LP_Solve

IE 311 - Operations Research I

Spring 2005

Solved Exercises on LP Modeling 1. [Problem 3 on page 64 in the 3rd edition of the textbook] Leary chemical manufactures three chemicals: A, B, and C. These

chemicals are produced via two production processes: 1 and 2. Running process 1 for an hour costs $4 and yields 3 units of A, 1 of B and 1 of C. Running process 2 for an hour costs $1 and produces 1 unit of A and one unit of B. To meet customer demands, at least 10 units of A, 5 of B and 3 units of C must be produced daily. Graphically determine a daily production plan that minimizes the cost of meeting Leary Chemical’s daily demands. Answer: Three types of chemicals are produced: A, B, and C. There are two types of processes (machines) available for production.

Process Cost per hour

Output (units of chemical)

A B C 1 4 3 1 1 2 1 1 1 0

Daily Demand 10 5 3

A. Decision Variables: 1X = Number of hours of process 1 used

2X = Number of hours of process 2 used B. Constraints: For chemical A: 103 21 ≥+ XX (I)

For chemical B: 521 ≥+ XX (II)

For chemical C: 31 ≥X (III)

Non-negativity constraints: 0, 21 ≥XX C. Objective function:

Z = Cost of meeting daily demands for chemicals = Cost of operating process 1 and 2

Minimize 21 4 XXZ +=

Process 1A

C

B Process 2 A

B

5

1 0

3 5

X 2

X 1

( I I )

( I ) ( I I I )

Z

F e a s i b l e R e g i o n

B

A

Point 1X 2X Z

A 5 0 (4) (5) + 0 = 20 B 3 2 (4) (3) + 2 = 14 Minimum

Point B is the optimal point. Processes 1 and 2 should be used 3 hours and 2 hours, respectively.

2. [Problem 6 on page 64 in the 3rd edition of the textbook] Farmer Jane owns 45 acres of land. She is going to plant each with wheat or corn. Each acre planted with wheat yields $200 profit; each with corn yields $300 profit. The labor and fertilizer used for each acre are given in the table below. One hundred workers and 120 tons of fertilizer are available. Use linear programming to determine how Jane maximizes profits from her land.

Crop

Wheat Corn Labor 3 worke rs 2 workers Fertilizer 2 tons 4 tons

Answer: A. Decision Variables: W = acres of land planted with wheat C = acres of land planted with corn B. Constraints:

1) Total acres of land used must equal to the land she owns. 45=+ WC

2) Maximum number of workers to be used is 100. 10023 ≤+ CW

3) Maximum tons of fertilizer is 120 tons. 12042 ≤+ CW

3) Non-negativity 0, ≥CW

C. Objective function:

Maximize WCZ 200 300 += 3. [Problem 3 on page 73 in the 3rd edition of the textbook] Peg and Al Fundy have a limited food budget, so Peg is trying to feed

the family as cheaply as possible. However, she still wants to make sure her fa mily members meet their daily nutritional requirements. Peg can buy two foods. Food 1 sells for $7 per pound and, each pound contains 3 units of vitamin A and 1 units of vitamin C. Food 2 sells for $1 per pound, and each pound contains 1 unit of each vitamin. Each day, the family needs at least 12 units of vitamin A and 6 units of vitamin C. (a) Verify that Peg should purchase 12 units of food 2 each day and thus oversatisfy the vitamin C requirement by 6 units. (b) Al has put his foot down and demanded that Peg fulfill the family’s daily nutritional requirement exactly by obtaining

precisely 12 units of vitamin A and 6 units of vitamin C. The optimal solution to new problem will involve ingesting less vitamin C, but it will be more expensive. Why?

Answer: (a) A. Decision Variables: 1X = pounds of food 1 purchased

2X = pounds of food 2 purchased B. Constraints:

1) Vitamin A requirement 123 21 ≥+ XX 2) Vitamin C requirement

621 ≥+ XX

3) Non-negativity 0, 21 ≥XX


Minimize 21 X 7 += XZ

Minimum is 12, so Peg should purchase 12 units of food 2 each day. From second constraint, 612021 >+=+ XX . Thus, vitamin C requirement is over satisfied by 12-6 = 6 units.

