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.
A
Process1
A
Process 2
B
B
C
Process
1
2
Constraints:
For chemical A:
For chemical B:
For chemical C:
Non-negativity constraints:
C.
4
1
Daily Demand
Output
(units of chemical)
A
B
C
3
1
1
1
1
0
10
5
3
X 1 = Number of hours of process 1 used
X 2 = Number of hours of process 2 used
A. Decision Variables:
B.
Cost per
hour
3 X1 + X 2 ≥ 10
X1 + X 2 ≥ 5
X1 ≥ 3
X1, X 2 ≥ 0
(I)
(II)
(III)
Objective function:
Z = Cost of meeting daily demands for chemicals = Cost of operating process 1 and 2
Minimize
Z = 4 X1 + X 2
( I I I )
( I )
X
2
Feasible Region
( I I )
10
5
Z
B
A
3
5
X
1
Point
X1
X2
Z
A
B
5
3
0
2
(4) (5) + 0 = 20
(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
Labor
Fertilizer
Answer:
A. Decision Variables:
Wheat
3 worke rs
2 tons
Corn
2 workers
4 tons
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.
C + W = 45
2) Maximum number of workers to be used is 100.
3W + 2C ≤ 100
3) Maximum tons of fertilizer is 120 tons.
2W + 4C ≤ 120
3) Non-negativity
W ,C ≥ 0
C. Objective function:
Maximize
3.
Z = 300C + 200W
[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:
X 1 = pounds of food 1 purchased
X 2 = pounds of food 2 purchased
B. Constraints:
1) Vitamin A requirement
3 X1 + X 2 ≥ 12
2) Vitamin C requirement
X1 + X 2 ≥ 6
3) Non-negativity
X 1, X 2 ≥ 0
C. Objective function:
Minimize
Z = 7 X1 + X 2
(I)
X2
(II)
12
C
( X1,
Point
Feasible Region
6
X2 )
Z
A
(6,0)
42
B
(3,3)
24
C
(0,12)
12
Z
B
A
4
X1
6
Minimum is 12, so Peg should purchase 12 units of food 2 each day. From second constraint,
Thus, vitamin C requirement is over satisfied by 12-6 = 6 units.
X1 + X 2 = 0 + 12 > 6 .
(b) The new problem becomes:
Minimize
s.t.
Z = 7 X1 + X 2
3 X 1 + X 2 = 12
X1 + X 2 = 6
X1 , X 2 ≥ 0
From the graph, it is obvious that the optimal solution is at point B. That is,
X1 = 3
*
and
X 2 = 3 . Then Z * = 24
*
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
6 A.M. to 12 P.M.
12 P.M. to 6 P.M.
15
5
12
6 P.M. to 12 A.M.
6
Answer:
Let us give numbers to the shifts as below:
1
12 A.M.
2
6 A.M.
Shifts
12P.M.
3
4
6P.M.
12 A.M.
X i = # of workers whose two shifts are consecutive and start at shift i (i = 1,2 ,3, 4)
X j = # of workers whose two shifts are not consecutive and start at shift j ( j = 1,2 )
A. Decision Variables:
B. Constraints:
1) Total # of workers in shift 1 is equal to 15.
X 4 + X 1 + Y1 = 15
2) Total # of workers in shift 2 is equal to 5.
X 1 + X 2 + Y2 = 5
3) Total # of workers in shift 3 is equal to 12.
X 2 + X 3 + Y1 = 12
4) Total # of workers in shift 4 is equal to 6.
X 3 + X 4 + Y2 = 6
5) Non-negativity
Xi ≥ 0
for
Y j ≥ 0 for
i = 1,2,3, 4
j = 1,2
C. Objective function:
Minimize
5.
Z = 12 ( X 1 + X 2 + X 3 + X 4 ) + 18 (Y1 + Y2 )
[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.
Outputs
Inputs
Gasoline
Oil1
Oil2
Production
Process
Heatingoil
Demand
(barrels per $ spent
for advertising)
5
Sales Price
($ per barrel)
Gasoline
Minimum
Average
Quality Level
8
Heating oil
6
10
20
Oil
Type
On hand Inventory
(Barrels)
Quality
Level
Product
1
5,000
10
2
10,000
5
A. Decision Variables:
25
X i , j = 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)
a j = dollar spent advertising product j ( j = 1,2 )
B. Constraints:
1) The barrels of each oil used to make products must not exceed its on hand inventory level.
