Consider the following LP model in standard form, with a row for the objective function Z.a) Put it into Canonical form ( or Simplex Tableau form) with basic variables X1, X2 , and X3.b) Determine the association BFS (Basic Feasible Solution) and the new formula for the objective function ZMinimize  10X1 + 4X2Sujbject to 3X1 + 2X2 - X3           = 607X1 + 2X2       - X4      = 843X1 + 6X2           -X5 = 72  X1,  X2,  X3 ,  X4 ,  X5 >= 0

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Asked Sep 13, 2019
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Consider the following LP model in standard form, with a row for the objective function Z.

a) Put it into Canonical form ( or Simplex Tableau form) with basic variables X1, X2 , and X3.

b) Determine the association BFS (Basic Feasible Solution) and the new formula for the objective function Z

Minimize  10X1 + 4X2

Sujbject to 

3X1 + 2X2 - X3           = 60

7X1 + 2X2       - X4      = 84

3X1 + 6X2           -X5 = 72

  X1,  X2,  X3 ,  X4 ,  X5 >= 0

 

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Expert Answer

Step 1

Tableau Form

The variables x3, x4 and x5 are having negative coefficients and hence they will get a negative value if they are made the basic variables, leading to infeasible solution. Hence, we cannot have x1, x2 and x3 as the basic variables in the starting tableau (they may enter the basic set in the subsequent steps). So we need to use artificial variables in the three equations to get a starting basic feasible solution and proceed with the Big-M method or 2 phase method for solving the LP. Using the artificial variables A1, A2 and A3 for the three equations and then including them in the objective function with large positive coefficient M (to penalize these artificial variables) we get the following format:

Minimize z = 10x1 + 4x2 +MA1 + MA2 + MA3

subject to

3X1 + 2X2 - X3 + A1          = 60

7X1 + 2X2       - X4 + A2     = 84

3X1 + 6X2           -X5 + A3 = 72

  X1,  X2,  X3 ,  X4 ,  X5,  A1, A2, A3 >= 0

Step 2

Artificial Variables

Hence the simplex tableau form is show below:

But the basic variables A1, A2 and A3 need to have 0 values in the z row as they are part of the starting basic feasible solution. Hence we need to substitute out the variables A1, A2 and A3 in the z-equation.

z = 10 x1 + 4x2 + MA1 + MA2 + MA3

A1 = 60 – 3x1 – 2x2 +x3       (from constraint 1)

A2  = 84 – 7x1 – 2x2 +x4      (from constraint 2)

A3 = 72 – 3x1 – 6x2 + x5      (from constraint 3)

Basic X1
RHS
X2
ХЗ
Х4
X5
A1
A2
АЗ
Z
-M
-M
-10
4
0
0
10
-M
A1
3
2
-1
0
0
1
0
60
A2
7
2
0
-1
1
84
АЗ
3
6
0
0
-1
0
0
1
72
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Image Transcriptionclose

Basic X1 RHS X2 ХЗ Х4 X5 A1 A2 АЗ Z -M -M -10 4 0 0 10 -M A1 3 2 -1 0 0 1 0 60 A2 7 2 0 -1 1 84 АЗ 3 6 0 0 -1 0 0 1 72

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Step 3

Modified z-equation

Therefore using these values of A1, A2 and A3 in z-equation we get

z = 10x1 + 4x2 +M(60-3x1-2x2+x3)+M(84-7x1-2x2+x4)+M(72-3x1-6x2+x5)

= 10x1 +4x2+60M-3Mx1-2Mx2+M...

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