TimeCost TradeOff Time Reduction Time Compression Time Shortening

Time-Cost Trade-Off (Time Reduction = Time Compression) (Time Shortening)

Reasons to Reduce Project Durations: ØTo avoid late penalties (or avoid damaging the company’s relationship) ØTo realize incentive pay (monetary incentives for timely or early competition of a project ØTo beat the competition to the market to fit within the contractually required time (influences Bid Price) ØTo free resources for use on other projects (complete a project early & move on to another project) ØTo reduce the indirect costs associated with running the project ØTo complete a project when weather conditions make it less expensive (Avoid temporary heating, avoid completing site work during raining season) ØMany things make crashing a way of life on some projects (i. e. last minutes changes in client specification, without permission to extend the project deadline by an appropriate increment) Shortening the duration is called project crashing

Methods to reduce durations 1. Overtime: Have the existing crew work overtime. This increase the labor costs due to increase pay rate and decrease productivity. 2. Hiring and/or Subcontracting: a) Bring in additional workers to enlarge crew size. This increases labor costs due to overcrowding and poor learning curve. b) Add subcontracted labor to the activity. This almost always increases the cost of an activity unless the subcontracted labor is far more efficient. 3. Use of advanced technology: Use better/more advanced equipment. This will usually increase costs due to rental and transport fees. If labor costs (per unit) are reduced, this could reduce costs.

How project cost is estimated? Practices to Estimate the Project Cost Ø Some company assigns Overhead office as % of direct cost. Ø Most companies don’t consider profit as a cost of the job. Ø Cost analysis are completed by using: o [Direct costs + Indirect costs + Company overhead] vs. Budgeted (estimated costs) Ø Similar to each activity, the project as a whole has an ideal (least expensive) duration.

How project cost is estimated?

How project is crashed? Specify 2 activity times and 2 costs o 1 st time/cost combination (D, CD)- called Normal [Normal – usual ‘average’ time, resources] o 2 nd time/cost combination (d, Cd)- called Crash [Expedite by applying additional resources]

General 2 Steps in Project Crashing • In order to speed up any project schedule with a minimum cost, you need to follow these two general steps: 1. Select the Critical Path Activities. 2. Within the critical path activities, select the lowest reduction cost rate “Cost Slope” to reduce it. • In following slides more details on these 2 General Steps

Steps in Project Crashing Basic Steps 1. Compute the crash cost per time period. If crash costs are linear over time: 2. Using current activity times, find the critical path and identify the critical activities

Steps in Project Crashing 3. If there is only one critical path, then select the activity on this critical path that (a) can still be crashed, and (b) has the smallest crash cost period. If there is more than one critical path, then select one activity from each critical path such that (a) each selected activity can still be crashed, and (b) the total crash cost of all selected activities is the smallest. Note that the same activity may be common to more than one critical path. 4. Update all activity times. If the desired due date has been reached, stop. If not, return to Step 2.

Example 1: Crashing The Project Critical Path for Milwaukee Paper (A, C, E, G, H) 0 2 2 4 4 3 7 C F 2 0 4 10 6 13 A 0 0 2 0 4 4 8 Start 0 0 13 2 15 H E 0 13 0 15 4 0 8 0 3 3 B 1 1 4 3 4 7 8 5 13 D G 4 1 8 8 0 13

Example 1: Crashing The Project CALCULATION OF CRAH COST/PERIOD Crash and Normal Times and Costs for Activity B Activity Cost Crash $34, 000 — $33, 000 — Crash Cost, Cd = $32, 000 — = $31, 000 — $30, 000 — Normal Cost, CD Crash Cost – Normal Cost Normal Time – Crash Time Crash Cost/Wk = $34, 000 – $30, 000 3– 1 $4, 000 = $2, 000/Wk 2 Wks Normal — | 1 Crash Time, d | 2 | 3 Normal Time, D Time (Weeks) Table 3. 16 (Frome Heizer/Render; Operation Management)

