ch 9
pert and cpm
Learning objectives:
1- Give a general description of/techniques.
2- Construct simple net work diagrams.
3- Analyze net works that have deterministic time.
4- Describe activity crashing “and solve simple problem.
Summary
Projects are comprised f a unique set of activities
established to achieve a given set of objectives during a
limited life span .
PERI/CPM are two commonly used techniques for
developing and monitoring projects.
Glossary
Activity
A task
CPM (critical Path method )
Pert/CPM were developed
Separately at about the
same time in the late
1950s. Both techniques
are concerned with
integration of a number of
different tasks.
Crashing
Accelerating a project by
speeding up those critical- path
activities that have lowest ratio
of incremental cost to
incremental time saved .
Critical Path
The path in a network diagram
that takes the longest time for
completion .
Critical Activity
An activity on the critical Path .
PERT program evaluation and Review Technique) See CPM
Pert / CPM
Program Evolution
and Review Technique
Critical path Method
Most projects have Certain elements in common
1. The involve Considerable Cost
2. Usually have a long time horizon
3. Involve a large number of activities that must carefully, planned
and performance guidelines.
Planning and Scheduling with Gantt charts :
Grantt chart is a popular tool for planning and Scheduling simple projects.
This tool enable a manager to initially Schedule project activities and
than monitor progress over time by comparing planned progress to
actual progress.
For example :Consider the case of bank that planned to great a new direct marketing
department .
He developed a list of the major activities that would be requires
Activity
time to accomplish
in week
1. Locate new facility
2. Interview staff
3.Hire + train staff
4. Select and order furniture
5. Remodel and intall phones
6. Receive Furniture and set up
7. Move in/ start up
8
4
9
6
11
3
1
week after
project start
immediate
immediate
4
8
8
14
19
prepare a Gantt chart for this project .
Gantt chart for bank case :
Activity
Start
2
4
6
8
start 2
4
6
8
10
12
14
16
18
20
1
2
3
4
5
6
7
10 12 14 16 18 20
PERT and
CPM
By using pert or cpm , managers are able to obtain :
1. A graphical display of project activities .
2. An estimate of how long the project will take .
3.An indication of which activities are the most critical to
timely completion of the project .
4. An indication of how long any activity can be delayed
without lengthening the project.
Note : pert : stressed probabilistic activity time estimates
because the environment in which it was developed was
typified by high uncertainty .
Cpm: originally made no provision for variable time estimates
The net work diagram
Recall the bank case:
4
2
6
1
5
3
Nodes: represent both beginning and endings of activities.
arrows : represent the project activities that lead from the
starting node to the finishing node.
Path: is a sequence of activities that lead from the starting
node to the finishing node .
from bank
Case digram
1-2–4-5–6
1 – 2- 5 – 5
1–3–5–6
Critical path : The longest path which referred to critical
activities .
Deterministic time Estimates
1
Deterministic time estimates:
Example :Given the information provided in the a companying net
work diagram , determine each of the following
1. The length of each path
2.The critical path
3. The expected length of the project
4. The amount of slack time for each path .
.
4
2
1
3
5
6
Solution :(The start and end “activities” have no time they merely
serve as reference points )
1. The path length are
A . 1- 2- 4- 5- 6
8 + 6 + 3 +1 = 18 weeks.
B . 1 -2 – 5- 6
8 + 11 + 1
= 20 weeks .
C. 1-3–5-6
4+9+1
= 14 weeks .
2 . The longest 20 week (1 -2 -5 -6 )
is the critical path .
3. The expected length of the project is equal to the length of
the critical path 20 weeks .
4. The amount of slack time for each path
path
length ( week)
1-2-4-5-6
8+6+3+1= 18
1- 2- 5 – 6
8+ 11+ 1 = 20
1- 3 – 5 – 6
4 + 9 +1 = 14
slack
20-18 = 2
20-20 = 0
20-14 = 6
Note :-
slack time = length of critical path – path length .
