Optimal Solutions for LP Problems: Cost Minimization Strategies

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

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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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Q 25. a. From the information of question assume: x1 = number of part-time employees beginning at 11:00 a.m. x2 = number of part-time employees beginning at 12:00 p.m. x3 = number of part-time employees beginning at 1:00 p.m. x4 = number of part-time employees beginning at 2:00 p.m. x5 = number of part-time employees beginning at 3:00 p.m. x6 = number of part-time employees beginning at 4:00 p.m. x7 = number of part-time employees beginning at 5:00 p.m. x8 = number of part-time employees beginning at 6:00 p.m. Each part-time employee assigned to a four-hour shift will be paid $7.60 (4 hours) = $30.40. Min 30.4x1 + 30.4x2 + 30.4x3 + 30.4x4 + 30.4x5 + 30.4x6 + 30.4x7 + 30.4x8 Part-Time Employees Needed S.t. x1 ≥ 8 11:00 a.m. x1 + x2 ≥ 8 12:00 p.m. x1 + x2 + x3 ≥ 7 1:00 p.m. x1 + x2 + x3 + x4 ≥ 1 2:00 p.m. x2 + x3 + x4 + x5 ≥ 2 3:00 p.m. x3 + x4 + x5 + x6 ≥ 1 4:00 p.m. x4 + x5 + x6 + x7 ≥ 5 5:00 p.m. x5 + x6 + x7 + x8 ≥ 10 6:00 p.m. x6 + x7 + x8 ≥ 10 7:00 p.m. x7 + x8 ≥ 6 8:00 p.m. x8 ≥ 6 9:00 p.m. xj ≥ 0 j = 1,2,...8 Full-time employees reduce the number of part-time employees needed. The optimal schedule calls for: 8 starting at 11:00 a.m. 2 starting at 3:00 p.m. 4 starting at 5:00 p.m. 6 starting at 6:00 p.m. b. Total daily salary cost = $608 There are 7 surplus employees scheduled from 2:00 - 3:00 p.m. and 4 from 9:00 - 10:00 p.m. Suggesting the desirability of rotating employees off sooner. c. Considering 3-hour shifts Let x denote 4-hour shifts and y denote 3-hour shifts where

y1 = number of part-time employees beginning at 11:00 a.m. y2 = number of part-time employees beginning at 12:00 p.m. y3 = number of part-time employees beginning at 1:00 p.m. y4 = number of part-time employees beginning at 2:00 p.m. y5 = number of part-time employees beginning at 3:00 p.m. y6 = number of part-time employees beginning at 4:00 p.m. y7 = number of part-time employees beginning at 5:00 p.m. y8 = number of part-time employees beginning at 6:00 p.m. y9 = number of part-time employees beginning at 7:00 p.m. Each part-time employee assigned to a three-hour shift will be paid $7.60 (3 hours) = $22.80 New objective function: Min ∑ j = 1 8 30.40 xj + Min ∑ i = 1 9 22.80 y i Each constraint must be modified with the addition of the yi variables. For instance, the first constraint becomes. x1 + y1 ≥ 8 And so on. Each yi appears in three constraints because each refers to a three hour shift. The optimal. Solution is shown below. x8 = 6 y1 = 8 y3 = 1 y5 = 1 y7 = 4 Optimal schedule for part-time employees: 4-Hour Shifts 3-Hour Shifts X 8 = 6 y1 = 8 y3 = 1 y5 = 1 y7 = 4 Total cost reduced to $501.60. Still have 20 part-time shifts, but 14 are 3-hour shifts. The surplus has been reduced by a total of 14 hours. Global optimal solution found. Objective value: 608.0000 Infeasibilities: 0.000000 Total solver iterations: 3 Elapsed runtime seconds: 0.13

Model Class: LP Total variables: 8 Nonlinear variables: 0 Integer variables: 0 Total constraints: 12 Nonlinear constraints: 0 Total nonzeros: 40 Nonlinear nonzeros: 0 Variable Value Reduced Cost X1 8.000000 0.000000 X2 1.000000 0.000000 X3 0.000000 0.000000 X4 1.000000 0.000000 X5 0.000000 0.000000 X6 4.000000 0.000000 X7 0.000000 0.000000 X8 6.000000 0.000000 Row Slack or Surplus Dual Price 1 608.0000 -1.000000 2 0.000000 -30.40000 3 1.000000 0.000000 4 2.000000 0.000000 5 9.000000 0.000000 6 0.000000 -30.40000 7 4.000000 0.000000 8 0.000000 0.000000 9 0.000000 0.000000 10 0.000000 -30.40000 11 0.000000 0.000000 12 0.000000 0.000000

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Case Problem: scheduling Workforce 1. Assume that: tij = number of temporary employees hired under option i (i = 1, 2, 3) in month j (j = 1 for January, j = 2 for February and so on). The following table depicts the decision variables used in this case problem. Option Jan Feb Mar. Apr. May June Option 1 t11 t12 t13 t14 t15 t16 Option 2 t21 t22 t23 t24 t25 - Option 3 t31 t32 t33 t34 - - Costs: Contract cost plus training cost Option Contract Cost Training Cost Training Cost Option 1 $2000 $875 $2875 Option 2 $4800 $875 $5675 Option 3 $7500 $875 $8375 Min. 2875 (t11 + t12 + t13 + t14 + t15 + t16) + 5675 (t21 + t22 + t23 + t24 + t25) + 8375 (t31 + t32 + t33 + t34) One constraint is required for each of the six months. Constraint 1: Need 10 additional employees in January Global optimal solution found. Objective value: 313525.0 Infeasibilities: 0.000000 Total solver iterations: 8 Elapsed runtime seconds: 0.15 Model Class: LP Total variables: 15 Nonlinear variables: 0 Integer variables: 0 Total constraints: 7 Nonlinear constraints: 0 t11 = number of temporary employees hired under Option 1 (one-month contract) in January t21 = number of temporary employees hired under Option 2 (two-month contract) in January t31 = number of temporary employees hired under Option 3 (three-month contract) in January

