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Operations Research Unit 11

Sikkim Manipal University 182

Unit 11 Project Scheduling and PERTCPM

Structure

11.1. Introduction

11.2. Basic difference between PERT and CPM

11.2.1 PERT

11.2.2 CPM

11.2.3 Project scheduling by PERTCPM

11.3. PERT / CPM network components and precedence relationship

11.3.1 Critical Path Calculations

11.3.2 Determination of the Critical Path

11.3.3 Determination of Floats

11.4. Project Management – PERT

11.5. Summary

Terminal Questions

Answers to SAQs and TQs

11.1. Introduction A project such as construction of a bridge, highway, power plant, repair and maintenance of an oil

refinery or an air plane design, development and marketing a new product, research and

development etc., may be defined as a collection of interrelated activities (or tasks) which must

be completed in a specified time according to a specified sequence and require resources such

as personnel, money, materials, facilities etc.

The growing complexities of today’s projects had demanded more systematic and more effective

planning techniques with the objective of optimizing the efficiency of executing the project.

Efficiency here implies effecting the utmost reduction in the time required to complete the project

while accounting for the economic feasibility of using available resources.

Project management has evolved as a new field with the development of two analytic techniques

for planning, scheduling and controlling projects. These are the Critical Path Method (CPM) and

the Project Evaluation and Review Technique (PERT). PERT and CPM are basically time

oriented methods in the sense that they both lead to the determination of a time schedule.



Learning Objectives:

After studying this unit, you should be able to understand the following

1. What is a project?

2. What is project management?

3. Application of PERT / CPM method to network analysis

11.2. Basic difference between PERT and CPM Though there are no essential differences between PERT and CPM as both of them share in

common the determination of a critical path and are based on the network representation of

activities and their scheduling that determines the most critical activities to be controlled so as to

meet the completion date of the project.

11.2.1 PERT 1. Since PERT was developed in connection with an R and D work, therefore it had to cope with

the uncertainties which are associated with R and D activities. In PERT, total project duration

is regarded as a random variable and therefore associated probabilities are calculated so as

to characterise it.

2. It is an eventoriented network because in the analysis of network emphasis is given an important

stages of completion of task rather than the activities required to be performed to reach to a

particular event or task.

3. PERT is normally used for projects involving activities of nonrepetitive nature in which time

estimates are uncertain.

4. It helps in pinpointing critical areas in a project so that necessary adjustment can be made to

meet the scheduled completion date of the project.

11.2.2 CPM 1. Since CPM was developed in connection with a construction project which consisted of

routine tasks whose resources requirement and duration was known with certainty, therefore

it is basically deterministic.

2. CPM is suitable for establishing a tradeoff for optimum balancing between schedule time and

cost of the project.

3. CPM is used for projects involving activities of repetitive nature.



11.2.3 Project scheduling by PERTCPM It consists of three basic phases: planning, scheduling and controlling.

1. Project Planning: The various steps involved during this phase are given below:

i) Identify various activities (task or work elements) to be performed in the project.

ii) Determining requirement of resources such as men, materials, machines etc., for carrying

out activities listed above.

iii) Estimating costs and times for various activities.

iv) Specifying the interrelationship among various activities.

v) Developing a network diagram showing the sequential interrelationships between the

various activities.

2. Scheduling: Once the planning phase is over, scheduling of the project, is when each of the activities required to be performed, is taken up. The various steps involved during this phase

are listed below:

1. Estimating the durations of activities, taking into considerations the resources required for

these execution in most economic manner.

2. Based on these time estimates, preparing a time chart showing the start and finish times

for each activity, and hence calculation of total project duration by applying network

analysis techniques such as forward (backward) pass and floats calculation; identifying

the critical path; carrying out resource smoothing (or levelling) exercise for critical or

scarce resources including recosting of the schedule taking into account resource

constraints.

3. Project Control: Project control refers to revaluating actual progress against the plan. If

significant differences are observed then rescheduling must be done to update

and revise the uncompleted part of the project.

