CCE 6100 HW4

WESTERN MICHIGAN UNIVERSITY CCE-6100-100 - Civil Systems Analysis HW.4 Done by: Aws Tarawneh

5. Kilgore’s Deli is a small delicatessen located near a major university. Kilgore does a large walk-in carry-out lunch business. The deli offers two luncheon chili specials, Wimpy and Dial 911. At the beginning of the day, Kilgore needs to decide how much of each special to make (he always sells out of whatever he makes). The profit on one serving of Wimpy is $0.45, on one serving of Dial 911, $0.58. Each serving of Wimpy requires 0.25 pound of beef, 0.25 cup of onions, and 5 ounces of Kilgore’s special sauce. Each serving of Dial 911 requires 0.25 pound of beef, 0.4 cup of onions, 2 ounces of Kilgore’s special sauce, and 5 ounces of hot sauce. Today, Kilgore has 20 pounds of beef, 15 cups of onions, 88 ounces of Kilgore’s special sauce, and 60 ounces of hot sauce on hand. A. Develop an LP model that will tell Kilgore how many servings of Wimpy and Dial 911 to make in order to maximize his profit today. W = Wimpy chili. D= Dial 911 chili. Max 0.45W + 0.58 D s.t. 0.25 W+0.25D<=20 0.25W+0.4D<=15 5W+2D<=88 5D<=60 D,W>0 Global optimal solution found. Objective value: 12.72000 Infeasibilities: 0.000000 Total solver iterations: 1 Model Class: LP Total variables: 2 Nonlinear variables: 0 Integer variables: 0 Total constraints: 5 Nonlinear constraints: 0 Total nonzeros: 9 Nonlinear nonzeros: 0 Variable Value Reduced Cost W 12.80000 0.000000 D 12.00000 0.000000 Row Slack or Surplus Dual Price 1 12.72000 1.000000 2 13.80000 0.000000 3 7.000000 0.000000 4 0.000000 0.9000000E-01 5 0.000000 0.8000000E-01

Min 12x11 + 13x12 + 14x13 + 10x21 + 11x22 + 10x23 s.t. x11 + x21 <= 600 x12 + x22 <= 1000 x13 + x23 <= 800 x11 + x12 + x13 = 1000 x21 + x22 + x23 = 800 x11>= 0 x12>= 0 x13>= 0 x21>= 0 x22>= 0 x23 >= 0 Global optimal solution found. Objective value: 20400.00 Infeasibilities: 0.000000 Total solver iterations: 4 Model Class: LP Total variables: 6 Nonlinear variables: 0 Integer variables: 0 Total constraints: 12 Nonlinear constraints: 0 Total nonzeros: 24 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 7 600.0000 0.000000 8 400.0000 0.000000 9 0.000000 0.000000 10 0.000000 0.000000 11 0.000000 0.000000 12 800.0000 0.000000 Component 1 will be bought from supplier 1 & 2 600 & 400 ,respectively Component 2 will be bought from supplier 3 X23=800 Purchase Cost = $20,400

17. MF= Manufactured Frame PF=Purchased Frame MS= Manufactured Support PS= Purchased Support MT= Manufactured Strap PT= Purchased Strap To produce one product we need 1 frame , 2 supports, 1 strap min 38MF+51PF+11.5MS+15PS+6.5MT+7.5PT s.t. MF+PF>=5000 MS+PS>=10000 MT+PT>=5000 3.5MF+1.3MS+0.8MT<=21000 2.2MF+1.7MS<=25200 3.1MF+2.6MS+1.7MT<=40800 Global optimal solution found. Objective value: 368076.9 Infeasibilities: 0.000000 Total solver iterations: 6 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 MF 5000.000 0.000000 PF 0.000000 3.576923 MS 2692.308 0.000000 PS 7307.692 0.000000 MT 0.000000 1.153846 PT 5000.000 0.000000 Row Slack or Surplus Dual Price 1 368076.9 -1.000000 2 0.000000 -47.42308 3 0.000000 -15.00000 4 0.000000 -7.500000 5 0.000000 2.692308 6 9623.077 0.000000 7 18300.00 0.000000

Variable Value Reduced Cost MF 2285.714 0.000000 PF 2714.286 0.000000 MS 10000.00 0.000000 PS 0.000000 0.9000000 MT 0.000000 0.6000000 PT 5000.000 0.000000 Row Slack or Surplus Dual Price 1 361500.0 -1.000000 2 0.000000 -45.00000 3 0.000000 -14.10000 4 0.000000 -7.500000 5 0.000000 2.000000 6 3171.429 0.000000 7 7714.286 0.000000 The purchased frames will be 2714.28 instead of 0 21. variables Bookshelf regular & overtime hours 1 st & 2 nd month production Floor regular & overtime hours 1 st & 2 nd month production BR1= Bookshelf regular hours 1 st month BR2= Bookshelf regular hours 2 nd month BO1= Bookshelf overtime hours 1st month BO2= Bookshelf overtime hours 2nd month FR1= Floor regular hours 1 st month FR2= Floor regular hours 2 nd month FO1= Floor overtime hours 1st month FO2= Floor overtime hours 2nd month Labor costs per unit BR 1&2=0.7*22=15.4 BO 1&2=0.7*33=23.1 FR 1&2=1*22=22 FO 1&2=1*33=33 Storage costs IB= inventory left from 1 st month for bookshelf units IF= inventory left from 1st month for floor units Raw material costs Bookshelf =$10 Floor=$12 Bookshelf & floor Costs = labor + raw material costs

