l12prob_plan2_04

l12prob_plan2_04 - Probabilistic Planning 2 Nathaniel...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Probabilistic Planning 2 Nathaniel Osgood 3-27-2004 Topics PERT (Cont'd) Review Merge node bias PNet refinement Monte Carlo Simulation approaches General Demo Process Interaction Activity Scanning PERT Basics Expresses uncertainty in activity duration Beta distribution assumed for activities Assume normally distributed project duration Project Duration Tends to be Normally Distributed (approx. sum of random variables) Assumes Independent Activity Durations - Not Always Satisfied 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 b-a s= 6 v=s 2 Recall: 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 Compute probability complete project by time t 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 Example-Standard Deviation Te = 11 S = V [C ] + V [ E ] + V [G ] 2 = 0.25 + 0.25 + 01111 . = 0.6111 S = 0.6111 = 0.7817 PERT Analysis-Probability 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 11.5 ) = P (T 11.5 ) - P (T 9 ) 11.5 - 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 Topics PERT (Cont'd) Review Merge node bias PNet refinement Monte Carlo Simulation approaches General Demo Process Interaction Activity Scanning Merge Node Bias Misleading to consider only variance from single predecessor for each node on critical path Early start of node depends on maximum of finish (or start) times of predecessors including non-critical! Basically ES = RV that is max of (non-iid) RVs Effect stronger if have More predecessors Predecessors with almost equal timing Less dependency among predecessors Consequence: Unrealistic optimism with respect to expected completion times, but especially variance Example Merge Node ES(C)=Max(EF(A), EF(B)) =10.777 =1.55 C A Late Finish: N(10,1) B Late Finish: N(9,3) Sample Problem Derived Parameters MEAN AND STANDARD DEVIATION OF THE CRITICAL AND NEAR CRITICAL PATHS FOR NETWORK TIME ESTIMATES PATH 0-3-7-8 (Critical Path) PATH 0-3-4-5-8 ACTIVITY 0-3 3-7 7-8 3-4 4-5 5-8 a 1 6 3.5 1 2 2 m 2 8 5 4 4 3 b 3 10 6.5 13 6 4 MEAN 2 8 5 VARIANCE 0.39 1.56 0.88 MEAN 2 VARIANCE 0.39 5 4 3 14.0 14.06 1.56 0.39 16.40 4.05 15.0 2.83 1.68 TOTALS* STANDARD DEVIATION - - * The mean and variance of the duration of a path is merely the sum of the means and variances of the activities along the path in question; the standard deviation of the path duration is then obtained as the square root of its variance. Impact of Multiple Paths Log Scale (maximum of times of both paths) Nave Approach Consider variance from all paths entering a merge node Assume Probability EF(i)<T = jPaths To(i) P(EF(j)<T) Recall PERT Analysis - ADG Path Te = 7 S = V [ A] + V [ D] + V [G ] = 0.25 + 0.25 + 011 . 2 = 0.6111 S = 0.6111 = 0.7817 PERT Analysis - ADG Path Probability of Ending before 10 10 - 7 P( T 10) = P z 0.7817 = P( z 38378) . = 0.9999 PERT Analysis - BG Path Te = 8 S = V [ B] + V [G ] 2 = 01111 + 01111 . . = 0.2222 S = 0.2222 = 0.4714 PERT Analysis - BG Path Probability of Ending before 10 10 - 8 P( T 10) = P z 0.4714 = P( z 4.2429) = 0.9999 PERT Analysis - ADG , BG and CEG Paths Probability of Ending before 10 P ( T 10) = P(TCEG 10) P(TADG 10) P(TBG 10) c = (01003)(09999)(09999) . . . = 01003 . = 10% PERT (cont): For the G finish within 10 days, all 3 paths must finish in 10 days or less (i.e. ADG and CEG and BG) Calculated as: P(T10)=P(ADG10)*P(CEG10)*P(BG10) What is wrong with this equation? The equation assumes the path durations are independent! This cannot be if there are shared activities between the paths. Example of Multiple Paths Dependent and Independent Activities with duration 2 have =.707 Activities with duration 4 have =1.414 PERT (cont): A Solution: Use either PNet Monte Carlo simulation PNet Aims at addressing merge node bias Basically works by Enumerate all paths P s.t. Dur(P)> Dur(crit path) Rank paths by decreasing duration (by decreasing naivelyestimated variance for ties) Compute linear correlation coefficient between paths Enter paths, eliminating any path whose correlation coefficient with a previously-entered path is > .5 P(T a ) = # remaining paths i =1 P ( pi T ) Validity of Beta distribution for activity durations Validity of central limit theorem for project duration Activity durations are not independent! PERT Disadvantages Take into consideration only critical path Not just sum of random variables -- have max. at joins Leads to overoptimism & underestimation of duration Multiple time estimates required to calibrate Can be time consuming Topics PERT (Cont'd) Review Merge node bias PNet refinement Monte Carlo Simulation approaches General Demo Process Interaction Activity Scanning Monte Carlo Simulation Characteristics Replaces analytic solution with raw computing power Avoids need to simplify to get analytic solution No need to assume functional form of activity/project distributions Used by Van Slyke (1963) Aimed at solving the merge bias problem in PERT Allows determining the criticality