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Planning of Barus & Holley Addition Procurement215 Site Preparation60 Pour Foundation31SP Erect Steel15P, PF Roof, Ext. Wall40ES Fabricate Glass50ES Int.

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Presentation on theme: "Planning of Barus & Holley Addition Procurement215 Site Preparation60 Pour Foundation31SP Erect Steel15P, PF Roof, Ext. Wall40ES Fabricate Glass50ES Int."— Presentation transcript:

1 Planning of Barus & Holley Addition Procurement215 Site Preparation60 Pour Foundation31SP Erect Steel15P, PF Roof, Ext. Wall40ES Fabricate Glass50ES Int Walls & Gl.85REW Landscaping55REW Acquire Furn.280 Install Furn.15IWG, AF, FG Activity Duration, Days Predecessors

2 Arrow Diagram (Network) 1 2 34 5 6 7 8 9 SP P PF ES REW FG IWG L AF IF Nodes (I) = EventsArrows (I,J) = Activities (Tasks)

3 Forward Pass: Find ES(I) 1 2 34 5 6 7 8 9 SP P PF ES REW FG IWG L AF IF 60 215 270 230 280 355 370 0 355 ES(I) = Earliest Start of Activities emanating from node I

4 Backward Pass: Find LC(I) 1 2 34 5 6 7 8 9 SP P PF ES REW FG IWG L AF IF 60 215 270 230 280 355 370 0 355 370 0 184 215 230 355 270 355 LC(I) = Latest Completion of Activities terminating at Node I.

5 Determining the Critical Path An Activity (I,J) is on the Critical Path if the following three conditions are met: ES(I) = LC(I) ES(J) = LC(J) ES(J) – ES(I) = LC(J) – LC(I) = D(I,J) where D(I,J) is the duration of Activity (I,J).

6 Identify Critical Path 1 2 34 5 6 7 8 9 SP P PF ES REW FG IWG L AF IF 60 215 270 230 280 355 370 0 355 370 0 184 215 230 355 270 355

7 Determining the Slack Times (Floats) The Slack Time, ST(I,J), of an Activity (I,J) is the difference between the maximum time available to perform the Activity (i.e. LC(J) – ES(I)) and its duration D(I,J); that is, ST(I,J) = LC(J) – ES(I) – D(I,J). For Activities on the Critical Path the Slack Time is zero.

8 Determine Slack Times 1 2 34 5 6 7 8 9 SP(124) P PF(124) ES REW FG(75) IWG L (45) AF(75) IF 60 215 270 230 280 355 370 0 355 370 0 184 215 230 355 270 355

9 Construct Time Chart 2004003001000 543 1 9 8 7 8 1 2 2 64 3 1 59 Elapsed Time, Days Critical TasksNon-Critical Tasks

10 Schedule Project 2004003001000 543 1 9 8 7 8 1 2 2 64 3 1 59 Elapsed Time, Days SP PF FG L AF

11 LINGO Model: Tasks MODEL: SETS: TASKS / START, PROCUREMENT, SITE_PREPARATION, POUR_FOUNDATION, ERECT_STEEL, ROOF_EXT_WALL, FABRICATE_GLASS, INT_WALLS_GLASS, LANDSCAPING, ACQUIRE_FURN, INSTALL_FURN, FINISH/: TIME, ES, LS, SLACK; PRED( TASKS, TASKS) / START, SITE_PREPARATION, START, PROCUREMENT, START, ACQUIRE_FURN, SITE_PREPARATION, POUR_FOUNDATION, PROCUREMENT, ERECT_STEEL, POUR_FOUNDATION, ERECT_STEEL, ERECT_STEEL, ROOF_EXT_WALL, ERECT_STEEL, FABRICATE_GLASS, ROOF_EXT_WALL, INT_WALLS_GLASS, FABRICATE_GLASS, INSTALL_FURN, ROOF_EXT_WALL, LANDSCAPING, INT_WALLS_GLASS, INSTALL_FURN, ACQUIRE_FURN, INSTALL_FURN, LANDSCAPING, FINISH, INSTALL_FURN, FINISH /; ENDSETS