A
Computing ALogrithm
Algorithm is used to develop four pieces of info. About the
net work
Activities :Es, earliest time the activity can start , assuming all preceding
activities start as early as possible
EF , earliest time the activity can finish
.
LS , latest time the activity can start and not delay the
project .
Once these values have been determined , they can be used
to find
1. Expected duration of the project
2. Slack time
3. Which activities are on the critical path .
Computing Es and EF Times
4
2
8
1
3
9
19
5
6
Computation of earliest starting and finished time is aided
by tow simple rules :
1. THE EF is equal to its Es plus it expected duration ,t:
?? EF=Es + t
2. For nodes with one entering arrow :
Es for activities at such nodes is equal to EF of entering
arrow. For nodes with Multiple entering arrows : equals
activities leaving such nodes equals largest EF of
entering arrows .
Compute Es and EF for our example :
Solution :
Assume an Es of o for activities without predecessors.
EF 1-2 = 0 +8=8
and EF 1-3=0+4=4
Likewise for the other activities .
T+Es
These results are Summarized in the table below :
Activity
Duration ,t
Es
EF
1-2
8
0
8
1-3
4
0
4
2-4
6
8
14
2-5
11
8
19
3-5
9
4
13
4-6
3
14
17
5-6
1
19
20
Note : last EF is the project duration , thus ,the expected
length of the project
is 20 week.
ES, EF
forward pass??
Computing LS and LF times
Computation of the Ls +LF times is aided by the use of tow
rules :
1.The L is equal to its LF minus its expected duration :
LS= LF -1
2. For nodes with one arrow :LF for arrows entering that node
equals the LS of the leaving arrows
For nodes with multiple leaving arrows.
LS, LF
back pass?
Compute LF, LS in our examples
Set LF of the last activity equal to the EF of that activity .
Thus
LF 5-6 =EF 5-6 = 20 weeks
LS 5-6 =LF 5-6 -1
= 20-1 -19
For LF 4-5 =LF 2-5 = LF 3 -5 =19
For the rest
LS 4-5 = 19-3 =16
LS 2-5 = 19 – 11= 18
LS 2-5 = 19 – 9 = 10
There are two arrow leaving Node 2 ,
2-4 with LS = 10 , 2-5 Ls = 8
The latest Finish for activity 2-1 thus become 8 which is the smallest Ls for
a leaving arrow the LF for 1-3 is equal to the LS for 3-5 .
LF 1-3 LS 3-5 =10
The LS for activity 1-3 is
LS 1-3 = 10 - 4 =6
The LS for activity 1-2 is LS 1-2 is
LS 1-2 =LF1-2 –t
=8-8=0
The LS , LF computation are Summarized
in the table below:
Activity
5-6
4-5
2-5
3-5
2-4
1-2
1-3
Duration
1
3
11
9
6
8
4
LF
LS
20
19
19
19
16
8
10
19
16
8
10
10
0
6
Computing Activity slack times
Activity slack = LS –ES or LF- ES
LS -ES
Activity
1-2
1-3
2-4
2-5
3-5
4-5
5-6
LS
0
6
10
8
10
16
19
ES
0
0
8
8
4
14
19
slack
0
6
2
0
6
2
0
Are all
Critical
Activities
Probabilistic time estimates
In the preceding example we assumed that activity times
were known and not subject to variation .
For other situations we need to use probabilistic approach
Probabilistic time estimates use three time estimates for
each activity Instead of one :
1- Optimistic time (a) length of time required under optimum
condition .
2- Pessimistic time (b) time under worse conditions
3- Most – likely time (m) most probable amount time .
Beta distribution is commonly , used to describe inherent
Variability in time estimates .
A Beta Distribution
Of special interest in network analysis
Are the Average or expected time
For each activity te and the variance of each activity time.
