Chap 9 Project Managementdocshare02.docshare.tips/files/20953/209533229.pdfCPM and PERT are special applications. Network It is the graphic representation of logically and sequentially

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CHAPTER 9

Project management

9.1 INTRODUCTION Network scheduling is a technique used for planning, and scheduling large projects in the field of

construction, maintenance, fabrication, purchasing computer system etc. The technique is a method

of minimizing the trouble spots such as production, delays and interruptions, by determining critical

factors and coordinating various parts of the overall job. There are two basic planning and control

technique that utilize a network to complete a predetermined project or schedule. These are Programme

Evaluation Review Technique (PERT) and Critical Path Method (CPM). A project is defined as a

combination of interrelated activities all of which must be executed in a certain order for its

completion. The work involved in a project can he divided into three phases corresponding to the

management functions of planning, scheduling and control.

Planning This phase involves setting the objectives of the project and the assumptions to be

made. Also it involves the listing of tasks or jobs that must be performed to complete a project

under consideration. In this phase, men, machines and materials required for the project in addition to

the estimates of costs and duration of the various activities of the project are also determined.

Scheduling This consists of laying the activities according to the precedence order and determining,

(i) the start and finish times for each activity

(ii) the critical path on which the activities require special attention and

(iii) the slack and float for the non-critical paths.

Controlling This phase is exercised after the planning and scheduling, which involves the

following:

(i) Making periodical progress reports

(ii) Reviewing the progress

(iii) Analyzing the status of the project and

(iv) Management decisions regarding updating, crashing and resource allocation etc.

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9.2 BASIC TERMS

To understand the network techniques one should be familiar with a few basic terms of which both

CPM and PERT are special applications.

Network It is the graphic representation of logically and sequentially connected arrows and nodes

representing activities and events of a project. Networks are also called arrow diagram.

Activity An activity represents some action and is a time consuming effort necessary to complete a

particular part of the overall project. Thus, each and every activity has a point of time where it

begins and a point where it ends.

It is represented in the network by an arrow as follows.

.

Here A is called the activity and i and j are start and end nodes.

Event The beginning and end points of an activity are called events or nodes. Event is a point in

the time and does not consume any resources. It is represented by a numbered circle. The head

event called the jth event has always a number higher than the tail event called the ith event.

Merge and burst events it is not necessary for an event to be the ending event of only one activity

but can be the ending event of two or more activities. Such event is defined as a Merge event.

If the event happens to be the beginning event of two or more activities it is defined as a Burst event

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Preceding, succeeding and concurrent activities

Activities which must be accomplished before a given event can occur are termed as preceding

activities.

Activities which cannot be accomplished until an event has occurred are termed as succeeding

activities.

Activities which can be accomplished concurrently are known as concurrent activities.

This classification is relative, which means that one activity can be preceding to a certain event, and the

same activity can be succeeding to some other event or it may be a concurrent activity with one or more

activities.

Dummy activity Certain activities which neither consumes time nor resources but are used simply

to represent a connection or a link between the events are known as dummies. It is shown in the

network by a dotted line. The purpose of introducing dummy activity is -

(i) to maintain uniqueness in the numbering system as every activity may have distinct set of

events by which the activity can be identified.

(ii) To maintain a proper logic in the network.

9.3 COMMON ERRORS Following are the three common errors in a network construction:

Looping (cycling) In a network diagram looping error is also known as cycling error. Drawing an

endless loop in a network is known as error of looping. A loop can be formed if an activity were

represented as going back in line.

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Dangling To disconnect an activity before the completion of all the activities in a network diagram is

known as Dangling.

Redundancy If a dummy activity is the only activity emanating from an event and which can be

eliminated is known as redundancy.

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9.4 RULES OF NETWORK CONSTRUCTION

There are number of rules in connection with the handling of events and activities of a project network

that should be followed.

(i) Try to avoid arrows which cross each other.

(ii) Use straight arrows.

(iii) No event can occur until every activity preceding, it has been completed.

(iv) An event cannot occur twice i.e. there must be no loops.

(v) An activity succeeding an event cannot be started until that event has occurred.

(vi) Use arrows from left to right. Avoid mixing two directions, vertical and standing arrows may be

used if necessary.

(vii) Dummies should be introduced if it is extremely necessary.

(viii) The network has only one entry point called the start event and one point of emergence called the

end or terminal event.

Numbering The Events

After the network is drawn in a logical sequence every event is assigned a number. The number

sequence must be such so as to reflect the flow of the network. In numbering the events the following

rules should be observed.

(i) Event numbers should be unique.

(ii) Event numbering should be carried out on a sequential basis from left to right.

(iii) The initial event which has all outgoing arrows with no incoming arrow is numbered as 1.

(iv) Delete all arrows emerging from all the numbered events. This will

create at least one new start event out of the preceeding events.

(v) Number all new start events 2,3 and so on. Repeat this process until all the terminal event without

any successor activity is reached, number the terminal node suitably.

Note: The head of an arrow should always bear a number higher than the one assigned to the tail of the

arrow.

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Construction of Network

EXAMPLE 9.1 Construct a network for the project whose activities and their precedence relationships

are as given in Table 9.1:

TABLE 9.1

Activities A B C D E F G H I

Immediate -Predecessor

__ A A __ D B,C,E F D G,H

Solution From the given constraints, it is clear that A, D are the starting activity and I the terminal

activity. B, C are starting with the same event and are both the predecessors of the activity F. Also E

has to be the predecessor of F. Hence, we have to introduce a dummy activity.

D1 is the dummy activity.

Finally we have the following network.

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Fig. 9.1

EXAMPLE 9.2 Construct a network for the project whose activities and their precedence relationships

are as given in Table 9.2.

TABLE 9.2

Activities A B C D E F G H I J K


__ __ __ A C B,D B,D E,F A G,H E,F

SOLUTION

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9.5 TIME ANALYSIS

Once the network of a project is constructed the time analysis of the network becomes essential for

planning various activities of the project. An activity time is a forecast of the time an activity is

expected to take from its starting point to its completion (under normal conditions).

