Lecture 2 PERT-CPM.pdf

Maruf Hossain Hredoy
BSc in CE,KUET
Lecturer
Department of Civil Engineering

WHAT IS A PROJECT?
A project is an assignment/task/job that has to be undertaken
and completed within a set time, budget, resources and
performance specifications designed to meet the needs of
stakeholder and beneficiaries
TECHNICAL ASPECTS TO PROJECT MANAGEMENT

TECHNICAL ASPECTS TO PROJECT MANAGEMENT
For example
The Canadian International Donor Agency (CIDA) has donated K7.5 million to
provide 3 bedrooms homes to 50 families living in the Kamanga informal
settlement. On 6 February 2004, the agency signed a contract with the
Department of Housing to implement the project. The following requirements,
amongst others were set in the contract:
➢ The 3 bedroomed houses must meet specifications in line with
government policy.
➢ In order to ensure sustainability and affordability for the 50 families, the
head of each of the
➢ 50 families must be given skills development training in small business
development and small business start-up. This is to ensure that the
families will be able to afford rentals, maintenance of the homes and to
expand their homes to accommodate the growth of the families in the
future.
➢ The project must be completed within three years and the handover of
the homes tothe 50 families must be a high profile public event.

From the example we see:
▪ a clear task - build 3 bedroomed homes for 50
families;
▪ a set time – within 3 years;
▪ a budget – K7.5 million;
▪ performance specifications to meet the
stakeholder needs – houses that meet the
specifications in line with government policy,
training for the head of each family;
▪ beneficiaries – 50 families;
▪ stakeholders – donor agency, Department of
Housing

Project stakeholders
▪ Project stakeholders are individuals and organisations who
are actively involved in the project, or whose interests may be
positively or negatively affected by the project.
Key stakeholders in the example above include:
➢ Project Manager - the individual responsible for managing the
project;
➢ Project beneficiaries – 50 families who are going to receive the
houses;
➢ Performing Organization – the Department of Housing
whose employees are most directly involved in doing the work
of the project;
➢ Sponsor – Canadian International Development Agency.

In addition to these there are many different typical stakeholders:
➢ Suppliers and contractors – e.g. Construction companies,
Skills development and education and training organisations,
legal firms, events management company;
➢ Project team members and their families;
➢ Government agencies – e.g. local municipality.
➢ Community representatives and organizations
Project stakeholders

The project life cycle typically passes through four stages, i.e. definition,
planning, execution, and delivery.
Definition stage – specifications of the project are defined, project
objectives are established, project teams are formed and major
responsibilities are assigned.
Planning stage – plans are developed to determine the project steps,
beneficiaries, timeframes, quality standards and budget.
Execution stage – the major portion of the project work takes place – both
physical and mental. Time, cost and specification measures are used for
control. The project managers have to ensure that the project is on
schedule within the budget and meeting specifications. They have to also
check if any changes are required.
Delivery stage – delivering the project product to the customer, may
involve customer training and transferring documents.
Project Life Cycle

the work of the project can be subdivided into smaller work
elements. The outcome of this process is called the work
breakdown structure.
work breakdown structure (WBS)
All the elements/steps that make up WBS are called work
packages. It is very useful as the structure clearly points to
what has to be done and in what sequence (order). It divides
the work and responsibility into individual work packages
which makes it easy for the project manager to manage and
monitor the implementation

work breakdown structure (WBS)

Project Bar Chart (Gantt Chart)

A convenient analytical and visual technique for managing the
projects :
PERT and CPM
PERT stands for Project Evaluation And Review Technique
CPM stands for Critical Path Method which was developed by.
PERT & CPM

CPM stands for Critical Path Method which was developed by
DuPont Company and applied first to the construction projects in the chemical
industry.
Though both PERT and CPM techniques have similarity in terms of
concepts, the basic difference is; CPM has single time estimate and PERT
has three time estimates for activities and uses probability theory to find
the chance of reaching the scheduled time.
Difference Between PERT & CPM

COMPONENTS of PERT/CPM NETWORK
PERT / CPM networks contain two majorcomponents
i. Activities, and
ii. Events
Activity: An activity represents an action and consumption of
resources (time, money, energy) required to complete a portion of a
project. Activity is represented byan arrow, .
Event: An event (or node) will always occur at the beginning and end of an
activity. The event has no resourcesand is represented byacircle. The ith event
and jth event are the tail event and head event respectively.

