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Dennis L. Bricker Dept. of Industrial Engineering The University-(Answered)

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Question

Hi, I would like assistance with the problem on page 10 in the attached file (Homework #5 Problem 1). Thank you!


Dennis L. Bricker

 

Dept. of Industrial Engineering

 

The University of Iowa

 


 

56:171 Operations Research

 

Homework #1 - Due Wednesday, August 30, 2000

 

In each case below, you must formulate a linear programming model that will solve the problem.

 

Be sure to define the meaning of your variables! Then use LINDO (or other appropriate

 

software) to find the optimal solution. State the optimal objective value, and describe in

 

?layman?s terms? the optimal decisions.

 

1. Walnut Orchard has two farms that grow wheat and corn. Because of differing soil

 

conditions, there are differences in the yields and costs of growing crops on the two farms.

 

The yields and costs are shown in the table below. Farm #1 has 100 acres available for

 

cultivation, while Farm #2 has 150 acres. The farm has contracted to grow 11,000 bushels of

 

corn and 6000 bushels of wheat. Determine a planting plan that will minimize the cost of

 

meeting these contracts.

 

Farm #1

 

Farm #2

 

Corn yield/acre

 

100 bushels

 

120 bushels

 

Cost/acre of corn

 

$90

 

$115

 

Wheat yield/acre

 

40 bushels

 

35 bushels

 

Cost/acre of wheat

 

$90

 

$80

 

Note: We are assuming that the costs and yields are known with certainty, which is not the

 

case in the "real world"!

 

2. A firm manufactures chicken feed by mixing three different ingredients. Each ingredient

 

contains four key nutrients: protein, fat, vitamin A, and vitamin B. The amount of each

 

nutrient contained in 1 kilogram of the three basic ingredients is summarized in the table

 

below:

 

Ingredient Protein (grams) Fat (grams) Vitamin A (units) Vitamin B (units)

 

1

 

25

 

11

 

235

 

12

 

2

 

45

 

10

 

160

 

6

 

3

 

32

 

7

 

190

 

10

 

The costs per kg of Ingredients 1, 2, and 3 are $0.55, $0.42, and $0.38, respectively. Each kg

 

of the feed must contain at least 35 grams of protein, a minimum of 8 grams (and a maximum

 

of 10 grams) of fat, at least 200 units of vitamin A and at least 10 units of vitamin B.

 

Formulate an LP model for finding the feed mix that has the minimum cost per kg.

 

--revised 8/28/00

 

3. ?Mama?s Kitchen? serves from 5:30 a.m. each morning until 1:30 p.m. in the afternoon.

 

Tables are set and cleared by busers working 4-hour shifts beginning on the hour from 5:00

 

a.m. through 10:00 a.m. Most are college students who hate to get up in the morning, so

 

Mama?s pays $9 per hour for the 5:00, 6:00, and 7:00 a.m. shifts, and $7.50 per hour for the

 

others. (That is, a person works a shift consisting of 4 consecutive hours, with the wages

 

equal to 4x$9 for the three early shifts, and 4x$7.50 for the 3 later shifts.) The manager

 

seeks a minimum cost staffing plan that will have at least the number of busers on duty each

 

hour as specified below:

 

5 am 6 am 7 am 8 am 9 am 10am 11am Noon 1 pm

 

#reqd

 

2

 

3

 

5

 

5

 

3

 

2

 

4

 

6

 

3

 


 

56:171 O.R. -- HW #1

 


 

08/28/00

 


 

page 1

 


 

56:171 Operations Research

 

Homework #2 -- Due Wednesday, Sept. 6

 

The Diet Problem. "The goal of the diet problem is to find the cheapest combination of foods

 

that will satisfy all the daily nutritional requirements of a person." Go to the URL:

 

http://www-fp.mcs.anl.gov/otc/Guide/CaseStudies/diet/index.html

 

and click on "Give it a try." Then on the next page select "Edit the constraints" and click on

 

"Go on" .

 

a. What are the restrictions on calories in the default set of requirements?

 

Go back to the previous page, where approximately 100 foods are listed for your selection. Choose "Default

 

requirements", and select 15 foods which you think would provide an economical menu meeting the requirements.

