4QA3 F12 Week 10 Lecture Notes

# Gandomi 35 r 335 200 140 440 300 200 starting

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Unformatted text preview: r = (335, 200, 140, 440, 300, 200) Starting in period 1: C(1) = 200 C(2) = [200 + (200)(.3)]/2 = 130 C(3) = [200 + (200)(.3) + (2)(140)(.3)]/3 = 114.67 C(4) = [344 + (440)(3)(.3)]/4 = 185 Stop. Starting in period 4: C(1) = 200 C(2) = [200 + (300)(.3)]/2 = 145 C(3) = [290 + (200)(2)(.3)]/3 = 136.67 Hence y1 = 335 + 200 + 140 = 675, y4 = 440 + 300 + 200 = 940 4QA3 F12 A. Gandomi 36 r = (335, 200, 140, 440, 300, 200) Starting in period 1: C (1) = 200/335= .597 Starting in period 5: C (1) = 200/300= .670 200 + (20 C (2) = 0)(.3) = .486 200 + (200 )(.3) (335 + 200 ) C (2 ) = = .520 (300 + 200 ) 200 + (200 )(.3) + (140 )(2 )(.3) C (3) = = .509 (335 + 200 + 140 ) Starting in period 3: C (1) = 200/140= 1.428 200 + ( 440 )(.3) = .572 (140 + 440 ) 200 + ( 440 )(.3) + (300 )(2 )(.3) C (3) = = .582 (140 + 440 + 300 ) Hence, y1 = 335+200 = 535, y3 = 140 + 440 = 580, y5 = 200+300 = 500 C (2 ) = 4QA3 F12 A. Gandomi 37 r = (335, 200, 140, 440, 300, 200) Starting in period 1: Periods Holding Cost 1 0 2 60 3 144 4 540 Starting in period 4: Periods Holding Cost 1 0 2 90 3 210 4QA3 F12 A. Gandomi 38 1. Lot- for- lot costs: 6 × 200 = \$1200 2. EOQ costs: 3 × 200 + (.3)[264 + 64+ 523 + 83 + 382 + 182] = 600 + 449.40 = \$1049.40 3. SM/PPB costs: 2 × 200+ (.3)[200 + 280 + 300 + 400] = 400 + 354 = \$754 4. LUC costs: 3 × 200 + (.3)[200 + 440 + 200] = 600 + 252 = \$852 ●  The Silver Meal and Part Period Balancing heuristics resulted in the same costs. 4QA3 F12 A. Gandomi 39 r = (335, 200, 140, 440, 300, 200) K = \$200 h = 0.30 ●  The resulting cij matrix for this problem is: 2 3 4 5 6 7 1 200 260 344 740 1100 1400 2 200 242 506 776 1016 3 200 332 512 692 4 200 290 410 5 200 260 6 200 4QA3 F12 A. Gandomi 40 f7 = 0 f6 = c67 = 200 at j = 7 f5 = min(c5 j + f j ) = min( 400, 260 ) = 260 at j = 7 j >5 f4 = min(c4 j + f j ) = min(200 + 260, 290 + 200, 410 ) = 410 at j >4 j=7 f3 = min(c3 j + f j ) = min(200 + 410, 332 + 260, 512 + 200, 692 ) = 592 at j >3 j=5 f2 = min(c2 j + f j ) = j >2 min(200 + 592, 242 + 410, 506 + 260, 776 + 200, 1016 ) = 652 at j=4 f1 = min(c1 j + f j ) = j >1 min(200 + 652, 260 + 592, 344 + 410, 740 + 260, 1100 + 200, 1400 ) = 754 4QA3 F12 A. Gandomi at j=4 41 ●  Uncertainty. MRP...
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