1. Chapter 3: Linear Programming Modeling Applications
Jason C. H. Chen, Ph.D.
Professor of MIS
School of Business Administration
Gonzaga University
Spokane, WA 99223

2. Linear Programming (LP) Can Be Used for Many Managerial Decisions:
1. Manufacturing applications
Product mix
Make-buy
2. Marketing applications
Media selection
Marketing research
3. Finance application
Portfolio selection
4. Transportation application and others
Shipping & transportation
Multiperiod scheduling

3. For a particular application we begin with
the problem scenario and data, then:
Define the decision variables
Formulate the LP model using the decision variables
Write the objective function equation
Write each of the constraint equations
Implement the model in Excel
Solve with Excel’s Solver

4. Manufacturing Applications Product Mix Problem: Fifth Avenue Industries
Produce 4 types of men's ties
Use 3 materials (limited resources)
Decision: How many of each type of tie to make per month?
Objective: Maximize profit

5. Resource Data Labor cost is $0.75 per tie Material Cost per yard
Yards available
per month
Silk
$20
1,000
Polyester $6
2,000
Cotton $9
1,250
Labor cost is $0.75 per tie

6. Product Data Type of Tie Silk Polyester Blend 1 Blend 2 Selling Price
(per tie)
$6.70
$3.55
$4.31
$4.81
Monthly Minimum
6,000
10,000
13,000
Monthly Maximum
7,000
14,000
16,000
8,500
Total material
(yards per tie)
0.125
0.08
0.10

7. Material Requirements (yards per tie)
Type of Tie
Silk
Polyester
Blend 1
(50/50)
Blend 2
(30/70)
0.125
0.08
0.05
0.03
Cotton
0.07
Total yards
0.125
0.08
0.10

8. Decision Variables S = number of silk ties to make per month
P = number of polyester ties to make per month
B1 = number of poly-cotton blend 1 ties to make per month
B2 = number of poly-cotton blend 2 ties to make per month

9. Profit Per Tie Calculation
(Selling price) – (material cost) –(labor cost)
Silk Tie
Profit = $6.70 – (0.125 yds)($20/yd) - $0.75
= $3.45 per tie

10. Material Limitations (in yards)
Objective Function (in $ of profit)
Max 3.45S P B B2
Subject to the constraints:
Material Limitations (in yards)
0.125S < 1,000 (silk)
0.08P B B2 < 2,000 (poly)
0.05B B2 < 1,250 (cotton)

11. Min and Max Number of Ties to Make
10,000 < P < 14,000
13,000 < B1 < 16,000
6,000 < B2 < 8,500
Finally nonnegativity S, P, B1, B2 > 0

12. LP Model for Product Mix Problem
Max 3.45S P B B2
Subject to the constraints:
0.125S < 1,000 (yards of silk)
0.08P B B2 < 2,000 (yards of poly)
0.05B B2 < 1,250 (yards of cotton)
6,000 < S < 7,000
10,000 < P < 14,000
13,000 < B1 < 16,000
6,000 < B2 < 8,500
S, P, B1, B2 > 0
Go to file 3-1.xls

13. Go to file 3-1.xls Fifth Avenue Industries S P B1 B2 All silk All poly
All silk
All poly
Blend-1
Blend-2
Number of units
7000.0
8500.0
Selling price
$6.70
$3.55
$4.31
$4.81
$192,614.75
Labor cost
$0.75
$31,668.75
Material cost
$2.50
$0.48
$0.81
$40,750.00
Profit
$3.45
$2.32
$2.81
$3.25
$120,196.00
Constraints:
Cost/Yd
Yards of silk
0.125
875.00
<=
1000
$20
Yards of polyester
0.08
0.05
0.03
2000 $6
Yards of cotton
0.07
1250 $9
Maximum all silk 1
7000
Maximum all poly
14000
Maximum blend-1
16000
Maximum blend-2
8500
Minimum all silk
>=
6000
Minimum all poly
10000
Minimum blend-1
13000
Minimum blend-2
LHS
Sign
RHS
Go to file 3-1.xls

