1. PERT CPM assumes duration of activity is known with certainty.
PERT assumes duration of activity as a random variable.
For an activity, PERT estimates three quantities:
a : optimistic duration under most favorable condition
b : pessimistic duration under least favorable condition
m : most likely value for activity’s duration

2. Probabilistic Estimates
Activity
start
Optimistic time
Most likely
time (mode)
Pessimistic
time to tp tm te

3. Let Tij be the duration of activity (i,j)
Let Tij be the duration of activity (i,j). PERT assumes that Tij follows a beta distribution and, mean and variance of Tij may be approximated by
: expected duration of activities on any path
: variance of duration of activities on any path

4. Example variance of PROJECT = variance of CRITICAL PATH
if more than one critical path, PROJECT VARIANCE=largest of CRITICAL
Example
Activity a b m
(1,2) 5 13 9
(1,3) 2 10 6
(3,5) 3 8
(3,4) 1 7
(4,5) 12
(5,6) 15

5. Project Diagram

7. Specified time – Path mean Path standard deviation
critical path = B-D-E-F. Thus,
E(CP) = = 38
var(CP) = = 7.22
standard deviation for CP is (7.22)1/2=2.69
What is the chance of completing project within 35 days ?
there is a 13% chance of completing project within 35 days.
Z =
Specified time – Path mean
Path standard deviation
Z indicates how many s.d the specified time is beyond the expected path duration.

8. PERT Example activity duration predecessor te A 2 / 3 / 6 weeks - 3.33
B 3 / 6 / 10 weeks A 6.17
C 1 / 1 / 2 week A 1.17
D 3 / 3 / 3 weeks B, C 3.00
CRITICAL PATH: A-B-D
EXPECTED DURATION: = 12.5
VARIANCE: { (6-2) / 6}^2 + {(10-3) / 6}^2 + {(3-3) / 6}^2 = 1.805
STD = 1.344
e.g.
if critical path expected = 9.5, and STD = 1.354
target=10 , Z = (10-9.5) / = .369
probability = .644

9. EXAMPLE

10. EXAMPLE (Cont.)
Identify all paths and their estimated completion times and variances.

11. Probability of Critical Path Being Completed in 39 Weeks or Less

12. PERT Example 1 2 4 6 7 3 5 9 8 Manual Testing Dummy System Training
System Testing
Orientation
Position recruiting
System development
Equipment testing and modification
Final debugging
System changeover
Job training
Equipment installation 25

13. Example (Cont.) Time estimates (wks) Mean Time Variance
Activity a b c t s2 26

14. Example (Cont.) Activity t s2 ES EF LS LF S 1 - 2 8 0.44 0 8 1 9 1 27

15. Example (Cont.) ( ) ( ) ( ) ( ) 1 2 4 6 7 3 5 9 8
( )
ES=8, EF=8
LS=9, LF=9
ES=6, EF=9
LS=6, LF=9
( )
ES=3, EF=5
LS=14, LF=16
ES=0, EF=3
LS=2, LF=5
ES=0, EF=6
LS=0, LF=6
ES=3, EF=7
LS=5, LF=9
ES=9, EF=13
LS=12, LF=16
LS=9, LF=16
ES=13, EF=13
LS=16 LF=16
ES=13, EF=25
LS=16 LF=25
ES=8, EF=13
LS=16 LF=21
( )
ES=0, EF=8
LS=1, LF=9
( )
ES=13, EF=25
LS=16 LF=25
s2 = s s s s2 79 = = weeks 28

16. . Determining Probability From Z Value s Z 0.00 0.01 … 0.09
What is the probability that project is completed within 30 weeks?
s2 = weeks, s = = 2.62 weeks
Z = x - m = = 1.91
P(Z < 1.91) =
2.62 s
P( x < 30 weeks) = = .
Z … 0.09 …
m = 25
Time (weeks)
x = 30 33

17. What is the probability that project will be completed within 22 weeks ?
Time (weeks)
x = 22
P( x< 22 weeks) =
P(Z< -1.14) = =
Z =
= -1.14 34

18. PERT Probability Example
A project has an expected completion time of 40 weeks, with a standard deviation of 5 weeks. What is the probability of finishing the project in 50 weeks or less? T
= 40 s
= 5 50 X
Normal Distribution Z = - 40 5 2 . m z
= 0
= 1
2.0
Standardized Normal Distribution 48

19. Obtaining the Probabilitym z
= 0 s Z
= 1
2.0
.00
.01
.02
0.0
.50000
.50399
.50798 :
.97725
.97784
.97831
2.1
.98214
.98257
.98300
Standardized Normal Probability Table (Portion)
Probabilities in body 49