60 SIMPLEX METHOD
60.1 INTRO RODU DUCT CTIION In the previous lesson, we have solved linear programming problems by graphical method. This method is useful when there are not more than two variables and the number of constraints is less. If we have a large number of either variables or constraints, it is still true that the optimal solution will be found at the vertex of the set of feasible solutions. In fact, we could find these vertices by writing all the equations corresponding to the inequalities of the problem and then proceeding to solve all possible combinations of these equations. We would of course, have to discard any solutions that are not feasible (because they don’t satisfy one or more of the constraints). Then we could evaluate the objective function at the remaining feasible solutions. After all this, we might discover that the problem has no optimal solution at all. But this would be a highly cumbersome procedure. If there were just four variables and 7 constraints, we would have to solve all possible combinations of 4 equations chosen from a set of 7 equations. Obviously, there would be 7c 4=35 solutions in all. Each of these solutions will then have to be tested for feasibility. So even for this relatively small number of variables and constraints, the work would be quite tedious. In the real world of applications, it is fairly common to encounter problems with hundreds, even thousands of variables and constraints. Of course such problems must be solved by
40 :: Mathematics
computer. Even so, choosing a more efficient problem solving strategy than the geometrical method might reduce computer’s running time from hours to seconds or for very large problems from days to hours. A more systematic approach would involve choosing a solution at one vertex of the feasible set, then moving from there to another vertex at which the objective function has a better value. This process is continued until the best possible value is found. One very efficient technique of doing this is simplex method. method .
60.2 O B JEC TI VE S After the study of this lesson you will be able to
define slack variables;
identify standard maximum problem;
solve the standard maximum problem; and
solve the linear programming problems for mixed constraints.
60.3 60 .3 SOLUTI SOLUTION ON OF THE LI LINEA NEAR R PROGR PROGRAMM AMMING ING PROBLEM BY SIMPLEX METHOD In this section we will solve the linear programming problem by simplex method. In linear programming problem we have two types of problems, that is, either to maximize the objective function or to minimize the objective function. We will consider these problems one by one. Let us consider first the problems in which the objective function is to be maximized. Example A : Solve the following linear programming problem. Max Ma x z = 3 x 1 + 4 x 2 subject to the conditions 2 x 1 + 4 x 2 ≤ 12 120 0 2 x 1 + 2 x 2 x 1
≤ 80
≥ 0, x 2
≥ 0
Simplex Method :: 41
Solution : The procedure for solving the given problem is illustrated in the following steps: Step I: Standard form of a maximum problem A linear programming problem in which the objective function is to be maximized is referred to as a maximum linear programming problem. problem. Such problems are said to be in standard form if following conditions are satisfied: (i) All the the variable variables s are non–ne non–negat gative. ive. (ii) All the other constraints constraints are written as a linear expression, expression, that is, less than or equal to a positive constant. Check whether the given problem is in standard form ? This is a maximum problem containing two variables x 1 an and d x 2. Since both the variables are non–negative and the other constraints are each written as linear expression less than or equal to a positive constant, therefore, we conclude that the maximum problem is in standard form. Steep II: Slack variables and starting simplex table In order to solve the maximum problem by simplex method, we need to do the following first (i) Int Introdu roductio ction n of slack slack variables variables,, and (ii) Constr Construction uction of the initial initial simplex table. table. In this problem we have the constraints as linear expressions less than or equal to some positive constants. That means there is a slack between the left and right sides of the inequalities. In order to take up the slack between the left and right sides of the problem, let us introduce the slack variables s 1, and s 2 which are greater than or equal to zero, such that 2 x 1+ 4 x 2+ s 1 = 120 2 x 1 + 2 x 2 + s 2 = 80 Furthermore, rewritten as
the
objective
function
z =
3 x 1 + 4 x 2 can be
–3 x 1 – 4 x 2 + z = 0 In effect, we have now replaced our original system of
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constraints and the objective function by a system of three equations in five unknowns as: 2 x 1+ 4 x 2+ s 1 = 120 2 x 1+ 2 x 2 + s 2 = 80 –3 x 1 – 4 x 2 + z = 0 Here, we have to find the particular solution ( x 1, x 2 , s 1, s 2 , z ) that gives the largest possible value for z . The augmented matrix for this system is x 1
x 2
s 1
s 2
z
2
4
1
0
0
120
2
2
0
1
0
80
–3
–4
0
0
1
0
This is called the initial (starting) simplex table and written as x 1
x 2
s 1
s 2
z
2
4
1
0
0
120
s 1
2
2
0
1
0
80
s 2
–3
–4
0
0
1
0
z
(1)) (1
Notice that the bottom row contains the negatives of the coefficients in the objective function and that we have set this row off from the rest of the matrix by a dotted line. From this point on, the simplex method consists of pivoting from one table to another until the optimal solution is found. What is pivoting ? Pivoting : To pivot a matrix about a given element, called the pivot element, element , is to apply row operations so that the pivot element is replaced by 1 and all other entries in the same column (called pivot column) become 0. More specifically, in the pivot row, divide each entry by the pivot element (we assume it is not 0). Obtain 0 elsewhere in the pivot column by performing row operations. How is the pivot element selected?
Simplex Method :: 43
Pivot element : The pivot element for the simplex method is found using the following rules: (i) The pivot column is selecte selected d by locating locating the most negative negative entry in the objective row. If all the entries in this column are negative, the problem is unbounded and there is no solution.. solution (ii) Divide each entry entry in the last column by the correspond corresponding ing entry (from the same row) in the pivot column. (Ignore the rows in which the pivot column entry is less than or equal to 0). The row in which the smallest positive ratio is obtained is the pivot row. row . The pivot element is the entry at the intersection of the pivot row and the pivot column. Note : The pivot element is never in the objective row. In this example, –4 is the most negative entry in the objective row, so column 2 is the pivot column.
