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

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

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

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

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

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

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

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

Material Limitations (in yards) 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 10,000 < P < 14,000 13,000 < B1 < 16,000 6,000 < B2 < 8,500 Finally nonnegativity S, P, B1, B2 > 0

LP Model for Product Mix Problem Max 3.45S + 2.32P + 2.81B1 + 3.25B2 Subject to the constraints: 0.125S < 1,000 (yards of silk) 0.08P + 0.05B1 + 0.03B2 < 2,000 (yards of poly) 0.05B1 + 0.07B2 < 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

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 13625.0 13100.0 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 2000.00 2000 $6 Yards of cotton 0.07 1250.00 1250 $9 Maximum all silk 1 7000.00 7000 Maximum all poly 13625.00 14000 Maximum blend-1 13100.00 16000 Maximum blend-2 8500.00 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

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

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

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

At Least 5 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 P < 25 N < 5 A < 20 Finally nonnegativity T, N, P, A > 0

LP Model for Media Selection Problem Objective Function Max 5000T + 8500N + 2400P + 2800A 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

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 67240.30 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

Finance application 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 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

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)

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

LP Model for Portfolio Selection Max 0.07T + 0.10B + 0.19G + 0.12P+ 0.08M + 0.14C Subject to the constraints: T + B + G + P + M + C < 5,000,000 (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

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

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

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

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

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

Minimum Num. Tellers Needed By Hour Time of Day F + P1 > 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 + P5 > 15 (3-4) F + P5 > 10 (4-5)

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

LP Model for Labor Planning Min 90 F + 28 (P1 + P2 + P3 + P4 + P5) Subject to the constraints: F + P1 > 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 + P5 > 15 (3-4) F + P5 > 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

Go to file 3-6.xls Hong Kong Bank F P1 P2 P3 P4 P5 FT tellers PT @9am   FT tellers PT @9am PT @10am PT @11am PT @Noon PT @1pm 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

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

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

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 0.448W4 + 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

LP Model for Vehicle Loading Objective Function Max 3.10W1 +3.20W2 +3.45W3 +4.15W4 +3.25W5+2.75W6 Subject to the constraints: W1 + W2 + W3 + W4 + W5 + W6 < 15,000 (Weight Limit) 0.125W1 + 0.064W2 + 0.144W3 +0.448W4 + 0.048W5 + 0.018W6 < 1300 (volume limit of truck) Pounds of Each Item Available W1 < 5000 (item 1 availability) W2 < 4500 (item 2 availability) W3 < 3000 (item 3 availability) W4 < 3500 (item 4 availability) W5 < 4000 (item 5 availability) W6 < 3500 (item 6 availability) Wi > 0, i=1,…,6 Go to file 3-7.xls

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 15000.00 <= 15000 Volume limit 0.125 0.064 0.144 0.448 0.048 0.018 1300.00 1300 Item 1 limit (pounds) 3037.38 5000 Item 2 limit (pounds) 4500.00 4500 Item 3 limit (pounds) 3000.00 3000 Item 4 limit (pounds) 3500 Item 5 limit (pounds) 4000.00 4000 Item 6 limit (pounds) LHS Sign RHS 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

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

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

Total Blend is 2 Ounces, or 0.125 Pounds 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

LP Model for a Blending Problem Objective Function Min 0.33A + 0.47B + 0.38C Subject to the constraints: 22A + 28B + 21C > 3 (protein units) 16A + 14B + 25C > 2 (riboflavin units) 8A + 7B + 9C > 1 (phosphorus units) 5A + 6C > 0.425 (magnesium units) A + B + C = 0.125 (lbs of total mix) A, B, C > 0 Go to file 3-9.xls

Whole Food Nutrition Center A 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

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 IAt = inventory of motor A at end of IBt = inventory of motor B at end of

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

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

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

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 (month 4) IB3 + PB4 – 1400 = IB4

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)

2240 < 1.3PA1 + 0.9PB1 < 2560 (month 1) 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

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 – IA1- = 800 (P&I balance month 1) PB1 – IB1 = 1000 IA1 + PA2 – IA2 = 700 (month 2) IB1 + PB2 – IB2 =1200 IA2 + PA3 – IA3 = 1000 (month 3) IB2 + PB3 –IB3 = 1400 IA3 + PA4 – IA4 = 1100 (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 < 1.3PA1 + 0.9PB1 < 2560 (range of labor hours 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) PAi, PBi, IAi, IBi > 0 Go to file 3-11.xls

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 1000.00 1000 GM3A Feb balance 700.00 700 GM3B Feb balance 1200.00 1200 GM3A Mar balance GM3B Mar balance 1400.00 1400 GM3A Apr balance 1100.00 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 2560.00 2560 Feb labor max Mar labor max 2355.00 Apr labor max Jan labor min >= 2240 Feb labor min Mar labor min Apr labor min LHS Sign RHS