Chapter 3
Linear Programming Modeling (Selected)
Applications:
With Computer Analyses in Excel

Learning Objectives
1. Model wide variety of linear programming
(LP) problems.
2. Understand major business application areas for LP problems: manufacturing, marketing, labor scheduling, blending, transportation, finance, and multi-period planning.
3. Gain experience in setting up and solving LP problems using Excel’s Solver.

Production Mix Problem
Fifth Avenue Industries
• Nationally known menswear manufacturer.
• Produces four varieties of neckties.
– All-silk tie.
– All-polyester tie.
– Two different polyester and cotton blends.
• Has fixed contracts with major department stores.
– Table 3.1 summarizes contract demand for products.

Material
Silk
Polyestar
Cotton
Production Mix Problem
Cost per yard
$20
$6
$9
Material available per month (yards)
1,000
3,000
1,600
Fifth Avenue uses a standard labor cost of $0.75 per tie (for any variety)

Production Mix Problem

Production Mix Problem
Fifth Avenue Industries
Each all-silk tie requires -
• Cost per tie = 0.125 yards of silk x $20 per yard =
$2.5
• Revenue per tie = $6.70 selling price per silk tie.
• Profit per tie = Revenue per tie - Cost per tie =
$6.70 - $2.5 – 0.75 = $3.45.
Profit for other three products -
• Profit per all-polyester tie = $2.32.
• Profit per Blend - 1 poly-cotton tie = $2.81
• profit per Blend - 2 poly-cotton tie = $3.25

Production Mix Problem
Fifth Avenue Industries
Objective: maximize profit menswear ties.
$3.45 S + $2.32P + $2.81 B
1
+ $3.25 B
2
Where:
S = number of all-silk ties produced per month.
P = number of polyester ties.
B
1
= number of Blend - 1 poly-cotton ties.
B
2
= number of Blend - 2 poly-cotton ties.

Production Mix Problem
Objective: maximize profit =
$3.45S + $2.32 P + $2.81 B
1
+ $3.25 B
2
Subject to:
0.125
S
1

800
0.08
P
2

0.05
B
1

0.03
B
2

3, 000
(Yards of silk)
(Yards of polyester)
0.05
B
1

0.07
B
2

1, 600
S

6, 000
(Yards of cotton)
(Contract minimum for all silk)
S

7, 000
P

10, 000
P

14, 000
B
1

13, 000
B
1

16, 000
B
2

6, 000
B
2

8,500
S P B B
1 2

0
(Contract minimum)
(Contract minimum for all polyester)
(Contract maximum)

0.125
S
1

800
0.08
P
2

0.05
B
1

0.03
B
2

3, 000
0.05
B
1

0.07
B
2

1, 600
S

S

P

10, 000 1
Subject to
P

14, 000
Constraints - Continued
B
1

13, 000
B
1

16, 000
B
2

6, 000
B
2

8, 500
S P B B
1 2

0
+ $4.00 B
2
(Contract minimum Blend 1)
(Contract maximum)
(Contract minimum Blend 2)
(Contract maximum)

Media Selection
Win Big Gambling Club promotes gambling junkets from a large Midwestern city to casinos in the
Bahamas.
• Club has budgeted up to $8,000 per week for local advertising.
– Money is to be allocated among four promotional media:
• TV spots,
• Newspaper ads, and
• Two types of radio advertisements.
• Win Big’s goal - reach largest possible high-potential audience through various media .

Media Selection
• Contract arrangements require at least five radio spots be placed each week.
• Management insists no more than $1,800 be spent on radio advertising each week.

Media Selection
Objective: maximize audience coverage=
5000T + 8500N + 2400P +2800A
T = number of 1-minute TV spots taken each week.
N = number of full-page daily newspaper ads taken each week.
P = number of 30-second prime-time radio spots taken each week.
A = number of 1-minute afternoon radio spots taken each week.

Media Selection
Objective: maximize audience coverage =
5000 T + 8500 N + 2400 P +2800 A
Subject to
T
N

12

5
P

25
A

20
800 T

925 N

290 P

380 A

$8, 000
P A 5
290 P

380 A

$1,800

Media Selection
The optimal solution found to be :
T = 1.97 television spots.
N = 5.00 newspaper ads.
P = 6.21 30-second prime time radio spots.
A = 0.00 1-minute afternoon radio spots.
This produces an audience exposure of 67,240 contacts.

