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PERT / CPM – Time-Cost Tradeoffs BSAD 30 Dave Novak Source: Anderson et al., 2015 Quantitative Methods for Business 13 th edition – some slides are directly from J. Loucks © 2013 Cengage Learning 1
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Overview Time-cost tradeoff Crashing activities Solving the crashing problem using LP 2
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Time-cost tradeoffs Investigating the trade-off between incurring higher costs to reduce the overall time it takes to complete the project Do to this, we must identify: Which activities should be targeted for reductions in completion time? What is the cost associated with reducing those activities? How will reducing the completion time of certain activities impact total project completion time and the critical path? 3
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Crashing activity times To reduce the overall time it takes to complete a project, the time it takes to complete select activities on the critical path must be reduced Activities can be shortened by adding resources (labor, equipment, etc.) for some additional cost The process of reducing activity times by adding resources is called crashing 4
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Crashing activity times To determine where and by how much to crash activity times, we need information about: 1) The “normal” cost of each activity under expected completion time 2) The shortest possible time for each activity that can be achieved by crashing 3) The cost for crashing each activity 5
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Crashing activity times In the Critical Path Method (CPM) approach, it is assumed that the normal time to complete an activity, t j, can be met at a normal cost, c j The activity, j, can be crashed to a reduced time, t j ’, for an increased cost, c j ’ So t j ≥ t j ’ and c j ≤ c j ’ 6
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Crashing activity times Notation 7
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Crashing activity times The maximum time each activity (j) can be reduced is M j M j = t j - t j ' The cost per unit reduction in activity (j), K j, is linear and can be calculated by: K j = (c j ' - c j ) / M j 8
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Example: Earthmover, Inc. 9 EarthMover is a manufacturer of road construction equipment including pavers, rollers, and graders The company is introducing a new line of loaders Management is concerned that the project might take longer than 26 weeks to complete without crashing some activities Assume that there is a huge financial penalty Earthmover must pay for a time overrun
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Example: Earthmover, Inc. 10 Immediate Completion Immediate Completion Activity Description Predecessors Time (wks) Activity Description Predecessors Time (wks) A Study Feasibility --- 6 A Study Feasibility --- 6 B Purchase Building A 4 B Purchase Building A 4 C Hire Project Leader A 3 C Hire Project Leader A 3 D Select Advertising Staff B 6 D Select Advertising Staff B 6 E Purchase Materials B 3 E Purchase Materials B 3 F Hire Manufacturing Staff B,C 10 F Hire Manufacturing Staff B,C 10 G Manufacture Prototype E,F 2 G Manufacture Prototype E,F 2 H Produce First 50 Units G 6 H Produce First 50 Units G 6 I Advertise Product D,G 8 I Advertise Product D,G 8
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Example: Earthmover, Inc. 11 Finish A. C. B. D. F. E. I. G. Start 6 4 3 6 10 3 2 8 H. 6
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Example: Earthmover, Inc. 12 Finish A. C. B. D. F. E. I. G. Start 6 4 3 6 10 3 2 8 H. 6 0 6 6 10 6 9 10 13 10 20 10 16 20 22 22 30 22 28
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Example: Earthmover, Inc. 13
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Example: Earthmover, Inc. 14 The completion time for this project using normal times is 30 weeks If we need to complete the project in 26 weeks, which activities should be crashed, and by how many weeks per activity to meet the 26 week deadline?
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Example: Crashing activity times We need the following information: Activity cost under the normal or expected activity time Time to complete the activity under maximum crashing (i.e., the shortest possible activity time) Activity cost under maximum crashing 15
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Example: Earthmover, Inc. 16
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Example: Crashing activity times Calculate crashing variables Activity A: t A = 6, t A ’ = 5, c A = $80,000, c A ’ = $100,000 M A = K A = 17
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Example: Crashing activity times Calculate crashing variables Activity C: t C = 3, t C ’ = 2, c C = $50,000, c C ’ = $100,000 M C = K C = 18
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Example: Crashing activity times Calculate crashing variables Activity F: t F = 10, t F ’ = 7, c F = $300,000, c F ’ = $480,000 M F = K F = 19
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Example: Earthmover, Inc. 20
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Example: Crashing activity times Keep in mind that the project is dynamic The effect of changing the start, finish, and completion time of activities in the project is going to “cascade” through the entire project The start and finish times of certain activities depends directly on the start and finish times of other activities As you crash activities along the current critical path, other activities may become critical (new critical paths may develop) 21
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Example: Crashing activity times You will need to check the critical path in the revised network as you go For a small network you can do this by trial and error, but the trial and error approach will not work for a larger network We can use Linear Programming (LP) to solve for which activities should be crashed and by how much 22
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Example: LP crashing We are trying to achieve a specific reduction in the project completion time by crashing activities in a least-cost manner 23
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Example: LP crashing 24 We need to set up constraints for each activities that allow us to introduce a slack variable for each activity If an activity has slack, then that particular activity does not need to start at the earliest start (ES)
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Example: LP crashing 25
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Example: LP crashing 26
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Example: LP objective function Y i is the amount of time activity (i) is crashed X i is the Early Finishing time of activity (i) Objective Function 27
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Example: LP constraints 2 different types of constraints 1. The time each activity can be crashed 2. Finish times for each activity based on preceding activities 28
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Example: LP constraints 29
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Example: LP constraints The amount of time each activity can be crashed 1) Y A ≤ 1 2) Y C ≤ 1 3) Y D ≤ 3 4) Y E ≤ 1 5) Y F ≤ 3 6) Y H ≤ 1 7) Y I ≤ 4 30
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Example: LP constraints Finish time constraints based on preceding activities 31 8) X A ≥ 0 + (6 – Y A )16) X G ≥ X F + (2 – Y G ) 9) X B ≥ X A + (4 – Y B )17) X H ≥ X G + (6 – Y H ) 10) X C ≥ X A + (3 – Y C )18) X I ≥ X D + (8 – Y I ) 11) X D ≥ X B + (6 – Y D )19) X I ≥ X G + (8 – Y I ) 12) X E ≥ X B + (3 – Y E )20) X H ≤ 26 13) X F ≥ X B + (10 – Y F )21) X I ≤ 26 14) X F ≥ X C + (10 – Y F ) 15) X G ≥ X E + (2 – Y G )X i, Y i ≥ 0 for all i
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Example: Earthmover, Inc. 33 Finish A. C. B. D. F. E. I. G. Start 5 4 3 6 7 3 2 8 H. 6
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Example: Earthmover, Inc. 34 Finish A. C. B. D. F. E. I. G. Start 5 4 3 6 7 3 2 8 H. 6 0 5 5 9 5 8 9 12 9 16 9 15 16 18 18 26 18 24
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Summary Time-cost tradeoff and crashing Example Using LP to select optimal activities to crash 35
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