Uploaded on

 

  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Be the first to comment
    Be the first to like this
No Downloads

Views

Total Views
80
On Slideshare
0
From Embeds
0
Number of Embeds
0

Actions

Shares
Downloads
0
Comments
0
Likes
0

Embeds 0

No embeds

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
    No notes for slide

Transcript

  • 1. Chapter 3: Linear Programming Modeling Applications © 2007 Pearson Education
  • 2. Linear Programming (LP) Can Be Used for Many Managerial Decisions: • • • • • • • Product mix Make-buy Media selection Marketing research Portfolio selection Shipping & transportation Multiperiod scheduling
  • 3. For a particular application we begin with the problem scenario and data, then: 1) Define the decision variables 2) Formulate the LP model using the decision variables • • Write the objective function equation Write each of the constraint equations 1) Implement the model in Excel 2) Solve with Excel’s Solver
  • 4. 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 Material Silk Yards available Cost per yard per month $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 Selling Price Polyester Blend 1 Blend 2 $6.70 $3.55 $4.31 $4.81 Monthly Minimum 6,000 10,000 13,000 6,000 Monthly Maximum 7,000 14,000 16,000 8,500 0.125 0.08 0.10 0.10 (per tie) Total material (yards per tie)
  • 7. Material Requirements (yards per tie) Type of Tie Material Silk Blend 1 Polyester (50/50) Blend 2 (30/70) Silk 0.125 0 0 0 Polyester 0 0.08 0.05 0.03 Cotton 0 0 0.05 0.07 0.125 0.08 0.10 0.10 Total yards
  • 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 Profit per tie = (Selling price) – (material cost) –(labor cost) Silk Tie Profit = $6.70 – (0.125 yds)($20/yd) - $0.75 = $3.45 per tie
  • 10. Objective Function (in $ of profit) Max 3.45S + 2.32P + 2.81B1 + 3.25B2 Subject to the constraints: Material Limitations (in yards) 0.125S < 1,000 (silk) 0.08P + 0.05B1 + 0.03B2 < 2,000 (poly) 0.05B1 + 0.07B2 < 1,250 (cotton)
  • 11. 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 file 3-1.xls
  • 12. 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
  • 13. Data Advertising Options Radio Radio TV Spot Newspaper (prime time) (afternoon) Audience Reached (per ad) 5,000 8,500 2,400 2,800 Cost (per ad) $800 $925 $290 $380 Max Ads Per week 12 5 25 20
  • 14. 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
  • 15. Objective Function (in num. audience reached) Max 5000T + 8500N + 2400P + 2800A Subject to the constraints: Budget is $8000 800T + 925N + 290P + 380A < 8000 At Least 5 Radio Spots per Week P+A>5
  • 16. No More Than $1800 per Week for Radio 290P + 380A < 1800 Max Number of Ads per Week T < 12 N< 5 Finally nonnegativity P < 25 A < 20 T, N, P, A > 0 Go to file 3-3.xls
  • 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 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 Investment
  • 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 + 0.10B + 0.19G + 0.12P + 0.08M + 0.14C 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 < 0.25 (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 > 0.30 (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 file 3-5.xls
  • 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. Time Period 9 – 10 10 – 11 11 – 12 Min Num. Tellers 10 12 14 12 – 1 1–2 2-3 3–4 4–5 16 18 17 15 10 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. 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. 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. 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 + P1 > 10 (9-10) F + P 1 + P2 > 12 (10-11) 0.5 F + P1 + P2 + P3 > 14 (11-12) 0.5 F + P1 + P2 + P3+ P4 > 16 (12-1) F + P 2 + P 3+ P 4 + P 5 > 18 (1-2) F + P 3+ P 4 + P 5 > 17 (2-3) F + P4 + P5 > 15 (3-4) F + P5 > 10 (4-5)
  • 32. Only 12 Full Time Tellers Available F < 12 finally nonnegativity: F, P1, P2, P3, P4, P5 > 0 Go to file 3-6.xls
  • 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 3500 $ / 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 3.10W1 + 3.20W2 + 3.45W3 + 4.15W4 + 3.25W5 + 2.75W6 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 0.125W1 + 0.064W2 + 0.144W3 + 0.448W4 + 0.048W5 + 0.018W6 < 1300 Pounds of Each Item Available W1 < 5000 W4 < 3500 W2 < 4500 W5 < 4000 W3 < 3000 W6 < 3500 Finally nonnegativity: Wi > 0, i=1,…,6 Go to file 3-7.xls
  • 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. A $ per pound Grain B C Minimum $0.33 $0.47 $0.38 Daily Requirement 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 0 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. Objective Function (in $ of cost) Min 0.33A + 0.47B + 0.38C Subject to the constraints: Total Blend is 2 Ounces, or 0.125 Pounds A + B + C = 0.125 (lbs)
  • 42. Minimum Nutritional Requirements 22A + 28B + 21C > 3 (protein) 16A + 14B + 25C > 2 (riboflavin) 8A + 7B + 9C > 1 (phosphorus) 5A + 6C > 0.425 (magnesium) Finally nonnegativity: A, B, C > 0 Go to file 3-9.xls
  • 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 month t (t=1,…,4) IAt = inventory of motor A at end of month t (t=1,…,4) IBt = inventory of motor B at end of month t (t=1,…,4)
  • 45. Sales Demand Data Motor A B Month 1 (January) 800 1000 2 (February) 700 1200 3 (March) 1000 1400 4 (April) 1100 1400
  • 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 most be between 2240 and 2560
  • 47. Inventory Data Motor A B Inventory cost $0.18 $0.13 (per motor per month) Beginning inventory (beginning of month 1) 0 0 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 PB2 + 6.6PB3 + 6.6PB4 + 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 IB3 + PB4 – 1400 = IB4 (month 4)
  • 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 + 0.9PB1 < 2560 (month 1) 2240 < 1.3PA2 + 0.9PB2 < 2560 (month 2) 2240 < 1.3PA3 + 0.9PB3 < 2560 (month 3) 2240 < 1.3PA4 + 0.9PB4 < 2560 (month 4) finally nonnegativity: PAi, PBi, IAi, IBi > 0 Go to file 3-11.xls