X1,1 + X1,2 ≤ 5,000 (for oil type 1)
X 2,1 + X 2 , 2 ≤ 10,000 (for oil type 2)
2) The minimum average quality level of each product must be satisfied.
10 X 1,1 + 5 X 2,1
Total Quality of oil used for gasoline
≥8 è
≥ 8 è 2 X 1 ,1 − 3 X 2 ,1 ≥ 0
X 1,1 + X 2,1
Total amount of gasoline produced
10 X 1, 2 + 5 X 2, 2
≥ 6 è 4 X 1, 2 − X 2 , 2 ≥ 0
X 1 ,2 + X 2 , 2
3) Production of each product must be equal to the demand created by advertising.
X 1 ,1 + X 2 ,1 = 5a1
(for gasoline)
X 1 ,2 + X 2 , 2 = 10a2
4) Non-negativity
X i, j ≥ 0
for
(for heating oil)
i = 1,2 ; j = 1,2
C. Objective function:
Z = Total sales revenue from gasoline + Total s ales revenue from eating oil – Total advertising cost
Maximize Z = 25 ( X 1,1 + X 2 ,1 ) + 20 ( X 1, 2 + X 2 , 2 ) − a1 − a 2
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
1
2
3
4
Answer:
A. Decision Variables:
Cost per lb
$8
$10
$11
$14
A
0.03
0.06
0.10
0.12
B
0.02
0.04
0.03
0.09
A = pounds (lb) of ingredient A used
B = pounds of ingredient B used
C
= pounds of ingredient C used
X i = pounds of chemical i used (i = 1,2,3,4)
B. Constraints:
1) Minimum ingredient A requirement for the drug (in weight)
A ≥ (0 .08 )(1,000 )
2) Minimum ingredient B requirement for the drug (in weight)
B ≥ (0 .04 )(1,000 )
3) Minimum ingredient C requirement for the drug (in weight)
C ≥ (0 .02 )(1,000 )
4) 100 lbs of chemical 2 should be used
X 2 ≥ 100
5) Chemical content
X 1 = 0.03 A + 0.02 B + 0.01C
X 2 = 0.06 A + 0.04 B + 0.01C
X 3 = 0.10 A + 0.03 B + 0.04C
X 4 = 0.12 A + 0.09B + 0.04C
6) Non-negativity
A, B, C ≥ 0
X i ≥ 0 for i = 1,2,3, 4
C. Objective function:
Minimize
Z = 8 X 1 + 10X 2 + 11 X 3 + 14X 4
C
0.01
0.01
0.04
0.04
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 $150 per vehicle is assessed. Each car gets 20 mpg and each truck gets
10 mpg. During each month, the vehicles produced by the company must average at least 16 mpg. Formulate an LP to meet
the demand and mileage requirements at minimum cost (include steel costs and holding costs).
Answer:
Let us draw the input-output diagram for the production process, and tabulate the given data.
steel for
trucks
steel for
cars
steel for
trucks
trucks in
inventory
100 trucks
trucks in
inventory
Month 1
Month 2
cars in
inventory
200 cars
400 trucks
800 cars
300 trucks
CUSTOMER
Vehicle
Truck
Car
Max # of vehicles that can be produced
Steel cost ($/ton)
Max amount of steel that can be purchased (ton)
Inventory holding cost ($/vehicle)
A. Decision Variables:
steel for
cars
cars in
inventory
300 cars
CUSTOMER
Monthly Demand
Month 1
Month 2
400
300
800
300
1,000
1,000
400
600
1,500
1,500
150
150
Steel Re quirement
(ton/vehicle)
2
1
Inventory on hand
100
200
S i = Steel bought during month i (i = 1,2 )
T i = trucks produced during month i (i = 1,2 )
IT i = trucks in inventory at the end of month i (i = 1,2 )
IC i = cars in inventory at the end of month i (i = 1,2)
B. Constraints:
1) Maximum amount of steel that can be purchased in each month is limited
S1 ≤ 1,500
S 2 ≤ 1,500
2) Steel used to produce vehicles in each month must be less than the amount of steel purchased in each month
2T1 + C1 ≤ S1
2T2 + C 2 ≤ S 2
3) Maximum number of vehicles produced in each month must be less than 1,000
T1 + C 1 ≤ 1,000
T2 + C 2 ≤ 1,000
4) Inventory balance constraints
For trucks
For cars
100 + T1 = 400 + IT1
IT1 + T1 = 300 + IT2
200 + C1 = 800 + IC1
IC 2 + C 2 = 300 + IC 2
T1 − IT1 = 300
è IT1 + T2 + IT2 = 300
è C1 − IC1 = 600
è IC 2 + C 2 − IC 2 = 300
è
5) Average mpg
20C1 + 10T1
≥ 16
C1 + T1
è
4C1 − 6T1 ≥ 0
20C 2 + 10T2
≥ 16
C 2 + T2
è
4C 2 − 6T2 ≥ 0
C. Objective function:
Z = Total Cost = Cost of steel + Inventory holding costs
Minimize
8.
