sol4 - Homework #4 solutions / IEOR4405 1 3.5. We solve...

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Homework #4 solutions / IEOR4405 1 3.5. We solve problem 3.4 initially and then infer the result for problem 3.5 We use Algorithm 3.2.1 for this problem. Notice that each of the functions h j ( C j ) is increasing in C j , and hence it is appropriate to use this algorithm. We do not have any precedence constraints, so J c = J 0 at all times. Also keep in mind that the algorithm decides on the jobs in backward order . These are the iterations of the algorithm: () J = ,J c = { 1 , 2 , 3 , 4 , 5 , 6 , 7 } . (a) We have j J c p j = 55 (this will be the completion time of the job that is about to be scheduled). The values of the h j (55) are: job j 1 2 3 4 5 6 7 h j (55) 165 77 3025 82.5 77.42 88 77 Jobs 2 and 7 have h j (55) minimum. So we can add either Job 2 or Job 7. Let’s choose Job 2 (choosing Job 7 gives an alternative schedule – see below). So schedule Job 2 last and delete it from J c . (b) J c = { 1 , 3 , 4 , 5 , 6 , 7 } . We have j J c p j = 47 and h j (47) are: job j 1 3 4 5 6 7 h j (47) 141 2209 70.5 76.86 75.2 65.8 So schedule Job 7 second to last and delete it from J c . (c) J c = { 1 , 3 , 4 , 5 , 6 } . We have j J c p j = 38 and h j (38) are: job j 1 3 4 5 6 h j (38) 114 1444 57 76.16 60.8 So schedule Job 4 third to last and delete it from J c . (d) J c = { 1 , 3 , 5 , 6 } . We have j J c p j = 31 and h j (31) are: job j 1 3 5 6 h j (31) 93 961 75.57 49.6 So schedule Job 6 fourth to last and delete it from J c . (e) J c = { 1 , 3 , 5 } . We have j J c p j = 22 and h j (22) are: job j 1 3 5 h j (22) 66 484 74.69 So schedule Job 1 fifth to last and delete it from J c . (f)
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This note was uploaded on 05/07/2011 for the course INDUSTRIAL IE 208 taught by Professor Serolbulkan during the Fall '11 term at Marmara Üniversitesi.

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sol4 - Homework #4 solutions / IEOR4405 1 3.5. We solve...

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