Dual Simplex Method Example Calculation
2
Question Maximize -3x 1 - 2x 2 Subject to -3x 1 - x 2 ≤-3 -4x 1 - 3x 2 ≤-6 x 1 + x 2 ≤ 3 Solution Starting Table C j -3 -2 0 0 0 x B C B A 1 A 2 A 3 A 4 A 5 Solution x 3 0 -3 -1 1 0 0 -3 x 4 0 -4 -3 0 1 0 -6 x 5 0 1 1 0 0 1 3 Z j 0 0 0 0 0 0 Z j - C j 3 2 0 0 0 Ratio -3/4 -2/3 0/0 0 0/0 x 4 leaves the basis and x 2 enters the basis Iteration 1 C j -3 -2 0 0 0 x B C B A 1 A 2 A 3 A 4 A 5 Solution x 3 0 -5/3 0 1 -1/3 0 -1 x 2 -2 4/3 1 0 -1/3 0 2 x 5 0 -1/3 0 0 1/3 1 1 Z j -8/3 -2 0 2/3 0 -4 Z j - C j 1/3 0 0 2/3 0 Ratio -1/5 0/0 0 -2 0/0 x 3 leaves the basis and x 1 enters the basis Iteration 2
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Example on Dual Simplex Method. Simplex Tableau included.
Transcript of Dual Simplex Method Example Calculation
QuestionMaximize −3x1 − 2x2
Subject to
−3x1 − x2 ≤ −3
−4x1 − 3x2 ≤ −6
x1 + x2 ≤ 3
Solution
Starting Table
Cj -3 -2 0 0 0xB CB A1 A2 A3 A4 A5 Solutionx3 0 -3 -1 1 0 0 -3x4 0 -4 -3 0 1 0 -6x5 0 1 1 0 0 1 3
Zj 0 0 0 0 0 0Zj − Cj 3 2 0 0 0Ratio -3/4 -2/3 0/0 0 0/0
x4 leaves the basis and x2 enters the basis
Iteration 1
Cj -3 -2 0 0 0xB CB A1 A2 A3 A4 A5 Solutionx3 0 -5/3 0 1 -1/3 0 -1x2 -2 4/3 1 0 -1/3 0 2x5 0 -1/3 0 0 1/3 1 1
Zj -8/3 -2 0 2/3 0 -4Zj − Cj 1/3 0 0 2/3 0Ratio -1/5 0/0 0 -2 0/0
x3 leaves the basis and x1 enters the basis
Iteration 2
Cj -3 -2 0 0 0xB CB A1 A2 A3 A4 A5 Solutionx1 -3 1 0 -3/5 1/5 0 3/5x2 -2 0 1 4/5 -3/5 0 6/5x5 0 0 0 -1/5 2/5 1 6/5
Zj -3 -2 1/5 3/5 0 -21/5Zj − Cj 0 0 1/5 3/5 0Ratio 0 0/0 -1/3 3 0/0
Optimal Solution Reached
The optimal Solution isx1 = 3/5x2 = 6/5
The optimal value isZmax = −21/5