14. Marketing applications Media Selection Problem: Win Big Gambling Club
Promote gambling trips to the Bahamas
Budget: $8,000 per week for advertising
Use 4 types of advertising
Decision: How many ads of each type?
Objective: Maximize audience reached

15. Data Audience Reached (per ad) 5,000 8,500 2,400 2,800 Cost $800 $925
Advertising Options
TV Spot
Newspaper
Radio
(prime time)
(afternoon)
Audience
Reached
(per ad)
5,000
8,500
2,400
2,800
Cost
$800
$925
$290
$380
Max Ads
Per week 12 5 25 20

16. Other Restrictions
Have at least 5 radio spots per week
Spend no more than $1800 on radio
Decision Variables
T = number of TV spots per week
N = number of newspaper ads per week
P = number of prime time radio spots per week
A = number of afternoon radio spots per week

17. At Least 5 Radio Spots per Week
Objective Function (in num. audience reached)
Max 5000T N P A
Subject to the constraints:
Budget is $8000
800T + 925N + 290P + 380A < 8000
At Least 5 Radio Spots per Week
P + A > 5

18. No More Than $1800 per Week for Radio 290P + 380A < 1800
Max Number of Ads per Week
T < 12 P < 25
N < 5 A < 20
Finally nonnegativity T, N, P, A > 0

19. LP Model for Media Selection Problem
Objective Function
Max 5000T N P A
Subject to the constraints:
800T + 925N + 290P + 380A < 8000
P + A > 5
290P + 380A < 1800
T < 12
P < 25
N < 5
A < 20
T, N, P, A > 0
Go to file 3-3.xls

20. Prime-time radio spots
Win Big Gambling Club T N P A
TV spots
Newspaper ads
Prime-time radio spots
Afternoon radio spots
Number of units
1.97
5.00
6.21
0.00
Audience
5000
8500
2400
2800
Constraints:
Maximum TV 1
<= 12
Maximum newspaper 5
Max prime-time radio 25
Max afternoon radio 20
Total budget
$800
$925
$290
$380
$8,000.00
$8,000
Maximum radio $
$1,800.00
$1,800
Minimum radio spots
>=
LHS
Sign
RHS
Go to file 3-3.xls

21. Finance application Portfolio Selection: International City Trust
Has $5 million to invest among 6 investments
Decision: How much to invest in each of investment options?
Objective: Maximize interest earned

22. Data Investment Interest Rate Risk Score Trade credits 7% 1.7
Corp. bonds
10%
1.2
Gold stocks
19%
3.7
Platinum stocks
12%
2.4
Mortgage securities 8%
2.0
Construction loans
14%
2.9

23. Constraints Invest up to $ 5 million
No more than 25% into any one investment
At least 30% into precious metals
At least 45% into trade credits and corporate bonds
Limit overall risk to no more than 2.0

24. Decision Variables T = $ invested in trade credit
B = $ invested in corporate bonds
G = $ invested gold stocks
P = $ invested in platinum stocks
M = $ invested in mortgage securities
C = $ invested in construction loans

25. Objective Function (in $ of interest earned)
Max 0.07T B G P
+ 0.08M C
Subject to the constraints:
Invest Up To $5 Million
T + B + G + P + M + C < 5,000,000

26. No More Than 25% Into Any One Investment
T < 0.25 (T + B + G + P + M + C)
B < 0.25 (T + B + G + P + M + C)
G < 0.25 (T + B + G + P + M + C)
P < 0.25 (T + B + G + P + M + C)
M < 0.25 (T + B + G + P + M + C)
C < 0.25 (T + B + G + P + M + C)

27. At Least 30% Into Precious Metals
G + P > 0.30 (T + B + G + P + M + C)
At Least 45% Into
Trade Credits And Corporate Bonds
T + B > 0.45 (T + B + G + P + M + C)