 

Then click on "Go on" again.

 

b. What is the minimum-cost menu meeting the nutritional requirements using the foods you indicated?

 

Indicate the solution in the left 2 columns of the table below.

 

Change the default upper limit on calories to 1500/day and solve the problem again. (Be sure that the lower bound ?

 

upper bound!)

 

c. What is the minimum-cost menu meeting the nutritional requirements using the foods you indicated?

 

Indicate the solution in the right 2 columns of the table below.

 

Quantity

 

(# servings)

 


 

Total Cost:

 


 

56:171 O.R. HW#2

 


 

Cost

 


 

$

 


 

Food

 

(& serving size)

 

1.

 

2.

 

3.

 

4.

 

5.

 

6.

 

7.

 

8.

 

9.

 

10.

 

11.

 

12.

 

13.

 

14.

 

15.

 

<><><><><><><>

 


 

Quantity

 

(# servings)

 


 

Cost

 


 

Total Cost:

 


 

$

 


 

page 1 of 1

 


 

56:171 Operations Research

 

Homework #3 -- Due Wednesday, Sept. 13

 

1. Simplex Algorithm: Use the simplex algorithm to find the optimal solution to the following LP:

 


 

Maximize z = 4 x1 + x2

 


 

2 x1 + x2 ? 9

 

x ? 5

 

2

 

subject to

 

x1 ? x 2 ? 4

 

x1 ? 0 , x 2 ? 0

 

Show the initial tableau, each intermediate tableau, and the final tableau. Explain how you have decided on the

 

location of each pivot and how you have decided to stop at the final tableau.

 

2. Below are several simplex tableaus. Assume that the objective in each case is to be minimized. Classify each

 

tableau by writing to the right of the tableau a letter A through G, according to the descriptions below. Also answer

 

the question accompanying each classification, if any.

 

(A) Nonoptimal, nondegenerate tableau with bounded solution. Circle a pivot element which would

 

improve the objective.

 

(B) Nonoptimal, degenerate tableau with bounded solution. Circle an appropriate pivot element. Would

 

the objective improve with this pivot?

 

(C) Unique nondegenerate optimum.

 

(D) Optimal tableau, with alternate optimum. State the values of the basic variables. Circle a pivot

 

element which would lead to another optimal basic solution. Which variable will enter the basis, and

 

at what value?

 

(E) Objective unbounded (below). Specify a variable which, when going to infinity, will make the

 

objective arbitrarily low.

 

(F) Tableau with infeasible basic solution.

 

Warning: Some of these classifications might be used for more than one tableau, while others might not

 

be used at all!

 

(i)

 


 

-z

 

1

 

0

 

0

 

0

 


 

(ii) -z

 

1

 

0

 

0

 

0

 

(iii)-z

 

1

 

0

 

0

 

0

 


 

X1

 

-3

 

0

 

-6

 

4

 


 

X2

 

0

 

0

 

0

 

1

 


 

X3

 

1

 

-4

 

3

 

2

 


 

X4

 

1

 

0

 

-2

 

-5

 


 

X5

 

0

 

0

 

1

 

0

 


 

X6

 

0

 

1

 

0

 

0

 


 

X7

 

2

 

0

 

2

 

1

 


 

X8

 

3

 

0

 

3

 

1

 


 

RHS

 


 

X1

 

3

 

0

 

-4

 

-6

 


 

X2

 

0

 

0

 

1

 

0

 


 

X3

 

-1

 

-4

 

2

 

3

 


 

X4

 

3

 

0

 

-5

 

-2

 


 

X5

 

0

 

0

 

0

 

1

 


 

X6

 

0

 

1

 

0

 

0

 


 

X7

 

2

 

3

 

-2

 

-4

 


 

X8

 

-2

 

0

 

1

 

3

 


 

RHS

 


 

X1

 

3

 

0

 

4

 

-6

 


 

X2

 

0

 

0

 

1

 

0

 


 

X3

 

1

 

-4

 

2

 

3

 


 

X4

 

1

 

0

 

-5

 

-2

 


 

X5

 

0

 

0

 

0

 

1

 


 

X6

 

0

 

1

 

0

 

0

 


 

