Access Millions of academic & study documents

Homework solutions for Management Science

Content type
User Generated
Subject
Business
Type
Homework
Showing Page:
1/3
Gabriel Erosa OMG 423 2/16/14
Module 4 Homework
3. a.
Y
Optimal Solution
2
4
6
8
10
2 4 6 8
10
X
X = 3, Y = 2
X = 2, Y = 3
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.

Sign up to view the full document!

lock_open Sign Up
Showing Page:
2/3
Gabriel Erosa OMG 423 2/16/14
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
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

Sign up to view the full document!

lock_open Sign Up
Showing Page:
3/3

Sign up to view the full document!

lock_open Sign Up
Unformatted Attachment Preview
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 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 ...
Purchase document to see full attachment
User generated content is uploaded by users for the purposes of learning and should be used following Studypool's honor code & terms of service.
Studypool
4.7
Indeed
4.5
Sitejabber
4.4