We shall use the following notation for basic scheduling computations.

(i, j) = Activity (i, j) with tail event i and head event j

Tij = Estimated completion time of activity (i, j)

ESij = Earliest starting time of activity (i, j)

EFij = Earliest finishing time of activity (i, j)

(LS)ii = Latest starting time of activity (i, j)

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(LF)ij = Latest finishing time of activity (i, j)

The basic scheduling computation can be put under the following three groups.

Forward Pass Computations (for earliest event time)

Before starting computations, the occurrence time of the initial network event is fixed. The forward

pass computation yields the earliest start and the earliest finish time for each activity (i, j) and

indirectly the earliest occurrence time for each event namely E. This consists of the following 3

steps.

Step 1 The computations begin from the start node and move towards the end node. Let zero be the

starting time for the project.

Step 2 Earliest starting time ( ) ijij EES = is the earliest possible time when an activity can begin

assuming that all of the predecessors are also started at their earliest starting time. Earliest finish time

of activity (i, j) is the earliest starting time + the activity time.

( ) ( ) ijijij tESEF +=

Step 3 Earliest event time for event j is the maximum of the earliest finish time of all the activities

ending at that event.

)tE(maxE ijiij +=

Backward Pass Computations (for latest allowable time)

The latest event time (L) indicates the time by which all activities entering into that event must be

completed without delaying the completion of the project. These can be calculated by reversing the

method of calculations used for the earliest event time. This is done in the following steps.

Step 1 For ending event assume E = L.

Step 2 Latest finish time for activity (i j) is the target time for completing the project

( ) ijij LLF =

Step 3 Latest starting time of the activity (i,j) =latest completion time of (i, j)-the activity time ijt

( ) ( ) ijijij tLFLS −=

ijj tL −=

Step 4 Latest event time for event i is the minimum of the latest start time of all activities originating

from toe event.

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)tL(minL ijjji −=

9.6 DETERMINATION OF FLOATS AND SLACK TIMES

Float is defined as the difference between the latest and the earliest activity time. Slack is defined as

the difference between the latest and the earliest event time. Hence the basic difference between the

slack and float is that slack is used for events only, whereas float is used for activities. There are

mainly three kinds of floats as given below.

Total float It refers to the amount of time by which the completion of an activity could be delayed

beyond the earliest expected completion time without affecting the overall project duration time.

Mathematically, the total float of an activity (i, j) is the difference between the latest start time and

the earliest start time of that activity.

Hence the total float for an activity (i, j) denoted by ijTF is calculated by the formula

=ijTF (Latest start-Earliest start) for activity (i, j)

where E, L are the earliest time and latest time for the tail event i and head event J and t is the normal

time for the activity (i, j) This is the most important type of float as it concerns with the overall project

duration.

Critical activity An activity is said to be critical if a delay in its .start will cause a further delay in the

completion of the entire project.

Critical path The sequence of critical activities in a network is called the critical path. It is the longest

path in the network from the starting event to the ending event and defines the minimum time required

to complete the project. In the network it is denoted by double line. This path identifies all the critical

activities of the project. Hence, for the activity (i, j) to lie on the critical path, following conditions

must be satisfied.

(a) ii LFES =

(b) jj LFES =

(c) ijjiji tLFLFESES −−=− .

Es., Es, are the earliest start, and finish time of the event i and j Lf , LF are the latest start, finish time

of the event i and j.

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9.7 CRITICAL PATH ANALYSIS The Critical path of a network gives the shortest time in which the whole project can be completed. It

is the chain of activities with the longest time durations. These activities are critical activities. They are

critical in the sense that delay in any of them results in the delay of the completion of the project.

There may be more than one critical path in a network and it is possible for the critical path to run

through a dummy. The critical path analysis consists of the following steps:

1. Calculate the time schedule for each activity : It involves the determination of the time by which

an activity must begin and the time before which it must be completed. The time schedule data for

each activity include the calculation of the earliest start, the earliest finish, the latest start, the latest

finish times and the float.

2. Calculate the time schedule for the completion of the entire project: It involves calculation of

project completion time and the probability of completion of theproject on or before that time.

3. Identify the critical activities and find the critical path : Critical activities are the once which must

be started and completed on schedule or else the project may get delayed. The path containing these

activities is the critical path and is the longest path in terms of duration.

EXAMPLE 9.3

Consider the network shown in Fig. 9.3 which consists of the following activities as shown in the Table 9.3

TABLE 9.3 Activities A B C D E F G H I J


__ __ A A B,C B,C E E D F,H,I

Duration(weeks) 15 15 3 5 8 12 1 14 3 14

Fig: 9.3

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The earliest start time (E) for an activity represents the time at which an activity can begin at the

earliest. It assumes that all the preceding activities start and finish at their earliest times. For instance

earliest start times of activities 1-2 and 1-3 are zero each or the earliest occurrence time of event 1 is

zero. Earliest start times of activities 2-3 and 2-5 or the earliest occurrence time of event 2 is obtained

by adding activity time t12 to earliest occurrence time of event 1 i.e., it is 0+15=15.

Next consider event 3. It can be reached directly from event 1 or via event 2, the times for the two

sequences being 15 and 15 + 3 = 18. Since event 3 can occur only when all the preceding activities and

events have taken place, its earliest occurrence time or the earliest start times of activities emanating

from even 3 is 18, the higher of the two values 15 and 18. This is represented by putting E3 = 18

around its node in the network. Likewise, the earliest occurrence time of each event can be determined

by proceeding progressively from left to right i.e., following the forward pass method according to the

following rule:

If only one activity converges on an event, its earliest start time E is given by E of the tail event of the

activity plus activity duration. If more than one activity converges on it, E's via all the paths would be

computed and the highest value chosen and put around the node.

The E's calculated for the problem at hand are shown in the network diagram.

