1. Project Management CPM, PERT, Crashing – An Illustrative Example
David S.W. Lai
Sept 19, 2013

2. The House Construction Problem
SEEM3530 Tutorial 2 Project Management
The House Construction Problem
The Build-Rite Construction Company has identified ten activities that take place in building a house. They are
Activity
Immediate Predecessors
Expected Time (days) 1
Walls and Ceiling 2 5
Foundation - 3
Roof Timbers 4
Roof Sheathing
Electrical Wiring 6
Roof Shingles 8 7
Exterior Siding
Windows 9
Paint
6, 7, 10 10
Inside Wall Board
8, 5
Modified from Moore and Weatherford, Decision Modelling, Pearson 2001.

3. SEEM3530 Tutorial 2 Project Management
Build-Rite’s engineers have calculated the cost of completing each activity. Their results are given below.
e.g. Cost for Activity 1
Activity
Normal
Time
Cost
Crash Time
Crash Cost 1 5 50 3 72 2 20 30 15 4 8 6 13 21 7 45 65 52 9 40 10 22 34

4. SEEM3530 Tutorial 2 Project Management
On the basis of company history, Build-Rite’s management has determined the following time estimates for each activity.
Activity
Optimistic Time ( a )
Most Probable Time ( m )
Pessimistic Time ( b ) 1 3 5 7 2 4 9 6 8 12 17 10

5. Questions Determine the slacks and the critical path.
How much would it cost to reduce the project duration by 7 days? 10 days?
What is the probability that all the activities on the current critical path(s) will be completed within 25 days?

6. Critical Path Method (CPM)
Step 1: Forward pass
Step 2: Backward pass
Step 3: Calculating
Identify the critical path(s).
interpret the meaning of slacks and critical activities.

7. Notations ES EF LS LF
ES: Earliest Start LS: Latest Start TS: Total Slack
EF: Earliest Finish LF: Latest Finish FS: Free Slack 2 3 8 3 4 6 2 3 5 3 4 9
END
Start 2 1 5 10 2 5 8 7

8. Step 1: Forward Pass 8 10 10 13 13 21 2 3 8 3 4 6 21 23 23 3 3 8 8 12 12 15 2 3 5 3 4 9
END
Start 2 1 5 10 8 10 10 15 2 5 8 7
Notations ES EF LS LF
ES: Earliest Start LS: Latest Start TS: Total Slack
EF: Earliest Finish LF: Latest Finish FS: Free Slack

9. SEEM3530 Tutorial 2 Project Management
Step 2: Backward Pass 8 10 8 10 10 13 10 13 13 21 13 21 2 3 8 3 4 6 21 23 21 23 23 23 3 3 3 8 3 8 8 12 8 12 14 18 12 15 18 21 12 15 2 3 5 3 4 9
END
Start 2 1 5 10 8 10 14 16 8 10 10 15 16 21 10 15 2 5 8 7
Notations ES EF LS LF
ES: Earliest Start LS: Latest Start TS: Total Slack
EF: Earliest Finish LF: Latest Finish FS: Free Slack

10. Step 3: Calculating Slacks
SEEM3530 Tutorial 2 Project Management
Step 3: Calculating Slacks
TS = 0
FS = 0
TS = 0
FS = 0
TS = 0
FS = 0
The slacks are equal to zero for all the critical activities. 8 10 8 10 10 13 10 13 13 21 13 21 8
TS = 0
FS = 0
TS = 0
FS = 0 2 3
TS = 0
FS = 0
TS = 0
FS = 0
TS = 0
FS = 0 3 4 6 21 23 21 23 23 23 3 3 3 8 3 8 8 12 8 12 14 18 12 15 18 21 12 15
TS = 6
FS = 6 2 3 5 3 4
TS = 6
FS = 0 9
END
Start 2 1 5 10
TS = 6
FS = 6 8 10 8 10 14 16
TS = 6
FS = 0 10 15 16 21 10 15 2 5 8 7
Notations ES EF LS LF
ES: Earliest Start LS: Latest Start TS: Total Slack
EF: Earliest Finish LF: Latest Finish FS: Free Slack

11. Crashing
In many cases, it is possible to reduce an activity’s duration by spending more money.
To investigate the tradeoff between project duration and project cost…

12. SEEM3530 Tutorial 2 Project Management
How much would it cost to reduce the project duration by 7 days? 10 days?
Activity
Normal
Time
Cost
Crash Time
Crash Cost 1 5 50 3 72 2 20 30 15 4 8 6 13 21 7 45 65 52 9 40 10 22 34
Max. Crash Days
Cost per Crash Day 2 11 1 10 15 6 - 4 5 7 12
When the task is performed in the normal way without extra resources….
The project cost is $288
The shortest possible project duration is 23 days
summing up the normal costs for all activities
can be determined using CPM

13. SEEM3530 Tutorial 2 Project Management
Project duration = 23
Project cost = 288
SEEM3530 Tutorial 2 Project Management
Crash activity 6 by 4 days for a cost of $8.
Critical activities to consider:
1, 2, 3, 4, 6
Crash Time
Cost per Crash Day 1 3 11 2 10 15 4 6 5 ∞ 7 8 9 12 2 3 8 3 4 6 2 3 5 4 3 9 2 1 5 10 2 5 8 7
We may crash an activity for multiple days only when the critical path(s) remain the same and the cost for crashing is linear.