28. Use a weighted average to calculate portfolio risk
Limit Overall Risk To No More Than 2.0
Use a weighted average to calculate portfolio risk
1.7T + 1.2B + 3.7G + 2.4P + 2.0M + 2.9C < 2.0
T + B + G + P + M + C OR
1.7T + 1.2B + 3.7G + 2.4P + 2.0M + 2.9C <
2.0 (T + B + G + P + M + C)
finally nonnegativity: T, B, G, P, M, C > 0

29. LP Model for Portfolio Selection
Max 0.07T B G P+ 0.08M C
Subject to the constraints:
T + B + G + P + M + C < 5,000, (total funds)
T < 0.25 (T + B + G + P + M + C) (Max trade credits)
B < 0.25 (T + B + G + P + M + C) (Max corp bonds)
G < 0.25 (T + B + G + P + M + C) (Max gold)
P < 0.25 (T + B + G + P + M + C) (Max platinum)
M < 0.25 (T + B + G + P + M + C) (Max mortgages)
C < 0.25 (T + B + G + P + M + C) (Max const loans)
1.7T + 1.2B + 3.7G + 2.4P + 2.0M + 2.9C < 2.0 (T + B + G + P + M + C) (Risk score)
G + P > 0.30 (T + B + G + P + M + C) (precious metal)
T + B > 0.45 (T + B + G + P + M + C) (Trade credits & bonds)
T, B, G, P, M, C > 0
Go to file 3-5.xls

30. Go to file 3-5.xls International City Trust T B G P M C Trade credits
Trade credits
Corp bonds
Gold
Platinum
Mortgages
Const loans
Dollars Invested
$1,250,000.00
$250,000.00
$500,000.00
Interest
0.07
0.10
0.19
0.12
0.08
0.14
$520,000.00
Constraints:
Total funds 1
$5,000,000.00
<=
$5,000,000
Max trade credits
$1,250,000
Max corp bonds
Max gold
Max platinum
Max mortgages
Max const loans
Risk score
1.7
1.2
3.7
2.4
2.0
2.9
10,000,000.00
10,000,000
Precious metals
$1,500,000.00
>=
$1,500,000
Trade credits & bonds
$2,500,000.00
$2,250,000
LHS
Sign
RHS
Go to file 3-5.xls

31. Employee Staffing Application Labor Planning: Hong Kong Bank
Number of tellers needed varies by time of day
Decision: How many tellers should begin work at various times of the day?
Objective: Minimize personnel cost

32. Time Period Min Num. Tellers 9 – 10 10 10 – 11 12 11 – 12 14 12 – 1 16
1 – 2 18
2 - 3 17
3 – 4 15
4 – 5
Total minimum daily requirement is 112 hours

33. Full Time Tellers
Work from 9 AM – 5 PM
Take a 1 hour lunch break, half at 11, the other half at noon
Cost $90 per day (salary & benefits)
Currently only 12 are available

34. Part Time Tellers
Work 4 consecutive hours (no lunch break)
Can begin work at 9, 10, 11, noon, or 1
Are paid $7 per hour ($28 per day)
Part time teller hours cannot exceed 50% of the day’s minimum requirement
(50% of 112 hours = 56 hours)

35. Decision Variables
F = num. of full time tellers (all work 9–5)
P1 = num. of part time tellers who work 9–1
P2 = num. of part time tellers who work 10–2
P3 = num. of part time tellers who work 11–3
P4 = num. of part time tellers who work 12–4
P5 = num. of part time tellers who work 1–5

36. Part Time Hours Cannot Exceed 56 Hours
Objective Function (in $ of personnel cost)
Min 90 F + 28 (P1 + P2 + P3 + P4 + P5)
Subject to the constraints:
Part Time Hours Cannot Exceed 56 Hours
4 (P1 + P2 + P3 + P4 + P5) < 56

37. Minimum Num. Tellers Needed By Hour
Time of Day
F + P > 10 (9-10)
F + P1 + P2 > 12 (10-11)
0.5 F + P1 + P2 + P3 > 14 (11-12)
0.5 F + P1 + P2 + P3+ P4 > 16 (12-1)
F + P2 + P3+ P4 + P5 > 18 (1-2)
F + P3+ P4 + P5 > 17 (2-3)
F + P4 + P > 15 (3-4)
F + P > 10 (4-5)

