• Share
  • Email
  • Embed
  • Like
  • Save
  • Private Content
Chapter 3
 

Chapter 3

on

  • 185 views

 

Statistics

Views

Total Views
185
Views on SlideShare
185
Embed Views
0

Actions

Likes
0
Downloads
0
Comments
0

0 Embeds 0

No embeds

Accessibility

Upload Details

Uploaded via as Microsoft PowerPoint

Usage Rights

© All Rights Reserved

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Processing…
Post Comment
Edit your comment

    Chapter 3 Chapter 3 Presentation Transcript

    • Chapter 3: Linear Programming Modeling Applications © 2007 Pearson Education
    • Linear Programming (LP) Can Be Used for Many Managerial Decisions: • • • • • • • Product mix Make-buy Media selection Marketing research Portfolio selection Shipping & transportation Multiperiod scheduling
    • 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
    • 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
    • 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
    • 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)
    • 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
    • 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
    • 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
    • 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)
    • 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
    • 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
    • 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
    • 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
    • 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
    • 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
    • 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
    • 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
    • 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
    • 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
    • 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
    • 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)
    • 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)
    • 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
    • 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
    • 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
    • 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
    • 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)
    • 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
    • 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
    • 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)
    • Only 12 Full Time Tellers Available F < 12 finally nonnegativity: F, P1, P2, P3, P4, P5 > 0 Go to file 3-6.xls
    • 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
    • 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
    • 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
    • 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
    • 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
    • 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
    • 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
    • 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
    • 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)
    • 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
    • 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
    • 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)
    • Sales Demand Data Motor A B Month 1 (January) 800 1000 2 (February) 700 1200 3 (March) 1000 1400 4 (April) 1100 1400
    • 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
    • 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
    • Production and Inventory Balance (inventory at end of previous period) + (production the period) - (sales this period) = (inventory at end of this period)
    • 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)
    • 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)
    • 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)
    • 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