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Explanation of Solution
Optimal solution:
Consider the following linear programing problem:
Subject to the constraints:
Use
Subject to the constraints:
Add slack variables s1,s2 and artificial variable a1 to get:
-(min w′ = -5x1+x2+a1)
Subject to the constraints:
Two Phase Method:
Phase I linear programming problem is,
Subject to the constraints:
The initial simplex table is given below:
w′ | x1 | x2 | a1 | s1 | s2 | rhs | basic variable | |
R0 | 1 | 0 | 0 | -1 | 0 | 0 | 0 | w′=0 |
R1 | 0 | 2 | 1 | 1 | 0 | 0 | 6 | a1=6 |
R2 | 0 | 1 | 1 | 0 | 1 | 0 | 4 | s1=4 |
R3 | 0 | 1 | 2 | 0 | 0 | 1 | 5 | s2=5 |
- Since, the basic variable a1 value in R0 is non-zero, therefore, do the transformations
w′ | x1 | x2 | a1 | s1 | s2 | rhs | basic variable | |
R0 | 1 | 0 | 0 | -1 | 0 | 0 | 0 | w′=0 |
R1 | 0 | 2 | 1 | 1 | 0 | 0 | 6 | a1=6 |
R2 | 0 | 1 | 1 | 0 | 1 | 0 | 4 | s1=4 |
R3 | 0 | 1 | 2 | 0 | 0 | 1 | 5 | s2=5 |
Since the highest positive entry 2 in R0 corresponds to x1, x1 enters the basis.
w′ | x1 | x2 | a1 | s1 | s2 | rhs | ratio | |
R0 | 1 | 2 | 1 | 0 | 0 | 0 | 6 | - |
R1 | 0 | 2 | 1 | 1 | 0 | 0 | 6 | 3* |
R2 | 0 | 1 | 1 | 0 | 1 | 0 | 4 | 4 |
R3 | 1 | 2 | 0 | 0 | 0 | 1 | 5 | 5 |
Apply the simplex method further:
w′ | x1 | x2 | a1 | s1 | s2 | rhs | basic variable | |
R0 | 1 | 0 | 0 | -1 | 0 | 0 | 0 | w′=0 |
R1 | 0 | 1 | 0 | 0 | 6 | x1 = 3 | ||
R2 | 0 | 0 | 1 | 0 | 1 | s1=1 | ||
R3 | 0 | 0 | 0 | 1 | 2 | s2=2 |
- Optimally reached for phase 1. Proceed to phase 2 with the actual objective function
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Operations Research : Applications and Algorithms
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- Operations Research : Applications and AlgorithmsComputer ScienceISBN:9780534380588Author:Wayne L. WinstonPublisher:Brooks Cole