X7

 

3

 

3

 

2

 

-4

 


 

X8

 

5

 

0

 

1

 

3

 


 

RHS

 


 

56:171 O.R. HW#3

 


 

-45

 

9

 

5

 

8

 


 

-45

 

9

 

0

 

5

 


 

-45

 

3

 

7

 

15

 


 

______

 


 

________

 


 

________

 


 

page 1 of 2

 


 

(iv) -z

 

1

 

0

 

0

 

0

 

(v)

 


 

-z

 

1

 

0

 

0

 

0

 


 

(vi) -z

 

1

 

0

 

0

 

0

 

(vii)-z

 

1

 

0

 

0

 

0

 

(viii)-z

 

1

 

0

 

0

 

0

 

(ix) -z

 

1

 

0

 

0

 

0

 


 

X1

 

3

 

0

 

4

 

-6

 


 

X2

 

0

 

0

 

1

 

0

 


 

X3

 

1

 

-1

 

-4

 

3

 


 

X4

 

-3

 

0

 

-5

 

-2

 


 

X5

 

0

 

0

 

0

 

1

 


 

X6

 

0

 

1

 

0

 

0

 


 

X7

 

2

 

3

 

2

 

-4

 


 

X8

 

0

 

0

 

1

 

3

 


 

RHS

 


 

X1

 

3

 

0

 

4

 

-6

 


 

X2

 

0

 

0

 

1

 

0

 


 

X3

 

0

 

-4

 

2

 

3

 


 

X4

 

1

 

0

 

-5

 

-2

 


 

X5

 

0

 

0

 

0

 

1

 


 

X6

 

0

 

1

 

0

 

0

 


 

X7

 

0

 

3

 

2

 

-4

 


 

X8

 

12

 

0

 

1

 

3

 


 

RHS

 


 

X1

 

3

 

0

 

-6

 

4

 


 

X2

 

0

 

0

 

0

 

1

 


 

X3

 

1

 

-4

 

3

 

2

 


 

X4

 

3

 

0

 

-2

 

-5

 


 

X5

 

0

 

0

 

1

 

0

 


 

X6

 

0

 

1

 

0

 

0

 


 

X7

 

2

 

3

 

-4

 

2

 


 

X8

 

0

 

0

 

3

 

1

 


 

RHS

 


 

X1

 

3

 

4

 

-6

 

0

 


 

X2

 

0

 

1

 

0

 

0

 


 

X3

 

1

 

2

 

3

 

-4

 


 

X4

 

1

 

-5

 

2

 

0

 


 

X5

 

0

 

0

 

1

 

0

 


 

X6

 

0

 

0

 

0

 

1

 


 

X7

 

-2

 

2

 

-4

 

3

 


 

X8

 

0

 

1

 

3

 

0

 


 

X1

 

2

 

0

 

6

 

4

 


 

X2

 

0

 

0

 

0

 

1

 


 

X3

 

-1

 

-4

 

3

 

2

 


 

X4

 

3

 

0

 

-2

 

-5

 


 

X5

 

0

 

0

 

1

 

0

 


 

X6

 

0

 

1

 

0

 

0

 


 

X7

 

2

 

3

 

-4

 

2

 


 

X8

 

0

 

0

 

3

 

1

 


 

RHS

 


 

X1

 

3

 

0

 

4

 

-6

 


 

X2

 

0

 

0

 

1

 

0

 


 

X3

 

1

 

-4

 

2

 

3

 


 

X4

 

4

 

0

 

-5

 

-2

 


 

X5

 

0

 

0

 

0

 

1

 


 

X6

 

0

 

1

 

0

 

0

 


 

X7

 

-2

 

-3

 

2

 

-4

 


 

X8

 

2

 

0

 

1

 

3

 


 

RHS

 


 

-45

 

9

 

3

 

5

 


 

-45

 

9

 

8

 

5

 


 

-45

 

9

 

5

 

8

 


 

________

 


 

______

 


 

________

 


 

RHS

 

-45

 

5

 

0

 

9

 


 

-45

 

9

 

5

 

8

 


 

-45

 

3

 

-8

 

15

 


 

________

 


 

________

 


 

________

 


 

3. LP Model Formulation (from Operations Research, by W. Winston (3rd edition), page 191): Carco uses robots

 

to manufacture cars. The following demands for cars must be met (not necessarily on time, but all demands

 

must be met by end of quarter 4):

 

Quarter #

 

1

 

2

 

3

 

4

 

Demand

 

600

 

800

 

500

 

400

 

At the beginning of the first quarter, Carco has two robots. Robots can be purchased at the beginning of each

 

quarter, but a maximum of two per quarter can be purchased. Each robot can build up to 200 cars per quarter.