12 LINGO Model: Data DATA: TIME = 0, 215, 60, 31, 15, 40, 50, 85, 55, 280, 15,0; ENDDATA

13 LINGO Model: Algorithm @FOR( TASKS( J)| J #GT# 1: ES( J) = @MAX( PRED( I, J): ES( I) + TIME( I))); @FOR( TASKS( I)| I #LT# LTASK: LS( I) = @MIN( PRED( I, J): LS( J) - TIME( I))); @FOR( TASKS( I): SLACK( I) = LS( I) - ES( I)); ES( 1) = 0; LTASK = @SIZE( TASKS); LS( LTASK) = ES( LTASK); END

14 LINGO Solution: Input Data Feasible solution found at step: 0 Variable Value LTASK 12.00000 TIME( START) 0.0000000 TIME( PROCUREMENT) 215.0000 TIME( SITE_PREPARATION) 60.00000 TIME( POUR_FOUNDATION) 31.00000 TIME( ERECT_STEEL) 15.00000 TIME( ROOF_EXT_WALL) 40.00000 TIME( FABRICATE_GLASS) 50.00000 TIME( INT_WALLS_GLASS) 85.0000 TIME( LANDSCAPING) 55.00000 TIME( ACQUIRE_FURN) 280.0000 TIME( INSTALL_FURN) 15.0000 TIME( FINISH) 0.0000000

15 LINGO Solution: Earliest Starts ES( START) 0.0000000 ES( PROCUREMENT) 0.0000000 ES( SITE_PREPARATION) 0.0000000 ES( POUR_FOUNDATION) 60.00000 ES( ERECT_STEEL) 215.0000 ES( ROOF_EXT_WALL) 230.0000 ES( FABRICATE_GLASS) 230.0000 ES( INT_WALLS_GLASS) 270.0000 ES( LANDSCAPING) 270.0000 ES( ACQUIRE_FURN) 0.0000000 ES( INSTALL_FURN) 355.0000 ES( FINISH) 370.0000

16 LINGO Solution: Latest Starts LS( START) 0.0000000 LS( PROCUREMENT) 0.0000000 LS( SITE_PREPARATION) 124.0000 LS( POUR_FOUNDATION) 184.0000 LS( ERECT_STEEL) 215.0000 LS( ROOF_EXT_WALL) 230.0000 LS( FABRICATE_GLASS) 305.0000 LS( INT_WALLS_GLASS) 270.0000 LS( LANDSCAPING) 315.0000 LS( ACQUIRE_FURN) 75.00000 LS( INSTALL_FURN) 355.0000 LS( FINISH) 370.0000 Latest Start is defined by LS(I,J) = LC(I,J) – D(I,J).

17 LINGO Solution: Slack Times SLACK( START) 0.0000000 SLACK( PROCUREMENT) 0.000000 SLACK( SITE_PREPARATION) 124.0000 SLACK( POUR_FOUNDATION) 124.0000 SLACK( ERECT_STEEL) 0.0000000 SLACK( ROOF_EXT_WALL) 0.0000000 SLACK( FABRICATE_GLASS) 75.00000 SLACK( INT_WALLS_GLASS) 0.0000000 SLACK( LANDSCAPING) 45.00000 SLACK( ACQUIRE_FURN) 75.00000 SLACK( INSTALL_FURN) 0.0000000 SLACK( FINISH) 0.0000000

18 CPM with Crashing: LINGO Model (a) MODEL: ! A CPM model with crashing; SETS: TASKS / START, PROCUREMENT, SITE_PREPARATION, POUR_FOUNDATION, ERECT_STEEL, ROOF_EXT_WALL, FABRICATE_GLASS, INT_WALLS_GLASS, LANDSCAPING, ACQUIRE_FURN, INSTALL_FURN, FINISH/: TIME, ! Normal time for task; TMIN, ! Min time at max crash; CCOST, ! Crash cost/unit time; EF, ! Earliest finish; CRASH; ! Amount of crashing;

