Post on 24-Dec-2015
Project ManagementCPM, PERT, Crashing – An Illustrative Example
David S.W. LaiSept 19, 2013
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.
The House Construction ProblemThe Build-Rite Construction Company has identified ten activities that take place in building a house. They are
Activity NormalTime
Normal Cost
Crash Time
Crash Cost
1 5 50 3 722 3 20 2 303 2 15 1 304 3 8 1 205 4 30 4 306 8 13 4 217 5 45 1 658 2 45 1 529 2 40 2 40
10 3 22 2 34
2 3 4 5 64050607080
Activity Time
Cost
e.g. Cost for Activity 1
Build-Rite’s engineers have calculated the cost of completing each activity. Their results are given below.
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
On the basis of company history, Build-Rite’s management has determined the following time estimates for each activity.
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 passStep 2: Backward passStep 3: Calculating
Identify the critical path(s).interpret the meaning of slacks and critical activities.
2 1
3
5
8 7
9 ENDStart
0 3 5
2
4
2
83
0
4
10
6
3
5
2
NotationsES EF
LS LF
ES: Earliest Start LS: Latest Start TS: Total SlackEF: Earliest Finish LF: Latest Finish FS: Free Slack
Step 1: Forward Pass
2 1
3
5
8 7
9 ENDStart
Notations
ES: Earliest Start LS: Latest Start TS: Total SlackEF: Earliest Finish LF: Latest Finish FS: Free Slack
0 3 5
2
4
2
83
0
4
10
6
3
5
2
0 0 0 3 3 8
8 10 10 13 13 21
21 23 23 23
8 12 12 15
8 10 10 15
ES EF
LS LF
2 1
3
5
8 7
9 ENDStart
Notations
ES: Earliest Start LS: Latest Start TS: Total SlackEF: Earliest Finish LF: Latest Finish FS: Free Slack
0 3 5
2
4
2
83
0
4
10
6
3
5
2
0 0 0 3 3 8
8 10 10 13 13 21
21 23 23 23
8 12 12 15
8 10 10 15
0 0
0 0
0 3
0 3
3 8
3 8
8 10
8 10
10 13
10 13
13 21
13 21
21 23
21 23
23 23
23 238 12
14 18
12 15
18 21
8 10
14 16
10 15
16 21
Step 2: Backward Pass
ES EF
LS LF
2 1
3
5
8 7
9 ENDStart
Notations
ES: Earliest Start LS: Latest Start TS: Total SlackEF: Earliest Finish LF: Latest Finish FS: Free Slack
0 3 5
2
4
2
83
0
4
10
6
3
5
2
0 0 0 3 3 8
8 10 10 13 13 21
21 23 23 23
8 12 12 15
8 10 10 15
0 0
0 0
0 3
0 3
3 8
3 8
8 10
8 10
10 13
10 13
13 21
13 21
21 23
21 23
23 23
23 238 12
14 18
12 15
18 21
8 10
14 16
10 15
16 21
TS = 0FS = 0
TS = 0FS = 0
TS = 0FS = 0
TS = 0FS = 0
TS = 0FS = 0
TS = 0FS = 0
TS = 0FS = 0
TS = 0FS = 0
TS = 6FS = 0
TS = 6FS = 6
TS = 6FS = 0
TS = 6FS = 6
The slacks are equal to zero for all the critical activities.
ES EF
LS LF
Step 3: Calculating Slacks
• 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…
Crashing
Max. Crash Days Cost per Crash Day
2 11
1 10
1 15
2 6
0 -
4 2
4 5
1 7
0 -
1 12
Activity NormalTime
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
How much would it cost to reduce the project duration by 7 days? 10 days?
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
2 1
3
5
8 7
93 5
2
4
2
83
4
10
6
3
5
2
Project duration = 23Project cost = 288 Crash activity 6 by 4
days for a cost of $8.
Crash Time
Cost per Crash Day
1 3 11
2 2 10
3 1 15
4 1 6
5 4 ∞
6 4 2
7 1 5
8 1 7
9 2 ∞10 2 12
Critical activities to consider:
1, 2, 3, 4, 6
We may crash an activity for multiple days only when the critical path(s) remain the same and the cost for crashing is linear.