The latest finish time (L) for an activity represents the latest by which an activity must be completed in

order that the project may not be delayed beyond its targeted completion time. This is calculated by

proceeding progressively from the end event to the start event. The L for the last event is assumed to be

equal to its E and the L's for the other events are computed by the following rule (using backward pass

method):

If only one activity emanates from an event, compute L by subtracting activity duration from L of

its head event. If more than one activity emanates from an event, compute L's via all the paths and

choose the smallest and put it around the event at hand.

The L's calculated for the problem at hand are shown in the network diagram.

Next, the earliest finish time (TEF) and the latest start time (TLS) for an activity are computed:

TEF = E + tij,

TLS = L – tij,

where tij if is the duration for activity i- j. Float (also called total float) for an activity is then

calculated:

F=L–TEF or F=TLS-E.

Float is, thus, the positive difference between the finish times or the positive difference between the

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start times. The following analysis table 9.4 is then compiled reference to the Fig. 9.3:

TABLE 9.4 Activity Duration Start time

Earliest Latest

Finish time

Earliest Latest

Total float

(weeks)

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

A(1-2) 15 0 0 15 15 0

B(1-3) 15 0 3 15 18 3

C(2-3) 3 15 15 18 18 0

D(2-5) 5 15 32 20 37 17

E(3-4) 8 18 18 26 26 0

F(3-6) 12 18 28 30 40 10

G(4-5) 1 26 36 27 37 10

H(4-6) 14 26 26 40 40 0

I(5-6) 3 27 37 30 40 10

J(6-7) 14 40 40 54 54 0

Columns 1 and 2 contain the activities and their durations is weeks. Under column 3 are noted the E's

for the tail events and under column 6 are noted the L's of the head events. Other columns are then

computed as follows:

column 4 = column 6 – column 2,

column 5 = column 3 + column 2,

and column 7= column 6 – column 5,

= column 4 – column 3.

As an example, consider activity 1-2. Its tail event 1 has E = 0. Put 0 against this activity under column

3. Its head event 2 has L = 15. Put 15 against it under column 6. Under column 4 note the value in

column 6 minus activity duration i.e., 15 – 15 = 0. Under column 5 note the value in column 3 plus

activity duration i.e., 0 + 15 = 15. Compute total float by subtracting column 5 from 6 or column 3

from 4. Total float for activity 1-2 is 0. Similarly, calculate total float for other activities. Critical path

is the path containing activities with zero float. These activities demand above normal attention with

no freedom of action. For the problem at hand it is 1-2-3-4-6-7 and is shown by double arrows in Fig.

1. The project duration is 54 weeks. Sometimes, there may be more than one critical path i.e., two or

more paths with the same maximum completion time. Non-critical activities have positive float (slack)

so that we may slacken while executing them and concentrate on the critical activities. While delay in

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any critical activity will delay the project completion, this may not be so with the non-critical

activities.

9.8 THE THREE FLOATS

Total float : It is the difference between the maximum time available to perform the activity and the

activity duration. The maximum time available for any activity is from the earliest start time to the

latest completion time. Thus for an activity i - j having duration tij,

Maximum time available = L – E.

Total Float = L – E – ty

= (L – tij) – E or L – (E + tij)

= TLS - E or L – TEF.

Thus the total float of an activity is the difference of its latest start and earliest start times or the

difference of its latest finish and earliest finish times. Total float represents the maximum time within

which an activity can be delayed without affecting the project completion time.

Free Float : It is that portion of the total float within which an activity can be manipulated without

affecting the floats of subsequent activities. It is computed by subtracting the head event slack from the

total float (TF). The head event slack is (L – E) of the event.

Free float (FF) of activity

i - j = T.F. – (L – E) of event j.

Independent Float : It is that portion of the total float within which an activity can be delayed for start

without affecting the floats of preceding activities. It is computed by subtracting the tail event slack

from the free float. If the result is negative, it is taken as zero.

Independent float of activity of i - j = F.F. – (L – E) of tail event j.

Apart from the above three floats, there is another float, namely the interfering float for the activities.

Subcritical Activity : Activity having next higher float than the critical activity is called the

subcritical activity and demands normal attention but allows some freedom of action. The path

connecting such activities is named as the subcritical path. A network may have more than one

subcritical path.

Supercritical Activity : An activity having negative float is called supercritical activity. Such an

activity demands very special attention and action. It results when activity duration is more than the

time available. Such negative float, though possible, indicates an abnormal situation requiring a

decision as to how to compress the activity. It can be done by employing more resources so as to make

the total float zero or positive. Compression of the network, however, involves an extra cost.

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Slack: It is the time by which occurrence of an event can be delayed. It is denoted by S and is the

difference between the latest occurrence time and earliest occurrence time of the event. i. e.,

S = L - E of the event

In the above discussion, the term float has been used in connection with the activities and slack for the

events. However, the two terms are being used interchangeably i.e., slack for the activities and float for

the events by some of the writers.

EXAMPLE 9.4

Tasks A, B, C, H, I constitute a project. The precedence relationships are A < D; A<E, B < F; D < F,

C < G, C < H; F < I, G < I

Draw a network to represent the project and find the minimum time of completion of the project when

time, in days, of each task is as follows:

Task A B C D E F G H I

Time 8 10 8 10 16 17 18 14 9

Also identify the critical path.

SOLUTION

Fig. 9.4(a)

The given precedence order reveals that there are no predecessors to activities A, B, and C and hence

they all start from the initial node. Similarly, there are no successor activities to activities E, H and I

and hence, they all merge into the end node of the project. The network obtained is shown in Fig.

9.4(a).

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The nodes of the network have been numbered by using the Fulkerson's rule. The activity. descriptions

and times are written along the activity arrows. To determine the minimum project completion time, let

event 1 occur at zero time. The earliest occurrence time (E) and the latest occurrence time (L) of each

event is then computed.

El = 0, E2 = El + t12 =0 + 8 = 8,

E3 = El + t13 = 0 + 8 = 8,

E4 = Max. [0 + 10, 8 + 10] = 18,

E5 = Max. [18 + 17, 8 + 18] = 35,

E6 = Max. [8 + 16, 35 + 9, 8+14]=44.

