Project Crashing

6/Allocating Resources to the Project

Solution to Project Crashing HW Problem 24 with Additional Instructions

24. This problem is just like 21, just with different information. This time we are not given Normal Costs. And we are given Crash Costs per week directly, instead of needing to calculate them. In answering the question, rather than give the new total cost, we will just give the additional crashing costs. See the spreadsheet screenshot on the next page for the Activity-on-Node network for this project.

The CP is:

C-F-H (duration: 23 weeks)

The other paths are:

A-D-H (12 weeks)

B-E-H (16 weeks)

C-G-I (18 weeks)

To get to 16 weeks, both the CP and C-G-I will need to be crashed. Begin with the CP by crashing H the maximum allowed three weeks for $30. Then move to the next cheapest activity to crash, which is F. Crash it four weeks for $80. That takes care of the CP. Now work on C-G-I. Crash activity I two weeks for $60. Total Crash Costs are $30 + $80 + $60 = $170. Use this approach to do homework problem 21 “by hand”—it will work. (But it didn’t completely work here—we can do better!)

For this problem, we missed $10 in savings. How? The two paths we needed to crash have a common activity, but our method led us to crash activities that were only on one path or the other. We didn’t get any synergies. It turns out we can save another $10 overall in crashing costs if instead of doing what we did, we crash the common activity C the allowed one week even though it is more expensive (which adds $40) and reduce our crashing of F and I by one week each (which subtracts $50). So our additional costs are only $160.

How would you have ever seen that?? As you can see, it doesn’t take a very complicated project to have the danger of this happening. What’s a project manager to do? Use Solver!

(If you don’t know how to use Excel Solver and are continuing on in the MPS program, you will learn how to use it in SCM 530 this July.)

This worksheet was used with Excel Solver to find the solution. Solver was set to minimize N11 by changing cells K4, K7, K9, and K10 subject to the constraints:

· J19 = 16, where the formula in J19 is “=max(J14:J17)”

· K4 ≥ L4

· K7 ≥ L7

· K9 ≥ L9

· K10 ≥ L10

You should be able to figure out what formulas need to go into cells K4, K7, K9, K10 and N4, N7, N9, N10 and N11. The other column K values are constants (or could just be left blank).

70