Probabilistic Time Estimates
• Optimistic time
– Time required under optimal conditions
• Pessimistic time
– Time required under worst conditions
• Most likely time
– Most probable length of time that will be
required
Variability in Activity Times
 CPM assumes we know a fixed time
estimate for each activity and there is no
variability in activity times
 PERT uses a probability distribution for
activity times to allow for variability
9B-3
Variability in Activity Times
 Three time estimates are required
– Optimistic time (a) – if everything goes
according to plan
– Most–likely time (m) – most realistic estimate
– Pessimistic time (b) – assuming very
unfavorable conditions
9B-4
Probabilistic Time Estimates
• Beta distribution
– A probability distribution traditionally used in PERT
Mean (expected time):
Variance:
a + 4m + b
6
t=
 =
2
b-a
6
2
where
a = optimistic estimate
m = most likely time estimate
b = pessimistic time estimate
9B-5
Why Use the Beta Distribution?
• Mean and variance can be estimated with 3
time estimates
• Continuous, but no predetermined shape
• Tradition
P(time)
P(time)
Examples of Beta Distributions
m
t
b
a
t
Time
Time
P(time)
a
a
m=t
Time
b
m
b
Project Network with Probabilistic Time
Estimates: Example
Lay foundation
Build house
4
2
7,8,12
Start
Finish work
11,12,20
7
1
3,4,7
9,12,15
Design house
and obtain
financing
3
3,4,5
6
5
2,4,5
Order and receive
Select paint
materials
2,4,6
Select carpet
9B-8
Activity Time Estimates
TIME ESTIMATES (WKS)
ACTIVITY
1
2
3
4
5
6
7
MEAN TIME
VARIANCE
a
m
b
t
б2
9
7
3
11
2
2
3
12
8
4
12
4
4
4
15
12
5
20
5
6
7
12
8.5
4
13.17
3.83
4
4.33
1.00
0.69
0.11
2.25
0.25
0.44
0.44
9B-9
Activity Early, Late Times,
and Slack
ACTIVITY
1
2
3
4
5
6
7
t
12
8.5
4
13.17
3.83
4
4.33
б
1.00
0.69
0.11
2.25
0.25
0.44
0.44
ES
EF
LS
LF
S
0
12
12
20.5
20.5
24.3
33.7
12
20.5
16
33.7
24.3
28.3
38
0
12
16.5
20.5
25.8
29.7
33.7
12
20.5
20.5
33.7
29.7
33.7
38
0
0
4.5
0
5.4
5.4
0
Earliest, Latest, and Slack
Critical Path
Lay foundation
Build house
Start
1
0
12
3
0
12
Design house
and obtain
financing
3
1
12
2
12
2
12
20.5
20.5
4
20.5 33.7
3
20.5 33.7
6
24.3 28.3
1
29.7 33.7
7
33.7 38
1
33.7 38
Finish work
16
16.5 20.5
Order and receive
materials
5
20.5 24.3
1
25.8 29.7
Select carpet
Select paint
9B-11
Total project variance
2
= б1 2 + б2 2 + б4 2 + б7 2
2
= 1.00 + 0.694 + 2.25 + 0.444
= 4.389 weeks
9B-12
Probabilistic Network Analysis
Determine probability that project is
completed within specified time
Z=
where
x-

 = tp = project mean time
 = project standard deviation
x = proposed project time
Z = number of standard deviations x
is from mean
9B-13
Key Assumptions
• Activity times are statistically independent
– Allows summation of individual expected activity
times and variances to get the overall projects
expected time and variance
• Network mean and variance normally
distributed
9B-14
Normal Distribution Of Project Time
Probability
Z
 = tp
x
Time
9B-15
House Example – 40 Weeks
What is the probability that the project is completed within 40 weeks?

P(x  40 weeks)
 = 38 x = 40
2
= 4.389 weeks

=

= 2.095 weeks
4.389
Z =
x-

= 40 - 38
2.095
= 0.9547
Time (weeks)
From Table A.1, (appendix A) a Z score of 0.9547 corresponds to a probability of
~0.3289. Thus P(40) = 0.3289 + 0.5000 = 0.8289
9B-16
Probabilistic Estimates
Beta Distribution
to
Activity
start
Optimistic
time
tm
te
Most likely
time (mode)
tp
Pessimistic
time
Expected Time
te
t
+
4t
+t
o
m
p
=
6
te = expected time
to = optimistic time
tm = most likely time
tp = pessimistic time
Variance

