IEOR150F10_SampleFinal_Solution

Now we apply e g q i k? q i h i q i 2 c i ? to find

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Now we apply e G ( Q ( i ) ) = Q ( i ) + h ( i ) Q ( i ) 2 + c ( i ) λ to find the total cost. This leads to e G ( Q (1) ) = \$5253 . 57 + \$5253 . 57 + \$1380000 = \$1390507 . 14 and e G ( Q (2) ) = \$3000 + \$8800 + \$1320000 = \$1331800 , which imply that the optimal order quantity is Q * = Q (2) = 4000 units and the corresponding total cost is e G * = e G ( Q (2) ) = \$1331800. 4. Let c = \$10 be the unit purchasing cost, K = \$150 be the ordering cost per order, I = 0 . 2 be the annual interest rate, h = cI = \$2 be the unit holding cost per year, and τ = 2 12 = 0 . 167 year be the order lead time. Since the monthly demand follows a normal distribution with mean 50 and standard deviation 10, let μ = 500 × 2 = 1000 units be the mean of the demand and σ = 2 × 100 2 = 141 . 42 units be the standard deviation of the demand during a lead time. Let D denote the demand during a lead time. Finally, let λ = 500 × 12 = 6000 units be the annual demand size. (a) To achieve a type-1 service level of α = 0 . 9, all we need is to choose the reorder point R to satisfy P ( D R ) α P D - μ σ R - μ σ · α Φ R - μ σ · α, where Φ( · ) is the cdf of the standard normal distribution. From Table A-4 we know Φ(1 . 28) = 0 . 8997 and Φ(1 . 29) = 0 . 9015, so choosing R 1 . 29 σ + μ = 1182 . 43 is good enough. To minimize our expected total cost, we choose R * = 1182 . 43 as the reorder point. For the order quantity Q , we may simply choose the EOQ quantity Q * = r 2 h = 948 . 68 . It follows that the expected annual holding and setup cost is h Q * 2 + R * - λτ · + Q * = \$2262 . 23 . 2

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(b) To achieve a type-2 service level of β = 0 . 99, we iteratively compute the order quantity Q and the reorder point R by the following formulas: Q i = n ( R i - 1 ) 1 - F ( R i - 1 ) + v u u t 2 h + " n ( R i - 1 ) 1 - F ( R i - 1 ) # 2 , n ( R i ) = Q i (1 - β ) , L ( z i ) = n ( R i ) σ , R i = μ + z i σ, where F ( x ) = P ( D x ) is the cdf of the demand during a lead time. Note that we initiate this process by setting n ( R - 1 ) = 0 and F ( R - 1 ) = 0 so that Q 0 = q 2 h , the regular EOQ quantity. Also note that we use Table A-4 in the textbook to look up z i from a given value of L ( z i ). The iterative calculation is summarized below. Iteration i 0 1 2 Q i 948.68 1021.17 1021.85 n ( R i ) 9.49 10.21 10.22 L ( z i ) 0.0671 0.0722 0.0723 z i 1.12 1.08 1.08 R i 1158.39 1152.74 1152.74 We stop at the end of iteration 2 since R 2 = R 1 and conclude that Q * = 1021 . 85 and R * = 1152 . 74. The expected annual holding and setup cost is h ˆ Q * 2 + R * - λτ ! + Q * = \$2208 . 08 .
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