BU375 Assignment 2

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Wilfrid Laurier University *

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375

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Industrial Engineering

Date

Apr 3, 2024

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docx

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4

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Assignment Questions: 1) (15 marks) For the “Normal” scenario, a. What is the critical path(s) and estimated completion time for this project? Please see chart above. All tasks with 0 slack will make up the critical path, or path A-C-D-G-K-L will be the critical path. Length it will take to complete the project is 27 weeks. b. What is the estimated project cost? Taking a sum of all activities required to finish the project will give the estimated project cost of $205,000 (sum of all normal costs above). As all paths lead to a shared goal of task L, all tasks must be completed in order for the project to be finished, meaning each tasks individual costs should be added to get the total cost.
c. What is the probability that the project can be completed in 26 weeks with this project plan? P(x < 26) = P(z < ((26 – 27) / 1.452966))) = P(z < -0.68825) = 24.51% Taking this equation to find a z-score of -0.688 gives us a probability the project will be completed in 26 weeks of 24.51%. 2) (20 marks) Suppose that the cost of being late to the market is estimated to be $30,000 per week if the project completion time exceeds 26 weeks. Determine the most “Cost-Effective” project completion time. a. Which task(s) would you recommend crashing (and how) to achieve this cheapest completion time and why?
Please see above with tasks highlighted in blue being on the critical path, and task C highlighted in purple as the crashed task (A was not included as it is not crashable). Only tasks on the critical path should be crashed. Multiple tasks have the same crash cost/week on the critical path. The next indicator on which task to crash is the one with the highest variance. Task C and G are again tied in variance. The next criteria is the task with the most predecessors and/or successors. Task C has two successors, aka two paths it effects, while G only has one successor and predecessor. This means that crashing task C will have a greater affect on the project as a whole. After this, crashing task G will cost 10,000, however will only reduce the expected cost ($30,000 * 75.44% - new probability of finishing in 26 weeks) by $7,630.54, meaning crashing the task costs the company $2,369.46 and does not make sense to do. b. What is the total cost of crashing the project to complete the project in 26 weeks and is it worth it and why?
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See crash costs/week. Crashing task C costs $10,000/week, and we will crash one week, making total crash costs $10,000, increasing total project costs to $215,000. c. What is the probability of completing the project in 26 weeks with this project plan? P(x < 26) = P(z < ((26 – 26) / 1.452966))) = P(z < 0) = 50.00% Because the mean is the same as the expected finish time, 50% of data points fall under the mean, meaning that chances of finishing in 26 weeks are 50%. As stated before, the next cheapest/most efficient crashing option’s cost outweighs the potential benefits, meaning 50% is the optimal probability to complete the project in the timeframe.