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  • 1. Time-Cost Trade-Off Time-Cost Trade-Off Duration and cost of activity depends on the amount an type of resources used Assuming more workers will normally reduce time and will change its cost This relationship between activity time and cost called time-cost Trade off04/11/2006 Emad Elbeltagi 2 1
  • 2. Time-Cost Trade-OffWhy do we need to reduce project time? Finish the project in a predefined deadline date Recover early delays to avoid liquidated damages Free key resources early for other projects Avoid adverse weather conditions that might affect productivity Receive an early completion-bonus Improve project cash flowscheduling techniques are duration04/11/2006 Emad Elbeltagi 3 Time-Cost Trade-OffWhat is the way to reduce activity duration? Working extended hours (Over time) Offering incentive payments to increase the productivity Using additional resources Using materials with faster installation methods Using alternate construction methods or sequence04/11/2006 Emad Elbeltagi 4 2
  • 3. Time-Cost Trade-OffActivity Time-Cost Relationship Decreasing activity duration will increase its cost Cost Crash duration & Crash cost Normal duration & Normal cost Time04/11/2006 Emad Elbeltagi 5 Time-Cost Trade-OffActivity Time-Cost Relationship Cost Crash duration & Crash cost Normal duration & Normal cost Time04/11/2006 Emad Elbeltagi 6 3
  • 4. Time-Cost Trade-OffActivity Time-Cost Relationship For simplicity, linear relationship is adopted Cost Crash duration & Crash cost Normal duration & Normal cost Time04/11/2006 Emad Elbeltagi 7 Time-Cost Trade-OffActivity Time-Cost Relationship (Example) Consider the following options Estimated daily Crew size production Crew formation (men) (square meter) 166 5 1 scaffold set, 2 labors, 2 carpenter, 1 foreman 204 6 2 scaffold set, 3 labors, 2 carpenter, 1 foreman 230 7 2 scaffold set, 3 labors, 3 carpenter, 1 foremanConsider the following rates: Labor LE 96/day;carpenter LE 128/day;foreman LE144/day andscaffolding LE60/day04/11/2006 Emad Elbeltagi 8 4
  • 5. Time-Cost Trade-OffActivity Time-Cost Relationship (Example) 8400 Sq meters of scaffolds Crew Duration (days) Cost (LE) size 5 50.6 (use 51) 51 x (1x60 + 2x96 + 2x128 + 1x144) = 33252 6 41.2 (use 42) 42 x (2x60 + 3x96 + 2x128 + 1x144) = 33936 7 36.5 (use 37) 37 x (2x60 + 3x96 + 3x128 + 1x144) = 3463204/11/2006 Emad Elbeltagi 9 Time-Cost Trade-OffActivity Time-Cost Relationship (Example) 34800 3 34600 34400 34200 Cost (LE) 34000 2 33800 33600 33400 1 33200 33000 30 35 40 45 50 55 Duration (days)04/11/2006 Emad Elbeltagi 10 5
  • 6. Time-Cost Trade-OffProject Time-Cost Relationship Project direct cost = summation of direct cost of individual activities Indirect cost comprises site and head office overheads Project indirect cost = daily indirect cost x project duration Direct cost increases as project duration decreases Indirect cost decreases s project duration decreases04/11/2006 Emad Elbeltagi 11 Time-Cost Trade-OffProject Time-Cost RelationshipProject cost Project duration04/11/2006 Emad Elbeltagi 12 6
  • 7. Time-Cost Trade-Off Shortening Project Duration (Optimum Duration) It might be necessary to shorten project duration to meet specific deadline Or to determine the optimum project duration that is correspond to the least project total cost Many approaches can be used Heuristic Cost-Slope method wii be applied 04/11/2006 Emad Elbeltagi 13 Time-Cost Trade-OffProcedure for Shortening Project Duration Draw the project network. Perform CPM calculations and identify the critical path, use normal durations and costs for all activities. Compute the cost slope for each activity from the following equation: cost slope = crash cost – normal cost / normal duration – crash duration 04/11/2006 Emad Elbeltagi 14 7