1 4
On dividing all the entries in the last column by the corresponding entry in column 2, we get 30 as the smallest ratio, so row 1 is the pivot row. Hence a 21 =4 is the pivot element. This is marked in the initial table (reproduced below) by a circle and the pivot column and the row are marked by arrows. x 1
→
x 2
s 1
s 2
z
2
4
1
0
0
1 20
s 1
12 0 ÷ 4=30
2
2
0
1
0
80
s 2
80 ÷ 2=40
–3
–4
0
0
1
0
z
↑ Divide R 1 (row1)by 4 and then apply the operations R 2– 2 R 1 , R 3 + 4 R 1 .
The new matrix becomes x 1
x 2
s 1
s 2
z
0
0
30
x 2
1
0
20
s 2
0
1
120
z
1 2
1
1
0
–1
0
−1 2
1
(2 )
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Notice that the identity matrix I 3 which now appears in column 2,4, and 5 helps us to solve for x 2, s 2 an and d z in terms of x 1 an and d s 1 . This is why we wrote x 2 , s 2 and z to the right of the matrix (2). (2) is called the second simplex table. Now, the question is when does this process end? This process ends when there are no negative entries in the objective row. In the second simplex table –1 is the most negative entry in the objective row, so column 1 is the pivot column. On dividing all the entries in the last column by the corresponding entry in column 1, we get 20 as the smallest ratio, so row 2 is the pivot row. Hence, a 21 =1 is the pivot element. This is marked by a circle in the 2nd simplex table (reproduced below) and the pivot column and row are marked by arrows. x 1
x 2
s 1
1 →
1
0
–1 ↑
0
−1 2
1
s 2
z
0
0
30
x 2
30 ÷
1
0
20
s 2
20 ÷ 1=20
0
1
120
z
Divide R 2 by 1 and then apply the operations R 1 –
1 2
=6 0
R 2, R 3+ R 2.
The new matrix becomes x 1
x 2
0
1
1
0
0
0
s 1
s 2
z
0
20
x 2
1
0
20
x 1
1
1
140
z
(3 )
In the above matrix we notice that there are no negative entries in the objective row. Hence, the optimal solution has been found. Therefore, Maximum z = 140 which occurs at x 1 =20, x 2 =20
.
Simplex Method :: 45
Check point 1: 1. Is the followin following g maximum maximum problem problem in the the standard standard form? Max Ma x z = 8 x 1 + x 2 subject to the constraints 3 x 1 + 4 x 2 ≥ 2 x 1 + x 2 ≤ 6 x 1 ≥ 0, x 2 ≥ 0
2. Write the the second si simplex mplex table table for the followin following g initial initial simplex table. The pivot element is circled. x 1
x 2
s 1
s 2
z
1
2
1
0
0
300
s 1
3
2
0
1
0
480
s 2
–1
–2
0
0
1
0
z
Example B : Solve the following linear programming problem by simplex method : Maximize z = 6 x 1 – 8 x 2 + x 3 subject to the constraints 3 x 1+ x 2 ≤ 10 4 x 1 – x 2 ≤ 5 x 1 + x 2 – x3 ≥ –3 x 1 ≥ 0, x 2 ≥ 0, x 3 ≥ 0
Solution : This is a maximum problem containing three variables x 1, x 2 and an d x 3. Each variable is non–negative. The given set of constraints contains x 1+ x 2 – x 3 ≥ –3, which is not a linear expression, that is, less than or equal to a positive constant. Thus, the maximum problem is not in standard form. Notice, however that by multiplying this constraint by –1, we gett ge – x x 1 – x 2 + x 3 ≤ 3
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which is in the desired form. Thus, although the maximum problem as stated is not in the standard form, it can easily be modified to the requirements of the standard form. Now, Max Ma x z = 6 x 1 – 8 x 2 + x 3 subject to the constraints 3 x 1 + x 2 ≤ 10 4 x 1 – x 2 ≤ 5 – x x 1 – x 2 + x 3 ≤ 3 x 1 ≥ 0, x 2 ≥ 0, x 3 ≥ 0
After adding the slack variables the systems of constraints takes the following form : 3 x 1 + x 2 + 0. x 3 + s 1
= 10
4 x 1 – x 2 + 0. x 3 + s 2
= 5
– x x 1 – x 2 + x 3 + s 3
( i)
= 3
x 1 ≥ 0, x 2 ≥ 0, x 3 ≥ 0, s 1 ≥ 0, s 2 ≥ 0, s 3 ≥ 0
and objective function can be rewritten as –6 x 1 + 8 x 2 – x 3 + z = 0 The starting simplex table for this system is
→
x 1
x 2
x 3
s 1
s 2
s 3
z
3
1
0
1
0
0
0
10
s 1
10 ÷ 3=
4
–1
0
0
1
0
0
5
s 2
5 ÷ 4=
–1
–1
1
0
0
1
0
3
s 3
–6 8 –1 0 0 0 1 0 (1)) (1 z ↑ The pivot column is found by locating the column containing smallest entry in the objective row (i.e., –6 in column 1). The pivot row is obtained by dividing each entry in the last column by the corresponding entry in the pivot column and selecting the smallest non-negative ratio (Remember, in finding the pivot element, we ignore rows in which the pivot column contains a negative number, in this case –1). In this case pivot row is second row and the pivot element is the circled element 4.
