SREENIVASA INSTITUTE OF TECHNOLOGY AND MANAGEMENT STUDIES (AUTONOMOUS)
Department of Computer Science and Engineering
GATE BITS SUBJECT-WISE
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Table of Contents 1. GATE Syllabus …………………………………………………………………………….. 3 2. Programming and Data Structures………………………………………………... 5 3. Design And Analysis Of Algorithms ……………………………………………………… 57 4. Compiler Design ……………………………………………………………………. 69 5. Operating System …………………………………………………………………… 90 6. Databases …………………………………………………………………………... 124 7. Software Engineering ………………………………………………………………. 154 8. Computer Networks………………………………………………………………… 160 9. Web technologies……………………………………………………………………. 179 10. Computer Organization….……………………………………………………. 185 11. Digital Logic ……………………………………………………………………. 237
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GATE SYLLABUS 1. COMPUTER SCIENCE AND INFORMATION TECHNOLOGY – CS & IT Engineering Mathematics
Mathematical Logic: Propositional Logic; First Order Logic. Probability: Conditional Probability; Mean, Median, Mode and Standard Deviation; Random Variables; Distributions; uniform, normal, exponential, Poisson, Binomial. Set Theory & Algebra: Sets; Relations; Functions; Groups; Partial Orders; Lattice; Boolean Algebra. Combinatorics: Permutations; Combinations; Counting; Summation; generating functions; recurrence relations; asymptotics. Graph Theory: Connectivity; spanning trees; Cut vertices & edges; covering; matching; independent sets; Colouring; Planarity; Isomorphism. Linear Algebra: Algebra of matrices, determinants, systems of linear equations, Eigen values and Eigen vectors. Numerical Methods: LU decomposition for systems of linear equations; numerical solutions of non-linear algebraic equations by Secant, Bisection and Newton-Raphson Methods; Numerical integration by trapezoidal and Simpson's rules. Calculus: Limit, Continuity & differentiability, Mean value Theorems, Theorems of integral calculus, evaluation of definite & improper integrals, Partial derivatives, Total derivatives, maxima & minima. GENERAL APTITUDE(GA): Verbal Ability: English grammar, sentence completion, verbal analogies, word groups, instructions, critical reasoning and verbal deduction. Computer Science and Information Technology Programming and Data Structures: Programming in C; Functions, Recursion, Parameter passing, Scope, Binding; Abstract data types, Arrays, Stacks, Queues, Linked Lists, Trees, Binary search trees, Binary heaps. Algorithms: Analysis, Asymptotic notation, Notions of space and time complexity, Worst and 3
average case analysis; Design: Greedy approach, Dynamic programming, Divide-and-conquer; Tree and graph traversals, Connected components, Spanning trees, Shortest paths; Hashing, Sorting, Searching. Asymptotic analysis (best, worst, average cases) of time and space, upper and lower bounds, Basic concepts of complexity classes P, NP, NP-hard, NP-complete. Compiler Design: Lexical analysis, Parsing, Syntax directed translation, Runtime environments, Intermediate and target code generation, Basics of code optimization. Operating System: Processes, Threads, Inter-process communication, Concurrency, Synchronization, Deadlock, CPU scheduling, Memory management and virtual memory, File systems, I/O systems, Protection and security. Databases: ER-model, Relational model (relational algebra, tuple calculus), Database design (integrity constraints, normal forms), Query languages (SQL), File structures (sequential files, indexing, B and B+ trees), Transactions and concurrency control. Information Systems and Software Engineering: information gathering, requirement and feasibility analysis, data flow diagrams, process specifications, input/output design, process life cycle, planning and managing the project, design, coding, testing, implementation, maintenance. Computer Networks: ISO/OSI stack, LAN technologies (Ethernet, Token ring), Flow and error control techniques, Routing algorithms, Congestion control, TCP/UDP and sockets, IP(v4), Application layer protocols (icmp, dns, smtp, pop, ftp, http); Basic concepts of hubs, switches, gateways, and routers. Network security basic concepts of public key and private key cryptography, digital signature, firewalls. Theory of Computation: Regular languages and finite automata, Context free languages and Push-down automata, Recursively enumerable sets and Turing machines, Undecidability. Digital Logic: Logic functions, Minimization, Design and synthesis of combinational and sequential circuits; Number representation and computer arithmetic (fixed and floating point). Computer Organization and Architecture: Machine instructions and addressing modes, ALU and data-path, CPU control design, Memory interface, I/O interface (Interrupt and DMA mode), Instruction pipelining, Cache and main memory, Secondary storage. Web technologies: HTML, XML, basic concepts of client-server computing.
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Programming and Data Structures 1. Which one of the following is the tightest upper bound that represents the time complexity of inserting an object into a binary search tree of n nodes? (A) O(1)
(B) O(log n)
(C) O(n)
(D) O(n log n)
(GATE 2013)
Answer: (C) Explanation: For skewed binary search tree on n nodes, the tightest upper bound to insert a node is O(n). 2. Which one of the following is the tightest upper bound that represents the number of swaps required to sort n numbers using selection sort? (A)O(log n)
(B) O(n)
(C) O(n log n)
(D) O(n2)
(GATE 2013)
Answer: (B) Explanation: The maximum number of swaps that takes place in selection sort on n numbers is n 3. Consider the following operation along with Enqueue and Dequeue operations on queues, where k is global parameters MultiDequeue(Q) { m=k while (Q is not empty) and (m >0) { Dequeue(Q) m=m-1 } } What is the worst case time complexity of a sequence of n queue operations on an initially empty queue? (A) ϴ(n)
(B) ϴ(n+k)
(C) ϴ(nk)
(D) ϴ(n2)
(GATE 2013)
Answer: (C) 4. The preorder traversal sequence of a binary search tree is 30, 20, 10, 15, 25, 23, 39, 35, 42. Which one of the following is the postorder traversal sequence of the same tree? (A) 10,20,15,23,25,35,42,39,30
(B) 15,10,25,23,20,42,35,39,30
(C) 15,20,10,23,25,42,35,39,30
(D) 15,10,23,25,20,35,42,39,30 5
Answer: (D)
Explanation: Preorder : 30,20,10,15,25,23,39,35,42 Inorder :10,15,20,23,25,30,35,39,42
5. What is the return value of f p,pif the value of p is initialized to 5 before the call? Note that the first parameter is passed by reference, whereas the second parameter is passed by value.
int f(int&x,int c) { c=c-1; if (c==0) return 1; x=x+1; return f(x,c)*x; } (A) 3024 Answer: (B)
(B) 6561
(C) 55440
(D) 161051
(GATE 2013)
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Explanation:
Common Data Questions 6 & 7: The procedure given below is required to find and replace certain characters inside an input character string supplied in array A. The characters to be replaced are supplied in array oldc, while their respective replacement characters are supplied in array newc. Array A has a fixed length of five characters, while arrays oldc and newc contain three characters each. However, the procedure is flawed
voidfind_and_replace( char *A, char *oldc, char *newc) { for (inti=0;i<5;i++) for (int j=0;j<3;j++) if (A[i]==oldc[j]) A[i]=newc[j]; }
The procedure is tested with the following four test cases 7
(1) oldc= "abc ", newc= "dab" (2) oldc= " cde", newc= "bcd" (3) oldc= "bca", newc= "cda" (4) oldc= "abc ", newc= "bac "
6. The tester now tests the program on all input strings of length five consisting of characters ‘a’, ‘b’, ‘c’, ‘d’ and ‘e’ with duplicates allowed. If the tester carries out this testing with the four test cases given above, how many test cases will be able to capture the flaw? (A) Only one (B) Only two (C) Only three (D) All four
(GATE 2013)
Answer: (B) Explanation: Flaw in this given procedure is that one character of Array ‘A’ can be replaced by more than one character of newc array, which should not be so.Test case (3) and (4) identifies this flaw as they are containing ‘oldc’ and ‘newc’ array characters arranged in specific manner. Following string can reflect flaw, if tested by test case (3).
Likewise single character ‘b’ in A is replaced by ‘c’ and then by ‘d’. Same way test case (4) can also catch the flaw. 7. If array A is made to hold the string “abcde”, which of the above four test cases will be successful in exposing the flaw in this procedure? (A) None
(B) 2 only
(C) 3 and 4 only
(D) 4 only
(GATE 2013)
Answer: (C) Explanation: Now for string “abcde” in array A, both test case (3) and (4) will be successful in finding the flaw, as explained in above question. 8. What will be output of the following program? 8
char inChar=’A’; switch(inChar) { case ‘A’: print(“Choice A\n”); case ‘B’: case ‘C’: printf(“Choice B”); case ‘D’: case ‘E’: default: printf(“No Choice”); }
(A) No Choice (B) Choice A (C) Choice A Choice B No Choice (D) Program gives no output as it is erroneous
(GATE 2012)
Answer: (C) Explanation:-Since there is no ‘break’ statement, the program executes all the subsequent case statements after printing “choice A”. 9.
Supposeacircularqueueofcapacity(n–1)elementsisimplementedwithan arrayofnelements.Assumethattheinsertionanddeletionoperationsarecarriedoutusi ngREARandFRONTasarrayindexvariables,respectively.Initially,REAR=FRONT=0.Thecond itionstodetectqueuefullandqueueemptyare (A)full:(REAR+1)modn==FRONT
empty:REAR==FRONT
(B)full:(REAR+1)mod n==FRONT empty:(FRONT+1)modn==REAR
(C)full:REAR==FRONT(D)full:(FRONT+1)mod n==REAR empty:(REAR+1)modn==FRONTempty:REAR==FRONT (GATE 2012)
Answer:(A) Explanation:Thecounterexamplefortheconditionfull:REAR=FRONTis InitiallywhentheQueueisemptyREAR=FRONT=0bywhichtheabovefullcondition issatisfiedwhichisfalse Thecounterexamplefortheconditionfull:(FRONT+1)modn=REARis InitiallywhentheQueueisemptyREAR=FRONT=0andletn=3,soafterinsertingone elementREAR=1andFRONT=0,atthispointtheconditionfullaboveissatisfied,buts tillthereisplacefor onemoreelementinQueue,sothisconditionisalsofalse Thecounterexamplefortheconditionempty:(REAR+1)modn=FRONTis InitiallywhentheQueueisemptyREAR=FRONT=0andletn=2,soafterinsertingone elementREAR=1andFRONT=0,atthispointtheconditionemptyaboveissatisfied,b utthequeueofcapacityn-1isfullhere 9
Thecounterexamplefortheconditionempty:(FRONT+1)modn=REARis InitiallywhentheQueueisemptyREAR=FRONT=0andletn=2,soafterinsertingone elementREAR=1andFRONT=0,atthispointtheconditionemptyaboveissatisfied,b utthequeueofcapacityn-1isfullhere
COMMON DATA QUESTIONS 10 & 11 Consider the following C program
(2012)
int a, b, c = 0; voidprtFun (void); int main () { staticint a = 1; /* line 1 */ prtFun(); a += 1; prtFun(); printf ( "\n %d %d " , a, b) ; } voidprtFun (void) { staticint a = 2; /* line 2 */ int b = 1; a += ++b; printf (" \n %d %d " , a, b); } 10.What output will be generated by the given code segment? (A) 3 1 (B) 4 2 (c) 4 2 (D) 3 1 41 61 62 52 42 61 20 52 Answer (C)
Explanation: ‘a’ and ‘b’ are global variable. prtFun() also has ‘a’ and ‘b’ as local variables. The local variables hide the globals (See Scope rules in C). When prtFun() is called first time, the local ‘b’ becomes 2 and local ‘a’ becomes 4. When prtFun() is called second time, same instance of local static ‘a’ is used and a new instance of ‘b’ is created because ‘a’ is static and ‘b’ is non-static. So ‘b’ becomes 2 again and ‘a’ becomes 6. main() also has its own local static variable named ‘a’ that hides the global ‘a’ in main. The printf() statement in main() accesses the local ‘a’ and prints its value. The same printf() statement accesses the global ‘b’ as there is no local variable named ‘b’ in main. Also, the defaut value of static and global int variables is 0. That is why the printf statement in main() prints 0 as value of b.
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11.What output will be generated by the given code d\segment if: (GATE 2012) Line 1 is replaced by “auto int a = 1;” Line 2 is replaced by “register int a = 2;” (A) 3 1 (B) 4 2 (C) 4 2 (D) 4 2 41 61 62 42 42 61 20 20 Answer (D) Explanation: If we replace line 1 by “auto int a = 1;” and line 2 by “register int a = 2;”, then ‘a’ becomes non-static in prtFun(). The output of first prtFun() remains same. But, the output of second prtFun() call is changed as a new instance of ‘a’ is created in second call. So “4 2″ is printed again. Finally, the printf() in main will print “2 0″. Making ‘a’ a register variable won’t change anything in output. 12. What does the following fragment of C-program print? char c[] = "GATE2011"; char *p =c; printf("%s", p + p[3] - p[1]) ; (A) GATE2011 (B) E2011 (C) 2011 (D) 011 Answer: (C) Explanation: See comments for explanation. char c[] = "GATE2011"; // p now has the base address string "GATE2011" char *p =c;
(GATE 2011)
// p[3] is 'E' and p[1] is 'A'. // p[3] - p[1] = ASCII value of 'E' - ASCII value of 'A' = 4 // So the expression p + p[3] - p[1] becomes p + 4 which is // base address of string "2011" printf("%s", p + p[3] - p[1]) ; 13. A max-heap is a heap where the value of each parent is greater than or equal to the value of its children. Which of the following is a max-heap? (GATE 2011)
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Answer: - (B) Explanation: - Heap is a complete binary tree
COMMON DATA QUESTIONS 14 & 15 Consider the following recursive C function that takes two arguments (2011) unsignedint foo(unsigned int n, unsigned int r) { if (n > 0) return (n%r + foo (n/r, r )); else return 0; } 14. What is the return value of the function foo when it is called as foo(513, 2)? (A) 9 (B) 8 (C) 5 (D) 2 (GATE 2011) Answer: (D)
Explanation: foo(513, 2) will return 1 + foo(256, 2). All subsequent recursive calls (including foo(256, 2)) will return 0 + foo(n/2, 2) except the last call foo(1, 2) . The last call foo(1, 2) returns 1. So, the value returned by foo(513, 2) is 1 + 0 + 0…. + 0 + 1. The function foo(n, 2) basically returns sum of bits (or count of set bits) in the number n.
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15. What is the return value of the function foo when it is called as foo(345, 10) ? (GATE 2011) (A) 345 (B) 12 (C) 5 (D) 3 Answer: (B) Explanation: The call foo(345, 10) returns sum of decimal digits (because r is 10) in the number n. Sum of digits for 345 is 3 + 4 + 5 = 12.
16. WhatisthevalueprintedbythefollowingCprogram? #include intf(int*a,intn)
(GATE 2010)
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{ if(n<=0)return0; elseif(*a% 2==0)return*a+ f(a+1,n-1); elsereturn*a-f(a+1,n-1); } intmain() { inta[]={12,7,13,4,11,6}; printf("%d",f(a,6)); return0; } 1)-9
2)5
3)15
4)19
Answer (C) Explanation: f() is a recursive function which adds f(a+1, n-1) to *a if *a is even. If *a is odd then f() subtracts f(a+1, n-1) from *a. See below recursion tree for execution of f(a, 6). . f(add(12), 6) /*Since 12 is first element. a contains address of 12 */ | | 12 + f(add(7), 5) /* Since 7 is the next element, a+1 contains address of 7 */ | | 7 - f(add(13), 4) | | 13 - f(add(4), 3) | | 4 + f(add(11), 2) | | 11 - f(add(6), 1) | | 6+0 So, the final returned value is 12 + (7 – (13 – (4 + (11 – (6 + 0))))) = 15
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17. The following C function takes a singly-linked list as input argument. It modified the list by moving the last element to the front of the list and returns the modified list. Some part of the code is left blank. (GATE 2010) typedefstruct node { int value; struct node *next } Node; Node *mode )_to_front(Node *head){ Node *p,*q; if((head==NULL) ||(head->next==NULL))return head; q=NULL;p=head; while(p->next!=NULL) { q=p; p=p->next; } ---------------------------------------return head; } Choose the correct alternative to replace the blank line. 1) q=NULL; p->next=head; head=p; 2) q->next=NULL; head=p; p->next=head; 3) head=p; p->next=q; q->next=NULL; 4) q->next=NULL; p->next=head; head=p; Answer: (D) Solution: Here the program wants to make the last node of the list, the first node. Here q is the second last node and p is the last node The second last node’s next should be now NULL so q->next=NULL. 15
p->next should below head node. sop->next=head Now the head node is p. So head = p. Hence (D) is correct option. 18. ThecyclomaticcomplexityofeachofthemodulesAandBshownbelowis10.Whatisthe cyclomaticcomplexityofthesequentialintegrationshownontherighthandside? (GATE 2010)
1)19
2)21
3)20
4)10
Answer: (1) Explanation: Cyclomatic complexity is defined as the no. of independent paths form begin to exit in a module. If some dependent sub modules are there in the module then the individual cyclomatic complexities are added & overall sum is reduced by 1. Here CC(complete) = CC(A) + CC(B) –1 So 10+10-1=19 Hence (1) is correct option. 19. What does the following program print? (GATE 2010) #include void f(int *p, int *q) { p = q; *p = 2;} inti = 0, j = 1; int main() { f(&i, &j); printf("%d %d \n", i, j); getchar(); return 0; } (A) 2 2 (B) 2 1 (C) 0 1 (D) 0 2 Answer: (D) 16
Explanation: See below f() with comments for explanation. /* p points to i and q points to j */ void f(int *p, int *q) { p = q; /* p also points to j now */ *p = 2; /* Value of j is changed to 2 now */ } 20.The following program is to be tested for statement coverage : (GATE 2010) begin if(a==b){S1;exit} else if(c==d){S2;} else {S3;exit;} S4; end The test cases T1, T2, T3, and T4 given below are expressed in terms of the properties satisfied by the values of variables a_b_cand d. The exact values are not given. T1: a,b,c and d are equal T2: a,b,c and d are all distinct T3: a=b and c!=d T4: a!=b and c=d Which of the test suites given below ensures coverage of statements S1, S2, S3 and S4? A) T1, T2, T3 B) T2, T4 C) T3, T4 D) T1, T2, T4 Answer: (D) Explanation: The following test cases covers statements. T1: all are equal S1 executed and then so no other execution. T2: all are distinct Only S3 executed T3: a=b & c!=d Only S1 executed T4: a!=b & c=d Only S2,S4only So to have all statements the option should be either T1,T2,T4or T2,T3,T4 Option (D) is T1,T2,T4 Hence (D) is correct option. Statement for Linked Answer Questions 21 & 22
(GATE 2010) 17
A has table of length 10 uses open addressing with hash function h(k)=k mod 10, and linear probing. After inserting 6 values into an empty has table, the table is as shown below. 0 1 2
42
3
23
4
34
5
52
6
46
7
33
8 9
21. Which one oft he following choices gives a possible order in which the key values could have been inserted in the table ? (A) 46, 42, 34, 52, 23, 33 (B) 34, 42, 23, 52, 33, 46 (C) 46, 34, 42, 23, 52, 33 (D) 42, 46, 33, 23, 34, 52 Answer: (C) Explanation: Here for hashing Linear probing is used, i.e. it finds the hash key value through hash function and maps the key on particular position In Hash table. In case of key has same hash address then it will find the next address then it will find the next empty position in the Has Table. Here we check all options: (A) Here 42 will be inserted at the 2nd position in the array next 52, also has same hash address 2. But it already occupied so it will search for the next free place which is 3rd position. So here 52 is misplaced and it is not possible key values. (B) Here 46 is misplaced so it is not possible value. (C) This is same as given hash table. So correct order is 46, 34, 42, 23, 52, 33 Hence (C) is correct option. 22. How many different insertion sequences of the key values using the same hash function and linear probing will result in the hash table shown above ? (A) 10 (B) 20 (C) 30 (D) 40 Answer: (C) Explanation: Here the given order of insertion is 46, 34, 42, 23, 52, 33 Here 42, 23, 34, 46 are inserted direct using hash function" But to insert 52 we have 6 vacant places." After insertion of 52 at any of the 6 places, we have 5 places"remaining for 33. So total combination. 6 * 5 = 30 possible ways Hence (C) is correct option. 18
23.Whatisthenumberofswapsrequiredtosortnelements usingselectionsort,intheworst case? (GATE 2009) A)θ(n) C)θ(n2 )
B)θ(nlogn) D)θ(n2logn)
Answer: (C)
24. Consider the program below: #include int fun(int n, int *f_p){ intt,f; if (n<=1){ *f_p=1 return 1; } t=fun(n-1,f_p); f=t+*f_p; *f_p=t; return f; } int main () { int x=15; printf(“%d\n”, fun(5,&x)); return 0; } The value printed is: A) 6 B) 8 C) 14 D) 15 Answer: B Explanation:
(GATE 2009)
Here the table column I, II, & III are during the forward calls of the recursive function fun &The part after arrow in column III is updated value during return calls & column IV & V are the returned values. 19
In the end 8 is returned so only this will be printed Hence (B) is correct option. 25. What is the maximum height of any AVL-tree with 7 nodes ? Assume that the height of a tree with a single node is 0. (GATE 2009) (A) 2 (B) 3 (C) 4 (D) 5 Answer: (B)
Explanation: AVL tree is a partially balanced tree with the weights assigned to thenodes can be only −1,0 or 1. This weight is assigned on the basis of difference of the no. of children in the left subtree& right subtree. If some other weight is there then we rotate the tree to balance it.
In this tree the height is 3 and it is Also AVL balanced so maximum height will be 3 Hence (B) is correct option. Statement for Linked Answer Question 26 & 27 (GATE 2009) Consider a binary max-heap implemented using an array 26.Which one of the follow9ng array represents a binary max-heap? (A) {25, 12, 16, 13, 10, 8, 14} (B) {25, 14, 13, 16, 10, 8, 12} (C) {25, 14, 16, 13, 10, 8, 12} (D) {25, 14, 12, 13, 10, 8, 16} Answer: (C) Explanation: If the value presented at any node is greater then all its children then the tree is called the max heap. Here we need to draw heap for all options
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Hence (C) is correct option. 27. What is the content of the array after two delete operations on the correct answer to the previous question? (GATE 2009) (A) {14, 13, 12, 10, 8} (B) {14, 12, 13, 8, 10} (C) {14, 13, 8, 12, 10} (D) {14, 13, 12, 8, 10} Answer: (D) Explanation:
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28. Which combination of the integer variables x,y, and z makes the variable a get the value 4 in the following expression? (GATE 2008) a = (x >y)?((x>z)?x:z): ((y >z)?y:z) (A) x= 3,y = 4,z = 2 (C) x= 6,y = 3,z = 5
(B) x = 6,y = 5,z = 3 (D) x = 5,y = 4,z = 5
Answer: (A) Explanation: a = (x >y)?((x>z)?x:z): ((y >z)?y:z) Expr 1?expr 2 : expr 3 ; Here Expr 1 is a comparison expression whose result may be true or false. If true returned the expr 2 is selected as choice otherwise expr3. Here we want 4 to be printed which is only in option (A) & (D) for y = 4 to be printed. x >y should be false since y is in true part so this expr should be true. So both the conditions are true in option (A) only so correct. We can check. x=3y=4z=2 a = (x >y)?((x >z)?x:z) ((y >z)?y:z) First we can check 3 >2 ?3 : 2 thus 3 is selected Then 4 >2 ?4 : 2 here 4 is selected Hence a = 3 > 4?3:4 = 4 Hence (A) is correct option. 30. What is printed by the following C program? int f(int x, int *py, int **ppz) { int y, z; **ppz += 1; z = **ppz;
(GATE 2008)
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*py += 2; y = *py; x += 3; return x + y + z; } void main() { int c, *b, **a; c = 4; b = &c; a = &b; printf( "%d", f(c,b,a)); getchar(); } (A) 18 (B) 19 (C) 21 (D) 22 Answer :(B) Explanation: /* Explanation for the answer */ /*below line changes value of c to 5. Note that x remains unaffected by this change as x is a copy of c and address of x is different from c*/ **ppz += 1 /* z is changed to 5*/ z = **ppz; /* changes c to 7, x is not changed */ *py += 2; /* y is changed to 7*/ y = *py; /* x is incremented by 3 */ x += 3; /* return 7 + 7 + 5*/ return x + y + z;
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31. Choose the correct option to fill ?1 and ?2 so that the program below prints an input string in reverse order. Assume that the input string is terminated by a newline character. (GATE 2008) void reverse(void) { int c; if (?1) reverse() ; ?2 } main() { printf ("Enter Text ") ; printf ("\n") ; reverse(); printf ("\n") ; } (A) ?1 is (getchar() != ’\n’) ?2 is getchar(c); (B) ?1 is (c = getchar() ) != ’\n’) ?2 is getchar(c); (C) ?1 is (c != ’\n’) ?2 is putchar(c); (D) ?1 is ((c = getchar()) != ’\n’) ?2 is putchar(c); Answer: (D) Explanation: getchar() is used to get the input character from the user and putchar() to print the entered character, but before printing reverse is called again and again until ‘\n’ is entered. When ‘\n’ is entered the functions from the function stack run putchar() statements one by one. Therefore, last entered character is printed first. You can try running below program void reverse(void); /* function prototype */ void reverse(void) { int c; if (((c = getchar()) != '\n')) reverse(); putchar(c); } main() { printf ("Enter Text ") ; printf ("\n") ; 24
reverse(); printf ("\n") ; getchar(); } 32. The following postfix expression with single digit operands in evaluated using a stack 823^/23*+ 51*− . Note that ^ is the exponentiation operator. The top two elements of the stack after the first* is evaluated are (GATE 2007) (A) 6, 1 (B) 5, 7 (C) 3, 2 (D) 1, 5 Answer: (B) Explanation: Given postfix expression is 823^/23*+ 51*− Scanning the string from left to right we push all the digits,& we get any operator we evaluate the operation between top 2 elements poped from stack. Then the result is pushed again into stack.
Stack contain 5, 7 Hence (B) is correct option 33.Consider the following C-function in which a[n] and b[n] are two sorted integer arrays and c[n+m] be another integer array.
(GATE 2006)
void xyz (int a[],int b[],int c[]){ inti, j, k; i=j=k=0; while((i
else c[k++]=b[j++]; } Which of the following condition (s) hold (s) after the termination of the while loop ? I
j
II
i
(A) only (I) (B) only (II) (C) either (I) or (II) but not both (D) neither (I) nor (II) Answer: (C) Explanation: While loop will terminate i>= n &j >= m program is to merge a &b arrays into C . While loop terminates after merging then either (I) or (II) should hold but not both at the same time. (I) says j
(GATE 2006)
void swap(int *px,int*py){ *px=*px−*py; *py=*px+*py; *px=*py−*px; } S1: will generate a compilation error S2: may generate a segmentation fault at runtime depending on the arguments passed S3: correctly implements the swap procedure for all input pointers referreing to integers stored in memory locations accessible to the process S4: implements the swap procedure correctly for some but not all valid input pointers 26
S5: may add or subtract integers and pointers (A) S1 (B) S2 and S3 (C) S2 and S4 (D) S2 and S5 Answer: (C) Explanation: Here pointers are used without initialization also the address pointed by then may be out of segment of program, so segmentation. Fault may be there so. S2 correct. Here no compiler error S1 false. Correctly done swap procedure but not all valid import pointers so S4 also true. S2 &S4 are correct. Hence (C) is correct option. Common Data Questions 35 & 36 (GATE 2006) A 3-ary max heap os like a binary max heap, but instead of 2 children, nodes have 3 children, A 3-ary heap can be represented by an array as follows: The root is stored in the first location, a [0], nodes in the next level, from left to right, is stored form a[1] to a[3]. The nodes from the second level of the tree from left to right are stored from a[4] location onward. An item x can be inserted into a 3-ary heap containing n items by placing x in the location a [n] and pushing it up the tree to satisfy the heap property. 35.Which one of the following is a valid sequence of elements in an array representing 2-ary max heap ? (A) 1, 3, 5, 6, 8, 9 (B) 9, 6, 3, 1, 8, 5 (C) 9, 3, 6, 8, 5, 1 (D) 9, 5, 6, 8, 3, 1 Answer: (D) Explanation:
27
Here in option (A), (B) and (C), value present at node is not greater then all its children. Hence (D) is correct option. 36.Suppose the elements 7, 2, 10, and 4 are inserted, in that order, into the valid 3-ary max heap found in the above question, Q. 38. Which on of the following is the sequence of items in the array representing the resultant heap ? (GATE 2006) (A) 10, 7, 9, 8, 3, 1, 5, 2, 6, 4 (B) 10, 9, 8, 7, 6, 5, 4, 3, 2, 1 (C) 10, 9, 4, 5, 7, 6, 8, 2, 1, 3 (D) 10, 8, 6, 9, 7, 2, 3, 4, 1, 5 Answer: (A) Explanation: Given heap is as follows
To add 7, 2, 10, 4 we add the node at the end of array
28
We keep if at right place in the heap tree. Compare elements with its parent node. Since 10 > 6 and 7 >5, we interchange
Since 10 >9, we interchange and we get
n/2=10/2=5 3 is at right position 4 is at right position Hence (A) is correct option. 37. What does the following C-statement declare?
(GATE 2005)
Int (*f) (int*) (A) A function that takes an integer pointer as argument and returns an integer (B) A function that takes an integer pointer as argument and returns an integer pointer (C) A pointer to a function that takes an integer pointer as argument an returns (D) A function that takes an integer pointer as argument returns a function pointer 29
Answer: (C) Explanation: Given statement int (*f) (int*) This is not the declaration of any function since the f has *(pointer symbol) before it. So f is a pointer to a function also the argument type is int *& the return type is int. So overall we can say that f is a pointer to a function that takes an integer pointer as argument and returns an integer. Hence (C) is correct option. 38.An Abstract Data type (ADT) is
(GATE 2005)
(A) same as an abstract class (B) a data type that cannot be instantiated (C) a data type for which only the operations defined on it can be used, but none else (D) all of the above Answer: (C) Explanation: Abstract Data type :- It is defined as a user defined data type, specified by keyword ‘abstract’ & defines the variables & functions, these operations can only use the variables of this data type. So option (C) which says that Abstract data type for which only operations defined on it can be used is correct. Eg.stack data type Here operations defined are push & pop. So we can apply only these 2 operations on it. Hence (C) is correct option. 39.A common property of logic programming languages and functional languages is (A) both are procedural language
(GATE 2005)
(B) both are based onλ−calculus (C) both are declarative (D) all of the above Answer: (D) 30
Explanation: λ-calculus" It provides the semantics for computation with functions so that properties of functional computation can be studied. Both the languages require declaration before use of any object. Both are procedural So option (D) is correct. Both the languages are based on λ calculus, procedural & declarative Hence (D) is correct option. 40.Which of the following are essential features of an object-oriented programming languages?
(GATE 2005)
1. Abstraction and encapsulation 2. Strictly-typedness 3. Type-safe property coupled with sub-type rule 4. Polymorphism in the presence of inheritance (A) 1 and 2 only
(B) 1 and 4 only
(C) 1, 2 and 4 only
(D) 1, 3 and 4 only
Answer: (B) Explanation: Object oriented programming languages necessarily have features like Abstraction Encapsulation, inheritance with polymorphism. but OOPL are also strongly-typed since there are restrictions on how operations involving values having different data types can be intermixed. Eg.two integers can be divided but one integer & one string can’t. Hence (B) is correct option. 41.A program P reads in 500 integers in the range (0, 100) representing the scores of 500 students. It then prints the frequency of each score above 50. What be the best way for P to store the frequencies?
(GATE 2005)
(A) An array of 50 numbers (B) An array of 100 numbers (C) An array of 500 numbers 31
(D) A dynamically allocated array of 550 numbers Answer: (A) Explanation: Here the no. readable are range 0 to 100 but the output of the program is interested in scores above 50 so there are 50 values (51 to 100) in this range. So only an array 50 integers required as we get any no. we increment the value stored at index. Array [x − 50] by 1. Hence (A ) is correct option. 42.Consider the following C-program (GATE 2005) double foo (double); /* Line 1*/ int main(){ doubleda_db; // input da db _foo(da); } double foo(double a){ returna; } The above code complied without any error or warning. If Line 1 is deleted, the above code will show (A) no compile warning or error (B) some complier-warning not leading to unitended results (C) Some complier-warning due to type-mismatch eventually leading to unitended results (D) Complier errors Answer: (C) Explanation: Here if line 1 which is prototype declaration of the function foo, in C compilation process this would give an compile warning , due to type-mismatch. Since then compiler won’t know what is the return type of foo. So unintended results may occur. Hence (C) is correct option. 43.Postorder traversal of a given binary search tree, T produces the following sequence of keys10, 9, 23, 22, 27, 25, 15, 50, 95, 60, 40, 29 Which one of the following sequences of keys can be the result of an inorder traversal of the tree T?
(GATE 2005) 32
(A) 9, 10, 15, 22, 23, 25, 27, 29, 40, 50, 60, 95 (B) 9, 10, 15, 22, 40, 50, 60, 95, 23, 25, 27, 29 (C) 29, 15, 9, 10, 25, 22, 23, 27, 40, 60, 50, 95 (D) 95, 50, 60, 40, 27, 23, 22, 25, 10, 0, 15, 29 Answer: (A) Explanation: When we are given any no elements & even any order (preorder or post order) & we need to calculate inorder, then inorder is simply sorted sequence of the elements. Here 9, 10, 15, 22, 23, 25, 27, 29, 40, 50, 60, 95 Hence (A) is correct option. 44.Assume that the operators +, -,# are left associative and ^ is right associative .The order of precedence (from highest to lowest) is ^,#, +, -. The postfix expression corresponding to the infix expression a + b#c - d ^ e ^ f is
(A) abc#+def ^^−
(B) abc#+de^f^
(C) ab+c#d−e^f^
(D) −+a#bc^^def
(GATE 2004)
Answer: (A) Explanation: Given expression a + b)c− d/e/f parenthesizing the expression as per given rules. = ((a + (b)c)) − (d/(e/f))) = ((a + (bc))) − (d/(ef/))) = ((abc)+) − (def//)) = (abc)+def//−) So option (A) is correct You can also solve using stack. Hence (A) is correct option. 45.Consider the following C program main ()
(GATE 2004) 33
{ int x, y, m, n; scanf(“%d%d”, &x,&y); /*Assume x>0 and y>0*/ m=x; n=y; while (m!=n) { if (m>n) m=m−n; else n=n−m; } printf(“%d”,n); } The program computers (A) x'y, using repeated subtraction (B) x mod y using repeated subtraction (C) the greatest common divisor of x and y (D) the least common multiple of x only Answer: (A) Explanation: Here if m >n then m = m − n m 1, E> 0). (GATE 2004) x=m; y=1; while (x−y>E) { x=(x+y)/2; y=m/x; } print (x); 34
(A) log m
(B) m2
(C) m1/2
(D) m1/3
Answer: (C) Explanation: Here we take let x = 16 Loop will stop when x − y = 0 or >0 Iteration
X
Y
1
(16+1)/2=8
16/8=2
2
(8+2)/2=5
16/5=3
3
(5+3)/2=4
16/4=4
Here X = Y Then take X. which is 4. (m)1/2 = 4 = (16)1/2 Hence (C) is correct option.
47.Choose the best matching between the programming styles in Group 1 and their characteristics in Group 2.
(GATE 2004)
Group-1
Group-2
P. Functional
1. Command-based, procedural
Q. Logic
2. Imperative, abstract data types
R. Object-oriented
3. Side-effect free, declarative,
S. Imperative
expression evaluation 4. Declarative, clausal representation, theorem proving
35
(A) P-2, Q-3, R-4, S-1
(B) P-4, Q-3, R-2, S-1
(C) P-3, Q-4, R-1, S-2
(D) P-3, Q-4, R-2, S-1
Answer: (D) Explanation: p. Functional Programming is declarative in nature, involves expression evaluation, & side effect free. q Logic is also declarative but involves theorem proving. r. Object oriented is imperative statement based & have abstract (general) data types. s Imperative :- The programs are made giving commands & follows definite procedure & sequence. Hence (D) is correct option. 48)What is printed by the print statements in the program P1 assuming call by reference parameter passing ? (2001) Program P1( ) { _ _ 1__ _ _ __ ____1(*)_ _r___ x; _r___ y; } func1(x,y,z) { y=y+4 z=x+y+z; } (A) 10, 3 (B) 31, 3 (C) 27, 7 (D) None of the above
36
SOLUTION Since the function fun 1 doesn’t return the values of x & y and x & y are not passed by reference. So in program P1( ) would print x = 10 & y = 3. So 10, 3 Hence (A) is correct option. 49)Consider the following three functions. (2001) [P1] int *g(void) { Int x=10; return (& x); } [P2] int *g(void) { int *px; *px=10; return px; } [P3] int *g(void) { int *px px=(int*)malloc (size of (int)); *px=10; return px; } Which of the above three functions are likely to cause problems with pointers ? (A) Only P3 (B) Only P1 and P3 (C) Only P1 and P2 (D) P1, P2 and P3
SOLUTION 37
P1 : Here the function is returning address of the variable x (& x ) but the return type is pointer to integer not address. So incorrect. P2 : *px = 0 directly assigned a value but still px doesn’t point to any memory location, so memory initialization or allocation should be done before. So incorrect. P3: Correction made in P2, memory pre allocated, So correct. Hence (C) is correct option. 50)Consider the following program (2001) Program P2 Var n:int: procedure W (var x:int) begin X=X+1 Print x; end Procedure D Begin var n:int; n=3; W(n); End Begin \\begin P2 n=10; D; end If the language has dynamic scooping and parameters are passed by reference, what will be printed by the program ? (A) 10 (B) 11 (C) 3 (D) None of the above
38
SOLUTION n = 10 given but not passed to D. In D, n = 3 & W(n) increments by 1. So n = n + 1 = 4. Hence (D) is correct option. 51)The results returned by function under value-result and reference parameter passing conventions. (2002) (A) Do not differ (B) Differ in the presence of loops (C) Differ in all cases (D) May differ in the presence of exception SOLUTION The results returned by function under value & reference parameter passing may differ in presence of loops. Hence (B) is correct option. 52)Consider the following declaration of a two-dimensional array in C. (2002) Char a[100][100] Assuming that the main memory is byte-addressable and that array is stored starting form memory address 0, the address of a [40] [50] is (A) 4040 (B) 4050 (C) 5040 (D) 5050 SOLUTION Char a[100] [100] 1 char require 1 byte Total required 10000 bytes. Memory format is byte addressable A[0]
…
a[0][50]
…
a[0][99]
.