2
(t
–
t
)
= p o
36
 = variance
to = optimistic time
tp = pessimistic time
Example 5
Optimistic
time
Most likely
time
Pessimistic
time
2-4-6
b
3-4-5
d
3-5-7
e
4-6-8
h
5-7-9
f
Example 5 Time Estimates
Tabc = 10.0
Tdef = 16.0
Tghi = 13.50
4.00
b
4.00
d
5.0
e
6.0
h
7.0
f
Path Probabilities
Z =
Specified time – Path mean
Path standard deviation
Z indicates how many standard deviations
of the path distribution the specified tine
is beyond the expected path duration.
Example 6
17
Weeks
1.00
a-b-c
Weeks
10.0
d-e-f
16.0
Weeks
1.00
g-h-i
13.5
Weeks
Determining the Project Schedule
Perform a Critical Path Analysis
Earliest start (ES) = earliest time at which an activity can start,
Activity
Description
Time (weeks)
assuming all predecessors have been
A
Build internal
components
2
completed
B finishModify
roof andtime
floorat which an activity can be3
Earliest
(EF) = earliest
finished
C
Construct
collection stack
2
D startPour
and at
install
frame
Latest
(LS) =concrete
latest time
which
an activity can start4so as
to not delay the completion
time of the4
E
Build high-temperature
burner
entire project
F
Install pollution
control system
3
Latest
(LF) = air
latest
time by
which an activity has to be
G finishInstall
pollution
device
5
finished so as to not delay the completion
H
Inspect and test
2
time of the entire project
Total Time (weeks)
25
Table 3.2
Forward Pass
Begin at starting event and work forward
Earliest Start Time Rule:


If an activity has only a single immediate predecessor, its ES
equals the EF of the predecessor
If an activity has multiple immediate predecessors, its ES is the
maximum of all the EF values of its predecessors
ES = Max {EF of all immediate predecessors}
Forward Pass
Begin at starting event and work forward
Earliest Finish Time Rule:

The earliest finish time (EF) of an activity is the sum of its
earliest start time (ES) and its activity time
EF = ES + Activity time
Backward Pass
Begin with the last event and work backwards
Latest Finish Time Rule:


If an activity is an immediate predecessor for just a single activity,
its LF equals the LS of the activity that immediately follows it
If an activity is an immediate predecessor to more than one activity,
its LF is the minimum of all LS values of all activities that
immediately follow it
LF = Min {LS of all immediate following activities}
Backward Pass
Begin with the last event and work backwards
Latest Start Time Rule:

The latest start time (LS) of an activity is the difference of its latest
finish time (LF) and its activity time
LS = LF – Activity time
Computing Slack Time
After computing the ES, EF, LS, and LF times for all
activities, compute the slack or free time for each
activity

