ms-51

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MS-51 1.

Solv Solve e th the e fo foll llow owin ing g LP prob proble lem m th thro roug ugh h SIM SIMPLEX PLEX meth method od Minimize Z = 2x1 + 5x2 Subject to x1 + x2 = 100 x1 ≤ 40 x2 ≥ 30 x1, x2 ≥ 0

Ans: Since the the given objectiv objective e function function is of minimisa minimisation tion type type we shall convert convert it in to a maximis maximisation ation type as follows: Maximise (-Z) = Maximise Z* = - 2x1 - 5 x2 Subject to the constraints x1 + x2 = 100 x1 ≤ 40 x2 ≥ 30 x1, x2 ≥ 0 By introducing the non-negative slack variable s1, surplus variable s2 and the artificial variablesR variables R1 and R2, the standard form of LPP becomes Maximise Z* = - 2x1 - 5 x2 + 0s1+ 0s2 –MR1 -MR2 Subject to the constraints x1 + x2 + R1 = 100 x1 + s1 = 40 x2 - s2 + R2 = 30 x1, x2, s1, s2 ≥ 0 Hence the basic feasible solution is given by s1 = 40, R1 = 100, R2 = 30 (x1= x2 = s2 =0, nonbasic). Initial Iteration:

CB YB -M R1 0 s1 -M R2 Z j* – C j

C j XB 100 40 30 -130M

-2 x1 1 1 0 -M + 2

-5 x2 1 0 (1) - 2M + 5

0 s1 0 1 0 0

0 s2 0 0 -1 M

-M R1 1 0 0 0

-M R2 0 0 1 0

Ø 100 -30

Since there are some Z j* – C j < 0, the current basic feasible solution is not optimal. The non-basic variable x2 enters into the basis and the basic variable R2 leaves the basis. First iteration C j XB 70 40 30 -70M-150

CB YB -M R1 0 s1 -5 x2 Z j* – C j

-2 x1 1 (1) 0 -M + 2

-5 x2 0 0 1 0

0 s1 0 1 0 0

0 s2 1 0 -1 -M + 5

-M R1 0 0 0 M

-M R2 -1 0 1 M-5

Ø 70 40 --

Since there are some Z j* – C j < 0, the current basic feasible solution is not optimal. The non-basic variable x1 enters into the basis and the basic variable s1 leaves the basis. Second iteration

CB -M -2 -5

YB R1 x1 x2

C j XB 30 40 30

-2 x1 0 1 0

-5 x2 0 0 1

0 s1 -1 1 0 1

0 s2 (1) 0 -1

-M R1 1 0 0

-M R2 0 0 1

Ø 30


Z j* – C j

-30M-230

0

0

M-2

-M + 5

0

M-5

Since there are some Z j* – C j < 0, the current basic feasible solution is not optimal. The non-basic variable S2 enters into the basis and the basic variable R1 leaves the basis. Third iteration CB YB 0 s1 -2 x1 -5 x2 Z j* – C j

C j XB 30 40 60 - 3 80

-2 x1 0 1 0 0

-5 x2 0 0 1 0

0 s1 -1 1 -1 3

0 s2 1 0 0 0

-M R1 1 0 1 M-5

-M R2 0 0 1 M-5

Ø 30 ---

Since all Z j* – C j ≥ 0, the current basic feasible solution is optimal. The optimal solution is given by Maximum Z* = -380, x1 = 40, x2 = 60. Minimum Z = 380, x1 = 40, x2 = 60.