…
…
…
.
.
.
.
.
A[50][0]
a[40][50]
. 39
. . A[99][0]
…
…
…
a[99][99]
100 bytes per row. I.e 40#100 = 4000 1 byte per column I. e. 50#1 = 50 Total 4050 Hence (B) is correct option. 53) Consider the following C function.(2003) fl oat f(fl oat x, int y){ fl oat p, s; int i; for (s=1, p=1, i=1, i
1x1+x 4 x3/3! x/4 x4/4! # = 1 + x + x2/2! + x3/3! + x4/4!
Loop
P
S
1
x
1+x
2
x)x/2 : 1 = x2/2 : 1
1 + x + x2/2
3
x2/2)x/3 = x3/3 : 2 : 1
1 + x + x2/2! + x3/3!
4
x3/3! x/4 x4/4!
1 + x + x2/2! + x3/3! + x4/4!
counter (i)
Thus it can be checked for every value. Here the assumption is that the value of y is very large so y " 3 So the series 1 + x + x2/2! + x3/3!................3 will have infinite terms & from our previous knowledge we know that this 3 series is expansion of ex (exponential series) so. 1 + x + x2/2! + x3/3!...........3 = ex Hence (B) is correct option. 54)Assume the following C variable declaration (2003) int)A[10], B[10][10]; Of the following expressions (1) A[2] (2) A[2][3] (3) B[1] 41
(4) B[2][3] Which will not give compile-time errors if used as left hand sides of assignment statements in a C program ? (A) 1, 2, and 4, only (B) 2, 3, and 4, only (C) 2 and 4 only
(D) 4 only
SOLUTION We have int )* which is an array of 10 integer value pointer whereas B[10] [10] is an array which stores 10#10 = 100 integers So let us try to solve it eliminating way. " Option 3 B[1] can’t be at the left side since it is 2D array so can’t use single index. We need not necessarily specify the size of first dimension for B[][3] " Option 4 B[2][3] is assignment to the array B value so possible. " Option 1 A [2] is also possible to assign some address os integer value " Option 2 this is some what tricky. Here A[2][3] becomes a 2D array if integer where A [2] means the 2nd integer in this array and A[2][3]o means 3rd integer in this row. eg. A[2] [3] = 5 means that at second row the third integer value is 5. Hence (*) Is correct option.
55)Let T(n) be the number of different binary search trees on n distinct elements. (2003) Then T[n] T(k 1)T(x) k1 n =− =/ , where x is (A) n − k + 1 (B) n − k (C) n − k − 1 (D) n − k − 2
SOLUTION
42
Binary search tree has a root node & its 2 subtrees. So for every node other than the leaves, all the elements smaller than the node are its left subtree & all the nodes which have value equal to or greater than that node are at right subtree. Here the given expression. T(n) T(k 1)T(X) K n 1 =− =/ Figure n(B) = no. of nodes in left subtree n(C) " no. of nodes in right subtree T(n) = n(B) + n(C) + 1 T(n) T(X)T(k 1) K n 1 =− =/ Expanding forT (k − 1) we get T(n) T(X) [T(0) T(1) T(2) .....T(n 1)] K n 1 =:++− = 144444444424444444443 / no. of nodes in left subtree denoted by K Total nodes = n So remaining node n − (k − 1) i.e nodes in the right subtree. So = n − k + 1 So overall we can say that the no. of different BST’s on n different 43
elements. T(n) T(n k 1)T(k 1) n k 1 = −+ − =/ Hence ( ) is correct option.
56) A data structure is required for storing a set of integers such that each of the following operations can be done is (logn) time, where n is the number of elements in the set. (2003) 1. Delection of the smallest element. 2. Insertion of an element if it is not already present in the set. Which of the following data structures can be used for this purpose ? (A) A heap can be used but not a balanced binary search tree (B) A balanced binary search tree can be used but not a heap (C) Both balanced binary search tree and heap can be used (D) Neither balanced binary search tree nor heap can be used SOLUTION Both the tasks can be performed by both the data structures but heap is a data structure where to perform these function every element has to be checked so O(n) complexity. But the balance binary search tree is efficient data structure since at every decision it selects one of its subtree to no. of elements to be checked are reduced by a factor of 1/2 every time. n/2! = x x = logn Hence (B) is correct option. 57) Let S be a stack of size n $ 1. Starting with the empty stack, suppose we push the first n natural numbers in sequence, and then perform n pop operations. Assume that Push and Pop operation take X seconds each , and Y seconds elapse between the end of the one such stack operation and the start of the next operation. For m $ 1, define the stack-life of mcs the time 44
elapsed from the end or Push (m) to the start of the pop operation that removes m from S . The average stack-life of an element of this stack is (2003) (A) n(X + Y) (B) 3Y + 2X (C) n(X + Y) − X (D) Y + 2X
SOLUTION Here each of PURSH & POP operation take X seconds & Y seconds are elapsed between two consecutive stack operations. m is the life time of element in stack. So m X is time for push. m X is time for pop. m Y is time for intermediate So total m(2X + Y) Average stack life ( ) m = m 2X + Y = 2X + Y = Y + 2X Hence (D) is correct option. 58) Consider the C program shown below. #include #defi ne print(x)printf(“%d”,x) int x; void Q (int z){ z+=x; print(z); } void p (int)y){ int x=)y+2; Page 13 www.gatehelp.com CS Topicwise 2001-2010 Programming & Data 45
Structure Q(x);)y=x-1; print(x); } main (void){ x=5; p(&x); print(x); } The output of this program is (A) 12 7 6 (B) 22 12 11 (C) 14 6 6 (D) 7 6 6 SOLUTION Figure Here X is the global variable so still 5. Figure Here this is global X whose xy has been changed to 6 so 6 is printed 12 66 Hence (A) is correct option. First x=5 Then by function p(&x) X =5+2=7 Then by function Q(x) z =z+x =7+5=12 Here x is global variable so still it is 5. Return to function p(&x) Y =7-1=6 print x =7 return to main Print x =6 Here this is global x whose *y ahs been changed to 6 so 6 is printed. 59) Consider the function - defined below.(2003) struct item { int data; struct item)next; }; 46
int f (struct item )p){ return ((p==NULL)||(p−>next==NULL)|| ((p−>data<=p−>next−>data)&& f(p−>next))); } For a given linked list p, the function f return 1 if and only if (A) the list is empty or has exactly one element (B) the elements in the list are sorted in non-decreasing order of data value (C) the elements in the list are sorted in non-increasing order of data value (D) not all elements in the list have the same data value SOLUTION Here the return 1 any 1 of the following should be correct. (A) P == NULL i.e the list is empty (ends) (B) P " next = NULL i.e have one element. (C) P " data <= p " next " data i.e the element is smaller than its next element also. This is true for whole list. Since &&f(p "next) is also there. So overall it gives that the elements should be in sorted order. Hence (B) is correct option. 60) The goal of structured programming is to (2004) (A) have well indented programs (B) be able to infer the flow of control from the compiled code (C) be able to infer the flow of control form the program text (D) avoid the use of GOTO statements SOLUTION Structured programming :- It is way of programming using the sub structure method, i.e splitting the programs into sub sections. Structured programming prevents confusing transfer of control of avoiding the use of GOTO statements. Hence (D) is correct option. 61) Consider the following C function (2004) void swap (int a, int b) {int temp; 47
temp =a; a =b; b =temp; } In the order to exchange the values of two variables x and y . (A) call swap (x,y) (B) call swap (&x,&y) (C) swap (x,y) cannot be used as it does not return any value (D) swap (x,y) cannot be used as the parameters are passed by value SOLUTION Here the function takes the arguments by value. " Option (A) sends parameter by value but only the local variable a & b will be exchanged but not the actual variables x & y so incorrect. " Option (B) is incorrect sending address of x & y . " Option (C) swap (x,y) is usable there is no need to return. " Option (D) is the opposite statement of option (A), it says that the values are passed by value so won’t swap so the option is correct. Hence (D) is correct option. 62) A single array A [1........MAXSIZE] is used to implement two stacks. The two stacks grow from opposite ends of the array. Variables top 1 and top 2 (top 1
2
3
4
5
6
7
8
9
10
48
Here the stack will be fuel if both top 1 & top 2 are at the adjacent index values i.e. their difference is 1. So top 1 = top 2 – 1 Here (D) is correct option. 63) The best data structure to check whether an arithmetic expression has balanced parenthesis is a (A) queue (B) stack (C) tree (D) list SOLUTION Balanced parenthesis in an equation are such that the no. of opening and closing parenthesis and in correct order should be there. We can check balancing using stack. When we get any opening parenthesis then we push that in the stack & if we get a closing one then we pop the stack. After the complete scanning of input string if stack is found empty then the arithmetic expression is balanced. Hence (B) is correct option. 63) Consider the following C function int f(int n) {static int i=1; if (n>=5) return n; n=n+i; i++; return f(n); } The value returned by f(1) is (A) 5 (B) 6 (C) 7 49
(D) 8 SOLUTION Here i is an static variable, so if it is once initialized it can’t be initialized again during its scope n is incremented by 1 & f(n) is called then. The final return is when n>=5 i.e. n returned then Step
Call
n
i
(1) 1
1
1
condition false n < 5
(2)
1+1=2
2
(3) 2
2
2
(4)
2+2=4
3
(5) 3
4
3
(6)
4+3=7
4
(7) 4
7
4
false n < 5
false n < 5
true return n = 7
So return value is 7. Hence (C) is correct option. 64) Consider the following program fragment for reversing the digits in a given integer to obtain a new integer. Let ....... n d d d. = 1 2 m int n, rev; rev=0; while(n>0){ rev=rev)10+n%10; n=n/10; } The loop invariant condition at the end of the ith iteration is (A) n = d1d2......dm−i and rev = dmdm−1......dm−i+1 (B) n = dm−i+1.....dm−1dm or rev = dm−i .....d2d1 (C) n =Y rev (D) n = d1d2....dm or rev = dm......d2d1
SOLUTION Here after every iteration one digit is reduced from n since n = n/10 so unit place is removed.
50
This unit place is then added into the previous reverse sum (rev) after multiplying rev by 10. So 1 digit is incremented every iteration. So at the ith iteration n should have m − i digits d1d2.....dm−i & rev have dmdm−1..........dm−i+1 i
n
rev
1
d1d2....dm−1 dm
2
d1d2......dm−2 dmdm−1
So on. Hence (A) is correct option. 65)Consider the following C program segment. (2003) char p[20]; char)s= “string”; int length=strlen(s); for (i=0;i
51
66) A circularly linked list is used to represent a Queue. A single variable p is used to access the Queue. To which node should p point such that both the operations enQueue and deQueue can be performed in constant time ? (2003) (A) rear node (B) front node (C) not possible with a single pointer (D) node next to front SOLUTION: Here due to circular connection the rear & front are connected. Here if we point P to rear the P "next point to front node & P " data will point to rear value while inserting at rear following sequence of operations done. These operation done is 0(1) time So constant complexity. Hence (A) is correct option. 67) Suppose the numbers 7, 5, 1, 8, 3, 6, 0, 9, 4, 2 are inserted in that order into an initially empty binary search tree. The binary search tree uses the usual ordering on natural numbers. What is the inorder transversal sequence of the resultant tree ?(2003) (A) 7 5 1 0 3 2 4 6 8 9 (B) 0 2 4 3 1 6 5 9 8 7 (C) 0 1 2 3 4 5 6 7 8 9 (D) 9 8 6 4 2 3 0 1 5 7 SOLUTION We can solve it in shortcut that the first given element in 7, so we need to choose that particular option in which 7 is at the right place i.e. all the elements on its left should be smaller than it & all the elements on the right should be equal & greater than it. So this rule is followed in option C only. The method to make BST for given inputs is 7, 5, 1, 8, 3, 6, 0, 9, 4, 2.
52
To make in order of a binary search tree. (i) Start with the root node. (ii) Scan its left subtree, (iii) If the node in subtree has any left child then store the node in stack & repeat this step for its left child unit no. left child of any node. (iv) If leaf reached then print the node & pop the stack, print the poped value. (v) Check its right subtree & repeat step (III) for it. (vi) When stack empty then stop
53
So here inorder is 0 1 2 3 4 5 6 7 8 9. Actually a fact can be remembered that inorder traversal of a BST leads to a sorted sequence of elements. Hence (C) is correct option 67) Consider the following 2-3-4 tree (i.e., B-tree with a minimum degree of two in which each data item is a letter. The usual alphabetical ordering of letters is used in constructing the tree.(2003) What is the result of inserting G in the below tree ?
(D) None of the above SOLUTION
54
2-3-4 B-tree means the min degree of a node is two & it can be max 4 So maximum of 3 elements can be there in a node.
Here in this node the no. of element >3. So we need a split or rotation. Since the adjacent child has no. of element n 2 2 # = 4 = 2 so we apply a right rotation. So here.
Hence (C) is correct option. 68) The following numbers are inserted into an empty binary search tree in the given order: 10, 1, 3, 5, 15, 12, 16. What is the height of the binary search tree (tree height is the maximum distance of a leaf node from the root) (2003) (A) 2 (B) 3 (C) 4 (D) 6 SOLUTION 55
Given are 10, 1, 3, 5, 15, 12, 16
The height of the leaf node (5) is high 3. Hence (B) is correct option.
56
DESIGN AND ANALYSIS OF ALGORITHMS
1.Consider a linked list of n elements. What is the time taken to insert an element after an element pointed by some pointer? A. O (1)
B. O (n)
C. O (log2 n)
D .O (n log2 n)
Answer A 2.An algorithm is made up of two independent time complexities f (n) and g (n). Then the complexities of the algorithm is in the order of A. f(n) x g(n) B. Max ( f(n),g(n))
C. Min (f(n),g(n))
D. f(n) + g(n)
Answer B 3.Two main measures for the efficiency of an algorithm are A. Processor and memory B. Complexity and capacity C. Time and space D. Data and space Answer C 4.The total number of comparisons in a bubble sort is A. 0(log n) B. 0(n log n) C. 0(n) D. None of the above Answer B 5.Time complexities of three algorithms are given. Which should execute the slowest for large values of N? A. ( 1 2 ) O N B. O(N)
C. O(log N)
D. None of these
Answer B 6.The upper bound of computing time of m coloring decision problem is A. O(nm)
B .O(nm)
C. O(nmn)
D. O(nmmn)
Answer C 7.The space factor when determining the efficiency of algorithm is measured by A. Counting the maximum memory needed by the algorithm B. Counting the minimum memory needed by the algorithm C. Counting the average memory needed by the algorithm D. Counting the maximum disk space needed by the algorithm 57
Answer A 8.If the address of A[1][1] and A[2][1] are 1000 and 1010 respectively and each element occupies 2 bytes then the array has been stored in _________ order. A. row major
B. column major
C. matix major D. none of these
Answer A 9.The time factor when determining the efficiency of algorithm is measured by A. Counting microseconds B. Counting the number of key operations C. Counting the number of statements D. Counting the kilobytes of algorithm Answer B 10.The Worst case occur in linear search algorithm when A. Item is somewhere in the middle of the array B. Item is not in the array at all C. Item is the last element in the array D. Item is the last element in the array or is not there at all Answer D 11.A list of n strings, each of length n, is sorted into lexicographic order using the merge-sort algorithm. The worst case running time of this computation is A. O (n log n) B. O (n2 log n) C. O (n2 + log n)
D. O (n2)
Answer A 12.Which of the following case does not exist in complexity theory A. Best case
B. Worst case C. Average case
D. Null case
Answer D 13.The minimum number of multiplications and additions required to evaluate the polynomial P = 4x3+3x2-15x+45 is A. 6 & 3
B. 4 & 2
C. 3 & 3
D. 8 & 3
Answer C 58
14.The concept of order Big O is important because A. It can be used to decide the best algorithm that solves a given problem B. It determines the maximum size of a problem that can be solved in a given given amount of time C. It is the lower bound of the growth rate of algorithm D. Both A and B Answer A 15.The worst case running time to search for an element in a balanced binary search tree with n2n elements is A. T(nlogn)
B. T(n2n)
C. T(n) D. T(logn)
Answer C 16.Which of the following sorting algorithm is of divide-and-conquer type? A. Bubble sort B. Insertion sort
C. Quick sort
D. All of above
Answer C 17.The quick sort algorithm exploit _________ design technique A. Greedy
B. Dynamic programming
C. Divide and Conquer D. Backtracking
Answer C 18.The number of distinct simple graphs with up to three nodes are A. 15
B. 10
C. 7
D. 9
Answer C 19.The number of unused pointers in a complete binary tree of depth 5 is A. 4
B. 8
C. 16
D. 32
Answer C 20.A given connected graph G is a Euler graph , if and only if all vertices of G are of A. Same degree
B .Even degree C .Odd degree D. Different degree
Answer B 21.What is the maximum number of nodes in a B-tree of order 10 of depth 3 (root at depth 0) ? A. 111
B.999
C. 9999
D. None of the above 59
Answer D 22.One can convert a binary tree into its mirror image by traversing it in A. Inorder
B. Preorder
C. Postorder
D. Any order
Answer C 23.Graphs are represented using A. Adjacency tree
B. Adjacency linked list C. Adjacency graph
D. Adjacency queue
Answer B 24.The data structure required for breadth first traversal on a graph is A. Queue
B. Stack
C. Array
D. Tree
Answer A 25.Number of edges of a complete binary tree with 16 leaf nodes are A. 14
B. 30
C. 32
D. 28
Answer B
26.Tree A. Is a bipartite graph B. With n node contains n-1 edges C. Is a connected graph D. All of these Answer D 27.If every node u in G is adjacent to every other node v in G, A graph is said to be A. Isolated B. Complete C. Finite D. Strongly Connected Answer B 28.Consider the following pseudo-code : If (A > B) and (C > D) then A=A+1 B=B+1 Endif The cyclomatic complexity of the pseudo-code is A. 2
B. 3
C. 4
D. 5
Answer D 29.Leaves of which of the following trees are at the same level ? 60
A. Binary tree B. B-tree
C. AVL-tree
D. Expression tree
Answer B 30.The Inorder traversal of the tree will yield a sorted listing of elements of tree in A. Binary tree B. Binary search tree
C. Heaps
D. None of the above
Answer B 31.One can make an exact replica of a Binary Search Tree by traversing it in A. Inorder
B. Preorder
C. Postorder
D. Any order
Answer B 32.Let A be an adjacency matrix of a graph G. The th ij entry in the matrix K A , gives A. The number of paths of length K from vertex Vi to vertex Vj. B. Shortest path of K edges from vertex Vi to vertex Vj. C. Length of a Eulerian path from vertex Vi to vertex Vj. D. Length of a Hamiltonian cycle from vertex Vi to vertex Vj. Answer B 33. A graph in which all nodes are of equal degree is called A. Multi graph B. Non regular graph C. Regular graph D. Complete graph Answer C 34. The time complexity to build a heap of n elements is A. 0(1)
B. 0(lgn)
C. 0(n)
D. 0(nlgn)
Answer D 35. Given a binary tree whose inorder and preorder traversal are given by Inorder : EICFBGDJHK Preorder : BCEIFDGHJK The post order traversal of the above binary tree is A. IEFCGJKHDB
B. IEFCJGKHDB
C. IEFCGKJHDB
D. IEFCGJKDBH
Answer A 36. The running time of the following sorting algorithm depends on whether the partitioning is balanced or unbalanced 61
A. Insertion sort B. Selection sort C. Quick sort D. Merge sort Answer C 37. In worst case Quick Sort has order A. O (n log n) B. O (n2/2)
C. O (log n)
D. O (n2/4)
Answer B 38. The sorting technique where array to be sorted is partitioned again and again in such a way that all elements less than or equal to partitioning element appear before it and those which are greater appear after it, is called A. Merge sort B. Quick sort
C. Selection sort
D. None of these
Answer B 39. The best average behaviour is shown by A. Quick Sort B. Merge Sort C. Insertion Sort D. Heap Sort Answer A 40. Quick sort is also known as A. Merge sort B. Heap sort C. Bubble sort D. None of these Answer D 41. Assuming P ? NP, which of the following is TRUE? A. NP-complete = NP B. NP-completenP=theta C. NP-hard = NP D. P = NP-complete Answer B 42. If there is an NP complete language L whose complement is in NP ,then complement of any language in NP is in A. P
B. NP C. Both A and B
D. None of these
Answer B 43. Both P and NP are closed under the operation of A. Union
B. Intersection C. Concatenation
D. Kleene
Answer D 44. If every node u in G is adjacent to every other node v in G, A graph is said to be
A.
Isolated 62
B. C. D.
Complete Finite Strongly Connected
Answer:B
45. Which of the following sorting algorithms has the lowest worst case complexity?
A. B. C. D.
Merge sort Bubble sort Quick sort Selection sort
46. Randomized quicksort is an extension chosen randomly. What is the worst numbers using randomized quicksort?
(a) O(n)
(b) O(n log n)
of quicksort where the pivot is case complexity of sorting n (c) O(n2)
(d) O(n!)
47. Level of any node of a tree is A. B. C. D.
Height of its left subtree minus height of its right subtree Height of its right subtree minus height of its left subtree Its distance from the root None of these
48. The total number of comparisons in a bubble sort is A. 0(log n) B. 0(n log n) C. 0(n) D. None of the above 49. Time complexities of three algorithms are given. Which should execute the slowest for large values of N? A. ( 1 2 ) O N B. O(N) C. O(log N) D. None of these 50. The quick sort algorithm exploit _________ design technique 63
A. B. C. D.
Greedy Dynamic programming Divide and Conquer Backtracking
51. A sort which relatively passes through a list to exchange the first element with any element less than it and then repeats with a new first element is called A. Insertion sort B. Selection sort C. Heap sort
Quick sort
52. The pre order and post order traversal of a Binary Tree generates the same output. The tree can have maximum A. Three nodes B. Two nodes C. One node D. Any number of nodes 53. A search technique where we keep expanding nodes with least accumulated cost so far is called A. B. C. D.
Hill climbing Branch and bound Best first Divide and conquer
54. The spanning tree of connected graph with 10 vertices contains A. 9 edges B. 11 edges C. 10 edges D.10 vertices 55. The post order traversal of a binary tree is DEBFCA. Find out the preorder traversal. A. ABFCDE B. ADBFEC C. ABDECF D. ABDCEF 56. Which of the following statements are TRUE? (1) The problem of determining whether there exists a cycle in an undirected graph is in P. (2) The problem of determining whether there exists a cycle in an undirected graph is in NP. 64
(3) If a problem A is NP-Complete, there exists a non-deterministic polynomial time algorithm to solve A. A. 1,2 and 3 B. 1 and 2 only C. 2 and 3 only D. 1 and 3 only 57. A binary tree can easily be converted into q 2-tree A. by replacing each empty sub tree by a new internal node B. by inserting an internal nodes for non-empty node C. by inserting an external nodes for non-empty node D. by replacing each empty sub tree by a new external node 58. Which of the following sorting procedures is the slowest? A. Quick sort B. Heap sort C. Shell sort D. Bubble sort 59. The pre-order and post order traversal of a Binary Tree generates the same output. The tree can have maximum A. B. C. D.
Three nodes Two nodes One node Any number of nodes
60.
A. A B. B C. C D. D
61. Two isomorphic graphs must have
A. B. C. D.
Equal number of vertices Same number of edges Same number of vertices All of the above
62. If each node in a tree has value greater than every value in its left subtree and has value less than every in the its right subtree ,the tree is called 65
A. Complete tree B.Full binary tree C. Binary search tree D. Threaded tree 63. A simple graph in which there exists an edge between pair of vertices is called A. Regular graph B. Planner graph C.Euler graph
D. Complete graph
64. The best average behaviour is shown by A. Quick sort B. Merge sort C.Insertion sort D. Heap sort 65. Which of the following sorting algorithm is of divide-and-conquer type? A. Bubble sort B. Insertion sort C. Quick sort D. All of above 66. The recurrence relation capturing the optimal execution time of the Towers of Hanoi problem with n discs is A. B. C. D.
T(n) = 2T(n - 2) + 2 T(n) = 2T(n - 1) + n T(n) = 2T(n/2) + 1 T(n) = 2T(n - 1) + 1
67.
A. A B. B C. C D. D 68. The goal of hashing is to produce a search that takes
A. B. C. D.
O(1) time O(n2 ) time O(log n ) time O(n log n ) time 69. One can make an exact replica of a Binary Search Tree by traversing it in 66
A. Inorder B. Preorder C. Postorder D. Any order 70. When converting binary tree into extended binary tree, all the original nodes in binary
tree are A. internal nodes on extended tree B. external nodes on extended tree C. vanished on extended tree D. None of above 71. The postfix form of A*B+C/D is A. *AB/CD+ C. A*BC+/D
B. AB*CD/+ D. ABCD+/*
72. For the bubble sort algorithm, what is the time complexity of the best/worst case?
(assume that the computation stops as soon as no more swaps in one pass) (a) (b) (c) (d)
best case: O(n) worst case: O(n*n) best case: O(n) worst case: O(n*log(n)) best case: O(n*log(n)) worst case: O(n*log(n)) best case: O(n*log(n)) worst case: O(n*n)
Answer : A 73. For the
quick sort algorithm, what is the time complexity of the best/worst case?
(a) best case: O(n) worst case: O(n*n) (b) best case: O(n) worst case: O(n*log(n)) (c) best case: O(n*log(n)) worst case: O(n*log(n)) (d) best case: O(n*log(n)) worst case: O(n*n) Answer : D 74. In
an arbitrary tree ( not a search tree) of order M. Its size is N, and its height is K. The computation time needed to find a data item on T is (a) O(K*K) (b) O(M*M) (c) O(N) (d) O(K) Answer : C
67
75. When we organize our data
as an ordered list, what is the time complexity of inserting/deleting a data item to/from the list? (a) O(length_of_list*length_of_list) (b) O(length_of_list) (c) O(log(length_of_list * length_of_list)) (d) O(1) Answer : B 76. Five statements
about B-trees are below. Four of them are correct. Which one is
INCORRECT? (a) All B-trees are also search trees (b) The word B-tree stands for balanced tree (c) The word B-tree also stands for binary tree (d) All leaves of a B-tree must be on the same level Answer : C 77. For any
B-tree of height H (H>1), after inserting a new key, is it possible for a key, K, which was located in a leaf-node to move up to the root in this regard which of the following is correct? (a) Can’t be defined without data (b) Never (c) Yes, only if H=2 (d) Yes Answer : D 78. When we say
the order of a tree is M, we mean
(a) Every non-leaf node must have M subtrees (b) Every non-leaf node must have M keys (c) Every non-leaf node can have at most M subtrees (d) Every non-leaf node can have at most M keys Answer : C 79. T
is a search tree of order M, its size is N, and its height is K. The computation time needed to INSERT/DELETE a data item on T is (a) O( 1 ) (b) O( M ) (c) O( Log K ) (d) O( K ) Answer : D
68
80. Suppose that
we have a data file containing records of famous people, and we need to build a hash table to find a record from the person's birthday. The size of the hash table is 4096. The following are hash functions which convert a birthday to an integer. Which of the following function is the best? (a) h1( day/month/year ) = day + month + year (b) h2( day/month/year ) = day + month*31 + year (c) h3( day/month/year ) = (day + month*31 + year*365) mod 4096 (d) h4( day/month/year ) = (day + month*31 + year*365) mod 4093 Answer : D
COMPILER DESIGN 1.Which of the following statements is false ? (A) An unambiguous grammar has same left most and right most derivation (B) An LL(1) parser is a top-down parser (C) LALR is more powerful than SLR (D) An ambiguous grammar can never be LR (K) for any k SOLUTION So (A) & (C) are, true. An ambiguous grammar can’t be LR (K) So option (A) is false since an unambiguous grammar has unique right most derivation & left most derivations but both are not same. Hence (A) is correct option 2. Dynamic linking can cause security concerns because (A) Security is dynamic (B) The path for searching dynamic libraries is not known till run time. (C) Linking is insecure (D) Cryptographic procedures are not available for dynamic linking SOLUTION Dynamic linking is type of linking in which libraries required by the program are linked during run time. But at this time cryptographic procedures are not available, so make this process insecure. Hence (D) is correct option. 3. Which of the following suffices to convert an arbitrary CFG to an LL(1) grammar? (A) Removing left recursion alone 69
(B) Factoring the grammar alone (C) Removing left recursion and factoring the grammar (D) None of this SOLUTION If a grammar has left recursion & left factoring then it is ambiguous. So to convert a CFG to LL(1) grammar both removal of left recursion & left factoring need to be done. Hence (C) is correct option. 4. Assume that the SLR parser for a grammar G has n1 states and the LALR parser for G has n2 states. The relationship between n1 and n2 is (A) n1 is necessarily less than n2 (B) n1 is necessarily equal to n2 (C) n1 is necessarily greater than n2 (D) None of the above SOLUTION SLR parsue is less range of context free languages than LALR but still both n1 & n2 are same for SLR & LALR respectively. Hence (B) is correct option. 5. In a bottom-up evaluation of a syntax directed definition, inherited attributes can (A) always be evaluated (B) be evaluated if the definition is L-attributed (C) be evaluated only if the definition has synthesized attributes (D) never be evaluated SOLUTION Every S (synthesized) -attributed definitions is L- attributed. So in a bottom-up evaluation of SDD inherited attributes can be evaluated only if the definition has synthesized attributes. Hence (C) is correct option. 6. Which of the following statements is FALSE? (A) In statically typed language, each variable in a program has a fixed type (B) In up-typed languages, values do not have any types (C) In dynamically typed languages, variables have no types (D) In all statically typed languages, each variable in a program is associated with values of only a single type during the execution of the program 70
SOLUTION (1) True for statically typed languages where each variable has fixed type. Similarly (4) is also correct. (2) True, in un-typed languages types of values are not defined. But option (C) is false, since in dynamically typed language variables have dynamically changing types but not that they have no type. Hence (C) is correct option. 7.Consider the grammar shown below S " | EtSS' | α S' " eS |! E"b In the predictive parse table M of this grammar the entities M(S) and M[S1,$] respectively are (A) {s' " eS }and{S' " ε} (B) {s' " eS}and{} (C) {s' " ε}and{S' " ε} (D){s' " eS , S ' " ε}and{ S' " ε} SOLUTION:D 8. Consider the grammar shown below. S"CC C " eC | d The grammar is (A) LL (1) (B) SLR (1) but not LL (1) (C) LALR (1) but not SLR (1) (D)
LR (1) but not LALR (1)
SOLUTION Given grammar S " CC C "cC d it can’t be LL since C " cC is recursive. LR(1) also known as CLR parser, and every CF 71
grammar is CLR grammar. So (A) is false but (C) & (D) can be correct. This grammar is CLR and also reducible to LALR without any conflicts. So (D) is false. Only need to check for SLR(1) or LR(0) This grammar is not SLR. Hence (C) is correct option
9. Consider the translation scheme shown below S " TR R "+ T {print (‘+’);}R | ε T " num {print (num.val);} Here num is a token that represents an integer and num. val represents the corresponding integer value. For an input string ‘9 + 5+ 2’, this translation scheme will print (A) 9 + 5 + 2
(B) 9 5 + 2 +
(C) 9 5 2 ++
(D) ++ 9 5 2
SOLUTION S " TR R "+ T {pr int(' + ');}R ε T " num{print(num.val);} Given string 9 + 5 + 2 S " TR
T + TR T+T+T 9+T+T 9+5 +T 9+5 +2
{print(+);} {print(+);} {print(9);} {print(5);} {print(2);}
So ++ 952 is printed Hence (D) is correct option. 10. Consider the syntax directed definition shown below S " id: = E " newtemp (); gen(t . place .place t}
. place;); 72
"
.place
.place;}
Here, gen is a function that generates the output code, and newtemp is a function that returns the name of a new temporary variable on every call. Assume that t1’s are the temporary variable names generated by newtemp. For the statement ‘X: = Y + Z ’, the 3-address code sequence generated by this definition is (A) X = Y + Z (B) t1 = Y + Z; X t1 (C) t1 = Y; t2 = t1 + Z; X = t2 (D) t1 = Y; t2 = Z; t3 + t2; X = t3 SOLUTION In 3-address code we use temporary variables to reduce complex instructions so here t1 = Y t2 = Z t 3 = t 1 + t2 x = t3 Hence (D) is correct option.
Solve the problems and choose the correct answers. The following program fragment is written in a programming language that allows variables and does not allow nested declarations of functions. global inti void int i print i print } main () { (i ) } 11. If the programming language uses static scoping and call by need parameter passing mechanism, the values printed by the above program are (A) 115, 220 (B) 25, 220 (C) 25, 15
(D) 115, 105
SOLUTION In static scoping the variables are initialized at compile time only 73
So i = 100 & j = 5 P (i + j) = P (100 + 5) = P(105) So x = 105 x + 10 = 105 + 10 = 115 So 115 & 105 will be printed. Hence (D) is correct option. 12. If the programming language uses dynamic scoping and call by name parameter passing mechanism, the values printed by the above program are (A) 115, 220 (B) 25, 220 (C) 25, 15 (D) 115, 105 SOLUTION In dynamic scoping, the local values are considered & variables are initialized at run time. Since x = i + j & in P (x) i = 200 & j = 20 x = 200 + 20 = 220 & printing (x + 10) x = i + j + 10 = 10 + 5 + 10 = 25 Hence (B) is correct option 13. Consider the following class definitions in a hypothetical object oriented language that supports inheritance and uses dynamic binding. The language should not be assumed to be either Java or C++, thought the syntax is similar
Now consider the following program fragment: P x =new Q(); Q y =new Q(); 74
P z =new Q(); x. f (1);((P) y). f (1);z.f(1); Here ((P) y) denotes a typecast of y to P. The output produced by executing the above program fragment will be (A) 1 2 1 (B) 2 1 1 (C) 2 1 2
(D) 2 2 2
SOLUTION 1.
Px = newQ();
2.
Qy = newQ();
3.
Pz = newQ();
4.
x : f(1); print 2 # i = 2
5.
((P) y) : f(1);
6.
z : f(1) print 2 # i = 2
but line 5. will print 2 because typecast to parent class can’t prevent over ridding. So function f(1) of class Q will be called not f(1) of class P . Hence (D) is correct option. 14. Which of the following is NOT an advantage of using shared, dynamically linked libraries as opposed to using statically linked libraries? (A) Smaller sizes of executable (B) Lesser overall page fault rate in the system (C) Faster program startup (D) Existing programs need not be re-linked to take advantage of newer versions of libraries SOLUTION The advantages of shared dynamically linked libraries include. (A) smaller size of executable since less data (B) lesser overall page fault rate. (C) No need for re-linking if newer versions of libraries are there. But since compilation time doesn’t include linking so a long linking time required during runtime in DLL ' s so slow startup. Hence (C) is correct option. 15. Which of the following grammar rules violate the requirements of an operator grammar? P, Q, R are non-terminals, and r, s, t are terminals 75
. (i) P " QR
(ii) P " Q s R
(iii) P " ε
(iv) P " Q t R r
(A) (i) only
(B) (i) and (iii) only
(C) (ii) and (iii) only (D) (iii) and (iv) only SOLUTION (I) P " QR is not possible since two NT should include one operator as Terminal. (II) Correct (III) Again incorrect. (IV) Correct. Hence (B) is correct option. 16. Consider a program P that consists of two source modules M1 and M2 a reference to a function contained in two different files. If M1 contains defined in M2, the reference will be resolved at (A) Edit-time (B) Compile-time (C) Link-time
(D) Load-time
SOLUTION The two modules needed to be linked since definition exist & M2 & M1 refers it. So during linking phase M1 links to M2. Hence (C) is correct option. 17. Consider the grammar rule E " E1 − E2 for arithmetic expressions. The code generated is targeted to a CPU having a single user register. The subtraction operation requires the first operand to be in the register. If E1 and E2 do not have any common sub expression, in order to get the shortest possible code (A) E1 should be evaluated first (B) E2 should be evaluated first (C) Evaluation of E1 and E2 should necessarily be interleaved (D) Order of evaluation of E1 and E2 is of no consequence
SOLUTION E1 is to be kept in accumulator & accumulator is required for operations to evaluate E2 also. So E2 should be evaluated first & then E1, so finally E1 will be in accumulator, otherwise need to use move & load instructions. 76
Hence (B) is correct option. 18. Consider the grammar with the following translation rules and E as the start symbol. E " E 1 #T "
value = .value * .value} .value = .value} .value = .value + .value} .value = .value}
"num .value =num.value} Compute E . value for the root of the parse tree for the expression: 2 # 3 # & 5 # 6 & 4. (A) 200 (B) 180 (C) 160
(D) 40
SOLUTION The parse tree would be.
Now we evaluate bottom up
77
LL " left to right left most derivation no ambignity should be there SLR or LR(0) L to R reverse right sentential form create LR(0) items. CLR or LR(1) create LR(1) items no bound LALR reduced CLR if while reducing any conflict found then not LALR
Hence (C) is correct option. 19. The grammar A " AA |( A)| ε is not suitable for predictive-parsing 78
because the grammar is (A) ambiguous
(B) Left-recursive
(C) right-recurisve
(D) an operator-grammar
SOLUTION The grammar is definitely left & right recursive but it is not suitable for predictive parsing because it is ambiguous. Hence (A) is correct option. 20. Consider the grammar E " E + n | E # n | n For a sentence n + n, the handles in the right-sentential form of the reduction are (A) n, E + n and E + n # (C) n , n + n and n + n # n
n
B) n , E + n and E + E # n (D) n , E + n and E # n
SOLUTION Given grammar E "E+n E "E#n E "n String = n + n # n Right sentential so right most non terminal will be used. E"E#n {E " E # n} E+n#n {E " E + n} n+n#n {E " n} So during reduction the order is reverse. So {E " n , E " E + n, E " E # n} Hence (D) is correct option. 21. Consider the grammar S " (S)| a Let the number of states in SLR(1), LR(1) and LALR(1) parsers for the grammar n1 n2 and n3 respectively. The following relationship holds good (A) n1 < n2 < n3
(B) n1 = n3 < n2
(C) n1 = n2 = n3
(D) n1 $ n3 $ n2
SOLUTION The no. of states for SLR(1) & LALR(1) are equal so n 1 = n3, but CLR(1) or LR(1) will have no. of states greater than LALR & LR(0) both. 79
Hence (B) is correct option. 22. Consider line number 3 of the following C-program.
