Module 4 Homework
b. The same extreme point, X = 3 and Y = 2, remains optimal. The value of the objective function becomes 6(3) + 12(2) = 42. The vertices of the feasible region are (0, 9), (2, 3), (3,2), and (9, 0). You really should confirm these algebraically by finding where the lines intersect. Then you need to evaluate the objective function at each of these points, and choose the point that gives the minimum value for the objective function.
c. A new extreme point, X = 2 and Y = 3, becomes optimal. The value of the objective function becomes 8(2) + 6(3) = 34. You use the same points for the alternative objective functions. It's the same feasibility region for all three objective functions, but they might be minimized at different vertices.
d. The objective coefficient range for variable X is 4 to 12. Since the change in part (b) is within this range, we know that the optimal solution, X = 3 and Y = 2, will not change. The objective coefficient range for variable Y is 8 to 24. Since the change in part (c) is outside this range, we have to re-solve the problem to find the new optimal solution. You will find the minimum to be x = 3 y = 2 minimum is 8(3) + 12(2) = 48 changing the x coefficient to 6 and graphing the new equation does not change the minimum location x = 3 y = 2 but does change the value 6(3) + 12(2) = 42.
7. a. U = 800
H = 1200
Estimated Annual Return= $8400 is the return (note their value and reduced
cost of 0).
b. Constraints 1 and 2. All funds available are being utilized and the maximum permissible risk is being incurred. There is no slack on constraint 1 and 2, so these are the binding constraints. This means that if the value of the constraint is changed, the solution changes.
The binding constraints are
25U + 50H ? 80,000 Funds available
0.50U + 0.25H ? 700 Risk maximum
It is easy to verify that these are equalities at U=800 H=1200