2. Ans: Ans:

Writ Write e th the eg gen ener eral aliz ized ed fro from m of of line linear ar pro progr gram ammi ming ng pro probl blem em wit with h mat matri rix x not notat atio ion. n. The line linear ar progr programm amming ing in invol volvin ving g more more than two two varia variable bles s may be be expres expressed sed as foll follows ows:: Maximise or Minimise Z = c1x1 + c2x2 + c3x3 + ---- + cnxn Subject to the constraints a11x1 + a12x2 + ---- + a1nxn ≤ or = or ≥ b1 a21x1 + a22x2 + ---- + a2nxn ≤ or = or ≥ b2 ----------

am1x1 + am2x2 + ---- + amnxn ≤ or = or ≥ bm and the non – negative restrictions x1, x2, x3, - - -, xn ≥ 0 The general linear programming problem can always be expressed in the following form Maximise Z = c1x1 + c2x2 + c3x3 + ---- + cnxn Subject to the constraints a11x1 + a12x2 + ---- + a1nxn ≤ b1 a21x1 + a22x2 + ---- + a2nxn ≤ b2 ----------

am1x1 + am2x2 + ---- + amnxn ≤ bm and the non – negative restrictions x1, x2, x3, - - -, xn ≥ 0 This form of LPP is called the canonical form of LPP can be expressed as : Maximise Subject to

A=

Z = CX (objective function) AX ≤ b (constraints) and X ≥ 0 (non – negative restrictions) restrictions) Where C = (c1 c2 c3 ----cn) a11

a12

a21 ------a m1

a22

a m2

----

a1n

----

a2n

----

x1 ,

X =

a mn

x2 , x1n

b1 b=

b2 -

bm

----- x----- x -----

3.

A manu manufa fact ctur urer er requ requir ires es 1500 15000 0 units units of a part part annua annuall lly y for asse assemb mbly ly.. Manuf Manufac actu ture rerr can produce this at the rate of 100 per day. Set up cost for each production run is Rs. 24/- . to hold 2


one one unit unit of th this is part part in inve invent ntor ory y will will cost cost Rs. Rs. 5/ 5/-- per per year year.. Dete Determ rmin ine e th the e opti optimu mum m manufacturing quantity and the optimum number of production run in any year. Ans: Ans:

Numb Number er of of item items s requ requir ired ed per per uni unitt time time R = 1500 15000 0 per per year year Set up cost C3 = Rs. 24/Cost of holding C1 = Rs. 5/Production rate K = 100 x 365 = 36500

Economic batch quantity =

= = 494.43 units =495 units

Economic run length t =

=

=

= 0.032962 years = 12.03 days

Optimum number of production run in any year = 30.34 4.

(a) (a) Dis Disti ting ngui uish sh betw betwee een n pur pure e and and mixe mixed d str strat ateg egy y for for a gam game. e.

Ans: Ans:

Pure Pure Strate Strategie gies: s: If a playe playerr known known exact exactly ly what what the the other other player player is going going to do do a dete determi rminis nistic tic situat situation ion is obt obtain ained ed and object objective ive function function is to maximi maximise se the gain. gain. Theref Therefore ore,, the pure pure strategy is a decision rule always to select a particular course of action. Zero sum games with two players are called rectangular games. In this case the loss (gain) of the player is exactly equal to the gain (loss) of the other. In pure strategies, Maximin = Minimax Mixed strategy: when maximin ≠ minimax, then pure strategy fails. Therefore, each player with certain probabilistic fixation. This type of strategy is called mixed strategy. If a player decides in advance to use all or some of his available courses of action in some fixed proportion, he is said to use mixed strategy. Thus a mixed strategy, after the pattern of play has become evident, is that the opponents are kept guessing as to what a player’s course of action will be. A player may be able to choose only a specified number of pure strategies, but has an infinite number of mixed strategies to choose.

4.

Ans: Ans:

(b) So Solve tth he ga game wh whose pa pay-off ma matrix is is 1

7

2

6

2

7

5

1

6

Give Given n prob proble lem m is is sol solve ved d by Domi Domina nanc nce e prop proper erty ty.. Row III is dominated by Row II and column I is dominated by column III, So, we delete Row III and column III

A1

B1

B2

1

7 3


A2

6

2

Using strategy for B as x, 1 – x We have , 1.x + 7(1 – x) = 6x + 2(1 – x) 7 – 6x = 4x + 2 10x = 5 x=

, and 1 – x =

Strat trateg egy y fo forr B: {

,

, 0}

Using strategy for A as y, 1 – y 1.y + 6(1 – y) = 7y + 2(1 – y) 6 – 5y = 5y + 2 7

y=

,1–y=

y=

, and 1 – y =

Strategy for A: {

,

, 0}

Value of the game = 1.x + 7(1 – x)

= 1. (

) + 7( )

=

= 5.