Simplex Method :: 47
Dividing R 2 by 4 and applying the operations R 1 – 3 R 2, R 3 + R 2 , R 4 + 6R 2 ,
we get the second simplex table as: x 1
x 2
x 3
s 1
0
1
0
0
s 2
s 3
z
0
0
0
0
−3
0
4
25
−1
1 →
−5
0
s 1
5
4
0
4
1
4 13 2
1
0
–1
0
4
x 1
(2)) (2
17
4
1
0
0
1
4
s 3 z
↑
By the same procedure as before, we determine the next pivot element to be 1 in the third column and third row (since we have to ignore the rows in which the pivot column entry is zero). Applying the operation R 4 + R3 , we get the third simplex table x 1
47 3 21 15 1 7 −1
x 2
4 4 22
x 3
s 1
0
1
0
0
1
0
0
0
s 3
z
0
0
0
0
1
0
1
1
−3
7
0
s 2
4
4
25 4
s 1
5
1 −5
0
1
4
4
4 17 4
x 1 x 3
(3)) (3
47
0
4
z
In the above matrix we see that there are no negative entries in the objective row. Hence the optimal solution is found. Therefore, Maximum z
=
5
x 1 =
4
17
,
x 2 = 0,
x 3 =
4
25
We also see that s 1 =
4
, s 2 = 0, s 3 = 0. These values can
be used to check that these are, in fact solutions of the system of equations (i)
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Check point 2 : Solve the following maximum problem by simplex method Maximize z = 6 x 1 + 8x 2 + x 3 subject to the constraints 3 x 1 + 5 x 2 + 3 x 3 ≤ 20 x 1 + 3x 2 + 2x 3
≤ 9
6 x 1 + 2 x 2 + 5 x 3 ≤ 30 x 1 ≥ 0, x 2 ≥ 0, x 3 ≥ 0
60.3.1 Problem of minimization Involving ‘ ≤ ’ sign In general, a minimum problem can problem by realizing that in order maximize – z z . That is in such cases function by –1 and convert it into solve it as discussed above.
be changed to a maximum to minimize z we must we multiply the objective a maximum problem and
Example C : Solve the following linear programming problem : Minimum z = x 1 – 3x 2 + 2x 3 subject to the constraints 3 x 1 – x 2 + 2 x 3 ≤ 7 –2 x 1 + 4 x 2 ≤ 12 –4 x 1 + 3 x 2 + 8 x 3 ≤ 10 x 1 ≥ 0, x 2 ≥ 0, x 3 ≥ 0
Solution : This is a problem of minimization in which all the constraints are written as a linear expression, that is, less than or equal to a positive constant. Therefore converting the objective function for maximization, we have Max z1 = – x x 1 + 3x 2 – 2x 3 , where z 1 = – z z After introducing the slack variable the problem can be expressed as 3 x 1 – x 2 + 2 x 3 + s 1
= 7
–2 – 2 x 1 + 4 x 2 + 0. x 3 + s 2
= 12
–4 – 4 x 1 + 3 x 2 + 8 x 3 + s 3
= 10
Simplex Method :: 49
x 1 ≥ 0, x 2 ≥ 0, x 3 ≥ 0, s 1 ≥ 0, s 2 ≥ 0, s 3 ≥ 0,
(i )
The objective function can be written as x 1 – 3 x 2 + 2 x 3 + z 1
= 0
The starting simplex table for this system is
→
x 1
x 2
x 3
s 1
s 2
s 3
z 1
3
–1
2
1
0
0
0
7
s 1
–2
4
0
0
1
0
0
12
s 2
–4
3
8
0
0
1
0
10
s 3
1
–3
2
0
0
0
1
0
z 1
12 ÷ 4= 3 10 10 ÷ 3= 3 (1)) (1
↑
1 4
The pivot column is found by locating the column containing the smallest entry in the objective row (–3 in column 2). The pivot row is obtained by dividing each entry in the last column by the corresponding entry in the pivot column (ignoring row in which the pivot column contains a negative number) and selecting the smallest non-negative ratio. Thus, the second row is the pivot row and the pivot element in that row is the circled element 4. Diving R 2 by 4 and applying the operations R 1+ R 2, R 3– 3 R 2, R 4+ 3 R 2 ,
we get the second simplex table as x 1
→
x 2
x 3
s 1
0
2
1
1
0
0
s 2
s 3
z 1
0
0
10
s 1
0
0
3
x 2
5 2
−1 2
1
−5 2
−1 2
4
(2)
−3
0
8
0
0
2
0
4 3 4
1
0
1
s 3
0
1
9
z
↑ Applying the same procedure, we determine the next pivot 5
element to be
2
in the first column and first row.
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Dividing R 1 by
and applying the operations 5
R 1 ,
R 2 +
R 3 +
2
R 1 ,
R 4 +
R 1
we get the third simplex table as x 1
x 2
1
0
0
1
x 3
s 1
s 2
4
2
1
5
5
10
1
6
5
20
s 3
z 1
0
0
4
x 1
0
0
5
x 2
(3)) (3 0
0
0
0
10
1
1
0
11
s 3
0
1
11
z 1
16 20
In the above matrix we see that there are no negative entries in the objective row. Hence, the optimal solution is found. Therefore, Maximum z 1 =1 =11 1 x 1= 4, x 2 = 5, x 3 = 0, s 1 = 0, s 2 = 0, s 3= 11
Hence the solution of the original problem is Min Mi n z = –11 x 1 = 4, x 2 = 5, x 3 = 0
We also see that s 1 = 0, s 2 = 0, s 3 = 11 can be used to check that these are, in fact solutions of the system of equation (i)
INTE X T QUESTIONS 60.1 1.