 

It costs $5000 to purchase a robot. Each quarter, a robot incurs $500 in maintenance costs (even if it is not

 

being used to build any cars). Robots can also be sold at the beginning of each quarter for $3000. At the end of

 

each quarter, a holding cost of $200 for each car in inventory is incurred. If any demand is backlogged, a cost

 

of $300 per car is incurred for each quarter the customer must wait. At the end of quarter 4, Carco must have at

 

least two robots.

 

a. Formulate an LP to minimize the total cost incurred in meeting the next four quarters' demands for cars. Be

 

sure to define your variables (including units) clearly! (Ignore any integer restrictions.)

 

b. Use LINDO (or other LP solver) to find the optimal solution and describe it briefly in "plain English". Are

 

integer numbers of robots bought & sold?

 


 

56:171 O.R. HW#3

 


 

page 2 of 2

 


 

56:171 Operations Research

 

Homework #4 -- Due Wednesday, Sept. 20

 

1. LP Duality: Write the dual of the following LP:

 


 

Min 3 x1 + 2 x2 ? 4 x3

 

5 x1 ? 7 x2 + x3 ?12

 


 

x1 ? x2 + 2 x3 = 18

 


 

subject to 2 x1 ? x3 ? 6

 

x + 2 x ?10

 

3

 

2

 

x j ? 0, j=1,2,3

 

2. Consider the following primal LP problem:

 


 

Max x1 + 2 x2 ? 9 x3 + 8 x4 ? 36 x5

 

2 x2 ? x3 + x4 ? 3x5 ? 40

 


 

subject to x1 ? x2 + 2 x4 ? 2 x5 ? 10

 

x ? 0, j=1,2,3,4,5

 

j

 

a. Write the dual LP problem

 

b. Sketch the feasible region of the dual LP in 2 dimensions, and use it to find the optimal solution.

 

c. Using complementary slackness conditions,

 

? write equations which must be satisfied by the optimal primal solution x*

 

? which primal variables must be zero?

 

d. Using the information in (c.), determine the optimal solution x*.

 

3. Sensitivity Analysis (based on LP model Homework #3 from Operations Research, by W. Winston (3rd edition),

 

page 191): Carco uses robots to manufacture cars. The following demands for cars must be met (not

 

necessarily on time, but all demands must be met by end of quarter 4):

 

Quarter #

 

1

 

2

 

3

 

4

 

Demand

 

600

 

800

 

500

 

400

 

At the beginning of the first quarter, Carco has two robots. Robots can be purchased at the beginning of each

 

quarter, but a maximum of two per quarter can be purchased. Each robot can build up to 200 cars per quarter.

 

It costs $5000 to purchase a robot. Each quarter, a robot incurs $500 in maintenance costs (even if it is not

 

being used to build any cars). Robots can also be sold at the beginning of each quarter for $3000. At the end of

 

each quarter, a holding cost of $200 for each car in inventory is incurred. If any demand is backlogged, a cost

 

of $300 per car is incurred for each quarter the customer must wait. At the end of quarter 4, Carco must have at

 

least two robots.

 

Decision Variables :

 

Rt : robots available during quarter t (after robots are bought or sold for the quarter)

 

Bt : robots bought during quarter t

 

St : robots sold during quarter t

 

It : cars in inventory at end of quarter t

 

Ct : cars produced during quarter t

 

Dt : backlogged demand for cars at end of quarter t

 

Using the LINDO output below, answer the following questions:

 

a. During the first quarter, a one-time offer of 20% discount on robots is offered. Will this change the

 

optimal solution shown below?

 

b. In the optimal solution, is any demand backlogged?