(b) The new problem becomes:

Minimize 21 X 7 += XZ

s.t. 123 21 =+ XX

621 =+ XX

0, 21 ≥XX

From the graph, it is obvious that the optimal solution is at point B. That is, 3*1 =X and 3*

2 =X . Then 24* =Z which is greater than 12 (i.e. it will be more expensive).

4. [Problem 5 on page 76 in the 3rd edition of the textbook] Each day, workers at the Gotham City Police Depart ment work two

6-hour shift chosen from 12 a.m. to 6 a.m., 6 a.m. to 12 p.m., 12 p.m. to 6 p.m., and 6 p.m. to 12 a.m. The number of workers are needed during each shift are shown at table below. Workers whose two shifts are consecutive are paid $12 per hour; workers whose shifts are not consecutive are paid $18 per hour. Formulate an LP that can be used to minimize the cost of meeting the daily work-force demands of the Gotham City Police Department.

Shift Workers Needed 12 A.M. to 6 A.M 15 6 A.M. to 12 P.M. 5 12 P.M. to 6 P.M. 12 6 P.M. to 12 A.M. 6

Answer: Let us give numbers to the shifts as below:

Point ( 1X , 2X ) Z

A (6,0) 42

B (3,3) 24

C (0,12) 12 6

12

4 6

X 2

X 1

( I I )

( I )

Z

Feasible Region

B

A

C

Shifts

12 A.M. 12 A.M.6 P.M.6 A.M. 12 P.M.

1 32 4

A. Decision Variables: iX = # of workers whose two shifts are consecutive and start at shift i )4,3,2,1( =i

jX = # of workers whose two shifts are not consecutive and start at shift j )2,1( =j

B. Constraints: 1) Total # of workers in shift 1 is equal to 15. 15114 =++ YXX 2) Total # of workers in shift 2 is equal to 5.

5221 =++ YXX 3) Total # of workers in shift 3 is equal to 12. 12132 =++ YXX

4) Total # of workers in shift 4 is equal to 6. 6243 =++ YXX

5) Non-negativity 0≥iX for 4,3,2,1=i

0≥jY for 2,1=j


Minimize )Y (18)X X (12 214321 +++++= YXXZ 5. [Problem 5 on page 91 in the 3rd edition of the textbook] Chandler Oil Company has 5,000 barrels of oil 1 and 10,000 barrels

of oil 2. The company sells two products: gasoline and heating oil. Both products are produced by combining oil 1 and oil 2. The quality level of each oil is as follows: oil 1, 10; oil 2, 5. Gasoline must have an average quality level of at least 8 and heating oil, at least 6. Demand for each product must be created by advertising. Each dollar spent advertising gasoline creates 5 barrels of demand and each spent on heating oil creates 10 barrels of demand. Gasoline is sold for $25 per barrel, heating oil for $20. Formulate an LP to help Chandler maximize profit. Assume that no oil of either type can be purchased. Answer: Let us draw the input-output diagram for the production process, and tabulate the given data.

A. Decision Variables: jiX , = barrels of oil i used to make product j ( i =1 is oil 1, i = 2 is oil 2, j =1 is gasoline,

j =2 is heating oil)

ja = dollar spent advertising product j )2,1( =j

B. Constraints:

1) The barrels of each oil used to make products must not exceed its on hand inventory level. 000,52,11,1 ≤+ XX (for oil type 1)

000,102,21,2 ≤+ XX (for oil type 2)

Oil Type

On hand Inventory (Barrels)

Quality Level Product

Minimum Average

Quality Level

Demand (barrels per $ spent

for advertising)

Sales Price ($ per barrel)

1 5,000 10 Gasoline 8 5 25

2 10,000 5 Heating oil 6 10 20

Oil 1Gasoline

Inputs

ProductionProcess

Heating oilOil 2

Outputs

2) The minimum average quality level of each product must be satisfied.

