IE383_Extra exercise for Final S09_Sol

# IE383_Extra exercise for Final S09_Sol - IE383 Extra...

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IE383 Extra Exercise for Final Exam Solution Problem 1 a) Minimize maximum tardiness G E A D C F H B I Completion Time: 3 7 12 22 29 37 40 42 48 Due Date: 4 8 15 20 28 32 37 44 45 Mean flow time = (3+7+12+22+29+37+40+42+48)/9 = 240/9 = 26.67 days Mean tardiness = (2+1+5+3+3)/9 = 14/9 = 1.56 days Maximum tardiness = 5 days b) Minimize mean flow time B G H E A I C F D Completion Time: 2 5 8 12 17 23 30 38 48 Due Date: 44 4 37 8 15 45 28 32 20 Mean flow time = (2+5+8+12+17+23+30+38+48)/9 = 183/9 = 20.33 days Mean tardiness = (1+4+2+2+6+28)/9 = 43/9 = 4.78 days Maximum tardiness = 28 days Problem 2 a) EDD E G A D C F H I B Completion Time: 4 7 12 22 29 37 40 46 48 Arrival Time: 0 2 0 9 21 7 15 25 41 Due Date: 8 4 15 20 28 32 37 45 44 Mean flow time = (4+5+12+13+8+30+25+21+7)/9 = 125/9 = 13.89 days Mean tardiness = (3+2+1+5+3+1+4)/9 = 19/9 = 2.11 days Maximum tardiness = 5 days b) SPT E G A F H C I D B Completion Time: 4 7 12 20 23 30 36 46 48 Arrival Time: 0 2 0 7 15 21 25 9 41 Due Date: 8 4 15 32 37 28 45 20 44 Mean flow time = (4+5+12+13+8+9+11+37+7)/9 = 106/9 = 11.78 days Mean tardiness = (3+2+26+4)/9 = 35/9 = 3.89 days Maximum tardiness = 26 days c) Critical ratio Job A B C D E F G H I Processing Time 5 2 7 10 4 8 3 3 6 Due Date 15 44 28 20 8 32 4 37 45 Arrival Time 0 41 21 9 0 7 2 15 25 Job: A E At time = 0, CR: 0.33 0.5 Select job E Job: A G At time = 4, CR: 0.46 inf Select job G

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IE383 Extra Exercise for Final Exam Solution Job: A F At time = 7, CR: 0.625 0.32 Select job A Job: D F At time = 12, CR: 1.25 0.4 Select job D Job: C F H At time = 22, CR: 1.17 0.8 0.2 Select job C
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## This note was uploaded on 09/23/2009 for the course IE 383 taught by Professor Leyla,o during the Spring '08 term at Purdue.

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IE383_Extra exercise for Final S09_Sol - IE383 Extra...

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