I de d e nt n t i fy fy t he he s ta t a nd n d a rd rd m ax a x i mu mu m li l i ne n e ar a r p ro r o g ra ra mm m m i ng ng problems from the following (give reason for your answers)
(a) Ma Maxim ximiz ize e z = 2 x 1 + x 2 subject to the constraints x 1 + x 2 ≤ 5
2 x 1 +3 x 2 ≤ 2 x 1 ≥ 0, x 2 ≥ 0
Simplex Method :: 51
(b) Maximize z = 3 x 1+ x 2 +x 3 subject to the constraints x 1 + x 2+ x 3 ≤ 6
2 x 1 + 3 x 2+ 4 x 3 ≤ 6 x 1 ≥ 0
(c) Maximize z = 2 x 1+x 2 subject to the constraints x 1 + x 2 ≥ –6
2 x 1 + x 2 ≤ –4 x 1 ≥ 0, x 2 ≥ 0
2. The follow following ing maximu maximum m problems problems are are not in stand standard ard form. form. Determine if the problem can be modified so as to be in standard form. (a) Ma Maxim ximize ize z = x 1+ x 2 subject to the constraints 3 x 1 – 4 x 2 ≤ –6 x 1 + x 2 ≤ 4 x 1 ≥ 0, x 2 ≥ 0
(b) Maximize z =2 x 1 + x 2+ 3 x 3 subject to the constraints – x x 1 + x 2 – x 3 ≥ –6 2 x 1 – 3 x 2 ≥ –1 –12 2 x 1 ≥ 0, x 2 ≥ 0, x 3 ≥ 0
3.
Write Writ e the the secon second d si simpl mplex ex tabl table e of the the foll followi owing ng sta start rting ing simplex table. The pivot element is circled. x 1
x 2
s 1
s 2
z
1
4
1
0
0
100
s 1
2
5
0
1
0
50
s 2
–2
–1
0
0
1
0
z
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4.
Find Fin d the the pivot pivot ele elemen mentt in the the foll followi owing ng simp simple lex x tabl table. e. x 1
x 2
x 3
s 1
s 2
s 3
z
1
2
4
1
0
0
0
24
s 1
2
–1
1
0
1
0
0
32
s 2
3
2
4
0
0
1
0
18
s 3
–1
–2
–3
0
0
0
1
0
z
60.4 T HE H E SI S I MP M P LE L E X M E T H O D CONSTRAINTS
WITH
M I X E D
So far we have solved the linear programming problems in the standard form. Recall that for a maximum problem in standard form each constraint must be of the form a 1x 1+ a 2 x 2 + ... + a n x n ≤ b 1 ,
b 1 > 0
That is, each is a linear expression less than or equal to a positive constant. What happens when the constraints are of any other form i.e., greater than or equal to, or equal to? Such problems are dealt by a method known as two phase method.. The method is illustrated in the following example. method Example D : Solve the following linear programming problem : Maximize z =20 x 1 + 15 x 2 subject to the constraints x 1 + x 2 ≥ 7
9 x 1 + 5x 2 ≤ 45 2 x 1+ x 2 ≥ 8 x 1 ≥ 0, x 2 ≥ 0
Solution : We first observe that this is a maximum problem which is not in standard form. Second, it cannot be modified so as to be in the standard form. In order to solve this problem the following steps be employed:
Simplex Method :: 53
Step I: Write each constraint except the non-negative constraints as an inequality with the variable on the left side of a ‘ ≤ ’ sign. To do this we merely multiply the first and third inequality by –1. The result is that the constraints become – x x 1 – x 2 ≤ –7 9 x 1 + 5x 2 ≤ 45 –2 x 1 – x 2 ≤ –8 Step II: Introduce non-negative variables (slack variables) on the left side of each inequality to form an equality. To do this, we will use the variables s 1 , s 2, s 3 to obtain – x x 1 – x 2+ s 1
= –7
9 x 1 + 5 x 2 + s 2
= 45
–2 x 1 – x 2 + s 3
= –8
x 1 ≥ 0, x 2 ≥ 0, x 3 ≥ 0, s 1 ≥ 0, s 2 ≥ 0, s 3 ≥ 0.
The objective function can be written as –20 x 1 –15 x 2 + z
= 0
Step III : Write the initial simplex table. x 1
x 2
s 1
s 2
s 3
z
–1
–1
1
0
0
0
–7
s 1
9
5
0
1
0
0
45
s 2
–2
–1
0
0
1
0
–8
s 3
–2 0
–1 5 –1
0
0
0
1
0
z
The initial table represents the solution x 1 = 0, x 2 = 0, x 3 = 0, s 1 = –7, s 2 = 45, s 3 = –8
This is not a feasible solution. The reason for this is the existence of the two negative constraints in the right hand column. That is, this table represents a solution that causes two of the variables to be negative, in violation of the non negativity requirement. Whenever this occurs, the simplex algorithm consist of two phases.