8510

1,21,1

1,21,1 ≥++

XXXX

è 8producedgasolineofamountTotal

gasolineforusedoilofQualityTotal≥ è 032 1,21,1 ≥− XX

6510

2,22,1

2,22,1 ≥+

+

XX

XX è 04 2,22,1 ≥− XX

3) Production of each product must be equal to the demand created by advertising. 11,21,1 5aXX =+ (for gasoline)

22,22,1 10aXX =+ (for heating oil)

4) Non-negativity 0, ≥jiX for 2,1=i ; 2,1=j


Z = Total sales revenue from gasoline + Total s ales revenue from eating oil – Total advertising cost

Maximize 212,22,11,21,1 )(20)(25 aaXXXXZ −−+++=

6. [Problem 11 on page 92 in the 3rd edition of the textbook] Eli Daisy produces the drug Rozac from four chemicals. Today they

must produce 1,000 lb of drug. The three active ingredients in Rozac are A , B, and C. By weight, at least 8% of Rozac must consist of A, at least 4% of B, and at least 2% of C. The cost per pound of each chemical and the amount of each ingredient in 1 lb of each chemical are given in table below. It is necessary that at least 100 lb of chemical 2 be used. Formulate an LP whose solution would determine the cheapest way of producing today’s batch of Rozac.

Chemical Cost per lb A B C

1 $8 0.03 0.02 0.01 2 $10 0.06 0.04 0.01 3 $11 0.10 0.03 0.04 4 $14 0.12 0.09 0.04

Answer: A. Decision Variables: A = pounds (lb) of ingredient A used B = pounds of ingredient B used C = pounds of ingredient C used iX = pounds of chemical i used )4,3,2,1( =i

B. Constraints: 1) Minimum ingredient A requirement for the drug (in weight) )000,1)(08.0(≥A 2) Minimum ingredient B requirement for the drug (in weight)

)000,1)(04.0(≥B 3) Minimum ingredient C requirement for the drug (in weight) )000,1)(02.0(≥C

4) 100 lbs of chemical 2 should be used 1002 ≥X

5) Chemical content CBAX 01.002.003.01 ++=

CBAX 01.004.006.02 ++=

CBAX 04.003.010.03 ++=

CBAX 04.009.012.04 ++= 6) Non-negativity

0,, ≥CBA

0≥iX for 4,3,2,1=i C. Objective function:

Minimize 4321 14X 1110X 8 +++= XXZ

7. [Problem 5 on page 104 in the 3rd edition of the textbook] During the next two months, General Cars must meet (on time) the

following dema nd for trucks and cars: month 1 – 400 trucks, 800 cars; month 2 – 300 trucks, 300 cars. During each month, at most 1000 vehicles can be produced. Each truck uses 2 tons of steel, and each car uses 1 ton of steel. During month 1, steel cost $400 per ton; during month 2, steel costs $600 per ton. At most 1500 tons of steel may be purchased at each month (steel may only be used during the month in which it is purchased). At the beginning of month 1, 100 trucks and 200 cars are in inventory. At the end of each month, a holding cost of $150 per vehicle is assessed. Each car gets 20 mpg and each truck gets 10 mpg. During each month, the vehicles produced by the company must average at least 16 mpg. Formulate an LP to meet the demand and mileage requirements at minimum cost (include steel costs and holding costs). Answer: Let us draw the input-output diagram for the production process, and tabulate the given data.

Monthly Demand Vehicle Month 1 Month 2

Steel Requirement (ton/vehicle)

Inventory on hand

Truck 400 300 2 100 Car 800 300 1 200 Max # of vehicles that can be produced 1,000 1,000 Steel cost ($/ton) 400 600 Max amount of steel that can be purchased (ton) 1,500 1,500 Inventory holding cost ($/vehicle) 150 150

A. Decision Variables: iS = Steel bought during month i )2,1( =i

iT = trucks produced during month i )2,1( =i

iIT = trucks in inventory at the end of month i )2,1( =i

iIC = cars in inventory at the end of month i )2,1( =i B. Constraints:

