Slide 1IntroductionTime-Cost Trade-offsReducing Project DurationSome Time-Cost ModelsSome Time-Cost ModelsSome Time-Cost ModelsConsider the following exampleFinding the Critical Path (Using Normal Times)Finding the Critical Path (A  B  E)Shortening the Project 1 dayReevaluate the Critical PathShortening the Project by an Additional DayIn summary…Critical Path for Crashed Schedule(Minimum) Cost of Crashing Project to Minimum Duration(Minimum) Cost of Crashing Project to Minimum Duration(Minimum) Cost of Crashing Project to Minimum Duration(Minimum) Cost of Crashing Project to Minimum Duration1BA 341 – Lean OperationsDr. Bogdan BichescuCost-Time Trade-offs andProject CrashingIntroductionRecall the project objectives:Performance (scope)CostTimeProject manager often forced to make trade-offs between project objectives, e.g., A delayed project must be expedited in order to be completed by deadlineResource constraints lead to changes in project scope and total durationTime-Cost Trade-offsWe will focus next on time-cost models, where the goal is to find either the most cost-effective approach to reducing the duration of a project, orthe project duration corresponding to minimum costIs there a link between activity duration and activity cost?Direct costs, e.g., overtime costs, hiring costs, leasing/buying resource costs, etc.Indirect costs, e.g., overhead, various fixed costsReducing Project DurationUsually done by adding manpower and/or resources to the project or working overtimeNote that doubling manpower does not reduce activity durations in half, e.g., Brook’s Law (IBM): Adding manpower to a late software project makes it even later! Fast-trackingRearrange the logic of project activities, such that (some) critical ones can be done in parallel, rather than sequentiallySome Time-Cost ModelsGoal: finding the most cost-effective approach to reduce project duration by x daysSteps1) Determine the project inputs:List of project activities and activity sequenceActivity durations and costsNormal Cost (NC) – the lowest expected activity costNormal Time (NT) – duration associated with NCCrash Time (CT) – the shortest possible activity timeCrash Cost (CC) – the cost associated with CTSome Time-Cost ModelsSteps:2) Find the cost per period (slope), to expedite each activity CTNTNCCCcrash time - timenormalcost normal -cost crash slopeSome Time-Cost ModelsSteps:3) Identify the critical path, using normal activity times4) Start crashing (reducing) the activities on the critical path, one activity at a time, one period (e.g., day) at a time, in increasing order of their slopes (the lower the slope, the cheaper the crashing)5) Reevaluate the critical path after each one-period crash; if new duration is satisfactory stop, otherwise, go back to step 4Consider the following exampleProject precedence table and crashing infoWant to shorten project by 2 days 3510, 453, 2BE3030, 1204, 1AD-20, 202, 2AC6020, 802, 1AB4040, 803, 2-A Slope ($/Day) Cost(NC, CC) Days(NT, CT)PredecessorActivityFinding the Critical Path (Using Normal Times)Start,00000A, 33030C, 28653B, 25353D, 48473E, 38585End,08888Earliest Start Time (ES)Latest Start Time (LS)Activity Name & DurationSlack=LS-ES=LF-EFFinding the Critical Path (A  B  E)Start,00000A, 33030C, 28653B, 25353D, 48473E, 38585End,08888Earliest Start Time (ES)Latest Start Time (LS)Activity Name & DurationSlack=LS-ES=LF-EFShortening the Project 1 dayIdentify the critical activity that is cheapest to crashA: slope = $40/dayB: slope = $60/dayE: slope = $35/dayProject duration is now 7 days, but is critical path the same?Activity E has the lowest slope  let’s crash E 1 day @ a cost of $35Reevaluate the Critical Path Start,00000A, 33030C, 27553B, 25353D, 47373E, 27575End,07777Earliest Start Time (ES)Earliest Finish Time (EF)Latest Start Time (LS)Activity Name & DurationSlack=LS-ES=LF-EFShortening the Project by an Additional DayIdentify the critical activities that are cheapest to crashA: slope = $40/dayB: slope = $60/dayE: cannot be crashed furtherD: slope = $30/dayThe cheapest alternative to further reduce project duration by 1 day is to crash A 1 day @ a cost of $40Now, choose between:• A @ $40, and• B & D @ $90In summary…The most cost-effective alternative to reduce the project duration by 2 days is:Crash E 1 day @ $35Crash A 1 day @ $40Final Cost: $75Can the project duration be further reduced?Yes, minimum project duration is 5 daysCrash B & D 1 day each, at a cost of $60 + $30 = $90See solved problems in the text for additional examplesCritical Path for Crashed ScheduleStart,00000A, 22020C, 26442B, 24242D, 46262E, 26464End,06666Earliest Start Time (ES)Latest Start Time (LS)Activity Name & DurationSlack=LS-ES=LF-EFLatest Finish Time (LF)(Minimum) Cost of Crashing Project to Minimum DurationStep 1- Replace all activity normal durations with their crashed durations Start,00000A, 22020C, 25342B, 13232D, 15432E, 25353End,05555(Minimum) Cost of Crashing Project to Minimum DurationStep 2- Identify the critical pathStart,00000A, 22020C, 25342B, 13232D, 15432E, 25353End,05555(Minimum) Cost of Crashing Project to Minimum DurationStep 3 – Relax non-critical activities in decreasing order of their slopes (i.e., what is the point of fully crashing non-critical activities?)Start,00000A, 22020C, 25342B, 13232D, 35432E, 25353End,05555(Minimum) Cost of Crashing Project to Minimum DurationFinally, the minimum cost is the sum of the:Cost of crashing the critical activities down to their minimum durations: $40 (A) + $60 (B) + $35 (E) = $135Cost of crashing activity D 1 day (from 4 to 3 days):$30$135 + $30 =