Time-Cost Trade-Off(Time Reduction = Time Compression) (Time Shortening) 13 June 2013GE404 ENGINEERING MANAGEMENT1TOPIC-6- TIME-COST TRADE OFFReasons to Reduce Project Durations:Shortening the duration is called project crashingTo avoid late penalties (or avoid damaging the companys relationship),To realize incentive pay, (monetary incentives for timely or early competition of a project,To beat the competition to the market, to fit within the contractually required time (influences Bid price) To free resources for use on other projects. (complete a project early & move on to another project)To reduce the indirect costs associated with running the project.To complete a project when weather conditions make it less expensive (Avoid temporary Heating, avoid completing site work during raining season).Many 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)13 June 2013GE404 ENGINEERING MANAGEMENT2TOPIC-6- TIME-COST TRADE OFFMethods to reduce durationsOvertime: Have the existing crew work overtime. This increase the labor costs due to increase pay rate and decrease productivity.Hiring and/or Subcontracting: Bring in additional workers to enlarge crew size. This increases labor costs due to overcrowding and poor learning curve. Add subcontracted labor to the activity. This almost always increases the cost of an activity unless the subcontracted labor is for more efficient.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 costSome company assigns Overhead office as % of direct cost.Most companies dont consider profit as a cost of the job. Cost analysis are completed by using: [Direct costs + Indirect costs + Company overhead] vs. Budgeted (estimated costs).Similar to each activity, the project as a whole has an ideal (Lease expensive) Duration.How project cost is estimated?

How project is crashed?Specify 2 activity times and 2 costs1st time/cost combination (D, CD)- called normal [Normal usual average time, resources]2nd time/cost combination (d, Cd)- called Crash [expedite by applying additional resources]

Steps in Project CrashingCompute the crash cost per time period. If crash costs are linear over time:

Using current activity times, find the critical path and identify the critical activities