1) Maximum amount of steel that can be purchased in each month is limited 500,11 ≤S

500,12 ≤S 2) Steel used to produce vehicles in each month must be less than the amount of steel purchased in each month

1112 SCT ≤+

2222 SCT ≤+ 3) Maximum number of vehicles produced in each month must be less than 1,000 000,111 ≤+ CT

000,122 ≤+ CT 4) Inventory balance constraints For trucks 11 400100 ITT +=+ è 30011 =− ITT

211 300 ITTIT +=+ è 300221 =++ ITTIT

For cars 11 800200 ICC +=+ è 60011 =− ICC

222 300 ICCIC +=+ è 300222 =−+ ICCIC

steel fortrucks

Month 1 Month 2

steel forcars

100 trucks

200 cars

800 cars400 trucks

trucks ininventory

cars ininventory

steel fortrucks

steel forcars

300 trucks 300 cars

trucks ininventory

cars ininventory

CUSTOMER CUSTOMER

5) Average mpg

161020

11

11 ≥++

TCTC

è 064 11 ≥− TC

161020

22

22 ≥++

TCTC

è 064 22 ≥− TC


Z = Total Cost = Cost of steel + Inventory holding costs

Minimize 212121 150150150150600400 ICICITITSSZ +++++= 8. [Problem 44 on page 119 in the 3rd edition of the textbook] A paper recycling plant process box board, tissue, paper,

newsprint, and book paper in to pulp that can be used to produce three grades of recycled paper (grades 1, 2, and 3). The prices per ton and pulp contents of the four inputs are shown in below table. Two methods, de-inking and asphalt dispersion, can be used to produce 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, leaving 90% of the original pulp. It costs $15 to apply asphalt dispersion to a ton of material. The asphalt dispersion removes 20% of the input’s pulp. At most 3,000 tons of input can be run through the asphalt dispersion process or 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, 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. Formulate an LP to minimize the cost of meeting the demands for pulp. Answer: Let us tabulate the given data.

Item Cost Pulp content Box board $5 15% Tissue paper $6 20% Newsprint $8 30% Book paper $10 40%

Method of Pulp Production

De-inking Asphalt Dispersion Cost($/tons of input) 20 15 % of input that is converted in to pulp 90 80 Max amount of input that can be processed 3,000 3,000

Pulp type can be used Product Box board Tissue Paper Newsprint Book Paper

Required amount of pulp for satisfying

demand Grade 1 X X 500 Grade 2 X X X 500 Grade 3 X X X 600

A. Decision Variables: iI = tons of input (raw material) i purchased ( i = 1, box board; i = 2, tissue paper; i = 3,

newsprint; i = 4, book paper)

iD = tons of input i sent through de- inking

iA = tons of input i sent through asphalt dispersion

iP = tons of type i pulp produced

jiU , = tons of type i pulp used for grade j paper ( j = 1, grade 1 paper; j = 2, grade 2

paper; j = 3, grade 3 paper)

B. Constraints: 1) Tons of each input sent through de- inking and asphalt dispersion processes can not exceed the purchased

amount 111 IAD ≤+

222 IAD ≤+

333 IAD ≤+

444 IAD ≤+ è (Total amount of book paper processed <= amount of book paper purchased) 2) Total amount of each pulp type must equal to its corresponding amount of purchase

111 )80.0)(15.0()90.0)(15.0( PAD =+ (Amount of box board pulp obtained by de- inking + amount of box board obtained by asphalt dispersion = total amount of box board pulp)

222 )80.0)(20.0()90.0)(20.0( PAD =+

333 )80.0)(30.0()90.0)(30.0( PAD =+

444 )80.0)(40.0()90.0)(40.0( PAD =+

3) Maximum amount of input that can be processed by each method of pulp production is limited. 000,34321 ≤+++ DDDD è For de- inking

000,34321 ≤+++ AAAA èFor asphalt dispersion 4) Inventory balance constraints 13,12,1 PUU ≤+ è For box board pulp