Example 1: Crashing The Project Time (Wks) Activity Normal A B C D E F G H 2 3 2 4 4 3 5 2 Crash 1 1 1 2 2 1 Cost ($) Normal 22, 000 30, 000 26, 000 48, 000 56, 000 30, 000 80, 000 16, 000 Crash 22, 750 34, 000 27, 000 49, 000 58, 000 30, 500 84, 500 19, 000 Crash Cost Critical Per Wk ($) Path? 750 2, 000 1, 000 500 1, 500 3, 000 Yes No Yes Table 3. 5 (Frome Heizer/Render; Operation Management)

Example 2: Crashing Project For the small project shown in the table, it is required to reduce the project duration. A) Reduce by 2 periods. B) Reduce by 5 periods. Activity Precedence Normal Crash Time, day Cost, $ A - 4 210 3 280 B - 8 400 6 560 C A 6 500 4 600 D A 9 540 7 600 E B, C 4 500 1 1100 F C 5 150 4 240 G E 3 150 H D, F 7 600 6 750 Σ 3050 4280

Example 2: Crashing Project 1 - develop network Activity Precedence A Normal Crash Time, day Cost, $ - 4 210 3 280 B - 8 400 6 560 C A 6 500 4 600 D A 9 540 7 600 E B, C 4 500 1 1100 F C 5 150 4 240 G E 3 150 H D, F 7 600 6 750 Σ 3050 4280 9 D 4 6 5 7 A C F H FINISH START 8 4 3 B E G

Example 2: Crashing Project 2 - calculate times, find critical path 9 4 13 D 6 0 4 4 4 A 0 0 0 START 0 0 0 6 2 15 10 4 0 15 F C 4 15 10 10 0 7 22 H 15 15 0 22 22 0 22 FINISH 0 8 8 10 B 7 7 4 14 14 Ø Project completion time = 22 working days Ø Critical Path: A, C, F, H. 15 5 17 22 0 22 G E 15 3 19 Activity Total float Free float 19 A 0 0 5 22 B C D 7 0 2 2 0 2 E F G H 5 0 0 0 5 0

Example 2: Crashing Project 3 - calculate cost slope Activity Precedence Time, day Cost, $ Cost Slope, $/day Normal Crash A - 4 210 3 280 70 B - 8 400 6 560 80 C A 6 500 4 600 50 D A 9 540 7 600 30 E B, C 4 500 1 1100 200 F C 5 150 4 240 90 G E 3 150 ** H D, F 7 600 6 750 150 Σ 3050 4280 Remarks 1 - [G can not expedite] 2 - lowest slope and can be expedited on critical path is activity (C) WITH 2 periods

Example 2: Crashing Project Reduce 2 periods of activity (C) with increase of cost (2*50) =100 ES Activity Precedence A LS Activity EF TF LF 9 4 Crash limit (d @ cost) D 4 0 4 4 4 A 0 0 0 START 0 0 13 0 4 0 8 4 0 Cost Slope, Time, day Cost, $ $/day - 4 210 3 280 70 B - 8 400 6 560 80 C A 6 500 4 600 50 D A 9 540 7 600 30 E B, C 4 500 1 1100 200 F C 5 150 4 240 90 G E 3 150 ** H D, F 7 600 6 750 150 Σ 8 5 13 13 F 8 Crash 3050 4280 13 C 4 Normal 8 0 7 20 H 13 13 0 20 20 FINISH 0 8 8 8 B 5 5 4 12 12 13 5 15 20 0 20 G E 13 3 17 Ø Project completion time = 20 working days Activity Total float Ø Critical Path: A, C, F, H. & A, D, H Free float 17 5 20 A 0 0 B C D 5 0 0 0 E F G H 5 0 0 0 5 0

