IEOR
IEOR150F10_SampleFinal_Solution

# 5 a the scheduling result is shown in table 1 the

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5. (a) The scheduling result is shown in Table 1. The total tardiness is 22 + 21 + 17 + 28 = 88. (b) The scheduling result is shown in Table 2. The total tardiness is 1+2+4+10+17 = 34. (c) The scheduling result is shown in Table 3. The total tardiness is 18 + 10 + 17 = 45. Job t i d i F i L i T i 1 5 14 5 -9 0 2 9 28 14 -14 0 3 11 32 25 -7 0 4 8 11 33 22 22 5 3 15 36 21 21 6 6 25 42 17 17 7 7 21 49 28 28 Table 1: FIFO Job t i d i F i L i T i 4 8 11 8 -3 0 1 5 14 13 -1 0 5 3 15 16 1 1 7 7 21 23 2 2 6 6 25 29 4 4 2 9 28 38 10 10 3 11 32 49 17 17 Table 2: EDD Job t i d i F i L i T i 5 3 15 3 -12 0 1 5 14 8 -6 0 6 6 25 14 -11 0 7 7 21 21 0 0 4 8 11 29 18 18 2 9 28 38 10 10 3 11 32 49 17 17 Table 3: SPT (d) We first schedule these jobs by EDD until we find the first tardy job, job 5 (cf. Table 4). We compare the processing time of jobs 4, 1, and 5 and find that job 4 has the longest processing time. We thus remove job 4 for a while. Now we apply EDD to the remaining six jobs (excluding job 4) until we find the first tardy job, job 2 (cf. Table 5). We compare the processing time of jobs 1, 5, 7, 6, and 2 and find that job 2 has the longest processing time. We thus remove job 2 for a while. 3

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Now we apply EDD to the remaining five jobs (excluding jobs 2 and 4) until we find a tardy job. Because there is no tardy job (cf. Table 6), we may append jobs 2 and 4 at the end. Because we want to minimize the maximum tardiness, we schedule job 4 first since its due time is earlier. The total tardiness is 29 + 21 = 50 (cf. Table 6). Job t i d i F i L i T i 4 8 11 8 -3 0 1 5 14 13 -1 0 5 3 15 16 1 1 Table 4: First round Job t i d i F i L i T i 1 5 14 5 -9 0 5 3 15 8 -7 0 7 7 21 15 -6 0 6 6 25 21 -4 0 2 9 28 30 2 2 Table 5: Second round Job t i d i F i L i T i 1 5 14 5 -9 0 5 3 15 8 -7 0 7 7 21 15 -6 0 6 6 25 21 -4 0 3 11 32 32 0 0 4 8 11 40 29 29 2 9 28 49 21 21 Table 6: Third round 6. (a) Let F 1 ( x ) = P ( D 1 x ) be the cdf of D 1 , then F 1 (0) = 0 . 5, F 1 (1) = 0 . 667, and F 1 (2) = 1. Because the critical ratio is 12 - 6 12 - 2 = 0 . 6, which is between F 1 (0) and F 1 (1), we round up and conclude that the optimal inventory level is q * 1 = 1. With one unit on hand, the expected sales quantity is h min { D, 1 } i = 0 × 1 2 + 1 × 1 6 + 1 × 1 3 = 0 . 5 . The expected sales revenue is thus 0 . 5 × \$10 = \$5. As the purchasing cost is \$4, the expected profit is \$5 - \$4 = \$1. (b) Let E = ( D 1 - q * 1 ) + be the excess demand not satisfied by retailer A and R = D 2 + E be the effective demand of retailer B. We need to characterize the cdf of R to solve this problem. Note that E may only be 0 or 1 and P ( E = 0) = P ( D 1 q * 1 ) = 2 3 and P ( E = 1) = P ( D 1 > q * 1 ) = 1 3 .
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