(Total amount of box board pulp used for product 2 and 3) 23,22,2 PUU ≤+ è For tissue paper pulp

33,31,3 PUU ≤+ è For newsprint pulp

42,41,4 PUU ≤+ è For book print pulp

5) Demand for each product (paper type) must be satisfied

5001,41,3 ≥+ UU è For grade 1 paper

5002,42,22,1 ≥++ UUU è For grade 2 paper

5003,33,23,1 ≥++ UUU è For grade 3 paper

6) All variables are non negative

0,, , ≥iiii PADI for 4,3,2,1=i

0, ≥jiU for 4,3,2,1=i ; 3,2,1=j C. Objective function:

Minimize 432143214321 151515152020202010865 AAAADDDDIIIIZ +++++++++++=

9. [Problem 3 on page 110 in the 3rd edition of the textbook] The IRS has determined that during each of the next twelve months

they will need the number of super computers given in table below. To meet these requirements the IRS rents supercomputers for a period of one, two, or three months. It costs $100 to rent a supercomputer for one month, $180 for two months, and $250 for three mo nths. At the beginning of month 1 the IRS has no supercomputers. Determine the rental plan that meets the next twelve months’ requirements at minimum cost. Note: you may assume that fractional rentals are okay. Thus if your solution says rent 140.6 computers for one month you can round this up or down without having much effect on total cost.

Month 1 2 3 4 5 6 7 8 9 10 11 12 Requirement 800 1,000 600 500 1,200 400 800 600 400 500 800 600

Answer:

A. Decision Variables: pmX , = the # of supercomputers rented at the beginning of month m for p number of periods

B. Constraints: 8003,12,11,1 ≥++ XXX

000,1)()( 3,22,21,23,12,1 ≥++++ XXXXX

600)()( 3,32,31,33,22,23,1 ≥+++++ XXXXXX

500)()( 3,42,41,43,32,33,2 ≥+++++ XXXXXX

200,1)()( 3,52,51,53,42,43,3 ≥+++++ XXXXXX

400)()( 3,62,61,63,52,53,4 ≥+++++ XXXXXX

800)()( 3,72,71,73,62,63,5 ≥+++++ XXXXXX

600)()( 3,82,81,83,72,73,6 ≥+++++ XXXXXX

400)()( 3,92,91,93,82,83,7 ≥+++++ XXXXXX

500)()( 3,102,101,103,92,93,8 ≥+++++ XXXXXX

800)()( 2,111,113,102,103,9 ≥++++ XXXXX

600)()()( 1,122,113,10 ≥++ XXX


Minimize

)(250

)(180

)(100

3,103,93,83,73,63,53,43,33,23,1

2,112,102,92,82,72,62,52,42,32,22,1

1,121,111,101,91,81,71,61,51,41,31,21,1

XXXXXXXXXX

XXXXXXXXXXX

XXXXXXXXXXXXZ

+++++++++

+++++++++++

++++++++++++=

10. [Problem 48 on page 120 in the 3rd edition of the textbook] Bank 24 is open 24 hours per day. Tellers work two consecutive 6-

hour shifts and are paid $10 per hour. The possible shifts are as follows: midnight – 6 a.m., 6 a.m. – noon, noon – 6 p.m., 6 p.m. – midnight. During each shift, the following numbers of customers enter the bank: midnight – 6 a.m.,100; 6 a.m. – noon, 200; noon – 6 p.m., 300; 6 p.m. – midnight, 200. Each teller can serve up to 50 customers per shift. To model customer a cost for customer impatience, we assume that any customer who is present at the end of a shift “costs” the bank $5. We assume that by midnight of each day, all customers must be served, so each day’s midnight – 6a.m. shift begins with zero customers in the bank. Formulate an LP that can be used to minimize the sum of the bank’s labor and customer impatience costs. Answer:

A. Decision Variables: iX = # of tellers beginning work on shift i

iY = # of customers unserved in shift i and to be served in the next shift

Shift (i ) Period 1 midnight – 6 A.M. 2 6 A.M. – noon 3 noon – 6 P.M. 4 6 P.M. – midnight

