Find All Optimal Solutions for a Linear Programming Problem (LP) using the Simplex Method
To find all optimal solutions for the given LP, we need to solve the LP problem using the simplex method.
The initial LP problem is as follows: Maximize z = x1 + x2 Subject to: x1 + x2 + x3 ≤ 1 x1 + 2*x3 ≤ 1 xi ≥ 0
We convert the inequalities into equations by introducing slack variables: x1 + x2 + x3 + s1 = 1 x1 + 2x3 + s2 = 1
The initial tableau is:
Basic Variables | x1 x2 x3 s1 s2 z |
-----------------------------------------------
-z | -1 -1 0 0 0 0 | 0
-----------------|-------------------------------
s1 | 1 1 1 1 0 0 | 1
-----------------|-------------------------------
s2 | 1 0 2 0 1 0 | 1
The pivot column is x2 (the most negative coefficient in the -z row). The pivot row is s1 (the minimum positive ratio of the right-hand side divided by the pivot column coefficient).
Dividing the right-hand side of each row by the pivot column coefficient: s1: 1/(-1) = -1 s2: 1/(-1) = -1
The next tableau is:
Basic Variables | x1 x2 x3 s1 s2 z |
-----------------------------------------------
-z | -1 0 1 1 1 0 | 1
-----------------|-------------------------------
x2 | 1 1 1 1 0 0 | 1
-----------------|-------------------------------
s2 | 0 0 1 1 1 0 | 0
The pivot column is x3 (the most negative coefficient in the -z row). The pivot row is s2 (the minimum positive ratio of the right-hand side divided by the pivot column coefficient).
Dividing the right-hand side of each row by the pivot column coefficient: s2: 1/(-1) = -1
The next tableau is:
Basic Variables | x1 x2 x3 s1 s2 z |
-----------------------------------------------
-z | -1 0 0 2 1 0 | 1
-----------------|-------------------------------
x2 | 1 1 0 1 1 0 | 1
-----------------|-------------------------------
x3 | 0 0 1 1 1 0 | 0
Since there are no negative coefficients in the -z row, we have reached an optimal solution. The optimal solution is: x1 = 0 x2 = 1 x3 = 0 z = 1
Therefore, the only optimal solution for the given LP problem is x1 = 0, x2 = 1, x3 = 0, z = 1.
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