Self Assessment Questions 1

Verify whether the following statements are True or False

1. Project consists of interrelated activities.

2. Project activities are to be completed in a specified time according to specified sequence.

3. PERT and CPM identifies non critical activities.

4. PERT is activity oriented network.



5. CPM is used for projects that are repetitive in nature.

11.3 PERT/CPM Network Components And Precedence Relationship

PERT/CPM networks consists of two major components as discussed below:

a) Events: An event represents a point in time that signifies the completion of some activities and the beginning of new ones. The beginning and end points of an activity are thus

described by 2 events usually known as the Tail and head events. Events are commonly

represented by circles (nodes) in the network diagram. They do not consume time and

Resource

b) Activities: Activities of the network represent project operations or task to be conducted. An arrow is commonly used to represent an activity, with its head indicating the direction of

progress in the project. Activities originating from a certain event cannot start until the

activities terminating at the same event have been completed. They consume time and

Resource.

Events in the network diagram are identified by numbers. Numbers are given to events such that

arrow head number must be greater than arrow tail number.

Activities are identified by the numbers of their starting (tail) event and ending (head) event. An

arrow (i .J) extended between two events, the tail event i represents the start of the activity and

the head event J represents the completion of the activity as shown in Fig. 9.1:

Fig. 9.1

J i

Starting event Completion Event

Activity



Figure 9.2 shows another example, where activities (1, 3) and (2, 3) must be completed before

activity (3, 4) can start.

Fig. 9.2

The rules for constructing the arrow diagram are as follows:

1. Each activity is represented by one and only one arrow in the network.

2. No two activities can be identified by the same head and tail events.

3. To ensure the correct precedence relationship in the arrow diagram, the following questions

must be answered as every activity is added to the network:

a) What activities must be completed immediately before these activity can start ?

b) What activities must follow this activity ?

c) What activity must occur concurrently with this activity ?

This rule is selfexplanatory. It actually allows for checking (and rechecking) the precedence

relationships as one progresses in the development of the network.

Example 1: Construct the arrow diagram comprising activities A, B, C …….. and L such that the following relationships are satisfied: 1) A, B and C the first activities of the project, can start simultaneously. 2) A and B precede D. 3) B precedes E, F and H. 4) F and C precede G. 5) E and H precede I and J. 6) C, D, F and J precede K. 7) K precedes L. 8) I, G and L are the terminal activities of the project.

1

2

3 4



Fig. 9.3

The dummy activities D1 and D2 are used (dotted lines) to establish correct precedence

relationships. D3 is used to identify activities E and H with unique end events. The events of the

project are numbered such that their ascending order indicates the direction of the progress in the

project. Note: A dummy activity in a project network analysis has zero duration.

11.3.1. Critical Path Calculations The application of PERT/CPM should ultimately yield a schedule specifying the start and

completion time of each activity. The arrow diagram is the first step towards achieving that goal.

The start and completion times are calculated directly on the arrow diagrams using simple

arithmetic. The end result is to classify the activities as critical or non critical. An activity is said

to be critical if a delay in the start of the course makes a delay in the completion time of the entire

project. A noncritical activity is such that the time between its earliest start and its latest

completion time is longer than its actual duration. A noncritical activity is said to have a slack or

float time.

11.3.2. Determination of the Critical Path A critical path defines a chain of critical activities that connects the start and end events of the

arrow diagram. In other words, the critical path identifies all the critical activities of the project.

The critical path calculations include two phases. The first phase is called the Forward Pass where all calculations begin from the start node and move to the end node. At each node a

number is computed representing the earliest occurrence time of the corresponding event. These

J



numbers are shown in squares . In forward pass we note the number of heads joining the event.

We take the maximum earliest timing through these heads.The second phase called the Backwards Pass, begins calculations from the “end” node and moves to the “start” node. The

number computed at each node is shown in a triangle ∆ near end point which represent the latest

occurrence time of the corresponding event. Consider the forward pass In backward pass we see

the number of tails and take minimum value through these tails.

Let ESi be the earliest start time of all the activities emanating from event i, i.e. ESi represents

the earliest occurrence time of event i, if i = 1 is the “start” event then conventionally, for the

critical path calculations, ESi = 0 , Let Dij be the duration of the activity (i, j). Then the forward

pass calculations are given by the formula:

ESi = maxi ESi+Dij, for all defined (i, j) activities with ESi=0. Thus in order to compute EsJ for

event j, Esi for the tail events of all the incoming activities (i, j) must be computed first.

With the computation of all ESj, the forward pass calculations are completed. The backward pass

starts from the “end” event. The objective of this phase to calculate LCi, the latest completion time for all the activities coming into the event i. Thus if i = n is the end event LCn = ESn initiates

the backward pass.

