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Here is the optimal tableau for a standard Max problem. zx1 x2 x3 24 81 82 83 rhs 1 0 5 3 0 6 0 1 .3 7.5 0 - .1 .2 0 0 28 360 0 -8 522 0 2700 0 6 12 1 60 0 0 -1/15-3 1 1/15 -1/10 0 2 Using that the dual solution y = CBy B-1 and finding B = (B-¹)-¹ we find the original CBV and rhs b. The allowable increase for b₂ is If b₂ is increased by 3 then, using Dual Theorem, the new value for * is If c₂ is increased by 10, then the new value for optimal > is i.e. if no change to BV, then just a change to profit on selling product 2. The original coefficients c₁ = =☐ a and c4 = 5 If c4 is changed to 512, then (first adjusting other columns of row0 by adding Delta times row belonging to x4 or using B-matrix method to update row0) the new optimal value, after doing more simplex algorithm, for > is

Here is the optimal tableau for a standard Max problem. zx1 x2 x3 24 81 82 83 rhs 1 0 5 3 0 6 0 1 .3 7.5 0 - .1 .2 0 0 28 360 0 -8 522 0 2700 0 6 12 1 60 0 0 -1/15-3 1 1/15 -1/10 0 2 Using that the dual solution y = CBy B-1 and finding B = (B-¹)-¹ we find the original CBV and rhs b. The allowable increase for b₂ is If b₂ is increased by 3 then, using Dual Theorem, the new value for * is If c₂ is increased by 10, then the new value for optimal > is i.e. if no change to BV, then just a change to profit on selling product 2. The original coefficients c₁ = =☐ a and c4 = 5 If c4 is changed to 512, then (first adjusting other columns of row0 by adding Delta times row belonging to x4 or using B-matrix method to update row0) the new optimal value, after doing more simplex algorithm, for > is

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter5: Inverse, Exponential, And Logarithmic Functions
Section5.6: Exponential And Logarithmic Equations
Problem 68E
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Here is the optimal tableau for a standard Max problem. zx1 x2 x3 24 81 82 83 rhs 1 0 5 3 0 6 0 1 .3 7.5 0 - .1 .2 0 0 28 360 0 -8 522 0 2700 0 6 12 1 60 0 0 -1/15-3 1 1/15 -1/10 0 2 Using that the dual solution y = CBy B-1 and finding B = (B-¹)-¹ we find the original CBV and rhs b. The allowable increase for b₂ is If b₂ is increased by 3 then, using Dual Theorem, the new value for * is If c₂ is increased by 10, then the new value for optimal > is i.e. if no change to BV, then just a change to profit on selling product 2. The original coefficients c₁ = =☐ a and c4 = 5 If c4 is changed to 512, then (first adjusting other columns of row0 by adding Delta times row belonging to x4 or using B-matrix method to update row0) the new optimal value, after doing more simplex algorithm, for > is
Transcribed Image Text:Here is the optimal tableau for a standard Max problem. zx1 x2 x3 24 81 82 83 rhs 1 0 5 3 0 6 0 1 .3 7.5 0 - .1 .2 0 0 28 360 0 -8 522 0 2700 0 6 12 1 60 0 0 -1/15-3 1 1/15 -1/10 0 2 Using that the dual solution y = CBy B-1 and finding B = (B-¹)-¹ we find the original CBV and rhs b. The allowable increase for b₂ is If b₂ is increased by 3 then, using Dual Theorem, the new value for * is If c₂ is increased by 10, then the new value for optimal > is i.e. if no change to BV, then just a change to profit on selling product 2. The original coefficients c₁ = =☐ a and c4 = 5 If c4 is changed to 512, then (first adjusting other columns of row0 by adding Delta times row belonging to x4 or using B-matrix method to update row0) the new optimal value, after doing more simplex algorithm, for > is
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