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**WOOD 492 MODELLING FOR DECISION SUPPORT**

Today we will talk about the format of the course and some introductions to operations research and its applications Lecture 23 CPM

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**Review Minimum spanning tree**

Goal: find the smallest network that has a path between each two nodes (application example: railway networks) Greedy algorithm solution : start from any node and in each step pick the closest unconnected node and add it to the network Critical path method (CPM) for time-cost trade-off Goal: find the optimal plan to expedite some activities within a project in order to minimize the costs while meeting the project deadline First, we visualize the activities within a project using a network (each activity is represented with a node in the network) Then we find the “critical path”, which is the longest path through the project network, going from start to finish nodes. Oct 31, 2012 Wood Saba Vahid

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**Example 15 – Project network**

“Reliable Constructions Co.” has identified the activities within a plant construction project The deadline is in 40 weeks Project budget is $5.4 million Based on activity information we create the project network How long will the project take? We need to find the paths through the project network and their lengths Path: one of the routes following the arcs from Start to Finish node Length: sum of the estimated durations of the activities along that path Example 15 Oct 31, 2012 Wood Saba Vahid

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**Example 15 - Critical Path Method**

There are six paths for this network Start-A-B-C-D-G-H-M-Finish 40 weeks Start-A-B-C-E-H-M-Finish 31 weeks Start-A-B-C-E-F-J-K-N-Finish 43 weeks Start-A-B-C-E-F-J-L-N-Finish 44 weeks Start-A-B-C-I-J-K-N-Finish 41 weeks Start-A-B-C-I-J-L-N-Finish 42 weeks Activities on each path are done sequentially (can not overlap), so the project length can not be shorter than the length of a given path, but it might be longer e.g. activity H has 2 predecessors (E, and G which is not on the path). Looking at the duration of activities after C, we see that activity E only takes 4 weeks, while activity D and G take 13 weeks combined. Since H needs to wait for both E and G to finish, the project will take much longer than the 31 weeks estimated length for the second path Critical Path Oct 31, 2012 Wood Saba Vahid

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**Critical Path Critical path: Start-A-B-C-E-F-J-L-N-Finish 44 weeks**

The longest path through the project network May have more than one critical path (same length) Activities on the longest path can be done sequentially without having to wait for activities on other paths, so the project length is equal to the length of this path All other paths will reach the Finish node sooner than the critical path Project length is 44 wks: longer than the 40 wks deadline The management should focus on reducing the length of activities on the critical path to finish the project by the deadline Which activities to choose? Oct 31, 2012 Wood Saba Vahid

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**CPM for time-cost trade-offs**

Crashing an activity: using costly measured to reduce its length CPM: determines how much (if any) to crash each activity to reduce the estimated duration of the project Time-cost graph : shows the relationship between time and cost of an activity in the normal and crash modes Assumptions: Maximum time reduction and associated crash costs For each activity can be estimated with certainty Partially crashing an activity is possible, the associated cost and time move along the line segment in the graph Time-cost trade-off data are required for each activity Cost Crash cost Crash Normal Normal cost Crash time Normal time Duration Oct 31, 2012 Wood Saba Vahid

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**Cost-time trade-off Example of cost-time trade-off for an activity**

Activity J (putting up the wallboard) It’s possible to reduce the duration of this activity by two weeks (through hiring temporary workers and using overtime) Normal time: 8 weeks Normal Cost: $430,000 Crash time: 6 weeks Crash Cost:$490,000 Maximum reduction in time=8 – 6 = 2 weeks Crash cost per week saved = (490,000 – 430,000)/2 = $30,000 So for each week the company saves in time, $30,000 are required in extra costs Example 15 Oct 31, 2012 Wood Saba Vahid

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**Which activities to crash?**

How to select the activities to crash, while minimizing the costs of doing so? Marginal Cost analysis Linear programming Marginal Cost Analysis select the longest path through the network Among the activities on that path, select the one with the lowest “crash cost per week” Reduce its length by one week if possible, if not move to the next lowest crash cost/week Review the length of all paths after this reduction and repeat steps 1 to 3 until you reach the desired length of the project Example 15 Oct 31, 2012 Wood Saba Vahid

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**Using LP to choose activities to crash**

The objective: minimize the total cost of crashing the activities Variables: xj : reduction in activity j duration by crashing (j=A,B,..,N) Constraints: Total project duration is less than the desired length The reductions can not be more than the maximum allowed reduction The reductions must be non-negative Precedence relationships between activities must hold (we need to define extra variables to write these constraints) Oct 31, 2012 Wood Saba Vahid

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**Precedence relationships**

Define new variables: Yj = start time of activity j (for j=B,C,…,N) The start time of each activity has to be after all its predecessors are finished, so Start time of activity j >= (start time + duration) of its immediate predecessor Duration of each activity j= its normal time – xj For example, immediate predecessor of activity F is E, so: YF >= YE + 4 – XE For activities with more than one predecessor, more inequalities are needed Oct 31, 2012 Wood Saba Vahid

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**The LP for cost-time trade-off, Example 15**

Minimize Z = Normal Cost + (crash cost per week of activity j ∗ reduction in activity j duration) Subject to: 𝑥 𝐴 ≤1, 𝑥 𝐵 ≤2,… (maximum reduction constraints) 𝑥 𝐴 ≥0, 𝑥 𝐵 ≥0,… (non-negativity constraints) 𝑦 𝐵 ≥0+2− 𝑥 𝐴 𝑦 𝐶 ≥ 𝑦 𝐵 +4− 𝑥 𝐵 𝑦 𝐹𝑖𝑛𝑖𝑠ℎ ≥ 𝑦 𝑀 +2 − 𝑥 𝑀 𝑦 𝐹𝑖𝑛𝑖𝑠ℎ ≥ 𝑦 𝑁 +6 − 𝑥 𝑁 …. (start time constraints/precedence) 𝑦 𝐹𝑖𝑛𝑖𝑠ℎ ≤40 (finish time constraint) Example 15 Oct 31, 2012 Wood Saba Vahid

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1 1 Slide © 2005 Thomson/South-Western Q 5 – 13 x 1 = the probability that Station A will take Sitcom Rerun x 2 = the probability that Station A will take.

1 1 Slide © 2005 Thomson/South-Western Q 5 – 13 x 1 = the probability that Station A will take Sitcom Rerun x 2 = the probability that Station A will take.

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