Question 17.SP.5: Costs for a project are $12,000 per week for as long as the ...
Costs for a project are $12,000 per week for as long as the project lasts. The project manager has supplied the cost and time information shown. Use the information to:
a. Determine an optimum crashing plan
b. Graph the total costs for the plan
| Activity | Crashing Potential (weeks) |
Cost per Week to Crash |
| a | 3 | $11,000 |
| b | 3 | 3,000 first week, |
| $ 4,000 others | ||
| c | 2 | 6,000 |
| d | 1 | 1,000 |
| e | 3 | 6,000 |
| f | 1 | 2,000 |
a. (1) Compute path lengths and identify the critical path:
| Path | Duration (weeks) |
| a–b | 24 (critical path) |
| c–d | 19 |
| e–f | 23 |
(2) Rank critical activities according to crash costs:
| Activity | Cost per Week to Crash |
| b | $ 3,000 |
| a | 11,000 |
Activity b should be shortened one week since it has the lower crashing cost. This would reduce indirect costs by $12,000 at a cost of $3,000, for a net savings of $9,000. At this point, paths a–b and e–f would both have a length of 23 weeks, so both would be critical.
(3) Rank activities by crashing costs on the two critical paths:
| Path | Activity | Cost per Week to Crash |
| a–b | b | $ 4,000 |
| a | 11,000 | |
| e–f | f | 2,000 |
| e | 6,000 |
Choose one activity (the least costly) on each path to crash: b on a–b and f on e–f, for a total cost of $4,000 + $2,000 = $6,000 and a net savings of $12,000 − $6,000 = $6,000.
(4) Check to see which path(s) might be critical: a–b and e–f would be 22 weeks in length, and c–d would still be 19 weeks.
(5) Rank activities on the critical paths:
| Path | Activity | Cost per Week to Crash |
| a–b | b | $ 4,000 |
| a | 11,000 | |
| e–f | e | 6,000 |
| f | (no further crashing possible) |
Crash b on path a–b and e on e–f for a cost of $4,000 + $6,000 = $10,000, for a net savings of $12,000 − $10,000 = $2,000.
(6) At this point, no further improvement is possible: paths a–b and e–f would be 21 weeks in length, and one activity from each path would have to be shortened. This would mean activity a at $11,000 and e at $6,000 for a total of $17,000, which exceeds the $12,000 potential savings in costs.
b. The following table summarizes the results, showing the length of the project after crashing n weeks:
| Path | n = 0 | 1 | 2 | 3 |
| a–b | 24 | 23 | 22 | 21 |
| c–d | 19 | 19 | 19 | 19 |
| e–f | 23 | 23 | 22 | 21 |
| Activity crashed | b | b,f | b,e | |
| Crashing costs ($000) | 3 | 6 | 10 |
A summary of costs for the preceding schedule would look like this:
| Project Length |
Cumulative Weeks Shortened |
Cumulative Crashing Costs ($000) |
Indirect Costs ($000) |
Total Costs ($000) |
| 24 | 0 | 0 | 24(12) = 288 | 288 |
| 23 | 1 | 3 | 23(12) = 276 | 279 |
| 22 | 2 | 3 + 6 = 9 | 22(12) = 264 | 273 |
| 21 | 3 | 9 + 10 = 19 | 21(12) = 252 | 271 |
| 20 | 4 | 19 + 17 = 36 | 20(12) = 240 | 276 |
The graph of total costs is as follows:
