Question 17.P.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:
(2) Rank critical activities according to crash costs:
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:
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:
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:
A summary of costs for the preceding schedule would look like this:
The graph of total costs is as follows:
| Path | Duration (weeks) |
| a-b | 24 (critical path) |
| c-d | 19 |
| e-f | 23 |
| Activity | Cost per Week to Crash |
| b | $ 3,000 |
| a | 11,000 |
| Path | Activity | Cost per Week to Crash |
| a-b | b | $ 4,000 |
| a | 11,000 | |
| e-f | f | 2,000 |
| e | 6,000 |
| 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) |
| 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 |
| 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 |
