091 Chp4 Case Solutions

CASE SOLUTION: SUMMER SPORTS CAMP AT STATE UNIVERSITY

Model Formulation: wi = new sheets purchased for week i (i = 1,2,...8) xi = sheets cleaned at laundry at end of week i yi = sheets cleaned by Mary’s friends at end of week i minimize Z = 10(w1 + w2 + w3 + w4 + w5 + w6 + w7 + w8) + 4(x1 + x2 + x3 + x4 + x5 + x6) + 2(y1 + y2 + y3 + y4 + y5) subject to w1 = 115 x1 + y1 = 115 w2 = 210 x2 + y2 = 210 w3 + .8x1 = 250 x3 + y3 = 250 w4 + .8x2 + .8y1 = 230 x4 + y4 = 230 w5 + .8x3 + .8y2 = 260 x5 + y5 = 260 w6 + .8x4 + .8y3 = 300 x6 = 300 w7 + .8x5 + .8y4 = 250 w8 + .8x6 +.8y5 ≥ 190 wi, xi, yi ≥ 0 Model Solution: w1 = 115 x1 = 0 y1 = 115 w2 = 210 x2 = 0 y2 = 210 w3 = 250 x3 = 52.5 y3 = 197.5 w4 = 138 x4 = 177.5 y4 = 52.5 w5 = 50 x5 = 260 y5 = 0 w6 = 0 x6 = 300 w7 = 0 w8 = 0 Z = $11,940

CASE SOLUTION: SPRING GARDEN TOOLS Model Formulation: Let i = 1 (trowel), 2 (hoe), 3 (rake), 4 (shovel) Ri = regular production of product i in stage 1 Si = subcontracted production of product i in stage 1 xi = overtime production of product i in stage 1 Ai = regular production of product i in stage 2 yi = overtime production of product i in stage 2 minimize Z = $6R1 + 10R2 + 8R3 + 10R4 + 7.2S1 + 12S2 + 9.6S3 + 12S4 + 6.2x1 + 10.7x2 +8.5x3 + 10.7x4 + 3A1 + 5A2 + 4A3 + 5A4 + 3.1y1 + 5.4y2 + 4.3y3

+ 5.4y4 subject to .04R1 + .17R2 + .06R3 + .12R4 ≤ 500 hrs. .04x1 + .17x2 + .06x3 + .12x4 ≤ 100 hrs. .05R1 + .14R2 + .14R4 ≤ 400 hrs. .05x1 + .14x2 + .14x4 ≤ 100 hrs. 1.2R1 + 1.6R2 + 2.1R3 + 2.4R4 + 1.2x1 + 1.6x2 + 2.1x3 + 2.4x4 ≤ 10,000 ft2 .06A1 + .13A2 + .05A3 + .10A4 ≤ 600 hrs. .06y1 + .13y2 + .05y3 + .10y4 ≤ 100 hrs. .05A1 + .21A2 + .02A3 + .10A4 ≤ 550 hrs. .05y1 + .21y2 + .02y3 + .10y4 ≤ 100 hrs. .03A1 + .15A2 + .04A3 + .15A4 ≤ 500 hrs. .03y1 + .15y2 + .04y3 + .15y4 ≤ 100 hrs. R1 + S1 + x1 = A1 + y1 R2 + S2 + x2 = A2+ y2 R3 + S3 + x3 = A3 + y3 R4 + S4 + x4 = A4 + y4 y1 + A1 = 1,800 y2 + A2 = 1,400 y3 + A3 = 1,600 y4 + A4 = 1,800 Ri, Si, xi, Ai, yi ≥ 0 61 Solution: R1 = 1,691.954 s1 = s2 = s3 = 0 R2 = 1,319.54 s2 = 866.6667 R3 = 1,600 y1 = 0 Z = $85,472.60 R4 = 933.33 y2 = 315.1514 A1 = 1,800 y3 = 0 A2 = 1,084.849 y4 = 388.1822 A3 = 1,600 x1 = 108.0457 A4 = 1,461.818 y2 = 80.4599 x3 = x4 = 0

CASE SOLUTION: SUSAN WONG’S PERSONAL BUDGETING MODEL (a) Let xij = amount invested for i months in month j where i = 1,3 and 7, and j = 1, 2, ..., 12. subject to Jan: –x11 – x31 – x71 + 2,450 + 3,800 = 2,750 Feb: x11 – x12 – x32 – x72 + 2,450 = 2,860 Mar: x12 – x13 – x33 – x73 + 2,450 = 2,335 Apr: x13 – x14 + x31 – x34 – x74 + 2,450 = 2,120 May: x14 – x15 + x32 – x35 – x75 + 2,450 = 1,205 June: x15 – x16 + x33 – x36 – x76 + 2,450 = 1,600 July: x16 – x17 + x34 – x37 – x77 + 2,450 = 3,050 Aug: x17 – x18 + x35 – x38 + x71 – x78 + 2,450 = 2,300 Sep: x18 – x19 + x36 – x39 + x72 – x79 + 2,450 = 1,975 Oct: x19 – x110 + x37 – x310 + x73 – x710 + 2,450 = 1,670 Nov: x110 – x111 + x38 – x311 + x74 – x711 + 2,450 = 2,710 Dec: x111 – x112 + x39 – x312 + x75 – x712 + 2,450 = 2,980 Investments (i = 1,3,7)

Month ( j) 1-month 3-month 7-month 1. January x11 = 410 x31 = 390 2. February 3. March x13 = 115 4. April x74 = 3535 5. May x75 = 1245 6. June x16 = 600 x36 = 250 7. July 8. August x18 = 150 9. September x39 = 875 10. October x110 = 780 11. November x111 = 4055 12. December x712 = 5645 Z = $844.60 earned in interest payments Max Z x x x j j j j j j

= + + = = =

 0 005  0 02  0 07

1

1 12 3 1 12 7 1 12

... (b)Using sensitivity analysis for the January constraint, the lower range for the right hand side is –410. Thus, Susan needs $710 out of her original $3,800 to make the model feasible (i.e., avoid an infeasible solution).

CASE SOLUTION: WALSH’S JUICE COMPANY xij = tons of unprocessed grape juice transported from vineyard i to plant j where i = n (New York), p (Pennsylvania), o (Ohio), and j = v (Virginia), m (Michigan), t (Tennessee), i (Indiana). yij = tons of grape juice processed into product i at plant j where i = j (juice), k (concentrate), l (jelly) minimize Z = $850xnv + 720xnm + 910xnt + 750xni + 970xpv + 790xpm + 1,050xpt + 880xpi + 900xov + 830xom + 780xot + 820xoi + 2,100yjv + 2,350yjm + 2,200yjt + 1,900yji + 4,100ykv + 4,300ykm + 3,950ykt + 3,900yki + 2,600ylv + 2,300ylm + 2,500ylt + 2,800yli subject to xnv + xnm + xnt + xni ≤ 1,400 xpv + xpm + xpt + xpi ≤ 1,100 xov + xom + xot + xoi ≤ 1,700

xnv + xpv + xov ≤ 1,200 xnm + xpm + xom ≤ 1,100 xnt + xpt + xot ≤ 1,400 xni + xpi + xoi ≤ 1,400 yjv + yjm + yjt + yji = 1,200 ykv + ykm + ykt + yki = 900 ylv + ylm + ylt + yli = 700 yjv + 2ykv + 1.5ylv = xnv + xpv + xov yjm + 2ykm + 1.5ylm = xnm + xpm + xom yjt + 2ykt + 1.5ylt = xnt + xpt + xot yji + 2yki + 1.5yli = xni + xpi + xoi xij, yij ≥ 0 Solution: xni = 1400 yji = 1200 xpm = 1100 ykv = 75 xov = 150 ykm = 25 xot = 1400 ykt = 700 yki = 100 ylm = 700 Z = $10,606,000 62

CASE SOLUTION: KING’S LANDING AMUSEMENT PARK xi = experienced employees in week i yi = new employees in week i zi = pool of employees available in week i minimize z = _yi subject to: 10yi + 30xi ≥ weekly hours needed, where i = 1, . . . , 16 30xi ≥ weekly hours needed, where i = 17, 18, 19, 20 x1 = 700 xi = .85xi–1 + yi, where i = 2, 3, . . . , 16 x17 = .25x16 xi = .90xi–1, where i = 18, 19, . . . , 10 yi ≤ zi, where i = 2, 3, . . . , 16 z1 = 1,500 z2 = 1,500 – y1 + 200 z3 = z2 – y2 + 200 z4 = z3 – y3 + 200 z5 = z4 – y4 + 200 z6 = z5 – y5 + 200 z7 = z6 – y6 + 200 z8 = z7 – y7 + 200 z9 = z8 – y8 + 100 z10 = z9 – y9 + 100 z11 = z10 – y10 + 100 z12 = z11 – y11 + 100 z13 = z12 – y12 + 100 z14 = z13 – y13 + 100 z14 = z14 – y14 + 100 z16 = z15 – y15 + 100 Solution: y1 = 100 x1 = 700.0 y2 = 171.5 x2 = 695.0

y3 = 213.2 x3 = 762.3 y4 = 417.6 x4 = 861.1 y5 = 54.3 x5 = 1148.6 y6 = 408.4 x6 = 1030.5 y7 = 147.0 x7 = 1284.3 y8 = 284.9 x8 = 1238.7 y9 = 189.7 x9 = 1336.8 y10 = 222.1 x10 = 1326.0 y11 = 152.4 x11 = 1349.2 y12 = 302.4 x12 = 1299.2 y13 = 279.9 x13 = 1406.7 y14 = 73.1 x14 = 1475.6 y15 = 517.8 x15 = 1327.4 y16 = 0 x16 = 1646.1 x17 = 411.5 x18 = 370.4 x19 = 333.3 x20 = 300.0 Z = 3,532 Note: This would logically be best solved as an integer programming model, however the model is so large it exceeds the capabilities of Excel to solve in a reasonable amount of time.