 


 

56:171 O.R. HW#4

 


 

Fall 2000

 


 

page 1 of 4

 


 

c. Suppose that the penalty for backlogging demand is $250 per month instead of $300. Will this change

 

the optimal solution? Note: this change applies to all quarters simultaneously!

 

d. If the demand in quarter #3 were to increase by 100 cars, what would be the change in the objective

 

function?

 

e. Suppose that we know in advance that demand for 10 cars must be backlogged in quarter #2. Using the

 

substitution rates found in the tableau, describe how this would change the optimal solution.

 

MIN

 


 

500 R1 + 500 R2 + 500 R3 + 500 R4 + 200 I1 + 200 I2 + 200 I3

 

+ 200 I4 + 5000 B1 + 5000 B2 + 5000 B3 + 5000 B4 - 3000 S1 - 3000 S2

 

- 3000 S3 - 3000 S4 + 300 D1 + 300 D2 + 300 D3 + 300 D4

 

SUBJECT TO

 

2)

 

R1 - B1 + S1 =

 

2

 

3) - R1 + R2 - B2 + S2 =

 

0

 

4) - R2 + R3 - B3 + S3 =

 

0

 

5) - R3 + R4 - B4 + S4 =

 

0

 

6)

 

I1 - D1 - C1 = - 600

 

7) - I1 + I2 + D1 - D2 - C2 = - 800

 

8) - I2 + I3 + D2 - D3 - C3 = - 500

 

9) - I3 + I4 + D3 - D4 - C4 = - 400

 

10)

 

R4 >=

 

2

 

11) - 200 R1 + C1 <=

 

0

 

12) - 200 R2 + C2 <=

 

0

 

13) - 200 R3 + C3 <=

 

0

 

14) - 200 R4 + C4 <=

 

0

 

15)

 

D4 =

 

0

 

END

 

SLB

 

R4

 

2.00000

 

SUB

 

B1

 

2.00000

 

SUB

 

B2

 

2.00000

 

SUB

 

B3

 

2.00000

 

SUB

 

B4

 

2.00000

 

OBJECTIVE FUNCTION VALUE

 

1)

 

9750.000

 

VARIABLE

 

R1

 

R2

 

R3

 

R4

 

I1

 

I2

 

I3

 

I4

 

B1

 

B2

 

B3

 

B4

 

S1

 

S2

 

S3

 

S4

 

D1

 

D2

 

D3

 

D4

 

C1

 

C2

 

C3

 

C4

 

ROW

 

2)

 

3)

 

4)

 

5)

 

6)

 

7)

 

8)

 

9)

 


 

56:171 O.R. HW#4

 


 

VALUE

 

3.000000

 

4.000000

 

2.500000

 

2.000000

 

0.000000

 

0.000000

 

0.000000

 

0.000000

 

1.000000

 

1.000000

 

0.000000

 

0.000000

 

0.000000

 

0.000000

 

1.500000

 

0.500000

 

0.000000

 

0.000000

 

0.000000

 

0.000000

 

600.000000

 

800.000000

 

500.000000

 

400.000000

 

SLACK OR SURPLUS

 

0.000000

 

0.000000

 

0.000000

 

0.000000

 

0.000000

 

0.000000

 

0.000000

 

0.000000

 


 

REDUCED COST

 

0.000000

 

0.000000

 

0.000000

 

3500.000000

 

190.000000

 

210.000000

 

202.500000

 

200.000000

 

0.000000

 

0.000000

 

2000.000000

 

2000.000000

 

2000.000000

 

2000.000000

 

0.000000

 

0.000000

 

310.000000

 

290.000000

 

297.500000

 

300.000000

 

0.000000

 

0.000000

 

0.000000

 

0.000000

 

DUAL PRICES

 

5000.000000

 

5000.000000

 

3000.000000

 

3000.000000

 

2.500000

 

12.500000

 

2.500000

 

0.000000

 


 

Fall 2000

 


 

page 2 of 4

 


 

10)

 

11)

 

12)

 

13)

 

14)

 

15)

 


 

0.000000

 

0.000000

 

0.000000

 

0.000000

 

0.000000

 

0.000000

 


 

0.000000

 

