Here is the optimal tableau for a Max LP 2 23 21 0 22 5 S2 10 1 S1 0 1 0 rhs 280 24 0 2 0 2 S3 10 8 - 4 1.5 0 0 0 1 2 8 - 2 1.25 0 0 0 1 0 .5 2 1. Using B-matrix method, the original rhs vector b had these coordinate values: b1 = ; b2 = ; and b3 = 2. If we increase the original c2 by 8 and update this (formerly) optimal tableau, the new rowo value for X2 is and so this tableau is not optimal using regular simplex, the new optimal values (enter as a/b rationals) are x1 = ; x2 = ; and x3 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: z S3 21 0 0 X2 5 23 0 $1 0 1 S2 10 1 0 10 rhs 340 36 2 0 2 -- 8 0 0 0 1 2 5/4 1 0 2 4 0 0 -1/2 3/2 20 - 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 =