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Expected length of a critical path:
The expected length of a sequence of independent activities is simply the sum of their separate expected lengths. This gives us the expected length of the entire project. We have to calculate the expected length te of every activity with the weights attached to the three time estimates and find the critical path in the manner described previously. The expected length of the entire project denoted by Te is the length of the critical path (i.e.) the sum of the, te's of all activities along the critical path.
In the same way, the variance of a sum of independent activity times is equal to the sum of their individual variances. Since Te is the is the sum of te's along the critical path, then variance of Te equals the sum of all the variances of the critical activities. The standard deviation of the expected project duration is the square root of the of the variance Te as calculated above.
At this juncture, consider the following example to illustrate the application of these formulae.
Example
Fig. 20
Activity Expected time (te) Std. deviation Variance 1-2 (1+4+7)/6 = 2 (7-1)/6 = 1 1 1-3 (1+16+7)/6 = 4 (7-1)/6 = 1 1 1-4 (2+8+8)/6 = 3 (8-2)/6 = 1 1 2-5 (1+4+1)/6 = 1 (1-1)/6 = 0 0 3-5 (2+20+14)/6 = 6 (14-2)/6 = 2 4 4-6 (2+20+8)/6 = 5 (8-2)/6 = 1 1 5-6 (3+24+15)/6 = 7 (15-3)/6 = 2 4 For each activity, the optimistic mostly likely and pessimistic time estimates are labeled in the same order. Using PERT formulae for te and St tabulate the results as above. To calculate the critical path, list all the three paths with their expected time of completion from figure 21.
1-2-5-6 = 10 1-3-5-6 = 17 1-4-6 = 8
Expected time and Critical path.
Fig. 21
The longest path is the critical path and it consists of activities 1-3, 3-5 and 5-6 with a length Te = 4 + 6 + 7 = 17. The variance of the critical path is then VT = Variance of 1-3 + variance of 3-5 + variance of 5-6.
VT 1 4 4 9
Standard deviation of duration of the critical path
VT 9 3
So far what we have done with PERT model is to recognize uncertainty by using three time estimates and these are reduced to a single time estimate for finding the critical path. It can be used to find early start-early finish programme, late start-late finish programme and slack. The variability of the time estimates for each activity is also reduced to a standard deviation and variance and this is used to find the standard deviation of expected completion time for the project.
Probability of completing a project with a given date
One may wonder how the calculations made to find Te, ST and VT are useful to a project manager. These parameters serve as a very useful tool for a project manager to estimate the probability of completing a project with a given date.
We have seen that the time required for an activity is uncertain and hence it is a random variable. Its average or expected value (te) can be estimated on the basis of an assumption regarding the type of probability distribution, and three points on this distribution namely optimistic, most likely and pessimistic time estimates.
We estimate the average, or expected, projected length, Te by adding the expected activity durations along the critical path. As the te's are all random variables, then so is Te. It is assumed that Te is to follow approximately a normal distribution. This gives the probability of completing a project with a given date.
Referring to the example illustrated above, the expected project length is 17 days and a standard deviation of 3 days. These are the two parameters useful to calculate the probability if the due date is to be met. For example,
(a) Draw the project network. (b) Calculate the length and variance of the critical path. (c) What the is approximate probability that the jobs on the critical path will be completed by the due date of 42 days? (d) What due date has about 90% chance of being met?
Solution:
Before proceeding to draw the project network, let us calculate the expected time of activity te, standard deviation and variance of the expected time of activity using
o m p e
t t t t
. ( ) / 6; Variance (. )^2 S D t (^) p to S D
Activity te (Days) S.D (Days) Variance 1 - 2 7 2 4 1 - 6 6 2 4 2 - 3 14 4 16 2 - 4 5 1 1 3 - 5 11 2 4 4 - 5 7 2 4 6 - 7 11 4 16 5 - 8 4 1 1 7 - 8 18 4 16
(a) Project Network: 3
14 11
2 5
7 5 7 4
1 4 8
6 11 18 6 7
Expected time and Critical path Fig. 24
(b) There are three paths:
1-2-3-5-8 = 36 days 1-2-4-5-8 = 23 days 1-6-7-8 = 35 days
1-2-3-5-8 is the longest path and hence the critical path.
Expected length of the critical path is 36 days. The variance for 1-2, 2-3, 3-5 and 5-8 are 4, 16, 4 and 1 respectively and variance of the projection duration is 25 and hence
Standard deviation of the project duration = 25 = 5 days.
(c) Due date = 42 days (T) Expected duration = 36 days (Te) and S.D = 5 days (o)
Z ( T Te ) / S D.. (42 36) / 5 1.
The area under the normal curve for Z = 1.2 is 0.3849. Therefore, the probability of completing the project in 42 days
= 0.5000 + 0. = 0. = 88.49%
Exercise
(1) A project has the following characteristics:
Activity Optimistic time Pessimistic time Most likely time 1-2 1 5 1. 2-3 1 3 2 2-4 1 5 3 3-5 3 5 4 4-5 2 4 3 4-6 3 7 5 5-7 4 6 5 6-7 6 8 7 7-8 2 6 4 7-9 5 8 6 8-10 1 3 2 9-10 3 7 5
Construct a network. Find the critical path and variance for each event. Find the project duration at 95% probability.