Total nonzeros: 43 Nonlinear nonzeros: 0 Variable Value Reduced Cost T11 0.000000 75.00000 T12 1.000000 0.000000 T13 0.000000 175.0000 T14 0.000000 75.00000 T15 6.000000 0.000000 T16 0.000000 175.0000 T21 3.000000 0.000000 T22 0.000000 100.0000 T23 0.000000 175.0000 T24 0.000000 0.000000 T25 0.000000 100.0000 T31 7.000000 0.000000 T32 12.00000 0.000000 T33 0.000000 0.000000 T34 14.00000 0.000000 Row Slack or Surplus Dual Price 1 313525.0 -1.000000 2 0.000000 -2800.000 3 0.000000 -2875.000 4 0.000000 -2700.000 5 0.000000 -2800.000 6 0.000000 -2875.000 7 0.000000 -2700.000 t11 + t21 + t31 = 10 Constraint 2: Need 23 additional employees in February t12, t22 and t32 are the number of temporary employees hired under Options 1, 2 and 3 in February. But, temporary employees hired under Option 2 or Option 3 in January will also be available to satisfy February needs. t21 + t31 + t12 + t22 + t32 = 23 Note: The following table shows the decision variables used in this constraint Option Jan Feb Mar. Apr. May June Option 1 - t12 - - - - Option 2 t21 t22 - - - - Option 3 t31 t32 - - - -

Constraint 3: Need 19 additional employees in March Option Jan Feb Mar. Apr. May June Option 1 - - t13 - - - Option 2 - t22 t23 - - - Option 3 t31 t32 t33 - - - t 31 + t 22 + t 32 + t 13 + t 23 + t 33 = 19 Constraint 4: Need 26 additional employees in April Option Jan Feb Mar. Apr. May June Option 1 - - - t14 - - Option 2 - - t23 t24 - - Option 3 - t32 t33 t34 - - t 32 + t 23 + t 33 + t 14 + t 24 + t 34 = 26 Constraint 5: Need 20 additional employees in May Option Jan Feb Mar. Apr. May June Option 1 - - - - t 15 - Option 2 - - - t 24 t 25 - Option 3 - - t33 t34 - - t 33 + t 24 + t 34 + t 15 + t 25 = 20. Constraint 6: Need 14 additional employees in June Option Jan Feb Mar. Apr. May June Option 1 - - - - - t 16 Option 2 - - - - t 25 - Option 3 - - - t34 - - t 34 + t 25 + t 16 = 14 Optimal Solution: Total Cost = $313,525 Option Jan Feb Mar. Apr. May June Option 1 0 1 0 0 6 0 Option 2 3 0 0 0 0 - Option 3 7 12 0 14 - -

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2. Option Number Hired Contract Cost Training Cost Total Cost Option Number Hired Contract Cost Training Cost Total Cost Option 1 7 $14,000 $6,125 $20,125 Option 2 3 $14,400 $2,625 $17,025 Option 3 33 $247,500 $28,875 $276,375 Total: $275,900 $37,625 $313,525 3. Hiring 10 full-time employees at the beginning of January will reduce the number of temporary employees needed each month by 10. Using the same linear programming model with the right-hand sides of 0, 13, 9, 16, 10 and 4, provides the following schedule for temporary employees: Option Jan Feb Mar. Apr. May June Option 1 0 4 0 0 3 0 Option 2 0 0 0 3 0 - Option 3 0 9 0 4 - - Option Number Hired Contract Cost Training Cost Total Cost Option Number Hired Contract Cost Training Cost Total Cost Option 1 7 $14,000 $6,125 $20,125 Option 2 3 $14,400 $2,625 $17,025 Option 3 13 $97,500 $11,375 $108,875 Total: 23 $146,025 Full-time employees cost: Training cost: 10($875) = $8,750 Salary: 10(6) (168) ($16.50) = $166,320 Total Cost = $146,025 + $8750 + $166,320 = $321,095 Hiring 10 full-time employees is $321,095 - $313,525 = $7,570 more expensive than using temporary employees. Do not hire the 10 full-time employees. Davis should continue to contract with Workforce to obtain temporary employees. 4. With the lower training costs, the costs per employee for each option are as follows: Option Cost Training Cost Total Cost Option 1 $2000 $700 $2700

Option 2 $4800 $700 $5500 Option 3 $7500 $700 $8200 Resolving the original linear programming model with the above costs indicates that Davis should hire all temporary employees on a one-month contract specifically to meet each month's employee needs. Thus, the monthly temporary hire schedule would be as follows: January - 10; February - 23; March - 19; April - 26; May - 20; and June - 14. The total cost of this strategy is $302,400. Note that if training costs were any lower, this would still be the optimal hiring strategy for Davis.

HW4

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