Identify the compiler’s response about this line while creating the object-module (A) No compilation error (B) Only a lexical error (C) Only syntactic errors (D) Both lexical and syntactic errors SOLUTION There are no lexical errors for C because all the wrong spelled keywords would be considered as identifiers until the syntax is checked. So the compiler would give syntax errors. Hence (C) is correct option. Data for Q. 23 & 24 are given below. Solve the problems and choose the correct answers. Consider the following expression grammar. The semantic rules for expression calculation are stared next to each grammar production. E " number E E # E
Eval E .val
number val E .VAL
E .val
E .val
E .VAL
E .val
; 23. The above grammar and the semantic rules are fed to a yacc tool (which is an LALR(1) parser generator) for parsing and evaluating arithmetic expressions. Which one of the following is true about the action of yacc for the given grammar? (A) It detects recursion and eliminates recursion (B) It detects reduce-reduce conflict, and resolves (C) It detects shift-reduce conflict, and resolves the conflict in favor of a shift over a reduce 80
action (D) It detects shift-reduce conflict, and resolves the conflict in favor of a reduce over a shift action SOLUTION Yace tool is used to create a LALR(1) parser. This parser can detect the conflicts but to resolve the conflicts it actually prefers shift over reduce action. Hence (C) is correct option. 24. Assume the conflicts part (a) of this question are resolved and an LALR(1) parser is generated for parsing arithmetic expressions as per the given grammar. Consider an expression 3 # 2 + 1. What precedence and associativity properties does the generated parser realize? (A) Equal precedence and left associativity; expression is evaluated to 7 (B) Equal precedence and right associativity, expression is evaluated to 9 (C) Precedence of 'x' is higher than that of ‘+’, and both operators are left associative; expression is evaluated to 7 (D) Precedence of ' # ' is higher than that of ‘#’, and both operators are left associative; expression is evaluated to 9
SOLUTION The grammar has equal precedence and it is also ambiguous. Since LALR(1) parser prefer shift over reduce so + operation will be executed here before ). 2 + 1 = 3 & 3 # 3 = 9 also the operators are right associative. Hence (B) is correct option. 25. Consider the following grammar. S "S*E S "E E"F+EE"F F " id Consider the following LR(0) items corresponding to the grammar above. (i)
S " S * .E
(ii)
E"F.+E
(iii)
E " F + .E
Given the items above, which two of them will appear in the same set in the canonical sets-ofitems for the grammar? 81
(A) (i) and (ii)
(B) (ii) and (iii)
(C) (i) and (iii)
(D) None of these
SOLUTION If S " S ): E is in LR(0) then E " F +: E will also be there because both of them has ' : ' before E . Hence (C) is correct option. 26. Consider the following grammar S " FR R " * S | ε F " id In the predictive parser table, M , of the grammar the entries M [ S, id] and M [ R,$] respectively (A) {S " FR} and {R " ε} (C) {S " FR} and {R " * S}
(B) {S " FR} and {} (D) {F " id} and {R " ε}
SOLUTION The predictive parser table is given as. Non Terminal
)
id
S
S " FR
F
F " id
R So at
R ") S R "! M [ S, id] = S {
M [ R,$] = {R Hence (A) is correct option.
$
R "! " FR} "!}
27. Consider the following translation scheme. S " ER R " * E{print{’ * ’);R | f E " F + E{print(’ + ’); | F F " (S) | id{print(id.value);} Here id is a taken that represents an integer and id. value represents the corresponding integer value. For an input ‘2 * 3 + 4’, this translation scheme prints (A) 2 * 3 + 4 (B) 2 * + 3 4 (C) 2 3 * 4 +
(D) 2 3 4 + * 82
SOLUTION Input string 2 ) 3 + 4 S " ER FR idR {print(2)} id)ER {print())} id) F+ER {print(+)}id) id + ER {print(3)} id) id ) id +id So 2 )+ 3 4 are printed Hence (B) is correct option. 28. Consider the following C code segment. for for if i #i } } } Which one to the following false? (A) The code contains loop-in variant computation (B) There is scope of common sub-expression elimination in this code (C) There is scope strength reduction in this code (D) There is scope of dead code elimination in this code
SOLUTION All the statements are true except option (D) since there is no dead code to get eliminated. Hence (D) is correct option. 29. Which one of the following grammars generates the language L = (a i b i | i ! j}?
(A)
S " AC | CB C " aCb | a | b A " aA | ε B " Bb | ε
(B) S " aS | Sb | a | b
(C)
S " ACCB
(D)
S " AC | CB 83
C " aCb |! A " aA |! B " Bb |!
C " aCb |! A " aA | a B " bB | b
SOLUTION The grammar S " AC CB C "aCb ! A "aA a B " bB b Consider string aaabb S " AC AaCb AaaCbb Aaabb aaabb But string aabb S " AC And this string is not derivable. Hence (D) is correct option. 30.In the correct grammar above, what is the length of the derivation (number of steps starting from S to generate the string al bm with l ! m? (A) max (l, m) + 2 (B) l+m+2 (C) l + m + 3
(D) max (l, m) + 3
SOLUTION It is very clear from the previous solution that the no. of steps required depend upon the no. of a' s & b ' s which ever is higher & exceeds by 2 due to S " AC CB & C "! So max(l , m) + 2 Hence (A) is correct option. 31. Which one of the following is a top-down parser? (A) Recursive descent parser (B) Operator precedence parser (C) An LR(k) parser
(D) An LALR(k) parser
SOLUTION Clearly LR & LALR are not top down they are bottom up passers. Also not operator precedence parser. ut yes recursive descent parser is top down parser. Starts from start symbol & derives the terminal string. Hence (A) is correct option. 84
32.Consider the grammar with non-terminals N = {S , C , S}, terminals T = {a, b , i , t, e}, with S as the start symbol, and the following of rules S " iCtSS1 | a S1 " eS | ε C"b The grammar is NOTLL(1) because: (A) It is left recursive (B) It is right recursive (C) It is ambiguous
(D) It is not context-free
SOLUTION The grammar has production S " iCtSS1 here the right hand side of grammar has the same symbol as left side. So the grammar is left recursive. The grammar is not ambiguous. Hence (A) is correct option. 33.Consider the following two statements: P: Every regular grammar is LL(1) Q: Every regular set has LR(1) grammar Which of the following is TRUE? (A) Both P and Q are true
(B) P is true and Q is false
(C) P is false and Q is true
(D) Both P and Q are false
SOLUTION LL(1) parsers can recognize the regular grammars also LL(1) is subset of LR(1) or CLR grammar so it also recognizes regular sets. So both accept regular grammar. 34. In a simplified computer the instructions are: OP R j , Ri − Performs Rj OP Ri and stores the result in register Ri OP m, Ri − Performs val OP Ri abd stores the result in Ri. value denotes the content of memory location m. MCVm, Ri −Moves the content off memory loction m to register Ri . MCVm, Ri , m −Moves the content of register Ri to memory location m.
The computer has only two registers, and OP is either ADD or SUB. Consider the following basic block: 85
t1 = a + b t2 = c + d t 3 = e − t2 t 4 = t 1 − t2 Assume that all operands are initially in memory. The final value of the computation should be in memory. What is the minimum number of MOV instructions in the code generated for this basic block? (A) 2
(B) 3
(C) 5
(D) 6
SOLUTION The operation sequence would be MOV a, R1 ADD b , R1 {R 1 = t1 MOV c , R2 ADD d, R2 {R 2 = t2 SUB e , R2 {t 3 = e − R 2 = R2 SUB R 1, R2 {R 2 = t4 MOV R 2, t4 {finally in memory Totally no. of move operation are 3 Hence (B) is correct option Data for Q. 35 & 36 are given below. Solve the problems and choose the correct answers. Consider the CFG with {S, A, B} as the non-terminal alphabet, {a, b} as the terminal alphabet, S as the start symbol and the following set of production rules
35.Which of the following strings is generated by the grammar? (A) aaaabb
(B) aabbbb
(C) aabbab
(D) abbbba
SOLUTION aabbab 86
S " aB " aaBB " aabSB " aabbAB " aabbab Hence (C) is correct option. 36.For the correct answer string to Q. 9 how many derivation trees are there? (A) 1
(B) 2
(C) 3
(D) 4
SOLUTION For the derivation two trees are possible
So due to ambiguity 2 trees are possible Hence (B) is correct option.
37. Which of the following describes a handle (as applicable to LR-parsing) appropriately? (A) It is the position in a sentential form where the next shift or reduce operation will occur (B) It is a non-terminal whose production will be used for reduction in the next step (C) It is a production that may be used for reduction in a future step along with a position in the sentential form where the next shift or reduce operation will occur. (D) It is the production p that will be used for reduction in the next step along with a position in the sentential form where the right hand side of the production may be found SOLUTION Handles are the part of sentential form, & they are identified as the right side of any given production which will be used for reduction in the net step. 87
Hence (D) is correct option. 38.Some code optimizations are carried out on the intermediate code because (A) They enhance the portability of the complier to other target processors (B) Program analysis is name accurate on intermediate code than on machine code (C) The information from data flow analysis cannot otherwise be used for optimization (D) The information from the front end cannot otherwise be used for optimization SOLUTION Code optimizations are carried out on the intermediate code because program analysis is more accurte on intermediate code than on machine code. Hence (B) is correct option. 39. Which of the following are true? (i) A programming language option does not permit global variables of any king and has no nesting of procedures/functions, but permits recursion can be implemented with static storage allocation (ii) Multi-level access link (or display) arrangement is needed to arrange activation recordsonly if the programming language being implemented has nesting of procedures/function (iii) Recursion in programming languages cannot be implemented with dynamic storage allocation (iv) Nesting of procedures/functions and recursion require a dynamic heap allocation scheme and cannot be implemented with a stack-based allocation scheme for activation records (v) Languages which permit a function to return a function as its result cannot be implemented with a stack-based storage allocation scheme for activation records (A) (ii) and (v) only
(B) (i), (iii) and (iv) only
(C) (i), (ii) and (v)
(D) (ii), (iii) and (v) only
SOLUTION I.
Statement is false since global variables are required for recursions with static storage. This is due to unavailability of stack in static storage. II. This is true III. In dynamic allocation heap structure is used, so it is false. IV. False since recursion can be implemented. V. Statement is completely true. So only II & V are true. Hence (A) is correct option. 88
40. An LALR(1) parser for a grammar can have shift-reduce (S-R) conflicts if and only if (A) The SLR(1) parser for G has S-R conflicts (B) The LR(1) parser for G has S-R conflicts (C) The LR(0) parser for G has S-R conflicts (D) The LALR(1) parser for G has reduce-reduce conflicts SOLUTION LALR parser is reduced form of CLR or LR(1) parser, LALR parser uses the LR(1) items of CLR parser & of any shift reduce conflicts are there then it is due to LR(1) parser. Hence (B) is correct option. 41. Which of the following statements are TRUE ? There exist parsing algorithms 3for some programming languages hose complex are less than θ(n ) II A programming language which allows recursion can be implemented with static storage allocation III No L-attributed definition can be evaluated in the framework of bottom-up parsing IV Code improving transformations can be performed at both source language and intermediate code level (A) I and II
(B) I and IV
(C) III and IV
(D) I, III and IV
SOLUTION I.
Statement is true since there are some parsers which take 0(n log2n) time for parsing. II. Completely false, since there is no use of stack which is required for recursion. III. False IV. True since both types of optimizations are applied Hence (B) is correct option. 42.What data structure in a complier is used for managing information about variables and their attributes? (A) Abstract syntax tree (B) Symbol table (C) Semantic stack
(D) Parse table
SOLUTION Symbol table is used for storing the information about variables and their attributes by 89
compiler. Hence (B) is correct option. 43. Which languages necessarily need heap allocation in the runtime environment ? (A) Those that support recursion (B) Those that use dynamic scoping (C) Those that allow dynamic data structure (D) Those that use global variables
SOLUTION Dynamic memory allocation is maintained by heap data structure. So to allow dynamic data structure heap is required. Hence (C) is correct option.
OPERATING SYSTEMS YEAR 2001 Question. 1 Which of the following statements is false ? (A)Virtual memory implements the translation of a program’s address space into physical memory address space. (B)Virtual memory allows each program to exceed the size of the primary memory. (C)Virtual memory increases the degree of multi-programming (D)Virtual memory reduces the context switching overhead.
SOLUTION Virtual memory enables a program to exceed the size of primary memory so it increases degree of multi-programming. Since data required by executing program is available here so context switching is reduced. But virtual memory doesn’t translate program’s address space into physical memory. Hence (A) is correct option. Question. 2
90
Consider a set of n tasks with known runtimes r1,r2,........rn to be run on a uniprocessor machine. Which of the following processor scheduling algorithms will result in the maximum throughput ? (A) Round-Robin (C) Highest-Response-Ratio-Next SOLUTION
(B) Shortest-Job-First (D) First-come-First-Served
Here the running times r1....rn are already known, single processor system. In this scenario, throughput i.e. CPU is maximum utilized in shortest job first scheduling. Hence (B) is correct option. Question. 3 Where does the swap space reside ? (A) RAM (C) ROM
(B) Disk (D) On-chip cache
SOLUTION Swap space is the memory space where the part of the program not currently in main memory for execution is stored, this program part can be swapped into memory when required. This space is generally in disk. Hence (B) is correct option. Question. 4 Consider a virtual memory system with FIFO page replacement policy. For an arbitrary page access pattern, increasing the number of page frames in main memory will. (A)Always decrease the number of page faults (B)Always increase the number of page faults (C)Sometimes increase the number of page faults (D)Never affect the number of page faults
SOLUTION During F1F0 page replacement policy, due to increase in the no. of page frames in memory should decrease the no. of page faults since more frames can be kept there. But due to Belady’s Anomaly after certain limit or in some page access instances no. of page faults are high. 91
Hence (C) is correct option. Question. 5 Consider a machine with 64 MB physical memory and a 32-bit virtual address space. If the page size is 4 KB, what is the approximate size of the page table ? (A) 16 MB (C) 2 MB
(B) 8 MB (D) 24 MB
SOLUTION Size of main memory = 64 MB Size of virtual memory = 232 B No. of pages = 232/212 = 220 pages Required 1 M enteries. But each entry has both a virtual address & corresponding physical address. Total bits in each entry = 32 +26 (Physical) = 58
= 58/8 = 8 Bytes. So total memory = 8*1 MB= 8 MB Hence (B) is correct option. Question. 6 Consider Peterson’s algorithm for mutual exclusion between two concurrent processes i and j. The program executed by process is shown below. repeat flag[i]=true; turn=j; while(p)do no-op; Enter critical section, perform actions, then exit critical section Flag[i]=false; Perform other non-critical section actions. Until false;
For the program to guarantee mutual exclusion, the predicate P in the while loop should be (A) flag [j]= true and turn =j (C) flag [i]=true and turn=j
(B) flag [j]=true and turn =j (D) flag [i]=true and turn=i
SOLUTION 92
While loop if true predicate then the program enters into critical region. This program enters into critical region of flag [i]=true act as semaphore, & true =j, the requirement of resource is by some other process. Hence (B) is correct option. YEAR 2002 Question. 7 Which of the following scheduling algorithms is non-preemptive ? (A) Round Robin (B) First-In First-Out (C)Multilevel Queue Scheduling (D)Multilevel Queue Scheduling with Feedback
SOLUTION Round robin is preemptive since processes are cycled for CPU time, & run for a particular time stamp in one cycle. Multilevel queue scheduling maintains various quenes, each having different priorities. But in FIFO scheme, only the process which enters once, would be completed first, so no. preemption. Hence (B) is correct option.Question. 8 The optimal page replacement algorithm will select the page that (A)Has not been used for the longest time in the past. (B)Will not be used for the longest time in the future. (C)Has been used least number of times. (D)Has been used most number of times
SOLUTION Optimal page replacement algorithm assumes that the pages that will come in future are already known, so replacement of the page which will not be used in future occurs. Hence (B) is correct option. Question. 9 Which combination of the following features will suffice to characterize an OS as a multiprogrammed OS ? More than one program may be loaded into main memory at the same time for execution. (B) If a program waits for certain events such as I/O, another program is immediately scheduled for execution. (C) If the execution of a program terminates, another program is immediately scheduled for execution. 93
(A) A (C) A and C
(B) A and B (D) A, B and C
SOLUTION Multi-programmed:- More than one program can run on single CPU, when one is blocked. (A)Is true and a characteristic of multi-programmed (B)Is true & also characterize a multi-programmed OS (C)Is true but no necessary for this type this happens in all OS, even in batch processor. Hence (B) is correct option.
Question. 10 In the index allocation scheme of blocks to a file, the maximum possible size of the file depends on (A)The size of the blocks, and the size of the address of the blocks (B)The number of blocks used for the index, and the size of the blocks. (C)The size of the blocks, the number of blocks used for the index, and the size of the address of the blocks. (D)None of the above.
SOLUTION When indexes are created, the maximum no. of blocks given to a file are totally dependent upon size of the index which tells how many blocks can be there, & size of each block. Hence (B) is correct option. YEAR 2003 Question. 11 Using a larger block size in a fixed block size file system leads to (A)better disk throughput but poorer disk space utilization (B)better disk throughput and better disk space utilization (C)poorer disk throughput but better disk space utilization (D)poorer disk throughput and poorer disk space utilization SOLUTION Using larger block size in a fixed block size system lead to poor disk space utilization due to data items which are very small comparable to block size cause fragmentation. But it leads to better disk through put since no. of blocks needs to fetch & replace become less. 94
Hence (A) is correct option. Question. 12 In a system with 32 bit virtual addresses and 1 KB page size, use of one-level page tables for virtual to physical address translation is not practical because of (A)the large amount of internal fragmentation (B)the large amount of external fragmentation (C)the large memory overhead in maintaining page tables (D)the large computation overhead in the translation process
SOLUTION 32 bit virtual address, i.e. 232 kB of virtual memory & 1 kB page size. So total pages = 232. So. we need to maintain a page table of 232 rows, this require 4 GB main memory which is quite impractical due to large memory overhead. Hence (C) is correct option. Question. 13 A uni-processor computer system only has two processes, both of which alternate 10 ms CPU bursts with 90 ms I/O bursts. Both the processes were created at nearly the same time. The I/O of both processes can proceed in parallel. Which of the following scheduling strategies will result in the least CPU utilizations (over a long period of time) for this system ? (A)First come first served scheduling (B)Shortest remaining time first scheduling (C)Static priority scheduling with different priorities for the two processes (D)Round robin scheduling with a time quantum of 5 ms.
SOLUTION There should be no doubt that round robin scheduling would lead to maximum CPU utilization, but since in FCFS one task would starve for a long time so min CPU utilization would be in this case. Hence (A) is correct option. Data for Q. 14 & 15 are given below. A processor uses 2-level page table fro virtual to physical address translation. Page table for both levels are stored in the main memory. Virtual and physical addresses are both 32 bits wide. The memory is byte addressable. For virtual to physical address translation, the 10 most 95
significant bits of the virtual address are used as index into the first level page table while the next 10 bits are used as index into the second level page table. The 12 least significant bits of the virtual address are used as offset within the page. Assume that the page table entries in both levels of page tables are 4 a bytes wide. Further, the processor has a translation look aside buffer(TLB), with a hit rate of 96%. The TLB caches recently used virtual page numbers and the corresponding physical page numbers. The processor also has a physically addressed cache with a bit ratio of 90%. Main memory access time is 10 ns, cache access time is 1 ns, and {LB access time is also 1ns. Question. 14 Assuming that no page faults occur, the average time taken to access a virtual address is approximately (to the nearest 0.5 ns) (A) 1.5 ns (C) 3 ns
(B) 2 ns (D) 4 ns
SOLUTION TLB is successfully 96% of total request & for remaining 4%. RAM is accessed twice. So average time taken. =0.96(1 +(0.9*1) +0.1*(1 +10))+0.04(21 +(0.9*0.1)) +0.1*(1 +10) =.96(1 +.9 +1.1) +0.4(21 +.09 +1.1) =.96*3 +0.4*23 =2.88 +0.92 =3.80≈4 ns (Nearest .5) Hence (D) is correct option.
Question. 15 Suppose a process has only the following pages in its virtual address space; two contiguous code pages starting at virtual address 0x0000000, two contiguous data pages starting at virtual address 0x00400000, and a stack page starting at virtual address 0xFFFFF000. The amount of memory required for storing the page tables of this process is (A)8 KB
(B) 12KB
(C) 16 KB
(D) 20KB
SOLUTION Total no. of pages required = 5 But due to 2 level page table = 2*4 Kb*2 =16 kB Hence (D) is correct option. Data for Q. 16 & 17 are given below.
96
Suppose we want to synchronize two concurrent processes P and Q using binary semaphores S and T. The code for the processes P and Q is shown below. Process P while(1) { W: print ‘0’; print ‘0’; X: }
Process Q: while(1) { Y: print ‘1’ print ‘1’ Z: }
Synchronization statements can be inserted only at points W, X, Y and Z. Question. 16 Which of the following will always lead to an output staring with ‘001100110011’? (A) P(S) at W, V(S) at X, P(T) at Y, V(T) at Z, S and T initially 1 (B) P(S) at W, V(T) at X, P(T) at Y, V(S) at Z, S initially 1, and T initially 0 (C)P(S) at W, V(T) at X, P(T) at Y, V(S) at Z, S and T initially 1 (D)P(S) at W, V(T) at X, P(T) at Y, V(S) at Z, S initially 1, and T initially 0
SOLUTION For output string 001100110011 alternatingly we require process P & Q to execute. For this to happen P(s) with S = 1 should be placed at W. At the same time P(T) with T = 0 will be at Y. At X we have V(T) which will have T = 1 so process Q starts. At the same time at Z we have V(s) which make S = 0 to stop process P. Hence (B) is correct option.
Question. 17 Which of the following will ensure that the output string never contains a substring of the form 0.1” or 10”1 where n is odd? (A)P(S) at W, V(S) at X, P(T) at Y, V(T) at Z, S and T initially 1 (B)P(S) at W, V(T) at X, P(T) at Y, V(S) at Z, S and T initially 1 (C)P(S) at W, V(S) at X, P(T) at Y, V(S) at Z, S initially 1 (D)(S) at W, V(T) at X, P(T) at Y, P(S) at Z, S and T initially 1
SOLUTION To ensure this condition that substring of form 01n0 or 10n1, where n is odd S should be initially 1, we will case only 1 semaphore S. So at W P(s), at X V(s) whereas at Y P(s), at Z V(s) 97
Hence (C) is correct option. YEAR 2004 Question. 18 Consider the following statements with respect to user-level threads and kernel-supported threads (i)Context which is faster with kernel-supported threads (ii)For user-level threads. a system call can block the entire process (iii)Kernel-supported threads can be scheduled independently (iv)User-level threads are transparent to the kernel Which of the above statements are true?
(A) (ii),(iii) and (iv) only (C) (i) and (iii) only
(B) (ii) and (iii) only (D) (i) and (ii) only
SOLUTION (I)It is false, context switch is not faster in support of kernel threads. (II)A system call can truly block the user level threads, since they don’t have permission to do that. (III)True since kernel supported threads have their independent memory & resources. (IV) False since user level threads might need support of kernel threads. Hence (B) is correct option.
Question. 19 Consider an operating system capable of loading and executing a single sequential user process at a time. The disk head scheduling algorithm used is First Come First Served (FCFS). If FCFS is replaced by shortest seek Time Fist (SSTF), claimed by the vendor to given 50% better beachmark results, what is the expected improvement in the I/O performance of user programs? (A) 50% (C) 25%
(B) 40% (D) 0%
SOLUTION I/O performance is not entirely dependent upon disk access, it has effect of various other devices, so using SSTF in place of FCFS may reduce disk access time but no improvement in the I/O is done. Hence (D) is correct option. Question. 20 The minimum number of page frames that must be allocated to a running process in a virtual memory environment is determined by (A)the instruction set architecture 98
(B)page size (C)physical memory size (D)number of processes in memory
SOLUTION Page frames are allocated in main memory, for virtual memory pages. This no. of page frames depends upon the instruction set architecture. Hence (A) is correct option. Question. 21 Consider the following set of processes, with the arrival times and the CPU-burst times given in milliseconds
Process P1 P2 P3 P4
Arrival Time 0 1 2 4
Burst Time 5 3 3 1
What is the average turnaround time for these processes with the preemptive shortest remaining processing time first (SRPT) algorithm? (A) 5.50 (C) 6.00
(B) 5.75 (D) 6.25
SOLUTION Then around time = (Submit time −finish time) Gantt chart for the scheduler.
T.A time for P1 = 12 −0 = 12 P2 = 4 −1 = 3 P3 = 8 −2 = 6 99
P4 = 5 −4 = 1 Total = 22 Average turnaround time = 22/4 = 5.5 Hence (A) is correct option.
Question. 22 Consider a system with a two-level paging scheme in which a regular memory access takes 150 nanoseconds, and servicing a page fault takes 8 milliseconds. An average instruction takes 100 nanoseconds of CPU time, and two memory accesses. The TLB hit ratio is 99%, and the page fault rate is one in every 10,000 instructions. What is the effective average instruction execution time? (A) 645 nanoseconds (C) 1215 nanoseconds
(B) 1050 nanoseconds (D) 1230 nanoseconds
SOLUTION Memory access time =0.90*150 +0.10*(150 +150) =135 +30 =165 ns The error rate = 1/10000 = 10−4 CPU burst time = 100 ns Total execution time =[100 +2[165] +10−4*8*106] =100 +330 +800 =100 +1130 =1230 ns Hence (D) is correct option.
Question. 23 Consider two processes P1 and P2 accessing the shared variables X and Y protected by two binary semaphores Sx and Sy respectively, both initialized to 1. P and V denote the usual semaphore operators, where P decrements the semaphore value, and V increments the semaphore value. The pseudo-code of P1 and P2 is as follows: P1: { while true do L1:……… L2:……… X = X+1; Y = Y-1; V(Sx); V(Sy); }
P 2: { while true do L3:……… L4:……… Y = Y+1; X = Y-1; V(Sy); V(Sx); } 100
In order to avoid deadlock, the correct operators at L1, L2, L3 and L4 are respectively (A) P(Sy),P(Sx);P(Sx),P(Sy) (C) P(Sx),P(Sx);P(Sy),P(Sy)
(B) P(Sx),P(Sy);P(Sy),P(Sx) (D) P(Sx),P(Sy);P(Sx),P(Sy)
SOLUTION Here semaphores are required to obtain mutual exclusion since both access X & Y . So at L1 P(Sx) which means now Sx = wait at L2P(Sy) Sy wait, this prevents process P2 to start access X &Y . V(Sx) & V(Sy) in the end of P1 makes Sx & Sy signal so that at L3 & L4 P(Sx) & P(Sy) can start. Hence (D) is correct option. Question. 24 A Unix-style I-node has 10 direct pointers and one single, one double and one triple indirect pointers. Disk block size is 1 Kbyte, disk block address is 32 bits, and 48-bit integers are used. What is the maximum possible file size? (A) 224 bytes
(B) 232 bytes
(C) 234 bytes
(D) 248 bytes
SOLUTION Size of 1 block = 1 kB Block addresses size 1 pointer size = 32 bit = 4 bytes. So, no. of pointers in= 210B/ 22B 1 block = 256 So direct pointer will have = 10*1kB = 10kB Double will have = 256*256*1 kB Triple will have = 256*256*256*1 kB = 28*28*28*210 B = 234 B Hence (C) is correct option. YEAR 2005
Question. 25 Suppose n processes, P1……Pn share m identical resource units, which can be reserved and released one at a time. The maximum resource requirement of process Pi is sp where si<0. Which one of the following is a sufficient condition for ensuring that deadlock does not occur? (A) i, s < m n (C) ∑ si < (m +n) i=1
(B) i,s,< n n (D) ∑ si < (m * n) i=1 101
SOLUTION For every Pi Si is maximum resource requirement where Si > 0. To allot resources to all processes without any deadlock situation is n ∑ si < (m +n) i=1 i.e. sum of all maximum resource requirement should be less than m +n. Hence (C) is correct option.
Question. 26 Consider the following code fragment: if (fork()==0) {a = a +5; printf("%d,%/n"a, and a);} else {a −5; printf("%d,%d/n",a,&a);} let u, v be the values printed by the parent process, and x,y be the values printed by the child process. Which one of the following is TRUE? (A) u = x+10 and v = y (B) u = x+10 and v≠y (C) u+10 = x and v = y (D) u+10 = x and v≠ y SOLUTION Initial value of a is let 10 and its address &a would be different for both parent & child process so. (A) & (B) are incorrect also parent process executes a = a −5 = 5 = u & child executes a = a +5 = 15 = x so u +10 = x Hence (D) is correct option. YEAR 2006 Question. 27 Consider three CPU-intensive processes, which require 10,20 and 30 time units and arrive at times 0,2, and 6, respectively. How many context switches are needed if the operating system 102
implements a shortes remaining time first scheduling algorithm? Do not count the context switches at time zero and at the end (A) 1
(B) 2
(C) 3
(D) 4
SOLUTION When CPU burst are given to another process, called context switching. The Gantt chart for shortest remaining time first is.
So there are two context & witches at T = 10 & T = 30 Hence (B) is correct option. Question. 28 The atomic feth-and-set x,y instruction unconditionally sets the memory location x to 1 and fetches the old value of x in y without allowing any intervening access to the memory location x. Consider the following implementation of P and V functions on a binary semaphore S. void p (binary_semaphore*S){ unsigned y; unsigned*x =& (S->value);} do { fetch-and-set x,y; } while(y); } void V (binary_semphore*S){ S_>value = 0; } Which one of the following is true? (A)The implementation may not work if context switching is disabled in P (B)Instead of using fetch-and-set, a pair of normal load/store can be used (C)The implementation of V is wrong (D)The code does not implement a binary semaphore 103
SOLUTION If there are more than two processes and context & switching processes is disabled in P then this implementation doesn’t work properly and can’t synchronize the processes. Hence (A) is correct option. Question. 29 A CPU generates 32-bit virtual addresses. The page size is 4 KB. The processor has a translation look-aside buffer (TLB) which can hold a total of 128 page table entries and is 4way set associative. The minimum size of the TLB tag is (A)11bits
(B) 13bits
(C) 15bits
(D) 20bits
SOLUTION TLB has 128 page table enteries, each page table would have. 64 bits i.e. 32 +32 virtual addresses. So total memory required. 27*26 But 4 way set associative. 27*26/22 = 211 So 11 bits are required. Hence (A) is correct option. Question. 30 A computer system supports 32-bit virtual addresses as well as 32-bit physical addresses, Since the virtual address space is of the same size as the physical address space, the operating system designers decide to get rid of the virtual entirely. Which one of the following is true? (A)Efficient implementation of multi-user support is no longer possible (B)The processor cache organization can be made more efficient now (C)Hardware support for memory management is no longer needed (D)CPU scheduling can be made more efficient now 104
SOLUTION Since both virtual and physical memory has 32 bit addresses, so there is no need for address translation hardware is required. Hence (C) is correct option. Question. 31 Consider three processes (process id 0,1,2, respectively) with compute time bursts 2,4, and 8 time units. All processes arrive at time zero. Consider the longest remaining time first (LRTF) scheduling algorithm. In LRTF ties are broken by giving priority to the process with the lowest process id. The average turn around time is (A) 13 units
(B) 14 units
(C) 15 units
(D) 16 units
SOLUTION Process id
0
1
2
CPU Burst
2
4
8
So we draw Gantt chart for scheduler
At t = 0 longest remaining time for P2 At t = 4 remaining times for both P1 & P2 is 4 so P1 is given priority. At t = 8 remaining time for P0P1 & P2 is 2. P0 is given priority & cyclically done till t = 14. Process P0 P1 P2
TAT 12 −0 = 12 13 −0 = 13 14 −0 = 14 39/3=13 units
Hence (A) is correct option. Question. 32 105
Consider three processes, all arriving at time zero, with total execution time of 10, 20 and 30 units, respectively. Each process spends the first 20% of execution time doing I/O, the next 70% of time doing computation, and the last 10% of time doing I/O again.The operating system uses a shortest remaining compute time first scheduling algorithm and schedules a new process either when the running process get blocked on I/O or when the running process finishes its compute burst. Assume that all I/O operations can be overlapped as much as possible. For what percentage of time does the CPU remain idle? (A) 0%
(B) 10.6%
(C) 30.0%
(D) 89.4%
SOLUTION Process 2 3
First I/O 2 4 6
Computation 7 14 21
Second I/O 1 2 3
Total 10 20 30
Since I/0 can be done parallely. Gantt chart
Total time taken = 51 units CPU has to wait = 6 units =(6/51)*100 =10.6% Hence (B) is correct option.
Question. 33 106
Consider the following snapshot of a system running n processes. Process i is holding xi instances of a resource R, for 1≤i≤n. Currently, all instances of R are occupied. Further, for all i, process i has placed a request for an additional y, instances while holding the xi instances it already has, There are exactly two processes p and q such that yp = yq = 0: Which one of the following can serve as a necessary condition to guarantee that the system is not approaching a deadlock? (A) min(xp,xq) < maxk≠p,q yk (B) xp+xq ≤ maxk≠p,q yk (C) min(xp,xq) < 1 (D) min(xp,xq)>1 SOLUTION Here option (B) is that min (xp,xq) < maxk≠ p,q yk Means the min no. of resources allocated should be less than the maximum number of resources required by any process other than p & q It prevent from deadlock. Hence (B) is correct option. Data for Q. 34 & 35 are given below. Barrier is a synchronization construct where a set of processes synchronizes globally i.e. each process in the set arrives at the barrier and waits for all others to arrive and then all processes leave the barrier. Let the number of processes in the set be three and S be a binary semaphore with the usual P and V functions. Consider the following C implementation of a barrier with line numbers shown on the left. Void barrier(void) { 1 :P(S) 2 :Process_arrived++; 3 :V (S) : 4 :while (process_arrived’=3); 5 :P(S); 6 :Process_left++; 7 :if(process_left==3) 8 :process_arrived=0; 9 :process_left+0; 10 :} 11 :V(S); } The variable process_arrived and process_left are shared among all processes and are initialized to zero. In a concurrent program all the three processes call the barrier function when they need to synchronize globally. Question. 34 107
The above implementation of barrier is incorrect. Which one of the following is true? (A)The barrier implementation is wrong due to the use of binary semaphore S (B)The barrier implementation may lead to a deadlock if two barrier invocations are used in immediate succession (C)Lines 6 to 10 need not be inside a critical section (D)The barrier implementation is correct if there are only two processes instead of three
SOLUTION This barrier implementation is to keep track of arrival & completion of processes in system, by incrementing no. of process arrived & left. This implementation may lead to deadlock if two barrier function’s invocations are used is immediate sessions. SinceV(s) in first invocation at line 3 removes 1-3 from critical section bring 5-11 in critical section at line 7. Hence (B) is correct option. Question. 35 Which one of the following rectifies the problem in the implementation? (A)lines 6 to 10 are simply replaced by process_arrived (B)At the beginning of the barrier the first process to enter the barrier waits until process_arrived becomes zero before proceeding to execute P(S) (C)Context switch is disabled at the beginning of the barrier and re-enabled at the end. (D)The variable process_left is made private instead of shared SOLUTION To rectify the barrier function at the beginning of the functions the first process which inters the function waits until process arrived becomes zero before executing P(s). This removes any deadlock situation. Hence (B) is correct option. YEAR 2007 Question. 36 Group-1 contains some CPU scheduling algorithms and group-2 contains some applications. 108
Match entries in Group-1 entries in Group-2 Group-1 P. Gang Scheduling Q. Rate Monotonic Scheduling R. Fair Share Scheduling (A)P-3 Q-2 R-1
(B) P-1 Q-2 R-3
Group-2 1.Guaranteed Scheduling 2. Real-time scheduling 3.Thread Scheduling
(C) P-2 Q-3 R-1
(D) P-1 Q-3 R-2
SOLUTION Rate monotonic scheduling is for real time processes, Gang scheduling is used to schedule group of program threads and fair share scheduling is a scheduling which guarantees a fair share of CPU burst to every competing process. Hence (A) is correct option. Question. 37 Consider the following statements about user level threads and kernel level threads. Which one of the following statements is FALSE? (A)Context switch time is longer for kernel level threads than for user level threads (B)User level threads do not need any hardware support (C)Related kernal level thread can be scheduled on different processors in a multiprocessor system (D) Blocking one kernel level thread blocks all related threads
SOLUTION Threading a method of executing small sub processes instead of single big process. This prevents thread from blocking. So blocking of one kernel level thread doesn’t block other related threads, they are unaffected. So (D) is false. Hence (D) is correct option. Question. 38 An operating system uses Shortest Remaining Time first (SRT) process scheduling algorithm. Consider the arrival times and execution times for the following processes Process P1 P2 P3 P4
Execution Time 20 25 10 15
Arrival Time 0 15 30 45 109
What is the total waiting time for process P2? (A) 5
(B) 15
(C) 40
(D) 55
SOLUTION Gantt chart for the processes.
P2 came at t = 15 Scheduled first time at t = 20 wait = 5. Wait between t = 30 & t = 40 due to short remaining time of new process P3. Wait now = 5 +10 = 15 Then complete at t = 55 Hence (B) is correct option.