38. Only 12 Full Time Tellers Available
finally nonnegativity: F, P1, P2, P3, P4, P5 > 0

39. LP Model for Labor Planning
Min 90 F + 28 (P1 + P2 + P3 + P4 + P5)
Subject to the constraints:
F + P > 10 (9-10)
F + P1 + P2 > 12 (10-11)
0.5 F + P1 + P2 + P > 14 (11-12)
0.5 F + P1 + P2 + P3+ P > 16 (12-1)
F + P2 + P3+ P4 + P5 > 18 (1-2)
F + P3+ P4 + P5 > 17 (2-3)
F + P4 + P5 > 15 (3-4)
F + P > 10 (4-5)
F < 12
4 (P1 + P2 + P3 + P4 + P5) < 56
F, P1, P2, P3, P4, P5 > 0
Go to file 3-6.xls

40. Go to file 3-6.xls Hong Kong Bank F P1 P2 P3 P4 P5 FT tellers PT @9am
FT tellers
Number of tellers
10.0
0.0
7.0
2.0
5.0
Cost
$90.00
$28.00
$1,292.00
Constraints:
9am-10am needs 1
>= 10
10am-11am needs
17.0 12
11am-Noon needs
0.5
14.0 14
Noon-1pm needs
19.0 16
1pm-2pm needs
24.0 18
2pm-3pm needs 17
3pm-4pm needs
15.0 15
4pm-5pm needs
Max full time
<=
Part-time limit 4
56.0 56
LHS
Sign
RHS
Go to file 3-6.xls

41. Transportation application and others Vehicle Loading: Goodman Shipping
How to load a truck subject to weight and volume limitations
Decision: How much of each of 6 items to load onto a truck?
Objective: Maximize the value shipped

42. Data
Item 1 2 3 4 5 6
Value
$15,500
$14,400
$10,350
$14,525
$13,000
$9,625
Pounds
5000
4500
3000
3500
4000
$ / lb
$3.10
$3.20
$3.45
$4.15
$3.25
$2.75
Cu. ft. per lb
0.125
0.064
0.144
0.448
0.048
0.018

43. Decision Variables
Wi = number of pounds of item i to load onto truck , (where i = 1,…,6)
Truck Capacity
15,000 pounds
1,300 cubic feet

44. Objective Function (in $ of load value)
Max 3.10W W W W W W6
Subject to the constraints:
Weight Limit Of 15,000 Pounds
W1 + W2 + W3 + W4 + W5 + W6 < 15,000

45. Volume Limit Of 1300 Cubic Feet 0.448W4 + 0.048W5 + 0.018W6 < 1300
Pounds of Each Item Available
W1 < 5000 W4 < 3500
W2 < W5 < 4000
W3 < 3000 W6 < 3500
Finally nonnegativity: Wi > 0, i=1,…,6

46. LP Model for Vehicle Loading
Objective Function
Max 3.10W W W W W5+2.75W6
Subject to the constraints:
W1 + W2 + W3 + W4 + W5 + W < 15,000 (Weight Limit)
0.125W W W3
+0.448W W W < 1300 (volume limit of truck)
Pounds of Each Item Available
W < 5000 (item 1 availability)
W < 4500 (item 2 availability)
W < 3000 (item 3 availability)
W < 3500 (item 4 availability)
W < 4000 (item 5 availability)
W < 3500 (item 6 availability)
Wi > 0, i=1,…,6
Go to file 3-7.xls

47. Go to file 3-7.xls Goodman Shipping W1 W2 W3 W4 W5 W6 Item 1 Item 2
Item 1
Item 2
Item 3
Item 4
Item 5
Item 6
Weight in pounds
3,037.38
4,500.00
3,000.00
0.00
4,000.00
462.62
Load value
$3.10
$3.20
$3.45
$4.15
$3.25
$2.75
$48,438.08
Constraints:
Weight limit 1
<=
15000
Volume limit
0.125
0.064
0.144
0.448
0.048
0.018
1300
Item 1 limit (pounds)
5000
Item 2 limit (pounds)
4500
Item 3 limit (pounds)
3000
Item 4 limit (pounds)
3500
Item 5 limit (pounds)
4000
Item 6 limit (pounds)
LHS
Sign
RHS
Go to file 3-7.xls