Merge and Burst Events
Oneor moreactivities can startand end simultaneouslyat an event .
Preceding and Succeeding Activities
Activities performed before given events are known as preceding activities, and
activities performed aftera given event are known as succeeding activities.
Activities A and B precedeactivities C and D respectively
.

Dummy Activity
An imaginary activity which does not consume any resource and time is called a
dummy activity. Dummy activities are simply used to represent a connection
between events in order to maintain a logic in the network. It is represented by a
dotted line in a network.

RULES IN CONSTRUCTING A NETWORK
1. No single activity can be represented more than once in a network.
The lengthof an arrow has nosignificance.
2.The event numbered 1 is the start event and an event with highest
number is the end event. Before an activity can be undertaken, all
activities preceding it must be completed. That is, the activities must
followa logical sequence (or – interrelationship) betweenactivities.
3.In assigning numbers to events, there should not be any duplication of
event numbers ina network.
4.Dummy activities must be used only if it is necessary to reduce the
complexityof a network.
5. A network should haveonlyonestarteventand oneend event.

Some conventions of network diagram are shown in Figure (a),
(b), (c), (d) below:

PROCEDURE FOR NUMBERING THE EVENTS
USING FULKERSON'S RULE
Step1: Numberthe startor initial eventas 1.
Step2: From event 1, strike off all outgoing activities. This would have
made one or more events as initial events (event which do not have
incoming activities). Numberthateventas 2.
Step3: Repeat step 2 for event 2, event 3 and till the end event. The end
event must have the highest number
Example 1:
Draw a network for a house construction project. The sequence of
activities with theirpredecessorsaregiven in Table , below.

The network diagram in Figure shows the procedure relationship between
the activities. Activity A (preparation of house plan), has a start event 1
as well as an ending event 2. Activity B (Construction of house) begins at
event 2 and ends at event 3. The activity B cannot start until activity A
has been completed. Activities C and D cannot begin until activity B
has been completed, but they can be performed simultaneously.
Similarly, activities E and F can start only after completion of
activities C and D respectively. Both activities E and F finish at the end of
event 6.
Example 2: Consider the project given in Table 8.2 and construct a
network diagram. Table : Sequenceof Activities for Building
Construction Project

Solution:
The activities C and D have a common predecessor A. The network
representation shown in Figure (a), (b) violates the rule that no two activities can
begin and end at the same events. It appears as if activity B is a predecessor of
activity C, which is not the case. To construct the network in a logical order, it is
necessaryto introduce adummy activityas shown in Figure .

Example 3:
Construct a network for a project whose activities and their predecessor relationship
aregiven in Table .
Solution: The network diagram for the given problem is shown in Figure
with activities A, B and C starting simultaneously.

Example 4: Draw a network diagram for
a project given in Table .
Solution: An activity network diagram describing the project is shown in
Figure , below:

CRITICAL PATH ANALYSIS
Thecritical path forany network is the longestpath through theentire network.
Since all activities must be completed to complete the entire project, the length
of the critical path is also the shortest time allowable for completion of the
project.
Thus if the project is to be completed in that shortest time, all activities on the
critical path must bestarted assoon as possible.
Theseactivitiesarecalled critical activities.
If the project has to be completed ahead of the schedule, then the time required
forat leastoneof thecritical activity must be reduced.
Further, any delay in completing the critical activities will increase the project
duration.