Slack is the length of time an activity can be delayed without
delaying the entire project
Slack = LS – ES
or
Slack = LF – EF
Computing Slack Time
Activity
A
B
C
D
E
F
G
H
Earliest
Start
ES
0
0
2
3
4
4
8
13
Earliest
Finish
EF
Latest
Start
LS
Latest
Finish
LF
Slack
LS – ES
2
3
4
7
8
7
13
15
0
1
2
4
4
10
8
13
2
4
4
8
8
13
13
15
0
1
0
1
0
6
0
0
On
Critical
Path
Yes
No
Yes
No
Yes
No
Yes
Yes
Variability in Activity Times
 CPM assumes we know a fixed time
estimate for each activity and there is no
variability in activity times
 PERT uses a probability distribution for
activity times to allow for variability
Variability in Activity Times
 Three time estimates are required
 Optimistic time (a) – if everything goes
according to plan
 Pessimistic time (b) – assuming very
unfavorable conditions
 Most likely time (m) – most realistic estimate
Variability in Activity Times
Estimate follows beta distribution
Expected time:
t = (a + 4m + b)/6
Variance of times:
v = [(b – a)/6]2
Variability in Activity Times
Probability
Estimate follows beta distribution
Expected time:
t = (a + 4m + b)/6
Probability
of
1
Variance
of
times:
in 100 of
Probability of 1
< a occurring
100 of > b
v = [(b − a)/6]2 inoccurring
Activity
Time
Optimistic Time
(a)
Most Likely Time
(m)
Pessimistic Time
(b)
Computing Variance
Optimistic
Most
Likely
Pessimistic
Expected
Time
Variance
Activity
a
m
b
t = (a + 4m + b)/6
[(b – a)/6]2
A
B
C
D
E
F
G
H
1
2
1
2
1
1
3
1
2
3
2
4
4
2
4
2
3
4
3
6
7
9
11
3
2
3
2
4
4
3
5
2
.11
.11
.11
.44
1.00
1.78
1.78
.11
Probability of Project Completion
Project variance is computed by
summing the variances of critical
activities
p2 = Project variance
= (variances of activities
on critical path)
Probability of Project Completion
Project variance is computed by
summing the variances of critical
Project variance
activities
2p = .11 + .11 + 1.00 + 1.78 + .11 = 3.11
Project standard deviation
p = Project variance
= 3.11 = 1.76 weeks
Probability of Project Completion
PERT makes two more assumptions:
 Total project completion times follow a
normal probability distribution
 Activity times are statistically independent
Probability of Project Completion
Standard deviation = 1.76 weeks
15 Weeks
(Expected Completion Time)
Probability of Project Completion
What is the probability this project can be
completed on or before the 16 week
deadline?
Z=
due
date
–
expected date
of completion
/p
= (16 wks – 15 wks)/1.76
= 0.57
Where Z is the number of standard deviations
the due date or target date lies from the
mean or expected date
Probability of Project Completion
From Appendix
What is the probability
this project.07can be.08
.00
.01
completed.1on .50000
or before
the 16 week
.50399
.52790 .53188
deadline? .2 .53983 .54380
.56749 .57142
due
.5
.6
expected date
Z .69146
= date .69497
−
of completion
.71566 /
.71904
p
.72575
.72907
.74857
= (16 wks − 15 wks)/1.76
= 0.57
.75175
Where Z is the number of standard deviations
the due date or target date lies from the
mean or expected date
Probability of Project Completion
Probability
(T ≤ 16 weeks)
is 71.57%
0.57 Standard deviations
15
Weeks
16
Weeks
Time
Determining Project Completion
Time
Probability of 0.99
Probability of 0.01
From Appendix I
0
2.33 Standard
deviations
Z
2.33
Variability of Completion Time
for Noncritical Paths
 Variability of times for activities on
noncritical paths must be considered
when finding the probability of finishing
in a specified time
 Variation in noncritical activity may cause
change in critical path
What Project Management Has
Provided So Far
 The project’s expected completion time is 15
weeks
 There is a 71.57% chance the equipment will be
in place by the 16 week deadline
 Five activities (A, C, E, G, and H) are on the
critical path
 Three activities (B, D, F) are not on the critical
path and have slack time
 A detailed schedule is available
Project Planning, Controlling
and Scheduling
Project Planning:
1.
2.
3.
4.
Setting goals.
Defining the project.
Tying needs into timed project activities.
Organizing the team.
Project Scheduling:
1. Tying resources to specific activities.
2. Relating activities to each other.
3. Updating and revising on regular basis.
Before Project
During Project
Project Controlling:
1. Monitoring resources, costs, quality and
budgets.
2. Revising and changing plans.
3. Shifting resources to meet demands.
Six Steps of PERT/CPM
1. Define the project and all of its significant activities or tasks
2. Develop the relationships among the activities and decide which
activities must precede others
3. Draw the network connecting all of the activities
4. Assign time and/or cost estimates to each activity
5. Compute the longest time path through the network; this is called the
critical path
6. Use the network to help plan, schedule, monitor, and control the
project