2.500000

 

12.500000

 

2.500000

 

0.000000

 

0.000000

 


 

RANGES IN WHICH THE BASIS IS UNCHANGED:

 

VARIABLE

 

R1

 

R2

 

R3

 

R4

 

I1

 

I2

 

I3

 

I4

 

B1

 

B2

 

B3

 

B4

 

S1

 

S2

 

S3

 

S4

 

D1

 

D2

 

D3

 

D4

 

C1

 

C2

 

C3

 

C4

 


 

ROW

 

2

 

3

 

4

 

5

 

6

 

7

 

8

 

9

 

10

 

11

 

12

 

13

 

14

 

15

 


 

CURRENT

 

COEF

 

500.000000

 

500.000000

 

500.000000

 

500.000000

 

200.000000

 

200.000000

 

200.000000

 

200.000000

 

5000.000000

 

5000.000000

 

5000.000000

 

5000.000000

 

-3000.000000

 

-3000.000000

 

-3000.000000

 

-3000.000000

 

300.000000

 

300.000000

 

300.000000

 

300.000000

 

0.000000

 

0.000000

 

0.000000

 

0.000000

 


 

CURRENT

 

RHS

 

2.000000

 

0.000000

 

0.000000

 

0.000000

 

-600.000000

 

-800.000000

 

-500.000000

 

-400.000000

 

2.000000

 

0.000000

 

0.000000

 

0.000000

 

0.000000

 

0.000000

 


 

OBJ COEFFICIENT RANGES

 

ALLOWABLE

 

ALLOWABLE

 

INCREASE

 

DECREASE

 

62000.000000

 

500.000000

 

38000.000000

 

2500.000000

 

42000.000000

 

500.000000

 

INFINITY

 

3500.000000

 

INFINITY

 

190.000000

 

INFINITY

 

210.000000

 

INFINITY

 

202.500000

 

INFINITY

 

200.000000

 

62000.000000

 

500.000000

 

500.000000

 

2000.000000

 

INFINITY

 

2000.000000

 

INFINITY

 

2000.000000

 

INFINITY

 

2000.000000

 

INFINITY

 

2000.000000

 

500.000000

 

2000.000000

 

3500.000000

 

500.000000

 

INFINITY

 

310.000000

 

INFINITY

 

290.000000

 

INFINITY

 

297.500000

 

INFINITY

 

300.000000

 

310.000000

 

190.000000

 

190.000000

 

210.000000

 

210.000000

 

202.500000

 

0.000000

 

17.500000

 

RIGHTHAND SIDE RANGES

 

ALLOWABLE

 

INCREASE

 

1.000000

 

1.000000

 

INFINITY

 

INFINITY

 

200.000000

 

200.000000

 

100.000000

 

0.000000

 

0.000000

 

200.000000

 

200.000000

 

100.000000

 

0.000000

 

0.000000

 


 

ALLOWABLE

 

DECREASE

 

1.000000

 

1.000000

 

1.500000

 

0.500000

 

200.000000

 

200.000000

 

300.000000

 

0.000000

 

INFINITY

 

200.000000

 

200.000000

 

300.000000

 

0.000000

 

0.000000

 


 

THE TABLEAU

 

ROW (BASIS)

 

1 ART

 

2

 

R1

 

3

 

R2

 

4

 

S3

 

5

 

R3

 

6

 

B1

 

7

 

B2

 

8

 

S4

 

9 ART

 

10 SLK

 

10

 

11

 

C1

 

12

 

C2

 

13

 

C3

 

14

 

C4

 


 

56:171 O.R. HW#4

 


 

R1

 

0.000

 

1.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 


 

R2

 

0.000

 

0.000

 

1.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 


 

R3

 

0.000

 

0.000

 

0.000

 

0.000

 

1.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 


 

Fall 2000

 


 

R4

 

3500.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

1.000

 

-200.000

 

-1.000

 

0.000

 

0.000

 

0.000

 

-200.000

 


 

I1

 

190.000

 

-0.005

 

0.005

 

0.005

 

0.000

 

-0.005

 

0.010

 

0.000

 

0.000

 

0.000

 

-1.000

 

1.000

 

0.000

 

0.000

 