The activity, which does not lie on the critical path, is called non-critical
activity.
These non-critical activities may havesome slack time.
The slack is the amount of time by which the start of an activity may be
delayed withoutaffecting theoverall completion timeof theproject.
Butacritical activity has noslack.
To reduce the overall project time, it would require more resources (at
extracost) toreduce the time taken by thecritical activities tocomplete.

Scheduling of Activities: Earliest Time (TE) and Latest
Time(TL)
Before the critical path in a network is determined, it is necessary to
find the earliest and latest time of each event to know the earliest
expected time (TE) at which the activities originating from the event
can be started and to know the latest allowable time (TL) at which
activities terminating at theeventcan becompleted.
Forward Pass Computations (tocalculate Earliest, TimeTE)
Step 1: Begin from the startevent and move towards theend event.
Step 2: Put TE = 0 for the startevent.
Step 3: Go to the next event (i.e node 2) if there is an incoming activity for event 2,
add calculateTE of previous event (i.e event 1) and activity time.
Note: If there are more than one incoming activities, calculate TE for all incoming
activities and take the maximum value. Thisvalue is the TE forevent 2.
Step 4: Repeat the same procedure from step 3 till theend event.

Backward Pass Computations (to calculate Latest Time TL)
Procedure :
Step 1: Begin from end event and move towards the start event. Assume that
thedirectionof arrows is reversed.
Step 2: Latest Time TL for the last event is the earliest time. TE of the last
event.
Step 3: Go to the next event, if there is an incoming activity, subtract the value
of TL of previous event from the activity duration time. The arrived value is
TL for that event. If there are more than one incoming activities, take the
minimum TE value.
Step 4: Repeatthesame procedurefrom step 2 till thestartevent.

DETERMINATION OF FLOAT AND SLACK TIMES
As discussed earlier, the non – critical activities have some slack or float.
The float of an activity is the amount of time available by which it is
possible to delay its completion time without extending the overall
projectcompletion time.
tij = duration of activity
TE = earliestexpected time
TL = latestallowable time
ESij = earliest start time of the activity
EFij = earliest finish time of theactivity
LSij = latest start time of the activity
LFij = latest finish time of theactivity
Total Float TFij: The total float of an activity is thedifference between the
lateststart time and theearlieststart time of thatactivity.
TFij = LS ij – ESij ....................(1)
or
TFij = (TL – TE) – tij …………..(ii)

Free Float FFij: The time by which the completion of an activity can be
delayed from itsearliestfinish timewithoutaffecting theearlieststart
timeof thesucceeding activity iscalled free float.
FF ij = (Ej – Ei) – tij ....................(3)
FFij = Total float – Head event slack
Independent Float IFij: The amount of time by which the start of an activity
can be delayed without affecting the earliest start time of any immediately
following activities, assuming that the preceding activity has finished at its
latest finish time.
IF ij = (Ej – Li) – tij ....................(4)
IFij = Free float – Tail event slack Wheretail event slack = Li – Ei
The negativevalueof independentfloat isconsidered to bezero.

Critical Path:
After determining the earliest and the latest scheduled times for
various activities, the minimum time required to complete the
project is calculated. In a network, among various paths, the
longest path which determines the total time duration of the
project is called the critical path. The following conditions must
besatisfied in locating thecritical path of a network.
Anactivity is said to becritical only if both theconditionsaresatisfied.
1. TL – TE = 0
2. TLj – tij – TEj = 0
Example :
A projectschedule has the following characteristicsas shown in Table
i. Construct PERT network.
ii. Compute TE and TL
for each activity.
iii. Find thecritical path.

