4th year (Production & Mechatronics)
Project Management (M1462)
Eng. Mohamed Ismail
Eng. Noah El-Zathry
Benha university
Faculty of engineering
Mechanical engineering Department
Semester 2 (2019/2020)
Sheet (4): PERT
The main objective in the analysis through PERT is to find out the completion for a particular
event within specified date. The PERT approach takes into account the uncertainties. The
three-time values are associated with each activity.
1. Optimistic time – It is the shortest possible time in which the activity can be finished. It
assumes that everything goes very well. This is denoted by t0.
2. Most likely time – It is the estimate of the normal time the activity would take under
normal conditions with normal delays. This is denoted by tm.
3. Pessimistic time – It represents the longest time the activity could take if everything goes
wrong. This is denoted by tp.
In PERT calculation, all values are used to obtain the percent expected value.
1. Expected time – It is the average time an activity will take if it were to be repeated on
large number of times and is based on the assumption that the activity time follows Beta
distribution, this is given by equations 1 and 2:
te =
(t0 +4tm + tp )
(1)
6
2. The variance for the activity is given by
σ2 = [
𝑡𝑝 −𝑡0 2
6
]
(2)
PERT solution steps
1. compute the expected duration of each activity te using eq (1).
2. compute the expected variance of each activity σ 2 using eq (2).
3. Draw the project network and identify the critical path as illustrated in the previous
section.
4. calculate the expected time and variance of the critical path by summing te and σ 2 for
each activity in the critical path. 𝑇𝑒 = 𝑡1 + 𝑡2 + ⋯ + 𝑡𝑘 and σ2𝑇 = σ12 + σ22 + ⋯ + σ2𝑘 .
5. to determine the probability of the project to be finished at a given time t, we can assume
that the total project follow normal distribution and use the following equation.
𝑥−𝜇
𝑡− 𝑇𝑒
𝑧=
=
σ
σ𝑇
*
Note: according to the central limit theorem, the sum of several independent activity duration
tends approximately to be normally distributed, with a mean equal to the sum of their
individual job times and variance equal to the sum of their individual activity variance.
1
(1)
(a)
(b)
By applying eq (1) and eq (2) for each activity we can compute t e and σ2 for each activity
te =
(t0 +4tm + tp )
6
and σ2 = [
𝑡𝑝 −𝑡0 2
6
] .
𝑇𝑒 = 𝑡1 + 𝑡2 + ⋯ + 𝑡𝑘
Path A-C-E: Te = 6 + 8.17 + 6.33 =20.5 days.
Path D-F-G: Te = 12 + 6 + 3.5 = 21.5 days.
Path B-H-I: Te = 8.5 + 4.17 +6.83 = 19.5 days.
So path D-F-G is the critical path with Te = 21.5 days.
For critical path: σ2𝑇 = σ12 + σ22 + ⋯ + σ2𝑘 = 36/36 + 4/36 + 25/36 = 65/36.
(c)
Revised D : te = (5 + 4*6 + 7) /6 = 6 days
7−5 2
Revised σ2 of D: [
6
] = 4/36.
2
𝑇𝑒 = 𝑡1 + 𝑡2 + ⋯ + 𝑡𝑘
Path C-E from start of the project: Te = 7 + (8.71 + 6.33) =21.5 days.
Path D-F-G from start of the project: Te = 7 + (6 + 6 + 3.5) = 22.5 days.
Path B-H-I from start of the project: Te = 7 + (4.17 +6.83) = 18 days.
So path D-F-G is the critical path with Te = 22.5 days.
For critical path: σ2𝑇 = σ12 + σ22 + ⋯ + σ2𝑘 = 4/36 + 4/36 + 25/36 = 33/36.
3
(5)
(a)
𝑇𝑒 = 𝑡1 + 𝑡2 + ⋯ + 𝑡𝑘 and σ2𝑇 = σ12 + σ22 + ⋯ + σ2𝑘
Path 1-2-3-8-11:
Te = 8 + 12 + 10.17 + 7.17 = 37.34 weeks.
σ2 = 0/36 + 4/36 + 9/36 + 25/36 = 38/36.
Path 1-2-4-6-11:
Te = 8 + 6 + 18.67 + 10.33 = 43 weeks.
σ2 = 0/36 + 4/36 + 144/36 + 36/36 = 184/36.
Path 1-2-4-7-11:
Te = 8 + 6 + 13 + 13 = 40 weeks.
σ2 = 0/36 + 4/36 + 0/36 + 25/36 = 29/36.
Path 1-2-5-9-10-11:
Te = 8 + 12 + 9.83 + 11 + 6 = 46.83 weeks.
σ2 = 0/36 + 16/36 + 25/36 + 4/36 + 0/36 = 45/36.
4