=4

Write short notes on the following (a) Goal programming (b) Dynamic programming (c) Simulation

Ans:

(a) Goal programming: all oth other er opt optimi imisat sation ion method methods s only only con consid sidere ered d one criter criterion ion or perfor performan mance ce measur measure e to def define ine the optimum. optimum. It is not possible possible to find find a soluti solution on tha thatt simultaneously minimise cost and maximises reliability and minimises energy utilisation. In many practical situations it would be desirable to achieve a solution that is best with respect to a number of different criteria. One way of treating multiple competing objectives is to select one criterion as primary and the remaining criteria as secondary. secondary. The primary criterion is used as an optimisation performance measure, while the secondary criteria are assigned acceptable minimum or maximum values and are treated as problem constraints. If careful considerations were are not given while selecting the acceptable levels, a feasible design that satisfies all the constraints may not exit. This problem is overcome by a technique called goal programming, which is fast becoming a practical method for handling multiple criteria. 4


In goal programm programming ing all the objectives objectives are assigned assigned target levels for achieveme achievement nt and a relative priority on achieving these levels. Goal programming treats these targets as goal to aspire for and not as absolute constraints. It then attempts to find an optimal solution that comes as “close as possible” to the targets in the order of specified priorities. The general goal programming model can be expressed as m Minimise Z = ∑ (wi+di+ + wi-di-) i=1 m Subject to Z = ∑ aij x j +di- - di+ = bi for all i i=1 x j , di- , di+ ≥ 0 for I and j (b) Dynamic programming: Dyna Dynami mic c prog progra ramm mmin ing g is a math mathem emat atic ical al te tech chniq nique ue of optimisation using multistage decision process. That is, the process in which a sequence of interrelated decisions has to be made. It provides a systematic procedure for determining the combination of decisions which maximise overall effectiveness. Classical mathematics has provided inadequate in handling many optimisation problems that involve large number of decision variables and /or large number of inequality constraints. Such type of problems is handle han dled d by the dyn dynam amic ic progra programm mming ing techni technique que.. The dyn dynami amic c progra programm mming ing techni technique que decompos deco mposes es the original original problem in ‘n’ variables in to ‘n’ sub problems problems (stages) (stages) each in one variable. The solution is obtained in an orderly manner by starting from the one stage to the next and is completed after the final stage is reached. In many situations we observed that the decision making process consists of selecting a combination of plans from a large number of alternative combination. Before making a decision, it is required that, all the decisions of combination are specified and the optimal policy can be selected only aft5er all the combinations are evaluated. The dynamic programming technique deal with such situations by dividing the given problem is to sub problems or stages. Only one stage is considered at a time and the various infeasible combinations are eliminated with the objective of reducing the volume of computations. The solution is obtained by moving from one stage to the next and is completed when the final stage is reached. (c) Simulation: Simulation is numerical technique for conducting experiments on a digital computer, which involves logical and mathematical relationships that interact to describe the behaviour and structure of a complex real-world system over extended periods of time. Simulation has often been described as the process of creating the essence of reality without ever ever actu actual ally ly at atta tain inin ing g th that at real realit ity y itse itself lf.. Simu Simula lati tion on will will invo involv lve e th the e cons constr truc ucti tion on,, experimentation and manipulation of a complex model on a digital computer. The techniques that will be described have also been implemented on modern microcomputers and elements of our discussion will directly address those possibilities. Although the simulation is sometimes viewed as “method of last resort”. Often to be employed when all else fails, recent advances in simulation methodologies, software availability and technical developments have made simulation one of the most widely used and accepted tools in systems analysis and operations research.

5

ms-51

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