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‘Crashing’ – reducing task durations by increased costs

‘Crashing’ – reducing task durations by increased costs. ( lecture ). 0. 0. 1. 0. 0. c. b. c. d. e. a. b. a. d. e. 0. 0. 4. 6. 6. 8. 0. 1. 0. 1. 0. 6. 11. 4. 10. 8. 4. 2. 5. 5. 2. 2. 2. 2. 4. 2. 4. 9. 6. 0. 0. 7. 11. 6. 4. 9. 11. 0. f. f. 11.

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‘Crashing’ – reducing task durations by increased costs

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  1. ‘Crashing’ – reducing task durations by increased costs (lecture)

  2. 0 0 1 0 0 c b c d e a b a d e 0 0 4 6 6 8 0 1 0 1 0 6 11 4 10 8 4 2 5 5 2 2 2 2 4 2 4 9 6 0 0 7 11 6 4 9 11 0 f f 11 0 14 3 3 11 14 Definition of crashing • Obtaining reduction in time at an increased cost (increasing the employed resources). • Cost-slope: the cost of reducing duration time by unit time. • Let’s see the following example:

  3. Procedure for crashing • Crash one time unit at a time • Only crashing critical path activities has any effect on TPT • Crash the that activity first that is the cheapest to reduce in time • Be aware of multiple critical paths • Stop crashing when: • the crash-time is reached at every activity, • benefits of possible crashing are lower than crashing costs.

  4. Crashing table • If the costs to reduce times are known, then a table can be set up showing the relative costs for the reduction in time of each activity by a constant amount. • Crash-time is the minimum duration of an activity. It is given by technical factors. Benefit of reducing TPTby one day: 400€/day

  5. Solution method • step: identify the critical activities • step: find the critical activity with cheapest crash cost, and if its cost slope is lower than the daily benefit from crashing, reduce its duration with one day. If there is no activity to crash, or it is too costly, stop crashing and go to step 4. • step: reidentify the critical path, and go back to step two. • step: identify the newest critical path, TPT and the total net benefit of crashing.

  6. Solution After crashing: • there are two critical paths • TPT is 10 days • total benefit of crashing is €890

  7. g b a c d e f 0 3 5 2 2 3 3 3 0 Example 2 (for individual work) • Identify the critical path and the TPT. 7

  8. 0 0 0 3 0 0 0 e f d g c b a 12 7 5 3 3 0 6 0 3 0 0 0 1 0 15 5 6 9 7 12 3 2 3 5 3 2 3 3 0 9 4 12 3 7 5 7 5 12 3 7 12 15 Example 2 (for individual work) Critcal: a-b-d-e-g TPT: 15 Using tbe table on the next slide, calculate the optimal TPT with crashing. 7

  9. Benefit of reducing TPTby one day: 1200€/day • What is the new TPT? • What is the total profit on crashing? 10 days €3000

  10. Reading • Textbook chapter 8. pp. 61-63.

  11. Thank you for listening

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