Question

original right-hand Table Q2 shows the final optimal maximization simplex tableau. The sides were 100 and 90 for the two constraints. Table Q2: final optimal maximization simplex tableau 0 0.12 8.4 16 ? 99.2 04 0.48 0.20 4.24 -.24 X3 -0.20 0.40 X1 G-2 i. ii. Replace the (?) sign with the correct value. What would the new solution be if there had been 150 units available in the first constraint? ii. What would the new solution be if there had been 70 units available in the second constraint?

Answer 1

i) From the given values, the completed tableau is following:

x1 CB 4 0 0 Solution Basis 0 0.48 0.12 -0.04 8.4 0.4 0.56 0.48 99.2 x1 0 16 Zj Cj - Zj 2 4.24 0.24 0 0 0.56 -0.48

ii)

Allowable increase for constraint 1 RH Side = -(16/-0.2) = 80

Allowable decrease for constraint 1 RH Side = 8.4/0.12 = 70

So, upper limit for constraint 1 RH side is 100+80 = 180

Hence, if there were 150 units available in the first constraint, then it is within the feasibility range and the dual price is applicable.

The new optimal value would increase by = shadow price * units increased = 0.56*(150-100) = 28

iii)

Allowable increase for constraint 2 RH Side = -(8.4/-0.04) = 210

Allowable decrease for constraint 2 RH Side = 16/0.4 = 40

So, lower limit for constraint 2 RH side is 90-40 = 50

Hence, if there were 90 units available in the second constraint, then it is within the feasibility range and the dual price is applicable.

The new optimal value would decrease by = shadow price * units decreased = 0.48*(90-70) = 9.6