The critical path is important since any delay in these activities can delay the
completion of the project
PERT/CPM
• Given the large number of tasks in a project, it is easy
to see why the following questions are important
1. When will the entire project be completed?
2. What are the critical activities or tasks in the project,
that is, the ones that will delay the entire project if
they are late?
3. Which are the non-critical activities, that is, the ones
that can run late without delaying the entire project’s
completion?
4. If there are three time estimates, what is the
probability that the project will be completed by a
specific date?
PERT/CPM
5. At any particular date, is the project on schedule,
behind schedule, or ahead of schedule?
6. On any given date, is the money spent equal to, less
than, or greater than the budgeted amount?
7. Are there enough resources available to finish the
project on time?
8. If the project is to be finished in a shorter amount of
time, what is the best way to accomplish this at the
least cost?
General Foundry Example of PERT/CPM
• General Foundry, Inc. has long been trying to avoid the
expense of installing air pollution control equipment
• The local environmental protection group has recently
given the foundry 16 weeks to install a complex air filter
system on its main smokestack
• General Foundry was warned that it will be forced to close
unless the device is installed in the allotted period
• They want to make sure that installation of the filtering
system progresses smoothly and on time
General Foundry Example of PERT/CPM
• Activities and immediate predecessors for General
Foundry
ACTIVITY
DESCRIPTION
IMMEDIATE
PREDECESSORS
A
Build internal components
—
B
Modify roof and floor
—
C
Construct collection stack
A
D
Pour concrete and install frame
B
E
Build high-temperature burner
C
F
Install control system
C
G
Install air pollution device
D, E
H
Inspect and test
F, G
Drawing the PERT/CPM Network
• There are two common techniques for drawing PERT networks
• Activity-on-node (AON) where the nodes represent activities
• Activity-on-arc (AOA) where the arcs are used to represent the
activities
• The AON approach is easier and more commonly found in software
packages
• One node represents the start of the project, one node for the end of
the project, and nodes for each of the activities
• The arcs are used to show the predecessors for each activity
General Foundry Example of PERT/CPM
• Network for General Foundry
A
C
F
Build Internal
Components
Construct
Collection Stack
Install Control
System
Start
E
H
Build Burner
Inspect
and Test
B
D
G
Modify Roof
and Floor
Pour Concrete
and Install Frame
Install Pollution
Device
Finish
Activity Times
• In some situations, activity times are known with certainty
• CPM assigns just one time estimate to each activity and this
is used to find the critical path
• In many projects there is uncertainty about activity times
• PERT employs a probability distribution based on three time
estimates for each activity
• A weighted average of these estimates is used for the time
estimate and this is used to determine the critical path
Activity Times
• The time estimates in PERT are
Optimistic time (a) = time an activity will take if everything goes
as well as possible. There should be only a
small probability (say, 1/100) of this
occurring.
Pessimistic time (b) = time an activity would take assuming very
unfavorable conditions. There should also
be only a small probability that the activity
will really take this long.
Most likely time (m) = most realistic time estimate to complete
the activity
Activity Times
• PERT often assumes time estimates follow a beta probability distribution
– The beta probability distribution is often used when there is no solid historical data
upon which to activity time base estimates
– Found to be appropriate in many cases for determining an expected value and
variance for activity completion times
Probability
Probability of 1 in 100 of
a Occurring
Probability of 1 in 100 of
b Occurring
Most
Optimistic
Time
Most
Likely
Time
(a)
(m)
Most
Pessimistic
Time
(b)
Activity Time
Activity Times
• To find the expected activity time (t), the beta distribution
weights the estimates as follows
a  4m  b
t=
6

To compute the dispersion or variance of activity completion time, we use the
formula
ba
Variance = 

 6 
2
Activity Times
• Time estimates (weeks) for General Foundry
ACTIVITY
OPTIMISTIC,
a
MOST
PROBABLE,
m
PESSIMISTIC,
b
A
1
2
3
2
4/36
B
2
3
4
3
4/36
C
1
2
3
2
4/36
D
2
4
6
4
16/36
E
1
4
7
4
36/36
F
1
2
9
3
64/36
G
3
4
11
5
64/36
H
1
2
3
2
4/36
EXPECTED
TIME,
t = [(a + 4m + b)/6]
25
VARIANCE,
[(b – a)/6]2
How to Find the Critical Path
• We accept the expected completion time for each
task as the actual time for now
• The total of 25 weeks in Table does not take into
account the obvious fact that some of the tasks
could be taking place at the same time
• To find out how long the project will take we
perform the critical path analysis for the network
• The critical path is the longest path through the
network
How to Find the Critical Path
• General Foundry’s network with expected activity times
A
2
C
2
F
E
Start
B
3
D
4
3
4
H
G
5
2
Finish
How to Find the Critical Path
•
1.
2.
3.
4.
To find the critical path, need to determine the following
quantities for each activity in the network
Earliest start time (ES): the earliest time an activity can begin
without violation of immediate predecessor requirements
Earliest finish time (EF): the earliest time at which an activity
can end
Latest start time (LS): the latest time an activity can begin
without delaying the entire project
Latest finish time (LF): the latest time an activity can end
without delaying the entire project
How to Find the Critical Path
• In the nodes, the activity time and the early and late start
and finish times are represented in the following manner
ACTIVITY
ES
LS

t
EF
LF
Earliest times are computed as
Earliest finish time = Earliest start time
+ Expected activity time
EF = ES + t
Earliest start = Largest of the earliest finish times of
immediate predecessors
ES = Largest EF of immediate predecessors
How to Find the Critical Path
• At the start of the project we set the time to zero
• Thus ES = 0 for both A and B
A
ES = 0
t =2
EF = 0 + 2 = 2
B
ES = 0
t =3
EF = 0 + 3 = 3
Start
How to Find the Critical Path
• General Foundry’s ES and EF times
A
0
2
2
C
2
2
4
F
4
E
4
Start
B
0
3
3
D
3
4
7
3
7
4
8
H
13
G
8
5
13
2
15
Finish
How to Find the Critical Path
• Latest times are computed as
Latest start time = Latest finish time
– Expected activity time
LS = LF – t
Latest finish time = Smallest of latest start times
for following activities
LF = Smallest LS of following activities