Similarly,

L6 = E6 = 44,

L5 = L6 — t56 = 44 – 9 = 35,

L4 = L5 — t45 = 35 – 17 = 18,

L3 = Min. [44 – 14, 35 – 18 = 17,

L2 = Min. [44 –. 16, 18 – 10] = 8,

L1 =Min. [8-8, 17-8, 18– 10] =0.

The E and L values for each event have been written along the nodes in Fig. 9.4(b).

Fig. 9.4(b)

The critical path is now determined by any of the following methods:

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Method 1. The network analysis table is compiled as shown in the Table 9.5.

TABLE 9.5 Activity Duration Start time

Earliest Latest Finish time Earliest Latest

Total float (weeks)

1-2 8 0 0 8 8 0 1-3 8 0 9 8 17 9 1-4 10 0 8 10 18 8 2-4 10 8 8 18 18 0 2-6 16 8 28 24 44 20 3-5 18 8 17 26 35 9 3-6 14 8 30 22 44 22 4-5 17 18 18 35 35 0 5-6 9 35 35 44 44 0

Activities 1-2, 2-4, 4-5 and 5-6 having zero float are the critical activities and 1-2-4-5-6 is the critical path.

Method 2. For identifying the critical path, the following conditions are checked. If an activity satisfies all the three conditions, it is critical.

(i). E = L for the tail event. (ii). E = L for the head event. (iii). Ej - Ei = Lj - Li = tij. Activities 1-2, 2-4, 4-5 and 5-6 satisfy these conditions. Other activities do not fulfill all the three

conditions. The critical path is, therefore, 1-2-4-5-6. Method 3. The various paths and their duration are:

Path Duration (days) 1-2-6 24

1-2-4-5-6 44 1-4-5-6 36 1-3-5-6 35

1-3-6 22 Path 1-2-4-5-6, the longest in time involving 44 days, is the critical path. It represented by

doublebold lines in Fig. 9.4(b).

EXAMPLE 9.5 A project consists of a series of tasks labelled A, B. H, I with the following constraints A< D,E; B,

D<F: C<G; B<H; F, G<I. (W<X, Y means X, and Y can't start until W is completed.) You are required

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to construct a network using this notation. Also find the minimum time of completion of the project

when the time of completion of each task is given as follows.

Task A B C D E F G H I

Time(days) 23 8 20 16 24 18 19 4 10

SOLUTION The given constraints can be given in the Table 9.5.

Activity A B C D E F G H I

Proceeding -- -- -- A A B,D C B G,F

Activity

To determine the minimum time of completion of the project, we compute ESi and LFj for each of

the tasks (i, j) of the project. The critical path calculations are as shown in the Table 9.6.

Critical path 1-2-3-5-6

TABLE 9.6

Actvity Normal time Earliest Latest Total floats

Start Finish Start Finish

A (1-2) 23 0 23 0 23 0

B (1-3) 8 0 8 31 39 8

C (1-4) 20 0 20 18 38 18

D (2-3) 16 23 39 23 39 0

E(2-6) 24 23 47 43 67 20

F( 3-5) 18 39 57 39 57 0

H( 3-6) 4 39 43 63 67 24

G( 4-5) 19 20 39 38 57 18

I(5-6) 10 57 67 57 67 0

The above table shows that the critical activities are 1-2, 2-3, 3-5, 5-7 as their total float is zero. Hence,

we have the critical path, 1-2-3-5-7 with the total project duration (the least possible time to complete

the entire project as 67 days.

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Fig. 9.5

EXAMPLE 9.6 A project schedule has the following characteristics:

Activity Time (weeks)

Activity Times (weeks)

1-2 4 5-6 4 1-3 1 5-7 8 2-4 1 6-8 1 3-4 1 7-8 2 3-5 6 8-10 5 4-9 5 9-10 7

(i) Construct the network. (ii) Compute E and L for each event, and (iii) Find the critical path.

SOLUTION The given data results in a network shown in Fig. 9.6. The figures along the arrows represent

the activity times.

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Fig. 9.6 The earliest occurrence time (E) and the latest occurrence time (L) of each event are now

computed by employing forward and backward pass calculations.

In forward pass computations, E values are represented in Fig.9.6 and in backward pass

computations, L values are also represented in Fig. 9.6. Network analysis table is given Table

9.7.

TABLE 9.7

Activity Duration Start time Earliest Latest

Finish time Earliest Latest

Total float (weeks)

1-2 4 0 5 4 9 5

1-3 1 0 0 1 1 0

2-4 1 4 9 5 10 5

3-4 1 1 9 2 10 8

3-5 6 1 1 7 7 0

4-9 5 5 10 10 15 5

5-6 4 7 12 11 16 5

5-7 8 7 7 15 15 0

6-8 1 11 16 12 17 5

7-8 2 15 15 17 17 0

22

8-10 5 17 17 22 22 0

9-10 7 10 15 17 22 5

Path 1-3-5-7-8-10 with project duration of 22 weeks is the critical path.

EXAMPLE 9.7

The utility data for a network are given below. Determine the total, free, and independent

floats and identify the critical path.

Activity 0-1 1-2 1-3 2-4 2-5 3-4 3-6 4-7 5-7 6-7

Duration 2 8 10 6 3 3 7 5 2 8

SOLUTION

The network diagram for the given project data is shown in Fig. 9.7. Activity durations are

written along the activity arrows.

The earliest start and latest finish times of the activities are computed by employing the forward

pass and backward pass calculations, as explained in example 9.2. These times are represented

in the network around the respective nodes.

The network analysis table is now constructed in Table 9.8.

Fig.9.7

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TABLE 9.8

Activity Durat

ion

Start time

Earliest Latest

Finish time

Earliest Latest

Float

Total Free Independent

0-1 2 0 0 2 2 0 0 0

1-2 8 2 8 10 16 6 0 0

1-3 10 2 2 12 12 0 0 0

2-4 6 10 16 16 22 6 0 –6 ≈ 0

2-5 3 10 22 13 25 12 0 –6 ≈ 0

3-4 3 12 19 15 22 7 1 1

3-6 7 12 12 19 19 0 0 0

4-7 5 16 22 21 27 6 6 0

5-7 2 13 25 15 27 12 12 0

6-7 8 19 19 27 27 0 0 0

Total float is the positive difference between latest and earliest finish times or latest and earliest

start times. For activity 1-2,

Total float (T.F.) = 16 – 10 = 8 – 2 = 6.