48. Blending Problem: Whole Food Nutrition Center
Making a natural cereal that satisfies minimum daily nutritional requirements
Decision: How much of each of 3 grains to include in the cereal?
Objective: Minimize cost of a 2 ounce serving of cereal

49. Minimum Daily Requirement
Grain
Minimum Daily Requirement A B C
$ per pound
$0.33
$0.47
$0.38
Protein per pound 22 28 21 3
Riboflavin per pound 16 14 25 2
Phosphorus per pound 8 7 9 1
Magnesium per pound 5 6
0.425

50. Decision Variables A = pounds of grain A to use
B = pounds of grain B to use
C = pounds of grain C to use
Note: grains will be blended to form a 2 ounce serving of cereal

51. Total Blend is 2 Ounces, or 0.125 Pounds
Objective Function (in $ of cost)
Min 0.33A B C
Subject to the constraints:
Total Blend is 2 Ounces, or Pounds
A + B + C = (lbs)

52. Minimum Nutritional Requirements
22A + 28B + 21C > 3 (protein)
16A + 14B + 25C > 2 (riboflavin)
8A + 7B + 9C > 1 (phosphorus)
5A C > (magnesium)
Finally nonnegativity: A, B, C > 0

53. LP Model for a Blending Problem
Objective Function
Min 0.33A B C
Subject to the constraints:
22A + 28B + 21C > 3 (protein units)
16A + 14B + 25C > 2 (riboflavin units)
8A + 7B + 9C > 1 (phosphorus units)
5A + 6C > (magnesium units)
A + B + C = (lbs of total mix)
A, B, C > 0
Go to file 3-9.xls

54. Whole Food Nutrition CenterA B C
Grain A
Grain B
Grain C
Number of pounds
0.025
0.050
Cost
$0.33
$0.47
$0.38
$0.05
Constraints:
Protein 22 28 21
3.00
>= 3
Riboflavin 16 14 25
2.35 2
Phosphorus 8 7 9
1.00 1
Magnesium 5 6
0.425
Total Mix
0.125 =
LHS
Sign
RHS
file 3-9.xls

55. Multiperiod Scheduling: Greenberg Motors
Need to schedule production of 2 electrical motors for each of the next 4 months
Decision: How many of each type of motor to make each month?
Objective: Minimize total production and inventory cost

56. Decision Variables PAt = number of motor A to produce in
month t (t=1,…,4)
PBt = number of motor B to produce in
IAt = inventory of motor A at end of
IBt = inventory of motor B at end of

57. Sales Demand Data Month Motor A B 1 (January) 800 1000 2 (February)
700
1200
3 (March)
1400
4 (April)
1100

58. Production Data Motor (values are per motor) A B Production cost $10$6
Labor hours
1.3
0.9
Production costs will be 10% higher in months 3 and 4
Monthly labor hours most be between
2240 and 2560

59. Max inventory is 3300 motors
Inventory Data
Motor A B
Inventory cost
(per motor per month)
$0.18
$0.13
Beginning inventory
(beginning of month 1)
Ending Inventory
(end of month 4)
450
300
Max inventory is 3300 motors

60. Production and Inventory Balance
(inventory at end of previous period)
+ (production the period)
- (sales this period)
= (inventory at end of this period)

61. Objective Function (in $ of cost) Min 10PA1 + 10PA2 + 11PA3 + 11PA4
+ 6PB1 + 6 PB PB PB4
+ 0.18(IA1 + IA2 + IA3 + IA4)
+ 0.13(IB1 + IB2 + IB3 + IB4)
Subject to the constraints:
(see next slide)

