PERT

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PERT
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What is PERT?

• Cases where activity durations are uncertain or
  subject to uncontrollable variation, exact
  project duration cannot be worked out. In this
  case use PERT
• PERT uses 3 estimates instead of one for each
  activity
• From these, the mean duration for each activity
  can be computed
• The statistical standard deviation of the
  activity duration can also be computed
• From the mean, we can compute (by theory of
  statistics)
• The standard deviation of the project duration
• The probabililty of the project being completed
  within certain times
• PERT suitable for RD projects

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Time Estimates

• For each activity, 3 estimates of the activity
  duration are needed
• aoptimistic time (if execution goes well)
• bpessimistic time (if execution goes badly)
• mmost likely time (if execution is normal)
• The mean
• Te(a4mb)/6
• The standard deviation which gives an indication
  of the possible spread of activity duration
• ?i(b-a)/6

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Mean and Standard Deviation of Project Duration

• Once the expected time te for all activities has
  been computed, proceed to use te in place of the
  single activity duration in CPM to work out the
  critical path and the project duration
• The resulting project duration is the mean
  project duration TE
• We also need to work out the standard deviation
  of the project duration ? as follows
• Project Duration ? ?(Summation of ?i2 f all the
  activities on the critical path)

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Probability of Different Project Durations

• From statistics, once we know the mean project
  duration, TE, and the standard deviation of the
  project duration, ? we can work out the
  probability that the project duration will be
  shorter than any specific time, T (i.e. the
  project will take T days or less) through the
  following formula
• Z(T- TE )/ ? , where Z is the quantity called
  the Normal variate
• Knowing Z, we can read off the probability from
  Normal Distribution Tables which are provided in
  nest slides

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Normal Distribution Table for Negative Values of Z
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Normal Distribution Table for Positive Values of Z
Z Probability --------------------- 0.0
0.5000 0.1 0.5398 0.2 0.5793 0.3 0.6179
0.4 0.6554 0.5 0.6915 0.6 0.7257 0.7
0.7580 0.8 0.7881 0.9 0.8159 1.0 0.8413
1.1 0.8643 1.2 0.8849 1.3 0.9032 1.4
0.9192 1.5 0.9332
Z Probability --------------------- 1.6
0.9452 1.7 0.9554 1.8 0.9641 1.9
0.9713 2.0 0.9772 2.1 0.9821 2.2
0.9861 2.3 0.9893 2.4 0.9918 2.5
0.9938 2.6 0.9953 2.7 0.9965 2.8
0.9974 2.9 0.9981 3.0 0.9987 gt3.0 1
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Example

• Consider a project with TE 5days and ?2
  days.If we wish to find out the probability that
  the project will take 7 days or less. Thus
  T7days. First, work out a value (calles the
  normal variate) Z, as follows
• Z(T- TE )/ ?(7-5)/21
• Read off the Normal Distribution Tables, the
  probability for Z1. We get the value 0.8413.
  Thus the probability that the project will take 7
  days or less is 0.8413
• If we need to find the probability that the
  project takes more than 7 days, we make use of
  the fact that
• Probability that project takes more than x days
  1-Probability that project takes x days or less
• Probability that project takes more than 7 days
  1-Probability that project takes 7 days or less
  1-0.84130.1587

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Interpolating from the Normal Distribution Table

• In the previous example, the Z value was 1.0
  and could be read off directly. If you had a
  value like 1.01, you could still round it off to
  1.0
• However there will be instances when you will get
  a value like 1.275, in which case you will need
  to interpolate from the table
• From the table Z11.2, P10.8849
• Z21.3, P20.9039
• Use the interpolation formula
• PP1Z-Z1 (P2-P1)
• Z2-Z1
• Therefore at Z1.275,
• P0.8849 1.275 -1.2 (0.9039-0.8849) 0.8992
• 1.3-1.2