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Step IV :Determine whether phase I or phase II applies. Phase I is used whenever negative entries appear in the right hand column. Phase II is used whenever all the entries in the right hand column are nonnegative. In determining whether phase I or phase II applies, the objective row is ignored. Step V : Select the pivot element Phase I: I: The pivot row is the row with the most negative value in the right hand side column. If all the entries in the pivot row are non-negative, there would be no feasible solution and hence the problem has no solution. For each column with a negative entry in the pivot row, find the ratio using the right hand side column entry as denominator. The largest of these is the pivot column and the point of intersection of the pivot row and pivot column would be the pivot element. For this we follow the pivoting strategy discussed in Phase II: II:For the preceeding sections. In the initial simplex table we notice that in the right hand side column two entries are negative. So we apply phase I. The pivot row is row 3 (–8 is the most negative value in the right hand side column). To find the pivot column 1 1 –2 –1 we find the ratios; i.e., = and = . The 4 –8 8 –8 1 largest is . Therefore, the pivot column is column 1. 4 The pivot element is –2 which is circled in the initial table (reproduced below). x 1
→
x 2
s 1
s 2
s 3
z
–1
–1
1
0
0
0
–7
9
5
0
1
0
0
45
–2
–1
0
0
1
0
–8
–20
–1 5
0
0
0
1
0
↑
Dividing R 3 by –2 and applying R 1 + R 3 , R 2 – 9R 3 , R 4 + 20 R 3
Simplex Method :: 55
we get the following table x 1
→
x 2
s 1
s 2
s 3
z
0
–1 2
1
0
–1 2
0
–3
0
1 2
0
1
9 2
0
9
1
1 2
0
0
–1 2
0
4
0
–5
0
0
–1 0
1
80
↑
Notice that in the right hand side column still one entry is negative; so apply the phase I again, as above. The pivot row is row 1, the pivot column is column 2 (or column 5). Here we take column 2 as pivot column for our convenience. 1
Dividing R 1 by – R 2 –
2
R 1, R 3 –
and applying R 1, R 4
+ 5 R 1
1 2
we get the following table.
→
x 1
x 2
0
1
0
s 1
s 2
s 3
z
–2
0
1
0
6
0
1
1
4
0
6
1
0
1
0
–1
0
1
0
0
–1 0
0
–5
1
110
↑
Observing that right hand side column has all the entries non-negative, so we will use phase II to obtain the optimal solution. Here pivot column is column 3 (being –10 as largest negative entry in the objective row) and R 3 is the pivot row (being the smallest non-negative ratio). The pivot element is circled and the pivot row and column are marked by arrows. Applying R 1 + 2 R 3 , R 2 – R 3 , R 4 + 10 R 3 , we get the following table.
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x 1
→
x 2
s 1
s 2
s 3
z
2
1
0
0
–1
0
8
–1
0
0
1
5
0
5
1
0
1
0
–1
0
1
10
0
0
0
–1 5
1
120
↑
Again apply the phase II and repeat the procedure as above. Dividing R 2 by 5 and applying the operations R 1+ R 2 , R 3 + R 2, R 4+ 15 R 2 ,
we get the following table x 1
x 2
s 1
s 2
s 3
z
9 5
1
0
1 5
0
0
9
x 2
–1 5
0
0
1 5
1
0
1
s 3
4 5
0
1
1 5
0
0
2
s 1
7
0
0
3
0
1
135
z
In the above matrix we see that there are no negative entries in the objective row. Hence, the optimal solution is found. Maximum z = 135 x 1= 0, x 2= 9, s 1= 2, s 2 = 0, s 3 = 1
Check point 3 : Solve the following linear programming problem: Maximize z = 3 x 1+ 4 x 2 subject to the constraints x 1+ x 2 ≤ 12
5 x 1 + 2 x 2 ≥ 36 7 x 1 + 4 x 2 ≥ 14 x 1 ≥ 0, x 2 ≥ 0, x 3 ≥ 0
Simplex Method :: 57
Example E : Solve the following linear programming problem Minimize z = 5 x 1+ 6 x 2 subject to the constraints x 1 + x 2 ≤ 10 x 1 +2 x 2 ≥ 12
2 x 1 + x 2 ≥ 12 x 1 ≥ 3 x 1 ≥ 0, x 2 ≥ 0
Solution : We change our problem from minimizing maximizing z 1 = – z z = – 5 x 1 – 6 x 2
z = 5 x 1 + 6 x 2 t o
Step I: Write each constraint with ‘ ≤ ’ , for this multiply second, third and fourth constraint by –1. i.e.,
x 1+ x 2 ≤ 10
– x –12 2 x 1 – 2 x 2 ≤ –1 –2 x 1 – x 2 ≤ –1 –12 2 – x x 1 ≤ –3 x 1 ≥ 0, x 2 ≥ 0
Step II: Introduce non-negative variables to form equalities x 1 + x 2 + s 1 =1 0
– x x 1 – 2 x 2 + s 2=–12 –2 x 1 – x 2 + s 3 =–12 – x x 1 + s 4 = –3 x 1 ≥ 0, x 2 ≥ 0, s 1 ≥ 0, s 2 ≥ 0, s 3 ≥ 0, s 4 ≥ 0 The objective function can be written as 5 x 1 + 6 x 2 + z 1 = 0
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Form the initial simplex table.
Step III:
→
x 1
x 2
s 1
s 2
s 3
s 4
z 1
1
1
1
0
0
0
0
10
s 1
–1
–2
0
1
0
0
0
–12
s 2
–2
–1
0
0
1
0
0
–12
s 3
–1
0
0
0
0
1
0
–3
s 4
5
6 ↑
0
0
0
0
1
0
z 1
Step IV: In the right hand side column three entries are negative. Apply phase I. pivot row is row row 2 (or 3) and for this row row the pivot pivot Step V V:: The pivot column is column 2. Pivoting as before, we get the following table (Dividing R 2 by –2 and applying the operation R 1 – R R 2, R 3+ R 2 , R 5 –6 R 2 ) x 1
s 3
s 4
z 1
1
0
0
0
4
s 1
1
0
0
0
0
6
x 2
0
0
1
0
0
–6
s 3
–1
0
0
0
0
1
0
–3
s 4
2
0
0
3
0
0
1
–3 6
z 1
→
x 2
s 1
0
s 2
Notice that still the right hand column has negative entries, so apply the phase I again. Pivot row is row 3 and for this row, the pivot column is column 1.