Question. 39 A virtual memory system uses first In First Out (FIFO) page replacement policy and allocates a fixed number of frames to a process. Consider the following statements: P:Increasing the number of page frames allocated to a process sometimes increases the page fault rate. Q:Some program do not exhibit locality of reference. Which one of the following is TRUE? (A)Both P and Q are ture, and Q is the reason for P (B)Both P and Q are true, but Q is not the reason for P (C)P is false, but Q is true (D)Both P and Q are false
SOLUTION Due to Belady’s Anomaly, increasing the number of page frames allocated to a process sometimes increase the page fault rate so P is true. Also some program do not exhibit locality of reference, so both are true but Q is not reason for P. Hence (B) is correct option. 110
Question. 40 A single processor system has three resource types X,Y, and Z, which are shared by three processes. There are 5 units of each resource type. Consider the following scenario, where the column alloc denotes the number of units of each resource type allocated to each process, and the column request denotes the number of units of each resource type requested by a process in order to complete execution. Which of these processes will finish LAST? alloc XYZ P0 121 P1 201 P2 221
request XYZ 103 012 120
(A)P0
(B)P1
(C)P2
(D)None of the above,
since the system is in a deadlock SOLUTION Initially. Process P0 P1 P2
Allow XYZ 121 121 221
Request/need XYZ 103 012 120
Available XYZ 012
Here looking at available resources, only need of P1 can be completed. So P1 will execute & free 2 0 1 so now available XYZ = 203. This is need for P0 so at second no. P0 will execute & free XYZ = 121 so available 3 2 4 & in the end need of P2 is fulfilled. Hence (C) is correct option. Question. 41 Two processes, P1 and P2, need to access a critical section of code. Consider the following synchronization construct used by the processes: /* P1 */ /*P2*/ while (true) { while (true) { wants1=true; wants2 = true; while(wants2==true); while (wants1 == true); /* Critical /* Critical 111
Section
Section
*/ */ wants 1 = false; wants 2 = false; } } /* Remainder section*/ /*Remainder section*/ Here, wants 1 and wants 2 are shared variables, Which are initialized to false. Which one of the following statements is TRUE about the above construct? (A)It does not ensure mutual exclusion. (B)It does not ensure bounded waiting. (C)It requires that processes enter the critical section in strict alternation. (D)It does not prevent deadlocks, but ensures mutual exclusion
SOLUTION If P1 make wants 1 = true then P2 goes in critical section & vice versa so both together are implementing mutual exclusion but. Since both are accessing wants 1 & wants 2 concurrently 4 wants 1 is first captured by P1 so P2 will wait & P2 captures want 2 so P1 will have to wait. So a definite deadlock. Hence (D) is correct option. Data for Q. 42 & 43 are given below. A process has been allocated 3 page frames. Assume that none of the pages of the process are available in the memory initially. The process makes the following sequence of page references (reference string): 1,2,1,3,7,4,5,6,3,1. Question. 42 If optimal page replacement policy is used, how many page faults occur for the above reference string? (A) 7
(B)8
(C) 9
(D)10
SOLUTION Reference string 1, 2, 1, 3, 7, 4, 5, 6, 3, 1 Using optimal replacement policy, we will replace that page in memory which either will not be used or latest used in future.
112
*denotes page fault So no. of page faults are 7. Hence (A) is correct option. Question. 43 Least Recently Used (LRU) page replacement policy is a practical approximation to optimal page replacement. For the above reference string, how many more page faults occur with LRU than with the optimal page replacement policy? (A) 0
(B)1
(C) 2
(D)3
SOLUTION Instead of optimal policy if we use LRU then we don’t have future knowledge but we replace that page in memory which has been recently used in past. Order 1, 2,1, 3, 7, 4, 5, 6, 3,1
*denotes page faults. In this case no. of page fault = 9 In optimal strategy = 7 Difference = 9 −7 = 2 Hence (C) is correct option. 113
YEAR 2008 Question. 44 Which of the following system calls results in the sending of SYN packets? (A) socket
(B)bind
(C) listen
(D)connect
SOLUTION SYN packets are used for synchronization between sender & receiver, these packets are sent by sender during connect system call for synchronous connection. Hence (D) is correct option. Question. 45 The data block of a very large file in the Unix file system are allocated using (A) Contiguous allocation (B) Linked allocation (C) indexed allocation (D) an extension of indexed allocation
SOLUTION Generally a large file system for UNIX OS use indexed allocation, but for very large systems an extension of indexed allocation i.e. ext 2, ext 3 are used. Hence (D) is correct option. Question. 46 The P and V operations on counting semaphores, where s is a counting ewmaphore, are defined as follows: P(s);s = s −1; If s < 0 then wait; V(s): s = s +1; If s <= 0 then wakeup a process waiting on s; Assume that Pb and Vb the wait and signal operations on binary semaphores are provided. Two binary semaphores Xb and Yb are used to implement the semaphore operations P(s) and V(s) as follows: P( s ): Pb( Xb ); s = s-1; if( s <0 ) { 114
Vb( Xb ); Pb( Yb ); } else Vb ( Xb ); V( s ): Pb( Xb ); s = s+1; if( s <= 0 ) Vb( Yb ); Vb( Xb ); The initial values of Xb and Yb are respectively (A) 0 and 0
(B)0 and 1
(C) 1 and 0
(D)1 and 1
SOLUTION Xb & Yb are binary semaphores used to implement mutual exclusion here. So when Xb is 1 Yb should be zero so only 1 code between the two could run. Since Pb(Xb) module implementing wait process so Xb should be 1. Pb(Yb) implementing signal s = s +1 So Yb should be 0 initially. Hence (C) is correct option. Question. 47 Which of the following statements about synchronous and asynchronous I/O is NOT true? (A) An ISR is invoked on completion of I/O in synchronous I/O but not in asynchronous I/O (B)In both synchronous and asynchronous I/O an ISR (Interrupt Serive Routine) is invoked after completion of the I/O (C) A process making a synchronous I/O cal waits until I/O is complete, but a process making an asynchronous I/O call does not wait for completion of the I/O (D) In the case of synchronous I/O, the process waiting for the completion of I/O is woken up by the ISR that is invoked afterr the completion of I/O
SOLUTION For the completion of I/0 an interrupt should be generated for CPU in case of both synchronous & asynchronous I/0 and this ISR call is before them. So CPU can switch to I/0. Hence (B) is correct option. Question. 48 Which of the following is NOT true of deadlock prevention and deadlock avoidance schemes? 115
(A) In deadlock prevention, the request for resources is always granted if the resulting state is safe (B) In deadlock avoidance, the request for resources is always granted if the resulting state is safe (C) Deadlock avoidance is less restrictive than deadlock prevention (D) Deadlock avoidance requires knowledge of resource requirements a priori
SOLUTION Both deadlock prevention, & avoidance allocate resources if the resulting state is safe state so option (A) & (B) are tree. The difference in both schemes is that in avoidance we know the requirement a prior. But Deadlock avoidance is less restrictive than prevention is false. Hence (C) is correct option. Question. 49 A process executes the following code for(i = 0;i < n;i ++) fork(); The total number of child processes created is (B)2n-1
(A) n
(C) 2n
(D)2n+1-1
SOLUTION The loop is called for n times. The first process is parent process so this should not be counted in child process. But after that every child process has its own child created so after every loop. 20,2',........2n total threads. But subtracting the parent. Hence (B) is correct option.
Question. 50 A processor uses 36 bit physical addresses and 32 bit virtual addresses, with a page frame size of 4 Kbytes. Each page table entry is of size 4 bytes. A three level page table is used for virtualto-physical address translation, where the virtual address is used as follows
bits 30-31 are used to index into the first level page table, bits 21-29 are used to index into second level page table bits 12-20 are used to index into third level page table bits 0-11 are used as offset within the page
The number of bits required for addressing the next level page table(or page frame) in the page table entry of the first, second and third level page table are respectively 116
(A) 20,20 and 20
(B)24,24 and 24
(C) 24,24 and 20
(D) 25,25 and 24
SOLUTION Total address single = 36 bit First level no. of bits = 2*12 = 24 bits Second level = 4*6 = 24 bits Third level = 8*3 = 24 bits Hence (B) is correct option.
YEAR 2009 Question. 51 Consider a system with 4 type of resources R1 (3 units), R2 (2 units), R3 (3 units), R4 (4units). A non-preemptive resource allocation policy is used. At any give instance, a request is not entertained if it cannot be completely satisfied. Three processes P1, P2, P3 request the resources as follows if executed independently.
Which one of the following statements is TRUE if all three processes run concurrently starting at time t = 0 ? (A)All processes will finish without any deadlock (B)Only P1 and P2 will be in deadlock (C)Only P1 and P3 will be in deadlock (D)All three processes will be in deadlock SOLUTION Total resources = 3 2 3 2 117
Time 0 1 2 3 4 5 6 7 8 9 10
Process 1Process 2Process 3Available 1234 1234 1234 1234 0200 0020 0001 3011 0210 0020 0001 3001 0210 0021 2001 1000 0210 0021 2001 1000 0210 1021 2001 0000 1110 1021 0001 1100 1110 1011 0001 1110 1100 1011 0101 1020 1100 0000 0111 2021 1102 0000 0000 2130 0000 0000 0000 3232
All process computed without deadlock Hence (A) is correct option. Question. 52 In which of the following page replacement policies, Belady’s anomaly may occur ? (A) FIFO
(B)Optimal
(C) LRU
(D)MRU
SOLUTION Belady’s anomaly is a properly of FIFO page replacement policy in which the no. of page faults increases even if the no. of page frames are increased in some cases. Hence (A) is correct option. Question. 53 The essential content(S) in each entry of a page table is/are (A)virtual page number (B)page frame number (C)Both virtual page number and page frame number (D)access right information
SOLUTION A page table is in main memory which mop a page frame to a virtual memory page. So essential content is page frame number Hence (B) is correct option.
118
Question. 54 Consider a disk system with 100 cylinders. The requests to access the cylinders occur in following sequence : 4, 34, 10, 7, 19, 73, 2, 15, 6, 20 Assuming that the head is currently at cylinder 50, what is the time taken to satisfy all requests if it takes 1 ms to move from one cylinder to adjacent one and shortest seek time first policy is used? (A) 95ms
(B) 119ms
(C) 233ms
(D) 276ms
SOLUTION Arranging in order.
Total = 16 +14 +1 +4 +5 +3 +1 +2 +2 +71 = 119 ms Hence (B) is correct option. Question. 55 In the following process state transition diagram for a uniprocessor system, assume that there are always some processes in the steady state : Now consider the following statements : I.
If a process makes a transition D, it would result in another process making transition A immediately 119
II. III. IV.
A process P2 in blocked state can make transition E while another process P1 is in running state The OS uses non-preemptive scheduling The OS uses non-preemptive scheduling
Which of the above statement are TRUE ? (A) I and II
(B) I and III
(C) II and III
(D) II and IV
SOLUTION IIs not necessary that a new process enters immediately. IIIs true different process can be in different states one is ready and another in running state. IIIIs also correct since if one blocked it stop running so other can run. IV Is not correct. Hence (C) is correct option.
Question. 56 The enter_CS( ) and leave_CS( ) functions to implement critical section of a process are realized using test and set instruction as follows : Void enter_cs(X) { while (test-and-set)(X)) : } Void leave_CS(X) { X=0; } In the above solution, X is a memory location associated with the CS and is initialized to 0. Now consider the following statements I. II. III. IV.
The above solution to CS problem is deadlock-free II The solution is starvation free The processes enter CS in FIFO order More than one process can enter CS at the same time
Which of the above statements are TRUE (A) I only
(B) I and II
(C) II and III
(D) IV only
SOLUTION I
Is true since X is initialized to zero, this cause only 1 process to enter into critical so no deadlock. 120
II & III
Are false, other process which doesn’t capture X will wait so ultimately starve.
IV
Since I true so it is false.
Hence (A) is correct option. Question. 57 A multilevel page table is preferred in comparison to a single level page table for translating virtual address to physical address because (A) It reduces the memory access time to read or write and memory location (B) It helps to reduce the size of page table needed to implement the virtual address space of a process (C) If is required by the translation look aside buffer (D) If helps to reduce the number of page faults in page replacement algorithms. SOLUTION Single level page table is not preferred since it requires the number of page table entries equal to the number of virtual memory addresses, but a multilevel page table has smaller number of entries so reduce the size of page table needed. Hence (B) is correct option. YEAR 2010 Question. 58 Consider the methods used by processes P1 and P2 for accessing their critical sections whenever needed, as given below. The initial values of shared boolean variables S1 and S2 are randomly assigned. Method used by P1 Method used by P2 While(S1= = S2) Critical Section S1= S2;
While(S1!= S2) Critical Section S2= not(S1);
While one of the following statements describes properties achieved? (A)Mutual exclusion nut nor progress (B)Progress but not mutual exclusion (C)Neither mutual exclusion nor progress 121
(D)Both mutual exclusion and progress
SOLUTION Method used by P1 & P2 enters into critical section when S1 = S2 & S1! = S2 respectively, since both are opposite conditions so mutual exclusion is there, but if P1's while loop true then it will always be true & in P2 if while loop true it would run infinitely so no progress. Hence (A) is correct option. Question. 59 A system uses FIFO policy for page replacement. It has 4 page frames with no pages loaded to begin with. The system first accesses 100 distinct pages in some order and then accesses the same 100 pages but now in the reverse order. How many page faults will occur ? (A) 196
(B) 192
(C) 197
(D) 195
SOLUTION In FIFO page replacement policy, the pages which entered first are replaced first so for first 100 accesses, 100 page faults will occur. Now in memory there are last 4 pages 97, 98, 99 & 100th page. So during reverse access there 4 pages would not create any page fault & other 97 page faults.
So total 100 +96 = 196 page faults. Hence (A) is correct option. Question. 60 Which of the following statements are true ? I. II. III.
Shortest remaining time first scheduling may cause starvation Preemptive scheduling may cause starvation Round robin in better than FCFS in terms of response time
(A) I only
(B) I and III only
(C) II and III only
(D) I, II and III
SOLUTION I.
SRIF scheduling may cause starvation since the processes which require large CPU burst periods would have to wait. 122
Generally preemptive scheduling doesn’t cause starvation but in some cases, it may cause
II.
starvation when one process doesn’t block due to I/O so others have to wait. III.
It is quite obvious that due to preemption involvement round orbit is better than FCFS, in
case of response time.
All are true Hence (D) is correct option. Question. 61 The following program consists of 3 concurrent processes and 3 binary semaphores. The semaphores are initialized as S0 = 1, S1 = 0, S2 = 0 Process P0
Process P1
Process P2
while( true ){ wait ( S0 ); print ‘0’ release ( S1 ); release ( S2 ); }
wait ( S1 ); release ( S0 );
wait ( S2 ); release ( S0 );
How many times will process P0 print ‘0’? (A) At least twice once
(B) Exactly twice
(C) Exactly thrice
(D) Exactly
SOLUTION Let us see what will happen initially S0 = 1, S1 = 0 S2 = 0 so P1 & P2 will wait & P0 runs & make S0 = 0 wait decrements semaphore by 1 & release increments if by 1. So after P0's 1st run S1 = 1 & S2 = 1 ‘0’ is printed Now P1 & P2 can run & make S1 = 0, S2 = 0 but both increments S0 = 1 again. So P0 again starts and print '0' so this would continue again & again. So '0' will be printed at least twice. Hence (A) is correct option.
Question. 62 A system has n resources R0..Rn-1, and k processes P0..Pk-1.The implementation of the resource request logic of each process Pi, is as follows: if (i%2==0){ 123
if(i
In which one of the following situations is a deadlock possible ? (A) n=40, k=26
(B) n=21, k=12
(C) n=20 , k=10
(D) n=41, k=19
SOLUTION Allotment of resources would be successful to all cases but in option (B) where n = 21 & k = 12. Let use take i = 10 So for P10 only R10 can be allotted whereas for P11 odd case (else). R21−11 i.e R10 can only be allotted. This is an conflict. Hence (B) is correct option.
DATABASE MANAGEMENT SYSTEMS 1) Which of the following statements are TRUE about an SQL query? (GATE 2012) P: An SQL query can contain a HAVING clause even if it does not a GROUP BY clause Q: An SQL query can contain a HAVING clause only if it has a GROUP BY clause R: All attributes used in the GROUP BY clause must appear in the SELECT clause S: Not all attributes used in the GROUP BY clause need to apper in the SELECT clause (A) P and R (B) P and S (C) Q and R (D) Q and S Answer (B) P is correct. HAVING clause can also be used with aggregate function. If we use a HAVING clause without a GROUP BY clause, the HAVING condition applies to all rows that satisfy the search condition. In other words, all rows that satisfy the search condition make up a single group. S is correct. To verify S, try following queries in SQL. CREATE TABLE temp ( id INT, name VARCHAR(100) ); INSERT INSERT INSERT INSERT
INTO INTO INTO INTO
temp temp temp temp
VALUES VALUES VALUES VALUES
(1, (2, (3, (4,
"abc"); "abc"); "bcd"); "cde");
124
SELECT Count(*) FROM temp GROUP BY name;
Output: count(*) -------2 1 1
2) Given the basic ER and relational models, which of the following is INCORRECT? (GATE 2012) (A) An attributes of an entity can have more that one value (B) An attribute of an entity can be composite (C) In a row of a relational table, an attribute can have more than one value (D) In a row of a relational table, an attribute can have exactly one value or a NULL value Answer (C) The term ‘entity’ belongs to ER model and the term ‘relational table’ belongs to relational model. A and B both are true. ER model supports both multivalued and composite attributes (C) is false and (D) is true. In Relation model, an entry in relational table can can have exactly one value or a NULL.
3) Suppose (A, B) and (C,D) are two relation schemas. Let r1 and r2 be the corresponding relation instances. B is a foreign key that refers to C in r2. If data in r1 and r2 satisfy referential integrity constraints, which of the following is ALWAYS TRUE? (GATE 2012)
Answer (A) B is a foreign key in r1 that refers to C in r2. r1 and r2 satisfy referential integrity constraints. So every value that exists in column B of r1 must also exist in column C of r2.
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4) Which of the following is TRUE? (GATE 2012) (A) Every relation in 2NF is also in BCNF (B) A relation R is in 3NF if every non-prime attribute of R is fully functionally dependent on every key of R (C) Every relation in BCNF is also in 3NF (D) No relation can be in both BCNF and 3NF Answer (C) BCNF is a stronger version 3NF. So every relation in BCNF will also be in 3NF. 5) Consider the following transactions with data items P and Q initialized to zero: T1: read (P) ; read (Q) ; if P = 0 then Q : = Q + 1 ; write (Q) ; T2: read (Q) ; read (P) ; if Q = 0 then P : = P + 1 ; write (P) ;
Any non-serial interleaving of T1 and T2 for concurrent execution leads to 2012) (A) A serializable schedule (B) A schedule that is not conflict serializable (C) A conflict serializable schedule (D) A schedule for which a precedence graph cannot be drawn
(GATE
Answer (B) Two or more actions are said to be in conflict if: 1) The actions belong to different transactions. 2) At least one of the actions is a write operation. 3) The actions access the same object (read or write). The schedules S1 and S2 are said to be conflict-equivalent if the following conditions are satisfied: 1) Both schedules S1 and S2 involve the same set of transactions (including ordering of actions within each transaction). 2) The order of each pair of conflicting actions in S1 and S2 are the same. A schedule is said to be conflict-serializable when the schedule is conflict-equivalent to one or more serial schedules. In the given scenario, there are two possible serial schedules: 1) T1 followed by T2 2) T2 followed by T1. In both of the serial schedules, one of the transactions reads the value written by other 126
transaction as a first step. Therefore, any non-serial interleaving of T1 and T2 will not be conflict serializable.
6) Consider the following relations A, B, C. How many tuples does the result of the following relational algebra expression contain? Assume that the schema of A U B is the same as that of A. (GATE 2012)
Table A Id Name Age ---------------12 Arun 60 15 Shreya 24 99 Rohit 11 Table B Id Name Age ---------------15 Shreya 24 25 Hari 40 98 Rohit 20 99 Rohit 11 Table C Id Phone Area ----------------10 2200 02 99 2100 01
(A) 7 (B) 4 (C) 5 (D) 9 Answer (A) Result of AUB will be following table Id Name Age ---------------12 Arun 60 15 Shreya 24 99 Rohit 11 25 Hari 40 98 Rohit 20 The result of given relational algebra expression will be
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Id Name Age Id Phone Area --------------------------------12 Arun 60 10 2200 02 15 Shreya 24 10 2200 02 99 Rohit 11 10 2200 02 25 Hari 40 10 2200 02 98 Rohit 20 10 2200 02 99 Rohit 11 99 2100 01 98 Rohit 20 99 2100 01
7) Consider the above tables A, B and C. How many tuples does the result of the following SQL query contains? (GATE 2012) SELECT A.id FROM A WHERE A.age > ALL (SELECT B.age FROM B WHERE B. name = "arun")
(A) 4 (B) 3 (C) 0 (D) 1 Answer (B) The meaning of “ALL” is the A.Age should be greater than all the values returned by the subquery. There is no entry with name “arun” in table B. So the subquery will return NULL. If a subquery returns NULL, then the condition becomes true for all rows of A So all rows of table A are selected. 8. Consider a relational table with a single record for each registered student with the following attributes. 1. Registration_Number:< Unique registration number for each registered student 2. UID: Unique Identity number, unique at the national level for each citizen 3. BankAccount_Number: Unique account number at the bank. A student can have multiple accounts or joint accounts. This attributes stores the primary account number 4. Name: Name of the Student 5. Hostel_Room: Room number of the hostel Which of the following options is INCORRECT? (GATE 2011) (A) BankAccount_Number is a candidate key (B) Registration_Number can be a primary key (C) UID is a candidate key if all students are from the same country (D) If S is a superkey such that S UID is NULL then S UID is also a superkey
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Answer (A) A Candidate Key value must uniquely identify the corresponding row in table. BankAccount_Number is not a candidate key. As per the question “A student can have multiple accounts or joint accounts. This attributes stores the primary account number”. If two students have a joint account and if the joint account is their primary account, then BankAccount_Number value cannot uniquely identify a row.
9) Consider a relational table r with sufficient number of records, having attributes A1, A2,…, An and let 1 <= p <= n. Two queries Q1 and Q2 are given below.
The database can be configured to do ordered indexing on Ap or hashing on Ap. Which of the following statements is TRUE? (GATE 2011) (A) Ordered indexing will always outperform hashing for both queries (B) Hashing will always outperform ordered indexing for both queries (C) Hashing will outperform ordered indexing on Q1, but not on Q2 (D) Hashing will outperform ordered indexing on Q2, but not on Q1. Answer (C) If record are accessed for a particular value from table, hashing will do better. If records are accessed in a range of values, ordered indexing will perform better. See this for more details.
10) Database table by name Loan_Records is given below. Borrower Ramesh Suresh Mahesh
Bank_Manager Sunderajan Ramgopal Sunderajan
Loan_Amount 10000.00 5000.00 7000.00
What is the output of the following SQL query?
(GATE 2011)
SELECT Count(*) FROM ( (SELECT Borrower, Bank_Manager FROM Loan_Records) AS S NATURAL JOIN (SELECT Bank_Manager, Loan_Amount FROM Loan_Records) AS T );
(A) 3 (B) 9 (C) 5 (D) 6 129
Answer (C) Following will be contents of temporary table S Borrower Bank_Manager -------------------------Ramesh Sunderajan Suresh Ramgqpal Mahesh Sunderjan
Following will be contents of temporary table T Bank_Manager Loan_Amount --------------------------Sunderajan 10000.00 Ramgopal 5000.00 Sunderjan 7000.00
Following will be the result of natural join of above two tables. The key thing to note is that the natural join happens on column name with same name which is Bank_Manager in the above example. “Sunderjan” appears two times in Bank_Manager column, so their will be four entries with Bank_Manager as “Sunderjan”. Borrower Bank_Manager Load_Amount -----------------------------------Ramesh Sunderajan 10000.00 Ramesh Sunderajan 7000.00 Suresh Ramgopal 5000.00 Mahesh Sunderajan 10000.00 Mahesh Sunderajan 7000.00
11) the table at any point in time. Using MX and MY, new records are inserted in the table 128 times with X and Y values being MX+1, 2*MY+1 respectively. It may be noted that each time after the insertion, values of MX and MY change. What will be the output of the following SQL query after the steps mentioned above are carried out? (GATE 2011) SELECT Y FROM T WHERE X=7;
(A) 127 (B) 255 (C) 129 (D) 257 Answer (A) X Y ------1 1
130
2 3 3 7 4 15 5 31 6 63 7 127 ...... ......
12) A relational schema for a train reservation database is given below. Passenger (pid, pname, age) Reservation (pid, class, tid) Table: Passenger pid pname age ----------------0 Sachin 65 1 Rahul 66 2 Sourav 67 3 Anil 69 Table : Reservation pid class tid --------------0 AC 8200 1 AC 8201 2 SC 8201 5 AC 8203 1 SC 8204 3 AC 8202
What pids are returned by the following SQL query for the above instance of the tables? (GATE 2010) SLECT pid FROM Reservation , WHERE class ‘AC’ AND EXISTS (SELECT * FROM Passenger WHERE age > 65 AND Passenger. pid = Reservation.pid)
(A) 1, 0 (B) 1, 2 (C) 1, 3 (S) 1, 5 Answer (C) When a subquery uses values from outer query, the subquery is called correlated subquery. The correlated subquery is evaluated once for each row processed by the outer query.
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The outer query selects 4 entries (with pids as 0, 1, 5, 3) from Reservation table. Out of these selected entries, the subquery returns Non-Null values only for 1 and 3.
13) Which of the following concurrency control protocols ensure both conflict serialzability and freedom from deadlock? I. 2-phase locking II. Time-stamp ordering (GATE 2010) (A) I only (B) II only (C) Both I and II (D) Neither I nor II Answer (B) 2 Phase Locking (2PL) is a concurrency control method that guarantees serializability. The protocol utilizes locks, applied by a transaction to data, which may block (interpreted as signals to stop) other transactions from accessing the same data during the transaction’s life. 2PL may be lead to deadlocks that result from the mutual blocking of two or more transactions. See the following situation, neither T3 nor T4 can make progress.
Timestamp-based concurrency control algorithm is a non-lock concurrency control method. In Timestamp based method, deadlock cannot occur as no transaction ever waits.
14) Consider the following schedule for transactions T1, T2 and T3:
Which one of the schedules below is the correct serialization of the above? 2010) (A)T1 T3 T2
(GATE
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(B)T2 (C)T2 (D)T3
T1 T3 T1
T3 T1 T2
Answer (A) T1 can complete before T2 and T3 as there is no conflict between Write(X) of T1 and the operations in T2 and T3 which occur before Write(X) of T1 in the above diagram. T3 should can complete before T2 as the Read(Y) of T3 doesn’t conflict with Read(Y) of T2. Similarly, Write(X) of T3 doesn’t conflict with Read(Y) and Write(Y) operations of T2. Another way to solve this question is to create a dependency graph and topologically sort the dependency graph. After topologically sorting, we can see the sequence T1, T3, T2.
15) The following functional dependencies hold for relations R(A, B, C) and S(B, D, E): B A, A C The relation R contains 200 tuples and the rel ation S contains 100 tuples. What is the maximum number of tuples possible in the natural join R S (R natural join S) (GATE 2010) (A) 100 (B) 200 (D) 300 (D) 2000 Answer (A) From the given set of functional dependencies, it can be observed that B is a candidate key of R. So all 200 values of B must be unique in R. There is no functional dependency given for S. To get the maximum number of tuples in output, there can be two possibilities for S. 1) All 100 values of B in S are same and there is an entry in R that matches with this value. In this case, we get 100 tuples in output. 2) All 100 values of B in S are different and these values are present in R also. In this case also, we get 100 tuples. 16) Consider two transactions T1 and T2, and four schedules S1, S2, S3, S4 of T1 and T2 as given below: T1 = R1[X] W1[X] W1[Y] T2 = R2[X] R2[Y] W2[Y] S1 = R1[X] R2[X] R2[Y] W1[X] W1[Y] W2[Y] S2 = R1[X] R2[X] R2[Y] W1[X] W2[Y] W1[Y] S3 = R1[X] W1[X] R2[X] W1[Y] R2[Y] W2[Y] S1 = R1[X] R2[Y]R2[X]W1[X] W1[Y] W2[Y] Which of the above schedules are conflict-serializable? (GATE: 2009) (A) S1 and S2 (B) S2 and S3 133
(C) S3 only (D) S4 only Answer (B) There can be two possible serial schedules T1 T2 and T2 T1. The serial schedule T1 T2 has the following sequence of operations R1[X] W1[X] W1[Y] R2[X] R2[Y] W2[Y] And the schedule T2 T1 has the following sequence of operations. R2[X] R2[Y] W2[Y] R1[X] W1[X] W1[Y] The Schedule S2 is conflict-equivalent to T2 T1 and S3 is conflict-equivalent to T1 T2.
17) Let R and S be relational schemes such that R={a,b,c} and S={c}. Now consider the following queries on the database:
IV) SELECT R.a, R.b FROM R,S WHERE R.c=S.c
Which of the above queries are equivalent? (A) I and II (B) I and III (C) II and IV (D) III and IV
(GATE: 2009)
Answer (A) I and II describe the division operator in Relational Algebra and Tuple Relational Calculus respectively. See Page 3 of this and slide numbers 9,10 of this for more details.
18) Consider the following relational schema: Suppliers(sid:integer, sname:string, city:string, street:string) Parts(pid:integer, pname:string, color:string) Catalog(sid:integer, pid:integer, cost:real)
Consider the following relational query on the above database: SELECT S.sname FROM Suppliers S WHERE S.sid NOT IN (SELECT C.sid FROM Catalog C WHERE C.pid NOT IN (SELECT P.pid
134
FROM Parts P WHERE P.color<> 'blue'))
Assume that relations corresponding to the above schema are not empty. Which one of the following is the correct interpretation of the above query? (GATE: 2009) (A) Find the names of all suppliers who have supplied a non-blue part. (B) Find the names of all suppliers who have not supplied a non-blue part. (C) Find the names of all suppliers who have supplied only blue parts. (D) Find the names of all suppliers who have not supplied only blue parts. Answer (B) The subquery “SELECT P.pid FROM Parts P WHERE P.color<> ‘blue’” gives pids of parts which are not blue. The bigger subquery “SELECT C.sid FROM Catalog C WHERE C.pid NOT IN (SELECT P.pid FROM Parts P WHERE P.color<> ‘blue’)” gives sids of all those suppliers who have supplied blue parts. The complete query gives the names of all suppliers who have not supplied a non-blue part
19) Assume that, in the suppliers relation above, each supplier and each street within a city has a unique name, and (sname, city) forms a candidate key. No other functional dependencies are implied other than those implied by primary and candidate keys. Which one of the following is TRUE about the above schema? (GATE: 2009) (A) The schema is in BCNF (B) The schema is in 3NF but not in BCNF (C) The schema is in 2NF but not in 3NF (D) The schema is not in 2NF Answer (A) The schema is in BCNF as all attributes depend only on a superkey (Note that primary and candidate keys are also superkeys). 20) Let R and S be two relations with the following schema R (P,Q,R1,R2,R3) S (P,Q,S1,S2) Where {P, Q} is the key for both schemas. Which of the following queries are equivalent? (GATE: 2008)
(A) Only I and II (B) Only I and III (C) Only I, II and III (D) Only I, III and IV 135
Answer (D) In I, Ps from natural join of R and S are selected. In III, all Ps from intersection of (P, Q) pairs present in R and S. IV is also equivalent to III because (R – (R – S)) = R S. II is not equivalent as it may also include Ps where Qs are not same in R and S.
21) Consider the following ER diagram.
(GATE: 2008)
The minimum number of tables needed to represent M, N, P, R1, R2 is (A) 2 (B) 3 (C) 4 (D) 5 Answer (A) Many-to-one and one-to-many relationship sets that are total on the many-side can be represented by adding an extra attribute to the “many” side, containing the primary key of the “one” side. Since R1 is many to one and participation of M is total, M and R1 can be combined to form the table {M1, M2, M3, P1}. N is a week entity set, so it can be combined with P.
22) Which of the following is a correct attribute set for one of the tables for the correct answer to the above question? (GATE: 2008) (A) {M1, M2, M3, P1} (B) {M1, P1, N1, N2} (C) {M1, P1, N1} (D) {M1, P1} Answer (A)
23) Consider the following relational schemes for a library database: Book (Title, Author, Catalog_no, Publisher, Year, Price) Collection (Title, Author, Catalog_no) with in the following functional dependencies: 136
I. Title Author --> Catalog_no II. Catalog_no --> Title Author Publisher Year III. Publisher Title Year --> Price
Assume {Author, Title} is the key for both schemes. Which of the following statements is true? (GATE: 2008) (A) Both Book and Collection are in BCNF (B) Both Book and Collection are in 3NF only (C) Book is in 2NF and Collection is in 3NF (D) Both Book and Collection are in 2NF only Answer (C) Table Collection is in BCNF as there is only one functional dependency “Title Author –> Catalog_no” and {Author, Title} is key for collection. Book is not in BCNF because Catalog_no is not a key and there is a functional dependency “Catalog_no –> Title Author Publisher Year”. Book is not in 3NF because non-prime attributes (Publisher Year) are transitively dependent on key [Title, Author]. Book is in 2NF because every non-prime attribute of the table is either dependent on the key [Title, Author], or on another non prime attribute. 24) Information about a collection of students is given by the relation studinfo(studId, name, sex). The relation enroll(studId, courseId) gives which student has enrolled for (or taken) that course(s). Assume that every course is taken by at least one male and at least one female student. What does the following relational algebra expression represent? (GATE: 2007)
(A) Courses in which all the female students are enrolled. (B) Courses in which a proper subset of female students are enrolled. (C) Courses in which only male students are enrolled. (D) None of the above Answer (B) The expression given in question does following steps in sequence. a) Select studids of all female students and selects all courseids of all courses. b) Then the query does a Cartesian Product of the above select two columns from different tables. c) Finally it subtracts enroll table from the result of above step (b). This will remove all the (studid, courseid) pairs which are present in enroll table. If all female students have registered in a courses, then this course will not be there in the subtracted result. So the complete expression returns courses in which a proper subset of female students are enrolled. studinfo table studid name
sex
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-----------------------1 a Male 2 c Female 3 d Female enroll table studid courseid -----------------1 1 2 1 3 1 2 2 3 3 3 2 Result of step b studid courseid --------------------2 1 2 2 2 3 3 1 3 2 3 3 Result of step c studid courseid ------------------2 3
25) Consider the relation employee(name, sex, supervisorName) with name as the key. supervisorName gives the name of the supervisor of the employee under consideration. What does the following Tuple Relational Calculus query produce? (GATE: 2007)
(A) Names of employees with a male supervisor. (B) Names of employees with no immediate male subordinates. (C) Names of employees with no immediate female subordinates. (D) Names of employees with a female supervisor. Answer (C) The query selects all those employees whose immediate subordinate is “male”. In other words, it selects names of employees with no immediate female subordinates
26) Consider the table employee(empId, name, department, salary) and the two queries 138
Q1 ,Q2 below. Assuming that department 5 has more than one employee, and we want to find the employees who get higher salary than anyone in the department 5, which one of the statements is TRUE for any arbitrary employee table? (GATE: 2007) Q1 : Select e.empId From employee e Where not exists (Select * From employee s where s.department = “5” and s.salary >=e.salary) Q2 : Select e.empId From employee e Where e.salary > Any (Select distinct salary From employee s Where s.department = “5”)
(A) Q1 is the correct query (B) Q2 is the correct query (C) Both Q1 and Q2 produce the same answer. (D) Neither Q1 nor Q2 is the correct query Answer (D) Consider the following example table. empid 1 2 3 4
name department salary a 4 90k b 5 30k c 5 50k d 5 80k
Q1 will give empid 1 Q2 will give empid 1, 3, 4 But the correct answer is 4
27) Which one of the following statements if FALSE? (GATE: 2007) (A) Any relation with two attributes is in BCNF (B) A relation in which every key has only one attribute is in 2NF (C) A prime attribute can be transitively dependent on a key in a 3 NF relation. (D) A prime attribute can be transitively dependent on a key in a BCNF relation. Answer (D)
28) Consider the following schedules involving two transactions. Which one of the following statements is TRUE? (GATE: 2007)
139
(A) Both S1 and S2 are conflict serializable. (B) S1 is conflict serializable and S2 is not conflict serializable. (C) S1 is not conflict serializable and S2 is conflict serializable. (D) Both S1 and S2 are not conflict serializable. Answer (C) S1 is not conflict serializable, but S2 is conflict serializable Schedule S1 T1 T2 --------------------r1(X) r1(Y) r2(X) r2(Y) w2(Y) w1(X) The schedule is neither conflict equivalent to T1T2, nor T2T1. Schedule S2 T1 T2 --------------------r1(X) r2(X) r2(Y) w2(Y) r1(Y) w1(X) The schedule is conflict equivalent to T2T1.
29) Consider the following log sequence of two transactions on a bank account, with initial balance 12000, that transfer 2000 to a mortgage payment and then apply a 5% interest. 1. 2. 3. 4. 5. 6. 7.
T1 T1 T1 T1 T2 T2 T2
start B old=12000 new=10000 M old=0 new=2000 commit start B old=10000 new=10500 commit
Suppose the database system crashes just before log record 7 is written. When the system is restarted, which one statement is true of the recovery procedure? (GATE: 2006)
(A) We must redo log record 6 to set B to 10500 (B) We must undo log record 6 to set B to 10000 and then redo log records 2 and 3 (C) We need not redo log records 2 and 3 because transaction T1 has committed (D) We can apply redo and undo operations in arbitrary order because they are idempotent. Answer (C) Once a transaction is committed, no need to redo or undo operations. 140
30) Consider the relation enrolled (student, course) in which (student, course) is the primary key, and the relation paid (student, amount) where student is the primary key. Assume no null values and no foreign keys or integrity constraints. Given the following four queries: Query1: select student from enrolled where student in (select student from paid) Query2: select student from paid where student in (select student from enrolled) Query3: select E.student from enrolled E, paid P where E.student = P.student Query4: select student from paid where exists (select * from enrolled where enrolled.student = paid.student)
Which one of the following statements is correct?
(GATE : 2006)
(A) All queries return identical row sets for any database (B) Query2 and Query4 return identical row sets for all databases but there exist databases for which Query1 and Query2 return different row sets. (C) There exist databases for which Query3 returns strictly fewer rows than Query2. (D) There exist databases for which Query4 will encounter an integrity violation at runtime. Answer (A) The output of Query2, Query3 and Query4 will be identical. Query1 may produce duplicate rows. But rowset produced by all of them will be same. Table enrolled student course ---------------abc c1 xyz c1 abc c2 pqr c1 Table paid student amount ----------------abc 20000 xyz 10000 rst 10000 Output of Query 1 abc abc xyz Output of Query 2 abc xyz Output of Query 3
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abc xyz Output of Query 4 abc xyz
31) Consider the relation enrolled(student, course) in which (student, course) is the primary key, and the relation paid(student, amount), where student is the primary key. Assume no null values and no foreign keys or integrity constraints. Assume that amounts 6000, 7000, 8000, 9000 and 10000 were each paid by 20% of the students. Consider these query plans (Plan 1 on left, Plan 2 on right) to “list all courses taken by students who have paid more than x”.
A disk seek takes 4ms, disk data transfer bandwidth is 300 MB/s and checking a tuple to see if amount is greater than x takes 10 micro-seconds. Which of the following statements is correct? (GATE: 2006) (A) Plan 1 and Plan 2 will not output identical row sets for all databases. (B) A course may be listed more than once in the output of Plan 1 for some databases (C) For x = 5000, Plan 1 executes faster than Plan 2 for all databases. (D) For x = 9000, Plan I executes slower than Plan 2 for all databases. Answer (C) Assuming that large enough memory is available for all data needed. Both plans need to load both tables courses and enrolled. So disk access time is same for both plans. Plan 2 does lesser number of comparisons compared to plan 1. 1) Join operation will require more comparisons as the second table will have more rows in plan 2 compared to plan 1. 2) The joined table of two tables will will have more rows, so more comparisons are needed to find amounts greater than x.
