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Q 11 . From the information of question: Let xij = units of component i purchased from supplier j Min 12x11 + 13x12 + 14x13 + 10x21 + 11x22 + 10x23 S.t. x11 + x12 + x13 = 1000 x21 + x22 + x23 = 800 x11 + x21 ≤ 600 x12 + x22 ≤ 1000 x13 + x23 ≤ 800 x11, x12, x13, x21, x22, x23 ≥ 0 By Lingo program: Global optimal solution found. Objective value: 20400.00 Infeasibilities: 0.000000 Total solver iterations: 4 Elapsed runtime seconds: 14.29 Model Class: LP Total variables: 6 Nonlinear variables: 0 Integer variables: 0 Total constraints: 6 Nonlinear constraints: 0 Total nonzeros: 18 Nonlinear nonzeros: 0 Variable Value Reduced Cost X11 600.0000 0.000000 X12 400.0000 0.000000 X13 0.000000 1.000000 X21 0.000000 1.000000 X22 0.000000 1.000000 X23 800.0000 0.000000 Row Slack or Surplus Dual Price 1 20400.00 -1.000000 2 0.000000 -13.00000 3 0.000000 -10.00000 4 0.000000 1.000000 5 600.0000 0.000000 6 0.000000 0.000000 1 2 3 Component 1 600 400 0 Component 2 0 0 800 Purchase Cost = $20,400

Q 17 a). from the information of question: Min 38FM + 51FP + 11.5SM + 15SP + 6.5TM + 7.5TP S.t. 3.5FM + 1.3SM + 0.8TM ≤ 21,000=350*60 2.2FM + 1.7SM ≤ 25,200=420*60 3.1FM + 2.6SM + 1.7TM ≤ 40,800=680*60 FM + FP ≥ 5,000 SM + SP ≥ 10,000 TM + TP ≥ 5,000 FM, FP, SM, SP, TM, TP ≥ 0. By Lingo program: Global optimal solution found. Objective value: 368076.9 Infeasibilities: 0.000000 Total solver iterations: 6 Elapsed runtime seconds: 0.13 Model Class: LP Total variables: 6 Nonlinear variables: 0 Integer variables: 0 Total constraints: 7 Nonlinear constraints: 0 Total nonzeros: 20 Nonlinear nonzeros: 0 Variable Value Reduced Cost FM 5000.000 0.000000 FP 0.000000 3.576923 SM 2692.308 0.000000 SP 7307.692 0.000000 TM 0.000000 1.153846 TP 5000.000 0.000000 Row Slack or Surplus Dual Price 1 368076.9 -1.000000 2 0.000000 2.692308 3 9623.077 0.000000 4 18300.00 0.000000 5 0.000000 -47.42308 6 0.000000 -15.00000 7 0.000000 -7.500000 FM = number of frames manufactured FP = number of frames purchased SM = number of supports manufactured SP = number of supports purchased TM = number of straps manufactured TP = number of straps purchased

Manufacture Purchase Frames 5000 0 Support 2692 7308 Strap 0 5000 b) Total Cost = 368,076.91$. c) Subtract values of slack variables from minutes available to determine minutes used. Divide by 60 to determine hours of production time used. Constraint 1 Cutting Slack = 0 350 hours used 2 Milling (25200 - 9623) / 60 = 259.62 hours 3 Shaping (40800 - 18300) / 60 = 375 hours d) Nothing, there are already more hours available than are being used. e) Yes. The current purchase price is $51.00 and the reduced cost of 3.577 indicates that for a purchase price below $47.423 the solution may improve. Resolving with the coefficient of FP = 45 shows that 2714 frames should be purchased. The optimal solution is as follows: OPTIMAL SOLUTION Objective Function Value = 361500.000 Variable Value Reduced Costs -------------- --------------- ------------------ FM 2285.714 0.000 FP 2714.286 0.000 SM 10000.000 0.000 SP 0.000 0.900 TM 0.000 0.600 TP 5000.000 0.000By Lingo program Constraint Slack/Surplus Dual Prices -------------- --------------- ------------------ 1 0.000 2.000 2 3171.429 0.000 3 7714.286 0.000 4 0.000 -45.000 5 0.000 -14.100 6 0.000 -7.500

Q 21 Decision variables: Regular Model Month 1 Month 2 Bookshelf B1R B2R Floor F1R F2R Decision variables: Overtime Model Month 1 Month 2 Bookshelf B1O B2O Floor F1O F2O Labor costs per unit Model Regular $ Overtime $ Bookshelf .7 (22) = 15.40 .7 (33) = 23.10 Floor 1(22) = 22 1 (33) = 33 Linear Programming Applications in Marketing, Finance and Operations Management IB = Month 1 ending inventory for bookshelf units. IF = Month 1 ending inventory for floor model. Objective function Min 15.40 B1R + 15.40 B2R + 22 F1R + 22 F2R + 23.10 B1O + 23.10 B2O + 33 F1O + 33 F2O + 10 B1R + 10 B2R + 12 F1R + 12 F2R + 10 B1O + 10 B2O + 12 F1O + 12 F2O + 5 IB + 5 IF Or (15.4+10) (22+12) (23.10+10) (33+12) Min 25.40 B1R + 25.40 B2R + 34 F1R + 34 F2R + 33.10 B1O + 33.10 B2O + 45 F1O + 45 F2O + 5 IB + 5 IF s.t. .7 B1R + 1 F1R ≤ 2400 Regular time: month 1 .7 B2R + 1 F2R ≤ 2400 Regular time: month 2 .7B1O + 1 F1O ≤ 1000 Overtime: month 1 .7B2O + 1 F2O ≤ 1000 Overtime: month 2 B1R + B1O - IB = 2100 Bookshelf: month 1 IB + B2R + B2O = 1200 Bookshelf: month 2 F1R + F1O - IF = 1500 Floor: month 1 IF + F2R + F2O = 2600 Floor: month 2 Global optimal solution found. Objective value: 241130.0 Infeasibilities: 0.000000 Total solver iterations: 8 Elapsed runtime seconds: 0.20

Model Class: LP Total variables: 10 Nonlinear variables: 0 Integer variables: 0 Total constraints: 9 Nonlinear constraints: 0 Total nonzeros: 30 Nonlinear nonzeros: 0 Variable Value Reduced Cost B1R 2100.000 0.000000 B2R 1200.000 0.000000 F1R 930.0000 0.000000 F2R 1560.000 0.000000 B1O 0.000000 0.000000 B2O 0.000000 0.000000 F1O 610.0000 0.000000 F2O 1000.000 0.000000 B1 0.000000 1.500000 F1 40.00000 0.000000 Row Slack or Surplus Dual Price 1 241130.0 -1.000000 2 0.000000 11.00000 3 0.000000 16.00000 4 390.0000 0.000000 5 0.000000 5.000000 6 0.000000 -33.10000 7 0.000000 -36.60000 8 0.000000 -45.00000 9 0.000000 -50.00000

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