# of customersentering the bank

in shift (i)

# of customersunserved andto be served in

shift (i+1)

i

# of customersserved in shift ( i )

# of customersleft unserved in

shift (i -1)

B. Constraints:

1005050 141 =++ YXX

1221 2005050 YYXX +=++

2332 3005050 YYXX +=++

343 1005050 YXX +=+

0≥iX and 0≥iY for 4,3,2,1=i C. Objective function:

Minimize 3214321 55560606060 YYYXXXXZ ++++++=

11. Linear programming models are used by many Wall Street firms to select a desirable bond portfolio. The following is a

simplified version of such model. Solodrex is considering investing in four bonds: $1,000,000 is available for investment. The expected annual return, the worst-case annual return on each bond, and the duration of each bond are given in the following table.

Bond Expected return Worst-case return Duration 1 13% 6% 3 2 8% 8% 4 3 12% 10% 7 4 14% 9% 9

The duration of a bond is measure of the bond’s sensitivity to interest rates. Solodex wants to maximize the expected return from its bond investments, subject to the following restrictions:

1. The worst-case return of the bond portfolio must be at least 8%. 2. The average duration of the portfolio must be at most 6. For example a portfolio that invested $ 600,000 in bond 1

and $400,000 in bond 4 would have an average duration of [(600,000) (3) + (400,000) (9)] / 1,000,000 = 5.4. 3. Because of the diversification requirements, at most 40% of the total amount invested can be invested in a single

bond. Formulate an LP model to help Solodex tom achieve its objective. Answer: A. Decision Variables: iX = the amount of money invested in bond i , for )4,3,2,1( =i B. Constraints:

1) )000,000,1)(08.0(09.01.008.006.0 4321 ≥+++ XXXX

2) 6000,000,1

9743 4321 ≤+++ XXXX

3) )000,000,1)(40.0(1 ≤X

)000,000,1)(40.0(2 ≤X

)000,000,1)(40.0(3 ≤X

)000,000,1)(40.0(4 ≤X

100

Y1

50(X1+X4 )

1 2 3 4

200 300 200

Y2 Y3

50(X1+X2 ) 50(X2+X3 ) 50(X3+X4 )

4) 0≥iX for 4,3,2,1=i C. Objective function:

Maximize 4321 14.012.008.013.0 XXXXZ +++=

12. An automobile company is planning its fall advertising campaign to unveil the new models for the coming year. The

marketing department has assembled the following data.

The company would like to limit their TV advertising expenses to $3 million and buy at least five prime time spots and at least four nonprime time spots. They would like to buy a minimum of 6 radio advertising units, and at least 9 advertising units in newspaper and magazines. They also want make sure that their message reaches at least 30 million youth viewers. It is required to devise an advertising campaign costing no more than $12 million that reaches as many viewers as possible, subject to these constraints. Ignoring any integer requirements of the variables, formulate this as a Linear Programming model. Answer: A. Decision Variables: 1X = # of prime time TV spots

2X = # of nonprime time TV spots

3X = # of radio advertising units

4X = # of newspapers and magazines advertising units B. Constraints:

1) TV advertising cost must be less than 3 million 000,000,3000,78000,100 21 ≤+ XX

2) Minimum required number of each advertising units

51 ≥X

42 ≥X

63 ≥X

94 ≥X

3) Total # of youth viewers must be greater than 30 million 304.05.15.2 4321 ≥+++ XXXX 4) Total advertising campaign budget limit 000,000,12000,20000,40000,78000,100 4321 ≤+++ XXXX

5) Non negativity constraints 0≥iX for 4,3,2,1=i


Z = the number of all viewers Maximize 4321 5.246 XXXXZ +++=

Viewers/spot (in millions) Medium Cost/spot

All viewers Youth TV-Prime time $100,000 6 2.5 TV-Nonprime time $78,000 4 1.5 Radio $40,000 2.5 1 Newspapers and magazines $20,000 1 0.4

13. A farm family owns 125 acres of land and has $40,000 in found available for investment. Its members can produce a total of

3,500 person-hours worth of labor during the winter months (mid-September to mid-May) and 4,000 person-hours during the summer. If any of these person-hours are not needed to farmer family, younger members of family will use their capacity to work on a neighboring farm for $5/hour during the winter months and $6/hours during the summer. Cash income may be obtained from three crops and two types of livestock: diary cows and lying hens. No investment founds are needed for corps. However, each cow will require an investment outlay of $1,200, and each hen will cost $9.

