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Unformatted text preview: IEOR 150, Fall 2010 Suggested Solution to Sample Midterm 2 1. Denote the DVD player, 32-inch flat screen TV, Blank DVDs (box of 50), Blank DVDs (box of 10), Compact stereo, and Telephone as items 1, 2, ..., and 6. Let K i , i , h i , c i , and w i be the setup cost per order, annual demand size, holding cost per unit per year, and purchasing cost per unit, and size per unit of item i , i = 1 , 2 ,..., 6. Let I , C , and W be the annual interest rate, total budget, and warehouse space. Note that h i = Ic i . (a) The first step is to calculate the EOQ quantity for each product. Let Q i be the EOQ quantity of product i , we have Q i = r 2 K i i h i for all i = 1 , 2 ,..., 6. These quantities are listed in Table 1. We then check whether the constraint 6 X i =1 c i Q i C is satisfied. The LHS is 288811.60, which is larger than C = 200000. Therefore, we need to do an adjustment. To determine which adjustment to do, we need to check whether c 1 h 1 = c 6 h 6 . Obvi- ously this is satisfied, so we only need to do a normalization to find the optimal order quantities. Let m = C 6 i =1 c i Q i = 0 . 692 and Q * i = m Q i for all i = 1 , 2 ,..., 6, we have Q * i as the optimal order quantity of item i . The values of Q * i s are listed in Table 1. Item 1 2 3 4 5 6 Q i 252.98 413.12 2065.59 3265.99 160.00 1239.35 c i Q i 50596.44 61967.73 61967.73 58787.75 40000.00 15491.93 Q * i = m Q i 175.19 286.08 1430.41 2261.67 110.80 858.24 Table 1: Results for Problem 1a (b) Similar to Part (a), first we use the EOQ values in Table 1 to check whether the constraint 6 X i =1 w i Q i W is satisfied. The LHS is 30758.72 (cf. Table 2), which is larger than W = 20000. Therefore, we need to do an adjustment. 1 Item 1 2 3 4 5 6 w i Q i 3035.79 7436.13 6196.77 6531.97 3840.00 3718.06 Table 2: Results for Problem 1b To determine which adjustment to do, we need to check whether w 1 h 1 = w 6 h 6 . Unfor- tunately, this is not satisfied, so we need to do the adjustment with . More precisely, we need to numerically search for the value of...
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