1. 1 6. T IME -C OST O PTIMIZATION Objective: The optimization of project duration and cost by an appropriate crashing of activities. Summary: 6.1 Finding the Minimum Project Duration and Corresponding Minimum Cost. 6.2 Sensitivity Analysis to Determine the Minimum Project Cost. 6.3Determining the Minimum Project Cost for a Target Project Duration.

2. 2 6.1 F INDING THE M INIMUM P ROJECT D URATION AND C ORRESPONDING M INIMUM C OST Find the minimum practicable project duration that can be achieved, and then find the corresponding minimum project cost. –Eg: highway construction, reduce project duration to minimize inconvenience to road users.

3. 3 Can control the duration of an activity by varying the type and numbers of resources used, and the number of hours they are employed. To reduce activity duration, or bring forward its completion date: –can add resources; –change to higher performance resources; –employ more hours per day (overtime); –increase the number of shifts; –increase the number of working days; this is termed Activity Crashing.

4. 4 Typically there is a time-cost trade-off in that reducing the duration of a task tends to: –increase direct costs (labor, equipment, materials, subcontractors). Why? However, reducing the project duration will tend to: –reduce indirect costs (site staff, head-office expenses, and penalty clauses). Also, increasing the duration of a task can: –increase direct costs by introducing idle time. There is a practical limit on how far an activity can be crashed. Why?

5. 5 In reducing the project duration, there is no point in crashing an activity to a point where it is no longer critical. Why? –Will not reduce project duration. –Likely to add to the costs. Also, where there is a choice, crash the activities that give rise to the least increase in costs.

6. 6 Fig. 6-1: Affect of Crashing an Activity on its Direct Costs (a) activity 1-2 time direct costs $ 3 4 5 900 500 normal duration crash duration (b) activity 2-3 time direct costs $ 4 5 6 7 1600 700 crash cost normal cost direct costs $ (c) activity 2-4 time 6 7 8 9 10 200 100 (d) activity 3-5 time direct costs $ 5 6 900 500 (e) activity 4-6 time direct costs $ 6 7 8 9 10 400 200 direct costs $ (f) activity 5-7 & 6-7 time 3 4 5 6 7 500 300 100 act 5-7 act 6-7

7. 7 Fig. 6-2: Initial Attempt at Reducing Project Duration (a) progress using normal activities for foundation operation continued... 5 7 10 0 6 0 5 7 12 3 4 5 6 7 0 5 12 15 25 32 27 25 15 5 0 TF = 0 TF = 3 TF = 0 TF = 9 TF = 3 TF = 0 TF = 2 TF = 0 Normal Duration = 32 Normal cost = 500 + 700 + 100 + 500 + 200 + 300 + 100 = $2,400

8. 8 Fig. 6-2: Initial Attempt at Reducing Project Duration (b) progress with all activities crashed 3 4 6 0 5 6 0 3 4 12 3 4 5 6 7 0 3 7 9 15 19 16 15 9 9 3 0 TF = 0 TF = 2 TF = 0 TF = 4 TF = 2 TF = 0 TF = 1 TF = 0 Note, activities use crashed durations Crashed Duration = 19 (down 13) Crashed cost = 900 + 1600 + 200 + 900 + 400 + 500 + 200 = $4,700 (an increase of $2,300)

9. 9 Fig. 6-3: Optimization of Cost by Relaxing Non-Critical Activities (a) first step in relaxing non-critical activities continued... Save costs by relaxing non-critical activities Start where the great- est savings can be made 3 4 6 0 6 6 0 3 4 12 3 4 5 6 7 0 3 7 9 15 19 16 15 9 9 3 0 TF = 0 TF = 2 TF = 0 TF = 3 TF = 2 TF = 0 TF = 1 TF = 0 relaxed by 1 Crashed Duration = 19 (unchanged) Crashed cost = 900 + 1600 + 200 + 500 + 400 + 500 + 200 = $4,300 (a saving of $400)

10. 10 Fig. 6-3: Optimization of Cost by Relaxing Non-Critical Activities (b) second step in relaxing non-critical activities continued... Select activity with next best cost savings 3 6 6 0 6 6 0 3 4 12 3 4 5 6 7 0 3 9 9 15 19 16 15 9 9 3 0 TF = 0 TF = 1 TF = 0 TF = 1 TF = 0 relaxed by 2 Crashed Duration = 19 (unchanged) Crashed cost = 900 + 1000 + 200 + 500 + 400 + 500 + 200 = $3,700 (a saving of $1,000 from complete crash)

