# Precedence Diagram Technique Precedence Networks Critical Path Analysis.

## Presentation on theme: "Precedence Diagram Technique Precedence Networks Critical Path Analysis."— Presentation transcript:

Precedence Diagram Technique Precedence Networks Critical Path Analysis

why do it? what are the inputs? the process using the outputs PERT and uncertainty

identify tasks sequence estimate resources analysis scheduling optimisation the context

what are the benefits? critical path non-critical activities

what are the inputs? tasks sequence/dependencies durations

rules and conventions just one start just one finish no danglers task number and/or name duration early start time late start time early finish time late finish time float slack

the process - forward pass Duration = 18 Task 06 2 Task 01 3 30 3 Task 04 6 3 9 Task 03 3 10 7 5 Task 08 2 7 5 Task 02 4 3 7 Task 09 1 18 17 Task 05 3 9 12 Task 07 5 12 17

the process - backward pass Task 06 2 3 5 13 15 10 Task 09 1 18 17 18 17 0 Task 07 5 12 17 12 0 Task 05 3 9 12 9 0 Task 04 6 3 9 9 3 0 Task 01 3 30 3 0 0 Task 08 2 7 5 171510 Task 03 3 10 7 12 9 2 Task 02 4 3 7 9 5 2

using the outputs Gantt Charts resource histograms resource smoothing optimising the schedule

PERT and uncertainty - 2 Task 01Task 02Task 03Task 04 4 2 5 4 The better estimate for Task 03 might be its PERT estimate, or Expected Value, EV) EV = O + 4L + P ___________ 6 O = Optimistic estimate (say 2) L = Most Likely (say 4) P = Pessimistic (say 12) In this example the EV = 6, which does, in fact change the critical path

PERT and uncertainty - 3 But how confident can we be in these results? An durations spread is the degree to which estimates of the duration differ from each other. If every estimate of duration were about equal, the estimate would have very little spread. There are many measures of spread. The distributions on this page have the same mean but differ in spread: the distribution on the bottom is more spread out.

PERT and uncertainty - 4 Standard deviation is used as a measure of spread. In a normal distribution, about 68% of estimates are within one standard deviation of the mean and about 95% of the estimates are within two standards deviations of the mean. EV = O + 4L + P ___________ 6 O = Optimistic estimate (say 2) L = Most Likely (say 4) P = Pessimistic (say 12) SD = (P – O)/6In our example SD = (12-2)/6 = 1.666 So, we could say that, for task 03: With 68% certainty, the duration will be between 2.34 and 5.66 (4 ± SD) With 95% certainty the duration will be between 0.68 and 7.32 (4 ± 2SD)