index of an activity (Proportion of runs in which the activity was in the critical path) Hundreds to thousands of simulations needed Monte Carlo Simulation Process Set the duration distribution for each activity No functional form of distribution assumed Could be joint distribution for multiple activities Iterate: for each "trial" ("realization") Sample random duration from each distributions Find critical path & durations with standard CPM Record these results Report recorded results Duration distribution Per-node criticality index (% runs where critical) Network Monte Carlo Simulation Example Optimistic Time, a 2 1 7 4 6 2 4 Statistics for E xample Activities Most Likely Pessimistic Expected Time, Time, Val u e, m b d 5 8 5 3 5 3 8 9 8 7 10 7 7 8 7 4 6 4 5 6 5 Stan d ard Deviation , s 1 0.66 0.33 1 0.33 0.66 0.33 Activity A B C D E F G Monte Carlo Simulation Example Run Number 1 2 3 4 5 6 7 8 9 10 Summary of Simulation Runs for Example Project Activity Duration Critical Completion B C D E F G Path Time 2.2 8.8 6.6 7.6 5.7 4.6 A-C-F-G 25.4 1.8 7.4 8.0 6.6 2.7 4.6 A-D-F-G 17.4 4.9 8.8 7.0 6.7 5.0 4.9 A-C-F-G 26.5 2.3 8.9 9.5 6.2 4.8 5.4 A-D-F-G 25.0 2.6 7.6 7.2 7.2 5.3 5.6 A-C-F-G 23.0 0.4 7.2 5.8 6.1 2.8 5.2 A-C-F-G 22.3 4.7 8.9 6.6 7.3 4.6 5.5 A-C-F-G 24.2 4.4 8.9 4.0 6.7 3.0 4.0 A-C-F-G 22.1 1.1 7.4 5.9 7.9 2.9 5.9 A-C-F-G 18.9 3.6 8.3 4.3 7.1 3.1 4.3 A-C-F-G 19.7 A 6.3 2.1 7.8 5.3 4.5 7.1 5.2 6.2 2.7 4.0 Project Duration Distribution Probability P( X ) = Number of Times Project Finished in Less Than or Equal to t weeks Total Number of Replications The Probability that the project ends in 20 weeks or less is P (X 20 )= 13 / 50 = 26 % Criticality Index Definition: Proportion of runs in which the activity was in the critical path PERT, CPM assume binary (either 100% or 0%) Helpful for prioritizing effort in Monitoring Controlling How Many Runs are Needed? Criticality Index p (particular node) Originally very conservative (10K runs) Empirical tests suggest 1000 runs adequate Estimate of confidence interval for criticality ^ (1- ) confidence interval=symmetric interval around p such that P(true value p is within that interval) is (1-)% ^ p - Z 2 ^ ^ p (1 - p) ^ , p + Z 2 n ^ ^ p (1 - p) n Consider a 95% confidence interval with 10% p 90%, ^ 400n 1000. Then with 95% confidence, p will be within 2%-5% of p How Many Runs are Needed? Mean Project Duration Must make assumptions regarding coefficient of variation = / (i.e. Std Dev/Mean) Z ^ u Basic formula Error % 100 2 n For Empirical range of CoV (5%..15%) Sample size 400: within .5% to 1.5% of true value Sample size 1000: within .3% to 1% of true value Note inverse-root relationship: Halving error requires increasing # of trials by a factor of 4! How Many Runs are Needed? Project Duration Standard Deviation Basic formula Error % 100 Z 2 2n ^ Sample size 400: within 7% of true value ^ Sample size 1000: within 4.38% of true value Inverse-root relationship again present Monte Carlo: Summary Conceptually simple Standard CPM used No need for special assumptions about functional form of distributions Provides criticality index (valuable prioritization) Scalable analysis quality (albeit with super-linear effort required to reduce error) Computationally expensive Estimation of duration distributions can be expensive Topics PERT (Cont'd) Review Merge node bias PNet refinement Monte Carlo Simulation approaches General Demo Process Interaction Activity Scanning (Dynamic) Simulation Approach CPM-Based methods use simple representations One-pass: No iteration Represented uncertainty only with respect to duration Explicitly representing process brings benefits Reasoning about process design Identifying emergent behavior (e.g. dynamic bottleneck) Simpler estimation of some uncertainties Must be clear about whether representations are just process-level or also project-level Detailed Representation Repetitive processes for which aggregate representation is not desirable Processes where static planning is not possible Repetitive processes for which # cycles unknown Scheduling and coordinating complex interactions (Large #s of brief interactions, dependent on timing) Cases where timing uncertainties change schedule Cases where individual timing component can be estimated, but where aggregate stats not known Examples of Repetitive Processes Earth moving Tunneling Hotel/Apartment/Dormitory construction Road/Bridge construction Plumbing and glazing in high-rise Topics PERT (Cont'd) Review Merge node bias PNet refinement Monte Carlo Simulation approaches General Demo Process Interaction Activity Scanning Simulation Example: Excavation and Transporting Given Front-end loader Output: ofront-end loader Instantaneous time between loads Trucks n vehicles nc ncsl se otrucks = = Capacity c d d d ( se + sl ) + Load time tl sl se Instantaneous dump time Fully loaded speed sl , empty speed se Distance to dumpsite d Nave productivity: min(ofront-end loader, otrucks) ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online