EMGT 501

HW #2Chapter 6 - SELF TEST 21

Chapter 6 - SELF TEST 22

Due Day: Sep 27

s.t.

634Max 321 xxx

20211

3012

1515.01

321

32

321

xxx

xx

xxx

0 , , 321 xxx

Ch. 6 – 21Consider the following linear programming problem:

a. Write the dual problem.b. Solve the dual.c. Use the dual solution to identify the optimal solution to

the original primal problem.d. Verify that the optimal values for the primal and dual

problems are equal.

Ch. 6 – 22A sales representative who sells two products is trying to determine the number of sales calls that should be made during the next month to promote each product. Based on past experience, representatives earn an average $10 commission for every call on product 1 and a $5 commission for every call on product 2. The company requires at least 20 calls per month for each product and not more than 100 calls per month on any one product. In addition, the sales representative spends about 3 hours on each call for product 1 and 1 hour on each call for product 2. If 175 selling hours are available next month, how many calls should be made for each of the two products to maximize the commission?

a. Formulate a linear program for this problem.b. Formulate and solve the dual problem.c. Use the final simplex tableau for the dual problem to

determine the optimal number of calls for the products. What is the maximum commission?

d. Interpret the values of the dual variables.

EMGT 501

HW #2

SolutionsChapter 6 - SELF TEST 21

Chapter 6 - SELF TEST 22

s.t.

u20u30u51Max 321

6u2uu

3uu2u0.5

4uu

321

321

31

0u ,u ,u 321

Ch. 6 – 21a.

0u ,5.0u ,4u

75z*3

*2

*1

*

1u

15

10

0

15

0

2u

30

01

0

30

0

3u

20

11/4

3/4

45/2

5/2

1s

0

1-1/4

-3/4

15/2

15/2

2s

0

01/2

-1/2

15

15

41/2

3/2

1530

0

2s

0

00

1

0

0

BCBasis

1u

2u

3s

75jz

jj cz

b.

c. From the zj values for the surplus variables we see that the optimal primal solution is x1=15/2, x2=15, and x3=0.

d. The optimal value for the dual is shown in part b to equal 75. Substituting x1=15/2 and x2=15 into the primal objective function, we find that it gives the same value.

4(15/2)+3(15)=75

Ch. 6 – 22a.

s.t.

x5x01Max 21

175xx3

100x

100x

20x

20x

21

2

1

2

1

0x ,x 21

s.t.

u175u100u100u20u20-Min 54321

5uuu-

10u3uu-

542

531

0u ,u ,u ,u ,u 54321

b.

The optimal solution to this problem is given by: u1=0, u2=0, u3=0, u4=5/3, and u5=10/3

c. The optimal number of calls is given by the negative of the dual prices for the dual: x1=25 and x2=100.Commission=$750.

d. u4=5/3: $1.67 commission increase for an additional call for product 2.u5=10/3: $3.33 commission increase for an additional hour of selling time per month.