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.) Make sure you do not check the Assume Linear Model box in Solver.

Example 15.3