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Unformatted text preview: Deterministic Planning II & Probabilistic Planning I
Nathaniel Osgood 3152004 Announcements
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Primavera tutorials Complicated scheduling case Problem set 4 (Scheduling; due Monday April 5) Wednesday guest lecture on behavioral managerial issues Recall: AON (PDM) Scheduling
Activities shown on nodes O(n) Forward/backward pass to determine ES/EF/LS/LF Multiple types of relationships
FS, SS, FF, SF No dummy arrows required Example Applications
TASK NO. 1 2 3 4 5 6 7 TASK NAME Place and Secure Trusses Install Roof Deck Apply Vapor Barrier Apply Roof Cladding Construct Roof Overhang Install Soffits Apply Flashing DURATION (in days) 2 7 2 2 4 4 6 Example Of CPM Algorithm Forward Pass Backwards Pass Recall: PDM Relationships
PDM Extends CPM to include
Multiple relationships (SS, SF, FF) beyond FS Lags (negative as "leads") Consider relationship XY with lag t between activities A and B
X, Y { S, F }, t Interpretation is that event Y of activity B can occur no earlier than t units after evnt X occurs for activity A Think of relationships as linking events Special relationships not needed in AOA
Can be placed directly between nodes Notation
Nodes are no longer simply vertices in graph
Arrow on left side of node indicates a start relationship Arrow on right side of node indicates finish relationship Nonplanar networks may require "jumps" PDM Activity Relationships
FinishtoStart Lead FinishtoStart Lag
LayOut & Excavate Install Fuel Tanks FS = 1
Pour 4thFloor Slab Remove 4th Floor Shoring FS = +14 Install Pipe Install Fuel Tanks StarttoStart Lag
Partially Adapted from Kellegeiros, 2003 Install Exterior Conduits SS = +1 PDM Activity Relationships (Cont'd) FinishtoFinish Lag
Excavate Trench Lay Pipe FF = +3 StarttoFinish Lag Install Wood Paneling & Base SF = +1
Install Carpeting SF '
Partially Adapted from Kellegeiros, 2003 SF '' PDM Caveats
Can have different semantics, but same result Asymmetries complicate reasoning Make sure you understand the meaning of relationships for the software you use! "Lag" and "Lead" lack standard definition May have different floats for same activity
Start float (LSES) Finish float (LFEF) Arises from successors for these events PDM Caveats II Critical Path
Choices impact critical path!
E.g. Finishtostart vs. Starttostart Think of critical path as running through events Tracing critical path can be difficult
Noncritical activity can have critical start/finish w/o splitting, can be counterintuitive (longer duration leads to shorter critical path!) Finishfinish constraints with leads can lead to "vanishing" critical path How critical path displayed depends on software "Vanishing Critical Path" Example of CounterIntuitive
The longer A20 is, the smaller the critical path duration and quicker can complete!
A30 FF2 A20 SS A10 Equivalent Timing Results Vs. Meaning is different Critical path may be different Reasoning about Relationships
Key Point: PDM relationships often represent relationships between particular parts of an activity. Think about
On what portion of an activity the other activity depends On how dependency would change if target activity duration changed If unclear, think about unbundling activity Multiple Relationships Vs. Asymmetries Vs. Bases for Formal Analysis
Legal SB A30 A20 SA Illegal 1 Method 1 Method 2 Distinguishing FF Interpretations NonBinding; A10 Time Unaffected Distinguishing FF Interpretations Binding; A20 Waits for A10 Activity Splitting I: NonSequential
Some algorithms allow division of an activity into two nonsequential pieces Advantages: Allows more flexibility in time, resource demands
Permits shorter critical paths Eliminates counterintuitive cases where prefer longer activity Allows predecessor activities connected via SF and SS relationships to begin Allows successor activities connected via SF or FF relationships to begin bulk of work early, and then just wait for event to finish Example of CounterIntuitive
The longer A20 is, the smaller the critical path duration and quicker can complete!
A30 FF2 A20 first SS A10 A20 second Example
Because of executive offices, can't finish carpeting until wood panelling starts Problem: Want carpenters for other work Answer: Split wood paneling
Do all carpeting except executive offices Allow carpenters to work on executive offices Finish carpeting work for the executive offices Carpenters back to finish job once available Activity Splitting 2: Pipelining
Turns monolithic tasks into subtasks that operate in parallel Typically increases resource demand Typically done manually (generally not enough information to permit automation) Often represent with SS constraint
Bubble Patterns when Splitting Tasks
Before P 10 Q 9 P1 5 After Q1 4 48 48 58 58 58 58 67 67 48 48 53 53 53 54 1 57 58 0 58 58 Q2 5 63 63 53 53 0 P2 5 58 58 0 Activity Windows
Mechanism for imposing time constraints on absolute activity times Can impose constraint for any of times
ES, EF, LS, LF By set WES=WLS, fix exact timing Particularly useful for timecritical milestones Forward Pass for node k (no splits; no leads)
INITIAL TIME WESk WEFk  Dk EFk = ESk + Dk EFp + FSpk ESp + SSpk EFp + FFpk  Dk ESp + SFpk  Dk ESk = Maxall p Moder Key factor: Cannot start until all predecessors ready! Must take maximum of predecessors' values Backward Pass (node k, no splitting; no leads)