19 CPM with Crashing: LINGO Model (b) ! Here are the precedence relations; PRED( TASKS, TASKS)/ START, SITE_PREPARATION, START, PROCUREMENT, START, ACQUIRE_FURN, SITE_PREPARATION, POUR_FOUNDATION, PROCUREMENT, ERECT_STEEL, POUR_FOUNDATION, ERECT_STEEL, ERECT_STEEL, ROOF_EXT_WALL, ERECT_STEEL, FABRICATE_GLASS, ROOF_EXT_WALL, INT_WALLS_GLASS, FABRICATE_GLASS, INSTALL_FURN, ROOF_EXT_WALL, LANDSCAPING, INT_WALLS_GLASS, INSTALL_FURN, ACQUIRE_FURN INSTALL_FURN, LANDSCAPING, FINISH, INSTALL_FURN, FINISH /; ENDSETS

20 CPM with Crashing: LINGO Model (c) DATA: TIME = 0, 215, 60, 31, 15, 40, 50, 85, 55, 280, 15, 0; ! Normal times; TMIN = 0, 195, 50, 28, 12, 35, 45, 50, 40, 240, 13, 0; ! Crash times; CCOST = 0, 2, 3, 3, 4, 4, 5, 3, 2, 1, 2, 0; ! Cost ($K)/day to crash; DUEDATE = 348; ! Project due date; ENDDATA

21 CPM with Crashing: LINGO Model (d) ! The crashing LP model; ! Define earliest finish, each predecessor of a task constrains when the earliest time the task can be completed. The earliest the preceding task can be finished plus the time required for the task minus any time that could be reduced by crashing this task.; @FOR( PRED( I, J): EF( J) >= EF( I) + TIME( J) - CRASH( J) );

22 CPM with Crashing: LINGO Model (e) ! For each task, the most it can be crashed is the regular time of that task minus minimum time for that task; @FOR( TASKS( J): CRASH( J) <= TIME( J) - TMIN( J) ); ! Meet the due date; ! This assumes that there is a single last task; EF( @SIZE( TASKS)) <= DUEDATE;

23 CPM with Crashing: LINGO Model (f) ! Minimize the sum of crash costs; MIN = @SUM( TASKS: CCOST * CRASH); END

24 CPM with Crashing: LINGO Solution EF( START) 0.0000000 2.000000 EF( PROCUREMENT) 195.0000 0.0000000 EF( SITE_PREPARATION) 60.00000 0.0000000 EF( POUR_FOUNDATION) 91.00000 0.0000000 EF( ERECT_STEEL) 210.0000 0.0000000 EF( ROOF_EXT_WALL) 250.0000 0.0000000 EF( FABRICATE_GLASS) 335.0000 0.0000000 EF( INT_WALLS_GLASS) 335.0000 0.0000000 EF( LANDSCAPING) 305.0000 0.0000000 EF( ACQUIRE_FURN) 280.0000 0.0000000 EF( INSTALL_FURN) 348.0000 0.0000000 EF( FINISH) 348.0000 0.0000000 Optimal solution found at step: 13 Objective value: 44.00000 Variable Value Reduced Cost

25 CPM with Crashing: LINGO Solution (Crash Times) Variable Value Reduced Cost CRASH( START) 0.0000000 0.0000000 CRASH( PROCUREMENT) 20.00000 0.0000000 CRASH( SITE_PREPARATION) 0.0000000 3.000000 CRASH( POUR_FOUNDATION) 0.0000000 3.000000 CRASH( ERECT_STEEL) 0.0000000 2.000000 CRASH( ROOF_EXT_WALL) 0.0000000 2.000000 CRASH( FABRICATE_GLASS) 0.0000000 5.000000 CRASH( INT_WALLS_GLASS) 0.0000000 1.000000 CRASH( LANDSCAPING) 0.0000000 2.000000 CRASH( ACQUIRE_FURN) 0.0000000 1.000000 CRASH( INSTALL_FURN) 2.000000 0.0000000 CRASH( FINISH) 0.0000000 0.0000000

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