Each cow will require 1.5 acres of land, 100 person-hours of work during the winter months, and other 50 person-hours during the summer. Each cow will produce a net annual cash income of $1,000 for the family. The corresponding figures for each hen are: no acreage, 0.6 person-hours during the winter, 0.3 more person-hours during the summer, and an annual net cash income of $5. The chicken house can accommodate a maximu m of 3,000 hens, and the size of the barn limits the herd to a maximum 32 cows. Estimated person-hours and income per acre planted in each of the three corps are as folows

Crop Soybeans Corn Oats Winter person-hours 20 35 10 Summer person-hours 50 75 40 Net annual cash income ($) 600 900 450

The family wishes to determine how much acreage should be planted in each of the corps and how many cows and hens should be kept to maximize its net cash income. Formulate the Linear Programming model for this problem. Answer: A. Decision Variables: SL = Amount of acres allocated for soybean production

cL = Amount of acres allocated for corn production

oL = Amount of acres allocated for oat production

C = # of cows purchased H = # of hens purchased W = Excess person-hours in the winter S = Excess person-hours in the summer B. Constraints:

1) Total land allocated for crop production and cows limited by the available land of 125 acres 1255.1 ≤+++ CLLL ocS

2) Total cost of purchasing cows and hens must be less than $40,000 000,409200,1 ≤+ HC

3) Limitations on the size of barn and chicken house

32≤C 000,3≤H

4) Labor limitations 500,36.0100103520 =+++++ WHCLLL ocS è For winters

000,49.0150407550 =+++++ SHCLLL ocS è For summers

4) Non negativity constraints 0,,,,,, ≥SWHCLLL ocS C. Objective function:

Z = total of crop income, income from animals, and income from neighbor farm Maximize SWHCLLLZ ocS 655000,1450900600 ++++++=

14. An investor has money-making activities A and B available at the beginning of each of the next five years (call them years 1 to 5). Each dollar invested in A at the beginning of a year returns $1.40 (a profit of $0.40) two years later (in time for immediate reinvestment). Each dollar invested in B at the beginning of a year returns $1.70 three years later. In addition, money-making activities C and D will each be available at one time in the future. Each dollar invested in C at the beginning of year 2 returns $1.90 at the end of year 5. Each dollar invested in D at the beginning of year 5 returns $1.30 at the end of year 5.

The investor begins with $60,000 and wishes to know which plan maximizes the amount of money that can be accumulated by the beginning of year 6. Formulate an LP model for this problem. Answer:

A. Decision Variables: tA = Amount of money invested to A in year t )4,3,2,1( =t

tB = Amount of money invested to B in year t )3,2,1( =t

2C = Amount of money invested to C in year 2

5D = Amount of money invested to C in year 2

tR = Amount of money not invested in year t )4,3,2,1( =t B. Constraints:

1) Equalities of years 000,60111 =++ RBA èYear 1

1222 RRBA =++ èYear 2 (Amount of money invested in year 2 +amount of money not invested in year 2 = amount of money left from year 1)

21333 4.1 RARBA +=++ èYear 3

31244 7.14.1 RBARA ++=+ èYear 4

4235 7.14.1 RBAD ++= èYear 5

4) Non negativity constraints 0≥tA for 4,3,2,1=t

0≥tB for 3,2,1=t

02 ≥C

05 ≥D

0≥tR for 4,3,2,1=t C. Objective function:

Z = Amount of money accumulated by the beginning of year 6

Maximize 5432 3.14.17.19.1 DABCZ +++=

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