In general for any node i, LCi = min LCjDij for all defined activities are calculated, which ends

the calculation of backward pass.

The critical path activities can now be identified by using the results of the forward and backward

passes. An activity (i, j) lies on the critical path if it satisfies the following conditions.

A) ESI = LCi

B) ESJ = LCJ

C) ESJESI = LCJLCI = DiJ

These conditions actually indicate that there is no float or slack time between the earliest stand

and the latest start of the activity. Thus the activity must critical. In the arrow diagram these are

characterised by numbers in and ∆ are the same at each of the head and tail events and the

difference between the number in (or ∆) at the head event and the number in (or ∆) at the tail

event in equal to the duration of the Activity.

Thus we will get a critical path, which is chain of connected activities, which spans the network form start to end.



Example 2: Consider a network which stands from node 1 and terminate at node 6, the time

required to perform each activity is indicated on the arrows.

Fig. 9.4

Let us start with forward pass with ESi = 0.

Since there is only one incoming activity (1, 2) to event 2 with D12 = 3.

ES2 = ES1+ DS2 = 0+3=3.

Let us consider the end 3, since there only one incoming activity (2, 3) to event 3, with D23 = 3.

ES3 = ES2+ D23 = 3+3 = 6.

To obtain ES4, since there are two activities A (3, 4) and (2,4 ) to the event 4 with D24 = 2 and D34

= 0.

ES4= maxi=2, 3 ESi + De4

= max ES2 +D24, ES3 + D34

= max 3+2, 6+0 = 6

Similary ES5 = 13 and ES6 = 19

Which completed first phase.

In the second phase we have

LC6= 19 = ES6

LC5 = 196 = 13

LC4= minJ = 5, 6 LCJ – D4J = 6

LC3 = 6, LC2 = 3 and LC1 = 0

∴ activities (1, 2), (2, 3) (3, 4) (4, 5) (5, 6) are critical and (2, 4) (4, 6), (3, 6), are noncritical.



Thus the activities (1, 2) (2, 3 ) (3, 4) (4, 5) and (5, 6) define the critical path which is the shortest

possible time to complete the project.

11.3.3. Determination of Floats Following the determination of the critical path, the floats for the noncritical activities must be

computed. Note that for the critical activities this float is zero. Before showing how floats are

determined, it is necessary to define two new times that are associated with each activity. There

are Latest Start (LS) and the Earliest Completion (EC) times, which are defined activity (i, J) by LSeJ= LCJ – DiJ

and ECeJ = ESi + DiJ

There are two important types of floats namely, Total Float (TF) and Free Float (FF). The total float TFiJ for activity (i, J) is the difference between the maximum time available to perform the activity (= LCJ – ESi) and its duration (= DiJ )

That is,

TFiJ = LCJ – ESI – DiJ = LCJ – ECiJ = LSiJ – ESi

The free float is defined by assuming that all the activities start as early as possible. In this case FFiJ for activity (i, J) is the excess of available time (= ESi – ESi) over its deviation (= DiJ ); that is, FFiJ = ESi – ESi = DiJ .

Note that only for a critical activity must have zero total float. The free float must also be zero when the total float is zero. The converse is not true, that is in the sense that a noncritical activity may have zero free floats.

Let us consider the example taken before the critical path calculations together with the floats for the noncritical activities can be summarized in the convenient form shown in the following table:



Activit y

(i J)

Duration DiJ

Earliest Latest Table Float TFiJ

Free Float FFiJ

Start ESi

Completio n

ECiJ

Start LSij

Completio n

∆LCJ

(1, 2) (2, 3) (2, 4) (3, 4) (3, 5) (3, 6) (4, 5) (4, 6) (5, 6)

3 3 2 0 3 2 7 5 6

0 3 3 6 6 6 6 6 13

3 6 5 6 9 8 13 11 19

0 3 4 6 10 17 6 14 13

3 6 6 6 13 19 13 19 19

0*0* 1 0* 4 11 0* 8 0*

0 0 1 0 4 11 0 8 0

Note: Total float = ESij = LFij ESij

Free float = Total float Head slack

* Critical activity *

Example 3: A project consists of a series of tasks A, B, C, – D, – E, F, G, H, I with the following relationships. (W < X,Y means X and Y cannot starts until W is completed, X,Y < W means W

cannot start until both X and Y are completed). With this notation construct the network diagram

having the following constraints A < D, E; B, D < F; C < G, B < H; F,G < I.