11. 11 Fig. 6-3: Optimization of Cost by Relaxing Non-Critical Activities (c) third step in relaxing non-critical activities Select activity with next best cost savings 3 6 6 0 6 6 0 4 4 12 3 4 5 6 7 0 3 9 9 15 19 15 9 9 3 0 TF = 0 relaxed by 1 Crashed Duration = 19 (unchanged) Crashed cost = 900 + 1000 + 200 + 500 + 400 + 400 + 200 = $3,600 (a saving of $1,100 from complete crash) Note, increase in number of critical activities

12. 12 6.2 S ENSITIVITY A NALYSIS TO D ETERMINE M INIMUM P ROJECT C OST Find the minimum project cost, and then find the corresponding minimum project duration. Two methods: –Start with normal activity network and gradually crash: Crash critical activities with smallest rate of change in cost. –Start with crashed activity network (after optimized for cost) and gradually relax: Relax activities with largest rate of change in cost.

13. 13 Fig. 6-4: Cost Senstivity Analysis by Relaxing Critical Activities (a) first step in cost sensitivity analysis continued... Start with crashed activity network optimized for cost P RIMAVERA BCN Mouse C rashing a ctivities !!! CRASH !!! 3 6 6 0 6 6 0 4 4 12 3 4 5 6 7 0 3 9 9 15 19 15 9 9 3 0 TF = 0 Project Duration = 19 days Direct Cost = 900 + 1000 + 200 + 500 + 400 + 400 + 200 = $3,600 Relax activities, slowly extending project duration (for all parallel paths). Which activity(ies) give the greatest cost reduction rate ? There are six alternatives ! 1) Relaxing activity 1-2 gives what rate ? $200/day 2) Relaxing activities 2-3 & 2-4 gives ? 300+25=$325/day 3) Relaxing activities 3-5 & 2-4 gives ? 0+25=$25/day activity 3-5 is already at its normal duration 4) Relaxing activities 3-5 & 4-6 gives ? 0+50=$50/day 5) Relaxing activities 3-5 & 6-7 gives ? 0+33.3=$33.3/day 6) Relaxing activities 5-7 & 6-7 gives ? 100+33.3=$133.3/day So, relax activities 2-3 & 2-4 by 1 day saving 300+25=$325 7 7 3 0 6 6 0 4 4 12 3 4 5 6 7 0 3 10 16 20 16 10 3 0 TF = 0 Project Duration = 20 days Direct Cost = 900 + 700 + 175 + 500 + 400 + 400 + 200 = $3,275

14. 14 Fig. 6-5: Senstivity of Costs to Varying the Project Duration 5000 4600 4200 3800 3400 3000 2600 2200 1800 1400 19 20 21 22 23 24 25 26 27 28 29 30 31 32 project duration (days) cost ($) = Indirect Costs @ $75/day = Direct Costs = Combined Costs

15. 15 Fig. 6-4: Cost Senstivity Analysis by Relaxing Critical Activities (b) second step in cost sensitivity analysis continued... 3 0 6 6 0 4 4 12 3 4 5 6 7 0 3 10 16 20 16 10 3 0 TF = 0 7 7 Project Duration = 20 days Direct Cost = 900 + 700 + 175 + 500 + 400 + 400 + 200 = $3,275 Relax activity 1-2 by 2 days saving 2x200=$400 5 3 0 6 6 0 4 4 12 3 4 5 6 7 0 5 12 18 22 18 12 5 0 TF = 0 7 7 Project Duration = 22 days Direct Cost = 500 + 700 + 175 + 500 + 400 + 400 + 200 = $2,875

16. 16 Fig. 6-5: Senstivity of Costs to Varying the Project Duration 5000 4600 4200 3800 3400 3000 2600 2200 1800 1400 19 20 21 22 23 24 25 26 27 28 29 30 31 32 project duration (days) cost ($) = Indirect Costs @ $75/day = Direct Costs = Combined Costs