Dividing R 3 by
and applying the operations 1
R 1 –
R 3,
R 2 –
2
R 3 , R 4 + R 3, R 5 – 2 R 3,
Simplex Method :: 59
we get the following table x 1
x 2
s 1
0
0
1
0
1
1
s 2
s 3
s 4
z 1
0
0
2
s 1
0
0
0
4
x 2
0
0
0
0
4
x 1
0
0
0
1
0
1
s 4
0
0
0
0
1
–4 4
z 1
1 3
Notice that all the entries in the last column are non-negative. So now we will use phase II in order to get the optimal solution.
1 4 7 −2 33
But here all the entries in the objective row are non-negative. Hence, this is the final table and the optimal solution is found. Since the maximum value of z 1 =–44, the minimum value of z=– z= – z z 1 =44. This occurs when =4,, x 2=4, s 1 =2, s4 =1 =1,, s 2=0 =0,, s 3 =0 x 1 =4 So far all our constraints were of the form ‘ ≤ ’ or ‘ ≥ ’ . What happens when one of the constraints is an equality? One way is to replace the equality constraint with the two constraints ‘ ≤ ’ and ‘ ≥ ’ . The process is illustrated in the next example. Example F : Minimize z = 7 x 1 + 5 x 2 + 6 x 3 subject to the constraints x 1 + x 2 + x 3 = 1 0 x 1 + 2 x 2 + 3x 3 ≤
19
2 x 1 + 3x 2 ≥ 21 x 1 ≥ 0, x 2 ≥ 0, x 3 ≥ 0
60 :: Mathematics
Solution : Here we wish to Maximize z 1 =–7 x 1 –5 x 2 –6 x 3 , where ( z 1=– z z ) subject to the constraints x 1+ x 2+ x 3 ≥ 10 x 1+ x 2+ x 3 ≤ 10 x 1+2 x 2+3 x 3 ≤ 19
2 x 1 +3 x 2 ≥ 21 x 1 ≥ 0, x 2 ≥ 0 x 3 ≥ 0
Step I: Write each constraint with ‘ ≤ ’ , for this multiply first and fourth constraint by –1. – x x 1 – x x 2 – x x 3 ≤ –10 x 1 + x 2+ x 3 ≤ 10 x 1 +2 x 2+3 x 3 ≤ 19
–2 x 1 –3 x 2 ≤ –2 –21 1 x 1 ≥ 0, x 2 ≥ 0, x 3 ≥ 0
Introduce duce non-negativ non-negative e variables variables to form equations equations.. Step II II:: Intro – x x 1 – x x 2 – x x 3 + s 1
= –10
x 1 + x 2 + x 3 + s 2
= 10
x 1 +2 x 2 +3 x 3 + s 3
= 19
–2 x 1 –3 x 2+ s 4
= –21
x 1 ≥ 0, x 2 ≥ 0, x 3 ≥ 0 s i ≥ 0, i = 1, 2, 3, 4.
The objective function can be written as 7 x 1+5 x 2 +6 x 3 + z 1 = 0 Form the initial simplex table. Step III: III:Form x 1
x 2
x 3
s 1
s 2
s 3
s 4
z 1
–1
–1
–1
1
0
0
0
0
–10
s 1
1
1
1
0
1
0
0
0
10
s 2
1
2
3
0
0
1
0
0
19
s 3
–2
–3
0
0
0
0
1
0
–21
s 4
7
5
6
0
0
0
0
1
0
z 1
Simplex Method :: 61
Step IV: Notice that in the right hand side column two entries are negative. Apply phase I. Step V: The pivot row is row 4 and pivot column is column 2. D i v i d i n g R 4 b y – 3 a n d a p p l y i n g t h e o p e r a t i o n s R 1+ R 4 , R 2 – R R 4 , R 3 –2 R 4 , R 5 –5 R 4 , we get the following table. x 1
2 3
x 2
x 3
s 1
s 2
s 3
s 4
0
–1
1
0
0
0
–3
s 1
0
1
0
1
0
0
3
s 2
0
3
0
0
1
0
5
s 3
1
0
0
0
0
0
7
x 2
0
6
0
0
0
1
–35
z 1
−1 3
z 1
Notice that one entry is still negative in the right hand column. So again apply phase I. 11 1 5 4 2 −1 3 3
Pivot row is row 1 and the pivot column is column 3. Dividing R 1 by –1 and applying the operations R 2–R 1, R 3– 3 R 1, R 5– 6 R 1 ,
we get the following table x 1
0 →
x 2
x 3
0
1
0
s 1
s 2
s 3
–1
0
0
0
1
1
0
0
0
3
0
1
0
0
0
0
6
s 4
z 1
0
3
x 3
0
0
s 2
1
0
–4
s 3
0
0
0
7
x 2
0
0
1
–5 3
z 1
0
↑ Notice that one entry is still negative in the right hand column. So again apply phase I.
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Pivot row is row 3, the pivot column is column 1. Pivot element is circled in the above table. Dividing R 3 by R 1 –
and applying the operations
R 4 –
R 3,
R 3,
R 5 –
R 3
we get the following table
→
x 1
x 2
x 3
0
0
1
0
0
0
1
0
0
0
1
0
0
s 1
s 2
s 3
s 4
0
z 1
0
2
x 3
0
0
s 2
0
0
3
x 1
0
0
0
5
x 2
0
0
1
–58
z 1
1
1
0
0
↑ Notice that all the entries in the right hand column are nonnegative. Therefore, we will apply phase II for finding the optimal solution.