Marketing Research Problem
Management Sciences Associates (MSA) handles consumer surveys. MSA has to determine, for a client, that it must fulfill several requirements in order to draw statistically valid conclusions on sensitive issue of new
U.S. immigration laws:
1 .
Survey at least 2,300 U.S. households.
2.
Survey at least 1,000 households whose heads are 30 years of age or younger.

Marketing Research Problem
3.
Survey at least 600 households whose heads are between 31 and 50 years of age.
4.
Ensure that at least 15% of those surveyed live in a state that borders on Mexico.
5. Ensure that at least 50 % of those surveyed who are 30 years of age or younger live in a state that does not border
Mexico *(missing)
5.
Ensure that no more than 20% of those surveyed who are 51 years of age or over live in a state that borders on Mexico.

Marketing Research Problem
Objective: minimize total interview costs =
$7.50 B
1
+ $6.80 B
2
+ $5.50 B
3
+
$6.90 N
1
+ $7.25 N
2
+ $6.10 N
3
B
1
= number 30 years or younger and live in border state.
B
2
= number 31-50 years and live in border state.
B
3
= number 51 years or older and live in border state.
N
1
= number 30 years or younger and do not live in border state.
N
2
= number 31-50 years and do not live in border state.
N
3
= number 51 years or older and do not live in border state.

Marketing Research Problem
Objective: minimize total interview costs =
$7.50 B
1
+ $6.80 B
2
+ $5.50 B
3
+
$6.90 N
1
+ $7.25 N
2
+ $6.10 N
3
Subject to
B
1

B
2

B
3

N
1

N
2

N
3

B
1

N
1

1, 000
2,300
B
2

N
2

600
B
1

B
2

B
3

B
3

0.2

0.15
B
3

N
3


B
1

B
2

B
3

N
1

N
2

N
3

B B B N N N
1 2 3 1 2 3

0

Marketing Research Problem
B
1
+ B
2
+ B
3

0.15( B
1
+ B
2
+ B
3
+ N
1
+ N
2
+ N
3
)
Rewritten as:
B
1
+ B
2
+ B
3
- 0.15( B
1
+ B
2
+ B
3
+ N
1
+ N
2
+ N
3
)

0
Simplifies to:
0.85
B
1
+ 0.85
B
2
+ 0.85
B
3
- 0.15
N
1
- 0.15
N
2
- 0.15
N
3

0
And
B
3

0.2( B
3
+ N
3
)
Rewritten as:
0.8B
3
- 0.2N
3
< 0

Marketing Research Problem
Optimal solution shows that it costs $15,166 and requires one to survey households as follows:
State borders Mexico and 31-50 years = 600
State borders Mexico and

51 years = 140
State not borders Mexico and

30 years = 1,000
State not borders Mexico and

51 years = 560

Employee Scheduling Application
Hong Kong Bank now employs 12 full-time tellers. Parttime employees (four hours per day) are available. They can start work anytime between 9 A.M. and 1 P.M.
Tellers requirements:

Employee Scheduling Application
Hong Kong Bank
Labor Constraints:
• Full-timers work from 9 A.M. to 5 P.M.
– Allowed 1 hour for lunch.
– Half of full-timers eat at 11 A.M. and other half at noon.
– Full-timers thus provide 35 hours per week of productive labor time.
• Part-time hours limited to a maximum of 50% of day’s total requirement.
Costs:
• Part-timers earn $7 per hour (or $28 per day) on average.
• Full-timers earn $90 per day in salary and benefits, on average.

Employee Scheduling Application
Hong Kong Bank
Decision Variables:
F = number of. full-time tellers
P
1
= part-timers starting at 9 A.M. (leaving at 1 P.M.)
P
2
= part-timers starting at 10 A.M. (leaving at 2 P.M.)
P
3
= part-timers starting at 11 A.M. (leaving at 3 P.M.)
P
4
= part-timers starting at noon (leaving at 4 P.M.)
P
5
= part-timers starting at 1 P.M. (leaving at 5 P.M.)