Similarly, for activity, say 2-5,

Total float = 25-13 = 22 -10 = 12 and so on.

Total float calculations are depicted in column 7 of table 4.

Free float of activity i - j = Total float – head event slack

= Total float – (L – E) of event j.

Thus free float of activity 0 – 1 =0–(L–E) of event 1,

= 0–(2-2)=0,

free float of activity 1 -2 =6–(16– 10)=6-6=0etc.

Free floats of various activities are calculated in column 8 of the network analysis table.

Independent float of activity i - j = Free float – tail event slack

= Free float – (L – E) of event i.

Thus independent float of activity 0 - 1 = 0 – (0 – 0) = 0,

independent float of activity I - 2= 0 – (2 – 2) = 0,

independent float of activity 2 – 4 = 0 – (16 – 10) = – 6≈ 0 and so on.

Independent floats of various activities are calculated in column 9 of the Table 9.8. If

independent float of an activity is negative, it is taken as zero.

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EXAMPLE 9.8

For the network given in Fig 9.8, determine the total, free, and independent floats for each

activity. Times for activities are in months.

Fig. 9.8

SOLUTION

The computations of earliest start, earliest finish, latest start and latest finish times along with

floats are given in Table 9.9.

TABLE 9.9

Activity Duratio

n

Start time

Earliest Latest

Finish time

Earliest Latest

Float

Total Free Independent

1-2 8 0 0 8 8 0 0 0

1-3 10 0 2 10 12 2 2 2

2-3 4 8 8 12 12 0 0 0

24 0 8 17 8 17 9 9 9

3-4 5 12 12 17 17 0 0 0

3-5 6 12 15 18 21 3 3 3

4-5 4 17 17 21 21 0 0 0

4-8 8 17 24 25 32 7 7 7

5.6 5 21 21 26 26 0 0 0

5-7 7 21 22 28 29 1 1 1

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6-7 3 26 26 29 29 0 0 0

6-8 5 26 27 31 32 1 1 1

7-8 3 29 29 32 32 0 0 0

Activities 1-2, 2-3, 3-4, 4-5, 5-6, 6-7 and 7-8 have zero float and hence are critical. The path 1-2-34-5-

6-7-8 is the critical path with the project duration of 32 months. Total, free and independent floats are

calculated as explained in example 9.4 and are represented in the last four columns of the Table 9.9.

EXAMPLE 9.9

Estimated times for the jobs of a project are given below:

Job: A B C D E F G H I J K L

Time: 13 5 8 10 9 7 7 12 8 9 4 17

(weeks)

The constraints governing the jobs are as fo1lows:

A and B are start jobs; A controls C, D and E; B controls F and J; G depends upon C; H depends on D;

E and F control I and L; K follows J; L is also controlled by K; G, H, I and L are the last jobs. Draw the

network, determine float for each activity, project duration and the critical path.

SOLUTION

The network obtained by using the given precedence relationship is shown in Fig. 9.10. Events have

been numbered using the Fulkerson's rule. Note that a dummy activity 6-8 has been included to draw

the network. The earliest start times and latest completion times of the activities can be computed by

using the forward and backward pass methods. Critical path is 1-2-6-8-9 and project duration is 39

weeks.

26

Fig. 9.10

9.9 COST ANALYSIS AND CRASHING THE NETWORK

9.9.1 Project Cost

Each activity of the project consumes some resources and hence has cost associated with it. In most

of the cases cost of an activity will vary to some extent with the amount of time consumed by the

activity. The cost of total project, which is the aggregate of the activity costs will also depend upon

the project duration time. Thus by increasing the costs, the project duration can be cut down to some

extent. The aim is always to strike a balance between the costs and time, and to obtain an optimum

project schedule. An optimum project schedule implies lowest possible cost and the associated time

for the project. The total cost of any project consists of the direct and indirect costs involved in its

execution.

9.9.2 Direct Cost

The cost is directly dependent upon the amount of resources involved in the execution of the

individual activities. This is the cost of the materials, equipment and labour required to perform the

activity. If the activity is performed by a subcontractor, the price of the subcontract will be the

activity direct cost. It can be seen from the direct cost-time relationship shown in Fig. 9.11 that the

activity duration can be varied by varying the cost. Point N, referred to as the 'Normal Point

27

corresponds to the normal activity time. If the duration is further increased, the decrease in cost is

insignificant. As the activity is compressed, the direct cost goes on increasing. If it is compressed

beyond C, the cost increases very rapidly for insignificant change in the activity duration. This point

is called 'crash point'. Crash time is thus the minimum activity duration to which an activity can be

compressed by increasing the resources and hence by increasing the direct cost.

To make the calculations simple, the direct cost-time curve is approximated to either a single

straight line or multi-straight lines (segments), as shown in Fig. 9.12. The slope of the line (or a

segment of line) gives the increase in direct cost per unit time for expediting the activity. This is

called cost slope.

For the straight line approximation in Fig. 9.12,

TC

TTCC

nc

nc∆∆

=−−

=slopecost

Fig. 9.11

The choice between the single straight line and segmented approximation depends upon the non-

linearity of the cost curve. Also the segmented approximation is adopted only when the activity can be

broken down into sub activities. Above all, it is the judgment of the executive whether to approximate

the curve by segments and go through involved calculations or to use a little rough single slope

approximation and save calculation work.

28

Fig. 9.12

9.9.3 Indirect Cost

Project indirect cost can further be sub-divided into two parts: fixed indirect cost and variable indirect

cost. The fixed indirect cost is due to the general and administrative expenses, licence fee, insurance

cost and taxes and does not depend upon the progress of the project. The variable indirect cost depends

upon the time consumed by the project and consists of overhead expenditure, suppervision, interest on

capital and depreciation, etc. It is assumed that the indirect cost increases linearly with time as shown in

figure 9.13.

