Question

(3) Kyle gave Cartman the tableau of a max-LP to solve using the simplex method. The LP had two variables (x1, 22) and two < constraints. Cartman decided to play a joke on Kyle, so he (a) changed the coefficient of zi in the first constraint from 6 to 9, and (b) solved the LP as a min-LP With these modifications, Cartman got the following"optimal" tableau after performing a single pivot. Find the correct optimal tableau that Kyle should have obtained.

Answer 1

1617181 911011 12113I A two variable max problem with S looks like Changed problem is で2x2 1242 2212-2 2212-2 x1,X2 2 0 x1,X2 2 0 a) The coefficient of x1 changed from 6 to 9, Thus original problem is b) Then solved the problem as min, thus Min c1x1 + C2X2 1242 2212-2 2 x1,X2 2 0 x1,X2 2 0 9 a1:2 21 a22 : b 2 0

9 a12 21 a22 : b 2 0 Taking 9 as pivot, divide the first row by 9, to get 1 a12/9b1/9 a22' a21 0 Then to make a21 zero, apply R2 a21R1, and to make c1 zero, apply R3 c1R1, which gives a12/9 a22 - a21a12/9i b2 - a21b1/9 0 This is the tableau got by Cartman Actually, it should have been 412b 21 a22 0

Taking 6 as pivot, divide the first row by 6, to get 1 a12/6 br/6] 21 a22 C1 C2 2 0 Then to make a21 zero, apply R2 - a21R1, and to make c1 zero, apply R3 + c1R1, which gives a12/6 a22 a21012/6 b116 [1