  • 8. Time-Cost Trade-OffProcedure for Shortening Project Duration Start by shortening the activity duration on the critical path which has the least cost slope and not been shortened to its crash duration. Reduce the duration of the critical activities with least cost slope until its crash duration is reached or until the critical path changes. When multiple critical paths are involved, the activity(ies) to be shorten is determined by comparing the cost slope of the activity which lies on all critical paths (if any) 04/11/2006 Emad Elbeltagi 15 Time-Cost Trade-OffProcedure for Shortening Project Duration Having shortened a critical path, you should adjust activities timings, and floats. The cost increase due to activity shortening is calculated as the cost slope multiplied by the time of time units shortened. Continue until no further shortening is possible, and then the crash point is reached. The results may be represented graphically by plotting project completion time against cumulative cost increase. 04/11/2006 Emad Elbeltagi 16 8
  • 9. Time-Cost Trade-Off Example Application Consider the following project Indirect cost LE 125/day Crash to 49 days Normal CrashActivity Preceded by Duration (day) Cost (LE) Duration (day) Cost (LE) A - 12 7000 10 7200 B A 8 5000 6 5300 C A 15 4000 12 4600 D B 23 5000 23 5000 E B 5 1000 4 1050 F C 5 3000 4 3300 G E, C 20 6000 15 6300 H F 13 2500 11 2580 I D, G, H 12 3000 10 3150 04/11/2006 Emad Elbeltagi 17 Time-Cost Trade-OffExample Application Total direct cost LE 36500 20 43 D (23) 24 47 0 12 12 20 20 25 27 47 47 59 A (12) B (8) E (5) G (20) I (12) 0 12 14 22 22 27 27 47 47 59 2@100 2@150 1@50 5@60 2@75 12 27 27 32 32 45 C (15) F (5) H (13) 12 27 29 34 34 47 3@200 1@300 2@40 04/11/2006 Emad Elbeltagi 18 9
  • 10. Time-Cost Trade-OffExample Application Activity G has the least cost slope 60 LE/day Crash G by 2 days, cost increase by 2 x 60 20 43 D (23) 22 45 0 12 12 20 20 25 27 45 45 57 A (12) B (8) E (5) G (18) I (12) 0 12 14 22 22 27 27 45 45 57 2@100 2@150 1@50 3@60 2@75 12 27 27 32 32 45 C (15) F (5) H (13) 12 27 27 32 32 45 3@200 1@300 2@40 04/11/2006 Emad Elbeltagi 19 Time-Cost Trade-OffExample Application Activity I has the least cost slope 75 LE/day Crash I by 2 days, cost increase by 2 x 75 20 43 D (23) 22 45 0 12 12 20 20 25 27 45 45 55 A (12) B (8) E (5) G (18) I (10) 0 12 14 22 22 27 27 45 45 55 2@100 2@150 1@50 3@60 12 27 27 32 32 45 C (15) F (5) H (13) 12 27 27 32 32 45 3@200 1@300 2@40 04/11/2006 Emad Elbeltagi 20 10
  • 11. Time-Cost Trade-OffExample Application Activity A has the least cost slope 100 LE/day Crash A by 2 days, cost increase by 2 x 100 18 41 D (23) 20 43 0 10 10 18 18 23 25 43 43 53 A (10) B (8) E (5) G (18) I (10) 0 10 12 20 20 25 25 43 43 53 2@150 1@50 3@60 10 25 25 30 30 43 C (15) F (5) H (13) 10 25 25 30 30 43 3@200 1@300 2@40 04/11/2006 Emad Elbeltagi 21 Time-Cost Trade-OffExample Application Activity H & G has the least cost slope 100 LE/day Crash H & G by 2 days, cost increase by 2 x 100 18 41 D (23) 18 41 0 10 10 18 18 23 25 41 41 51 A (10) B (8) E (5) G (16) I (10) 0 10 10 18 20 25 25 41 41 51 2@150 1@50 1@60 10 25 25 30 30 41 C (15) F (5) H (11) 10 25 25 30 30 41 3@200 1@300 04/11/2006 Emad Elbeltagi 22 11
  • 12. Time-Cost Trade-OffExample Application Activity C & B has the least cost slope 350 LE/day Crash H & G by 2 days, cost increase by 2 x 350 16 39 D (23) 16 390 10 10 16 16 21 23 39 39 49 A (10) B (6) E (5) G (16) I (10)0 10 10 16 18 23 23 39 39 49 1@50 1@60 10 23 23 28 28 39 C (13) F (5) H (11) 10 3 23 28 28 39 1@200 1@300 04/11/2006 Emad Elbeltagi 23 Time-Cost Trade-Off Example Application Duratio Direct cost X 1000 LE Indirect cost x 1000 LE Total cost x 1000 LE n 59 36.50 7.375 43.875 57 36.62 7.125 43.745 55 36.77 6.875 43.645 53 36.97 6.625 43.595 51 37.17 6.375 43.545 49 37.87 6.125 43.995 04/11/2006 Emad Elbeltagi 24 12
  • 13. Time-Cost Trade-OffExample Application 50 40 LE x 1000 30 20 10 0 48 50 52 54 56 58 60 Time (days)04/11/2006 Emad Elbeltagi 25 13