The document discusses graphs, networks, project scheduling, critical path method (CPM), program evaluation and review technique (PERT), activity-on-node (AON) and activity-on-arc (AOA) representations, activity slack, floats, crashing activities to reduce project duration, and performing time-cost tradeoffs to optimize project costs. It provides definitions of key terms like vertices, edges, predecessors, successors, critical and non-critical activities, earliest start times, latest completion times, total float and free float. It also includes an example problem demonstrating how to construct a network diagram, find the critical path, compute activity times, and determine the optimal crash time and costs to meet a deadline.

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

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

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 - - -

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 - - -

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 - - -

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

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

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