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

2. 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: a 0 0 4 b 2 c 2 e 2 d 5 f 3 a 0 40 0 4 4 0 b 4 60 4 6 2 0 c 6 81 7 9 2 0 e 8 101 9 11 2 1 d 6 0 6 5 0 f 140 11 14 3 0

3. Procedure for crashing 1.Crash one time unit at a time 2.Only crashing critical path activities has any effect on TPT 3.Crash the that activity first that is the cheapest to reduce in time 4.Be aware of multiple critical paths 5.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. Activity (label) Duration (day) Float (day) Crash time Cost-slope (€/day) a402100 b202150 c211110 d503200 e211160 f302500 Benefit of reducing TPT by one day: 400 €/day

5. Solution method 1.step: identify the critical activities 2.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. 3.step: reidentify the critical path, and go back to step two. 4.step: identify the newest critical path, TPT and the total net benefit of crashing.

6. Solution Path durations Path / activity crached normalstep 1step 2step 3step 4step 5 Cost: Cumulated net benefit: Path durations Path / activity crashed normalstep 1step 2step 3step 4step 5 –aadd, cnone a-b-c-e-f131211 10– a-b-d-f1413121110– Cost:–100 200310– Cumulated net benefit: –300600800890– After crashing: – there are two critical paths – TPT is 10 days – total benefit of crashing is €890

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

8. 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. b 3 50 3 5 2 0 a 0 30 0 3 3 0 c 3 61 4 7 3 0 d 5 70 5 7 2 0 e 7 120 7 5 0 f 6 93 9 3 3 g 150 12 15 3 0 7

9. Activity (label) Normal duration (day) Float (day) Crash time Cost- slope (€/day) a31500 b21550 c21150 d53900 e54400 f32100 g33200 Benefit of reducing TPT by one day: 1200 €/day What is the new TPT? What is the total profit on crashing? 10 days €3000 Activity (label) Normal duration (day) Float (day) Crash time Cost- slope (€/day) a301500 b201550 c211150 d503900 e504400 f332100 g303200

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

11. Thank you for listening