EMGT 501

HW #1Chapter 2 - SELF TEST 18

Chapter 2 - SELF TEST 20

Chapter 3 - SELF TEST 28

Chapter 4 - SELF TEST 3

Chapter 5 - SELF TEST 6

Due Day: Sep 13

s.t.

14Max 21 xx

1022

1223

30210

21

21

21

xx

xx

xx

0 , 21 xx

Ch. 2 – 18For the linear program

a. Write this linear program in standard form.b. Find the optimal solution using the graphical solution

procedure.c. What are the values of the three slack variables at the

optimal solution?

Ch. 2 – 20Embassy Motorcycle (EM) manufactures two lightweight motorcycles designed for easy handling and safety. The EZ-Rider model has a new engine and a low profile that make it easy to balance. The Lady-Sport model is slightly larger, uses a more traditional engine, and is specifically designed to appeal to women riders. Embassy produces the engines for both models at its Des Moines, Iowa, plant. Each EZ-Rider engine requires 6 hours of manufacturing time and each Lady-Sport engine requires 3 hours of manufacturing time. The Des Moines plant has 2100 hours of engine manufacturing time available for the next production period. Embassy’s motorcycle frame supplier can supply as many EZ-Rider frames as needed.

However, the Lady-Sport frame is more complex and the supplier can provide only up to 280 Lady-Sport frames for the next production period. Final assembly and testing requires 2 hours for each EZ-Rider model and 2.5 hours for each Lady-Sport model. A maximum of 1000 hours of assembly and testing time are available for the next production period. The company’s accounting department projects a profit contribution of $2400 for each EZ-Rider produced and $1800 for each Lady-Sport produced.

a. Formulate a linear programming model that can be used to determine the number of units of each model that should be produced in order to maximize the total contribution to profit.

b. Find the optimal solution using the graphical solution procedure.

c. Which constraints are binding.

Ch. 3 – 28National Insurance Associates carries an investment portfolio of stocks, bonds, and other investment alternatives. Currently $200,000 of funds are available and must be considered for new investment opportunities. The four stock options National is considering and the relevant financial data are as follows:

Stock A B C D

Price per share $100 $50 $80 $40Annual rate of return 0.12 0.08 0.06 0.10Risk measure per dollar invested 0.10 0.07 0.05 0.08

The risk measure indicates the relative uncertainty associated with the stock in terms of its realizing the projected annual return; higher values indicate greater risk. The risk measures are provided by the firm’s top financial advisor.

National’s top management has stipulated the following investment guidelines: the annual rate of return for the portfolio must be at least 9% and no one stock can account for more than 50% of the total dollar investment.

a. Use linear programming to develop an investment portfolio that minimizes risk.

b. If the firm ignores risk and uses a maximum return-on-investment strategy, what is the investment portfolio?

c. What is the dollar difference between the portfolios in parts (a) and (b)? Why might the company prefer the solution developed in part (a)?

Ch. 4 – 3The employee credit union at State University is planning the allocation of funds for the coming year. The credit union makes four types of loans to its members. In addition, the credit union invests in risk-free securities to stabilize income. The various revenue-producing investments together with annual rates of return are as follows:

Type of Loan/Investment Annual Rate of Return (%) Automobile loans 8 Furniture loans 10 Other secured loans 11 Signature loans 12 Risk-free securities 9

The credit union will have $2,000,000 available for investment during the coming year. State laws and credit union policies impose the following restrictions on the composition of the loans and investments.

• Risk-free securities may not exceed 30% of the total funds available for investment.• Signature loans may not exceed 10% of the funds invested in all loans (automobile, furniture, other secured, and signature loans).• Furniture loans plus other secured loans may not exceed the automobile loans• Other secured loans plus signature loans may not exceed the funds invested in risk-free securities.

How should the $2,000,000 be allocated to each of the loan/investment alternatives to maximize total annual return? What is the projected total annual return?

Ch. 5 – 6

5203

20 1 2 0

25 0 1-1/2

0100

0010

0001

403015

1x 2x 3x 1s 2s 3s

BcBasis

jj zc jz

a. Complete the initial tableau.b. Write the problem in tableau form.c. What is the initial basis? Does this basis correspond to

the origin? Explain.d. What is the value of the objective function at this initial

solution?

e. For the next iteration, which variable should enter the basis, and which variable should leave the basis?f. How many units of the entering variable will be in the next solution? Before making this first iteration, what do you think will be the value of the objective function after the first iteration?g. Find the optimal solution using the simplex method.

EMGT 501

HW #1 SolutionsChapter 2 - SELF TEST 18

Chapter 2 - SELF TEST 20

Chapter 3 - SELF TEST 28

Chapter 4 - SELF TEST 3

Chapter 5 - SELF TEST 6

Ch. 2 – 18(a)

Max 4x1 + 1x2 + 0s1 + 0s2 + 0s3

s.t.