 

I2

 

210.000

 

0.000

 

-0.005

 

-0.010

 

0.005

 

0.000

 

-0.005

 

0.005

 

0.000

 

0.000

 

0.000

 

-1.000

 

1.000

 

0.000

 


 

page 3 of 4

 


 

15 ART

 


 

0.000

 


 

0.000

 


 

0.000

 


 

0.000

 


 

0.000

 


 

0.000

 


 

B2

 

B3

 

B4

 

S1

 

0.000 2000.000 2000.000 2000.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

-1.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

-1.000

 

1.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

-1.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 


 

ROW

 

1

 

2

 

3

 

4

 

5

 

6

 

7

 

8

 

9

 

10

 

11

 

12

 

13

 

14

 

15

 


 

I3

 

202.500

 

0.000

 

0.000

 

0.005

 

-0.005

 

0.000

 

0.000

 

-0.005

 

-1.000

 

0.000

 

0.000

 

0.000

 

-1.000

 

0.000

 

0.000

 


 

I4

 

200.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

1.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 


 

B1

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

1.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 


 

ROW

 

1

 

2

 

3

 

4

 

5

 

6

 

7

 

8

 

9

 

10

 

11

 

12

 

13

 

14

 

15

 


 

S2

 

2000.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

-1.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 


 

S3

 

0.000

 

0.000

 

0.000

 

1.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 


 

S4

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

1.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 


 

D1

 

310.000

 

0.005

 

-0.005

 

-0.005

 

0.000

 

0.005

 

-0.010

 

0.000

 

0.000

 

0.000

 

1.000

 

-1.000

 

0.000

 

0.000

 

0.000

 


 

D2

 

290.000

 

0.000

 

0.005

 

0.010

 

-0.005

 

0.000

 

0.005

 

-0.005

 

0.000

 

0.000

 

0.000

 

1.000

 

-1.000

 

0.000

 

0.000

 


 

D3

 

297.500

 

0.000

 

0.000

 

-0.005

 

0.005

 

0.000

 

0.000

 

0.005

 

1.000

 

0.000

 

0.000

 

0.000

 

1.000

 

0.000

 

0.000

 


 

D4

 

300.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

-1.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

1.000

 


 

ROW

 

1

 

2

 

3

 

4

 

5

 

6

 

7

 

8

 

9

 

10

 

11

 

12

 

13

 

14

 

15

 


 

C1

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

1.000

 

0.000

 

0.000

 

0.000

 

0.000

 


 

C2

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

1.000

 

0.000

 

0.000

 

0.000

 


 

C3

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

1.000

 

0.000

 

0.000

 


 

C4

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

1.000

 

0.000

 


 

SLK

 

10

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

1.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 


 

SLK

 

11

 

2.500

 

-0.005

 

0.000

 

0.000

 

0.000

 

-0.005

 

0.005

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 


 

SLK

 

12

 

12.500

 

0.000

 

-0.005

 

-0.005

 

0.000

 

0.000

 

-0.005

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 


 

ROW

 

1

 

2

 

3

 

4

 

5

 

6

 

7

 

8

 

9

 

10

 

11

 

12

 

13

 

14

 

15

 


 

SLK

 

13

 

2.500

 

0.000

 

0.000

 

0.005

 

-0.005

 

0.000

 

0.000

 

-0.005

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 

0.000

 


 

56:171 O.R. HW#4

 


 

SLK

 

14

 

0.000 -9750.000

 

0.000

 

3.000

 

0.000

 

4.000

 

0.000

 

1.500

 

0.000

 

2.500

 

0.000

 

1.000

 

0.000

 

1.000

 

0.000

 

0.500

 

1.000

 

0.000

 

0.000

 

0.000

 

0.000

 

600.000

 

0.000

 

800.000

 

0.000

 

500.000

 

1.000

 

400.000

 

0.000

 

0.000

 


 

Fall 2000

 


 

page 4 of 4

 


 

56:171 Operations Research

 

Homework #5 -- Fall 2000

 

1. Linear Programming sensitivity. A paper recycling plant processes box board, tissue

 

paper, newsprint, and book paper into pulp that can be used to produce three grades of