Example 2: Crashing Project Reduce 1 period of activity (A) with increase of cost =70 9 3 Activity Precedence A 12 D 3 0 3 3 3 A 0 0 0 START 0 0 0 4 0 7 3 0 Cost Slope, Time, day Cost, $ $/day - 4 210 3 280 70 B - 8 400 6 560 80 C A 6 500 4 600 50 D A 9 540 7 600 30 E B, C 4 500 1 1100 200 F C 5 150 4 240 90 G E 3 150 ** H D, F 7 600 6 750 150 Σ 7 5 12 12 F 7 Crash 3050 4280 12 C 3 Normal 7 0 7 19 H 12 12 0 19 19 0 19 FINISH 0 8 8 7 B 8 4 4 11 11 12 5 14 19 0 19 G E 12 3 16 Ø Project completion time = 19 working days Activity Total float Ø Critical Path: A, C, F, H. & A, D, H Free float 16 5 19 A 0 0 B C D 4 0 0 0 E F G H 5 0 0 0 5 0

Example 2: Crashing Project Activity Precedence A Reduce farther 1 period of 2 activities (D, F) with increase of cost (30+90) =120 3 8 11 D 3 0 3 3 3 A 0 0 0 START 0 0 0 4 0 3 0 Crash Cost Slope, Time, day Cost, $ $/day - 4 210 3 280 70 B - 8 400 6 560 80 C A 6 500 4 600 50 D A 9 540 7 600 30 E B, C 4 500 1 1100 200 F C 5 150 4 240 90 G E 3 150 ** H D, F 7 600 6 750 150 Σ 3050 4280 11 7 7 4 11 11 F C 3 Normal 7 7 0 7 18 H 11 11 0 18 18 0 18 FINISH 0 8 8 7 B 3 3 4 11 11 11 Ø Project completion time = 19 working days Ø Critical Path: A, C, F, H. & A, D, H 14 15 15 4 Activity Total float Free float A 0 0 B C D 3 0 0 0 4 18 0 18 G E 11 3 18 E F G H 4 0 0 0 4 0

Example 2: Crashing Project Activity Precedence A Reduce farther 1 period of activity (H) with increase of cost =150 3 8 11 D 3 0 3 3 3 A 0 0 0 START 0 0 0 4 0 3 0 Crash Cost Slope, Time, day Cost, $ $/day - 4 210 3 280 70 B - 8 400 6 560 80 C A 6 500 4 600 50 D A 9 540 7 600 30 E B, C 4 500 1 1100 200 F C 5 150 4 240 90 G E 3 150 ** H D, F 7 600 6 750 150 Σ 3050 4280 11 7 7 4 11 11 F C 3 Normal 7 7 0 6 17 H 11 11 0 17 17 0 17 FINISH 0 8 8 7 B 2 2 4 11 11 10 Ø Project completion time = 19 working days Ø Critical Path: A, C, F, H. & A, D, H 14 14 14 3 Activity Total float Free float A 0 0 B C D 2 0 0 0 3 17 0 17 G E 10 3 17 E F G H 3 0 0 0 3 0

Example 2: Crashing Project Reduction from 22 to 17 increase to 3490 Cycle Activity 0 - 1 C Time cost 2 Total cost Duration 3050 22 3150 20 100 2 A 1 70 3220 19 3 D, F 1 30+90 3340 18 4 H 1 150 3490 17 Activity Precedence A Normal Crash Time, day Cost, $ $/day - 4 210 3 280 70 B - 8 400 6 560 80 C A 6 500 4 600 50 D A 9 540 7 600 30 E B, C 4 500 1 1100 200 F C 5 150 4 240 90 G E 3 150 ** H D, F 7 600 6 750 150 Σ 3050 3550 3450 Cost 3350 3250 3150 3050 2950 16 17 18 19 20 Project Duration Cost Slope, 21 22 23 4280