The sum of the direct and indirect cost gives the total project cost. As the direct cost decreases with

time and indirect cost increases with time, the total project cost curve will have a point where the total

cost will be minimum (Fig. 9.14). The time corresponding to this point is called the optimum duration

and the cost, the optimum cost.

29

9.10 Crashing (Contracting or Compressing) the Network Often there may be compelling reasons to complete the project earlier than would be the case with the

duration of the critical path computed on the basis of normal activity times. Examples are war

situations or market introduction of a new product. In such situations, our concern is to find the project

cost if some or all the activities are crashed. From the knowledge of the normal direct cost and crashed

direct cost, the cost slope for each activity can be determined. If the indirect cost/time is known, the

total cost of the project can be found by adding the direct and indirect costs.

The normal time (tn) for the completion of the project will be the sum of the normal time durations of

the critical activities and the normal direct cost (Cn) of the project will be the sum of the normal cost of

all the activities since each and every activity has to be executed to complete the project.

The minimum time (tc) that the project will take for its completion will be sum of the crashed time

durations of the activities along the critical path. If all the activities (critical as well as noncritical) are

crashed, the cost will be very high without any additional advantage over and above the one obtained

by crashing only the critical activities. Therefore, the non-critical activities need not be expedited since

their crashing is not going to decrease the project duration further.

However, in the process of crashing the critical activities, it may so happen that some of the non-

critical activities become critical. It is, therefore, essential to proceed step by step, crashing one activity

at a time and examine whether any other non-critical activity has also become critical. Obviously, first

the critical activity with minimum cost slope should be selected for crashing and so on till all the

critical activities are fully crashed or no further crashing is possible.

At each step of crashing, the direct cost is calculated; it would naturally be higher than the normal

30

direct cost as it. includes the additional crash cost too. However, the indirect cost will reduce as the

project duration has been reduced. The total cost is then found at each step. A table or a curve may be

constructed between the total cost and time. It is found that as the project duration is decreased, the

total cost may also decrease, reaching the minimum value, beyond which it may increase. This

minimum cost is called the optimum project cost and the corresponding time, the optimum project

time. The crashing may be further continued to get the minimum or crash project time and the

corresponding crash cost.

EXAMPLE 9.10

List of activities for erecting a canteen in a factory is given in Table 9.10 with other relevant details.

Job A must precede all others while job E must follow others. Apart from this, jobs can run

concurrently.

TABLE 9.10

Code Job description Normal Crash Duration Cost Duration Cost (days) (Rs.) (days) (Rs.) A Lay foundation and build walls 5 3,000 4 4,000

B Tile roofing 6 1,200 2 2,000

C Instal electricity 4 1,000 3 1,800

D Instal plumbing 5 1,200 3 2,000

E Connect services to finish 3 1,600 3 1,600

Total cost 8,000

(i) Draw the network and identify the critical path.

(ii) Crash the network fully to find out minimum duration.

(iii) If indirect costs are Rs. 300 per day, determine time-cost trade offer the project.

SOLUTION

First, the cost slope for each activity and the normal direct cost of the project is calculated. This

is shown in the table below.

Job A (1-2) B (2-3) C (2-5) D (2-4) E (5-6)

Cost slope 1,000(1) 200(4) 800(1) 400(3) __

(Rs./ day)

Next, the network is drawn and the critical path is determined. An activity (i - j) is critical only if

31

E = L for its tail event, E = L for its head event and Ej – Ei= Lj – Li = tij. Critical path is shown in Fig.

9.15.

Fig. 9.15

(i) Critical path is 1-2-3-5-6.

Normal duration = 14 days. Normal cost = Rs. (8,000 + 300 x 14) = Rs. 12,200.

(ii) Represent the network on time-scaled diagram. This is shown in Fig. 9.15. Activities along

the critical path are arranged along the horizontal line 1-2-3-5-6.

Early start times for the activities have been assumed. Dotted lines show the floats for the non-critical

activities.

Project length can be reduced by crashing the critical activities. This will involve extra direct-cost.

While crashing a critical activity, it is possible that some associated non-critics activity (or

activities) may also have to be crashed. If float is available no extra cost for such activity (or

activities) will be involved; if not, the cost of crashing such activity (activities) will also be

considered and the total cost will be calculated. Then the critical activity involving the least

associated cost will be selected for crashing.

While crashing a critical activity, only those non-critical activities will require crashing which start

from the tail event of the critical activity under consideration or earlier and finish at the head event

of the critical activity or later than that. This rule may be kept in mind while crashing the

activities.

32

Fig. 9.16 Crash activity B (2-3), C (2-5) and D (2-4) by one day each. it is crashed by one day. The various alternative critical activities and their crash costs are given in Table 9.11.

TABLE 9.11 Activity Cost (Rs.) A (1-2) 1,000(1) B (2-3) 200(4) C(2-5) 800(1) D(2-4) 400(2) E(5-6) 0

Since activity 2-3 has the lower associated crash cost of Rs. 200 per day, Crash cost = Rs. 200, project duration = 13 days. The network is shown in Fig. 9.17.

33

Fig. 9.18

Crash activities B, C and D by one day each. The various activities and their crash costs are given below.

Activity Cost (Rs)

A (1 -2) 1,000

B (2-3) 200

D (2-4) 400

C (2-5) Nil

Total 600

Crash cost = Rs. (200 + 600) = Rs. 800, project duration = 12 days.

The network is shown in Fig. 9.19.

34

Fig. 9.19

Crash activity A (1-2) by one day.

The various activities and their crash costs are given below.

Activity Cost (Rs)

B (2-3) 200

D (2-4) 400

C (2-5) 800

Total 1,400

Crash cost = Rs. (1,000 + 800) Rs. 1,800, project duration = 11 days.

The network is shown in Fig. 9.20

Crash activities B, C and D by one day each.