Objective function Min 25.4BR1+25.4BR2+34FR1+34FR2+33.1BO1+33.1BO2+45FO1+45FO2+5IB+5IF s.t. 0.7BR1+1FR1 <= 2400 0.7BR2+1FR2 <= 2400 0.7 BO1+FO1<=1000 0.7 BO2+FO2<=1000 BR1+BO1-IB=2100 IB+BR2+BO2=1200 FR1+FO1-IF=1500 IF+FR2+FO2=2600 Global optimal solution found. Objective value: 241130.0 Infeasibilities: 0.000000 Total solver iterations: 8 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 BR1 2100.000 0.000000 BR2 1200.000 0.000000 FR1 930.0000 0.000000 FR2 1560.000 0.000000 BO1 0.000000 0.000000 BO2 0.000000 0.000000 FO1 610.0000 0.000000 FO2 1000.000 0.000000 IB 0.000000 1.500000 IF 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

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 The optimal solution schedule calls for 8 starting at 11:00 AM 1 starting at 12:00 PM 1 starting at 2:00 PM 4 starting at 4:00 PM 6 starting at 6:00 PM b. What is the total payroll for the part-time employees? How many part-time shifts are needed? Use the surplus variables to comment on the desirability of scheduling at least some of the part- time employees for 3-hour shifts 16 part time shifts Total daily salary cost = $608 There are 9 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. Assume that part-time employees can be assigned either a 3-hour or a 4-hour shift. Develop a minimum-cost schedule for the part-time employees. How many part-time shifts are needed, and what is the cost savings compared to the previous schedule? Considering 3-hour shifts Let x denote 4-hour shifts and y denote 3-hour shifts where 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

Min 30.4x1 + 30.4x2 + 30.4x3 + 30.4x4 + 30.4x5 + 30.4x6 + 30.4x7 + 30.4x8+22.80y1 +22.80y2 +22.80y3 +22.80y4 +22.80y5 +22.80y6 +22.80y7 +22.80y8 +22.80y9 S.t. x1+y1 ≥ 8 11:00 a.m. 1 full time employee x1 + x2+y1+y2 ≥ 8 12:00 p.m. 1 full time employee x1 + x2 + x3 +y1+y2+y3≥ 7 1:00 p.m. 2 full time employee x1 + x2 + x3 + x4 +y2+y3+y4≥ 1 2:00 p.m. 2 full time employee x2 + x3 + x4 + x5 +y3+y4+y5≥ 2 3:00 p.m. 1 full time employee x3 + x4 + x5 + x6 +y4+y5+y6≥ 1 4:00 p.m. 2 full time employee x4 + x5 + x6 + x7 +y5+y6+y7≥ 5 5:00 p.m. 1 full time employee x5 + x6 + x7 + x8+y6+y7+y8 ≥ 10 6:00 p.m. 2 full time employee x6 + x7 + x8+y7+y8+y9 ≥ 10 7:00 p.m. 2 full time employee x7 + x8 +y8+y9≥ 6 8:00 p.m. 1 full time employee x8+y9 ≥ 6 9:00 p.m. 1 full time employee Global optimal solution found. Objective value: 501.6000 Infeasibilities: 0.000000 Total solver iterations: 9 Model Class: LP Total variables: 17 Nonlinear variables: 0 Integer variables: 0 Total constraints: 12 Nonlinear constraints: 0 Total nonzeros: 76 Nonlinear nonzeros: 0 Variable Value Reduced Cost X1 0.000000 0.000000 X2 0.000000 7.600000 X3 0.000000 7.600000 X4 0.000000 0.000000 X5 0.000000 0.000000 X6 0.000000 7.600000 X7 0.000000 7.600000 X8 6.000000 0.000000 Y1 8.000000 0.000000 Y2 0.000000 15.20000 Y3 1.000000 0.000000 Y4 0.000000 0.000000 Y5 1.000000 0.000000 Y6 0.000000 7.600000 Y7 4.000000 0.000000 Y8 0.000000 7.600000 Y9 0.000000 0.000000 Row Slack or Surplus Dual Price 1 501.6000 -1.000000 2 0.000000 -22.80000 3 0.000000 0.000000 4 2.000000 0.000000 5 0.000000 -7.600000 6 0.000000 -15.20000

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 Total 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 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 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

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

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 - - 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.