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ISyE 3104: Introduction to Supply Chain Modeling: Manufacturing and Warehousing Instructor : Spyros Reveliotis Summer 2006 Solutions for Homework #3 ISYE 3104 Spring 2006 Homework 7 Solution Section 5.2 2. Deterministic variation is systematic variation in demand that can be predicted in advance. This variation is typical in the demand of seasonal products, i.e., products that tend to present the main volume of their demand at certain periods of a regular cycle. Examples of such seasonal products are products pertaining to summer recreational activities or toys that are consumed primarily during the Christmas season. Aggregate and production planning problems to be discussed in the next part of the course seek to characterize and accommodate this systematic variation of the demand in the most efficient / cost-effective manner. On the other hand, random variation is the unpredictable component in the variation of the experienced demand, and it is typically absorbed through safety stock. Section 5.3 7. If he only keeps track of the number of sales, he has no way to accurately estimate the demand since demand = sales + lost sales. He would need some way to gauge the lost sales. One method would be to increase his supply for a period of time so that he would be able to meet all demand. An alternative approach would be to consult some published statistics on this demand. 8. a) c0 = 0.08 - 0.03 = 0.05 cu = 0.35 - 0.08 = 0.27 0.27 0.84375 Critical ratio = 0.05 0.27 From the given distribution, we have: Q 0 5 10 15 20 f(Q) 0.05 0.1 0.1 0.2 0.25 F(Q) 0.05 0.15 0.25 0.45 0.7 25 30 35 0.15 0.1 0.05 0.85 0.95 1 0.8438 Since the critical ratio falls between 20 and 25, the optimal is Q = 25 bagels. b) The answers should be close since the given distribution appears to be close to the normal. The quality of approximation could be assessed more systematically by using the relevant tests from your statistics (or simulation) classes. c) xf ( x) (0)(0.05) (5)(0.10) ... (35)(0.05) 18 2 x 2 f ( x) 2 402.5 (18) 2 78.5 78.5 8.86 The z value corresponding to a critical ratio of 0.84375 is 1.01. Hence, Q* = z + = (8.86)(1.01) + 18 = 26.95 ~ 27 2 ISYE 3104 Spring 2006 Homework 7 Solution 10. a) A period is three months. Holding cost per year is $500, which means that the cost for a 3-month period is: 500/4 = $125 = c0 cu = 250 (emergency shipment cost) 250 0.667 z 0.44 Critical ratio = 250 125 Hence, Q* = z + = (6)(0.44) + 60 = 62.64 ~ 63 cars. b) cu = 150 c0 = 125 150 0.5454 z 0.11 150 125 Hence, Q* = z + = (6)(0.11) + 60 = 60.66 ~ 61 cars. Critical ratio = c) This corresponds to an infinite horizon problem with lost sales. In this case, cu = lost profit = $3,500 c0 = holding cost = 125 3500 0.9655 z 1.82 Critical ratio = 3500 125 which gives Q* = (6)(1.82) + 60 = 70.92 ~ 71 cars. Section 5.5 14. a) Note: We assume 4 weeks/month and 48 weeks/year. Monthly demand is normal (= 28= 8) = 14 weeks = 3.5 months => Lead time demand is normally distributed with = 28(3.5) = 98, = (8) 3.5 15 h = Ic = (0.3)(6) = $1.8 = (28)(12) = 336 units per year p = 10 K = 15 = 98 3 ISYE 3104 Spring 2006 Homework 7 Solution 2(336)(15) 75 1.8 (75)(1.8) F ( R1 ) 0.04 (10)(336) z 1.75 R1 z 124, n( R1 ) L( z ) 0.2426 Q0 2(336) (15 (10)(0.2426)) 81 1.8 (81)(1.8) F ( R2 ) 0.0434 (10)(336) z 1.71 R2 z 124 Since R2 = R1, we stop. Conclude that (Q,R) = (81,124) Q1 b) S = R - = 124 - 98 = 26 units 15. a) EOQ = 75 (from 14(a)) Type 1 service of 90% => F(R) = 0.10, z = 1.28 R = z = (15)(1.28) + 98 = 117 (Q,R) = (75,117) b) Find Type II service level achieved in part (1) n( R ) L( z ) (15)(0.0475) 1 0.0095 Q Q 75 0.9905 (99.05% service level) So it is actually much higher than the 90% fill rate that he wanted. Section 5.6 22. a) F(Q) = 0.95. The corresponding z value is z = 1.645, giving Qz= (6)(1.645) + 60 = 69.87 ≈ 70. b) n(Q) = (1-) = (0.05)(60) = 3. L(z) = n(Q)/ = 3/6 = 0.5 z = -0.19 Qz= 58.86 ≈ 59. 4