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Graph ,[object Object],[object Object]
Digraph & Network ,[object Object],[object Object],[object Object]
Project Scheduling ,[object Object],[object Object],[object Object],[object Object],[object Object]
Project Network ,[object Object],[object Object],[object Object],[object Object],[object Object]
CPM/PERT ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
CPM/PERT ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
CPM/PERT ,[object Object],[object Object],[object Object],[object Object]
Activity Slack (Float) ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Float  … ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Problem ,[object Object],F and G are the terminal activities of the project Activity Predecessor(s) Time (days) A - 3 B - 4 C A,B 5 D B 6 E D 7 F C,E 8 G D 9
Problem … ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Project Network 10 10 1 2 3 4 5 6 A C E B D F G 3 5 7 4 6 8 9 0 0 4 4 17 17 25 25 12 4 4, 3 9, 0 12, 4 17, 9 10, 16 25, 19
Computations (Example) Critical path is 1    2    4    5    6 Activity (i,j) Dur. D i, j ES i LC j ES j Total Float LC j  – ES i  – D i,j Criti-cal Free Float ES j  – ES i  – D i,j A (1,3) 3 0 12 4 12 – 0 – 3 = 9 - 4 – 0 – 3 = 1 B (1,2) 4 0 4 4 4 – 0 – 4 = 0 Yes 4 – 0 – 4 = 0 C (3.5) 5 4 17 17 17 – 4 – 5 = 8 - 17 – 4 – 5 = 8 D (2,4) 6 4 10 10 10 – 4 – 6 = 0 Yes 10 – 4 – 6 = 0 E (4,5) 7 10 17 17 17 – 10 – 7 = 0 Yes 17 – 10 – 7 = 0 F (5,6) 8 17 25 25 25 – 17 – 8 = 0 Yes 25 – 17 – 8 = 0 G (4,6) 9 10 25 25 25 – 10 – 9 = 6 - 25 – 10 – 9 = 6
Project Cost Optimization ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Project Cost Optimization contd… ,[object Object],[object Object]
Project Cost Optimization contd… ,[object Object],[object Object],[object Object],[object Object],[object Object]
Problem Consider the following arrow diagram with activity times given in days: The normal and crash data for this project are as follows: 1 2 3 4 A B C D 4 6 8 10 Activity Normal Time Crash Time Normal Cost Crash Cost A 4 3 80 105 B 6 4 180 250 C 8 5 200 320 D 10 6 350 530
Problem  … ,[object Object],[object Object],[object Object],1 2 3 4 A B C D 4 6 8 10 6 6 0 0 10, 0 4, 14 24 24 14 14 {FF=10}
Problem  … ,[object Object],[object Object],[object Object],[object Object],Activity (i,j) Crash Limit (D – D ’ ) Crash Cost/day A(1, 3) 4 – 3 = 1 (105-80)/(4-3)=25 B(1, 2) 6 – 4 = 2 (250-180)/(6-4)=35 C(2, 3) 8 – 5 = 3 (320-200)/(8-5)=40 D(3, 4) 10 – 6 = 4 (530-350)/(10-6)=45
Problem  … From the critical path calculations, we have the following information: Since the critical activity B has the lowest “crash cost per day”, it becomes the first candidate for crash.  The length by which B can be reduced is found as follows: Reduction limit = min(crash limit, +ve FF limit) = min (2,10) = 2  Hence crash activity B by 2 days Acitivity (i,j) A(1, 3) B(1, 2) C(2, 3) D(3, 4) Critical - Yes Yes yes Free Float (FF) 10 - - -
Problem  … ,[object Object],[object Object],[object Object],1 2 3 4 A B C D 4 6 8 10 4 4 0 0 8, 0 4, 12 22 22 12 12 {FF=8} 4
Problem  … Since the crash limit for critical activity B is reached, consider critical activity C with the next lowest “crash cost per day” for crash.  The length by which C can be reduced is found as follows: Reduction limit = min(crash limit, +ve FF limit) = min(3,8)=3  Hence crash  activity C  by 3 days Acitivity (i,j) A(1, 3) B(1, 2) C(2, 3) D(3, 4) Critical - Yes Yes yes Free Float (FF) 8 - - -