Find also the minimum time of completion of the project, the critical path, and the total floats of each task, when the time (in days) of completion of each task is as follows: Task: A B C D E F G H I Time: 23 8 20 16 24 18 19 4 10

Fig. 9.5



ES1 = 0, ES2 = 20, ES3 = 23, ES4 = 59, ES5 = 39, ES6 = 57, ES7 = 67

Activity (iJ)

Durati on DiJ

Earliest Latest Table Float TFiJ

Free Float FFiJ

Start ESe

Finish Eeij

Start LJDiJ

Finish LJ

(1, 2) (1, 3) (1, 4) (2, 5) (3, 4) (3, 7) (4, 5) (4, 6) (5, 6) (5, 7) (6, 7)

20 23 8 19 16 24 0 18 0 4 10

0 0 0 20 23 23 39 39 39 39 37

20 23 8 39 39 47 39 57 39 43 67

18 0 31 38 23 43 57 39 57 63 57

38 23 39 57 39 67 57 57 57 67 67

180*31 180*20 100*18 240*

0 0 31 0 0 20 0 0 18 24 0

Critical path is 1 – 3 – 4 – 6 – 7.

Self Assessment Questions 2 Fill in the blanks

1. Events do not consume ________ and _________. 2. Arrow’s head number is _________ than its tail number. 3. Dummy activity in network is introduced t keep proper _________ relationship. 4. Critical path calculation include both _________ and _________.

11.4 Project Management – PERT

Probability and Cost Consideration in Project Scheduling The analysis in CPM does not take into the case where time estimates for the different activities

are probabilistic. Also it does not consider explicitly the cost of schedules. Here we will consider

both probability and cost aspects in project scheduling.

Probability considerations are incorporated in project scheduling by assuming that the time

estimate for each activity is based on 3 different values. They are –

a = The optimistsic time, which will be required if the execution of the project goes extremely

well.

b = The pessimistic time, which will be required if everything goes bad.



m = The most likely time, which will be required if execution is normal.

The most likely estimate m need not coincide with midpoint 2 b a + of a and b. Then the

expected duration of each activity D can be obtained as the mean of 2 b a + and 2 m. i.e.

6 m 4 b a

3

m 2 2 b a

D + +

= +

+

= .

This estimate can be used to study the single estimate D in the critical path calculation.

The variance of each activity denoted by V is defined by

variance V = 2

6 a b

− .

The earliest expected times for the node i denoted by E(µi) for each node i is obtained by

taking the sum of expected times of all activities leading to the node i, when more than one

activity leads to a node i, then greatest of all E(µi) is chosen . Let µi be the earliest occurrence

time of the event i, we can consider µi as a random variable. Assuming that all activities of the

network are statistical independent, we can calculate the mean and the variance of the µi as

follows Eµi = ESI and Varµi = ∑k

k V . Where K defines the activities along the largest path

leading to i.

For the latest expected time, we consider the last node. Now for each path move backwords,

substituting the eJ D for each activity (ij). Thus we have E(LJ) = E(µa) and E(µi) = L(LJ) – iJ D if only

one path events from J to i or it is the minimum of E[LJ) – iJ D ] for all J for which the activities (i, j)

is defined.

Note: The probability distribution of times for completing an event can be approximated by the

normal distribution due to central limit theorem.

Since µi represents the earliest occurrence time, event will meet a certain schedule time STi

(specified by an analyst) with probability

Pr (µi ≤ STi) = Pr

µ

µ − ≤

µ

µ − µ

) ( V

) ( E ST

) ( V

) ( E

i

i i

i

i i

= Pr (Z ≤ Ki)



where Z ∼N(01) and Ki = ) ( V ) ( E ST

i

i i µ

µ − . It is common practice to compute the probability that event i

will occur no later than its LCe such probability will then represent the chance that the succeeding

events will occur within the (ESe, LCe) duration.

Example 4: A project is represented by the network shown below and has the following data.

Task A B C D E F G H I Optimistic Time 5 18 26 16 15 6 7 7 3 Pessimistic Time 10 22 40 20 25 12 12 9 5 Most Likely Time 8 20 33 18 20 9 10 8 4

Determine the following:

a) Expected task time and their variance.

b) The earliest and latest expected times to reach each event.

c) The critical path.