Crash cost = Rs. (1,400 + 1,800) = Rs. 3,200, project duration = 10 days.

35

Fig. 9.20

Fig. 9.21

The network is shown 9.21. All actvities are fully crashed. Thus minimum project duration is

10 days. The complete crasging of the network from 14 days to 10 days.

The above results are summarized in the Table 9.12.

.

36

TABLE 9.12

Duration(days) Normal direct

cost (Rs.)

Crash cost

(Rs)

Indirect cost

(Rs)

Total cost

(Rs)

14 8,000 0 4,200 12,200

13 8,000 200 3,900 12,100

12 8,000 800 12,400

11 8,000 1,800 3,300 13,100

10 8,000 3,200 3,000 14,200

Thus time-cost trade off exists at 13 days when B is crashed by I day and the tots minimum cost

is Rs. 12,100.

EXAMPLE 9.11 The Table 9.13 gives the activities of a constriction project and other relevant information.

(a) What is the normal project length and the minimum project length?

(b) Determine the minimum crashing costs of schedule ranging from normal length down to and

including the minimum length schedule.

TABLE 9.13

Activity i- j

Normal duration (days)

Crash duration (days)

Cost of crashing (Rs per day)

1-2 9 6 20 1-3 8 5 25 1-4 15 10 30 2-4 5 3 10 3-4 10 6 15 4-5 2 1 40 (c) What is the optimal length schedule duration of each job for your solution?

Overhead of the project is Rs. 60 per day.

SOLUTION

37

Since the overhead cost is given the cost is indirect. The project duration can by reducing the cost

associated with it.

The critical path comprises the 1-3, 3-4 and 4-5 with the normal duration as 20 days.

The total cost associated with the project is 20 X 60 = Rs. 1,200.

The cost slope is given in the data. We reduce the project duration by crashing the activity lying on the critical path. The critical activity 3-4 has the minimum cost slope. It is Ranked I, so this activity is crashed first for 4 days. The other paths of the network are: Normal duration

Crash activity Crash cost Overhead cost Total Cost

20 - - 20 x 60 = 1200 1200 19 3-4 (1) 1 x 15 = 15 19 x 60 = 1140 1155 18 3-4 (2) 15 +1 x15 = 30 18 x 60 = 1080 1110 17 3-4 (3) 30 + 1 x 15 = 45 17 x 60 = 1020 1065 16 4-5 45 + 1 x 40 = 85 16 x 60 = 960 1045

Activity

Cost of crashing

1-2 20 (3) III 1-3 25 (3) IV 1-4 30 (5) V 2-4 10 (2) I 3-4 15 (4) II 4-5 40 (1) VI

38

(1-3-4-5-1-4-5) 15 3-4 (1-3-4-5) 85 + 15 + 30 15 x 60 = 900 1-4 ( 1-4-5) +10=140 2-4 (1-2-4-5) 14 1-3 (1-3-4-5) 140+25+30 14 x 60 = 840 1045 1-4 (1-4-5) +10=205 2-4 (1-2-4-5) 13 1-3 ( 1-3-4-5) 205+25+30 13 x 60= 780 1060 1-4 (1-4-5 ) +20 = 280 1-2 (1-2-4-5) After 17 days, we get parallel paths namely, 1-3-4-5 and 1-4-5. We crash the activity 4-5 for one day at

the rate of Rs. 40 per day.

After 16 days we get all the critical paths. As three is no common activity between them, we crash the

activity 3-4 for the path 1-3-4-5, 1-4 for the path 1-4-5 and 2-4 for the path 1-2-4-5.

As the activity 3-4 in the path 1-3-4-5 is in crash time, we crash the activity 1-3 for 1 day at the rate of

Rs. 25 per day. The optimum duration is 15 days as it gives the minimum cost of Rs. 1040.

To find the minimum duration we crash further.

Normal duration

Crash activity Crash cost Overhead cost Total Cost

12 1-3 (1-3-4-5) 1-4(1-4-5) 1-2(1-2-4-5)

280+25+30+20=355 12 x 60 = 720 1075

The minimum duration is 12 days and no more crashing is possible as all activities in the path 1-3-4-5

are in crash time.

Optimum duration = 15 days with total cost associated as Rs 1040

Minimum duration = 12 days with total cost associated as Rs 1075.

Note: From the above problem we observe that the optimum duration and minimum duration are not

same. Optimum duration refers to the which yields the minimum total cost where as minimum duration

is the one in which no more crashing is possible beyond that duration.

EXAMPLE 9.12 The table below provides cost and estimates for a seven activity project.

Activity Time estimate

(weeks) Direct cost estimate

(Rs. 1000) Normal Crash Normal Crash A (1-2) 2 1 10 15

1040

39

B (1-3) 8 5 15 21 C (2-4) 4 3 20 24 D (3-4) 1 1 7 7 E (3-5) 2 1 8 15 F (4-6) 5 3 10 16 G (5-6) 6 2 12 36 (a) Draw the project network corresponding to normal time.

(b) Determine the critical path and the normal duration and cost of the project.

(c) Crash the activities so that the project completion time reduces to 9 weeks.

SOLUTION

As the problem involves direct cost we expect that the project duration can be reduced with increase in

total cost. First we draw the network.

Critical path 1-3-5-6; Normal duration = 16 weeks

Total cost = RS. 82,000

The calculation of cost slope and crashing number of days are shown in the table below:

Activity Slope (Rs. 1000 ) 1-2 5 (1) IV 1-3 2 (3) I 2-4 4 (1) III 3-4 0 3-5 7 (1) VI 4-6 3 (2) II 5-6 6 (4) V

The different paths of the network are

First crashing We crash the activity 1-3 as it is the critical activity with minimum rank. We crash this

activity for 3 weeks at the rate of Rs. 2 (1000) per day.

Project duration reduced to 16-3 = 13 weeks

Total cost = 82 + 3x 2 = 88 (RS. 1000)

Second Crashing Next we crash the activity 5-6 for 2 weeks at the rate of Rs. 6 (1000) per week as this

activity has the next minimum rank.


