# Notes

```Project Management
CPM, PERT, Crashing – An Illustrative Example
David S.W. Lai
Sept 19, 2013
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
2
Foundation
-
3
3
Roof Timbers
1
2
4
Roof Sheathing
3
3
5
Electrical Wiring
1
4
6
Roof Shingles
4
8
7
Exterior Siding
8
5
8
Windows
1
2
9
Paint
6, 7, 10
2
10
Inside Wall Board
8, 5
3
Modified from Moore and Weatherford, Decision Modelling, Pearson 2001.
Build-Rite’s engineers have calculated the cost of completing each
activity. Their results are given below.
Normal
Time
Normal
Cost
Crash
Time
Crash
Cost
1
5
50
3
72
2
3
20
2
30
3
2
15
1
30
4
3
8
1
20
5
4
30
4
30
6
8
13
4
21
7
5
45
1
65
8
2
45
1
52
9
2
40
2
40
10
3
22
2
34
e.g. Cost for Activity 1
Cost
Activity
80
70
60
50
40
2
3
4
5
Activity Time
6
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
2
3
4
3
1
2
3
4
1
2
9
5
4
4
4
6
4
8
12
7
1
3
17
8
1
2
3
9
2
2
2
10
2
3
4
Questions
1) Determine the slacks and the critical path.
2) How much would it cost to reduce the project duration by 7
days? 10 days?
3) What is the probability that all the activities on the current
critical path(s) will be completed within 25 days?
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.
Notations
ES: Earliest Start LS: Latest Start
EF: Earliest Finish LF: Latest Finish
ES EF
TS: Total Slack
FS: Free Slack
LS LF
2
3
0
Start
3
2
5
1
8
3
4
6
3
4
5
10
2
8
5
7
2
9
0
END
Step 1: Forward Pass
8
10
10
2
3
0
0
0
0
3
3
2
Start
3
8
12
2
8
Notations
EF
LS
LF
ES: Earliest Start LS: Latest Start
EF: Earliest Finish LF: Latest Finish
6
12
TS: Total Slack
FS: Free Slack
21
23
23
15
2
9
10
10
21
8
3
4
5
8
ES
13
3
4
8
5
1
13
10
15
5
7
0
END
23
Step 2: Backward Pass
8
10
10
13
13
21
8
10
10
13
13
21
2
3
0
0
0
3
3
8
0
0
0
3
3
8
0
3
2
Start
12
12
15
14
18
18
21
5
1
3
4
5
10
8
10
10
15
14
16
16
21
Notations
EF
LS
LF
ES: Earliest Start LS: Latest Start
EF: Earliest Finish LF: Latest Finish
6
8
2
8
ES
8
3
4
TS: Total Slack
FS: Free Slack
5
7
21
23
23
23
21
23
23
23
2
9
0
END
Step 3: Calculating Slacks
The slacks are equal to zero
for all the critical activities.
TS = 0
FS = 0
TS = 0
FS = 0
TS = 0
FS = 0
0
0
0
3
3
8
0
0
0
3
3
8
0
3
2
Start
TS = 0
FS = 0
TS = 0
FS = 0
TS = 0
FS = 0
8
10
10
13
13
21
8
10
10
13
13
21
2
3
12
12
15
14
18
18
21
5
1
4
5
TS = 6
FS = 0
8
10
14
16
TS = 6
FS = 0
Notations
EF
LS
LF
ES: Earliest Start LS: Latest Start
EF: Earliest Finish LF: Latest Finish
6
8
2
8
ES
8
3
4
TS: Total Slack
FS: Free Slack
TS = 6
FS = 6
3
10
10
15
16
21
5
7
TS = 6
FS = 6
TS = 0
FS = 0
TS = 0
FS = 0
21
23
23
23
21
23
23
23
2
9
0
END
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…
How much would it cost to reduce the project duration by 7
days? 10 days?
Activity
Normal
Time
Normal
Cost
Crash
Time
Crash
Cost
Max. Crash Days
Cost per Crash
Day
1
5
50
3
72
2
11
2
3
20
2
30
1
10
3
2
15
1
30
1
15
4
3
8
1
20
2
6
5
4
30
4
30
0
-
6
8
13
4
21
4
2
7
5
45
1
65
4
5
8
2
45
1
52
1
7
9
2
40
2
40
0
-
10
3
22
2
34
1
12
When the task is performed in the normal way without extra resources….
summing up the normal costs for all
• The project cost is \$288 activities
• The shortest possible project duration is 23 days can be determined using CPM
Project duration = 23
Project cost = 288
Critical activities to
consider:
1, 2, 3, 4, 6
Crash
Time
Cost per
Crash Day
1
3
11
2
2
10
3
1
15
4
1
6
3
5
5
4
∞
2
1
6
4
2
7
1
5
8
1
7
9
2
10
2
Crash activity 6 by 4
days for a cost of \$8.
2
3
8
3
4
6
2
4
3
5
10
∞
2
5
12
8
7
9
We may crash an activity for multiple days only when the critical
path(s) remain the same and the cost for crashing is linear.