Pivot column is column 7 and pivot row is row 1. Pivot element is circled in the above table. Dividing R 1 , by R 3 –
R 1,
and applying the operations
R 4 +
R 1 ,
+
R 1 ,
s 2
s 3
s 4
z 1
R 5
we get the following table x 1
x 2
x 3
s 1
0
0
4
–1
0
1
1
0
8
s 4
0
0
0
1
1
0
0
0
0
s 2
1
0
–1
–2
0
–1
0
0
1
x 1
0
1
2
1
0
1
0
0
9
x 2
0
0
3
9
0
2
0
1
– 52
z 1
Simplex Method :: 63
Notice that all the entries in the objective row are non-negative. Hence, the optimal solution is found. Since the ma x imum i mum z 1= –52, so minimum z = 52. This occurs when x 1=1 =1,, x 2=9, x 3=0 =0,, s 1=0 =0,, s 2 =0, s 4 =8 Example G : Solve the following linear programming problem Minimum z = x 1+x 2 subject to the constraints 2 x 1+ x 2 ≥ 4 x 1 +7 x 2 ≥ 7 x 1 ≥ 0, x 2 ≥ 0
Solution: We change our problem from minimizing z = x 1 +x 2 to maximizing z 1 = – z z = – x x 1–x 2 Step I: Now, we write all the constraints with sign ‘ ≤ ’ and for this multiply each constraint (except the non-negative constraint) by –1. The constraints become –2 x 1 – x x 2 ≤ –4 – x x 1 –7 x 2 ≤ –7 x 1 ≥ 0, x 2 ≥ 0
Step II: Introduce non-negative variables to form equations. –2 x 1 – x x 2+ s 1 = –4 – x x 1 –7 x 2+ s 2 = –7 x 1 ≥ 0, x 2 ≥ 0, s 1 ≥ 0, s 2 ≥ 0
The objective function can be written as x 1 + x 2+ z 1 = 0
Form the initial simplex table Step III: III:Form
→
x 1
x 2
s 1
s 2
z 1
–2
–1
1
0
0
–4
s 1
–1
–7
0
1
0
–7
s 2
1
1
0
0
1
0
z 1
↑
64 :: Mathematics
Step IV: Notice that in the right hand side column, the entries have negative sign. Phase I applies here. Step V: Pivot row is R 2 and pivot column is column 2. The pivot element is circled in the above table. Dividing row 2 by –7 and applying the operations R 1 + R 2 ,
R 3 – R 2,
we get the following table x 1
→
x 2
s 1
s 2
z 1
–13 7
0
1
–1 7
0
–3
s 1
1 7
1
0
–1 – 1 7
0
1
x 2
6 7
0
0
1 7
1
–1
z 1
↑
Step IV: Notice that one entry in the right hand side column is negative. Phase I applies here. row is row 1 and pivot pivot column column is column column 1 and Step V V:: Pivot row the pivot element is circled. Dividing R 2 –
by
R 1
and
R 1 , R 3 –
applying
the
operations
R 1,
we get the following table x 1
x 2
1
0
0
0
s 1
−7
s 2
z 1
0
x 1
1
0
x 2
0
1
z 1
13
Notice that all the entries in the right hand column are nonnegative. So we will apply phase II to obtain the optimal solution. Here we see that all the entries in the objective row are non -negative. Hence, the optimal solution is found.
Simplex Method :: 65
Max. z 1=
, so min. z =
This occurs at x 1 =
,
.
and x 2=
, s 1=0 =0,, s 2 =0
INTE X T QUESTIONS 60.2 1. Solve the following linear programming problems (a) Ma Maxim ximiz ize e z =3 =3 x 1 +4 x 2 subject to the constraints – x x 1 – x x 2 ≥ –12 5 x 1 +2 x 2 ≥ 36 7 x 1 +4 x 2 ≥ 14 x 1 ≥ 0, x 2 ≥ 0
(b) Mi Mini nimiz mize e z =4 =4 x 1 +2 x 2 subject to the constraints 10 21 31 −31 13 13
3 x 1 +x 2 ≥ 27 – x –21 1 x 1 – x x 2 ≤ –2 x 1 +2 x 2 ≥ 30 x 1 ≥ 0, x 2 ≥ 0
(c) Maximize z =3 =3 x 1+2 x 2 subject to the constraints 2 x 1+x 2 ≤ 4 x 1 + x 2 ≤ 3 x 1 ≥ 0, x 2 ≥ 0
WHAT YOU HAVE LEARNT We use simplex method to find the optimal solution of a linear programming problem when the given problem has a large number of either variables or constraints. We follow the following steps to find the optimal solution: 1. Write the problem in the standard form. 2. Introduce slack variables on the left side of each inequality
66 :: Mathematics
to form an equality. 3. Wri Write te the the initia initiall simplex simplex table table.. 4. Determine whether phase I or phase II applies. Phase I: Whenever negative entries appear in the right hand column. Phase II: Whenever all the entries in the right hand column are non– negative. 5. Phase I (i)
Selectt the Selec the pivot pivot elem elemen ent. t. If If all the the eleme element nts s in the the pivot pivot row are non–negative, there would be no feasible solution.
(ii)
Do all all the the row row operat operations ions to form form a new new simplex simplex tab table le and repeat (i). This process ends when there are no negative entries in the right hand column.
(iii) Now use use phase phase II to to get the the optima optimall solution solution.. Phase II ( i) i)
S el el ec ec t t he he pi pi vo vo t e le le me me nt nt . If all the entries in this column are negative, the problem is unbounded and there is no solution.
(ii)
Do all all the the row row operat operations ions to form form a new new simplex simplex tab table le and repeat step (i) . This process ends when all the entries in the objective row are non–negative.
