WOOD 492 MODELLING FOR DECISION SUPPORT

1. WOOD 492 MODELLING FOR DECISION SUPPORT Lecture 23 CPM

2. 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. Wood 492 - Saba Vahid

3. 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 Wood 492 - Saba Vahid

4. 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-Finish44 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 Wood 492 - Saba Vahid

5. Critical Path • Critical path: Start-A-B-C-E-F-J-L-N-Finish44 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? Wood 492 - Saba Vahid

6. 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 Crash cost Normal Normal cost Duration Crash time Normal time Wood 492 - Saba Vahid

7. 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 Wood 492 - Saba Vahid

8. 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 Wood 492 - Saba Vahid

9. 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) Wood 492 - Saba Vahid

10. 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 Wood 492 - Saba Vahid

11. The LP for cost-time trade-off, Example 15 • Minimize Z = Normal Cost + • Subject to: • (maximum reduction constraints) • , ,… (non-negativity constraints) …. (start time constraints/precedence) • (finish time constraint) Example 15 Wood 492 - Saba Vahid