Here is the optimal tableau for a Max LP 2 82 83 rhs 1 10 10...

 

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Here is the optimal tableau for a Max LP 2 82 83 rhs 1 10 10 280 0 0 1 2 -8 24 0 1 0 2 4 8 0 1 0 0 -.5 1.5 2 1. Using B-matrix method, the original rhs vector b had these coordinate values: ; and b3 b1 = F1 0 0 0 x1 X2 5 <-2 2 1.25 2. If we increase the original c2 by 8 and update this (formerly) optimal tableau. #3. 81 0 0 the new row0 value for x2 is : b2- and so this tableau is not optimal using regular simplex, the new optimal values are 2 1 x2 = 3. After increasing the original value of b2 by two more than its allowable increase, we find that we have the following (B-matrix method) updated tableau: and x3- 21 X2 23 81 83 rhs 82 10 10 340 0 5 0 0 0 0 - 2 0 1 2 -8 36 0 0 - 2 1 0 2 -4 20 0 1 5/4 0 0 -1/2 3/2 - 1 we now have to use Dual Simplex (that should have been a question) to get the new optimal tableau with new optimal value for z = Here is the optimal tableau for a Max LP 2 82 83 rhs 1 10 10 280 0 0 1 2 -8 24 0 1 0 2 4 8 0 1 0 0 -.5 1.5 2 1. Using B-matrix method, the original rhs vector b had these coordinate values: ; and b3 b1 = F1 0 0 0 x1 X2 5 <-2 2 1.25 2. If we increase the original c2 by 8 and update this (formerly) optimal tableau. #3. 81 0 0 the new row0 value for x2 is : b2- and so this tableau is not optimal using regular simplex, the new optimal values are 2 1 x2 = 3. After increasing the original value of b2 by two more than its allowable increase, we find that we have the following (B-matrix method) updated tableau: and x3- 21 X2 23 81 83 rhs 82 10 10 340 0 5 0 0 0 0 - 2 0 1 2 -8 36 0 0 - 2 1 0 2 -4 20 0 1 5/4 0 0 -1/2 3/2 - 1 we now have to use Dual Simplex (that should have been a question) to get the new optimal tableau with new optimal value for z =

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To use the Dual Simplex method to solve the new LP problem we start by constructing the dual problem The dual problem is Minimize w 10y1 5y2 280y3Subj... View the full answer