Z = 400S1 + 600S 2 + 150IT1 + 150IT 2 + 150IC1 + 150IC 2
[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
Box board
Tissue paper
Newsprint
Book paper
Cost
$5
$6
$8
$10
Pulp content
15%
20%
30%
40%
Method of Pulp Production
De-inking
Asphalt Dispersion
20
15
90
80
3,000
3,000
Cost($/tons of input)
% of input that is converted in to pulp
Max amount of input that can be processed
Pulp type can be used
Product
Box board
Grade 1
Grade 2
Grade 3
X
X
A. Decision Variables:
Tissue Paper
Newsprint
Book Paper
X
X
X
X
X
X
I i = tons of input (raw material) i
Required amount of
pulp for satisfying
demand
500
500
600
purchased ( i = 1, box board;
i = 2, tissue paper; i = 3,
newsprint; i = 4, book paper)
Di = tons of input i sent through de- inking
Ai = tons of input i sent through asphalt dispersion
Pi = tons of type i pulp produced
U i , j = 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
D1 + A1 ≤ I 1
D2 + A2 ≤ I 2
D3 + A3 ≤ I 3
D4 + A4 ≤ I 4 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
(0.15)( 0.90) D1 + ( 0.15)(0.80) A1 = P1
(Amount of box board pulp obtained by de- inking + amount of box board obtained by asphalt
dispersion = total amount of box board pulp)
(0.20)(0.90) D 2 + (0.20)( 0.80) A2 = P2
(0.30)( 0.90 ) D3 + (0.30)( 0.80 ) A3 = P3
(0.40)(0.90) D 4 + (0.40)( 0.80) A4 = P4
3) Maximum amount of input that can be processed by each method of pulp production is limited.
D1 + D 2 + D3 + D 4 ≤ 3,000
è For de- inking
A1 + A2 + A3 + A4 ≤ 3,000
èFor asphalt dispersion
4) Inventory balance constraints
U 1, 2 + U 1 ,3 ≤ P1
è For box board pulp
(Total amount of box board pulp used for product 2 and 3)
U 2, 2 + U 2, 3 ≤ P2
è For tissue paper pulp
U 3,1 + U 3, 3 ≤ P3
è For newsprint pulp
U 4 ,1 + U 4 , 2 ≤ P4
è For book print pulp
5) Demand for each product (paper type) must be satisfied
U 3,1 + U 4,1 ≥ 500
è For grade 1 paper
U 1, 2 + U 2 , 2 + U 4 , 2 ≥ 500
è For grade 2 paper
U 1, 3 + U 2, 3 + U 3, 3 ≥ 500
è For grade 3 paper
6) All variables are non negative
I i , Di , Ai , Pi ≥ 0 for i = 1,2 ,3, 4
U i , j ≥ 0 for i = 1,2 ,3, 4 ; j = 1,2,3
C. Objective function:
Minimize
9.