TERMINAL QUESTIONS 1. Solve the following following linear program programming ming problem problem by simplex method. (a) Ma Maxim ximize ize z =5 =5 x 1 +3 x 2 subject to the constraints 3 x 1 +5 x 2 ≤ 15 5 x 1 +2 x 2 ≤ 10 x 1 ≥ 0, x 2 ≥ 0
Simplex Method :: 67
(b) Maximize z = 3 x 1+ 2 x 2 – x 3 subject to the constraints x 1 + 3 x 2 + x 3 ≤ 9
2 x 1 + 3 x 2 – x 3 ≥ 2 3 x 1 – 2 x 2 + x 3 ≥ 5 x 1 ≥ 0, x 2 ≥ 0, x 3 ≥ 0
(c) Minimize z =6 x 1 + 8x 2 + x 3 subject to the constraints 3 x 1 + 5 x 2 + 3 x 3 ≥ 20 x 1 + 3 x 2 + 2x 3 ≥ 9
6 x 1 + 2 x 2 + 5 x 3 ≥ 30 x 1 + x 2 + x 3 ≤ 10 x 1 ≥ 0, x 2 ≥ 0, x 3 ≥ 0
(d) Min Minimiz imize e z = 2 x 1 + x 2 subject to the constraints 3 x 1 + x 2 = 3 4 x 1 + 3 x 2 ≥ 6 x 1 + 2 x 2 ≤ 4 x 1 ≥ 0, x 2 ≥ 0
(e) Minimize z = 2 x 1+ 9 x 2 + x 3 subject to the constraints x 1 + 4 x 2 + 2 x 3 ≥ 5
3 x 1 + x 2 + 2 x 3 ≥ 4 x 1 ≥ 0, x 2 ≥ 0, x 3 ≥ 0
(f) Maximize z =2 x 1 – x 2+ x 3 subject to the constraints x 1 + x 2 – 3 x 3 ≤ 8
4 x 1 – x 2 + x 3 ≥ 2 2 x 1 + 3 x 2 – x 3 ≥ 4 x 1 ≥ 0, x 2 ≥ 0, x 3 ≥ 0
68 :: Mathematics
(g)) Ma (g Maxim ximize ize z = 2 x 1+ x 2 subject to the constraints 3 x 1 + 5 x 2 ≤ 15 6 x 1 + 2 x 2 ≤ 24 x 1 ≥ 0, x 2 ≥ 0
(h) Max Maximiz imize e z = 3 x 1+ 2 x 2 subject to the constraints x 1 + x 2 ≤ 4 x 1 – x 2 ≤ 2 x 1 ≥ 0, x 2 ≥ 0
( i) i) M ax ax im im iz iz e z =2 x 1 + 4x 2 + x 3 subject to the constraints x 1 + 2 x 2 ≤ 4
2 x 1 + 2x 2 ≤ 3 x 2 + 4x 3 ≤ 3 x 1 ≥ 0, x 2 ≥ 0, x 3 ≥ 0
( j) j) M an an im im iz iz e z = 4 x 1 + 8 x 2 + 3 x 3 subject to the constraints x 1 + x 2 ≥ 2
2 x 1+ x 2 ≤ 5 x 1 ≥ 0, x 2 ≥ 0, x 3 ≥ 0
ANSWERS TO CHECK POINTS Check point 1 : 1.
2.
No t i n the standard form x 1
x 2
s 1
s 2
z
1 2
1
1 2
0
0
150
x 2
2
0
–1
1
0
180
s 2
0
0
1
0
1
300
z
Simplex Method :: 69
Check point 2: Max Ma x z =
,
x 1 =
,
x 2
=
,
x 3
= 0
Check point 3: Max z = 44, x 1 = 4, x 2 = 8
ANSWERS TO INTEXT QUESTIONS 60.1 1. (a) Standa Standard rd form. form. (b) Not in standar standard d form. form. (c) Not in stan standard dard form. form. 2. (a) Can Cannot not be be modifie modified d so as to be in in the the standa standard rd form. form. (b) Can be modif modified ied to Maximize z = 2 x 1 + x 2 + 3 x 3 subject to the constraints x 1 – x 2 + x 3 ≤ 6
–2 x 1 + 3 x 2 ≤ 12
1 55 5 2 75 3 4 −8
x 1 ≥ 0, x 2 ≥ 0 x 3 ≥ 0
12 45 5 2
3.
x 1
x 2
s 1
s 2
z
0
1
0
60
s 1
1
0
0
10
x 2
0
0
1
10
z
4. Pivot column column is column column 3, pivot pivot row is is row 3 and and pivot element is 4(a 33 ). 60.2 1 . ( a ) Ma x . z = 44,
x 1 = 4,
x 2 = 8
( b) b) M in in z = 48, x 1 = 3, x 2 = 18 ( c ) M ax z = 7, x 1 = 1,
x 2 = 2
70 :: Mathematics
ANSWERS TO TERMINAL QUESTIONS 1 . ( a ) Max
x 1 =
,
( b) b) M ax ax z = 27, x 1 = 9,
x 2 = 0,
x 3 = 0
( c ) M i n z =
x 2 = 0,
x 3 =
z
=
,
, x 1 = 0,
( d) d) M in in z =
(e ) Mi n
,
z =
x 1 =
,
,
x 2
=
x 2 =
x 1 = 0,
x 2 = 0,
x 3 =
(f) The problem problem is unbounded unbounded and there there is no no solution. solution.
( g) g) M ax ax . z =
,
x 1 =
,
x 2 =
(h)) Ma (h Max. x. z = 11,
x 1 = 3,
x 2 = 1
( i ) M a x z =
x 1 = 0,
x 2 =
(j) Min z = 10,
,
x 1 =
,
x 2 = 0,
, x 3 =
x 3 = 0.