2 1
3
5
8 7
93 5
2
4
2
43
4
10
6
3
5
2
Project duration = 19Project cost = 288+8
Crash activity 4 by 2 days for a cost of $12.
Crash Time
Cost per Crash Day
1 3 11
2 2 10
3 1 15
4 1 6
5 4 ∞
6 4 2
7 1 5
8 1 7
9 2 ∞10 2 12
Critical activities to consider:1, 2, 3, 4
2 1
3
5
8 7
93 5
2
4
2
41
4
10
6
3
5
2
Project duration = 17Project cost = 288+8+12
Crash Time
Cost per Crash Day
1 3 11
2 2 10
3 1 15
4 1 6
5 4 ∞
6 4 2
7 1 5
8 1 7
9 2 ∞10 2 12
Crash activity 2 by 1 day for a cost of $10.
There are multiple critical paths.
Critical activities to consider:
1, 2, 3, 7, 8, 10
• Crash 3, 8 and 10?• Crash 3, 7 and 10?• Crash 1?
2 1
3
5
8 7
93 5
2
4
2
41
4
10
6
3
5
2
Project duration = 16Project cost = 288+8+12+10
Crash Time
Cost per Crash Day
1 3 11
2 2 10
3 1 15
4 1 6
5 4 ∞
6 4 2
7 1 5
8 1 7
9 2 ∞10 2 12
Crash activity 1 by 2 day for a cost of $22.
• Crash 3, 8 and 10?• Crash 3, 7 and 10?
Critical activities to consider:
1, 3, 7, 8, 10
2 1
3
5
8 7
93 5
2
4
2
41
4
10
6
3
5
2
Project duration = 14Project cost = 288+8+12+10+22
Crash Time
Cost per Crash Day
1 3 11
2 2 10
3 1 15
4 1 6
5 4 ∞
6 4 2
7 1 5
8 1 7
9 2 ∞10 2 12
Critical activities to consider:3, 7, 8, 10
Option 2: Crash activity
3, 7 and 10 by 1 day.The cost is 15+5+12 = 32
Option 1: Crash activity
3, 8 and 10 by 1 day.The cost is 15+7+12 = 34
2 1
3
5
8 7
93 5
1
4
2
41
4
10
6
2
4
2
Project duration = 13Project 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
5 4 ∞
6 4 2
7 1 5
8 1 7
9 2 ∞10 2 12
Critical activities to consider:
7, 8
Crashing 7, 8 or both 7 and 8 will not change the project duration.
Time-Cost Trade-Off
Project Duration
ProjectCost
13 18 23280290300310320330340350360370380
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 2 3 43 Roof Timbers 1 2 34 Roof Sheathing 1 2 95 Electrical Wiring 4 4 4
6 Roof Shingles 4 8 127 Exterior Siding 1 3 178 Windows 1 2 39 Paint 2 2 2
10 Inside Wall Board 2 3 4
Beta Distribution
The graph is taken form http://www.isixsigma.com
Task 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
Expected Activity TimeVariance
Expected Activity
Time
Variance
5 0.444
3 0.111
2 0.111
3 1.778
4 0.000
8 1.778
5 7.111
2 0.1112 0.000
3 0.111
What is the probability that all the activities on the current critical path(s) will be completed within 25 days?
2 1
3
5
8 7
93 5
2
4
2
83
4
10
6
3
5
2
Task Expected Activity
Time
StandardDeviation
1 5 0.667
2 3 0.333
3 2 0.333
4 3 1.333
5 4 0
6 8 1.333
7 5 2.667
8 2 0.333
9 2 0
10 3 0.333
• 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)
Task 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
What is the probability that all the activities on the current critical path(s) will be completed within 25 days?
12 14 16 18 20 22 24 26 28 30 320.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00
If we plot for all T =12,13,…,32,
The project duration estimates could be more complicated when the effect of the other paths on the project duration become significant.
Project Duration
Probability
Questions?