10x1 + 2x2 + 1s1 = 30

3x1 + 2x2 + 1s2 = 12

2x1 + 2x2 + 1s3 = 10

x1, x2, s1, s2, s3 0

Ch. 2 – 18(b)

x2

x10 2 4 6 8 10

2

4

6

8

10

12

14

Optimal Solution

x1 = 18/7, x2 = 15/7, Value = 87/7

(c)s1 = 0, s2 = 0, s3 = 4/7

Ch. 2 – 20(a)

Let E = number of units of the EZ-Rider produced L = number of units of the Lady-Sport produced

Max 2400E + 1800L

s.t.

6E + 3L 2100 Engine time

L 280 Lady-Sport maximum

2E + 2.5L 1000 Assembly and testing

E, L 0

Ch. 2 – 20(b)

0

L

Profit = $960,000

Optimal Solution

100

200

300

400

500

600

700

100 200 300 400 500E

Engine Manufacturing Time

Frames for Lady-Sport

Assembly and Testing

E = 250, L = 200

Number of Lady-Sport Produced

Num

ber

of E

Z-R

ider

Pro

duce

d

Ch. 2 – 20(c)

The binding constraints are the manufacturing time and the assembly and testing time.

Ch. 3 – 28(a)

Let A = number of shares of stock AB = number of shares of stock BC = number of shares of stock CD = number of shares of stock D

To get data on a per share basis multiply price by rate of return or risk measure value.

Min 10A + 3.5B + 4C + 3.2D

s.t.

100A + 50B + 80C + 40D = 200,000

12A + 4B + 4.8C + 4D 18,000 (9% of 200,00)

100A 100,000

50B 100,000

80C 100,000

40D 100,000

A, B, C, D 0

Solution: A = 333.3, B = 0, C = 833.3, D = 2500Risk: 14,666.7Return: 18,000 (9%) from constraint 2

Ch. 3 – 28(b) Max 12A + 4B + 4.8C +

4D

s.t.

100A + 50B + 80C +

40D

= 200,000

100A 100,000

50B 100,000

80C 100,000

40D

100,000

A, B, C, D 0

Solution: A = 1000, B = 0, C = 0, D = 2500Risk: 10A + 3.5B + 4C + 3.2D = 18,000Return: 22,000 (11%)

Ch. 3 – 28(c)

The return in part (b) is $4,000 or 2% greater, but the risk index has increased by 3,333.

Obtaining a reasonable return with a lower risk is a preferred strategy in many financial firms. The more speculative, higher return investments are not always preferred because of their associated higher risk.

Ch. 4 – 3 x1 = $ automobile loansx2 = $ furniture loansx3 = $ other secured loansx4 = $ signature loansx5 = $ "risk free" securities

Max 0.08x1 + 0.10x2 + 0.11x3 + 0.12x4 + 0.09x5

s.t.

x5 600,000 [1]

x4 0.10(x1 + x2 + x3 + x4)

or -0.10x1 - 0.10x2 - 0.10x3 + 0.90x4 0 [2]

x2 + x3 x1

or - x1 + x2 + x3 0 [3]

x3 + x4 x5

or + x3 + x4 - x5 0 [4]

x1 + x2 + x3 + x4 + x5 = 2,000,000 [5]

x1, x2, x3, x4, x5 0

Automobile Loans (x1) = $630,000

Furniture Loans (x2) = $170,000

Other Secured Loans (x3) = $460,000

Signature Loans (x4) = $140,000

Risk Free Loans (x5) = $600,000

Solution

Annual Return $188,800 (9.44%)

Ch. 5 – 6(a)

x 3

25

0

1

-1/2

-25

s 1

0

1

0

0

0

x 1

5

2

0

3

-5

x 2

1

2

20

0

-20

Basis

s 1

s 2

s 3

c B

0

0

0

Z -c j + z j

s 2

0

0

1

0

0

s 3

0

0

0

1

0

40

30

15

0

Ch. 5 – 6(b)

Max 5x1 + 20x2 + 25x3 + 0s1 + 0s2 + 0s3

s.t.

2x1 + 1x2 + 1s1 = 40

2x2 + 1x3 + 1s2 = 30

3x1 -

1/2x3

+ 1s3 = 15

x1, x2, x3, s1, s2, s3, 0.

Ch. 5 – 6(c) The original basis consists of s1, s2, and s3. It is the

origin since the nonbasic variables are x1, x2, and x3 and are all zero.

(d) 0

(e)x3 enters because it has the largest negative zj - cj and s2 will leave because row 2 has the only positive coefficient.

(f) 30; objective function value is 30 times 25 or 750.

(g) Optimal Solution: x1 = 10 s1 = 20x2 = 0 s2 = 0x3 = 30 s3 = 0z = 800.