TERMINAL TIME WLFk WLSk + Dk LSs  FSks LFs  FFks LSs  SSks + Dk LFs  SFks + Dk LFk = Minall s , LSk = LFk  Dk Moder Key factor: Must finish in time for all successors to start in time! Otherwise will delay project completion time => Must take min of successors' values. Dealing with Leads
Dealing with leads ("negative lags") requires more general algorithm
Two O(n) passes may no longer be sufficient Be careful about meaning! Basic approach: Convert AON into AOAlike form
Start/Finish Nodes explicit for every activity Very helpful for thinking through meaning Use Dijkstra's algorithm to solve O(VlgV+E) Example Translated Diagram
8 Si 1 PS 3 4 i 5 Fi 6 PF Fj 7 Sj 2 j Unified Algorithm
Calculations for the Unified Network Model with Negative Link Durations
Forward Pass: Step 1: Set PL(i) =  and TL(i) =  , where Set TL(PS) = 0 where PS = the project start node PL(i) = the maximim distance from PS to node i TL(i) = the maximim distance from PS to node i found at intermediate stages Step 2: Select node i for which TL(i) is the maximum among all nodes. Set PL(i) = TL(i) and TL(i) =  For each link originating at node i, If PL(j) =  and PL(i) + D(i, j) > TL(j), then set TL(j) = PL(i) + D(i, j). If PL(j) =  and PL(i) + D(i, j) < TL(j), then do not change the labels on j. If PL(j) >  and PL(i) + D(i, j) > PL(j), then set PL(j) =  and TL(j) = PL(i) + D(i, j). If PL(j) >  and PL(i) + D(i, j) > PL(j), then do not change the label on j. Step 3: Step 4: Backward Pass: Repeat application of the algorithm with the following changes: 1. Reverse each link direction. 2. Start with the project finish node PF with TL(PF) = 0. 3. At the end of step 3, set the latest event time, L(i) = E(PF)  PL(i) for all nodes i. Repeat step 2 until PL(PF) >  is a number larger than any link duration. , where PF is the project finish node. Set the earliest event time for each node, E(i) = PL(i). Schedules exhibit much uncertainty Motivations for Dealing with Uncertainty Weather occurrences, design duration, productivity, delivery times, subcontractor quality, regulatory changes, etc... Clients, community may want to know milestones, finish date with confidence
Tenant movein dates Traffic planning Event planning Reasoning about schedule constraints such as weather, seasonal traffic Extensions may be much worse than early completion Case 1: Logan International Terminal
Firm date required by
Vendors Airlines Sought probabilistic scheduling to quantify uncertainty Case 2: Philadelphia Children's Hospital
Described in "Modern Steel Construction" article (posted on STELLAR site) Accounting for
Emergency helicopter usage Patient area activity Limited time windows for work (24am) Contingencies regarding telephone switch relocation Informal Ways of Handling Uncertainty
Most common: Ignore!
Assume expected duration Hope errors cancel Apply contingency factors "What if" scenario analysis to examine
Optimistic scenario Most likely scenario Pessimistic scenario Program Evaluation and Review Technique (PERT)
Developed by US Navy, BoozAllen Hamilton and Lockheed Corporation Polaris Missile/Submarine (1958) Captures probabilistic activity durations Allows analytic solution for Schedule duration Schedule variance PERT Basics
Beta Distribution for Activity Duration Assume normally distributed project duration
Project Duration Tends to be Normally Distributed (approx. sum of random variables) Assumes Independent Activity Durations  Not Always Satisfied Activity Duration Frequency Beta Distribution Three Cases of Beta Distribution Beta vs. Normal Can guarantee Beta nonnegative Normal Distribution Assumed for Schedule Stochastic Approach
Optimistic Most Likely (mode not mean) Pessimistic Expected Duration Variance Standard Deviation
d=
_ a m b
1 1 a + 4m + b 2m + (a + b ) = 3 2 6 ba s= 6 v=s 2 Steps in PERT Analysis
For each activity k
Obtain ak, mk (mode) and bk Compute expected activity duration (mean) dk=te Compute activity variance vk=s2 Compute expected project duration D=Te using standard CPM algorithm Compute Project Variance V=S2 as sum of critical path activity variance (this assumes independence!)
In case of multiple critical paths use the one with the largest variance Calculate probability of completing the project
Assuming project duration normally distributed PERT Example
Calculated
Activity Predecessor A B C D A E C F A G B,D,E a 1 5 2 1 4 3 1 m 2 6 4 3 5 4 2 b 4 7 5 4 7 5 3 d 2.17 6.00 3.83 2.83 5.17 4.00 2.00 v 0.25 0.11 0.25 0.25 0.25 0.11 0.11 Activity on Node Example Forward Pass Backward Pass PERT ExampleStandard Deviation Te = 11 S = V [C ] + V [ E ] + V [G ]
2 = 0.25 + 0.25 + 01111 . = 0.6111 S = 0.6111 = 0.7817 PERT AnalysisProbability of Ending before 10 (Critical Path Only)
P(T Td ) = P(T 10) 10  Te = P z S 10  11 = P z 0.7817 = P( z 12793) . = 1  P( z 12793) . = 1  0.8997 = 01003 . = 10% PERT Analysis  Probability of Ending before 13 (Critical Path Only) 13  11 P( T 13) = P z 07817 . = P( z 25585) . = 09948 . PERT Analysis  Probability of Ending between 9 and 11.5(CP Only)
P(TL T TU ) = P( 9 T 15) . = P( T 115)  P( T 9) . 115  11 9  11 = P z  P z 0.7817 0.7817 = P( z 0.6396)  P( z 2.5585) = P( z 0.6396)  [1  P( z 2.5585) ] = 0.7389  [1  0.9948] = 0.7389  0.0052 = 0.7337 ...
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This note was uploaded on 10/29/2008 for the course PM 1040 taught by Professor Dr.nathanielosgood during the Spring '04 term at MIT.
 Spring '04
 DR.NATHANIELOSGOOD

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