(i) From thedatagiven in the problem, theactivity network is constructed as
shown in Figure given below

(ii) To determine the critical path, compute the earliest time TE and latest
time TL for each of the activity of the project. The calculations of TE and
TLareas follows:,
TocalculateTE forall activities
TE1 = 0
TE2 = TE1 + t1, 2 = 0 + 4 = 4
TE3 = TE1 + t1, 3 = 0 + 1 =1
TE4 = max (TE2 + t2, 4 and TE3 + t3, 4)
= max (4 + 1 and 1 + 1) = max (5, 2)
= 5 days
TE5 = TE3 + t3, 6 = 1 + 6 = 7
TE6 = TE5 + t5, 6 = 7 + 4 = 11
TE7 = TE5 + t5, 7 = 7 + 8 = 15
TE8 = max (TE6 + t6, 8 and TE7 + t7, 8)
= max (11 + 1 and 15 + 2) = max (12, 17)
= 17 days
TE9 = TE4 + t4, 9 = 5 + 5 = 10
TE10 = max (TE9 + t9, 10 and TE8 + t8, 10)
= max (10 + 7 and 17 + 5) = max (17, 22)
= 22 days
TocalculateTL forall activities
TL10 = TE10 = 22
TL9 = TE10 – t9,10 = 22 – 7 = 15
TL8 = TE10 – t8, 10 = 22 – 5 = 17
TL7 = TE8 – t7, 8 = 17 – 2 = 15
TL6 = TE8 – t6, 8 = 17 – 1 = 16
TL5 = min (TE6 – t5, 6 and TE7 – t5, 7)
= min (16 – 4 and 15 –8) = min (12, 7)
= 7 days
TL4 = TL9 – t4, 9 = 15 – 5 =10
TL3 = min (TL4 – t3, 4 and TL5 – t3, 5 )
= min (10 – 1 and 7 – 6) = min (9, 1)
= 1 day
TL2 = TL4 – t2, 4 = 10 – 1 = 9
TL1 = Min (TL2 – t1, 2 and TL3 – t1, 3)
= Min (9 – 4 and 1 – 1) = 0

(iii) From the Table , we observe that the
activities 1 – 3, 3 – 5, 5 – 7,7 – 8 and 8 – 10 are
critical activities as their floats are zero.

PROJECT EVALUATION REVIEW TECHNIQUE, (PERT)
In the critical path method, the time estimates are assumed to be
known with certainty. In certain projects like research and
development, new product introductions, it is difficult to estimate
the timeof variousactivities.
Hence PERT is used in such projects with a probabilistic method using three time
estimates foran activity, ratherthan a singleestimate, asshown in Figure 8.22.
Optimistic time tO:
It is the shortest time taken to complete the
activity. It means that if everything goes well
then there is more chance of completing the
activitywithin this time.
Most likelytime tm:
It is the normal time taken to complete an activity,
if the activity were frequently repeated under the
sameconditions.
Pessimistic time tp:
It is the longest time that an activity would take to
complete. It is the worst time estimate that an
activity would take if unexpected problems are
faced.

Taking all these time estimates into consideration, the expected time
of an activity is arrived at.
Theaverage or mean (ta) valueof
theactivityduration isgiven by,
Thevarianceof theactivity time
iscalculated using the formula,
Probability for Project Duration
The probability of completing the
project within the scheduled time (Ts) or
contracted time may be obtained by
using the standard normal deviate
where Te is the expected time of project
completion.
Probability of completing the project
within thescheduled time is,

Example
An R & D project has a list of tasks to be performed whose timeestimates are
given in theTable , as follows.
a. Draw the project network.
b. Find thecritical path.
c. Find the probability that the project is completed in 19 days. If the
probability is less than 20%, find the probability of completing it in 24
days.

Time expected for each activity is
calculated using the formula (5):
Similarly, the expected time is
calculated for all the activities.
The variance of activity time is
calculated using the formula (6).
Similarly, variances of all the
activitiesarecalculated.

Construct a network diagram:
calculate the time earliest (TE) and
time Latest (TL) forall theactivities.
From the network diagram Figure , thecritical path is identified as
1-4, 4-6, 6-7, with a projectduration of 22 days.

