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.

Theory of Simplex Method

For any LP problem with n decision variables, each CPF (Corner Point Feasible) solution lies at the intersection of n constraint boundaries; i.e., the simultaneous solution of a system of n constraint boundary equations.

,53 21 xxZ

1x

0,0

1823

21

21

xx

xx122 2 x4

and

Max

s.t.

A two-variable linear programming problem

0

x

bAx

cx

,

0

0

0

0,,,,,, 2

1

2

1

21

nn

n

b

b

b

b

x

x

x

xcccc

Max

s.t.

mnmm

n

n

aaa

aaa

aaa

A

21

22221

11211

Original Form Augmented Form

Max s.t. bAX

0x

Maxs.t.

cX Z

0,0

0

001

S

S

S

XX

bIXAXZ

XcXZ

Matrix Form

b

X

X

Z

IA

c

S

00

0

1

(1)

(2)

Matrix Form (2) is

Max

s.t.

bIXAX

XcXZ

Z

S

S

00

(3)

orMax

s.t.

bXA

XcZ

Z

ˆˆ

0ˆˆ(4)

where

IAAX

XXcc

S

,ˆˆ0,ˆ

matrix nonbasic a :N

matrix basic a :B

)X (to tscoefficien

objective ingcorrespond theseof vector a :c

)X (to tscoefficien

objective ingcorrespond theseof vector a :c

variablesbasicnon of vector a :X

variablesbasic of vector a :X

N

N

B

B

N

B

Max

s.t.

bNXBX

XcXcZ

Z

NB

NNBB

0

Then, we have

(5)

(6)

where

NBAcccX

XX NB

N

B ,ˆ,ˆˆ

Eq. (6) becomes

bBNXBX NB11

Putting Eq. (7) into (5), we have

0)( 11 NNNB XcNXBbBcZ

(7)

(8)So,

bBcXcNBcXZ BNNBB11 )(0 (9)

become (9) Eq. and (7) Eq. ,0 Currently, NX

bBcZbBX BB11 , (10)

bB

bBc

bB

Bc

X

Z BB

B1

1

1

1 0

0

1

Eq. (10) can be expressed by

(11)

From Eq. (2),

bB

bBc

X

X

Z

BAB

BccABc

bB

bBc

X

X

Z

IA

c

B

Bc

X

Z

B

S

BB

B

S

B

B

1

1

11

11

1

1

1

1

0

1

0

01

0

1

(12)

Thus, initial and later simplex tableau are

IterationBVZOriginalVariables

SlackVariables

RHS

Z1 -c 0 00

BX0 A I b

IterationBVZOriginalVariables

SlackVariables

RHS

Z 1 cABcB 1 1BcB bBcB1

AnyBX 0 AB1 1B bB1

1. Initialization:

Same as for the original simplex method.

2. Iteration:

Step 1

Determine the entering basic variable:

Same as for the Simplex method.

The Overall Procedure

Step 2

Determine the leaving basic variable:

Same as for the original simplex method,

except calculate only the numbers required to

do this [the coefficients of the entering basic

variable in every equation but Eq. (0), and

then, for each strictly positive coefficient, the

right-hand side of that equation].

Step 3

Determine the new BF solution:

Derive and set

3. Optimality test:

Same as for the original simplex method, except

calculate only the numbers required to do this test,

i.e., the coefficients of the nonbasic variables in

Eq. (0).

1B .1bBxB

Fundamental Insight

Z

Z

RHS

Row0

Row1~N

1BcB

1B

bBcB1

bB 1

X

BX

SX

1

0 AB 1

cABcB 1

0

5x4x3x

RightSideBVIteration 3x 4x 5x2x1x

18

12

4

)(,

100

010

001

)(,

2

2

0

3

0

111 bBbBIA

-3 -5 0 0 0 0 1 0 1 0 0 4 0 2 0 1 0 12 3 2 0 0 1 18

Coefficient of:

,0,0,0,5,3,

5

4

3

SB cc

x

x

x

x

BVIteration

1

5x2x3x 1 0 1 0 0 4

0 1 0 0 6

3 0 0 -1 1 6

,

110

00

001

,0,5,0

21 ,

6

6

4

18

12

4

110

00

001

1B bB 1

Bc

21

21

RightSide3x 4x 5x2x1x

Coefficient of:

0 0 0

BVIteration

25

1

5x2x3x 1 0 1 0 0 4

0 1 0 0 6

3 0 0 -1 1 62

1

RightSide3x 4x 5x2x1x

Coefficient of:

-3 0 0 0

0

0

1

0

110

00

001

0,5,0

33

3

0

1

110

00

001

0,5,0

2

1

21

11 cZ 111 caBcB

44 cZ 441 caBcB 2

5

BVIteration

25

1

5x2x3x 1 0 1 0 0 4

0 1 0 0 6

3 0 0 -1 1 62

1

RightSide3x 4x 5x2x1x

Coefficient of:

-3 0 0 0

0

1

0

3

0

1

2

2

0

3

0

1

110

00

001

AB 12

1

BVIteration

25

1

5x2x3x 1 0 1 0 0 4

0 1 0 0 6

3 0 0 -1 1 62

1

RightSide3x 4x 5x2x1x

Coefficient of:

-3 0 0 0 30

30

6

6

4

0,5,01

bBcB

so

BVIteration

25

1

5x2x3x 1 0 1 0 0 4

0 1 0 0 6

3 0 0 -1 1 62

1

RightSide3x 4x 5x2x1x

Coefficient of:

-3 0 0 0 30

minimum

The most negative coefficient

23

6

41

4

4

6

BVIteration

2

1x2x3x 0 0 1 2

0 1 0 0 6

1 0 0 2

21

RightSide3x 4x 5x2x1x

Coefficient of:

0 0 0

,

0

00

1

,3,5,0Bc

21 ,

2

6

2

18

12

4

0

00

1

2

13

13

1

31

31

31

31

31

31

1B bB 1

31

31

31

31

BVIteration

2

1x2x3x 0 0 1 2

0 1 0 0 6

1 0 0 2

21

RightSide3x 4x 5x2x1x

Coefficient of:

0 0 0 1

31

31

31

31

10

1

0

0

0

0

0

1

3,5,0

230

0

1

0

0

0

0

1

0,5,0

44

144 caBccZ B

551

55 caBccZ B

23

21

31

31

31

31

21

31

31

31

31

BVIteration

2

1x2x3x 0 0 1 2

0 1 0 0 6

1 0 0 2

21

RightSide3x 4x 5x2x1x

Coefficient of:

0 0 0 1 36

31

31

31

31

36

2

6

2

3,5,01

bBcB

so

23