The expected time is computed as a weighted average
Of three time estimates :
te = a + 4m + b
Standard deviation of each activity
Time is estimated 1/6 of difference between
the pessimistic and optimistic time estimate .
The variance is found by squiring the standard deviation , thus :
b  a
2
6 =
 6

2
or
(b  a ) 2
36
Size of the variance reflects the degree of uncertainty associated with an
activity time
The larger the variance , the greater the uncertainty
It is also desirable to compute the standard deviation of the expected
Time for each path .
6path =
4(variancesofactivitiesonpath
Go to example
Example:
The net work diagram for a shown in the accompanying figure
, with three time estimates for each activity,
Activity times are in month , do the following :
Compute the expected time for each activity and the
.1
expected duration for each path
2. Identify the Critical Path .
3. Compute the variance for each activity and the variance
m
b
a
for each path .
Solution:
1. path
a -b - c
d –e -f
g- h –i
activity
times
amb
te =a+4m=b
6
a
b
c
1 3 4
2 4 6
2 3 5
2.83
4.00
3.17
d
e
f
3 45
3 5 7
5 7 9
4.00
5.00
7.00
path
total
=10.00
=16.00
largest critical path
g
2 3 6
3.33
h
4 6 8
6.00
13.50
i
3 4 6
4.17
2.* critical path is d- h -i which has the largest path total
16.00 weeks .
Which mean the expected duration of the project .
3.Variance :
Path
activity
amb
52 act= (b-a)2
52 path
path
36
a-b – c
D- e-f
g- h- i
a
b
c
1 3 4
2 4 6
2 35
(4-1)2/36=9/36
(6-2)2/36=16/36
(5-2)2/36= 9/36
34/63=.944
.97
d
e
f
34 5
357
5 7 9
(5-3)2/36= 4/36
(7-3)2/36= 16/36
(9-5)2/36= 16/36
36/36=1.00
1.00
g
h
i
236
468
346
(6-2)2/36=16/36
(8-4)2/36=16/36
(6-3)2/36= 9/36
41/36=1.39
1.07
The previous Solution enable a manager to make
probabilistic estimates of the project completion time ,
such as
1. The probability that the project will completed with
months of start ?
2. The probability that the project will take longer than
specific months more than it’s completion
Go to example
The next example illustrates the use of normal
distribution to determine the probabilities for
various completion time .
Before we look at that example , it is important
to make note of two points
1. Related to independence, it is assumed that
path duration times are independent of each
other
2.A project is not completed until all of it’s
activities have been completed (included those
on the critical path).
Example :using the information from the preceding example ,
Answer the following question :
1. Can the paths be considered independent ? Why ?
2. Determine the probability that the project will be
completed within it months of it’s start ?
3.Determine the probability that the project will be
completed with 15 months of it’s start.
4. What is the probability that the project will not be
completed within 15 months of it’s start ?
Solution :-
1. Yes , the paths can be considered Independent because
no activity is on information than on path , and we have
information that would suggest that any activity times
are interrelated .
2. In order to answer Q’s of this nature , we must take into
account the degree to witch the path distribution the
specified completion time. See the figures
path
17
months
100%
A–b–c
10.0
months
D–e–f
16.0
months
99.95
q–h–i
13.5
months
The shaded portion of each distribution corresponds to the probability
the part will be completed within the specified time , observe that
paths:
A -b - c and g -h -i are far enough to the left of the
specified time that is highly likely that both will finished
by month 17 but that critical path overlaps the specified
completion time.
Hence , we need only consider the distribution of path
d- e - f in assessing the probability of completion by month
17.to do so, we must first compute the value of Z using
the relationship :
Z=
specified time – Expected time
path standard deviation
In our example :expected time for path d – e – f
is 16.0 ,
Z = 17-16 =+1.00
1.00
From table B with Z = +1.00
Normal distribution
.8413
Probability 84% to finish the project within 17 months
3.The question illustrates how to handle a problem in which more
than one of the distribution over laps the specified time
See below?
