When discussing parametric analysis in Section 3.8, we considered reallocating floor space for a…
When discussing parametric analysis in Section 3.8, we considered reallocating floor space for a trailer-production problem to trade off woodworking capacity for metalworking capacity. The trade-off was governed by the parameter θ in the following tableau:
We found that the following tableau was optimal for values of θ in the range 0 ≤ θ ≤ 4:
The optimal solution at θ = 4 is x1 = 56, x2 = x3 = x4 = x5 = 0, and z = 336. At θ = 5, we found that the optimal basic variables are x1 = 52, x2 = 1.5, and that the optimal objective value is z = 348 − 3(5) = 333. Therefore, as θ increases from 4 to 5,
Δz = 333 − 336 = −3.
The previous optimal tableau tells us that for 0 ≤ θ ≤ 4, woodworking capacity is worth $11/day and metalworking capacity is worth $0.50/day. Increasing θ from 4 to 5 increases woodworking capacity by one day. Using the prices $11/day and $0.50/day, the change in the optimal objective value should be:
Δ z = 11(1) + 0.50(−1) = 10.50.
We have obtained two different values for the change in the optimal objective value, Δ z = −3 and Δ z = 10.50. Reconcile the difference between these values.