Project duration = 19
Project cost = 288+8
Critical activities to
consider:
1, 2, 3, 4
Crash
Time
Cost per
Crash Day
1
3
11
2
2
10
3
1
15
4
1
6
3
5
5
4
∞
2
1
6
4
2
7
1
5
8
1
7
9
2
10
2
Crash activity 4 by
2 days for a cost of
\$12.
2
3
4
3
4
6
2
4
3
5
10
∞
2
5
12
8
7
9
Project duration = 17
Project cost = 288+8+12
Crash
Time
Cost per
Crash Day
1
3
11
2
2
10
3
1
15
4
1
6
3
5
5
4
∞
2
1
6
4
2
7
1
5
8
1
7
9
2
10
2
Crash activity
2 by 1 day for
a cost of \$10.
Critical activities to
consider:
1, 2, 3, 7, 8, 10
There are multiple
critical paths.
2
1
4
3
4
6
2
4
3
5
10
∞
2
5
12
8
7
9
• Crash 3, 8 and 10?
• Crash 3, 7 and 10?
• Crash 1?
Project duration = 16
Project cost = 288+8+12+10
Crash
Time
Cost per
Crash Day
1
3
11
2
2
10
3
1
15
4
1
6
3
5
5
4
∞
2
1
6
4
2
7
1
5
8
1
7
9
2
10
2
Crash activity 1
by 2 day for a
cost of \$22.
Critical activities to
consider:
1, 3, 7, 8, 10
2
1
4
3
4
6
2
4
3
5
10
∞
2
5
12
8
7
9
• Crash 3, 8 and 10?
• Crash 3, 7 and 10?
Option 1: Crash
activity
Project duration = 14
Project cost = 288+8+12+10+22
3, 8 and 10
Critical activities to
consider:
3, 7, 8, 10
Crash
Time
Cost per
Crash Day
1
3
11
2
2
10
3
1
15
4
1
6
3
5
5
4
∞
2
1
6
4
2
7
1
5
8
1
7
9
2
∞
2
5
10
2
12
8
7
by 1 day.
The cost is
15+7+12 = 34
2
1
4
3
4
6
2
4
3
5
10
9
Option 2: Crash
activity
3, 7 and 10
by 1 day.
The cost is
15+5+12 = 32
Project duration = 13
Project cost = 288+8+12+10+22+32 = 372
Crash
Time
Cost per
Crash Day
1
3
11
2
2
10
3
1
15
4
1
6
3
5
5
4
∞
2
1
6
4
2
7
1
5
8
1
7
9
2
10
2
Critical activities to
consider:
7, 8
1
1
4
3
4
6
2
4
2
5
10
∞
2
4
12
8
7
9
Crashing 7, 8 or both 7 and 8 will not
change the project duration.
380
360
Project
Cost
340
320
300
280
13
18
Project Duration
23
What is the probability that all the activities on the current
critical path(s) will be completed within 25 days?
Activity
Optimistic
Time
(a)
Most
Probable Time
(m)
Pessimistic
Time
(b)
1
Walls and Ceiling
3
5
7
2
Foundation
2
3
4
3
Roof Timbers
1
2
3
4
Roof Sheathing
1
2
9
5
Electrical Wiring
4
4
4
6
Roof Shingles
4
8
12
7
Exterior Siding
1
3
17
8
Windows
1
2
3
9
Paint
2
2
2
10
Inside Wall Board
2
3
4
Beta Distribution
The graph is taken form http://www.isixsigma.com
Expected Activity Time
Variance
Optimistic
Time
(a)
Most
Probable
Time ( m )
Pessimistic Expected
Time
Activity
(b)
Time
1
3
5
7
5
0.444
2
2
3
4
3
0.111
3
1
2
3
2
0.111
4
1
2
9
3
1.778
5
4
4
4
4
0.000
6
4
8
12
8
1.778
7
1
3
17
5
7.111
8
1
2
3
9
2
2
2
2
2
0.111
0.000
10
2
3
4
3
0.111
Variance
What is the probability that all the activities on the
current critical path(s) will be completed within 25 days?
Expected
Activity
Time
Standard
Deviation
1
5
0.667
2
3
0.333
3
2
0.333
4
3
1.333
3
5
2
1
2
3
8
3
4
6
2
4
3
5
10
5
4
0
6
8
1.333
7
5
2.667
2
8
5
2
0.333
8
9
2
0
7
10
3
0.333
9
What is the probability that all the activities on the
current critical path(s) will be completed within 25 days?
Expected
Activity
Time
Variance
1
5
0.444
2
3
0.111
3
2
0.111
4
3
1.778
5
4
0.000
6
8
1.778
7
5
7.111
8
2
0.111
9
2
0.000
10
3
0.111
• Assume that the activity times are
independent random variables.
• The expected project duration is
E(X) = 5 + 3 + 2 + 3 + 8 + 2 = 23 days
• The corresponding variance is
V(X) = 0.444+0.111+0.111+1.778+1.778+0.000
= 4.222
• Assume that the project duration is normally
distributed (Based on the Central Limit
Theorem)
If we plot
for all T =12,13,…,32,
1.00
0.80
Probability
0.60
0.40
0.20
0.00
12
14
16
18
20
22
24
26
28
30
Project Duration
The project duration estimates could be more complicated when the effect of
the other paths on the project duration become significant.
32
Questions?
```