15 months
100%
A–b–c
10.0
months
.1587
D–e–f
16.0
months
15 months
.9192
Q–h–i
13.5
months
Note that d –e – f and g – h –i paths overlaps month 15.
This means that both paths have the potential for delaying the
project beyond 15 month.
Back to Z equation
Path Z= 15-expected path duration
from normal
path standard deviation
distribution table,
B1,
P(Z)
A –b - c 15-10 = +5.15
1.00
.97
D –e - f 15-16 =-1.00
.1587
1.00
G –h - i 15-13.5 =+1.40
.9192
1.07
Note :Any path with a Z of more than +2.5 is assigned a
probability of 1.00 .
The joint probability of finishing before month 15 is the
product of these probabilities
(1.00) ( .1587) (.9192) = .1459 ??
Time – Cost Trade – OFFs – Crashing
The desire to shorten the length of a project merely reflects o
attempt to reduce the indirect cost associated with running
the project .
Such as : Facilities of equipment cost , supervision and labor
and personal costs.
To avoid late penalties.
To crashing desirable , a manager needs the following info
1. Regular time and crash time estimates for each activity .
2. Regular cost and crash cost estimates for each activity
3. A list of activities that are on the critical path .
The general procedure for crashing is:
1. Obtain estimates of regular and crash time and costs
for each activity .
2.Determine the lengths of all paths and path slack times
3. Determine which activities are on the critical path .
4. Crash critical activities , in order of increasing costs do
not exceed benefits .
Example :Using the info . Below , develop an optimum time- cost
solution , Assume that indirect project costs are $1000per
day .
Cost per day
Activity
Normal time
crash time
to crash
1-2
6
6
2-5
10
8
$500
1-3
5
4
300
3-4
4
1
700
4-5
9
7
600
5-6
2
1
800
Solution:2
6
Start
10
2
1
5
5
9
4
3
6
4
1.Determine which activities are on the critical path it’s
length , and the length of the other path :
path
length
1-2-5-6
18
1-3-4-5-6
20
critical path
End
2. Rank the critical path activities in order of lost wet
crashing cost and determine the umber of days each can
be crashed
cost per day
available
activity
to crash
days
1-3
$300
1
4-5
600
2
3-4
700
3
5-6
800
1
3.Begin shorting the project , one day at a time , and chek
after each reducing to see which path is critical .( After
a certain point, another path may equal the length of the
shortened critical path ) thus.
4.A shorten Activity 1-3 one day at cost of $ 300, the length
of the critical path now become 19 days.
b. Activity 1-3 cannot be shortened any more shorten
activity 4-5 one day of a cost of $600the length of path
1-3-4-5-6 now become 18 days which is the same as the
length of path 1-2-5-6
c. Since the paths are now both critical , further
improvements will necessitate shorting one activity on
each.
The remaining points for crashing and their costs one:
Path
1-2-5-6
activity
1-2
2-5
5-6
1-3-4-5-6
crash cost per day
no reduction possible
$500
800
1-3
3-4
4-5
5-6
no further reduction possible
$700
600
800
Activity 5-6 would not be advantageous because it has highest crashing
cost . How ever activity 5-6 is on paths and he project ?
* The option of shortening the least expensive activity on each path
would cost $500 for 2-5 and $600 for 4-5 *
Thus , shorten Activity 5-6 by one day the project duration is now 17
days .
d. At this point , no additional improvement is feasible . The cost to crash
Activity 4-5 is $600 for a total of $1.100 and that would exceed the
project costs of $1000 per day .
E. The crashing sequence is Summarized below
length after
Path
crashing n days
n=0
1
2
3
1-2-5-6
18
18
18
17
1-3-4-5-6
20
19
18
17
Activity crashed
1-3
4-5 5-6
Cost
$300
$600
$800