Problem  … ,[object Object],[object Object],[object Object],1 2 3 4 A B C D 4 6 8 10 4 4 0 0 5, 0 4, 9 19 19 9 9 {FF=5} 4 5
Problem  … Since the crash limit for critical activity C is reached, consider critical activity D with the next lowest “crash cost per day” for crash.  Although we can reduce D by 4 days, it is only necessary to reduce it by 1 day to reach our project completion goal of 18 days. Acitivity (i,j) A(1, 3) B(1, 2) C(2, 3) D(3, 4) Critical - Yes Yes yes Free Float (FF) 5 - - -
Problem  … ,[object Object],[object Object],[object Object],1 2 3 4 A B C D 4 6 8 9 4 4 0 0 5, 0 4, 9 18 18 9 9 {FF=5} 4 5
The Algorithm ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Problem  … ,[object Object],[object Object],Activity Predecessor(s) Normal Time Normal Cost Cr. Cost per day Crash Time A - 4 400 125 3 B A 5 800 200 4 C A 4 520 150 2 D B 3 600 225 2 E C 3 255 100 2 F B, E 4 600 190 2
Problem  … 1 2 3 5 4 6 A B C E D F 4 5 4 3 3 4 0,0 4,4 8,8 11,11 9,11 15,15 FF = 3 FF = 2 FF = 0 ,[object Object],[object Object],[object Object]
Problem  … Activity (i,j) Crash Limit (D – D ’ ) Crash Cost A(1, 2) 4 – 3 = 1 125 B(2,3) 5 – 4 = 1 200 C(2, 4) 4 – 2 = 2 150 D(3, 6) 3 – 2 = 1 225 E(4, 5) 3 – 2 = 1 100 F(5, 6) 4 – 2 = 2 190
Problem  … The two lowest “crash cost per day” critical activities E and A have crash limits  of 1 day each Reduction limit = min(crash limit, FF limit) = min(2,2)=2  Hence crash  activity E and A  by 1 day each Acitivity (i,j) A(1, 2) B(2,3) C(2, 4) D(3, 6) E(4, 5) F(5, 6) Dummy (3,5) Critical yes - Yes - yes yes - (FF) - 0 - 3 - 2
Problem  … ,[object Object],[object Object],[object Object],1 2 3 5 4 6 A B C E D F 4 5 4 3 3 4 0,0 3,3 8,9 13,13 FF = 2 FF = 1 FF = 0 3 2 7,7 9,9
Problem  … The next lowest “crash cost per day” critical activity is C The length by which C can be reduced is found as follows: Reduction limit = min(crash limit, FF limit) = min(2,1)=1  Hence crash  activity C  by 1 day Acitivity (i,j) A(1, 2) B(2,3) C(2, 4) D(3, 6) E(4, 5) F(5, 6) Dummy (3,5) Critical yes - Yes - yes yes - (FF) - 0 - 2 - 1
Problem  … ,[object Object],[object Object],[object Object],1 2 3 5 4 6 A B C E D F 4 5 3 3 3 4 0,0 3,3 8,8 12,12 FF = 1 FF = 0 3 2 6,6 8,8 4
Problem  … A and E have reached their crash limits while C has 1 day remaining.  As there are two critical paths, the possible crashes are shown below: Acitivity (i,j) A(1, 2) B(2,3) C(2, 4) D(3, 6) E(4, 5) F(5, 6) Dummy (3,5) Critical yes yes Yes - yes yes yes (FF) - - - 1 - - Activity B, C F Crash cost/day 200, 150 190 Remaining crash limit 1, 1 1
Problem  … ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

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Critical Path Method

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  • 12. Project Network 10 10 1 2 3 4 5 6 A C E B D F G 3 5 7 4 6 8 9 0 0 4 4 17 17 25 25 12 4 4, 3 9, 0 12, 4 17, 9 10, 16 25, 19