Example 2: Crashing Project Accelerating the Critical and Noncritical path Normal Activity Precedence Crash Cost Slope, Time, day Cost, $ $/day A - 4 210 3 280 70 B - 8 400 6 560 80 C A 6 500 4 600 50 D A 9 540 7 600 30 E B, C 4 500 1 1100 200 F C 5 150 4 240 90 G E 3 150 ** H D, F 7 600 6 750 150 Σ 3050 4280

How project is crashed? Network Compression Algorithm Ø Determine Normal project duration, and cost. Ø Identify Normal duration Critical Path. Ø For large network, using CRITICALITY THEOREM to eliminate the noncritical paths that do not need to be crashed. Eliminate Activities with having “TF > the required project reduction time”. Ø Compute the cost slop

How project is crashed? Network Compression Algorithm Ø Shortening the CRITICAL ACTIVITIES beginning with the activity having the lowest cost-slope Ø Determine the compression limit (Nil) Ø Organize the data as in the following table:

How project is crashed? Network Compression Algorithm Ø Update the project network Ø When a new Critical Path is formed: - Shorten the combination of activity which Falls on Both Critical Paths, OR - Shorten one activity from each of the critical paths. Use the combined cost of shortening both activities when determining if it is cost-effective to shorten the project. Ø At each shortening cycle, compute the new project duration and project cost Ø Continue until no further shortening is possible

How project is crashed? Network Compression Algorithm Ø Tabulate and Plot the Indirect project Cost on the same timecost graph Ø Add direct and indirect cost to find the project cost at each duration. Ø Use the total project cost-time curve to find the optimum time.

Example 3 38470

Example 3 We have one critical path, A-C-G- I. Either crash A at cost SR 100/week or crash C at cost SR 200/week or crash G at cost SR 60/week or crash I at cost SR 75/week. 20 23 43 D 24 4 47 0 0 12 12 A 0 0 12 2@100 Cycle # 0 1 12 8 20 B 14 2 22 2@150 20 5 25 E 22 2 27 1@50 27 20 47 G 27 0 47 5@60 12 15 27 C 12 0 27 3@200 27 5 32 F 29 2 34 1@300 32 13 45 H 34 2 47 2@40 Activity to Can Be Shortened G 4 5 Nil 2 Weeks Cost per Shortened Week 2 60 47 12 59 I 47 0 59 2@75 2 ES D EF Activity LS TF LF Crash limit Cost for Project Total Cost Cycle Duration 59 36, 500 120 57 36, 620

Example 3 - 20 23 43 D 22 2 45 0 0 12 12 A 0 0 12 2@100 Cycle # 0 1 2 Now we have two critical paths: A-C-F-H-I and A-C-G- I. Either crash A at cost SR 100/week or crash C at cost SR 200/week or crash I at cost SR 75/week or crash F and G at cost SR 360/week or crash H and G at cost SR 100/week. 2 12 8 20 B 14 2 22 2@150 20 5 25 E 22 2 27 1@50 27 18 45 G 27 0 47 3@60 12 15 27 C 12 0 27 3@200 27 5 32 F 27 0 32 1@300 32 13 45 H 32 0 45 2@40 Activity to Can Be Shortened G 5 I 2 Nil 2 Weeks Cost per Shortened Week 2 60 2 75 45 12 57 I 45 0 55 2@75 ES D EF Activity LS TF LF Crash limit Cost for Project Total Cost Cycle Duration 36, 500 59 120 36, 620 57 150 36, 770 55

Example 3 - 20 23 43 D 22 2 45 0 0 12 12 A 0 0 12 2@100 Cycle # 0 1 2 3 Now we have two critical paths: A-C-F-H-I and A-C-G- I. Either crash A at cost SR 100/week or crash C at cost SR 200/week or crash F and G at cost SR 360/week or crash H and G at cost SR 100/week. 2 12 8 20 B 14 2 22 2@150 20 5 25 E 22 2 27 1@50 27 18 45 G 27 0 45 3@60 12 15 27 C 12 0 27 3@200 27 5 32 F 27 0 32 1@300 32 13 45 H 32 0 45 2@40 Activity to Can Be Shortened G 5 I 2 A 2 Nil 2 Weeks Cost per Shortened Week 2 60 2 75 2 100 45 10 55 I 45 0 55 0 ES D EF Activity LS TF LF Crash limit Cost for Project Total Cost Cycle Duration 59 36, 500 120 57 36, 620 150 55 36, 770 200 53 36, 970