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32) The following functional dependencies are given: AB
CD, AF
D, DE
F, C
G , F
Which one of the following options is false? (A)CF+ = {ACDEFG} (B)BG+ = {ABCDG} (C)AF+ = {ACDEFG} (D)AB+ = {ABCDFG}
E, G
A
(GATE: 2006)
Answer (C) Closure of AF or AF+ = {ADEF}, closure of AF doesn’t contain C and G. Option (D) also looks correct. AB+ = {ABCDG}, closure of AB doesn’t contain F. 33) Which one of the following statements about normal forms is FALSE? (GATE 2005) (a) BCNF is stricter than 3NF (b) Lossless, dependency-preserving decomposition into 3NF is always possible (c) Lossless, dependency-preserving decomposition into BCNF is always possible (d) Any relation with two attributes is in BCNF Answer (c) It is not always possible to decompose a table in BCNF and preserve dependencies. For example, a set of functional dependencies {AB –> C, C –> B} cannot be decomposed in BCNF. 34) The following table has two attributes A and C where A is the primary key and C is the foreign key referencing A with on-delete cascade. A C ----2 4 3 4 4 3 5 2 7 2 9 5 6 4
The set of all tuples that must be additionally deleted to preserve referential integrity when the tuple (2,4) is deleted is: (GATE 2005) (a) (3,4) and (6,4) (b) (5,2) and (7,2) (c) (5,2), (7,2) and (9,5) (d) (3,4), (4,3) and (6,4) Answer (C) When (2,4) is deleted. Since C is a foreign key referring A with delete on cascade, all entries 143
with value 2 in C must be deleted. So (5, 2) and (7, 2) are deleted. As a result of this 5 and 7 are deleted from A which causes (9, 5) to be deleted.
35) The relation book (title, price) contains the titles and prices of different books. Assuming that no two books have the same price, what does the following SQL query list? (GATE 2005) select title from book as B where (select count(*) from book as T where T.price > B.price) < 5
(a) Titles of the four most expensive books (b) Title of the fifth most inexpensive book (c) Title of the fifth most expensive book (d) Titles of the five most expensive books Answer (d) When a subquery uses values from outer query, the subquery is called correlated subquery. The correlated subquery is evaluated once for each row processed by the outer query. The outer query selects all titles from book table. For every selected book, the subquery returns count of those books which are more expensive than the selected book. The where clause of outer query will be true for 5 most expensive book. For example count (*) will be 0 for the most expensive book and count(*) will be 1 for second most expensive book. 36) Let r be a relation instance with schema R = (A, B, C, D). We define r1 = ‘select A,B,C from r’ and r2 = ‘select A, D from r’. Let s = r1 * r2 where * denotes natural join. Given that the decomposition of r into r1 and r2 is lossy, which one of the following is TRUE? (GATE 2005) (a) s is subset of r (b) r U s = r (c) r is a subset of s (d) r * s = s Answer (c) Consider the following example with lossy decomposition of r into r1 and r2. We can see that r is a subset of s. Table r A B C D --------------------------1 10 100 1000 1 20 200 1000 1 20 200 1001 Table r1
144
A B C -----------------1 10 100 1 20 200 Table r2 A D ----------1 1000 1 1001 Table s (natural join of r1 and r2) A B C D --------------------------1 10 100 1000 1 20 200 1000 1 20 100 1001 1 20 200 1001
37) Let E1 and E2 be two entities in an E/R diagram with simple single-valued attributes. R1 and R2 are two relationships between E1 and E2, where R1 is one-to-many and R2 is many-to-many. R1 and R2 do not have any attributes of their own. What is the minimum number of tables required to represent this situation in the relational model? (GATE 2005) (a) 2 (b) 3 (c) 4 (d) 5 Answer (c) The situation given can be expressed with following sample data. E1 a b c E2 x y z R1 E1 a a b R2 E1
E2 x y z
E2
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a a b
x y y
38) Consider a relation scheme R = (A, B, C, D, E, H) on which the following functional dependencies hold: {A–>B, BC–> D, E–>C, D–>A}. What are the candidate keys of R? (GATE 2005) (a) AE, BE (b) AE, BE, DE (c) AEH, BEH, BCH (d) AEH, BEH, DEH Answer (d) A set of attributes S is candidate key of relation R if the closure of S is all attributes of R and there is no subset of S whose closure is all attributes of R. Closure of AEH, i.e. AEH+ = {ABCDEH} Closure of BEH, i.e. BEH+ = {ABCDEH} Closure of DEH, i.e. DEH+ = {ABCDEH}
39) Let R1 (A, B, C) and R2 (D, E) be two relation schema, where the primary keys are shown underlined, and let C be a foreign key in R1 referring to R2. Suppose there is no violation of the above referential integrity constraint in the corresponding relation instances r1 and r2 . Which one of the following relational algebra expressions would necessarily produce an empty relation ? (GATE:2004) 1) ΠD (r2 ) - ΠC (r1 ) 2) ΠC (r1 ) - ΠD (r2 ) 3) ΠD (r1
C 1 D r2 )
4) ΠC (r1
C = Dr2)
Answer (2)
Explanation: C is an attribute in R1 but D is a key in K 2 . So consider ΠC (r1) − ΠD (r2) So the result of this query would be all those tuples which are in ΠC (r1) but not in ΠD (r2).Since D is a key so it has all the possible values of C . So difference would always be 146
empty.Hence( 2 ) is correct.
40) Consider the following relation schema pertaining to a students database: Student (rollno, name, address) Enroll (rollno, courseno, coursename) where the primary keys are shown underlined. The number of tuples in the Student and Enroll tables are 120 and 8 respectively. What are the maximum and minimum number of tuples that can be present in (Student * Enroll), where '*' denotes natural join ? (GATE: 2004) 1) 8, 8 2) 120, 8 3) 960, 8 4) 960, 120 Answer (1) Explanation: The boundary cases are when either all the tuples of Enroll table belong to one roll no. , so there can be at most 8 roll no & courses no. combinations or the other case is when all the tuples belong to different roll no. this also has 8 tuples. So (8,8) = (max, min) Hence (A) is correct option
41) Which of the following scenarios may lead to an irrecoverable error in a database system ? (GATE: 2003) 1) A transaction writes a data item after it is read by an uncommitted transaction 2) A transaction reads a data item after it is read by an uncommitted transaction 3) A transaction reads a data item after it is written by a committed transaction 4) A transaction reads a data item after it is written by an uncommitted transaction Answer (4)
147
42) The following SQL query select distinct al, a2,........., an from r1 , r2,........, rm where P For an arbitrary predicate P, this query is equivalent to which of the following relational algebra expressions GATE 2003 1. 2. 3. 4. Answer (1)
43) Relation R with an associated set of functional dependencies, F, is decomposed into BCNF. The redundancy (arising out of functional dependencies) in the resulting set of relations is (GATE: 2002) (a) Zero (b) More than zero but less than that of an equivalent 3NF decomposition (c) Proportional to the size of F+ (d) Indetermine Ans: solution (b) Explanation: Redundancy in BCNF is low when compared to 3NF. 44) With regard to the expressive power of the formal relational query languages, which of the following statements is true? (GATE: 2002) (a) Relational algebra is more powerful than relational calculus (b) Relational algebra has the same power as relational calculus. 148
(c) Relational algebra has the same power as safe relational calculus. (d) None of the above SOLUTION
Expressive power is the capacity of formal query languages to express various query statements, so relational algebra is as powerful as relational calculus only if calculus is safe relational calculus.Hence (C) is correct option.
45) Consider a relation geq which represents “greater than or equal to”, that is, (x,y)∈ geq only if y≥x. create table geq ( Ib integer not null ub integer not null primary key 1b foreign key (ub) references geq on delete cascade) Which of the following is possible if a tuple (x,y) is deleted? (GATE:2001) (a) A tuple (z,w) with z > y is deleted (b) A tuple (z,w) with z > x is deleted (c) A tuple (z,w) with w < x is deleted (d) The deletion of (x,y) is prohibited SOLUTION
Tuple (x, y) is deleted. Here y $x lb is primary key ub is foreign key Since y refer to same key z . & in table A x a primary key is deleted so all enteries from table B will also be deleted. And also all enteries referencing it in y will also be deleted.Hence (C) is correct option.
46) Suppose the adjacency relation of vertices in a graph is represented in a table Adj (X, Y).Which of the following queries cannot be expressed by a relational algebra expression of constant length ? (A) List all vertices adjacent to a given vertex. (B) List all vertices which have self loops (C) List all vertices which belong to cycles of less than three vertices (D) List all vertices reachable from a given vertex 149
SOLUTION
The database contains the adjacency list of the graph. So relation algebra with face problems when while calculating the length of cycle self loops come, then the query would execute in one tuple only. Hence (C) is correct option. 47) Given the relations employee (name, salary, deptno) department (deptno, deptname, address) Which of the following queries cannot be expressed using the basic relational algebra operations (σ,π,×, ,∪ ,∩,−)? (GATE:2000) (a) Department address of every employee (b) Employees whose name is the same as their department name (c) The sum of all employees' salaries (d) All employees of a given department Answer: (c) 48) Given relations r(w,x) and s(y,z), the result of select distinct w,x from r, s is guaranteed to be same as r, provided (GATE:2000) (a) r has no duplicates and s is non-empty (b) r and s have no duplicates (c) s has no duplicates and r is non-empty (d) r and s have the same number of tuples Answer: (A) 49) In SQL, relations can c o n t a i n null values, and c o m p a r i s o n s with null v a l u e s are treated as unknown. Suppose all c o m p a r i s o n s with a null value are treated as false. Which of the following pairs is not equivalent? (GATE:2000) (a) x = 5 not (not (x = 5) (b) x = 5 x > 4 and x < 6, where x is an integer (c) x ≠ 5 not (x = 5) (d) None of the above 150
Answer: (B) 50) Consider the join of a relation R with a relation S. If R has m tuples and S has n tuples then the maximum and minimum sizes of the join respectively are (GATE:1999) (a) m + n and 0 (b) mn and 0 (c) m + n and |m – n| (d) mn and m + n
151
52) Consider the schema R = (S T U V) and the dependencies S € T, T € U. U € V and V € S. Let R = (R1 and R2) be a decomposition such that R1 ∩ R2 = ∅ . The decomposition is (GATE:1999) (a) not in 2NF
(b) in 2NF but not 3NF (c) in 3NF
but not in 2NF
(d) in both 2NF and 3NF
53) The minimum number of record movements required to merge five files A (with 10 records), B (with 20 records), C (with 15 records), D (with 5 records) and E (with 25 records) is: (GATE:1999) (a) 165 (b) 90 (c) 75 (d) 65 54) Which of the following is/are correct?
(GATE:1999)
(a) An SQL query automatically eliminates duplicates (b) An SQL query will not work if there are no indexes on the relations (c) SQL permits attribute names to be repeated in the same relation (d) None of the above Answer: (d)
55) Consider the set of relations EMP (Employee-no. Dept-no, Employee-name, Salary) DEPT (Dept-no. Dept-name, Location) Write an SQL query to: (a) Find all employee names who work in departments located at ‘Calcutta’ and whose salary is greater than Rs.50,000. (b) Calculate, for each department number, the number of employees with a salary greater than Rs.1,00,000. (GATE:1999)
56) Which normal form is considered adequate for normal relational database Design? (GATE:1998) (a) 2 NF
(b) 5 NF
(c) 4 NF
(d) 3 NF
Ans: solution (d) Explanation: 152
A relational database table is often described as "normalized" if it is in the Third Normal Form because most of the 3NF tables are free of insertion, update, and deletion anomalies. 57) There are five records in a database. Name
Age
Occupatio Categor n CO y A
Rama
27
Abdul
22
N
A
Jeniffe
28
EN
B
r
32
G
D
Maya
24
DO
C
There is an index file Devassociated with thisC and it contains the values 1,3,2,5 and 4.Which one of the fields is the index built from? (GATE:1998) SER (a) Age Answer: (C)
(b) Name MUS
(c) Occupation
(d) Category
58) Suppose we have a database consisting of the following three relations. FREQUENTS (student, parlor) giving the parlors each student visits. SERVES (parlor, ice-cream) indicating what kind of ice-creams each parlor serves. LIKES (student, ice-cream) indicating what ice-creams each student likes. (Assume that each student likes at least one ice-cream and frequents at least one parlor) Express the following in SQL: Print the students that frequent at least one parlor that serves some ice- cream that they like. (GATE:1998) 59) AB+ -tree index is to be built on the Name attribute of the relation STUDENT . Assume that all student names are of length 8 bytes, disk blocks are of size 512 bytes, and index pointers are of size 4 bytes. Given this scenario, what would be the best choice of the degree (i.e. the number of pointers per node) of the B+ -tree ? (A) 16 (B) 42 (C) 43 (D) 44 SOLUTION
Size of 1 record of index = 8 + 4 = 12 bytes. Let no. of pointers required = P No. of index values per block = P – 1 So
(P − 1) 8 + 4P= 512 12P= 520 153
P , 44 60) Let r and s be two relations over the relation schemes R and S respectively, and let A be an attribute in R . Then the relational algebra expression σA = a ]r A s g is always equal to : (A) σA = a ]r g (B) r (C) σA = a ]r A s g (D) None of the above SOLUTION
Given query performs natural join between r & s , & then project attribute A where A = a . This same result is produced by the query σA = a (r) A s . This query selects attribute A = a from r & then performs join operation results are same. Hence (C) is correct option.
INFORMATIONS SYSTEMS SOFTWARE ENGINEERING
1.The coupling between different modules of a software is categorized as follows: I. Content coupling V. Data coupling
II. Common coupling
III. Control coupling
IV. Stamp coupling
Coupling between modules can be ranked in the order of strongest(least desirable) to weakest (most desirable) as follows: 1) I-II-III-IV-V 2) V-IV-III-II-I 3) I-III-V-II-IV 4) IV-II-V-III-I
Solution:1
2.
154
Solution: B
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3.
Solution: A
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4.
Solution: D
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5. In the system concepts, the term integration ? a. implies structure and order 155
b. refers to the manner in which each component functions with other components of the system c. means that parts of computer system depends on one another d. refers to the holism of systems Solution: D
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6.What is software? a. Set of computer programs, procedures and possibly associated document concerned with the operation of data processing. b. A set of compiler instructions. c. A mathematical formulae d. All of the above e. None of the above Solution:A
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7. Which is the last step in classic life cycle paradigm? a. System engineering b. Analysis c. Design d. Coding e. Maintenance. Solution:E
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8. Which of the following is closer to machine Code a. Assembly language b. Machine language c. High level language d. All of the above e. None of the above Solution:A
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9. The following are properties of Modularity except 156
a. It implement a single independent function b. It performs a single logical task. c. It has a single entry and exit point. d. It is entirely constructed of modules. e. None of the above. Solution:E
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10. The following are characteristics of software expects a. It is developed or engineered. b. Software does not wear out c. Software are custom made d. Software have don’t have spare parts instead it has backup e. Software consists of physical devices. Solution:E
gate-2011
11. Software genetic development process contains three genetic phrases namely a. Definition, development, maintenance. b. Coding, design, Software engineering c. Software engineering, Definition, Coding d. Design, Coding, Development e. Development, Definition, Testing Solution:A
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12. Which of the following translators convert high-level language on statement-by-statement basis? a. Compiler b. Machine level language converter c. Interpreter. d. Assembler e. None of the above 157
Solution:C
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13. Which of the following is not an example of Prototype in engineering paradigm? a. Paper prototype. b. Existing prototype. c. Working prototype. d. Software prototype. e. Engineering prototype. Solution:A
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14. Which of the following Construct in formal model in software engineering execute each statement in succession. a. Selection Construct. b. Sequence Construct. (Correct) c. Iteration Construct. d. Business Construct. e. Statement Construct. Solution:B
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15.What is software engineering? a. Set of computer programs, procedures and possibly associated document concerned with the operation of data processing. b. Software engineering is Design, Coding, Development c. Software engineering implement a single independent function d. Software engineering is the establishment and use of sound engineering practice in order to produce economical and reliable software that will perform efficiently on real machine (Correct) e.Software engineering is a step that encompasses the method, tools, procedure used in software Solution:D
gate-2012
16. The most important feature of spiral model is (A) requirement analysis. (B) risk management. (C) quality management. (D) configuration management. 158
Ans: B 17. The worst type of coupling is (A) Data coupling. (B) control coupling. (C) stamp coupling. (D) content coupling. Ans: D 18. IEEE 830-1993 is a IEEE recommended standard for (A) Software requirement specification. (B) Software design. (C)Testing. (D) Both (A) and (B) Ans: A 19. One of the fault base testing techniques is (A) unit testing. (B) beta testing. (C) Stress testing. (D) mutation testing. Ans: D 20. Changes made to an information system to add the desired but not necessarily the required features is called (A) Preventative maintenance. (B) Adaptive maintenance. (C) Corrective maintenance. (D) Perfective maintenance. Ans: D 21. All the modules of the system are integrated and tested as complete system in the case of (A) Bottom up testing (B) Top-down testing (C) Sandwich testing (D) Big-Bang testing Ans: D 22. SRS is also known as specification of Ans: D (A) White box testing (B) Stress testing (C) Integrated testing (D) Black box testing 23. The model in which the requirements are implemented by category is (A) Evolutionary Development Model (B) Waterfall Model 159
(C) Prototyping (D) Iterative Enhancement Model Ans: A 24. SRD stands for (A) Software requirements definition (B) Structured requirements definition (C) Software requirements diagram (D) Structured requirements diagram Ans: B
public key cryptography, X adds a digital signature σ to message M, encrypts , and sends it to Y, where it is decrypted. Which one of the following sequences of keys is used for the operations? 25. Using
(A) Encryption: X’s private key followed by Y’s private key; Decryption: X’s public key followed by Y’s public key (B) Encryption: X’s private key followed by Y’s public key; Decryption: X’s public key followed by Y’s private key (C) Encryption: X’s public key followed by Y’s private key; Decryption: Y’s public key followed by X’s private key (D) Encryption: X’s private key followed by Y’s public key; Decryption: Y’s private key followed by X’s public key Ans:D
2013
COMPUTER NETWORKS 1. The transport layer protocols used for real time multimedia, file transfer, DNS and email, respectively are GATE 2013 (A) TCP, UDP, UDP and TCP (B) UDP, TCP, TCP and UDP (C) UDP, TCP, UDP and TCP (D) TCP, UDP, TCP and UDP ANS: C SOL: Multimedia can be unreliable but has to be fast so UDP, File transfer has to be secure & reliable so uses TCP, DNS can be both TCP and UDP, E mail uses TCP for reliability. 2. Using public key cryptography, X adds a digital signature σ to message M, encrypts , and sends it to Y, where it is decrypted. Which one of the following sequences of keys is used for the operations? GATE 2013 160
(A) Encryption: X’s private key followed by Y’s private key; Decryption: X’s public key followed by Y’s public key (B) Encryption: X’s private key followed by Y’s public key; Decryption: X’s public key followed by Y’s private key (C) Encryption: X’s public key followed by Y’s private key; Decryption: Y’s public key followed by X’s private key (D) Encryption: X’s private key followed by Y’s public key; Decryption: Y’s private key followed by X’s public key ANS: D 3. Assume that source S and destination D are connected through two intermediate routers labeled R. Determine how many times each packet has to visit the network layer and the data link layer during a transmission from S to D. GATE 2013
(A) Network layer – 4 times and Data link layer – 4 times (B) Network layer – 4 times and Data link layer – 3 times (C) Network layer – 4 times and Data link layer – 6 times (D) Network layer – 2 times and Data link layer – 6 times ANS: C SOL: Therefore, Network layer – 4 times Data link layer – 6 times
4. Determine the maximum length of the cable (in km) for transmitting data at a rate of 500 Mbps in an Ethernet LAN with frames of size 10,000 bits. Assume the signal speed in the cable to be 2,00,000 km/s. GATE 2013 (A) 1 (B) 2 (C) 2.5 (D) 5 ANS: B SOL: Propagation time = Transmission time + Collision signal time 𝐹𝑎𝑟𝑚𝑒 𝑠𝑖𝑧𝑒 𝐿𝑒𝑛𝑔𝑡ℎ 𝐿𝑒𝑛𝑔𝑡ℎ = + 𝑃𝑟𝑜𝑝𝑎𝑔𝑎𝑡𝑖𝑜𝑛 𝑡𝑖𝑚𝑒 𝑆𝑖𝑔𝑛𝑎𝑙 𝑠𝑝𝑒𝑒𝑑 𝑆𝑖𝑔𝑛𝑎𝑙 𝑠𝑝𝑒𝑒𝑑 10000 𝑏𝑖𝑡 2 ∗ 𝑙𝑒𝑛𝑔𝑡ℎ = 500 ∗ 106 𝑏𝑖𝑡𝑠/ 𝑠𝑒𝑐 2 ∗ 105 𝑘𝑚/𝑠𝑒𝑐 161
Length = 2 km 5. In an IPv4 datagram, the M bit is 0, the value of HLEN is 10, the value of total length is 400 and the fragment offset value is 300. The position of the datagram, the sequence numbers of the first and the last bytes of the payload, respectively are GATE 2013 (A) Last fragment, 2400 and 2789 (B) First fragment, 2400 and 2759 (C) Last fragment, 2400 and 2759 (D) Middle fragment, 300 and 689 ANS: C SOL: Since M bit is 0, so there is no fragment after this fragment. Hence this fragment is the last fragment. Now, HLEN defines the length of header in datagram. Since Hlen is 10 so, size of header is 10 * 4 = 40 B Length of data = Total length – Header length = 400 – 40 = 360 B Now, fragment offset of data in original datagram is measured in units of 8 B. so to find first Byte of this fragment, First byte/8 = fragment offset First byte = 300 * 8 = 2400 B and since length of data is 360 B. So, last byte on this datagram will be 2759 6. In the IPv4 addressing format, the number of networks allowed under Class C addresses is GATE 2012 (A) 214 (B) 27 (C) 221 (D) 224 ANS: C SOL: For class C address, size of network field is 24 bits. But first 3 bits are fixed as 110; hence total number of networks possible is 221. 7. Which of the following transport layer protocols is used to support electronic mail? GATE 2012 (A) SMTP (B) IP (C) TCP (D) UDP ANS:C SOL: E-mail uses SMTP in application layer to transfer mail. And SMTP uses TCP to transfer data in transport layer. 8. The protocol data unit (PDU) for the application layer in the Internet stack is GATE 2012 (A) Segment (B) Datagram (C) Message (D) Frame ANS: C SOL: Protocol Data Unit (PDU) Application layer – Message Transport layer – Segment Network layer – Datagram Data Link layer – Frame Ans = Message 162
9. Consider an instance of TCP’s Additive Increase Multiplicative Decrease (AIMD) algorithm where the window size at the start of the slow start phase is 2 MSS and the threshold at the start of the first transmission is 8 MSS. Assume that a timeout occurs during the fifth transmission. Find the congestion window size at the end of the tenth transmission. GATE 2012 (A) 8 MSS (B) 14 MSS (C) 7 MSS (D) 12 MSS ANS: C SOL: Given threshold = 8 Time = 1, during first transmission, window size = 2 (slow start phase) Time = 2, congestion window size = 4 (double the no. of acknowledgments) Time = 3, congestion window size is = 8 Time = 4, congestion window size = 9, after threshold (increase by one addictive increase) Time = 5, transmits 10 MSS, but time out occurs congestion windw size = 10 Hence threshold = (congestion window size)/2=10/2 = 5 Time = 6, transmits 2 Time = 7, transmits 4 Time = 8, transmits 5(threshold is 5) Time = 9, transmits 6 Time = 10, transmits 7 During 10th transmission, it transmits 7 segments hence at the end of the 10th transmission the size of congestion window is 7 MSS. 10. Consider a source computer (S) transmitting a file of size 106 bits to a destination computer (D) over a network of two routers (R1 and R2) and three links (L1, L2, and L3). L1 connects S to R1; L2 connects R1 to R2; and L3 connects R2 to D. Let each link be of length 100km. Assume signals travel over each line at a speed of 108 meters per second. Assume that the link bandwidth on each link is 1Mbps. Let the file be broken down i n t o 1000 p a c k e t s each of size 1000 bits. Find the total sum o f transmission and p r o p a g a t i o n delays i n transmitting the file from S to D? GATE 2012 (A) 1005ms ANS: A SOL:
(B) 1010ms L1
S
(C) 3000ms L2
R1
R2
(D) 3003ms L3 D
Transmission delay for 1st packet from each of S, R1 and R2 will take 1 ms Propagation delay on each link l1,l2 and l3 for one packet is 1ms Therefore the sum of transmission delay and propagation delay on each link for one packet is 2ms. The first packet reaches the destination at 6th ms The second packet reaches the destination at 7th ms So, inductively we can say that 1000th packet reaches the destination at 1005th ms.
163
11. Consider the directed graph shown in the figure below. There are multiple shortest paths between vertices S and T. Which one will be reported by Dijkstra’s shortest path algorithm? Assume that, in any iteration, the shortest path to a vertex v is updated only when a strictly shortest path to v is discovered.
GATE 2012 (A) SDT (B) SBDT (C) SACDT (D) SACET ANS: D SOL: Let d[v] represent the shortest path distance computed from ‘S’ Initially d[S] = 0, d[A] = ∞, d[B] = ∞, - - - - -, d[T] = ∞ and let P[v] represent the predecessor of v in the shortest path from ‘S’ to ‘v’ and let P[v] = 1 denote that currently predecessor of ‘v’ has not been computed → Let Q be the set of vertices for which shortest path distance has not been computed → Let W be the set of vertices for which shortest path distance has not been computed → So initially, Q = {S, A, B, C, D, E, F, G, T}, W = ϕ We will use the following procedure Repeat until Q is empty{ 1. u = choose a vertex from Q with minimum d[u] value 2. Q = Q – u 3. update all the adjacent vertices of u 4. W = W U{u} } d[S] = 0, d[A] = ∞, d[B] = ∞, . . . . , d[T] = ∞ Iteration 1: Step 1: u = S Step 2: Q = {A, B, C, D, E, F, G, T} Step 3: final values after adjustment d[S] = 0, d[A] = 4, d[B] = 3, d[C] = ∞, d[D] = 7, d[E] = ∞ - - -, d[T] = ∞ P[A] = S, P[B] = S, P[C] = 1, P[D] = S, P[E] = 1 - - - , P[T] = 1 Step 4: W = {S} Iteration 2: Step 1: u = S Step 2: Q = {A, C, D, E, F, G, T} Step 3: final values after adjustment d[S] = 0, d[A] = 4, d[B] = 3, d[C] = ∞, d[D] = 7, d[E] = ∞ - - -, d[T] = ∞ P[A] = S, P[B] = S, P[C] = 1, P[D] = S, P[E] = 1 - - - , P[T] = 1 164
Step 4: W = {S, B} Iteration 3: Step 1: u = A Step 2: Q = {C, D, E, F, G, T} Step 3: final values after adjustment d[S] = 0, d[A] = 4, d[B] = 3, d[C] = 5, d[D] = 7, d[E] = ∞ - - -, d[T] = ∞ P[A] = S, P[B] = S, P[C] = A, P[D] = S, P[E] = 1 - - - , P[T] = 1 Step 4: W = {S, B, A} Iteration 4: Step 1: u = C Step 2: Q = {D, E, F, G, T} Step 3: final values after adjustment d[S] = 0, d[A] = 4, d[B] = 3, d[C] = 5, d[D] = 7, d[E] = 6, - - -, d[T] = ∞ P[A] = S, P[B] = S, P[C] = A, P[D] = S, P[E] = C, - - - , P[T] = 1 Step 4: W = {S, B, A, C} Iteration 5: Step 1: u = E Step 2: Q = {D, F, G, T} Step 3: final values after adjustment d[S] = 0, d[A] = 4, d[B] = 3, d[C] = 5, d[D] = 7, d[E] = 6, d[F] = ∞, d[G] = 8, d[T] = 10 P[A] = S, P[B] = S, P[C] = A, P[D] = S, P[E] = C, P[F] = 1, P[G] = E, P[T] = E Step 4: W = {S, B, A, C, E} After iteration 5, we can observe that P[T] = E, P[E] = C, P[C] = A, P[A] = S, So the shortest path from S to T is SACET 12. An Internet Service Provider (ISP) has the following chunk of CIDR-based IP addresses available with it: 245.248.128.0/20. The ISP wants to give half of this chunk of addresses to Organization. A, and a quarter to Organization B, while retaining the remaining with itself. Which of the following is a valid allocation of addresses to A and B? GATE 2012 (A) 245.248.136.0/21 and 245.248.128.0/22 (B) 245.248.128.0/21 and 245.248.128.0/22 (C) 245.248.132.0/22 and 245.248.132.0/21 (D) 245.248.136.0/24 and 245.248.132.0/21 ANS: A SOL:
th
Since half of 4096 host addresses must be given to organization A, we can set 12 bit to 1 and include that bit into network part of organization A, so the valid allocation of addresses to A is 245.248.136.0/21
165
th
th
Now for organization B, 12 bit is set to ‘0’ but since we need only half of 2048 addresses, 13 bit can be set to ‘0’ and include that bit into network part of organization B so the valid allocation of addresses to B is 245.248.128.0/22 13. Consider different activities related to email. m1: Send an email from a mail client to a mail server m2: Download an email from mailbox server to a mail client m3: Checking email in a web browser Which is the application level protocol used in each activity? (A) m1: HTTP m2: SMTP m3: POP
GATE 2011
(B) m1: SMTP m2: FTP m3: HTTP
(C) m1: SMTP m2: POP m3: HTTP (D) m1: POP m2: SMTP m3: IMAP ANS: C SOL: Mail client uses SMTP (Simple Mail Transfer Protocol) to send mail. (The client need not be web based. So, HTTP may not be involved here). POP (Post Office Protocol) is used to retrieve mail from mail server. HTTP (Hypertext transfer protocol) is used to transfer a HTML page containing the mail message that can be viewed on a web browser.
14. A layer-4 firewall (a device that can look at all protocol headers up to the transport layer) CANNOT GATE 2011 (A) Block entire HTTP traffic during 9:00PM and 5:00AM (B) Block all ICMP traffic (C) Stop incoming traffic from a specific IP address but allow outgoing traffic to the same IP address (D) Block TCP traffic from a specific user on a multi-user system during 9:00PM and 5:00Am ANS: A Statement for Linked Questions 15 and 16 Consider a network with five nodes, N1 to N5, as shown below.
166
The network uses a Distance Vector Routing protocol. Once the routes have stabilized, the distance vectors at different nodes are as following. N1: (0, 1, 7, 8, 4) N2: (1, 0, 6, 7, 3) N3: (7, 6, 0, 2, 6) N4: (8, 7, 2, 0, 4) N5: (4, 3, 6, 4, 0) Each distance vector is the distance of the best known path at the instance to nodes, N1 to N5, where the distance to itself is 0. Also, all links are symmetric and the cost is identical in both directions. In each round, all nodes exchange their distance vectors with their respective neighbors. Then all nodes update their distance vectors. In between two rounds, any change in cost of a link will cause the two incident nodes to change only that entry in their distance vectors. 15. The cost of link N2-N3 reduces to 2(in both directions). After the next round of updates, what will be the new distance vector at node, N3? GATE 2011 (A) (3, 2, 0, 2, 5) ANS: A
(B) (3, 2, 0, 2, 6)
(C) (7, 2, 0, 2, 5)
(D) (7, 2, 0, 2, 6)
16. After the update in the previous question, the link N1-N2 goes down. N2 will reflect this change immediately in its distance vector as cost, ∞. After the NEXT ROUND of update, what will be the cost to N1 in the distance vector of N3? GATE 2011 (A) 3 ANS: C
(B) 9
(C) 10
(D) ∞
17. One of the header fields in an IP datagram is the Time to Live (TTL) field. Which of the following statements best explains the need for this field? GATE 2010 (A) It can be used to prioritize packets (B) It can be used to reduce delays (C) It can be used to optimize throughput (D) It can be used to prevent packet looping ANS: D SOL: Whenever Time to live field reaches ‘0’ we discard the packet, so that we can prevent it from looping. 18. Which one of the following is not a client server application? (A) Internet chat ANS: D
(B) Web browsing
(C) E-mail
GATE 2010 (D) Ping
167
Statement for Linked Answer Questions: 19 & 20 Consider a network with 6 routers R1 to R6 connected with links having weights as shown in the following diagram
19. All the routers use the distance vector based routing algorithm to update their routing tables. Each router starts with its routing table initialized to contain an entry for each neighbor with the weight of the respective connecting link. After all the routing tables stabilize, how many links in the network will never be used for carrying any data? GATE 2010 (A) 4 (B) 3 (C) 2 (D) 1 ANS: D SOL: In Distance vector, the Router will update its routing tables by exchanging the information from all its neighbors. After all the routing tables stabilize the routing Table for ‘R1’ will not have any entry to Router R6., so that link will not be used. So one link. 20. Suppose the weights of all unused links in the previous question are changed to 2 and the distance vector algorithm is used again until all routing tables stabilize. How many links will now remain unused? GATE 2010 (A) 0 ANS: A
(B) 1
(C) 2
(D) 3
21. Which of the following statement(s) is / are correct regarding Bellman-Ford shortest path algorithm? P. Always finds a negative weighted cycle, if one exists. Q. Finds whether any negative weighted cycle is reachable from the source. GATE 2009 (A) P only (B) Q only (C) both P and Q (D) Neither P nor Q ANS: B SOL: The algorithm identifies a negative weight cycle iff it is reachable from Source.
168
22. Consider the following graph:
Which one of the following is NOT the sequence of edges added to the minimum spanning tree using Kruskal’s algorithm? GATE 2009 (A) (b,e) (e,f) (a,c) (b,c) (f,g) (c,d) (B) (b,e) (e,f) (a,c) (f,g) (b,c) (c,d) (C) (b,e) (a,c) (e,f) (b,c) (f,g) (c,d) (D) (b,e) (e,f) (b,c) (a,c) (f,g) (c,d) ANS: D SOL: Weight of edge (a,c) is less than (b,c) . So it cannot come after (b,c) 23. In the RSA public key cryptosystem, the private and public keys are (e, n) and (d, n) respectively, where n=p*q and p and q are large primes. Besides, n is public and p and q are private. Let M be an integer such that 0
ed ≡ 1 mod n ed ≡ 1 mod Φ(n) M' = Me mod Φ(n) M = (M')d mod Φ(n) Which of the above equations correctly represent RSA cryptosystem? (A) I and II ANS: B
(B) I and III
(C) II and IV
GATE 2009 (D) III and IV
Statement for Linked Answer Questions: 24 & 25 Frames of 1000 bits are sent over a 106 bps duplex link between two hosts. The propagation time is 25ms. Frames are to be transmitted into this link to maximally pack them in transit (within the link).
169
24. What is the minimum number of bits (i) that will be required to represent the sequence numbers distinctly? Assume that no time gap needs to be given between transmission of two frames. GATE 2009 (A) i=2 (B) i=3 (C) i=4 (D) i=5 ANS: D SOL: The transmission time for a frame is 1000/1Mbps = 1 ms .As the propagation time is 25 ms, the sender can transmit 25 packets before the first packet reaches the destination. Therefore the number of bits required to represent 25 packets is 5. 25. Suppose that the sliding window protocol is used with the sender window size of 2i, where i is the number of bits identified in the earlier part and acknowledgements are always piggy backed. After sending 2i frames, what is the minimum time the sender will have to wait before starting transmission of the next frame? (Identify the closest choice ignoring the frame processing time.) GATE 2009 (A) 16ms (B) 18ms (C) 20ms (D) 22ms ANS: B SOL: Sliding window size is 32 as i=5. The sender can expect an Ack after one RTT. Here Round trip time is 50ms. Therefore the sender has to wait at least 50-32= 18ms before transmission of the next frame. 26. What is the maximum size of data that the application layer can pass on to the TCP layer below? GATE 2008 (A) Any size (B) 216 bytes-size of TCP header (C) 216 bytes (D) 1500 bytes ANS: A SOL: Application layer can pass any length data. TCP layer will divide that data into frames. 27. Which of the following system calls results in the sending of SYN packets? GATE 2008 (A) socket (B) bind (C) listen (D) connect ANS: D SOL: In the process of establishing a connection between two endpoints, the user process on active end point invokes the connect() system call. The active end point then sends a SYN packet. The passive end point invokes an accept() system call and sends ACK to the other system then the connection is established. 28. Dijkstra’s single source shortest path algorithm when run from vertex ‘a’ in the following graph, computes the correct shortest path distance to
170
(A) only vertex a (C) only vertices a, b, c, d ANS: D
(B) only vertices a, e, f, g, h (D) all the vertices
SOL: Even though the graph has negative weights, it correctly computes the shortest path to all the vertices. There will not be any problem with the Dijkstra's algorithm operating on negative edge weights as long as the shortest path distance computed for the currently removed vertex is the actual shortest path distance. 29. In the slow start phase of the TCP congestion control algorithm, the size of the congestion window GATE 2008 (A) Does not increase (C) Increases quadratically ANS: B
(B) increases linearly (D) increases exponentially
30. If a class B network on the Internet has a subnet mask of 255.255.248.0, what is the maximum number of hosts per subnet? GATE 2008 (A) 1022 ANS: C SOL:
(B) 1023
(C) 2046
(D) 2047
Number of bits for subnet mask = 21 Number of bits for host = 11 Number of hosts = 211-2 = 2046
31. A computer on a 10Mbps network is regulated by a token bucket. The token bucket is filled at a rate of 2Mbps. It is initially filled to capacity with 16Megabits. What is the maximum duration for which the computer can transmit at the full 10Mbps? GATE 2008 (A) 1.6 seconds (B) 2 seconds (C) 5 seconds (D) 8 seconds ANS: B SOL: If the capacity of the token bucket is C bytes, Token arrival rate is R bytes/sec, and the Maximum possible transmission rate is M bytes/sec then the time(S) in seconds it is possible to transmit is S = C/(M-R) seconds , so 16/(10-2) = 2 seconds. 32. In Ethernet when Manchester encoding is used, the bit rate is:
GATE 2007 171
(A) Half the baud rate. (B) Twice the baud rate. (C) Same as the baud rate. (D) None of the above ANS: A SOL: In Ethernet when Manchester encoding is used, the bit rate is half of the baud rate. 33. Which one of the following uses UDP as the transport protocol?
