528 words - 3 pages

ï»¿Module 4 Homework

3. a.

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 ...view middle of the document...

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

c.

Constraint

Dual Values

Funds Avail.

0.09

Risk Max

1.33

U.S. Oil Max

0

The dual values are 0.093333 (clearly, really 7/75) for constraint 1 and 1.33333 (really, 4/3) for constraint 2. This means that a change of 1 in the constraints yields this change in the objective function. Note how 7/75*80000+4/3*700 = 8400

d. No, the optimal solution does not call for investing the maximum amount in U.S. Oil. It would not be beneficial, as we are at the optimum solution. From 0.50U + 0.25H ? 700, an increase of 1 in U will mean that H will have to decrease by 2 (.5*1+.25*-2=0)

Then, the objective function will change by 3U + 5H = 3(1) +5(-2) = 3 - 10 = -7; this is a decrease of 7.

Note that there is a typo above in constraint 2 (I corrected it above).

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