The probability of completing the project within 19 days is given by, P (Z< Z0)
To find Z0 ,
we know, P (Z <Z Network Model 0) = 0.5 – z (1.3416) (from normal tables, z (1.3416) = 0.4099)
= 0.5 – 0.4099
= 0.0901
= 9.01% Thus, the probabilityof completing the R & D project in 19 days is 9.01%.
Since the probability of completing the project in 19 days is less than 20% As in
question, we find the probability of completing it in 24 days.

COSTANALYSIS
The two importantcomponents of anyactivityare thecostand time. Cost is
directly proportional to time and viceversa.
For example, in constructing a shopping complex, the expected time of completion
can be calculated using the time estimates of various activities. But if the
construction has to be finished earlier, it requires additional cost to complete the
project. We need to arrive at a time/cost trade-off between total cost of project and
total time required tocomplete it.
Normal time:
Normal time is the time required to complete
the activity at normal conditions and cost.
Crash time:
Crash time is the shortest possible activity
time; crashing more than the normal time
will increase thedirect cost.
Cost Slope
Cost slope is the increase in cost per unit of
time saved by crashing. A linear cost curve
is shown in Figure 8.27.

Example
An activity takes 4 days to complete at a normal cost of Rs. 500.00. If it is possible
to complete the activity in 2 days with an additional cost of Rs. 700.00, what is the
incremental costof theactivity?
Incremental Costor Cost Slope
It means, if oneday is reduced we have tospend Rs. 100/- extra perday.
ProjectCrashing
Procedure for crashing
Step1: Draw the network diagram and mark the Normal time and Crash time.
Step2: Calculate TE and TL for all the activities.
Step3: Find the critical path and other paths.
Step 4: Find the slope for all activities and rank them in ascending order.

Step 5: Establish a tabular column with required field.
Step 6: Select the lowest ranked activity; check whether it is a critical activity. If
so,crash theactivity, elsego to the next highestranked activity.
Note: The critical path must remain critical while crashing.
Step 7: Calculate the total cost of project for each crashing
Step 8: Repeat Step 6 until all the activities in the critical path are fully
crashed.
Example
The following Table
8.13 gives the activities
of a construction
projectand otherdata.
If the indirect cost is Rs. 20 perday, crash theactivities to find the
minimum duration of the projectand the projectcostassociated.

Solution
From thedata provided in the table, draw the network diagram (Figure 8.28)
and find thecritical path.
From the diagram, we observe that the
critical path is 1-2-5 with projectduration of
14 days
Thecost slope forall activities and their rank
is calculated as shown in Table 8.14

The available paths of the network are listed down in Table 8.15
indicating the sequenceof crashing (see Figure 8.29).
The sequence of crashing and
the total cost involved is given
in Table 8.16 Initial direct cost
= sum of all normal costs given
= Rs. 490.00

It is not possible to crash more than 10 days, as all the activities in
the critical path are fully crashed. Hence the minimum project
duration is 10 dayswith the total costof Rs. 970.00.
Activity
Crashed
Project
Duration
Critical Path Direct Cost in (Rs.) Indirect
Cost in
(Rs.)
Total
Cost in
(Rs.)
- 14 1-2-5 490 14 x 20 =
280
770
1 – 2(2)
2 – 5(2)
2 – 4(1)
3 – 4(2)
10 1 – 2 – 5
1 – 3 – 4 – 5
1 – 2 – 4 – 5
490 + (2 x 15) + (2 x
100) + (1 x 10) + (2 x
20) = 770
10 x 20 =
200
970

a. Draw the project networkdiagram.
b. Calculate the lengthand variance of thecritical path.
c. What is the probabilitythat the jobson thecritical path can be
completed in 41 days?

Lecture 2 PERT-CPM.pdf

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Lecture 2 PERT-CPM.pdf