GATE 2007
(A) HTTP (B) Telnet (C) DNS (D) SMTP ANS: C SOL: DNS queries are normally short and they need fast responses. Therefore UDP is a better option as a transport protocol for DNS.
34. There are n stations in a slotted LAN. Each station attempts to transmit with a probability p in each time slot. What is the probability that ONLY one station transmits in a given time slot? GATE 2007 (A) np(1-p)n-1 (B) (1-p)n-1 (C) p(1-p)n-1 (D) 1-(1-p)n-1 ANS: A SOL: The probability that only one station transmits in a given slot is, probability that station 1 transmits, stations 2 to n are not tranmitting + station 2 is transmitting and station 1, stations 3 to n are not transmitting + ............... which is, p(1-p)(1-P)....... + (1-p)p(1-p)(1-p) ........ + (1-p)(1-p)p(1-p) ......... + ....... ntimes = p(1-p)(n-1) + p(1-p)(n-1) + p(1-p)(n-1) + p(1-p)(n-1) + p(1-p)(n-1) + ...... n times = np(1-p)(n-1) 35. In a token ring network the transmission speed is 7 10 bps and the propagation speed is 200 metres/Rs. The 1-bit delay in this network is equivalent to: GATE 2007 (A) 500 metres of cable. (B) 200 metres of cable. (C) 20 metres of cable. (D) 50 metres of cable. ANS: C SOL: The time taken to transmit 1 bit is 0.1micro second. Propagation speed is 200meters/microsecond.Therefore 1 bit delay implies 20 metres. 36. The address of a class B host is to be split into subnets with a 6-bit subnet number. What is the maximum number of subnets and the maximum number of hosts in each subnet? (A) 62 subnets and 262142 hosts. (B) 64 subnets and 262142 hosts. (C) 62 subnets and 1022 hosts. (D) 64 subnets and 1024 hosts. ANS: C SOL: Maximum number of subnets is 26-2 =62. Maximum number of hosts is 210-2 = 1022. Actually at present , subnets with addresses all 0's and all 1's can also be used. 37. The message 11001001 is to be transmitted using the CRC polynomial 3 x + 1 to protect it from errors. The message that should be transmitted is: GATE 2007 172
(A) 11001001000 (B) 11001001011 (C) 11001010 (D) 110010010011 ANS: B SOL: The divisor is 1001. After dividing the given data 11001001 by 1001, the remainder is 011 which is the CRC. Therefore the transmitted data is, data+CRC which is 11001001011. 38. The distance between two stations M and N is L kilometers. All frames are K bits long. The propagation delay per kilometer is t seconds. Let R bits/second be the channel capacity. Assuming that processing delay is negligible, the minimum number of bits for the sequence number field in a frame for maximum utilization, when the sliding window protocol is used, is:
(A) ⌈log 2
2𝐿𝑡𝑅+2𝐾
(C) ⌈log 2 ANS: A
𝐾 2𝐿𝑡𝑅+𝐾 𝐾
⌉
⌉
(B) ⌈log 2 (D) ⌈log 2
2𝐿𝑡𝑅+𝐾 2𝐾
2𝐿𝑡𝑅 𝐾
⌉
⌉
SOL: The distance between the stations is L kilometers. Propagation delay per kilometer is t seconds. Therefore the total propagation delay is Lt seconds. The round trip time is 2*propagation delay = 2Lt seconds. Maximum utilization can be achieved by transmitting data for the whole round trip time. The size of data that can be transmitted for Round trip time is Time * bandwidth = 2LtR. other than round trip time a packet is transmitted, and an ack. is also transmitted when the packet is received. The size in bits is 2k. Therefore total size in bits is 2LtR + 2k. Number of packets is (2LtR+2k)/k. Number of bits required to represent these packets is log(Number of packets). 39. Match the following: (P) SMTP (Q) BGP (R) TCP (S) PPP
(1) Application layer (2) Transport layer (3) Data link layer (4) Network layer (5) Physical layer
GATE 2007
(A) P - 2 Q - 1 R - 3 S – 5 (B) P - 1 Q - 4 R - 2 S - 3 (C) P - 1 Q - 4 R - 2 S – 5 (D) P - 2 Q - 4 R - 1 S – 3 ANS: B SOL: SMTP is an application layer protocol. TCP is the transport layer protocol. BGP is network layer protocol and PPP is the data link layer protocol.
173
40. For which one of the following reasons does Internet Protocol (IP) use the time-to- live (TTL) field in the IP datagram header? GATE 2006 (A) Ensure packets reach destination within that time (B) Discard packets that reach later than that time (C) Prevent packets from looping indefinitely (D) Limit the time for which a packet gets queued in intermediate routers. ANS: C SOL: Time to live indicates the time or maximum number of hops the packet is allowed to make before it is discarded. Normally it is set to twice the maximum length path from source to destination. 41. Station A uses 32 byte packets to transmit messages to Station B using a sliding window protocol. The round trip delay between A and B is 80 milliseconds and the bottleneck bandwidth on the path between A and B is 128 kbps. What is the optimal window size that A should use? GATE 2006 (A) 20 (B) 40 (C) 160 ANS: B SOL: The packet size is 32 bytes = 32*8 bits.
(D) 320
Round trip time is 80ms. Bandwidth is 128kbps. Therefore in 1 RTT, the source a transmit 128kbps * 80ms bits of data. The data possible to be transmitted divided by the packet size gives the window size, which is (128k * 80 ms)/(32*8) = 40. 42. Two computers C1 and C2 are configured as follows. C1 has IP address 203.197.2.53 and net mask 255.255.128.0. C2 has IP address 203.197.75.201 and net mask 255.255.192.0. which one of the following statements is true? GATE 2006 (A) C1 and C2 both assume they are on the same network (B) C2 assumes C1 is on same network, but C1 assumes C2 is on a different network (C) C1 assumes C2 is on same network, but C2 assumes C1 is on a different network (D) C1 and C2 both assume they are on different networks. ANS: C SOL: IP address of C1 is 203.197.2.53 and Subnet mask is 255.255.128.0 ending gives the network id which is 203.197.0.0. When C1 sees the ipaddress 203.197.75.201, to find the network id it will and with its subnet mask, which gives 203.197.0.0. So C1 assumes that C2 is on the same network with C1. Similarly, IP address of C2 is 203.197.75.201, subnet mask is 255.255.192.0 ending gives the network id which is 203.197.64.0. 174
When this computer looks at IP address of C1, to find the network id, it will and with its network mask giving 203.197.0.0. Therefore C1 assumes that C2 is on the same network with C2, but C2 assumes C1 is on a different network. 43. Station A needs to send a message consisting of 9 packets to Station B using a sliding window (window size 3) and go-back-n error control strategy. All packets are ready and immediately available for transmission. If every 5th packet that A transmits gets lost (but no acks from B ever get lost), then what is the number of packets that A will transmit for sending the message to B? GATE 2006 (A) 12
(B) 14
(C) 16
(D) 18
ANS: C
SOL:
Assume that correctly transmitted packet is acknowledged at the same time. When a packet is lost, the receiver waits for certain duration but the sender can send up to its window size. The total number of packets sent is 16. 44. Packets of the same session may be routed through different paths in: (a) TCP, but not UDP (b) TCP and UDP (c) UDP, but not TCP (d) Neither TCP nor UDP ANS: B SOL: Packet is the Network layer Protocol Data Unit (PDU). TCP and UDP are Transport layer protocols. Packets of same session may be routed through different routes. Most networks don’t use 175
static routing, but use some form of adaptive routing where the paths used to route two packets for same session may be different due to congestion on some link, or some other reason. 45. The address resolution protocol (ARP) is used for:
GATE 2005
(a) Finding the IP address from the DNS (b) Finding the IP address of the default gateway (c) Finding the IP address that corresponds to a MAC address (d) Finding the MAC address that corresponds to an IP address ANS: D 46. The maximum window size for data transmission using the selective reject protocol with n-bit frame sequence numbers is: GATE 2005 (a) 2n
(b) 2n - 1
(c) 2n - 1
(d) 2n-2
ANS: B 47. In a network of LANs connected by bridges, packets are sent from one LAN to another through intermediate bridges. Since more than one path may exist between two LANs, packets may have to be routed through multiple bridges. Why is the spanning tree algorithm used for bridge-routing? GATE 2005 (a) For shortest path routing between LANs (b) For avoiding loops in the routing paths (c) For fault tolerance (d) For minimizing collisions ANS: B 48. An organization has a class B network and wishes to form subnets for 64 departments. The subnet mask would be: GATE 2005 (a) 255.255.0.0 (b) 255.255.64.0 (c) 255.255.128.0 (d) 255.255.252.0 ANS: D SOL: The size of network ID is 16 bit in class B networks. So bits after 16th bit must be used to create 64 departments. Total 6 bits are needed to identify 64 different departments. Therefore, subnet mask will be 255.255.252.0. 49. Suppose the round trip propagation delay for a 10 Mbps Ethernet having 48-bit jamming signal is 46.4 ms. The minimum frame size is: GATE 2005 (a) 94 ANS: C SOL:
(b) 416
(c) 464
(d) 512
Transmission Speed = 10Mbps. Round trip propagation delay = 46.4 ms The minimum frame size = (Round Trip Propagation Delay) * (Transmission Speed) 176
= 10*(10^6)*46.4*(10^-3) = 464 * 10^3 = 464 Kbit. The concept behind the above formula is collision detection. Consider a situation where a node A wants to send a frame to another node B. When Node A begins transmitting, the signal must propagate the network length. In the worst-case collision scenario, Node B begins to transmit just before the signal for Node A’s frame reaches it. The collision signal of Node A and Node B’s frame must travel back to Node A for Node A to detect that a collision has occurred. The time it takes for a signal to propagate from one end of the network to the other is known as the propagation delay. In this worst-case collision scenario, the time that it takes for Node A to detect that its frame has been collided with is twice the propagation delay. Node A’s frame must travel all the way to Node B, and then the collision signal must travel all the way from Node B back to Node A. This time is known as the slot time. An Ethernet node must be transmitting a frame for the slot time for a collision with that frame to be detected. This is the reason for the minimum Ethernet frame size.
50. Choose the best matching between Group 1 and Group 2 Group –1 Group – 2 P. Data link layer
1. Ensures reliable transport of data over a physical Point-to-point link
Q. Network layer
2. Encodes/decodes data for physical transmission
R. Transport layer
3.Allows end-to-end communication between two processes 4. Routes data from one network node to the next (b) P – 2, Q – 4, R - 1 (d) P – 1, Q – 3, R – 2
(a) P – 1, Q – 4, R - 3 (c) P – 2, Q – 3, R - 1 ANS: A SOL: Transport layer is responsible for end to end communication, creation of sockets.Network layer routes the data from one node to other, till it reach to destination. Datalink layer ensures reliable data transfer by error correction, duplication check ordered delivery etc.
51. Which of the following is NOT true with respect to a transparent bridge and a router? (a) Both bridge and router selectively forward data packets (b) A bridge uses IP addresses while a router uses MAC addresses (c) A bridge builds up its routing table by inspecting incoming packets (d) A router can connect between a LAN and a WAN ANS: B 177
SOL: Bridge is the device which work at data link layer whereas router works at network layer. Both selectively forward packets, build routing table & connect between LAN & WAN but since bridge works at data link it uses MAC addresses to route whereas router uses IP addresses. 52. The routing table of a router is shown below:
On which interface will the router forward packets addressed to destinations 128.75.43.16 and 192.12.17.10 respectively?
(a) Eth1 and Eth2
(b) Eth0 and Eth2
(c) Eth0 and Eth3 ANS: C
(d) Eth1 and Eth3
SOL: Given IP Address 128.75.43.16. (1) Eth 0 128.75.43.0. (2) Mask 255.255.255.0. Equation (1) & (2) both are of same network. 192.12.17.10. (1) Eth3 192.12.17.5 (2) Mask 255.255.255.255 Equation (1) & (2) both are of same network. 53. Which of the following assertions is FALSE about the Internet Protocol (IP)? GATE 2003 (A) It is possible for a computer to have multiple IP addresses (B) IP packets from the same source to the same destination can take different routes in the network (C) IP ensures that a packet is discarded if it is unable to reach its destination within a given number of hops (D) The packet source cannot set the route of an outgoing packets; the route is determined only by the routing tables in the routers on the way ANS: A SOL: Internet protocol ensures that a packet is forwarded if it is nable to reach its 178
destination within a given no. of hops. One computer can have multiple IP addresses also packets having same source & destination can take different routes. Source doesn’t decide where to route the packet, but it is decided by the routing tables at intermediate routers.
54. Which of the following functionalities must be implemented by a transport protocol over and above the network protocol?GATE 2003 (A) Recovery from packet losses (B) Detection of duplicate packets (C) Packet delivery in the correct order (D) End to end connectivity ANS: D SOL: Transport protocols are mainly for providing end to end connections by making sockets. Recovery from packet loss & delivery in correct order, duplication is checked by Data link layer.
WEB TECHNOLOGIES 1.
DMSP stands for A. Distributed Mail System Protocol C. Distributed Message System Protocol
B. Distributed Message System Pool D. Distributed Mai l System Pool
Answer: Option A Explanation: DMSP stands for a “ Distributed Mai l system Protocol .
2.
The term byte stuffing refers to: A. Data stuffing used with character oriented hardware. C. data stuffing used with both (A) and(B)
B. Data stuffing used with bit oriented hardware. D. Data stuffing used with byte oriented hardware.
Answer & Explanation Answer: Option A Explanation: 179
The term byte stuffing refers to data stuffing used with character -oriented hardware
3.
In 32bit IP Addressing scheme all 1's represent A. this computer. C. limited broadcast. Answer & Explanation
B. directed broadcast. D. loop back.
Answer: Option C Explanation: In 32 bit IP Addressing scheme all 1s represent limited broadcast.
4.
Which Layer is not present in TCP/ IP model? A. Application Layer C. Transport Layer Answer & Explanation
B. Internet Layer D. Presentation Layer
Answer: Option D Explanation: Presentation layer is not present in TCP/ IP Model.
5.
All exceptions in Java are subclasses of built in class called A. Exception B. Error. D. Raise. C. Throwable. Answer & Explanation Answer: Option C Explanation: All exception in Java are subclasses of built in class called Throwable. 6.
Unlike Ipv4, Ipv6 does not include the following field in the base header A. Next Header field. B. Field for Fragmentation information D. Kind field. C. Flow Label. Answer & Explanation 180
Answer: Option B Explanation: Unlike Ipv4, Ipv6 does not include the Field for Fragmentation information in the base header. 7.
Let most segment of a name inn DNS represents A. Individual Network. C. Domain name Answer & Explanation
B. Individual computer. D. Network t ype.
Answer: Option B Explanation: Left Most segment of a name in DNS represents- Individual computer
8.
A header in CGI script can specify A. format of the document. C. (A) and (B) both Answer & Explanation
B. new location of the document. D. start of the document.
Answer: Option A Explanation: A header in CGI script can specify - Format of the document & New location of the document. 9.
Address 192.5.48.3 belongs to A. class A. C. class C. Answer & Explanation
B. class B. D. class D.
Answer: Option C Explanation: Address 192.5.48.3 belongs to class C. 10. FDDI (Fiber Distributed Data Interconnect) is an example of A. token ring. B. token bus D. multipoint network. C. star topology 181
Answer & Explanation Answer: Option A Explanation: FDDI is an example of token ring. 11. CIDR stands for A. Classified Internet Domain Routing C. Classless Internet Domain Routing Answer & Explanation
B. Classless Inter Domain Routing D. Classified Inter Domain Routing
Answer: Option B Explanation: The total number of class of IP addresses are 5. 12. A Network uses a star topology if A. Computers are arranged in a closed loop. C. All computers attach to a single long cable.
B. All computers attach to a central point. D. Computers attach to multiple hierarchical cables.
Answer & Explanation Answer: Option B Explanation: A Network uses a star topology if all computers attach to a central point. 13. In IP addressing scheme, class used for multicasting is: A. Class A B. Class B D. Class D C. Class C Answer & Explanation Answer: Option D Explanation: In IP addressing scheme, class used for multicasting is class D. 14. The total number of class of IP address are A. 3.
B. 4. 182
C. 5. Answer & Explanation
D. 9.
Answer: Option C Explanation: 15. In TCP protocol header 'checksum' is of___________ A. 8 bits C. 32 bits Answer & Explanation
B. 16 bits D. 64 bis
Answer: Option B Explanation: In TCP protocol header checksum is of 16 bits. 16. Error detecting method that can detect more errors without increasing additional information in each packet is A. checksum B. even parity mechanism D. odd parity mechanism. C. CRC Answer & Explanation Answer: Option C Explanation: Error detecting method that can detect more errors without increasing additional information in each packet is CRC. 17. Parent class of all Java classes is A. java.lang.system C. java.lang.class Answer & Explanation
B. java.lang.object D. java.lang.reflect.object
Answer: Option B Explanation: Parent class of all Java classes is java.lang.object. 18. Hardware that calculates CRC(Cyclic Redundancy Check) uses: A. Shift register B. Xor unit 183
C. Both (A)and (B) Answer & Explanation
D. Instruction register
Answer: Option B Explanation: Hardware that calculates CRC uses shift register and Xor unit. 19. Except ions of type error inn JAVA are handled by A. User program C. Operating system kerne Answer & Explanation
B. Java run time environment D. Interrupt
Answer: Option B Explanation: Exceptions of type error in JAVA are handled by JAVA run time environment. 20. MTU is specified by A. IP Datagram size C. TCP Segment size Answer & Explanation
B. Hardware technology D. None of the above.
Answer: Option B Explanation: MTU is specified by hardware technology. 21. Consider the HTML t able definition given below: (GATE 2009) < table border=1> | ab | cd |
| ef | gh |
| ik |
184
The number of rows in each column and the number of columns in each row are: A. (2,2,3) and (2,3,2) B. (2,2,3) and (2,2,3) D. (2,3,2) and (2,2,3) C. (2,3,2) and (2,3,2) Answer & Explanation Answer: Option C Explanation: Here two td command used in the first tr command and three td command used in second tr command, So required rows in each column is <2,3,2> 22. HTML (HyperText Markup Language) has language elements which permit certain actions other than describing the structure of the web document. Which one of the following actions is NOT supported by pure HTML (without any server or client side scripting) pages?(GATE 2011) A. Embed web objects from different B. Refresh the page automatically after a sites into the same page specified interval D. Display the client time as C. Automatically redirect to another page upon download part of the page Answer & Explanation Answer: Option D Explanation: As per Theory.
COMPUTER ORGANIZATION
YEAR 2001 Question. 1 More than one word are put in one cache block to (E) Exploit the temporal locality of reference in a program (F) Exploit the spatial locality of reference in a program (G) Reduce the miss penalty (H) None of the above
185
SOLUTION Cache is the small memory which has a very less access time. So it is used for temporal locality of reference whereas virtual memory is for spatial locality of reference. Hence (A) is correct option.
Question. 2 A low memory can be connected to 8085 by using (A) INTER
(B) RESET IN
(C) HOLD
(D) READY
SOLUTION memory can be connected to 8085 by using READY signal. If READY is set then communication is possible.Hence (D) is correct option. Question. 3 Suppose a processor does not have any stack pointer register. Which of the following statements is true ? (A) It cannot have subroutine call instruction (B) It can have subroutine call instruction, but no nested subroutine calls. (C) Nested subroutine calls are possible, but interrupts are not. (D) All sequences of subroutine calls and also interrupts are possible SOLUTION Stack pointer register holds the address of top of stack, which is the location of memory at which the CPU should resume its execution after servicing some interrupt or subroutine call. So if SP register not available then no subroutine call instructions are possible. Hence (A) is correct option. Question. 4 A processor needs software interrupt to (E) Test the interrupt system of the processor. (F) Implement co-routines. (G) Obtain system services which need execution of privileged instructions. (H) Return from subroutine. SOLUTION 186
A CPU needs software interrupt to obtain system services which need execution of privileged instructions. Hence (C) is correct opton. Question. 5 A CPU has two modes-privileged and non-privileged. In order to change the mode from privileged to non-privileged. (A) A hardware interrupt is needed. (E) A software interrupt is needed. (F) A privileged instruction (which does not generate an interrupt) is needed. (G) A non-privileged instruction (Which does not generate an interrupt) is needed. SOLUTION A software interrupt is initiated by some program module which need some CPU services, at that time the two modes can be interchanged. Hence (B) is correct option. Question. 6 The process of assigning load addresses to the various parts of the program and adjusting the code and date in the program to reflect the assigned addresses is called (A) Assembly (B) Parsing (C) Relocation
(D) Symbol resolution
SOLUTION Load addresses are assigned to various parts of the program, the program can be loaded at any location in memory. This location is added to all addresses in the code, to get correct references. This makes a code re-locatable. Hence (C) is correct option. Question. 7 Which of the following requires a device driver ? (A) Register (B) Cache (C) Main memory
(D) Disk
SOLUTION Device driver is the program which co-ordinates with CPU to regulate the devices. Register, cache & 187
main memory are directly connected to CPU. So only Disk from given options require device drivers. Hence (D) is correct option. Question. 8 Which is the most appropriate match for the items in the first column with the items in the second column (X.) Indirect Addressing
(I.) Array implementation
(Y.) Indexed Addressing
(II.) Writing re-locatable code
(Z.) Base Register Addressing (III.) Passing array as parameter (A) (X, III) (Y, I) (Z, II)
(B) (X, II) (Y, III) (Z, I)
(C) (X, III) (Y, II) (Z, I)
(D) (X, I) (Y, III) (Z, II)
SOLUTION Indexed addressing is used for array implementation where each element has indexes. Base register is used to re-locatable code, where starts from base address & then all local addresses as added to base address. Indirect addressing is done when array is passed as parameter only name is passed. Hence (A) is correct option. Question. 9 Consider the following data path of a simple non-pilelined CPU. The registers A, B, A1, A2, MDR the bus and the ALU are 8-bit wide. SP and MAR are 16-bit registers. The MUX is of size 8 X (2:1) and the DEMUX is of size 8 X (1:2). Each memory operation takes 2 CPU clock cycles and uses MAR (Memory Address Register) and MDR (Memory Date Register). SP can be decremented locally.
The CPU instruction “push r”, where = A or B, has the specification M [ SP] !r SP ! SP − 1 188
How many CPU clock cycles are needed to execute the “push r” instruction ? (A) 2
(B) 3
(C) 4
(D) 5
SOLUTION Push ‘r’ Consist of following operations M [ SP] !r SP ! SP − 1 ‘r’ is stored at memory at address stack pointer currently is, this take 2 clock cycles. SP is then decremented to point to next top of stack. So total cycles = 3 Hence (B) is correct option. Question. 10 Which of the following does not interrupt a running process ? (A) A device
(B) Timer
(C) Scheduler process
(D) Power failure
SOLUTION A device can request interrupt service. A timer when finishes or power failure causes a running process to stop. But a scheduler process doesn’t do this. Hence (C) is correct option. YEAR 2002 Question. 11 A device employing INTR line for device interrupt puts the CALL instruction on the data bus while (A) INTA is active
(B) HOLD is active
(C) READY is active
(D) None of the above
LUTION INTR is a signal which if enabled then microprocessor has interrupt enabled it receives high INR signal & activates INTA signal, so another request can’t be accepted till CPU is busy in servicing interrupt. Hence (A) is correct option. 189
Question. 12 In 8085 which of the following modifies the program counter ? (E) Only PCHL instruction (F) Only ADD instructions (G) Only JMP and CALL instructions (H) All instructions SOLUTION Program counter is the register which has the next location of the program to be executed next. JMP & CALL changes the value of PC. PCHL instruction copies content of registers H & L to PC. ADD instruction after completion increments program counter. So program counter is modified in all cases. Hence (D) is correct option. Question. 13 In serial data transmission, every byte of data is padded with a ‘0’ in the beginning and one or two ‘1’s at the end of byte because (E) Receiver is to be synchronized for byte reception (F) Receiver recovers lost ‘0’s and ‘1’ from these padded bits (G) Padded bits are useful in parity computation. (H) None of the above SOLUTION In serial data transmission the sender & receiver needs to be synchronized with each other. Receiver should know when 1 byte of data has been sent. 0 & 1’s which are padded tell the receiver to synchronize. Hence (A) is correct option. uestion. 14 Which of the following is not a form of memory ? (A) Instruction cache (B) Instruction register (C) Instruction opcode
(D) Translation-a-side buffer
SOLUTION 190
Instruction register stores instruction, look-a-side buffer & instruction cache are also memory. But instruction opcodes are the opcodes related to an instruction which are not part of memory hierarchy. Hence (C) is correct option. Question. 15 In the C language (3) At most one activation record exists between the current activation record and the activation record for the main. (4) The number of activation records between the current activation record and the activation record for the main depends on the actual function calling sequence. (5) The visibility of global variables depends on the actual function calling sequence. (6) Recursion requires the activation record for the recursive function to be saved on a different stack before the recursive fraction can be called. SOLUTION Activation record is the contiguous memory locations where the data needed by the program is kept so at most one activation record exist between current activation record & the record for the main. Hence (A) is correct option. Question. 16 In the absolute the addressing mode (E) The operand is inside the instruction (F) The address of the operand is inside the instruction (G) The register containing the address of the operand is specified nside the instruction (D) The location of the operand is implicit SOLUTION In absolute addressing mode, no need of giving operand, the operand are implicit, instruction itself has knowledge of operands. Hence (D) is correct option. Question. 17 The performance of a pipelined processor suffers if 191
(E) The pipelined stages have different delays (F) Consecutive instructions are dependent on each other (G) The pipeline stages share hardware resources (H) All the above SOLUTION Pipelining is a method to execute a program breaking it in several independent sequence of stages. In that case pipeline stages can’t have different delays, no dependency among consecutive instructions & sharing of hardware resources shouldn’t be there. So option (D) is true Hence (D) is correct option. Question. 18 Horizontal microprogramming t
Does not require use of signal decoders
t
Results in larger sized microinstructions than vertical microprogramming
t
Uses one bit for each control signal
t
All of the above
SOLUTION
192
In horizontal microprogramming the instruction size is not large, & no decoding is required. But 1 bit is used for all control signals. Hence (C) is correct option. YEAR 2003 Question. 19 For a pipelined CPU with a single ALU, consider the following situations 1. The j + 1 − st instruction uses the result of j − th instruction as an operand 2. The execution of a conditional jump instruction 3. The j − th and j + 1 − st instructions require the ALU at the same time Which of the above can cause a hazard? (A) 1 and 2 only (B) 2 and 3 only (C) 3 only
(D) All the three
SOLUTION Case 1 is here of data dependency, this can’t be safe with single ALU so read after write. Case 2 Conditional jumps are always hazardous they create conditional dependency in pipeline Case 3 This is write after read problem or concurrency dependency so hazardous All the three are hazardous. Hence (D) is correct option. Question. 20 Consider an array multiplier for multiplying two n bit numbers. If each gate in the circuit has a unit delay, the total delay of the multiplier is (A) Θ(1) (B) Θ(log n) (C) Θ(n) (D) Θ(n2) SOLUTION The no. of gates used in n bit array multiplier (n X n) is 2n − 1. So. if every single gate takes unit delay, then total delay 0(2n − 1) = 0(n) It is of linear order Hence (C) is correct option. 193
Question. 21 Consider the ALU shown below
If the operands are in 2’s complement representation, which of the following operations can be performed by suitably setting the control lines K and C0 only (+ and - denote addition and subtraction respectively)? (A) A + B, and A − B,but notA + 1 (B) A + B,and A + 1,but notA − B (C) A + B,but not A − B,orA + 1 (D) A + B,and A − B,andA + 1 SOLUTION This is the ckt to add two numbers in 2’s complement form. K & C0 are set to 1. So A + B & A − B using bit adders can be done. Also since C0 = 1 & in case B 0, B1........ all are 0 then it gives A + 1. Hence (D) is correct option.
Data for Q. 22 & 23 are given below. Consider the following assembly language program for a hypothetical processor. A,B and C are 8 bit registers. The meanings of various instructions are shown as comments. MO V B, # 0 ; B!0 MO V C, # 8 ; C!8 Z: CMP C, # 0 ; compare C with 0 JZX ; jump to X if zero flag is set SUB C, # 1
; C!C−1
194
RRC A, # 1
; right rotate A through carry ; by one bit. Thus: if the ; initial values of A and the ; carry flag are a7 ....... a0 and c0 ; respectively, their values ; after the execution of this ; instruction will be c0 a7 .....a1 ; and a0 respectively. ;jump to Y if carry flag is set ; jump to Z ; B!B+1 ; jump to Z
JCY JMP Z Y: ADD B, # 1 JMP Z X: Question. 22
If the initial value of register A is A0, the value of register B after the program execution will be (A) the number of 0 bits in A0 (B) the number of 1 bits in A (C) A0 (D) 8 SOLUTION Here value of B incremented by 1 only if carry flag is 1, carry is filled using right rotation, so B will store the no. of is in A0. Hence (B) is correct option. Question. 23 Which of the following instructions when inserted at location X will ensure that the value of register A after program execution is the same as its initial value? (A) RRC A,# 1 (B) NOP
;
no operation
(C) LRC A, # 1
;
left rotate A through carry flag by one bit
(D) ADD A, # 1 195
SOLUTION In the end of program execution to check whether both initial and final value of register A is A0, we need to right rotate register A through carry by one bit. Hence (A) is correct option. YEAR 2004 Question. 24 Which of the following addressing modes are suitable for program relocation at run time? 1. Absolute addressing 2. Based addressing 3. Relative addressing 4. Indirect addressing (A) 1 and 4 (C) 2 and 3
(B) 1 and 2 (D) 1,2 and 4
SOLUTION Program relocation at run time transfers complete block to some memory locations. This require as base address and block should be relatively addressed through this base address. This require both based addressing and relative addressing mode. Hence (C) is correct option. Question. 25 Consider a multiplexer with X and Y as data inputs and Z as control input.Z = 0 selects input X , and Z =1 selects input Y . What are the connection required to realize the 2-variable Boolean function f = T + R, without using any additional hardware? (A) R to X, 1 to Y, T to Z
(B) T to X, R to Y, T to Z
(C) T to X, R to Y, 0 to Z
(D) R to X, 0 to Y, T to Z
196
SOLUTIO N
We require f = T + R We have MUX equation f = Z' x + zy Now if we make following ckt
Truth table R T F
Z
0 0
0
0
0 1
1
1
1 0
1
0
1 1
1
1
So X = R Y = 1 Z = T f
=T'R+T = (T + T ')(T + R)
f
=T+R
Hence (A) is correct option.
Data for Q. 26 & 27 are given below. Consider the following program segment for a hypothetical CPU having 197
three user registers R1,R2 and R3.
198
Instruction
Operation
Instruction Size (in words)
MOV R1,5000
;R1←Memory[5000]
2
MOV R2,R3
;R2←R2+R3
1
ADD R2,R3
;R2←R2+R3
1
MOV 6000,R2
;Memory[6000]←R2
2
HALT
;Machine halts
1
Question. 26 Consider that the memory is byte addressable with size 32 bits, and the program has been loaded starting from memory location 1000 (decimal). If an interrupt occurs while the CPU has been halted after executing the HALT instruction, the return address (in decimal) saved in the stack will be (A) 1007
(B) 1020
(C) 1024
(D) 1028
SOLUTION Byte addressable so 1 word require 4 bytes. Instruction no.
Size
Address range
1
2
1000-1007
2
1
1008-1011
3
1
1012-1015
4
2
1016-1023
5
1
1024-1027
Next location 1028. CPU has executed the HALT instruction so next time the CPU will resume at next location i.e. 1028 which would be at the top of stack. Hence (D) is correct option. Question. 27 Let the clock cycles required for various operations be as follows: Register to/from memory transfer:
3 clock cycles
199
ADD with both operands in register: 1 clock cycle Instruction fetch and decode:
2 clock cycles per word
The total number of clock cycles required to execute the program is (A) 29
(B) 24
(C) 23
(D) 20
SOLUTION The clock cycles are per block so if an instruction size is 2 then it requires twice no. of clock cycles. Instruction No. Size
No. of clock cycles
1
2
3X2+2
8
2
1
1X3+2
5
3
1
1(add only)
1
4
2
3X2+2
8
5
1
2(fetch & decode)
2
Total
24
Hence (B) is correct option. Question. 28 Consider a small two-way set-associative cache memory, consisting of four blocks. For choosing the block to be replaced, use the least recently used (LRU) scheme. The number of cache misses for the following sequence of block addresses is 8, 12,0, 12,8 (A) 2
(B) 3
(C) 4
(D) 5
SOLUTION
200
After than 12 & 8 are referred but this does not cause any miss So no. of miss = 3 This stars ()) shows the misses. Hence (B) is correct option. Question. 29 The microinstructions stored in the control memory of a processor have a width of 26 bits. Each microinstruction is divided into three fields: a micro-operation field of 13 bits, a next address field (X), and a MUX select field (Y). There are 8 status bits in the inputs of the MUX.
How many bits are there in the X and Y fields, and what is the size of the control memory in number of words? (A) 10, 3, 1024
(B) 8, 5, 256
(C) 5, 8, 2048
(D) 10, 3, 512
SOLUTION MUX has 8 states bits as input lines so we require 3 select inputs to select & input lines. No. of bits in control memory next address field = 26 − 13 − 3 = 10 10 bit addressing, we have 210 memory size. So X, Y size = 10,3,1024 Hence (A) is correct option. Question. 30 A hard disk with a transfer rate of 10 M bytes/second is constantly
201
transferring data to memory using DMA. The processor runs at 600 MHz. and takes 300 and 900 clock cycles to initiate and complete DMA transfer respectively. If the size of the transfer is 20 Kbytes, what is the percentage of processor time consumed for the transfer operation? (A) 5.0% (B) 1.0% (C) 0.5%
(D) 0.1%
SOLUTION Transfer rate = 10 MB ps Data = 20 KB Time = 20 X210 = 2 X 10−3 10 X 220 = 2 ms Processor speed = 600 MHz = 600 cycles/sec. Cycles required by CPU = 300 + 900 For DMA = 1200 1200 So time = = .002 ms 6 600 X 10 %=
.002
X 100
2 = 0.1% Hence (D) is correct. Question. 31 A 4-stage pipeline has the stage delays as 150, 120, 160 and 140 nanoseconds respectively. Registers that are used between the stages have a delay of 5 nanoseconds each. Assuming constant clocking rate, the total time taken to process 1000 data items on this pipeline will be (A) 120.4 microseconds (B) 160.5 microseconds (C) 165.5 microseconds
(D) 590.0 microseconds
SOLUTION Delay = 5 ns/stage Total delay in pipline. 202
= 150 + 120 + 160 + 140 = 570 Delay due to 4 stages. Stage 1 delay 1 stage 2 delay 2 stage 3 stage 4 5 X 3 = 15 Total = 570 + 15 = 585 Total time
= 1000 data items 585 ns = 165.5 microseconds. Hence (C) is correct option
YEAR 2005 Question. 32 Which one of the following is true for a CPU having a single interrupt request line and a single interrupt grant line? (A) Neither vectored interrupt nor multiple interrupting devices are possible (B) Vectored interrupts are not possible but multiple interrupting devices are possible (C) vectored interrupts and multiple interrupting devices are both possible (D) vectored interrupt is possible but multiple interrupting devices are not possible SOLUTION CPU has single interrupt request and grant line
Here multiple request can be given to CPU but CPU interrupts only for highest priority interrupt so option (A) & (D) are wrong. But here in case of single interrupt lines definitely vectored interrupts are not possible. Hence (B) is correct option.
203
Question. 33 Normally user programs are prevented from handing I/O directly by I/O instructions in them. For CPUs having explicit I/O instructions, such I/O protection is ensured by having the I/O instructions privileged. In a CPU with memory mapped I/O, there is no explicit I/O instruction. Which one of the following is true for a CPU with memory mapped I/O? (E) I/O protection is ensured by operating system routine(s) (F) I/O protection is ensured by a hardware trap (G) I/O protection is ensured during system configuration (H) I/O protection is not possible SOLUTION In memory mapped I/0 the complete system (memory + I/0 ports) hold the same set of addresses. They are considered to be the part of that memory only. This management is done by OS only. Hence (A) is correct option. Question. 34 What is the swap apace in the disk used for? (E) Saving temporary html pages (F) Saving process data (G) Storing the super-block (H) Storing device drivers SOLUTION Swap space is the memory pre allowed to store process’s data. This can be compared with virtual memory. The data required to complete process is kept here. Hence (B) is correct option. Question. 35 Increasing the RAM of a computer typically improves performance because (E) Virtual memory increases (F) Larger RAMs are faster 204
(C) Fewer page faults occur (D) Fewer segmentation faults occur SOLUTION Due to increase in RAM size all the pages required by CPU are available in RAM so page fault chance are less, so virtual memory access chances are less and latency is reduced for secondary memory. Hence (C) is correct option. Question. 36 Consider a three word machine instruction ADD A [R0],@B The first operand (destination) “A [R0]” uses indexed addressing mode with R0 as the index register. The second operand (source) "@B" uses indirect addressing mode. A and B are memory addresses residing at the second and the third words, respectively. The first word of the instruction specifies the opcode, the index register designation and the source and destination addressing modes. During execution of ADD instruction, the two operands are added and stored in the destination (first operand). The number of memory cycles needed during the execution cycle of the instruction is (A)
3
(B) 4
(C)
5
(D) 6
SOLUTION ADD A [ R 0],@B This is instruction has 3 computational parts. ADD instruction requires 1 machine cycle, A [ R0] here R0 is index register which has starting address of index then this index has the block address. This whole operation require 3 machine cycles. Now @ B is indirect addressing. This takes 2 machine cycles. So overall 1 + 3 + 2 = 6 machine cycles. Hence (D) is correct option.
205
Question. 37 Match List-I with List-II and select the correct answer using the codes given below the lists: List-I
List-II
A. A [1] = B [j];
1. Indirect addressing
B. while [* A ++];
2. Indexed addressing
C. int temp=*x ;
3. Auto increment
Codes: A
B
C
(A)
3
2
1
(B)
1
3
2
(C)
2
3
1
S
123
SOLUTION LIST-I (iv) A [1] = B [ J];
2
(v) While [) A ++]
3.
Auto increment the memory locations is A are automatically incriminated.
1.
