IEOR150F10_SampleFinal_Solution

IEOR150F10_SampleFinal_Solution - IEOR 150, Fall 2010...

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IEOR 150, Fall 2010 Suggested Solution to Sample Final 1. Let c = $15 be the unit purchasing cost, K = $50 be the ordering cost per order, I = 0 . 3 be the annual interest rate, h = cI = $4 . 5 be the unit holding cost per year, p = $40 be the shortage cost per unit, and τ = 2 12 = 0 . 167 year be the order lead time. Since the monthly demand for the filter follows a normal distribution with mean 1800 and standard deviation 400, let μ = 1800 × 2 = 3600 units be the mean of the demand and σ = 2 × 400 2 = 565 . 69 units be the standard deviation of the demand during a lead time. Finally, let λ = 1800 × 12 = 21600 units be the annual demand size. Because we use the EOQ quantity as the order quantity, we have Q * = r 2 h = 692 . 82 as the order quantity. Then we compute R * = F - 1 1 - Q * h · = 5119 . 81 , z = R * - μ σ = 2 . 69 , and n ( R * ) = σL ( z ) = 0 . 62 , where the value of L ( z ) is looked up from Table A-4 in the textbook. R * is the optimal reorder level. It then follows that the safety stock level is R * - λτ = 1519 . 81 and the expected annual stock-out penalty cost is λn ( R * ) Q * = $776. 2. We have the production rate P = 250 , 000 units per year, demand rate λ = 120 , 000 units per year, setup cost K = $3 , 000 per production run, production cost c = $16 per unit, annual interest rate I = 0 . 25, holding cost h = cI = $16 × 0 . 25 = $4 per unit per year, and the effective holding cost h 0 = h (1 - λ P ) = $2 . 08 per unit per year. (a) The optimal production lot size is Q * = r 2 h 0 = 18605 . 21 units . (b) Note that P = 250 , 000 and λ = 120 , 000, so the proportion of uptime is λ P = 48% and the proportion of downtime is P - λ P = 52%. (c) The total holding and setup cost is G * = 2 Kλh 0 = $38698 . 84 . Therefore, the annual holding cost and the annual setup cost are both G * 2 = $19349 . 42. 3. Let range 0 be [0 , 2000), range 1 be [2000 , 4000), and range 2 be [4000 , ). Let c (0) = $25, c (1) = $23, and c (1) = $22 be the unit purchasing cost for ranges 0, 1, and 2, respectively. It follows that the unit holding costs are h (0) = Ic (0) = $5, h (1) = Ic (1) = $4 . 6, and h (0) = Ic (2) = $4 . 4. For an all-units quantity discount, we may directly apply ¯ Q ( i ) = r 2 h ( i ) , i ∈ { 0 , 1 , 2 } , 1
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to find the (unadjusted) EOQ quantity for each range. Here we have ¯ Q (0) = 2190 . 89 units , ¯ Q (1) = 2284 . 16 units , and ¯ Q (2) = 2335 . 50 units . Note that ¯ Q (1) is realizable (feasible), so we may ignore ¯ Q (0) . We then adjust ¯ Q (1) and ¯ Q (2) to Q (1) = 2284 . 16 units and Q (2) = 4000 units .
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IEOR150F10_SampleFinal_Solution - IEOR 150, Fall 2010...

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