Example 3 - 18 23 41 D 20 2 43 0 0 10 10 A 0 0 10 0 Cycle # 0 1 2 3 4 Activity to Shorten G I A H, G Now we have two critical paths: A-C-F-H-I and A-C-G- I. Either crash C at cost SR 200/week or crash F and G at cost SR 360/week or crash H and G at cost SR 100/week. 2 10 8 18 B 12 2 20 2@150 18 5 23 E 20 2 25 1@50 25 18 43 G 25 0 43 3@60 10 15 25 C 10 0 25 3@200 25 5 30 F 25 0 30 1@300 30 13 43 H 30 0 43 2@40 Can Be Shortened 5 2 2 2 Nil 2 2 Weeks Cost per Shortened Week 2 60 2 75 2 100 2 60+40 43 10 53 I 43 0 53 0 ES D EF Activity LS TF LF Crash limit Cost for Project Total Cost Cycle Duration 36, 500 59 120 36, 620 57 150 36, 770 55 200 36, 970 53 200 37, 170 51

- Example 3 Now we have three critical paths; A-C-F-H-I, A-C-G- I, and A-B-D-I. Either crash C and B at cost SR 350/week or crash F, G and B at cost SR 510/week. 18 23 41 D 18 0 41 0 0 10 10 A 0 0 18 5 23 E 20 2 25 1@50 25 16 41 G 25 0 41 1@60 25 5 30 F 25 0 30 1@300 30 11 41 H 30 0 41 0 10 8 18 B 10 0 18 2@150 0 10 15 25 C 10 0 25 3@200 Cycle # 0 1 2 3 4 5 Activity to Can Be Shortened G 5 I 2 A 2 G, H 2 B, C 2 Nil 2 2 Weeks Cost per Shortened Week 2 60 2 75 2 100 2 60+40 2 150+200 Cost for Cycle 120 150 200 700 41 10 51 I 41 0 51 0 ES D EF Activity LS TF LF Crash limit Project Total Cost Duration 59 36, 500 57 36, 620 55 36, 770 53 36, 970 51 37, 170 49 37, 870

Example 3 – Final Results 16 23 39 D 16 0 39 0 0 10 10 A 0 0 10 6 16 B 10 0 16 5 21 E 18 2 23 1@50 23 16 39 G 23 0 39 1@60 10 13 23 C 10 0 23 1@200 23 5 28 F 23 0 28 1@300 28 11 39 H 28 0 39 0 Cycle # 0 1 2 3 4 5 Project Direct Cost Duration 59 36500 57 36620 55 36770 53 36970 51 37170 49 37870 39 10 49 I 39 0 49 0 ES D EF Activity LS TF LF Crash limit Indirect Cost Total Cost 7375 7125 6875 6625 6375 6125 43875 43745 43645 43595 43545 43995

Example 3 – Final Results Project Optimal Duration

CASE WORK Data on small maintenance project is given as below: On completion, the project will give a return of SR 110/day. Using time-cost trade-off method, how much would you like to compress the project for maximizing the return (ignore the Indirect cost effect)? Show all calculations. Activity A B C D E F G H I Depends on A B B C D, E F, G Normal Time Cost 6 days SR 700 4 days 400 5 days 650 8 days 625 10 days 200 7 days 500 3 days 600 6 days 300 7 days 350 Time 4 days 5 days 7 days 5 days 3 days 5 days 4 days Crash Cost SR 800 400 700 350 700 600 425