Z = 5I 1 + 6I 2 + 8I 3 + 10 I 4 + 20 D1 + 20 D 2 + 20 D 3 + 20 D4 + 15 A1 + 15 A2 + 15 A3 + 15 A4
[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
Requirement
1
800
2
1,000
3
600
4
500
5
1,200
6
400
7
800
8
600
9
400
10
500
11
800
12
600
Answer:
A. Decision Variables:
X m , p = the # of supercomputers rented at the beginning of month m for p number of periods
B. Constraints:
X 1 ,1 + X 1, 2 + X 1, 3 ≥ 800
( X 1, 2 + X 1,3 ) + ( X 2 ,1 + X 2 , 2 + X 2 , 3 ) ≥ 1,000
X 1 ,3 + ( X 2 , 2 + X 2, 3 ) + ( X 3,1 + X 3 , 2 + X 3 ,3 ) ≥ 600
X 2, 3 + ( X 3, 2 + X 3,3 ) + ( X 4,1 + X 4, 2 + X 4,3 ) ≥ 500
X 3 ,3 + ( X 4 , 2 + X 4, 3 ) + ( X 5 ,1 + X 5 , 2 + X 5 ,3 ) ≥ 1,200
X 4 , 3 + ( X 5 , 2 + X 5 , 3 ) + ( X 6,1 + X 6 , 2 + X 6 , 3 ) ≥ 400
X 5 , 3 + ( X 6 , 2 + X 6 , 3 ) + ( X 7 ,1 + X 7 ,2 + X 7 , 3 ) ≥ 800
X 6, 3 + ( X 7, 2 + X 7,3 ) + ( X 8,1 + X 8, 2 + X 8,3 ) ≥ 600
X 7 , 3 + ( X 8 , 2 + X 8 ,3 ) + ( X 9 ,1 + X 9 ,2 + X 9 ,3 ) ≥ 400
X 8 ,3 + ( X 9 ,2 + X 9 ,3 ) + ( X 10 ,1 + X 10 , 2 + X 10 ,3 ) ≥ 500
X 9, 3 + ( X 10 , 2 + X 10 ,3 ) + ( X 11,1 + X 11, 2 ) ≥ 800
( X 10, 3 ) + ( X 11, 2 ) + ( X 12 ,1 ) ≥ 600
C. Objective function:
Minimize
Z = 100( X 1,1 + X 2,1 + X 3,1 + X 4,1 + X 5,1 + X 6 ,1 + X 7 ,1 + X 8,1 + X 9,1 + X 10,1 + X 11,1 + X 12 ,1 ) +
180( X 1,2 + X 2,2 + X 3,2 + X 4 ,2 + X 5,2 + X 6 ,2 + X 7, 2 + X 8, 2 + X 9 ,2 + X 10 ,2 + X 11,2 ) +
250( X 1,3 + X 2,3 + X 3,3 + X 4 ,3 + X 5,3 + X 6,3 + X 7,3 + X 8 ,3 + X 9,3 + X 10 ,3 )
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 6hour 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:
Shift ( i )
1
2
3
4
Period
midnight – 6 A.M.
6 A.M. – noon
noon – 6 P.M.
6 P.M. – midnight
# of customers
entering the bank
in shift (i)
# of customers
left unserved in
shift (i -1)
i
# of customers
served in shift ( i )
A. Decision Variables:
X i = # of tellers beginning work on shift i
Yi = # of customers unserved in shift i and to be served in the next shift
# of customers
unserved and
to be served in
shift (i+1)
100
1
200
Y1
50(X1+X4 )
2
50(X1+X2 )
300
Y2
3
200
Y3
50(X 2+X3 )
4
50(X3 +X4 )
B. Constraints:
50 X 1 + 50 X 4 + Y1 = 100
50 X 1 + 50 X 2 + Y2 = 200 + Y1
50 X 2 + 50 X 3 + Y3 = 300 + Y2
50 X 3 + 50 X 4 = 100 + Y3
Xi ≥ 0
and
Yi ≥ 0
for
i = 1,2,3, 4
C. Objective function:
Minimize
Z = 60 X 1 + 60 X 2 + 60 X 3 + 60 X 4 + 5Y1 + 5Y2 + 5Y3
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:
B. Constraints:
X i = the amount of money invested in bond i , for (i = 1,2,3,4)
1)
0.06 X 1 + 0.08 X 2 + 0.1X 3 + 0.09 X 4 ≥ (0.08)(1,000 ,000 )
2)
3X1 + 4 X 2 + 7X 3 + 9 X4
≤6
1,000 ,000
3)
X 1 ≤ ( 0.40)(1,000,000)
X 2 ≤ (0.40)(1,000,000)
X 3 ≤ (0.40 )(1,000 ,000 )
X 4 ≤ (0.40)(1,000,000)
4)
X i ≥ 0 for i = 1,2 ,3, 4
C. Objective function:
Z = 0.13 X 1 + 0.08 X 2 + 0.12 X 3 + 0.14 X 4
Maximize
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.