Basic StepsSteps in Project CrashingIf 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 per 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.Update all activity times. If the desired due date has been reached, stop. If not, return to Step 2.Example1: Crashing The ProjectCritical Path for Milwaukee Paper (A, C, E, G, H)Start0A2B3C2F3D4E4G5H200022447480337131315800021424848813101313415000000116Example1: Crashing The ProjectTable 3.16 (Frome Heizer/Render; Operation Management) CALCULATION OF CRAH COST/PERIOD Crash and Normal Times and Costs for Activity B|||123Time (Weeks)$34,000 $33,000 $32,000 $31,000 $30,000 Activity CostCrashNormalCrash Time, dNormal Time, DCrash Cost, CdNormal Cost, CDCrash Cost/Wk =Crash Cost Normal CostNormal Time Crash Time=$34,000 $30,0003 1= = $2,000/Wk$4,0002 WksExample1: Crashing The ProjectTime (Wks)Cost ($)Crash CostCriticalActivityNormalCrashNormalCrashPer Wk ($)Path?A2122,00022,750750YesB3130,00034,0002,000NoC2126,00027,0001,000YesD4248,00049,0001,000NoE4256,00058,0001,000YesF3230,00030,500500NoG5280,00084,5001,500YesH2116,00019,0003,000YesTable 3.5 (Frome Heizer/Render; Operation Management) Example2: Crashing Project (123)ActivityPrecedenceNormalCrashTime, dayCost, $Time, dayCost, $A-42103280B-84006560CA65004600DA95407600EB,C450011100FC51504240GE31503150HD,F7600675030504280For the small project shown in the table, it is required reduce the project duration. Reduce by 2 periods.Reduce by 5 periods. Example2: Crashing Project (123)1- develop networkA4C6B8E4G3F5H7STARTFINISHD9ActivityPrecedenceNormalCrashTime, dayCost, $Time, dayCost, $A-42103280B-84006560CA65004600DA95407600EB,C450011100FC51504240GE31503150HD,F7600675030504280Example2: Crashing Project (123)A044004C46104010B0887715E1041415519G1431719522F1051510015H1572215022000000START2202222022FINISHD49136215ActivityABCDEFGHTotal float07025050Free float02020050Project completion time = 22 working daysCritical Path: A, C, F, H.2- calculate times, find critical pathExample2: Crashing Project (123)ActivityPrecedenceNormalCrashTime, dayCost, $Time, dayCost, $A-42103280B-84006560CA65004600DA95407600EB,C450011100FC51504240GE31503150HD,F7600675030504280Cost Slope, $/day7080503020090**150Remarks1- [G can not expedite]2- lowest slope and can be expedited on critical path is activity (C) WITH 2 periods3- calculate cost slopeExample2: Crashing Project (123)A044004C448408B0885513E841213517G1231517520F85138013H1372013020000000START2002020020FINISHD49134013ActivityABCDEFGHTotal float05005050Free float00000050Project completion time = 20 working daysCritical Path: A, C, F, H. & A, D, HReduce 2 periods of activity (C) with increase of cost (2*50) =100ActivityPrecedenceNormalCrashCost Slope, Time, dayCost, $Time, dayCost, $$/dayA-4210328070B-8400656080CA6500460050DA9540760030EB,C450011100200FC5150424090GE31503150**HD,F7600675015030504280ESLSEFTFLFActivityCrash limit (d @ cost)Example2: Crashing Project (123)A033003C347307B0888412E741112516G1131416519F75127012H1271912019000000START1901919019FINISHD39123012ActivityABCDEFGHTotal float04005050Free float00000050Project completion time = 19 working daysCritical Path: A, C, F, H. & A, D, HReduce 1 period of activity (A) with increase of cost =70ActivityPrecedenceNormalCrashCost Slope, Time, dayCost, $Time, dayCost, $$/dayA-4210328070B-8400656080CA6500460050DA9540760030EB,C450011100200FC5150424090GE31503150**HD,F7600675015030504280Example2: Crashing Project (123)ActivityABCDEFGHTotal float03004040Free float00000040Project completion time = 19 working daysCritical Path: A, C, F, H. & A, D, HReduce farther 1 period of 2 activities (D,F) with increase of cost (30+90) =120ActivityPrecedenceNormalCrashCost Slope, Time, dayCost, $Time, dayCost, $$/dayA-4210328070B-8400656080CA6500460050DA9540760030EB,C450011100200FC5150424090GE31503150**HD,F7600675015030504280A033003C347307B0883311E741111415G1131415418F74117011H1171811018000000START1801818018FINISHD38113011Example2: Crashing Project (123)ActivityABCDEFGHTotal float02003030Free float00000030Project completion time = 19 working daysCritical Path: A, C, F, H. & A, D, HReduce farther 1 period of activity (H) with increase of cost =150ActivityPrecedenceNormalCrashCost Slope, Time, dayCost, $Time, dayCost, $$/dayA-4210328070B-8400656080CA6500460050DA9540760030EB,C450011100200FC5150424090GE31503150**HD,F7600675015030504280A033003C347307B0882210E741110314G1131414317F74117011H1161711017000000START1701717017FINISHD38113011Example2: Crashing Project (123)Reduction from 22 to 17 increase to 3490ActivityPrecedenceNormalCrashCost Slope, Time, dayCost, $Time, dayCost, $$/dayA-4210328070B-8400656080CA6500460050DA9540760030EB,C450011100200FC5150424090GE31503150**HD,F7600675015030504280CycleActivityTimecostTotal costDuration 0-3050221C21003150202A1703220193D,F130+903340184H1150349017Example2: Crashing Project (123)Accelerating the Critical and Noncritical path

ActivityPrecedenceNormalCrashCost Slope, Time, dayCost, $Time, dayCost, $$/dayA-4210328070B-8400656080CA6500460050DA9540760030EB,C450011100200FC5150424090GE31503150**HD,F7600675015030504280Determine Normal project duration, and cost.How project is crashed?Network Compression Algorithm

Identify Normal duration Critical Path.For large network, using CRITICALITY THEORM to eliminate the noncritical paths that do not need to be crashed.Compute the cost slopHow project is crashed?Network Compression Algorithmshortening the CRITICAL ACTIVITIES beginning with the activity having the lowest cost-slope

Organize the data as in the following table:

Determine the compression limit (Nil) How project is crashed?Network Compression AlgorithmUpdate the project networkWhen a new Critical path is formed:Shorten the combination of activity which Falls on Both Critical Paths, ORShorten 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.At each shortening cycle, compute the new project duration and project costContinue until no further shortening is possible How project is crashed?Network Compression AlgorithmUse the total project cost-time curve to find the optimum time.Tabulate and Plot the Indirect project Cost on the same time-cost graph

Add direct and indirect cost to find the project cost at each duration.Example 3

38470Example 3

Example 3

Example 3

Example 3

Example 3

Example 3

Project Optimal Duration

Example 3CASE WORK

13 June 2013GE404 ENGINEERING MANAGEMENT34TOPIC-6- TIME-COST TRADE OFF