  • 13. Computations (Example) Critical path is 1  2  4  5  6 Activity (i,j) Dur. D i, j ES i LC j ES j Total Float LC j – ES i – D i,j Criti-cal Free Float ES j – ES i – D i,j A (1,3) 3 0 12 4 12 – 0 – 3 = 9 - 4 – 0 – 3 = 1 B (1,2) 4 0 4 4 4 – 0 – 4 = 0 Yes 4 – 0 – 4 = 0 C (3.5) 5 4 17 17 17 – 4 – 5 = 8 - 17 – 4 – 5 = 8 D (2,4) 6 4 10 10 10 – 4 – 6 = 0 Yes 10 – 4 – 6 = 0 E (4,5) 7 10 17 17 17 – 10 – 7 = 0 Yes 17 – 10 – 7 = 0 F (5,6) 8 17 25 25 25 – 17 – 8 = 0 Yes 25 – 17 – 8 = 0 G (4,6) 9 10 25 25 25 – 10 – 9 = 6 - 25 – 10 – 9 = 6
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  • 17. Problem Consider the following arrow diagram with activity times given in days: The normal and crash data for this project are as follows: 1 2 3 4 A B C D 4 6 8 10 Activity Normal Time Crash Time Normal Cost Crash Cost A 4 3 80 105 B 6 4 180 250 C 8 5 200 320 D 10 6 350 530
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  • 20. Problem … From the critical path calculations, we have the following information: Since the critical activity B has the lowest “crash cost per day”, it becomes the first candidate for crash. The length by which B can be reduced is found as follows: Reduction limit = min(crash limit, +ve FF limit) = min (2,10) = 2 Hence crash activity B by 2 days Acitivity (i,j) A(1, 3) B(1, 2) C(2, 3) D(3, 4) Critical - Yes Yes yes Free Float (FF) 10 - - -
  • 21.
  • 22. Problem … Since the crash limit for critical activity B is reached, consider critical activity C with the next lowest “crash cost per day” for crash. The length by which C can be reduced is found as follows: Reduction limit = min(crash limit, +ve FF limit) = min(3,8)=3 Hence crash activity C by 3 days Acitivity (i,j) A(1, 3) B(1, 2) C(2, 3) D(3, 4) Critical - Yes Yes yes Free Float (FF) 8 - - -
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  • 24. Problem … Since the crash limit for critical activity C is reached, consider critical activity D with the next lowest “crash cost per day” for crash. Although we can reduce D by 4 days, it is only necessary to reduce it by 1 day to reach our project completion goal of 18 days. Acitivity (i,j) A(1, 3) B(1, 2) C(2, 3) D(3, 4) Critical - Yes Yes yes Free Float (FF) 5 - - -
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  • 29. Problem … Activity (i,j) Crash Limit (D – D ’ ) Crash Cost A(1, 2) 4 – 3 = 1 125 B(2,3) 5 – 4 = 1 200 C(2, 4) 4 – 2 = 2 150 D(3, 6) 3 – 2 = 1 225 E(4, 5) 3 – 2 = 1 100 F(5, 6) 4 – 2 = 2 190
  • 30. Problem … The two lowest “crash cost per day” critical activities E and A have crash limits of 1 day each Reduction limit = min(crash limit, FF limit) = min(2,2)=2 Hence crash activity E and A by 1 day each Acitivity (i,j) A(1, 2) B(2,3) C(2, 4) D(3, 6) E(4, 5) F(5, 6) Dummy (3,5) Critical yes - Yes - yes yes - (FF) - 0 - 3 - 2
  • 31.
  • 32. Problem … The next lowest “crash cost per day” critical activity is C The length by which C can be reduced is found as follows: Reduction limit = min(crash limit, FF limit) = min(2,1)=1 Hence crash activity C by 1 day Acitivity (i,j) A(1, 2) B(2,3) C(2, 4) D(3, 6) E(4, 5) F(5, 6) Dummy (3,5) Critical yes - Yes - yes yes - (FF) - 0 - 2 - 1
  • 33.
  • 34. Problem … A and E have reached their crash limits while C has 1 day remaining. As there are two critical paths, the possible crashes are shown below: Acitivity (i,j) A(1, 2) B(2,3) C(2, 4) D(3, 6) E(4, 5) F(5, 6) Dummy (3,5) Critical yes yes Yes - yes yes yes (FF) - - - 1 - - Activity B, C F Crash cost/day 200, 150 190 Remaining crash limit 1, 1 1
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