For activity H
LS = LF – t = 15 – 2 = 13 weeks
How to Find the Critical Path
• General Foundry’s LS and LF times
A
0
0
2
2
2
C
2
2
2
4
4
F
4
10
E
4
4
Start
B
0
1
3
3
4
D
3
4
4
7
8
3
7
13
4
8
8
H
13
13
G
8
8
5
13
13
2
15
15
Finish
How to Find the Critical Path
• Once ES, LS, EF, and LF have been determined, it is a simple
matter to find the amount of slack time that each activity
has
Slack = LS – ES, or Slack = LF – EF
• From Table we see activities A, C, E, G, and H have no slack
time
• These are called critical activities and they are said to be on
the critical path
• The total project completion time is 15 weeks
• Industrial managers call this a boundary timetable
How to Find the Critical Path
• General Foundry’s schedule and slack times
ACTIVITY
EARLIEST
START,
ES
EARLIEST
FINISH,
EF
LATEST
START,
LS
LATEST
FINISH,
LF
A
0
2
0
2
0
Yes
B
0
3
1
4
1
No
C
2
4
2
4
0
Yes
D
3
7
4
8
1
No
E
4
8
4
8
0
Yes
F
4
7
10
13
6
No
G
8
13
8
13
0
Yes
H
13
15
13
15
0
Yes
SLACK,
LS – ES
ON
CRITICAL
PATH?
How to Find the Critical Path
• General Foundry’s critical path
A
0
0
2
2
2
C
2
2
2
4
4
F
4
10
E
4
4
Start
B
0
1
3
3
4
D
3
4
4
7
8
3
7
13
4
8
8
H
13
13
G
8
8
5
13
13
2
15
15
Finish
Probability of Project Completion
• The critical path analysis helped determine the expected
project completion time of 15 weeks
• But variation in activities on the critical path can affect
overall project completion, and this is a major concern
• If the project is not complete in 16 weeks, the foundry will
have to close
• PERT uses the variance of critical path activities to help
determine the variance of the overall project
Project variance =
∑
variances of activities on the
critical path
Probability of Project Completion
• From Table we know that

ACTIVITY
VARIANCE
A
4/36
B
4/36
C
4/36
D
16/36
E
36/36
F
64/36
G
64/36
H
4/36
Hence, the project variance is
Project variance = 4/36 + 4/36 + 36/36 + 64/36 + 4/36 = 112/36 = 3.111
Probability of Project Completion
• We know the standard deviation is just the square root of
the variance, so
Project standard deviation =  T = Project variance
= 3.11 = 1.76 weeks

We assume activity times are independent and total project completion time
is normally distributed
Probability of Project Completion
• The project’s expected completion date is 15 weeks.
– Assume that the total project completion time follows a normal
probability distribution
– Chart tells us that there is a 50% chance of completing the entire project
in less than 15 weeks and a 50% chance it will exceed 15 weeks
Standard Deviation = 1.76 Weeks
15 Weeks
(Expected Completion Time)
Probability of Project Completion
• The standard normal equation can be applied as follows
Z=
Due date  Expected date of completion
T
16 weeks  15 weeks
=
= 0.57
1.76 weeks

From Appendix we find the probability of 0.71566 associated with this Z value
 That means there is a 71.6% probability this project can be completed in 16
weeks or less
Probability of Project Completion
• Probability of General Foundry meeting the 16-week
deadline
Expected Time is 15 Weeks
0.57 Standard Deviations
Probability
(T ≤ 16 Weeks)
is 71.6%
15
Weeks
16
Weeks
Time
Probability of Project Completion
From Appendix
What is the probability
this project.07can be.08
.00
.01
completed.1on .50000
or before
the 16 week
.50399
.52790 .53188
deadline? .2 .53983 .54380
.56749 .57142
due
.5
.6
expected date
Z .69146
= date .69497
−
of completion
.71566 /
.71904
p
.72575
.72907
.74857
= (16 wks − 15 wks)/1.76
= 0.57
.75175
Where Z is the number of standard deviations
the due date or target date lies from the
mean or expected date