In the crashing model in Example 15.3, we assumed that the cost per day crashed is constant. This is often unrealistic. For example, it might cost $300 to decrease the duration of an activity from 10...


In the crashing model in Example 15.3, we assumed that the cost per day crashed is constant. This is often unrealistic. For example, it might cost $300 to decrease the duration of an activity from 10 days to 9 days, but it might cost $450 to reduce it from 9 days to 8 days. One possible way to model this is to assume that the crashing cost, c(d), for reducing the duration by d days is a quadratic: c(d) = cd2 for some constant c
 0. This function produces the “increasing cost per day” behavior frequently seen. To try it out in Example 15.3, suppose the crashing cost for activity H, wiring offices, exhibits this quadratic behavior, with c = 300. Then, for example, the cost of reducing the duration of activity H from 12 days to 9 days is c(3) = 300(3)2 = $2700. Modify the Project Crashing Linear.xlsx model to accommodate this quadratic function, and then optimize to meet a deadline of 54 days. (You can still assume that activity H can be crashed by a maximum of 4 days.) Now you must use GRG Nonlinear Solver.


EXAMPLE 15.3 MEETING A DEADLINE FOR THE LAN PROJECT


From the CPM calculations in Example 15.1, the insurance company knows that if the LAN activities continue to take as long as listed in Table 15.2, the entire project will take 62 working days to complete. However, the project manager is under pressure to finish the job in 56 working days. He estimates that each activity could be crashed by a certain amount at a certain cost. Specifically, he estimates the cost per day of activity time reduction and the maximum possible days of reduction for each activity, as shown in Table 15.5. For example, activity A’s duration could be reduced from 10 days to 9 days at cost $600, or it could be reduced from 10 days to 8 days at cost $1200. (It is even possible to have a fractional reduction, such as from 10 days to 8.5 days at cost $900.) On the other hand, note that three of the activities cannot be crashed at all, probably due to technical considerations. How can the deadline be met at minimum cost?


Objective To use a Solver model to decide how much to crash each activity so that the deadline is met at minimum cost.


WHERE DO THE NUMBERS COME FROM?


The numbers in Table 15.5 are not necessarily easy to obtain. The project manager probably has some idea of the minimum possible time to perform any activity, regardless of the amount spent. For example, wiring offices takes a minimal amount of time, regardless of how many people are working on it. He probably also has a good idea of what it would take to expedite any activity—extra workers, for example—and the corresponding cost.

Dec 06, 2021
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