 

recycled paper (grades 1, 2, and 3). The prices per ton and the pulp contents of the four

 

inputs are:

 

Input

 

Cost

 

Pulp

 

type

 

$/ton

 

content

 

Box board

 

5

 

15%

 

Tissue paper

 

6

 

20%

 

Newsprint

 

8

 

30%

 

Book paper

 

10

 

40%

 

Two methods, de-inking and asphalt dispersion, can be used to process the four inputs into

 

pulp. It costs $20 to de-ink a ton of any input. The process of de-inking removes 10% of

 

the input's pulp. It costs $15 to apply asphalt dispersion to a ton of material. The asphalt

 

dispersion process removes 20% of the input's pulp. At most 3000 tons of input can be run

 

through the asphalt dispersion process or the de-inking process. Grade 1 paper can only be

 

produced with newsprint or book paper pulp; grade 2 paper, only with book paper, tissue

 

paper, or box board pulp; and grade 3 paper, only with newsprint, tissue paper, or box board

 

pulp. To meet its current demands, the company needs 500 tons of pulp for grade 1 paper,

 

500 tons of pulp for grade 2 paper, and 600 tons of pulp for grade 3 paper. The LP model

 

below was formulated to minimize the cost of meeting the demands for pulp.

 

Hint:: you may wish to define the variables

 

BOX = tons of purchased boxboard

 

TISS = tons of purchased tissue

 

NEWS = tons of purchased newsprint

 

BOOK = tons of purchased book paper

 

BOX1 = tons of boxboard sent through de-inking

 

TISS1 = tons of tissue sent through de-inking

 

NEWS1 = tons of newsprint sent through de-inking

 

BOOK1 = tons of book paper sent through de-inking

 

BOX2 = tons of boxboard sent through asphalt dispersion

 

TISS2 = tons of tissue sent through asphalt dispersion

 

NEWS2 = tons of newsprint sent through asphalt dispersion

 

BOOK2 = tons of book paper sent through asphalt dispersion

 

PBOX = tons of pulp recovered from boxboard

 

PTISS = tons of pulp recovered from tissue

 

PNEWS= tons of pulp recovered from newsprint

 

PBOOK = tons of pulp recovered from book paper

 

PBOX1 = tons of boxboard pulp used for grade 1 paper,

 

PBOX2 = tons of boxboard pulp used for grade 2 paper, etc.

 

...

 

PBOOK3 = tons of book paper pulp used for grade 3 paper.

 

The LP model using these variables is:

 

MIN

 


 

5 BOX +6 TISS +8 NEWS +10 BOOK +20 BOX1 +20 TISS1 +20 NEWS1

 

+20 BOOK1 +15 BOX2 +15 TISS2 +15 NEWS2 +15 BOOK2

 

SUBJECT TO

 


 

56:171 O.R. HW#5

 


 

Fall 2000

 


 

page 1 of 6

 


 

2)

 

3)

 

4)

 

5)

 

6)

 

7)

 

8)

 

9)

 

10)

 

11)

 

12)

 

13)

 

14)

 

15)

 

16)

 

17)

 

18)

 


 

-

 


 

-

 


 

BOX + BOX1 + BOX2 <=

 

0

 

TISS + TISS1 + TISS2 <=

 

0

 

NEWS + NEWS1 + NEWS2 <=

 

0

 

BOOK + BOOK1 + BOOK2 <=

 

0

 

0.135 BOX1 + 0.12 BOX2 - PBOX =

 

0

 

0.18 TISS1 + 0.16 TISS2 - PTISS =

 

0

 

0.27 NEWS1 + 0.24 NEWS2 - PNEWS =

 

0

 

0.36 BOOK1 + 0.32 BOOK2 - PBOOK =

 

0

 

PBOX + PBOX2 + PBOX3 <=

 

0

 

PTISS + PTISS2 + PTISS3 <=

 

0

 

PNEWS + PNEWS1 + PNEWS3 <=

 

0

 

PBOOK + PBOOK1 + PBOOK2 <=

 

0

 

PNEWS1 + PBOOK1 >=

 

500

 

PBOX2 + PTISS2 + P

 

Paper#9209391 | Written in 27-Jul-2016

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