Employee Scheduling Application
Objective: minimize total daily labor cost
$90 F + $28 ( P
1
+ P
2
+ P
3
+ P
4
+ P
5
)
Subject to
F

P
1

10
F

P
1

P
2

12
0.5
F

P
1

P
2

P
3

15
14
0.5
F

P
1

P
2

P
3

P
4

F

P
2

P
3

P
4

P
5

F

P
3

P
4

P
5

17
18
F

P
4

P
5

F

P
5

10
F

12
16
(9 A.M. - 10 A.M. needs)
(10 A.M. - 11 A.M. needs)
(11 A.M. - noon needs)
(noon - 1 P.M. needs)
(1 P.M. - 2 P.M. needs)
(2 P.M. - 3 P.M. needs)
(3 P.M. - 4 P.M. needs)
(4 P.M. - 5 P.M. needs)
(full-time tellers available)

Employee Scheduling Application
Constraints (Continued):
• Part-time worker hours cannot exceed 50% total hours required each day, which is sum of tellers needed each hour.
4( P
1
P
2
P
3
P
4
P
5
)
       
Simplifying yields,
4 P
1

4 P
2

4 P
3

4 P
4
F P P P P P
1 2 3 4 5

0

4 P
5

0.050(112)

Employee Scheduling Application
• Excel entries for model reveal optimal solution.
– Employ 10 full-time tellers.
– 7 part-time tellers at 10 A.M.
– 2 part-time tellers at 11 A.M.
– 5 part-time tellers at noon.
– Total cost of $1,292 per day.
• There are several alternate optimal solutions.

Employee Scheduling Application
• There are several alternate optimal solutions.
– In practice sequence in which constraints are listed in model may affect specific solution found.
– One alternate solution.
• Employ 10 full-time tellers.
• 6 part-time tellers at 9 A.M.
• 1 part-time teller at 10 A.M.
• 2 part-time teller at 11 A.M.
• 5 part-time tellers at noon.
• Total cost of this policy is also $1,292.

Ingredient Blending Applications
Diet Problems
Diet problem involves specifying a food or food ingredient combination that satisfies stated nutritional requirements at minimum cost.
• Whole Food Nutrition Center uses three bulk grains to blend natural cereal that sells by the pound.
• Each 2-ounce serving of cereal, when taken with 1.2 cup of whole milk, meets an average adult’s minimum daily requirement for protein, riboflavin, phosphorus, and magnesium.

Ingredient Blending Applications
Whole Food Nutrition Center
Diet Problems
• Minimum adult daily requirement:
– Protein 3 units.
– Riboflavin 2 units.
– Phosphorus 1 unit.
– Magnesium 0.425 unit.
• Select blend of grains to meet USRDA at minimum cost.

Ingredient Blending Applications
Decision Variables:
A = pounds of grain in one 2-ounce cereal serving.
B = pounds of grain in one 2-ounce cereal serving.
C = pounds of grain in one 2-ounce cereal serving.

Ingredient Blending Applications
Objective: minimize total cost of mixing 2ounce serving =
$0.33 A + $0.47 B + $0.38 C
Subject to
22 A

28 B

21 C

3
16 A

14 B

25 C

2
8 A

7 B

9 C

1
5 A

0 B

6 C

0.425
(Protein units)
(Riboflavin units)
(Phosphorous units)
(Magnesium units)
, ,

0
0.125
(Total mix 2 ounces or
0.125 pound)

Multi-period Applications
• Most challenging application of LP is modeling multi-period scenarios.
• Situations where decision maker has to determine optimal decisions for several periods (weeks, months, etc.).
• These problems especially difficult because decision choices in later periods are directly dependent on decisions made in earlier periods.

Multi-period Applications
Production Scheduling
Greenberg Motors, Inc., manufactures two different electrical motors for sale under contract to Drexel Corp.
Model GM3A is found in many Drexel food processors, and model GM3B is used in assembly of blenders.
•
Production planning must consider four factors:
– Desirability of producing same number of each motor each month.
– Necessity to reduce inventory carrying, or holding, costs.
– Warehouse limitations cannot be exceeded.
– Company’s no-layoff policy.

Multi-period Applications: Production
Scheduling
Greenberg Motors, Inc.
•
Scheduled Orders.
•
Decision Variables.
P
Ai
= number of model GM3A motors produced in month i (i =1,2,3,4 for January–April).
P
Bi
= number of model GM3B motors produced in month i (i=1,2,3,4 for January–April).

