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Solve this LP max z = 2x12x2 + 3x3 s.t. 2x1 + x2 + x3 x1 - x2 + x3 1, 2, 3 urs ≤1 ≤0 Set up the initial simplex tableau and then enter the value of 3 in the optimal solution. The value of 3 in optimal solution is z x1 x2 x3 x3" s1 s2 rhs watch out that your second tableau does not have negatives in rhs.

Solve this LP max z = 2x12x2 + 3x3 s.t. 2x1 + x2 + x3 x1 - x2 + x3 1, 2, 3 urs ≤1 ≤0 Set up the initial simplex tableau and then enter the value of 3 in the optimal solution. The value of 3 in optimal solution is z x1 x2 x3 x3" s1 s2 rhs watch out that your second tableau does not have negatives in rhs.

Practical Management Science
Practical Management Science
6th Edition
ISBN:9781337406659
Author:WINSTON, Wayne L.
Publisher:WINSTON, Wayne L.
Chapter4: Linear Programming Models
Section: Chapter Questions
Problem 75P
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Solve this LP max z = 2x1 − 2x2 + 3x3 s.t. 2x1 + x2 + x3 X1 X₂ X3 1, 2 > 0, 3 urs <1 ≤0 Set up the initial simplex tableau and then enter the value of 3 in the optimal solution. The value of 3 in optimal solution is Z watch out that your second tableau does not have negatives in rhs. x1 x2 x3 x3" s1 s2 00000 rhs
Transcribed Image Text:Solve this LP max z = 2x1 − 2x2 + 3x3 s.t. 2x1 + x2 + x3 X1 X₂ X3 1, 2 > 0, 3 urs <1 ≤0 Set up the initial simplex tableau and then enter the value of 3 in the optimal solution. The value of 3 in optimal solution is Z watch out that your second tableau does not have negatives in rhs. x1 x2 x3 x3" s1 s2 00000 rhs
Expert Solution
Step 1

To set up the initial simplex tableau, we first convert the maximization problem into a minimization problem by multiplying both sides of the objective function by -1:

minimize:

-2x1+2x2-3x3

subject to:

2x1+2x2+3x3 <= 1

 

x1 <= 0  x2 <= 0  x3 <= 0

 

We introduce slack variables s1 and s2 to convert the inequality constraint into an equality constraint:

2x1 + 2x2 + 3x3 + s1 = 1 x1 + s2 = 0 x2 = 0 x3 = 0

The initial simplex tableau is:

 

z  x2  x2 x3 s1 s2 RHS
1 2 2 3 1 0 1 
0 1 0 0 1 0 0
0 0 1 0 0 1 0
0 0 0 1 0 0 0
-21 22 0 -3 0 0 0

 

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