1. Linear Programming Models: Graphical and Computer Methods

2. Using Excel’s Solver for LP
Flair Furniture Example
Other Examples
Examples provided in the Excel LPExamplesFile

3. Using EXCEL Solver Setup the problem Determine the variables
Determine the solution cells
Objective function coefficients
Define the objective function
Constraint coefficiants
Coefficient equations (Left-Hand Side, LHS)
Coefficient signs
Coefficient constraint values (RHS)

4. the Flair Furniture Example:
Max 7T + 5C (profit)
Subject to the constraints:
3T + 4C < 2400 (carpentry hrs)
2T + 1C < 1000 (painting hrs)
C < (max # chairs)
T > (min # tables)
T, C > 0 (nonnegativity)
Go to LPExamplesFile.xls

5. Flair Furniture EXCEL Ex.
See the example in the Excel LPExamplesFile.

6. Linear Programming Modeling Applications

7. Linear Programming (LP) Can Be Used for Many Managerial Decisions:
Product mix
Make-buy
Media selection
Marketing research
Portfolio selection
Labor planning
Shipping & transportation
Blending
Multiperiod scheduling

8. 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

9. 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

10. 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

11. 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

12. 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

13. 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

14. 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

15. 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)

16. Min and Max Number of Ties to Make 6,000 < S < 7,000 10,000 < P < 14,000 13,000 < B1 < 16,000 6,000 < B2 < 8,500 Finally nonnegativity S, P, B1, B2 > 0 Go to Fifth Ave Industries in LPExamplesFile

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

18. 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

19. 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

20. 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

21. 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

22. No More Than 25% Into Any One Investment T < 0
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)

23. At Least 30% Into Precious Metals G + P > 0
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)

24. 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
Go to International City in LPExamplesFile

25. 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

26. Total minimum daily requirement is 112 hours
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

27. 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

28. Work 4 consecutive hours (no lunch break)
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)

29. Decision Variables F = num
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

30. 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

31. 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)

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

33. 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

34. 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

35. 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

36. 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

37. 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
Go to Goodman Shipping in LPExamplesFile

38. 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

39. 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

40. 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

41. 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)

42. 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
Go to Whole Foods in LPExamplesFile

43. 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

44. 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

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

46. 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 must be between
2240 and 2560

47. 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

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

49. 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)

50. Production & Inventory Balance 0 + PA1 – 800 = IA1 (month 1) 0 + PB1 – 1000 = IB1 IA1 + PA2 – 700 = 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

51. 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)

52. Range of Labor Hours 2240 < 1. 3PA1 + 0
Range of Labor Hours 2240 < 1.3PA PB1 < 2560 (month 1) 2240 < 1.3PA PB2 < 2560 (month 2) 2240 < 1.3PA PB3 < 2560 (month 3) 2240 < 1.3PA PB4 < 2560 (month 4) finally nonnegativity: PAi, PBi, IAi, IBi > 0 Try to do this one on your own.