Medium
Cost/spot
TV-Prime time
TV-Nonprime time
Radio
Newspapers and magazines
$100,000
$78,000
$40,000
$20,000
Viewers/spot
(in millions)
All viewers
Youth
6
2.5
4
1.5
2.5
1
1
0.4
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:
X 1 = # of prime time TV spots
X 2 = # of nonprime time TV spots
X 3 = # of radio advertising units
X 4 = # of newspapers and magazines advertising units
B. Constraints:
1) TV advertising cost must be less than 3 million
100,000X 1 + 78,000X 2 ≤ 3,000,000
2) Minimum required number of each advertising units
X1 ≥ 5
X2 ≥ 4
X3 ≥ 6
X4 ≥ 9
3) Total # of youth viewers must be greater than 30 million
2.5 X 1 + 1.5 X 2 + X 3 + 0.4 X 4 ≥ 30
4) Total advertising campaign budget limit
100 ,000 X 1 + 78,000 X 2 + 40,000 X 3 + 20,000 X 4 ≤ 12,000 ,000
5) Non negativity constraints
X i ≥ 0 for i = 1,2 ,3, 4
C. Objective function:
Z = the number of all viewers
Maximize Z = 6 X 1 + 4 X 2 + 2.5 X 3 + X 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 $5. The chicken house can accommodate a maximu m of 3,000 hens, and the size of the barn limits the herd to a maximum
32 cows.
Estimated person-hours and income per acre planted in each of the three corps are as folows
Crop
Winter person-hours
Summer person-hours
Net annual cash income ($)
Soybeans
20
50
600
Corn
35
75
900
Oats
10
40
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:
LS = Amount of acres allocated for soybean production
L c = Amount of acres allocated for corn production
L o = 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
LS + Lc + Lo + 1.5C ≤ 125
2) Total cost of purchasing cows and hens must be less than $40,000
1,200 C + 9 H ≤ 40 ,000
3) Limitations on the size of barn and chicken house
C ≤ 32
H ≤ 3,000
4) Labor limitations
20 LS + 35 L c + 10 L o + 100 C + 0.6 H + W = 3,500 è For winters
50 L S + 75 Lc + 40 Lo + 150 C + 0.9H + S = 4,000 è For summers
4) Non negativity constraints
LS , Lc , Lo , C, H ,W , S ≥ 0
C. Objective function:
Z = total of crop income, income from animals, and income from neighbor farm
Maximize Z = 600 L S + 900 Lc + 450 Lo + 1,000 C + 5H + 5W + 6S
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:
At = Amount of money invested to A in year t (t = 1,2,3,4)
Bt = Amount of money invested to B in year t (t = 1, 2,3)
C 2 = Amount of money invested to C in year 2
D5 = Amount of money invested to C in year 2
Rt = Amount of money not invested in year t (t = 1,2,3,4)
B. Constraints:
1) Equalities of years
A1 + B1 + R1 = 60,000
A2 + B2 + R2 = R1
èYear 1
èYear 2
(Amount of money invested in year 2 +amount of money not invested in year 2 = amount of money left
from year 1)
A3 + B3 + R3 = 1.4 A1 + R 2
èYear 3
A4 + R 4 = 1.4 A2 + 1.7B1 + R3 èYear 4
D5 = 1.4 A3 + 1.7 B 2 + R4
èYear 5
4) Non negativity constraints
At ≥ 0 for t = 1, 2,3,4
Bt ≥ 0 for t = 1,2 ,3
C2 ≥ 0
D5 ≥ 0
Rt ≥ 0 for t = 1, 2,3,4
C. Objective function:
Z = Amount of money accumulated by the beginning of year 6
Maximize
Z = 1.9C2 + 1.7 B3 + 1.4 A4 + 1.3D5