62. Production & Inventory Balance
0 + PA1 – = IA1 (month 1)
0 + PB1 – 1000 = IB1
IA1 + PA2 – = IA2 (month 2)
IB1 + PB2 – 1200 = IB2
IA2 + PA3 – 1000 = IA3 (month 3)
IB2 + PB3 – 1400 = IB3
IA3 + PA4 – 1100 = IA4 (month 4)
IB3 + PB4 – 1400 = IB4

63. Maximum Inventory level
Ending Inventory
IA4 = 450
IB4 = 300
Maximum Inventory level
IA1 + IB1 < 3300 (month 1)
IA2 + IB2 < 3300 (month 2)
IA3 + IB3 < 3300 (month 3)
IA4 + IB4 < 3300 (month 4)

64. 2240 < 1.3PA1 + 0.9PB1 < 2560 (month 1)
Range of Labor Hours
2240 < 1.3PA PB1 < (month 1)
2240 < 1.3PA PB2 < (month 2)
2240 < 1.3PA PB3 < (month 3)
2240 < 1.3PA PB4 < (month 4)
finally nonnegativity: PAi, PBi, IAi, IBi > 0
Go to file 3-11.xls

65. LP model for a Multiperiod Scheduling
Min 10PA1+10PA2+11PA3+11PA4+6PB1+6 PB2+6.6PB3+6.6PB4+0.18(IA1+IA2+IA3+ IA4)+0.13(IB1+IB2+IB3+IB4)
Subject to the constraints:
PA1 – IA = 800 (P&I balance month 1)
PB1 – IB1 = 1000
IA1 + PA2 – IA2 = (month 2)
IB1 + PB2 – IB2 =1200
IA2 + PA3 – IA3 = (month 3)
IB2 + PB3 –IB3 = 1400
IA3 + PA4 – IA4 = (month 4)
IB3 + PB4 – IB4 = 1400
IA4 = 450 (Ending Inventory)
IB4 = 300 (Ending Inventory)
IA1 + IB1 < 3300 (maximal inventory level month 1)
IA2 + IB2 < 3300 (month 2)
IA3 + IB3 < 3300 (month 3)
IA4 + IB4 < 3300 (month 4)
2240 < 1.3PA PB1 < (range of labor hours month 1)
2240 < 1.3PA PB2 < (month 2)
2240 < 1.3PA PB3 < (month 3)
2240 < 1.3PA PB4 < (month 4)
PAi, PBi, IAi, IBi > 0
Go to file 3-11.xls

66. Greenberg Motors
PA1
IA1
PA2
IA2
PA3
IA3
PA4
IA4
PB1
IB1
PB2
IB2
PB3
IB3
PB4
IB4
GM3A Jan prod
GM3A Jan inv
GM3A Feb prod
GM3A Feb inv
GM3A Mar prod
GM3A Mar inv
GM3A Apr prod
GM3A Apr inv
GM3B Jan prod
GM3B Jan inv
GM3B Feb prod
GM3B Feb inv
GM3B Mar prod
GM3B Mar inv
GM3B Apr prod
GM3B Apr inv
Number of Units
1,276.92
476.92
1,138.46
915.38
842.31
757.69
792.31
450.00
1,000.00
0.00
1,200.00
1,400.00
1,700.00
300.00
Cost
$10.00
$0.18
$11.00
$6.00
$0.13
$6.60
$76,301.62
Constraints:
GM3A Jan balance 1 -1
800.00 =
800
GM3B Jan balance
1000
GM3A Feb balance
700.00
700
GM3B Feb balance
1200
GM3A Mar balance
GM3B Mar balance
1400
GM3A Apr balance
1100
GM3B Apr balance
GM3A Apr Inventory
450
GM3B Apr Inventory
300
Jan storage cap
<=
3300
Feb storage cap
Mar storage cap
Apr storage cap
750.00
Jan labor max
1.3
0.9
2560
Feb labor max
Mar labor max
Apr labor max
Jan labor min
>=
2240
Feb labor min
Mar labor min
Apr labor min
LHS
Sign
RHS