17. 17 Fig. 6-4: Cost Senstivity Analysis by Relaxing Critical Activities (c) third step in cost sensitivity analysis continued... 5 0 6 6 0 4 4 12 3 4 5 6 7 0 5 12 18 22 18 12 5 0 TF = 0 7 7 Project Duration = 22 days Direct Cost = 500 + 700 + 175 + 500 + 400 + 400 + 200 = $2,875 Relax activities 5-7 & 6-7 by 1 day saving 100+33.3=$133.3 5 5 5 0 6 6 0 4 4 12 3 4 5 6 7 0 5 12 18 23 18 12 5 0 TF = 0 7 7 Project Duration = 23 days Direct Cost = 500 + 700 + 175 + 500 + 400 + 300 + 166.7 = $2,741.7

18. 18 Fig. 6-5: Senstivity of Costs to Varying the Project Duration 5000 4600 4200 3800 3400 3000 2600 2200 1800 1400 19 20 21 22 23 24 25 26 27 28 29 30 31 32 project duration (days) cost ($) = Indirect Costs @ $75/day = Direct Costs = Combined Costs

19. 19 Fig. 6-4: Cost Senstivity Analysis by Relaxing Critical Activities (d) fourth step in cost sensitivity analysis continued... 5 0 6 6 0 5 5 12 3 4 5 6 7 0 5 12 18 23 18 12 5 0 TF = 0 7 7 Project Duration = 23 days Direct Cost = 500 + 700 + 175 + 500 + 400 + 300 + 166.7 = $2,741.7 Relax activity 4-6 by 4 days saving 4x50 = $200 10 5 0 6 6 0 5 5 12 3 4 5 6 7 0 5 12 22 27 22 12 5 0 TF = 0 TF = 4 TF = 0 7 7 Project Duration = 27 days Direct Cost = 500 + 700 + 175 + 500 + 200 + 300 + 166.7 = $2,541.7

20. 20 Fig. 6-5: Senstivity of Costs to Varying the Project Duration 5000 4600 4200 3800 3400 3000 2600 2200 1800 1400 19 20 21 22 23 24 25 26 27 28 29 30 31 32 project duration (days) cost ($) = Indirect Costs @ $75/day = Direct Costs = Combined Costs

21. 21 Fig. 6-4: Cost Senstivity Analysis by Relaxing Critical Activities (e) fifth step in cost sensitivity analysis 5 0 6 10 0 5 5 12 3 4 5 6 7 0 5 12 22 27 22 12 5 0 TF = 0 TF = 4 TF = 0 7 7 Project Duration = 27 days Direct Cost = 500 + 700 + 175 + 500 + 200 + 300 + 166.7 = $2,541.7 Relax activity 6-7 by 2 days saving 2x33.3 = $66.7 7 5 0 6 10 0 5 5 12 3 4 5 6 7 0 5 12 22 29 24 22 12 5 0 TF = 0 TF = 6 TF = 0 TF = 2 TF = 0 7 7 0 Project Duration = 29 days Direct Cost = 500 + 700 + 175 + 500 + 200 + 300 + 100 = $2,475

22. 22 Fig. 6-5: Senstivity of Costs to Varying the Project Duration 5000 4600 4200 3800 3400 3000 2600 2200 1800 1400 19 20 21 22 23 24 25 26 27 28 29 30 31 32 project duration (days) cost ($) = Indirect Costs @ $75/day = Direct Costs = Combined Costs Finally, relaxing activity 2-4 by 3 days takes us back to the normal network. Optimum combined costs @ 23 days

23. 23 6.3 D ETERMINING M INIMUM P ROJECT C OST FOR A T ARGET D URATION Find the minimum project cost for a given target project duration. –Eg: speed-up of the project to meet the contractual target date.

24. 24 Fig. 6-5: Senstivity of Costs to Varying the Project Duration 5000 4600 4200 3800 3400 3000 2600 2200 1800 1400 19 20 21 22 23 24 25 26 27 28 29 30 31 32 project duration (days) cost ($) = Combined Costs If target project duration = 20 days; use network in Fig. 6-4a. If target project duration = 21 days; use network in Fig 6-4b, relaxing act 1-2 by 1 day only to 6 days If target project duration = 27 days; might use network relaxed to 23 days (Fig. 6-4c)