Indirect addressing here temp is assigned the value of int type stored at the address contained in X
(vi) int temp=) X
LIST-II Indexed addressing here the indexing is used
A2 B3 C1 Hence (C) is correct option. Question. 38
206
Consider a direct mapped cache of size 32 KB with block size 32 bytes. The CPU generates 32 bit addresses. The number of bits needed for cache indexing and the number of tag bits are respectively (A) 10,17 (B) 10,22 (C) 15,17
(D) 5,17
SOLUTION Cache is direct mapped. Size of Cache = 32 KB = 25 X 210 B = 215 Bytes. Require 15 bits for cache addressing so CPU address has tag and index No. of tag bits = 32 − 15 = 17 From 15 Cache addressing bits consist of blocks & words. Each block has 32 words (bytes) So require 5 bit. Index = block + word Block = 15 − 5 = 10 So, 10, 17 Hence (A) is correct option. Question. 39 A 5 stage pipelined CPU has the following sequence of stages IF-Instruction fetch from instruction memory. RD-Instruction decode and register read, EX- Execute:ALU operation for data and address computation, MA-Data memory access-for write access the register read at RD stage it used, WB-register write back. Consider the following sequence of instruction: I 1 : LR0,Locl; R0 <= M[Locl] I 2 AR0,R0; R0 <= R0 + R0 I 3 AR2,R0; R2 <= R2 − R0 Let each stage take one clock cycle. What is the number of clock cycles taken to complete the above sequence of instruction starting from the fetch of I1? 207
(A) 8
(B) 10
(C) 12
(D) 15
SOLUTION Order of phase in instruction cycle. IF " A 3" E X MA " WB 1 2
3
4
5
6
7
8
9
10
R 0 ! M [ LOC] IF RD EX MA WB R 0 ! R 0 + R0
IF
RD
EX MA WB
R 2 ! R 2 − R0 R 0 ! R 0 + R0 R 2 ! R 2 − R0 R 0 ! R 0 + R0
IF
RD
EX MA WB
can’t start before 3 since R0 has not been read by I1. can’t start before till 5th cycle
6th
since I2 has not executed
Total cycles = 10 Hence (B) is correct option. Question. 40 A device with data transfer rate 10 KB/sec is connected to a CPU. Data is transferred byte-wise. Let the interrupt overhead be 4 μ sec. The byte transfer time between the device interface register and CPU or memory is negligible. What is the minimum performance gain of operating the device under interrupt mode over operating it under program controlled mode? (A) 15 (B) 25 (C) 35
(D) 45
SOLUTION Data transfer rate = 10000 B/sec Total data = 25 X 103 Rate = 104 Performance gain = 10254 X 100 = 25% 208
Hence (B) is correct option. Question. 41 Consider a disk drive with the following specification 16 surfaces, 512 tracks/surface, 512 sectors/track, 1 KB/sector, rotation speed 3000 rpm. The disk is operated in cycle stealing mode whereby whenever one byte word is ready it is sent to memory; similarly, for writing, the disk interface reads a 4 byte word from the memory in each DMA cycle. Memory cycle time is 40 nsec. The maximum percentage of time that the CPU gets blocked during DMA operation is (A) 10
(B) 25
(C) 40
(D) 50
SOLUTION Disk revolutions = 3000 PM or 50 RPS At a time can read in One revolution = 512 KB = 219 ) Tracks read/sec 22 50 = 50 X 217 per sec. Interrupt = .2621 sec Percentage gain = .26211 X 100 , 26% Hence (B) is correct option.
Data for Q. 42 & 43 are given below Consider the following data path of a CPU
209
The, ALU, the bus and all the registers in the data path are of identical size. All operations including incrementation of the PC and the GPRs are to be carried out in the ALU. Two clock cycle are needed for memory read operation-the first one for loading data from the memory but into the MDR. Question. 42 The instruction “add R0,R1” has the register transfer in terpretation R0 <= R0 + R1. The minimum number of clock cycles needed for execution cycle of this instruction is (A) 2 (C) 4
(B) 3 (D) 5
SOLUTION R0 ! R 0 + R1 First cycle require to fetch operands two cycles required for this. The next cycle required to use ALU to perform ADD operation. So total cycles required = 3 Hence (D) is correct option. Question. 43 The instruction “call Rn, sub” is a two word instruction. Assuming that PC is incremented during the fetch cycle of the first word of the instruction, its register transfer interpretation is Rn <= PC = 1; PC <= M [PC]; 210
The minimum number of CPU clock cycles needed during the execution cycle of this instruction is (A) 2 (B) 3 (C) 4
(D) 5
SOLUTION Rn ! PC + 1 PC = M [ PC] Program outer is itself a register so incremented in 1 cycle. Now fetching the memory at PC & the value of at address stored in PC takes 2 cycles. So total 1 + 2 = 3 cycles. Hence (B) is correct option. Question. 44 A CPU has 24-bit instructions. A program starts at address 300(in decimal). Which one of the following is a legal program counter (all values in decimal)? (A) 400 (B) 500 (C) 600
(D) 700
SOLUTION Size of instruction = 24 bits. Since each instruction require 243 = 3 bytes & start address is 300 so the address for this range can be multiple of 3 only so 600. Hence (C) is correct option. YEAR 2006 Question. 45 A CPU has a cache with block size 64 bytes. The main memory has k banks, each bank being c bytes wide. Consecutive c-bute chunks are mapped on consecutive banks with warp-around. All the k banks can be accessed in parallel, but two accesses to the same bank must be serialized. A cache block access may involve multiple iterations of parallel bank accesses depending on the amount of data obtained by accessing all the k banks in parallel. Each iteration requires decoding
211
the bank numbers to be accessed in parallel and this takes k /2 ns. The latency of one bank access is 80 ns. If c = 2 and k=24, then latency of retrieving a cache block starting at address zero from main memory is (A) 92 ns
(B) 104 ns (D) 184 ns
(C) 172 ns SOLUTION Size of Cache block = 64 B No. of main memory banks K = 24 Size of each bank C = 2 bytes. So time taken for < access. T = decoding time + latency time T = K/2 + latency = 12 + 80 = 92 ns. But since C = 2 for accesses. 2 X 92 = 189 ns. Hence (D) is correct option Question. 46 A CPU has five-stages pipeline and runs at 1GHz frequency. Instruction fetch happens in the first stage of the pipeline. A conditional branch instruction computes the target address and evaluates the condition in the third stage of the pipeline. The processor stops fetching new instructions following a conditional branch until the branch outcome is known. A program executes 109 instructions out of which 20% are conditional branches. If each instruction takes one cycle to complete on average, then total execution time of the program is (A) 1.0 second (B) 1.2 seconds (C) 1.4 seconds
(D) 11.6 seconds
SOLUTION Given that 80% of 109 instruction require single cycle i.e. no conditional branching & for 20% an extra cycle required. Time taken by 1 cycle = 10−9 sec. 212
Total time = 10−9 b 10080 X 109 + 10020 X 2 X 109l = 10−9 X 109 b 54 + 25 l =
6
5
= 1.2 seconds.
Hence (B) is correct option. Question. 47 Consider a new instruction named branch-on-bit-set (mnemonic bbs). The instruction “bbs reg, pos, labbel” jumps to label if bit in position pos of register operand reg is one. a register is 32 bits wide and the bits are numbered 0 to 31, bit in position 0 being the least significant. Consider the following emulation of this instruction on a processor that does not have bbs implemented. temp!reg and mask Branch to label if temp is non-zero The variable temp is a temporary register. For correct emulation the variable mask must be generated by (A) mask! 0x1 << pos
(B) musk! 0x ffffffff >> pos
(C) mask! pos
(D) msdk! 0xf
SOLUTION Given instruction bbs reg, pos, Label Here pos bit decided whether to jump to label. So all other bits in temp set to 0. Temp ! reg and mask. So of temp is not zero branch to label. So shifting left over. Mask ! 0 X 1 << pos Hence (D) is correct option. Data for Q. 48 & Q. 49 are given below. Solve the problem and choose the correct answers. Consider two cache organizations: The first one is 32 KB 2-way set associative with 32-bytes block size. The second one is of the same
213
size but direct mapped. The size of an address is 32 bits in both cases A2to-1 multiplexes has latency of 0.6 ns where a k-bit comparator has a latency of k /10ns. The hit latency of the set associative organization is h1 while that of the direct mapped one is h2. Question. 48 The value of h1 is (A) 2.4ns
(B) 2.3ns
(C) 1.8ns
(D) 1.7ns
SOLUTION 2 way set-associative Cache. Size 32 KB 2 way so 16 KB sets. Require 14 bits for 214 B Block size = 32 Byte. = 5 bits. No. of blocks14 − 5 = 9 Tag index
h1 = 1810 + 0.6 ns = 2.4 ns. Hence (A) is correct option. Question. 49 The value of h2 is (A) 2.4ns
(B) 2.3ns 214
(C) 1.8ns
(D) 1.7ns
SOLUTION Similarly to previous question. The CPU address is same but Direct coaching require for 32 KB 15 bits. Which would be 10 + 5 = 17 17 10 5 h2 = 1710 + 0.6 = 2.3 ns Hence (B) is correct option. Data for Q. 50 & Q. 51 are given below. A CPU has a 32 KB direct mapped cache with 128-byte block size. Suppose A is a two dimensional array of size 512X512 with elements that occupy 8-bytes each. Consider the following two C code segments, P1 and P2, P1 : for (i=0;i<512;i++) { for (j=0;j<512;j++) { x+=A[i][j]; } } P2 : for (i=0;i<512;i++) { for (j=0;j<512;j++) { {x+=A[j][i];} } } P1 and P2are executed independently with the same initial state, namely, the array A is not in the cache and i, j, x are in registers. Let the number of cache misses experienced by P1 be M1 and that for P2 be M2. Question. 50 The value of M1 is
215
(A) 0
(B) 2048
(C) 16384
(D) 262144
SOLUTION Given loop P1 accesses array A row wise & P2 access column wise. M1 = ? Cache Capacity = 215 B. 1 element = 23 B Total elements 512 X 512 Total data = 512 X 512 X 8 B = 221 B Block size = 128 B 1 block can have = 1288 = 16 elements 512 X 512 16 So total blocks require = = 1638 blocks Since the memory is initially empty so all blocks are required at least once. So, M1 = 16384 Hence (C) is correct option. Question. 51 The value of the ratio M1/M2 is (A) 0
(B) 1/16
(C) 1/8
(D) −16
SOLUTION Now
M2
=?
In the case (P2 loop) the array is accessed column wise, so even the block brought for A [0][0] − A[0][15] would not be used for second column wise access i.e. A[1][0] So new block need to swap, similarly for A[3][0] & So on. This would continue for every element, since memory is contiguous. So M2 = 512 X 512 = 262144
216
&
M1 = 16384 = 1 M2 262144 16
Hence (B) is correct option. YEAR 2007 Question. 52 Consider a 4-way set associative cache consisting of 120 lines with a line size of 64 words. The CPU generates a 20-bit address of a word in main memory. The number of bits in the TAG, LINE and WORD fields are respectively (A) 9,6,5
(B) 7,7,6
(C) 7,5,8
(D) 9,5,6
SOLUTION 4 way set associative cache Size = 128 X 64 words 128 X 64 = 32 X 64 4 But for 4 sets For 32 lines we require 5 bits. For 64 words we require 6 bits. Indexing = 6 + 5 = 11 bits. Tag = 20 − 11 = 9 bits 9
5 6
Hence (D ) is correct option. Question. 53 Consider a disk pack with 16 surfaces, 128 tracks per surface and 256 sectors per track. 512 bytes of data are stored in a bit serial manner in a sector. The capacity of the disk pack and the number of bits required to specify a particular sector in the disk are respectively (A) 256 Mbytes, 19 bits
(B) 256 Mbyte, 28 bits
(C) 512 Mbytes, 20 bits
(D) 64 Gbyte, 28 bits
217
SOLUTION Surface = 6 Tracks = 16 X 128 Sectors = 16 X 128 X 256 = 2 4 X 2 7 X 28 = 219 So 19 lines are required to address all sectors. Bytes = 219 X 512 B = 219 X 29 B = 228 = 256 MB Hence (A) is correct option. Question. 54 Consider a pipelined processor with the following four stages IF: Instruction Fetch ID: Instruction Decode and Operand Fetch EX: Execute WB: Write Bank The IF, ID and WB stages take one clock cycle each to complete the operation. The number of clock cycles for the EX stage depends on the instruction. The ADD and SUB instructions need 1 clock cycle and the MUL instruction need 3 clock cycles in the EX stage. Operand forwarding is used in the pipelined processor. What is the number of clock cycles taken to complete the following sequence of instructions? ADD MUL
R2, R4,
R1, R3,
R0 R2
R2!R1+R0 R4!R3*R2
SUB (A) 7
R6,
R5,
R4
R6!R5−R4 (B) 8
(C) 10
(D) 14
SOLUTION Order of instruction cycle-phases. IF " ID " EX " WB We have 3 instructions. 1 R 2 ! R 1 ! R0 R 4 ! R 3 ! R2 R 6 ! R 5 ! R4
IF
2
3
4
5
6
7
8
ID EX WB IF
ID
EX
EX EX WB
IF
ID
-
-
EX
WB 218
Represent wait in pipeline due to result dependently. Clock cycles require = 8 Hence (B) is correct option.
Data for Q. 55, 56 & 57 are given below. Consider the following program segment. Here R1, R2 and R3 are the general purpose registers. Instruction MOV R1,(3000) LOOP: MOV R2,R1 ADD R2,R1 MOV (R3),R2 INC R3 DEC R1 BNZ LOOP HALT
Operation R1!M[3000] R2!M[R3] R2!R1+R2 M(R3]!R2 R3!R3+1 R1!R1-1 Branch on not zero Stop
Instruction size (no. of words) 2 1 1 1 1 1 2
Assume that the content of memory location 3000 is 10 and the content of the register R3 is 2000. The content of each of the memory locations from 2000 to 2010 is 100. The program is loaded from the memory location 100. All the numbers are in decided. Question. 55 Assume that the memory is word addressable. The number of memory references for accessing the data in executing the program completely is (A) 10
(B) 11
(C) 20
(D) 21
219
SOLUTION 1st memory reference R1 ! M 3000
and then in the loop which
^h runs for 10 times there are 2 memory reference every iteration. 10 X 2 = 20 Instruction Total
Words Location 20 + 1 = 21 R2 ! M [ R3] M [R3] ! R2 Hence (D) is correct option Question. 56 Assume that the memory is word addressable. After the execution of this program, the content of memory location 2010 is (A) 100
(B) 101
(C) 102
(D) 110
SOLUTION Program stores results from 2000 to 2010. It stores 110, 109, 108......100 at 2010 location. DEC R1 Hence (A) is correct option. Question. 57 Assume that the memory is byte addressable and the word size is 32 bits. If an interrupt occurs during the execution of the instruction “INC R3”, what return address will be pushed on to the stack? (A) 1005
(B) 1020
(C) 1024
(D) 1040
SOLUTION Now byte addressable memory so 1 word i.e. 4 bytes require 4 addresses. MOV (R3), R2
1
1016-1019
INC R3
1
1020-1023
DEC R1
1
1024-1027
220
MOV R1, (3000)
2
1000-1007
MOV R2, R1
1
1008-1011
ADD R2, R1
1
1012-1015
Interrupt occurs during execution of INC R3, So CPU will complete the execution of this instruction and then Push the next address 1024 to the stack, so after interrupt service the program can be resumed from next instruction. Hence (C) is correct option. Data for Q. 58 & Q. 59 are given below. Consider a machine with a byte addressable main memory of 216 bytes. Assume that a direct mapped data cache consisting of 32 lines of 64 bytes each is used in the system. A 50 X 50 two-dimensional array of bytes is stored in the main memory stating from memory location 1100H.Assume that the data cache is initially empty. The complete array is accessed twice. Assume that the contents of the data cache do not change in between the two accesses. Question. 58 How many data cache misses will occur in total? (A) 48 (B) 50 (C) 56
(D) 59
SOLUTION Size of main memory 216 bytes. Size of Cache = 32 X 64 B = 211 B Size of array = 2500 B Array is stored in main memory but cache is empty. Size of Cache = 2048 B So no. of page faults = 2500 − 2048 = 452 For second access = 452 X 2 = 904 Total = 1356 Hence (C) is correct option.
221
Question. 59 Which of the following lines of the data cache will be replaced by new blocks in accessing the array (A) line 4 to line 11 (B) line 4 to line 12 (C) line 0 to line 7
(D) line 0 to line 8
SOLUTION No of page faults = 452 One line has 64 B So the line at which these page faults will finish. = 45264 , 7 So 0 to 7 line Hence (C) is correct option. YEAR 2008 Question. 60 For a magnetic disk with concentric circular track, the latency is not linearly proportional to the seek distance due to (A) non-uniform distribution of requests (B) arm starting and stopping inertia (C) higher capacity of tracks on the periphery of the platter (D) use of unfair arm scheduling policies. SOLUTION Tracks on magnetic disks are concentric a seek is from me sector to other may or mayn’t be in different tracks. This seek distance is not proportional to latency since the tracks at periphery has higher diameter so high in capacity to store data. Hence (C) is correct option. Question. 61 Which of the following is/are true of the auto increment addressing mode? 1. It is useful in creating self relocating code 2. If it is included in an Instruction Set Architecture, then an 222
additional ALU is required for effective address calculation 3. The amount of increment depends on the size of the data item accessed. (A) 1 only (B) 2 only (C) 3 only
(D) 2 and 3 only
SOLUTION In auto increment addressing mode the address where next data block to be stored is generated automatically depending upon the size of single data item required to store. So statement 3 is correct. Statement says that this mode is used for self relocating code, but this is false since self relocating code, takes always some address in memory. Statement 2 is also incorrect since no additional ALV is required. Hence (C) is correct option. Question. 62 Which of the following must be true for the RFE (Return from Expectation) instruction on a general purpose processor. 1. It must be a trap instruction 2. It must be a privileged instruction 3. An exception can not be allowed to occur during execution of an RFE instruction. (A) 1 only (B) 2 only (C) 1 and 2 only
(D) 1, 2 and 3 only
SOLUTION RFE (Return From Exception) is a privileged trap trap instruction which is executed when exception occurs, so an exception is not allowed to execute. Hence (D) is correct option. Question. 63 For inclusion to hold between two cache level L1 and L2 in a multilevel cache hierarchy, which of the following are necessary? 1. L1 must be a write-through cache 2. L2 must be write-through cache
223
3. The associativity of L2 must be greater that of L1 4. The L2 cache must be at least as large as the L1 cache (A) 4 only (B) 1 and 2 only (C) 1, 2 and 4 only
(D) 1, 2, 3 and 4
SOLUTION Level 1 (L1) & Level 2 (L2) cache are placed between CPV & they can be both write through cache but this is not necessary. Associativity has no dependence but L2 cache must be at least as large as L1 cache, since all the words in L1 are also is L2. Hence (A) is correct option. Question. 64 Which of the following are NOT true in a pipe lined processor? 1. 2. 3.
Bypassing can handle alll Raw hazards. Register renaming can eliminate all register carried WAR hazards. Control hazard penalties can be eliminated by dynamic branch prediction. (A) 1 and 2 only (B) 1 and 3 only (C) 2 and 3 only
(D) 1,2 and 3
SOLUTION In a pipelined processor by passing can’t handle all the row hazards. Registers carried WAR doesn’t have register naming as proper solution. And control hazard penalties are eliminated by delayed branching not by dynamic branch prediction. So all are false. Hence (D) is correct option. Question. 65 The use of multiple register windows with overlap causes a reduction in the number of memory accesses for 1. Function locals and parameters 2. Register saves and restores
224
3. Instruction fetches (A) 1 only
(B) 2 only
(C) 3 only
(D) 1,2 and 3
SOLUTION Multiple register windows with overlap causes a reduction in the number of memory accesses for instruction fetching. Hence (C) is correct option. Question. 66 In an instruction execution pipeline, the earliest that the data TLB (Translation Look aside Buffer) can be accessed is (A) before effective address calculation has started (B) during effective address calculation (C) after effective address calculation has completed (D) after data cache lookup has completed SOLUTION TLB is used during effective address calculation in an instruction execution pipeline. Hence (B) is correct option. Data for Q. 67, 68 & 69 are given below. Consider a machine a 2-way set associative data cache of size 64 kbytes and block size 16 bytes. The cache is managed using 32 bit virtual addressed and the page size is 4 kbytes. A program to be run on this machine begins as follows: Double APR[1024]]1024] int i,j ; /*Initalize array APR to 0.0*/ for (i = 0;i < 1024;i ++) for (j = 0;k < 1024;j ++) APR[i][j] = 0.0; The size of double 8 bytes. Array APR is in memory stating at the beginning of virtual page 0 # FF000 and stored in row major order. The cache is initially empty and no pre-fetching is done. The only
225
data memory references made by the program are those to array APR. Question. 67 The total size of the tags in the cache directory is (A) 32 kbits (B) 34 kbits (C) 64 kbits
(D) 68 kbits
SOLUTION Virtual (CPU) address has = 32 bits 2 way set associative cache size = 64 KB Size of 1 set = 32 KB Require 15 bits for indexing. So Tag = 32 − 15 = 17 Size of block = 16 bytes = 4 bits are required Index = block + word Block = 15 − 4 = 11
17
11 4
CPV address Size of tags = There are 2 bytes of tags in every set of cache. So total = 17 X 2 X 1024 = 34 KB. Hence (B) is correct option. 17
Question. 68 Which of the following array elements has the same cache index as APR [0][0]? (A) APR[0][4] (B) APR[4][0] (C) APR[0][5]
(D) APR[5][0]
SOLUTION Elements stored in row major order. Two elements should have same cache index (15 bits) & their tags may be different (17 bits). SoAPR[%][ %] the MSB 17 bits will be changed. 226
APR[%][ %] APR[%][1]............... APR[2][%]................................ APR[4][%]................................ So on. This is virtual memory storage. So 15 LSB of APR [%][ %] & APR [%][ %] are same so same index APR [%] & APR [4] 17 MSB are different so tags differ. Hence (B) is correct option. Question. 69 The cache hit ratio for this initialization loop is (A) 0%
(B) 25%
(C) 50%
(D) 75%
SOLUTION No. of hits Cache hit ratio = Total accesses 1024 = 1024 + 1024
= 1 = 0.5 2
or = 50% Hence (C) is correct option.
Data for Q. 70 & 71 are given below. Delayed branching can help in the handling of control hazardous Question. 70 For all delayed conditional branch instruction, irrespective of weather the condition evato true or false, A (A) the instruction following the conditional branch instruction in memory is executed (B) the first instruction in the fall through path is executed (C) the first instruction in the taken path is executed (D) the branch takes longer to execute that any other instruction 227
SOLUTION Delayed branching for conditional instructions, irrespective of whether the condition evaluates to true or false, the first instruction in the fall through path would be executed this prevent from hazardous control transfer. Hence (B) is correct option. Question. 71 The following code is to run on a pipe lined processor with one branch delay slot 11: ADD R2 ! R7+R8 12: SUB R4 ! R5− R6 13: ADD R1 ! R2+ R3 14: STORE Memory [R4] ! R1 BRANCH to Label if R1==0 Which of the instruction 11,12,13 or 14 can legitimately occupy the delay slot without any other program modification? (A) 11
(B) 12
(C) 13
(D) 14
SOLUTION In pipelining result of 1 instruction is used for the next in pipeline. Delay slot will be occupied by the next instruction in the fall through path. The branching instruction R1 == 0 goto Label X. So in delay slot I2 will be there. Hence (B) is correct option. YEAR 2009 Question. 72 How many 32KX1 RAM chips are needed to provide a memory capacity of 356-bytes ? (A) 8
(B) 32
(C) 64
(D) 128 228
SOLUTION Memory capacity of 1 RAM = 32 K bits Total Memory required = 256 K bytes 3 No. of RAM = 256 X K X 2 X bits 32 X K X bits Chips required = 28 X25 23 = 26 = 64 Hence (C) is correct option. Question. 73 A CPU generally handles are interrupt by executing an interrupt service routine (A) As soon as an interrupt is raised (B) By checking the interrupt register at the end of fetch cycle (C) By checking the interrupt register after finishing the execution of the current instruction (D) By checking the interrupt register at fixed time intervals SOLUTION An interrupt is a signal delivered to CPU, which tells to stop its normal service routine & execute interrupt service routine. This interrupt service routine is checked as soon as CPU receives the interrupt but since CPU working unit is an instruction so CPU can switch to ISR only after execution of current instruction. Hence (C) is correct option. Question. 74 Consider a 4 stage pipeline processor. The number of cycles needed by the four instructions 11, 12, 13, 14 in stages S1, S2, S3, S4 is shown below:
I1 I2 I3
S1 2 2 2
S2 1 3 1
S3 1 2 1
S4 1 2 3
229
I4
1
2
2
2
What is the number of cycles needed to execute the following loop? (A) 16
for (i = 1to 2){I1; I2; I3; I4;} (B) 23
(C) 28
(D) 30
SOLUTION We can see a single iteration of given for loop according to the cycles required. Cycle
S1
S2
1
I1
2
I1
3
I2
I1
4
I3
I2
5
I3
I2
6
I4
I2
S3
S4
Completion
I1
I1
I1
7
I3
I2
8
I4
I2
9
I4
I3
I2
10
I4
I2
11
I4
I3
12
I3
13
I3
14
I4
15
I4
I2
I3 I4
No. of cycle of 2 iteration = 2 X 15 = 30 Hence (D) is correct option. Question. 75 Consider a 4 way set associative cache (initially empty) with total 16 cache blocks. The main memory consists of 256 blocks and the request for memory blocks is in the following order : 230
0, 255, 1, 4, 3, 8, 133, 159, 216, 129, 63, 8, 48, 32, 73, 92, 155 Which one of the following memory block will NOT be in the cache if LRU replacement policy is used ? (A) 3
(B) 8
(C) 129
(D) 216
SOLUTION 4 way set associative so 16 block will be divided in 4 sets of 4 blocks each. We apply (Address mod 4) function to decide set. 0
48
0
mod4 = 0
)
4
32
255
mod4 = 3
)
8
8
1
mod4 = 1
)
216
92
4
mod4 = 0
)
1
1
3
mod4 = 3
)
133
133
8
mod4 = 0
)
129 73
129 73
133 159
mod4 = 1 mod4 = 3
) )
216
mod4 = 0
)
129
mod4 = 1
)
63
mod4 = 3
)
8
mod4 = 0
)
Set 0
Set 1
Set 2
255
155
98
mod4 = 0
)
3
3
32
mod4 = 0
)
159
159
73
mod4 = 1
)
63
63
92
mod4 = 0
)
155
mod4 = 3
)
Set 3
All ) are misses S1 is the first stage & S2 is second. In the second stage 216 is not present in Cache Hence (D) is correct option. Common Data for Question 76 & 77 A hard disk has 63 sectors per track, 10 platters each with 2 recording 231
surfaces and 1000 cylinders. The address of a sector is given as a triple c , h, s , where c is the cylinder number, h is the surface number and s is the sector number. Thus, the 0th sector is addressed as 0,0,0 , the 1st sector as 0,0,1 , and so on. Question. 76 The address
400,16,29 , corresponds to sector number:
(A) 505035
(B) 505036
(C) 505037
(D) 505038
SOLUTION Each cylinder has 10 platters or 20 recording surfaces or 63 X 20 sector = 1260 sectors. Each recording surface has 63 sectors. So < 400,16,297 =< c , h, s > 400 X 1260 + 16 X 63 + 29 50503 7 Hence (C) is correct option. Question. 77 The address of 1039th sector is (A) 0,15,31
(B)
0,16,30
(C)
(D)
0,17,31
0,16,31
SOLUTION 1 cylinder has 1260 sectors So address for 1039th sector. C=0 Each surface has 63 sectors. Total surface here = 103963 = 16 surfaces Remainder here 31 sectors So < 0, 16, 31 >. Hence (C) is correct option. 232
YEAR 2010 Question. 78 A main memory unit with a capacity of 4 megabytes is build using 1M X 1 − bit DRAM chips. Each DRAM chip has 1K rows of cells with 1 K cells in each row. The time taken for a single refresh operation is 100 nanoseconds. The time required to perform one refresh operation on all the cells in the memory unit is (A) 100 nanoseconds (B) 100)210nanoseconds (C) 100)220nanoseconds (D) 3200)220nanoseconds SOLUTION Size of main memory = 4 MB 1 DRAM size = 1 Mb 4XMX8Xb No. of chips required = 1XMXb = 32 1 DRAM has 1 K rows 1 ROW has 1 K cells Total cells in 1 DRAM = K2 = 220 In 32 DRAM = 32 X 220 Cells 1 cell refresh take 100 ns. So total refresh time = 32 X 100 X 220 ns = 3200 X 220 ns. Hence (D) is correct option. Question. 79 A-5 stage pipelined processor has Instruction Fetch. (IF), Instruction Decode (ID), Operand Fetch (OF), Perform Operation (PO) and Write Operand (WO) stages. The IF, ID, OF and WO stages take 1 clock cycle each for any instruction. The PO stage takes 1 clock cycle for ADD and SUB instruction. The PO stage takes 1 stake clock cycle for ADD and SUB instructions 3 clock cycles for MUL instruction, and 6 clock cycles for DIV instruction respectively. Operand forwarding is used in the pipeline. What is the number of clock cycles needed to execute the following sequence of instructions ?
233
Instruction Meaning of instruction I0 : MUL R 2, R 0, R1
R 2 ! R 0)R1
I1: DIV R 5, R 3, R4
R 5 ! R 3 /R4
I2 : ADD R 2, R 5, R2
R 2 ! R 5 + R2
I3 : SUB R 5, R 2, R6
R 5 ! R 2 − R6
(A) 13
(B) 15
(C) 17
(D) 19
SOLUTION The order of operations IF " ID " OF " PO " WO Figure Here
A = (R 3 /R 4) + R 2, R6
So we can see that all the instruction can be executed in 17 clock cycles using piplining. Hence (C) is correct option. Question. 80 The program below uses six temporary variables a, b, c , d, e, f a =1 b = 10 c = 20 d=a+b e=c+d f = c + e b=c+e e = b + f d=5+e Assuming that all operations take their operands from register, what is the minimum number of registers needed to execute this program without spilling ? (A) 2 (B) 3 (C) 4
(D) 6 234
SOLUTION Replacement R1
R2
R3
a
b
c
d
b
c
d
e
c
f
e
c
f
b
c
f
e
c
f e d So all the operations done using 3 registers only. Hence (B) is correct option. Common Data for Questions 81 & 82 A computer system has an L1 and L2 cache, an L2 cache, and a main memory unit connected as shown below. The block size in L1 cache is 4 words. The block size is L2 cache is 16 words. The memory access times are 2 nanoseconds, 20 nanoseconds and 200 nanoseconds for L1 cache, L2 cache and main memory unity respectively.
Question. 81 When there is a miss in L1 cache and a hit in L2 cache, a block is transferred form L2 cache to L1 cache. What is the time taken for this transfer ? (A) 2 nanoseconds
(B) 20 nanoseconds
(C) 22 nanoseconds
(D) 88 nanoseconds
SOLUTION Each block is L2 Cache is 4 times L1 Cache. So far 1 block miss in L1 Cache the block from L2 to L1 will be transferred, but L2 block has size 16 words & L1 data bus of 4 words, so 4L2 & 4L1 access are
235
required. 4 X 2 + 4 X 20 8 + 80 88 ns Hence (D) is correct option. Question. 82 When there is a miss in both L1 cache and L2 cache, first a block is transferred from memory to L2 cache, and then a block is transferred from L2 cache to L1 cache. What is the total time taken for these transfers ? (A) 222 nanoseconds
(B) 888 nanoseconds
(C) 902 nanoseconds
(D) 968 nanoseconds
SOLUTION Miss in both L1 & L2. Cause main memory to transfer that block in both cache. 1 block of Main memory has 16 words but data bus of L2 has only 4 words. So 4 access of Main memory & 4 access of L2 Cache required to update L2 4 X 20 + 4 X 200 80 + 800 = 880 ns Now L2 updates L1, this takes 4 access to L1 & 4 access to L2 4 X 2 + 4 X 20 8 + 80 = 88 ns Total time = 880 + 88 = 968 ns Hence (D) is correct option. YEAR 2013 92.
Consider an instruction pipeline with five stages without any branch prediction: Fetch Instruction (FI), Decode Instruction (DI), Fetch Operand (FO), Execute Instruction (EI) and Write Operand (WO). The stage delays for FI, DI, FO, EI and WO are 5 ns, 7 ns, 10 ns, 8 ns and 6 ns, respectively. There are intermediate storage buffers after each stage and the delay of each buffer is 1 ns. A program consisting of 12 instructions I1 ,I2 ,I3 ,......I12 is executed in this pipelined processor. Instruction I4 is the only branch instruction and its branch target is I9 . If the branch is taken during the execution of this program, the time (in ns) needed to complete the program is (A) 132 (B) 165 (C) 176 (D) 328 236
Ans: (C) Exp:
Total clock slots taken are 16. Each slot will take maximum of {5, 7, 10, 8 ,7} =10. Hence
total
slots
for
all
the
instructions
=
DIGITAL DESIGN
YEAR 2001
Question. 1 Given the following Karnaugh map, which one of the following represents the minimal sum-of-Products of the map ?
(A) xy + y' z
(B) wx' y' + xy + xz
(C) w' x + y' z + xy
(D) xz + y
237
SOLUTION
There are 2 quads. y' z + yx So xy + y' z Hence (A) is correct option.
Question. 2 Consider the following circuit with initial state Q 0 = Q1 = 0. The D flip-flops are positive edged triggered and have set up times 20 nanosecond and hold times 0.
Consider the following timing diagrams of X and C ; the clock of C $ 40 nanosecond. Which one is the correct plot of Y
238
SOLUTION Consider the following circuit with initial state Q 0 = Q1 = 0. The D flipflops are positive edged triggered and have set up times 20 nanosecond and hold times 0. Consider the following timing diagrams of X and C ; the clock period of C $ 40 nanosecond. Which one is the correct plot of Y ? Figure
Question. 3 The 2’s complement representation of (− 539)10 is hexadecimal is (A) ABE
(B) DBC
(C) DE5
(D) 9E7
SOLUTION Binary of 539 = 1000011011 239
Binary :0010 0001 1011 2's comp :1101 1110 0101 Hexadecimal D E 5 (DES)16 Hence (C) is correct option. Question. 4 Consider the circuit shown below. The output of a 2:1 Mux is given by the function (ac ' + bc).
Which of the following is true ? (A) f = x1' + x2
(B) f = x1' x2 + x1x2'
(C) f = x1x2 + x1' x2'
(D) f = x1 + x2
SOLUTION Output of any 2:1 MUX = ac' + bc Here output of MUX 1. g = ax 1' + bx1 Output of MVX 2 (I) = gx 2' + x 1 x2 (J) = (ax 1' + bx 1) x 2' + x 1 x2 (K) = ax 1' x 2' + bx 1 x 2' + x 1 x2 Given a = 1, b = 0 (L) = x 1' x 2' + x 1 x2 Hence (C) is correct option. Question. 5 Consider the circuit given below the initial state Q 0 = 1,Q 1 = Q2 = 0. The state of the circuit is given by the value 4Q 2 + 2Q 1 + Q0
240
Which one of the following is the correct state sequence of the circuit ? (A) 1, 3, 4, 6, 7, 5, 2
(B) 1, 2, 5, 3, 7, 6, 4
(C) 1, 2, 7, 3, 5, 6, 4
(D) 1, 6, 5, 7, 2, 3, 5
SOLUTION
Initially
Q0
Q1
Q2
Value 4Q 2 + 2Q 1 + Q0
Clk 1
1 0
0 1
0 0
1 2
2
1
0
1
5
3
1
1
0
3
4
1
1
1
7
5
0
1
1
6
6
0
0
1
4
Hence (B) is correct option.
YEAR 2002
Question. 6 Minimum sum of product expression for f (w, x, y, z) shown in Karnaughmap below is 241
(A) xz + y' z
(B) xz' + zx'
(C) x' y + zx'
(D) None of the above
SOLUTION
There are 2 quads possible xz' + x' z Hence (B) is correct option. Question. 7 The decimal value of 0.25 (E) is equivalent to the binary value 0.1 (F) is equivalent to the binary value 0.01 (G) is equivalent to the binary value 0.00111..... (H) cannot be represented precisely in binary. 242
SOLUTION Given decimal no. 0.25 Binary = ? .25 # 2 = .5 .5 # 2 = 1 (.01)2 Hence (B) is correct option. Question. 8 The 2’s complement represent representation of the decimal value −15 is (A) 1111
(B) 11111
(C) 111111
(D) 10001
SOLUTION Given (− 15)10 Binary of 15 = (01111)2 2’s complement of 15 would represent (− 15). 01111 (10001)2 Hence (D) is correct option. Question. 9 Sign extension is a step in (I) floating point multiplication (J) signed 16 bit integer addition (K) arithmetic left shift (L) converting a signed integer from one size to another. SOLUTION Sign extension is the operation in computer arithmetic of increasing no. of bits of a binary no., while preserving sign and value done by appending MSB’s. In the floating point multiplication to bring the no. in desired no. of significant digits sign extension is done. Hence (A) is correct option. 243
Question. 10 In 2’s complement addition, overflow (H) Relational algebra is more powerful than relational calculus (I) Relational algebra has the same power as relational calculus. (J) Relational algebra has the same power as safe relational calculus. (K) None of the above. SOLUTION In 2’s complement addition, overflow occurs when the carries from sign bit & previous bit doesn’t match. So overflow can’t occur when a positive value is added to some negative value. Hence (B) is correct option. Question. 11 Consider the following logic circuit whose inputs are functions f1, f2, f3 and output is f
Given that f1(x, y, z) = Σ(0,1,3,5) f2(x, y, z) = Σ(6,7), and f (x, y, z) = Σ(1,4,5) f3 is (A) Σ(1,4,5)
(B) Σ(6,7)
(C) Σ(0,1,3,5)
(D) None of the above
SOLUTION f1 (x, y, z) = Σ(0,1,3,5)
244
= x' y' + y' z + x' z f2 (x, y, z) = Σ(6,7)
= xy f (x, y, z) = Σ(1,4,5)
= xy' + y' z f (x, y, z) = f1 f2 : f3 (I) f1 : f2 + f3 (J)xy (x' y' + y' z + x' z) + (xy' + y' z) f3 = xy' z + xy' z' + xy' z + x' y' z f3 = Σ(1,4,5) Hence (A) is correct option. Question. 12 Consider the following multiplexor where 10, 11, 12, 13 are four data input lines selected by two address line combinations A1A0 = 00,01,10,11 respectively and f is the output of the multiplexor. EN is the Enable input.
