1. ENM448-Project Planning and Management
Examples

2. Preliminaries for CPM Forward Pass—Earliest Times
How soon can the activity start? (early start—ES)
How soon can the activity finish? (early finish—EF)
How soon can the project be finished? (expected time—TE)
Backward Pass—Latest Times
How late can the activity start? (late start—LS)
How late can the activity finish? (late finish—LF)
Which activities represent the critical path (CP)? This is the longest path in the network which, when delayed, will delay the project.
How long can the activity be delayed? (slack or float—SL)

3. Forward Pass—Earliest Times
Add activity times along each path in the network (ES + Duration = EF).
Carry the early finish (EF) to the next activity where it becomes its early start (ES) unless…
The next succeeding activity is a merge activity, in which case the largest EF of all preceding activities is selected.

4. Backward Pass—Latest Times
Subtract activity times along each path in the network (LF - Duration = LS).
Carry the late start (LS) to the next activity where it becomes its late finish (LF) unless
The next succeeding activity is a burst activity, in which case the smallest LF of all preceding activities is selected.

5. Problem 1 (CPM)
You are completing a group term paper. Given the project network that follows, complete the forward and backward pass, compute activity slack, and identify the critical path. Use this information to create a Gantt chart for the project. Be sure to show slack for noncritical activities.
Activity
Description
Duration
Predecessor A
Identify topic 1
None B
Research topic 5 C
Draft paper 3 D
Edit paper 2 E
Create graphics F
References G
Final draft
D,E,F

6. Solution
Critical Path Network

7. Solution (cont’d)

8. Problem 2 (CPM) A.Construct a precedence diagram.
B. On the diagram, compute the four schedule dates (ESD, EFD, LSD, LFD) and the four floats (TF, FF, INTF, and IDF) for each activity, and the lag for each link.
C. Identify the critical path.

10. Preliminaries for PERT

11. Problem 3 (PERT)

12. a)Determine the expected value and the variance of the completion time for each activity. b) Use the expected times from (a) to find the critical path. c) Assuming that the normal distribution applies, determine the probability that the critical path will take between 18 and 26 days to complete. d) How much time must be allowed to achieve a 90% probability of timely completion?

13. Solution

14. Solution (cont’d)
b) By using the expected time (mean) of each activity, we find that the critical path is A-C-D.
Remark: For this simple project, we can find the longest path (A-C-D) has the largest expected time, which is the critical path.
The mean critical path duration is
μ=  = 24.
The variance of the critical path duration is the sum of the variances along the path:  
σ2cp = ( ) / 36 = 264/36
so that the standard deviation is readily computed as
σcp =

15. Solution (cont’d)
c) The interval probability may be computed as the difference between two cumulative probabilities as follows: 
P(18 ≤ t  ≤  26) = P(t  ≤  26) - P(t  ≤  18).
Two separate z computations are required. First at 26 we have 
 z26= (26-24) / 2.708=0.739
Then by looking up the normal table with z26=0.739, we have one result that
P(t ≤ 26) = 0.770

17. Solution (cont’d)
Secondly, at 18 we have z18= (18-24) / 2.708= and P(t ≤ 18) = Combining these two results yields the desired probability as P(18 ≤ t ≤ 26) = =

18. Solution (cont’d)
d) For 90% probability, we must pick a z value corresponding to 90% of the area under the normal curve, 50% left of mean and 40% right of mean, so z = Then solving for t we have t = *2.708 = days.

19. Problem 3 (PERT)
a) Determine the expected value and the variance of the completion time for each activity. b) Use the expected times from (a) to find the critical path. c) What is the probability the project will be completed before a scheduled time of 67?

20. Solution a)
Activity times and variances

21. Solution (cont’d) b)

22. Solution (cont’d)

23. Problem 5 (Crashing/Reducing Project Duration)
Use the information contained below to compress one time unit per move using the least cost method. Reduce the schedule until you reach the crash point of the network. For each move identify what activity(s) was crashed, the adjusted total cost, and explain your choice if you have to choose between activities that cost the same.
If the indirect cost for each duration are $1,500 for 17 weeks, $1, for 16 weeks, $1,400 for 15 weeks, $1,350 for 14 weeks, $1,300 for 13 weeks, $1,250 for 12 weeks, $1,200 for 11 weeks, and $1,150 for 10 weeks, what is the optimum cost-time schedule for the project? What is the cost?

24. Problem 5 (Crashing/Reducing Project Duration)

25. Solution

26. Solution (cont’d)

27. Solution (cont’d)

28. Solution (cont’d)