Project Management assignment on CPM

Résumé du document

Demand at the Winder Widget Company (W2C) has recently decreased due to the economic slowdown. In response, the managers of W2C have shut down one of their production lines. However, management is concerned about how long it would take to restart the line when (and if) demand increases. They have come to you for help.

After completing a careful study of the production process, you have determined from your work breakdown structure [...]

Extraits

[...] Here are the results summarized in the table: Table 2 presents the Earliest Starting and Finish Time, the Latest Starting and Finish Time, the slack times and the critical tasks: The model to find the Latest Starting Time and Finish Time is the following: The constraints are on all the Latest Finish Times to END). The Latest Starting Time is calculated by subtracting the Latest Finish Time and the Duration. LS = Latest Starting Time, LF = Latest Finish Time Duration ( LF(X) - Duration(X) = LS(X) We want to maximize the sum of the finish times We have to initialize the LS(END) = EF(END) because we want the Total time of the project to be the same in both cases (Earliest and Latest). [...]

[...] Therefore, we can't change the early starting time of D by changing its duration. ii) If we want to change the early start time of we have to change the early Finish time of G and thus change the early start time of G. ( D must become a critical task: if the duration of D becomes 8 current duration + slack the early start time of I becomes 23. iii) In this case (see above), we also change the earliest possible completion time of the project, which becomes 29. [...]

[...] We compress G. First, what happens if we compress the task with the lowest compressing cost, which is on the critical path? ( START-B-C-G-END becomes also a critical path! So if we want to have an earliest starting time of 15 weeks at we have to compress tasks on both paths, which raise the costs. How much does it cost? Compress F by 2 weeks ( F lasts 3 weeks Compress E by 1 week ( E lasts 5 weeks Total of 15 weeks to get to we have to reduce C Compress C by two weeks ( C lasts 5 weeks ( Total cost of 30$ The other solution is to reduce G by two weeks and reduce F by one week: Compress G by two weeks ( 10+10=20$, G lasts 8 weeks Compress F by 1 week ( F lasts 4 weeks ( Total cost of 25$ The second solution is the best because it's the cheapest to achieve the goal of reducing the project by 3 weeks. [...]

[...] The earliest finish time found in question is 28 weeks. For each week over 20, we would get a penalty which is a relatively high cost compared to the cost of compressing tasks (highest for 11$). If we don't compress any task, we have to pay 9*8=72$ penalty. On the other side, reducing tasks to go under 20 weeks can be interesting only if the bonus fee is higher than the reducing cost. ( But is it possible to go under 20 weeks anyway? [...]

[...] Exercise II: In and the tasks A,B,C and D follow a Finish-to-Start relationship In the tasks A,B,C and D follow a Start-to-Start relationship This case can be summarized by the following table: The critical path is clearly START-B-D-END. Only the tasks A and C have a total slack time of 3. If all the tasks take a lag of one day, here is how it influences the earliest/latest starting/finishing times: Since all the links take a lag, no critical task changes, i.e. the critical path does not change. The slack times of tasks A and C do not change either. [...]