14. SEEM3530 Tutorial 2 Project Management
Project duration = 19
Project cost = 288+8
SEEM3530 Tutorial 2 Project Management
Crash activity 4 by 2 days for a cost of $12.
Critical activities to consider:
1, 2, 3, 4
Crash Time
Cost per Crash Day 1 3 11 2 10 15 4 6 5 ∞ 7 8 9 12 2 3 4 3 4 6 2 3 5 4 3 9 2 1 5 10 2 5 8 7

15. SEEM3530 Tutorial 2 Project Management
Project duration = 17
Project cost =
SEEM3530 Tutorial 2 Project Management
Critical activities to consider:
1, 2, 3, 7, 8, 10
There are multiple critical paths.
Crash Time
Cost per Crash Day 1 3 11 2 10 15 4 6 5 ∞ 7 8 9 12
Crash activity 2 by 1 day for a cost of $10. 2 1 4 3 4 6 2 3 5 4 3 9 2 1 5 10 2 5 8 7
Crash 3, 8 and 10?
Crash 3, 7 and 10?
Crash 1?

16. SEEM3530 Tutorial 2 Project Management
Project duration = 16
Project cost =
SEEM3530 Tutorial 2 Project Management
Critical activities to consider:
1, 3, 7, 8, 10
Crash Time
Cost per Crash Day 1 3 11 2 10 15 4 6 5 ∞ 7 8 9 12
Crash activity 1 by 2 day for a cost of $22. 2 1 4 3 4 6 2 3 5 4 3 9 2 1 5 10 2 5 8 7
Crash 3, 8 and 10?
Crash 3, 7 and 10?

17. SEEM3530 Tutorial 2 Project Management
Project duration = 14
Project cost =
SEEM3530 Tutorial 2 Project Management
Option 1: Crash activity
3, 8 and 10
by 1 day.
The cost is = 34
Critical activities to consider:
3, 7, 8, 10
Crash Time
Cost per Crash Day 1 3 11 2 10 15 4 6 5 ∞ 7 8 9 12 2 1 4 3 4 6 2 3 5 4 3 9 2 1 5 10
Option 2: Crash activity
3, 7 and 10
by 1 day.
The cost is = 32 2 5 8 7

18. SEEM3530 Tutorial 2 Project Management
Project duration = 13
Project cost = = 372
SEEM3530 Tutorial 2 Project Management
Critical activities to consider:
7, 8
Crash Time
Cost per Crash Day 1 3 11 2 10 15 4 6 5 ∞ 7 8 9 12 1 1 4 3 4 6 2 3 5 4 2 9 2 1 5 10 2 4 8 7
Crashing 7, 8 or both 7 and 8 will not change the project duration.

19. SEEM3530 Tutorial 2 Project Management
Time-Cost Trade-Off
Project
Cost
Project Duration

20. What is the probability that all the activities on the current critical path(s) will be completed within 25 days?
Task
Activity
Optimistic Time
( a )
Most Probable Time ( m )
Pessimistic Time
( b ) 1
Walls and Ceiling 3 5 7 2
Foundation 4
Roof Timbers
Roof Sheathing 9
Electrical Wiring 6
Roof Shingles 8 12
Exterior Siding 17
Windows
Paint 10
Inside Wall Board

21. SEEM3530 Tutorial 2 Project Management
Beta Distribution
The graph is taken form

22. Expected Activity Time Variance
Task
Optimistic Time
( a )
Most Probable Time ( m )
Pessimistic Time
( b ) 1 3 5 7 2 4 9 6 8 12 17 10
Expected Activity Time
Variance 5
0.444 3
0.111 2
1.778 4
0.000 8
7.111

23. Expected Activity Time
What is the probability that all the activities on the current critical path(s) will be completed within 25 days?
Task
Expected Activity Time
Standard
Deviation 1 5
0.667 2 3
0.333 4
1.333 6 8 7
2.667 9 10 2 3 8 3 4 6 2 3 5 4 3 9 2 1 5 10 2 5 8 7

24. Expected Activity Time
What is the probability that all the activities on the current critical path(s) will be completed within 25 days?
Assume that the activity times are independent random variables.
The expected project duration is
E(X) = = 23 days
The corresponding variance is
V(X) =
= 4.222
Assume that the project duration is normally distributed (Based on the Central Limit Theorem)
Task
Expected Activity Time
Variance 1 5
0.444 2 3
0.111 4
1.778
0.000 6 8 7
7.111 9 10

25. SEEM3530 Tutorial 2 Project Management
If we plot for all T =12,13,…,32,
Probability
Project Duration
The project duration estimates could be more complicated when the effect of the other paths on the project duration become significant.

26. SEEM3530 Tutorial 2 Project Management
Questions?