Fig. 9.6

d) The probability of an event occurring at the proposed completion data if the original contract

time of completing the project is 41.5 weeks.

e) The duration of the project that will have 96% channel of being completed.

Solution: a) Using the formula we can calculate expected activity times and variance in the following table

2

6 a b V ) m 4 b a (

6 1 D

− = + + =



b)

Activity a B m v 12 13 14 25 26 36 47 57 67

5 18 26 16 15 6 7 7 3

10 22 40 20 25 12 12 9 5

8 20 33 18 20 9 10 8 4

78 2000 330 180 200 90 98 80 40

0.696 0.444 5.429 0.443 2.780 1.000 0.694 0.111 0.111

Forward Pass: E1 = 0 E2 = 7.8 E3 = 20 E4 = 33 E5 = 258 E6 = 29 E7 = 42.8

Backward Pass: L7 = 42.8 L6 = 38.8 L5 = 34.8 L4 = 33.0 L3 = 29.8 L2 = 16.8 L1 = 0.

The Evalues and Lvalues are shown in Fig.

Fig. 9.7 c) The critical path is shown by thick line in fig. The critical path is 147 and the earliest

completion time for the project is 42.8 weeks.

d) The last event 7 will occur only after 42.8 weeks. For this we require only the duration of

critical activities. This will help us in calculating the standard duration of the last event.

Expected length of critical path = 33+9.8 = 42.8

Variance of article path length = 5.429+0.694 = 6.123



Probability of meeting the schedule time is given by Pi (Z ≤ Ki) =

Pi (Z – 0.52) = 0.30 (From normal distribution table)

Thus the probability that the project can be completed in less than or equal to 41.5 weeks is

0.30. In other words probably that the project will get delayed beyond 41.5 weeks is 0.70.

e) Given that P (Z ≤ Ki) = 0.95. But Z0.9S = 1.6 u, from normal distribution table. Then 1.6 u =

or 47 . 2

8 . 42 ST u 6 . 1 is

) ( V

) ( E ST i

i

i i − =

µ

µ −

Sji = 1.64×2.47+42.8 = 46.85 weeks.

Self Assessment Questions 3

I) True or False 1) In a project network, a sequence of activities may form a loop.

2) A critical activity must have its total and free floats equal to zero.

3) A noncritical activity cannot have zero total float.

4) The critical path of project network represents the minimum duration needed to complete

the network.

5) A network may include more than one critical path.

11.5. Summary Critical Path computations are quite simple, yet they provide valuable information that simplifies the scheduling of complex projects. The result is that PERTCPM techniques enjoy tremendous popularity among practitioners in the field. The usefulness of the techniques is further enhanced by the availability of specialized computer systems for executing, analyzing and controlling network projects.

Terminal Questions 1. Write down the basic difference between PERT and CPM. 2. Explain Project Management (PERT) 3. A project has 10 activities. The following table shows the information about the activities.

Activity Preceding activity Duration in weeks A 6 B 3 C A 5 D A 4



E A 3 F C 3 G D 5 H B, D, E 5 I H 2 J I, G, F 3

a. Draw the network b. Find the project duration c. Identify the CPM d. Prepare the schedule

4. A small project consisting of eight activities has the following characteristics: TIME ESTIMATES (IN WEEKS)

Activity Preceding activity Most optimistic time (a)

Most likely time (m)

Most pessimistic time (b)

A None 2 4 12 B None 10 12 26 C A 8 9 10

D A 10 15 20 E A 7 7.5 11 F B, C 9 9 9 G D 3 3.5 7

H E, F, G 5 5 5

a. Draw the PERT network for the project. b. Determine the critical path. c. If a 30week deadline is imposed, what is the probability that the project will be finished within

the time limit? d. If the project manager wants to be 99% sure that the project is completed on this schedule

date, how many weeks before that date should he start the project work?

Answers To Self Assessment Questions Self Assessment Questions 1 1. True 2. True 3. True 4. False 5. True

Self Assessment Questions 2 1. Time, resource 2. Greater than



3. Precedence 4. Forward pass & backward pass

Self Assessment Questions 3 1) False 2) True 3) True 4) True 5) False

Answer for Terminal Questions 1. Refer to Section 11.2

2. Refer to Section 11.4

3. b) 20 weeks c) A – D – H – I J

4. b) A – D – G – H c) 0.6591 d) 34.7 weeks