245
The function f (x, y, z) implemented by the above circuit is (A) xyz'
(B) xy + z
(C) x + y
(D) None of the above
SOLUTION A1 A0
EN (MUX) work
0
0
1
do not
0
1
0
(MUX) Work
1
0
1
do not
1 1 0 So MUX is ENABLED only if A0 = 0 So output should have Z'. Consider xyz' option (A) A, A0 = 1 0 gives correct answer. Hence (A) is correct option. Question. 13 Let f (A, B) = A' + B. Simplified expression for function f (f (x + y, y), z) is (A) x' + z (B) xyz (C) xy' + z
(D) None of the above
SOLUTION f (x + y, y) = (x + y)' + y & x + y + y f (f (x + y, y), z) = x + y + y + z & (x + y : y ) + z [(x + y) : y ] + z
246
[xy + yy ] + z xy + z Hence (C) is correct option. Question. 14 What are the states of the Auxiliary Carry (AC) and Carry Flag (CY) after executing the following 8085 program ? MIV
H,
5DH
MIV
L,
6BH
MOV A,
H
ADD L (A) AC = 0 and CY = 0
(B) AC = 1 and CY = 1
(C) AC = 1 and CY = 0
(D) AC = 0 and CY = 1
SOLUTION Program is to add 2 nos kept in H & L, result of addition is stored in A. (5D) 16 + (6B)16 &
is the carry so CY = 0 (I) is auxillary carry AC = 1 Hence (C) is correct option. Question. 15 The finite state machine described by the following state diagram x y and x stands for with A as starting state, where an arc label is 1-bit input and y stands for 2-bit output. 247
(A) Outputs the sum of the present and the previous bits of the input. (B) Outputs 01 whenever the input sequence contains 11 (C) Outputs 00 whenever the input sequence contains 10 (D) None of the above. SOLUTION Previous input 0(A)
Present i/p 0(A)
Output 00
0(A)
1(B)
01
1(B) 1(B)
0(A) 1(C)
01 10
1(C)
1(C)
10
1(C)
0(A)
01
So output is always sum of the present and previous bits of input. Hence (A) is correct option. YEAR 2003 Question. 16 Assuming all numbers are in 2’s complement representation, which of the following number is divisible by 11111011? (A) 11100111
(B) 11100100
(C) 11010111
(D) 11011011
SOLUTION
248
We can’t judge the no’s in 2’s complement first we need to convert them in decimal Given no. 11111011"00000101=5 11100111 " 00011001 = 25
(A) 11100100 " 00011100 = 28 (B) 11010111 " 00101001 = 41 (C) 11011011 " 00100101 = 37 From all only option (A) is divisible by 5. Shortcut : To convert 2’s complement no. directly into original binary, we should complement all the digits from MSB till the last one (1). Keep the last 1 from the LSB as it is. Observe in the example. Question. 17 The following is a scheme for floating point number representation using 16 bits.
Let s, c and m be the number represented in binary in the sign, exponent, and mantissa fields respectively. Then the flouting point number represented id 2 − 9 e−31 , if the exponent 111111 0 otherwise )(− 1) (1 + m # 2 )2 What is the maximum difference between two successive real numbers representable in this system? (A) 2−40 (C) 222
(B) 2−9 (D) 231
SOLUTION e has 6 bits so max value can be 26 − 1 = 63 when e = 111111 But given e =Y 111111 So max e = 62 = 111110 Two consecutive number will have same exponent but difference in mantissa by 1. Difference would be (− 1)2 (1 + (m + 1)2− 9)262 − 31 − (− 1)2 (1 + m # 2− 9)262 − 31 231 # 2−9 = 222 Hence (C) is correct option. Question. 18 A 1-input, 2-output synchronous sequential circuit behaves as follows. 249
Let zk , nk denote the number of 0’s and 1’s respectively in initial k bits of the input (zk + nk = k). The circuit outputs 00 until one of the following conditions holds. 1. nk − nk = 2. In this case, the output at the k -th and all subsequency clock ticks is 10. 2. nk − zk = 2. In this case, the output at the k -th and all subsequent clock ticks is 01. What in the minimum number of states required in the state transition graph of the above circuit? (A) 5 (B) 6 (C) 7
(D) 8
SOLUTION The sequential circuit has 3 variables to decide the state in which input & 2 inputs are present. Output for particular inputs decide states. i/p
op 1
op 2
State
0
0
0
Intial
0
0
1
nK−zK=2
0
1
0
zK−nK=2
0
1
1
Not applicable
1
0
0
Initial
1
0
1
nK−zK=2
1
1
0
zK−nK=2
1
1
1
is correct
Using 3 bits we require 23 − 1 = 7 states here. Hence (C) is correct option. Question. 19 The literal count of a boolean expression is the sum of the number of times each literal appears in the expression. For example, the literal count of (xy + xz) is 4. What are the minimum possible literal counts of the product-of-sum and sum-of-product representations respectively of the function given by the following karnaugh map? Here,
denotes “don’t care”
250
(A) (11,9)
(B) (9,13)
(C) (9,10)
(D) (11,11)
SOLUTION Considering product of sum & sum of product separately.
Sum of product = wy + w' y' + z' wx' + xyz' 12 34 567 8910
Product of sum = (y' + z')(z' + y)(w' + z') (x + z + w) Lateral count =9
Literal count =10 Hence (C) is correct option. In SOP the K-map is solved for 1 & POS K-map solved for 0 Question. 20 Consider the following circuit composed of XOR gates and non-inverting buffers.
The non-inverting buffers have delays δ1 = 2ns and δ2 = 4ns as shown in the figure. both XOR gates and al wires have zero delay. Assume that all gate inputs, outputs and 251
wires are stable at logic level 0. If the following waveform is applied at input. A, how many transition (s) (change of logic levels) occur (s) at B during the interval from 0 to 10 ns?
(A) 1 (C) 3
(B) 2 (D) 4
SOLUTION Due to delays S1 = 2 & S2 = 4 the transitions would occur at time 1, 2 & 4. Time Input (A)
Output (B)
0
1
0
I
1
1
0
Transition
II
2
1
0
Transition
III
4
0
1
Transition
So total 3 transitions Hence (C) is correct option. YEAR 2004 Question. 21 The Boolean function x' y' + xy + x' y is equivalent to
252
(A) x' + y'
(B) x + y
(C) x + y'
(D) x' + y
SOLUTION x' y' + xy + xy' x'(y + y') + xy (A + A') = 1 x' + xy (A + AB) = (A + A) : (A + B) (x' + x) : (x' + y) 1 : (x' + y) x' + y Hence (D) is correct option. Question. 22 In an SR latch made by cross-coupling two NAND gates, if both S and R inputs are set to 0, then it will result in (A) Q = 0,Q' = 1 (B) Q = 1,Q' = 0 (C) Q = 1,Q' = 1
(D) Indeterminate states
SOLUTION SR latch both S and R when 0 leads to invalid state.
Transition table for SR flip flop. S R Q(Next state) ii 0 Invalid so Q=Q’=1 change 0 1 1 1 0 0 1 1 previous state For S=R=0 Q=Q’=1 253
Hence (C) is correct option.
254
Question. 23 If 73x (in base-x number system) is equal to 54, (in base-y number system), the possible values of x and y are (A) 8, 16
(B) 10, 12
(C) 9, 13
(D) 8, 11
SOLUTION (73)x = (54)y 7x + 3 = 5y + 4 (x', y) 7x + 3 8, 16 59 10, 12 73 9, 13 64 8, 11 59 Hence (D) is correct option.
5y + 4 84 64 69 59
Question. 24 What is the result of evaluating the following two expressions using three-digit floating point arithmetic with rounding? (113.+−111.)+7.51 113.+(−111.+7.51) (A) 9.51 and 10.0 respectively
(B) 10.0 and 9.51 respectively
(C) 9.51 and 9.51 respectively
(D) 10.0 and 10.0 respectively
SOLUTION Expression 1 (113.0 + (− 111.) + 7.51 (113.0 − 111.0) + 7.51 2.0 + 7.51 9.51 10 rounded off Expression 2 113.0 + (− 111.0 + 7.51) 113.0 + (− 103.49) 113.0 − 103.00 10.0 rounded off Hence (D) is correct option 255
Question. 25 A circuit outputs a digit in the form of 4 bits. 0 is represented by 0000, 1 by 0001,...9 by 1001. A combinational circuit is to be diesigned which takes these 4 bits as input and outputs 1 if the digit $ 5, and 0 otherwise. If only AND, OR and NOT gates may be used, what is the minimum number of gates required? (A) 2 (B) 3 (C) 4
(D) 5
SOLUTION CKT takes 4 bits as the input so K-Map will have 4 variable so 16 options are available.
1 digit = 5 0 otherwise Here for 0 to 4 we have 0 output, from 5 to 9 1 output & for 10 to 15 don’t care. 1 octed & 2 pounds. a + bd + bc a + b (d + c) Two OR gates One AND gate Total 3 Hence (B) is correct option. Question. 26 Which are the essential prime implicates of the following Boolean function? f (a, b , c ) = a' c + ac ' + b ' c 256
(A) a' c and ac'
(B) a' c and b ' c
257
(C) a' c only
(D) ac' and bc'
SOLUTION f (a, b , c) = a' c + ac ' + b ' c Making min terms a' bc + a' b' c + abc ' + ab' c ' + a' b' c + ab' c Since b ' c gives no new term. So a' c & ac' are only essential prime implicants. Solution detailed method Tabulation method Since b ' c gives no new term. So a' c & ac' are only essential prime implicants. Solution detailed method Tabulation method f (a, b , c ) = /m(1,3,5,6,4) Figure Figure 3 & 6 have only 1 cross they are in a' c & ac' Question. 27 Consider the partial implementation fo a 2-bit counter using T flip flops following the sequence 0-2-31-0, as shown below
To complete the circuit, the input X should be (A) Q2'
(B) Q2 + Q1
(C) (Q1 5 Q2)'
(D) Q1 5 Q2
SOLUTION Counter counts the no. of signal inversion change of states. Sequence input is 0 − 2 − 3 − 1 − 0
258
Binary 00 − 10 − 11 − 01 − 00 to generate signals if we XOR gate then it outputs 1 if both are different. So output sequence would be. 0 − 1 − 0 − 1 − 0 & the sequence would be counted. So. X = Q 1 5 Q2 Hence (D) is correct option. Question. 28 A 4-bit carry look ahead adder, which adds two 4-bit numbers, is designed using AND, OR, NOT, NAND, NOR gates only. Assuming that all the inputs are available in both complemented and uncompensated forms and the delay of each gate is one time unit, what is the overall propagation delay of the adder? Assume that the carry network has been implemented using two-level AND-OR logic. (A) 4 time units
(B) 6 time units
(C) 10 time units
(D) 12 time units
SOLUTION Carry of any higher order bit is dependent upon previous order bit addition generated carry. C out = g 0 + p0 C in P3 P2 P1 P0 g 3 g 2 g 1 g0 c 3 c 2 c 1 c0 c3 g 3 + P3 g 2 + P3 P2 g 1 + | P3 P2 P1 g 0 + P3 P2 P2 P0 C in This is 4 bit look ahead adder equation total gate delay = 1+1+2+2 = 6 Hence (B) is correct option. Question. 29 Let A = 11111010 and B 0000 1010 be two 8-bit 2’s complement numbers. Their product in 2’s complement is (A) 1100 0100
(B) 1001 1100
(C) 1010 0101
(D) 1101 0101
SOLUTION A and B are in 2’s complement form. 259
A = 11111010 Binary = 00000110 = 6 2’s complement represent −ve number So A =− 6 B = 00001010 MSB is 0 so +ve no. decimal 10. B = 10 A # B =− 6 # 10 =− 60 Binary of 60 = 00111100 2’s complement 11000100 Hence (A) is correct option.
YEAR 2005
Question. 30 Consider the following circuit.
Which one of the following is TRUE? (A) f is independent of X
(B) f is independent of Y
(C) f is independent of Z
(D) None of X, Y, Z is redundant
260
SOLUTION
f =X:Y:Y:Z = X:Y+Y:Z = X:Y+Y:Z For redundant check we need to draw K map to min terms. XY (Z + Z ) + (X + X ) : YZ XYZ + XY Z + XYZ + X YZ
XY + YZ + XZ Hence (D) is correct option. Question. 31 The range of integers that can be represented by an a bit 2’s complement number system is (A) − 2 n − 1 to (2n−1 − 1)
(B) − (2 n − 1 − 1) to (2n−1 − 1)
(C) −2n − 1 to 2n−1
(D) − (2 n − 1 + 1) to (2n−1 − 1)
SOLUTION n bit 2’s complement system must have corresponding bit binary system. But to implement +ve & −ve nos. Both we require MSB to be sign bit. So maximum magnitude can be 2n−1 − 1 suppose we take n = 4. 261
Using 4 bits. 1 1 1 1,. . . . . .0 0 0 0, . . . . .01 1 1 −7 +7 This would be the range. So − (2 n−1 − 1) to + (2 n−1 − 1) Hence (B) is correct option. Question. 32 The hexadecimal representation of 6578 is (A) 1AF (B) D78 (C) D71
(D) 32F
SOLUTION (657)8 = (?)16 Making binary S S= 0 0 010101111 (IAF) 16 A BBC Hence (A) is correct option. Question. 33 The switching expression corresponding to f (A, B, C , D) = /(1,4,5,9,11,12) is (A) BC' D' + A' C ' D + AB' D
(B) ABC' + ACF + B' C ' D
(C) ACD ' + A' BC' + AC' D'
(D) A' BD + ACD ' + BCD'
SOLUTION f (A, B, C , D) = /(1,4,5,9,11,12) Drawing K map for min terms.
262
BC D + A C D + AB D So min terms are BC D + A CD + AB D Hence (A) is correct option. Question. 34 Consider the following circuit involving a positive edge triggered D -FF.
Consider the following timing diagram. Let Ai represent the logic level on the line A in the i − th clock period.
Let A represent the complement of A. The correct output sequence on Y over the clock perids 1 through 5 is (A) A0 A1 A1 ' A3 A4 (B) A0 A1 A2 ' A3 A4 (C) A1 A2 A2 ' A3 A4 (D) A1 A2 ' A3 A4 A5 SOLUTION We need to calculate equation for D input. (D) = (A i X')' − (X' Q')' =Ai+X+X+Q (E) =Ai'+X+Q Drawing truth table for ckt Clock X Q0 = 0 Q1 = 1 Ai
Y
263
0
1
0
1
A0'
A0'
1
1
0
1
A1'
A0'
2
0
0
1
A2'
A1'
3
1
0
1
A3'
A1'
4
1
0
1
A4 '
A3'
5
0
0
1
A5'
A4 '
Hence (A) is correct option. Question. 35 The following diagram represents a finite state machine which takes as input a binary number from the least significant bit
Which one of the following is TRUE? (J) It computes 1’s complement of the input number (K) It computes 2’s complement of the input number (L) It increments the input number (M) It decrements the input number SOLUTION The transition table for the diagram Present state
Input
Next state
Output
Q0
0
Q0
0
Q0
1
Q1
1
Q1
0
Q1
1
Q1
1
Q1
0 264
So the FSM takes input from LSB side it doesn’t change state till the first 1 comes from LSB side, after that it complement all the bits. This is logic for 2’s complement. Hence (B) is correct option. Question. 36 Consider the following circuit
The flip-flops are positive edge triggered DFFs. Each state is designated as a two bit string Q0, Q1. Let the initial state be 00. The state transition sequence is (A) 00 " 11 " 01 (B) 00 " 11 A BBBBBBBC A BBBC (C) 00 " 10 " 01 " 11 (D) 00 " 11 " 01 " 10 A BBBBBBBBBBC A BBBBBBBBBBC SOLUTION Truth table for DFF CP
D
Qn+1
Action
0
X
Qn
No change
1
0
0
Reset
1
1
1
Set
D here AX + X' Q' Truth table for ckt
and so on. 265
Hence (D) is correct option. Data for Q. 37 & 38 are given below. Solve the problems and choose the correct answers. Consider the following floating point format
Mantissa is a pure fraction is sign-magnitude form. Question. 37 The decimal number 0.239 # 213 has the following hexadecimal representation without normalization and rounding off (A) 0D 24
(B) 0D 4D
(C) 4D 0D
(D) 4D 3D
SOLUTION Sign bit 0 Exponent = 13 Excess 64 = 13 + 64 = 77 = 1001101 Binary of 239
266
We have 8 bits for Mantissa 0 0 1 1 1 1 0 1 So the floating point format. 0
1001101
0100
1101
4
D
00111101 0011
1101
3
D
Hence (D) is correct option. Question. 38 The normalized representation for the above format is specified as follows. The mantissa has an implicit 1 preceding the binary (radix) point. Assume that only 0’s are padded in while shifting a field. The normalized representation of the above number (0.239 # 213) is (A) 0A 20 (B) 11 34 (C) 4D D0
(D) 4A E8
SOLUTION Given no. .239 # 213 Normalized form of binary. Binary " .239 = (00111101)2 Normalized = 1.11101 # 210 Proceeding implicit 1 So 8 bit mantissa 11101000S padding
267
Excess 64 exponent 1 0 0 1 0 1 0 = 74 Sign bit = 0 Floating
Point
Format
0
1001010
11101000
0100
1010
4
A
1110 E
1000 8
GAE8 Hence (D) is correct option.
268
YEAR 2006 Question. 39 You are given a free running clock with a duty cycle of 50% and a digital waveform f which changes only at the negative edge of the clock. Which one of the following circuits (using clocked D flip flops) will delay the phase of f by 180c ?
SOLUTION We require phase shift of 180 in f In ckt (B) the negation of signal f & clock delays signal f by 180. 269
So phase shift occurs. Hence (B) is correct option. Question. 40 Consider the circuit below. Which one of the following options correctly represents f (x, y, z)?
− − (A) xz + xy + yz −− (C) xz + xy + yz
− − (B) xz + xy + yz − − (D) xz + xy + yz
SOLUTION MVXI Selects X when Z = 0 Y' when Z = 0 MVX II Selects (XZ' + Y' Z) when y = 0 X when y = 0 so (XZ' + YZ) Y' + XY Simplifying = xz' y' + zy' y' + xy = xz' y' + xy (z + z') + zy' = xz' y' + xyz + xyz' + zy'(x + x') = xz' y' + xyz + xyz' + xy' z + x' y' z = y' z + xy' z + xyz' + xyz + xyz'[a + a = a] = y' z + xz'(y + y') + xy (z + z') = y' z + xz' + xy 270
Hence (A) is correct option. Question. 41 Given two three bit numbers a2 a1 a0 and b2 b1 b0 and c, the carry in, the function that represents the carry generate function when these two numbers are added is (E) a2 b2 + a1 a1 b1 + a2 a1 a0 b0 + a2 a0 b1 b0 + a1 b2 b1 + a1 a0 b2 b0 + a0 b2 b1 b0 (F) a2 b2 + a2 b1 b0 + a2 a1 b1 b0 + a1 a0 b21 b1 + a1 a0 b2 + a1 a0 b2 b0 + a2 a0 b1 b0 (G) a2 + b2 + (a2 5 b2)[a1 + b1 + (a1 5 b1)(a0 + b0)] (H) a2 b2 + a2 a1 b1 + a2 a1 a0 b0 + a2 a0 b1 b0 + a1 b2 b1 a1 a0 b2 b0 + a0 b2 b1 b0 SOLUTION a2 a1 a0 b2 b1 b0 C 0
0
0
0
0
0
0
0
0
1
0
0
1
1
0
1
0
0
1
0
0
1
0
0
1
1
1
0
1
1
0
0
1
0
0
1
0
1
1
1
1
1
1
0
1 1 0 1 1 1 Case I These are the possible value of a 2 a 1 a0 & b 2 b 1 b0 when a2 = 1 c=1 Case II b2 = 1 c = 1 so a 2 + b2 Case III If any 1 of a2 or b2 is 1 a 2 5 b2 then if a1 = 1 c = 1 b1 = 1 c = 1 so a 2 5 b 2 [ a 1 + b1] Case IV If any of a2 or b2 is 1 & any of a2 or b1 is 1 then if a0 = 1 c = 1 or if b0 = 1 then c = 1 so overall. a 2 + b 2 + [(a 2 5 b 2){a 1 + b 1 + (a 1 5 b 1)(a 0 + b0)}] Hence (C) is correct option. Question. 42 Consider a boolean function f (w, x, y, z). Suppose that exactly one of its inputs is allowed to change at a time. If the function happens to be true for two input vectors i1 +< w1, x1, y1, x1 > and i2 +< w2, x2, y2, z2 >
271
, we would like the function to remain true as the input changes from i1 to i2 (i1 and i2 differ in exactly one bit position), without becoming false momentarily. Let f (w, x, y, z) = /(5,, 711,12,13,15). Which of the following cube covers of f will ensure that the required property is satisfied? (A) wxz, wxy, xyz, xyz, wyz
(B) wxy, wxz, wyz
(C) wxyz , xz, wxyz
(D) wzy, wyz, wxz, wwxz , xyz, xyz
SOLUTION Given function f (w, x, y, z) = Σ(5,7,11,12,13,15) draw K-map of the above function.
1 quad = xz = xz (y + y') = xzy + xy' z 2 pairs = wxy' + wyz xyz + xy' z + wxy' + wyz Hence (A) is correct option. Question. 43 We consider addition of two 2’s complement numbers bn − 1 bn −2 .....b0 and an − 1 an −2 ....a0. A binary adder for adding unsigned binary numbers is used to add the two numbers. The sum is denoted by cn − 1 cn −2 ....c0 and the carryout by cout . Which one of the following options correctly identifies the overflow condition?
(A) cout (an − 1 5 bn−1)
(B) an − 1 bn − cn − an − 1 bn − 1 +cn−1 1 1
(C) cout 5 cn−1
(D) an − 1 5 bn − 1 5 cn−1
SOLUTION 272
Binary adder generates C out only if 1 C in
1
0 1
0a
1
1 1
1b
0
1 1
1 C out 1 1 1 C out in this case is Cn−1 generated carry. C in is Cn−2 So b ' n − 1 a' n − 1 c n − 2 + b n − 1 a n − 1 c'n−2 f = Cout 5 Cn−1 Hence (C) is correct option. Question. 44 Consider number represented in 4-bit gray code. Let h3 h2 h1 h0 be the gray code representation of a number n and let g3 g2 g1 g0 be the gray code of (n + 1) (modulo 16) value of the number. Which one of the following functions is correct? E g0 (h1 h2 h1 h0) = /(1,2,3,6,10,13,14,15) E g1 (h1 h2 h1 h0) = /(4,9,10,11,12,,13,14,15) E g2 (h1 h2 h1 h0) = /(2,4, 5,6,7,12,,13,15) E g3 (h1 h2 h1 h0) = /(0,1,6,7, 10,11,12,,13,) SOLUTION g3g2g1 g0
Binary h
h 3 h 2 h 1 h0 (n + 1) mod 16
0000
0
0000
1
0001
0001
1
0001
2
0011
0010
2
0011
3
0010
0011
3
0010
4
0110
0100
4
0110
5
0111
0101
5
0111
6
0101
0110
6
0101
7
0100
0111
7
0100
8
1100
1000
8
1100
9
1101
273
1001
9
1101
10
1111
1010
10
1111
11
1110
1011
11
1110
12
1010
1100
12
1010
13
1011
1101
13
1011
14
1001
1110
14
1001
15
1000
1111 15 1000 0 0000 This gives the solution option (B) g 1 (h 3, h 2, h 1, h0) = /(4,9,10,11,12,13,14,15) YEAR 2007 Question. 45 What is the maximum functions involving n Boolean variables? (A) n2 (C) 22n
number of different Boolean (B) 2n (D) 2n2
SOLUTION Each boolean variable can have values 0 or 1, so for expression involving n boolean variables will have terms 2n . These 2n terms need to be arranged in different manner and nos., suppose 2n = M. So this arrangement would take 2M ways or 22n ways. Hence (C) is correct option. Question. 46 How many 3-to-8 line decoders with an enable input are needed to construct a 6-to-64 line decoder without using any other logic gates? (A) 7 (B) 8 (C) 9
(D) 10
SOLUTION
274
Total output lines required = 64 We need to use 3 to 8 decoders. So decoders required 648 = 8 decoders for output. But we need one more decoder i.e for combining result. 8 + 1 = 9 decoders. Hence (C) is correct option. Question. 47 Consider the following Boolean function of four variables f (w, x,, y, z) = /(1,3,4,6,9,11,12,14) The function is (I) independent of one variable (J) independent of two variables (K) independent of three variables (L) dependent on all the variables SOLUTION f (w, x, y, z) = /m(1,3,4,6,9,11,12,14)
2 qlead 1st qlead xz' x' z 2nd qlead xz' + x' z xz' + x' z So independent of 2 variables. Hence (B) is correct option. Question. 48 Let f (w, x, y, z) = /(0,4,5,7,8,9,13,15). Which of the following expressions are NOT equivalent to f ? 275
(A) x' y' z + w' xy' + wy' z + xz
(B) w' y' x' + wx' y' + xz
(C) w' y' z' + wx' y' + xyz + xy' z
(D) x' y' z + wx' y' + w' y
276
SOLUTION f (w, x, y, z) = /m(0,4,5,7,8,9,13,15) Drawing K-map.
xz + w' y' z' + wx' y' Hence (B) is correct option. Question. 49 Define the connective* for the boolean variable X and Y as: X * Y = XY + X' Y' Let Z = X * Z Consider the following expression P, Q and R. P: X = Y * ZQ: Y = X * Z R: X * Y * Z = 1 Which of the following is TRUE? (A) only P and Qare valid
(B) Only Qand Rare valid
(C) Only P and Rare valid
(D) AllP, Q,Rare valid
SOLUTION Given Z = X ) Z & XZ + X' Z' P: X = Y ) Z (I) YZ + Y' Z' (J)Y (XZ + X' Z') + Y' Z' (K) XYZ + X' YZ' + Y' Z' (L) XYZ + X' YZ' + XY' Z' + X' Y' Z' valid. Q: Y = X ) Z (M) XZ + X' Z' (N) X (XZ + X' Z') + X' Z' 277
(G) XZ + X' Z' (H) X (Y + Y') Z + X'(Y + Y') Z' (I)XYZ + XY' Z + X' YZ' + X' Y' Z' valid R : X ) Y ) Z = 1 (XY + X' Y') ) Z & (XZ + X' Y') Z + (XY + X' Y') Z' (E)XYZ + X' Y' Z + [(XY : X' Y') Z'] (F) XYZ + X' Y' Z + [(X + Y ) : (X + Y)] Z' (G) XYZ + X' Y' Z + X' YZ' + XY' Z' =Y 1 So invalid Hence (A) is correct option. Question. 50 Suppose only one multiplexer and one inverter are allowed to be used to implement any Boolean function of nvariables. What is the minimum size of the multiplexer needed? (A) 2n line to 1
line
(B) 2n+1 line to 1 line
(C) 2n−1 line to
1 line
(D) 2n−2 line to 1 line
SOLUTION To select 2n lines we need a select function with n bits. Here with n variables we have (n − 1) select bits thus 2n−1 data lines. So MUX has 2n−1 lines to 1. Hence (C) is correct option. Question. 51 In a look-ahead carry generator, the carry generate function Gi and the carry propagate function Pi for inputs, Ai and Bi are given by Pi = Ai 5 Bi and Gi = Ai Bi The expressions for the sum bit S and carry bit Ci+1 of the look ahead carry adder are given by Si + Pi 5 Ci and Ci + 1Gi + Pi Ci , Where C0 is the input carry. Consider a two-level logic implementation of the look-ahead carry generator.. Assume that all Pi and Gi are available for the carry generator circuit and that the AND and OR gates can have any number of inputs. The number of AND gates and OR gates needed to implement the look-ahead carry generator for a 4-bit adder with S3, S2, S1, S0 and C4 as its outputs are respectively
278
(A) 6,3
(B) 10,4
(C) 6,4
(D) 10,5
SOLUTION The equation for 4 bit carry look ahead adder is Cout g 3 + p 3 g 2 + p 3 p 2 g 1 + p 3 p 2 p 1 g 0 + p 3 p 2 p 1 p 0 Cin Here no. of AND gates = 10 OR gates = 4 Hence (B) is correct option. Question. 52 The control signal functions of 4-bit binary counter are given below (where X is “don’t care”) Clear Clock Load Count Function 1 0
X X
X 0
X 0
Clear to 0 No change
0
-
1
X
Load input
0
-
0
1
Count next
The counter is connected as follows
279
Assume that the counter and gate delays are negligible. If the counter starts at 0, then it cycles through the following sequence (A) 0,3,4
(B) 0,3,4,5
(C) 0,1,2,3,4
(D) 0,1,2,3,4,5
SOLUTION From the truth table for the counter ckt we can see that when counter = 1. & load = 0, count next is the function. So it would count from 0 to 4 & then clear to 0 & again start if clock input is increasing. Hence (C) is correct option. YEAR 2008 Question. 53 In the IEEE floating point representation the hexadecimal value 0x00000000 corresponds to (A) the normalized value 2−127 (C) the normalized value +0
(B) the normalized value 2−126 (D) the special value +0
SOLUTION This 0X00000000 hexadecimal value can be converted into 32 bit binary. 0000 0000 0000 0000 0000 0000 0000 0000 0 # 2c This is representation in IEEE floating point format. Case for special +0. Hence (D) is correct option. Question. 54 In the karnaugh map shown below, X denoted a don’t care term. What is the nominal form of the function represented by the karnaugh map
280
−− −− (A) b .d + a. d −− −−− (C) b .d + a. b. d
−− −− −−− (B) a. b + b . d + a.b . d −− −− −− (D) a. b + b . d + a. d
SOLUTION Given K-map is
quad 1 " a b quad 2 " b d pair 1 " a c d So a b + b d + a d (c + c) a b + b d + a d Hence (D) is correct option. Question. 55 Let a denote number system radix. The only value(s) of r that satisfy the equation 121 + 11, is/are (A) decimal 10 (C) decimal 10 and 11 SOLUTION (121)r = (11)r If r = 10 it is true it can’t be 2 = 11 It is true for 3 to 10. So it is true for r > 2 Hence (D) is correct option.
(B) decimal 11 (D) any value> 2 since bit value can’t be 2 then. It is not true for r
Question. 56 Give f1, f3 and f in canonical sum of products form (in decimal) for the circuit 281
f1 = /m(4,5,6,7,8) f3 = /m(1,6,15) f = /m(1,6,8,15) Then f2 is (A) /m(4,6)
(B) /m(4,8)
(C) /m(6,8)
(D) /m(4,6,8)
SOLUTION
Given f = Σm(1,6,8,15) f3 = Σm(1,6,15) So output 1,6,8,15 here 1,6,15 can come form f3. Since the final gate is OR gate so from f1 AND f2 no minterm except 1,6,8,15 should come. f1 = Σm(4,5,6,7,8) f2 can be Σm(6,8) So Since 4,5, & 7 should no 7 come here. Hence (C) is correct option. Question. 57 − − −− − If P , Q, R are Boolean variables, (P + Q) (P.Q + P. R) (P. R + Q) simplifies to − − (B) P. R (A) P. Q
282
− (C) P. Q + R
− (D) P. R + Q
SOLUTION S (P + Q ) : (P Q + PR) : (P R + Q) S (PPQ + PPR + PQ Q + PQR)(P R + Q) S (P Q + PR + PQ + PQR)(P R + Q) S (PQ + PR + PQR)(P R + Q) S [ PQ (1 + R ) + PR](P R + Q) S P (Q + R)(P R + Q) S (P P R + PQ )(Q + R) S PQ : (Q + R) S PQ + PQR S PQ (1 + R) S PQ Hence (A) is correct option. YEAR 2009 Question. 58 (1217)8 is equivalent to (A) (1217)16
(B) (028F)16
(C) (2297)10
(D) (0B17)16
SOLUTION (vii) (1217)8 (viii) 001010001111SSS (ix) (028F)1 6 Hence (B) is correct option. Question. 59 What is the minimum number of gates required to implement the Boolean function (AB + C) if we have to use only 2-input NOR gates ? (A) 2
(B) 3
(C) 4
(D) 5 283
SOLUTION AB+C implementation through NOR gate (X + Y ) We require one AND gate & 1 OR gate
AND gate & OR gate can be implemented by NOR gate. 2 A+C+B+C 2 A+C:B+C 2 (A + C ) : (B + C) 2 C + AB So we require & NOR gates. Hence (B) is correct option. YEAR 2010 Question. 60 The minterm expansion of f (P, Q, R) = PQ + QR + PR is (A) m 2 + m 4 + m 6 + m1
(B) m 0 + m 1 + m 3 + m5
(C) m 0 + m 1 + m 6 + m1
(D) m 2 + m 3 + m 4 + m5
SOLUTION Given expression is f (P, Q, R) = PQ + QR + PR For min term expansion we add the remaining variables in the expression. 2. PQ (R + R ) + (P + P ) QR + P (Q + Q ) R 3. PQR + PQR + PQR + PQR + PQR + PQR 4. PQR + PQR + PQR + PQ R 5. m 7 + m 6 + m 2 + m4 6. 111 + 110 + 010 + 100 So = m 2 + m 4 + m 6 + m7 284
Hence (A) is correct option. Question. 61 P is a 16-bit signed integer. The 2’s complement representation of P is (F 87B)16. The 2’s complement representation of 8)P is (A) (C 3D8)16
(B) (187B)16
(C) (F878)16
(D) (987B)16
SOLUTIO N P’s 2’s complement. = (F87B)16 Is complement= F 87B − 1 = (F87A)16 In base 16 complement is done by subtracting from 15 i.e F. = = (0785)16 (0000 01111000 0101)2 1 # 2c + 1 # 2 2 + 1 # 2 7 + 1 # 2 8 + 1 # 2 9 + 1 # 210 = = 1925 8 # P = 8 # 1925 = 15400 Its binary 0011110000101000SSSS For hexadecimal make pairs of 4 i.e. (3C28)16 2’s complement P = 1100 0011 1101 1000 C 3 D 8 2’s complement of P = (C 3D8)16 Hence (A) is correct option. Question. 62 The Boolean expression for the output f of the multiplexer shown below is
(A) P 5 Q 5 R
(B) P 5 Q 5 R 285
(C) P + Q + R
(D) P + Q + R
SOLUTION S1 & so are the select bits which are used to select any 1 of the 4 inputs. Selection table S 1 (P) S 0 (Q) Input 0
0
0R
0
1
1R
1
0
2R
1
1
3R
The expression has 3 variables So K-map
This is K-map for P 5 Q 5 R i.e PQ R + P Q R + PQR + P QR Hence (B) is correct option. Question. 64 In the sequential circuit shown below, if the initial value of the output Q 1 Q0 is 00, what are the next four values of Q 1 Q0 ?
286
(A) 11, 10, 01, 00
(B) 10, 11, 01, 00
(C) 10, 00, 01, 11
(D) 11, 10, 00, 01
SOLUTION There are 2 T-toggle flip flops in the ckt. Truth table for TFF. CP
T
Qn+1
0
X
Qn
Qn previous state
1
0
Qn
CP clock pulse
1
1
Qn
Qn+1 next state T toggle input
Since initially Q, Q0 = 00, so during 1st clock cycle both T & clock signals in ckt are 1. After Q0 = 1 this fed to 2nd TFF which invert previous state Q1 = 1 so Q 1 Q0 = 11 11 when fed to next cycle clock = 1 so Q0 = 0 Q1 = 1 since no inversion Q, Q0 = 10 In next cycle clock = 1 Q0 = 1 inverse, Q1 = 0 in the end Q 1 Q0 = 00 So order 11, 10, 01, 00 Hence (A) is correct option. Which one of the following circuits is NOT equivalent to a 2-input XNOR (exclusive NOR) gate? (A)
(B) 287
(C)
(D)
Answer: - (D) Exp: - All options except option ‘D’ gives EX-NOR gates 67.
The minimum number of D flip-flops needed to design a mod-258 counter is (A) 9 (B) 8 (C) 512 (D) 258 Answer: - (A) Exp: - 2n ≥ 258 ⇒ Common Data Questions: 50 & 51 Consider the following circuit involving three D-type flip-flops used in a certain type of counter configuration.
P D
Q
Clock Q Q D
Q
Clock Q R D
Q
Clock Q
288
68. If all the flip-flops were reset to 0 at power on, what is the total number of distinct outputs (states) represented by PQR generated by the counter? (A) 3 (B) 4 (C) 5 (D) 6 Answer: - (B) Exp: CLOCK
Inputs
Outputs
D2 D3
D1
R
P
Q
R
1
0
1
0
0
1
0
2
0
1
1
0
1
1
3
1
0
0
1
0
0
4
0
0
0
0
0
0
So Total number of distinct outputs is 4
69.
If at some instance prior to the occurrence of the clock edge, P. Q and R have a value 0, 1 and 0 respectively, what shall be the value of PQR after the clock edge? (A) 000 (B) 001 (C) 010 (D) 011 Answer: - (D) Exp: -From the Table Shown in the explanation of question 50, if first state is 010 next State is 011
289
YEAR 2013 73.
Which one of the following expressions does NOT represent exclusive NOR of x and y? (B) x ⊕ y'
Ans:
(C) x'⊕ y
(D) x'⊕ y'
(D)
Exp: ⊕ ⊕ ⊕
⊕y
74. In the following truth table, V = 1 if and only if the input is valid. Inputs
Outputs
D0
D1
D2
D3
X0
X1
V
0 1
0 0
0 0
0 0
X 0
X 0
0 1
0
1
0
0
1
1
1
X
1
0
0
1
X
X
X
1
1
1
1
What function does the truth table represent? (A) Priority encoder (B) Decoder (C) Multiplexer (D) Demultiplexer Ans: (A) Exp: 4 to 2 priority encoder. 75. The smallest integer than can be represented byan 8-bit number in 2’s complement form is (A) -256 (B) -128 (C) -127 (D) 0 Ans: (B) Exp:
− 28 −1
−128. Range is -2(n-1) to +2(n-1)-1
76. A RAM chip has a capacity of 1024 words of 8 bi
ts each
8
. The number of 290
decoders
Ans: Exp.
(A) 4 (B)
with
enable
(B) 5
line (C) 6
needed
to
construct
a
(D) 7
16 Number of chips required =
2 [16 chips vertically with each
having 2 chips horizontally] So to select one chip out of 16 vertical chips, we need 4 x 16 decoder. Available decoder is – 2 x 4 decoder To be constructed is 4 x 16 decoder
So we need 5, 2 x 4 decoder in total to construct 4 x 16 decoder.
291
