Project Management CPM PERT Crashing An Illustrative Example

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.

Activity 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 6 8 13 4 21 7 5 45 1 65 8 2 45 1 52 9 2 40 10 3 22 2 34 Cost Build-Rite’s engineers have calculated the cost of completing each activity. Their results are given below. e. g. Cost for Activity 1 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 EF ES: Earliest Start LS: Latest Start EF: Earliest Finish LF: Latest Finish LS LF 2 3 0 Start 3 2 5 1 TS: Total Slack FS: Free Slack 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 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 2 3 0 0 0 3 3 8 0 3 2 Start 5 1 12 12 15 14 18 18 21 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 2 9 0 END

Step 3: Calculating Slacks The slacks are equal to zero for all the critical activities. TS = 0 FS = 0 0 3 3 8 0 0 0 3 3 8 0 3 2 Start 5 1 TS = 0 FS = 0 8 10 10 13 13 21 2 3 12 12 15 14 18 18 21 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 3 TS = 6 FS = 6 10 10 15 16 21 5 7 TS = 6 FS = 6 TS = 0 FS = 0 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 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 0 - 10 3 22 2 34 1 12 When the task is performed in the normal way without extra resources…. • The project cost is $288 summing up the normal costs for all 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 ∞ 6 4 2 2 1 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 ∞ 6 4 2 2 1 7 1 5 8 1 7 9 2 10 2 Crash activity 4 by 2 days for a cost of $12. 2 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 ∞ 6 4 2 2 1 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 ∞ 6 4 2 2 1 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 ∞ 6 4 2 2 1 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 ∞ 6 4 2 2 1 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.

Time-Cost Trade-Off 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? Task 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 Task 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 0. 000 6 4 8 12 8 1. 778 7 1 3 17 5 7. 111 8 1 2 3 9 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? Task Expected Activity Time Standard Deviation 1 5 0. 667 2 3 0. 333 3 2 0. 333 4 3 5 2 1 2 3 8 3 4 6 2 4 3 5 10 3 1. 333 5 4 0 6 8 1. 333 7 5 2. 667 2 5 8 2 0. 333 8 9 7 2 0 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? 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 • 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+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?