Multi-period Applications: Production
Scheduling
Greenberg Motors, Inc.
• Associated Costs
.
• Production costs now -
– GM3A is $10.
– GM3B is $6.
• Prices will increase in March to $11 and $6.60, respectively.
• Production Cost =
$10 P
A1
+ $10 P
A2
+ $11 P
A3
+ $11 P
A4
+ $6 P
B1
+
$6 P
B2
+ $6.60 P
B3
+ $6.60 P
B4

Multi-period Applications: Production
Scheduling
Greenberg Motors, Inc.
I
Ai
= level of on-hand inventory for GM3A at end of month i (i=1,2,3,4 for January–April)
P
Bi
= level of on-hand inventory for GM3B at end of month i (i=1,2,3,4 for January–April)
• Carrying Cost
– GM3A is $0.18 per month.
– GM3B is $0.13 per month.
Cost of carrying inventory =
$0.18 I
A1
+ $0.18 I
A2
+ $0.18 I
A3
+ $0.18 I
A4
+
$0.13 I
B1
+ $0.13 I
B2
+ $0.13 I
B3
+ $0.13 I
B4

Multi-period Applications: Production
Scheduling
Greenberg Motors, Inc.
Objective: minimize total costs = period production costs
+ period inventory costs =
$10 P
A1
+ $10 P
A2
+ $11 P
A3
+ $11 P
A4
+ $6 P
B1
+ $6 P
B2
+ $6.60 P
B3
+ $6.60 P
B4
+ $0.18 I
A1
+ $0.18 I
A2
+ $0.18 I
A3
+ $0.18 I
A4
+ $0.13 I
B1
+ $0.13 I
B2
+ $0.13 I
B3
+ $0.13 I
B4

Multi-period Applications:Production
Scheduling
• Inventory at end of month is:
• January’s demand for GM3As is 800 and for GM3Bs is
1,000, write relation as:
I
A1
= 0 + P
A1
- 800
I
B1
= 0 + P
B1
- 1000
• Rewrite as:
P
A1
- I
A1
= 800
P
B1
- I
B1
= 1000

Multi-period Applications: Production
Scheduling
Greenberg Motors, Inc.
P
A 2

I
A 1

I
A 2

P
B 2

I
B 1

I
B 2

700
1, 200
P
A 3

I
A 2

I
A 3

1, 000
P
B 3

I
B 2

I
B 3

1, 400
P
A 4

I
A 3

I
A 4

1,100
P
B 4

I
B 3

I
B 4

1, 400
February GM3A demand
February GM3B demand
March GM3A demand
March GM3B demand
April GM3A demand
April GM3B demand
• If Greenberg wants additional 450 GM3As and 300 GM3Bs at end of April, add constraints:
I

A 4
450 and I

B 4
300

Multi-period Applications:Production
Scheduling
• Warehouse space constraints:
I

I

A 1 B 1
I
A 2

I
B 2

3, 300
3, 300
I
A 3

I
B 3

3, 300
I
A 4

I
B 4

3, 300

Multi-period Applications: Production
Scheduling
• Labor Constraints:
1.3
1.3
P
P
A 1
A 1
1.3
P
A 2
1.3
P
A 2
1.3
1.3
P
P
A 3
A 3
1.3
P
A 4
1.3
P
A 4

0.9
P
B 1

0.9
P
B 1

0.9
P
B 2

0.9
P
B 2

0.9
P
B 3

0.9
P
B 3






2, 240
2, 560
2, 240
2, 560
2, 240
2, 560

0.9
P
B 4

2, 240

0.9
P
B 4

2, 560
(January min. hours)
(January max. hours)
(February min. hours)
(February max. hours)
(March min. hours)
(March max. hours)
(April min. hours)
(April max. hours)

Multi-period Applications:Production
Scheduling

• Continued discussion of LP models.
• More experience in formulating and solving problems from variety of disciplines and applications:
– Marketing, manufacturing, employee scheduling,
– Finance, transportation, ingredient blending, and
– Multi-period planning.
• Illustrated setup and solution of models using Excel’s
Solver add-in.