Total cost = 88 + 12 x 6 = Rs. 100 (1000)

40

Third Crashing After 11 days we get all the paths critical. As three is no activity in common, we crash

the activity 5-6 for 2 weeks in the paths 1-3-5-6 crash the activity 4-6 for 2 weeks which is common to

the paths 1-3-4-6 and 1-2-4-6.


Total cost = 100+2x6+2x3=118 (RS. 100)

The project duration can’t be reduced beyond 9 weeks as all the activities are in crash time. Hence, the

optimum duration is 9 weeks with the total cost associated Rs. 118 (1000).

9.11 PROGRAMME EVALUATION AND REVIEW TECHNIQUE (PERT) The network methods discussed so far may be termed as deterministic, since estimated activity

times are assumed to be known with certainty. However, in research project or design of gear box or a

new machine, various activities are based on judgment. It is difficult to obtain a reliable time

estimate due to the changing technology. Time values are subject to chance variations. For such cases

where the activities are non-deterministic in nature, PERT was developed. Hence, PERT is a

probabilistic method where the activity times are represented by a probability distribution. This

probability distribution of activity times is based upon three different time estimates made for each

activity. These are as follows.

(i) Optimistic time estimate

(ii) Most likely time estimate

(iii) Pessimistic time estimate

Optimistic time estimate It is the smallest time taken to complete the activity if everything goes on well.

There is very little chance that activity can be done in time less than the optimistic time. It is denoted

by to. or a.

Most likely time estimate It refers to the estimate of the normal time the activity would take. This

assumes normal delays. It is the mode of the probability distribution. It is denoted by tm.

Pessimistic time estimate It is the longest time that an activity would take if everything goes wrong. It

is denoted by tp or b. These three time values are shown in the following figure 1.

41

Fig. 1 Time distribution curve

From these three time estimates, we have to calculate the expected time of an activity. It is given by the

weighted average of the three time estimates

6tt4t pmo ++

=µ

Where pt , ot and mt are pessimistic, optimistic and most likely time estimates respectively.

Variance of the activity is given by 2

op2

6tt

−=σ

The range for the time estimates is from pt to ot . The most likely time will be anywhere in the range

from pt to ot . The expected project completion time is iµ∑ , where iµ is the expected duration of

the ith critical activity. The variance of the project completion time is ∑σ2 where i2σ is the

variance of the ith critical activity in the critical path.

As a part of statistical analysis, we may be interested in knowing the probability of completing the

project on or before a given due date (C) or we may be interested in knowing the expected project

completion time if the probability of completing the project is given. For this, the beta distribution

is approximated to standard normal distribution whose statistic is given below:

σµ−

=xz

where x is the actual project completion time, µ is the expected project completion time (sum of

expected durations of the critical activities) and σ is the standard deviation of the expected project

completion time (square root of the sum of the variances of all the critical activities). Therefore

( )CxP ≤ ) represents the probability that the project will be completed on or before the C time u This

42

can be converted into the standard normal statistic z as:

σµ−

≤=

σµ−

≤σµ− CzPCxP

Example 9.13 Consider Table 1 summarizing the details of a project involving 11 activities

Table 1 Details of Project with 11 Activities

b

Activity Predecessor(s) Duration (weeks)

ot mt pt

A __ 6 7 8

B __ 1 2 9

C __ 1 4 7

D A 1 2 3

E A, B 1 2 9

F C 1 5 9

G C 2 2 8

H E,F 4 4 4

1 E, F 4 4 10

J D, H 2 5 14

K I,G 2 5 8

(a) Construct the project network.

(b) Find the expected duration and variance of each activity.

(c) Find the critical path and the expected project completion time.

(d) What is the probability of completing the project on or before 25 weeks?

(e) If the probability of completing the project is 0.84, find the expected project completion t

Solution (a) The project network is shown in Figure 1.

43

Figure 1 Network for Example 1.

(b) The expected duration and variance of each activity are shown in Table 1.

Table 2 Computations of Expected Duration and Variance

Activity

Duration (weeks) Mean duration

Variance ot mt pt

A 6 7 8 7 0.11 B 1 2 9 3 1.78

C 1 4 7 4 1.00 D 1 2 3 2 0.11 E 1 2 9 3 1.78

F 1 5 9 5 1.78 G 2 2 8 3 1.00 H 4 4 4 4 0.00 1 4 4 10 5 1.00 J 2 5 14 6 4.00 K 2 2 8 3 1.00

(c) The calculations of critical path based on expected durations are summarized in Figure 2. The critical path is A-DI –E-F-J and the corresponding project completion time is 20 weeks.

44

*D1 — Dummy activity

Figure 2 Network for Example 1

(d) The sum of the variances of all the activities on the critical path is:

0.11 + 1.78 + 0.00 + 4.00 = 5.89 weeks.

Therefore 43.289.5 ==σ weeks. Also

( ) =≤ 25xP ( ) .9803.006.2zP43.2

2025xP =≤=

−

≤σµ−

This value is obtained from standard normal table. Therefore, the probability of completing

the project on or before 25 weeks is 0.9803.

(e)We also have ( ) 84.0CxP =≤ . Therefore,

84.043.220CzP

84.0CxP

=

−

≤

=

σµ−

≤σµ−

From the standard normal table, the value of z is 0.99, when the cumulative probability is 0.84.

Therefore,

weeks4.22Cor99.043.220C

==−

45

The project will be completed in 22.41 weeks (approximately 23 weeks) if the probability of

completing the project is 0.84.

Example 2 Consider the data of a project summarized in Table 3.

Table 3 Data of Example 2


ot

mt pt

A _ _ 4 4 10

B _ _ 1 2 B

C _ _ 2 5 14

D A

1

4 7

E A 1 2 3

F A 1 5 9

G B , C 1 2 9

H C 4 4 4

I D 2 2 8

J E , G 6 7 8


(b) Find the expected duration and the variance of each activity.



Solution

46

*D1 — Dummy activity

Figure 3 Network for Example 2

(a) The mean and variance of each activity in the project are shown in Table 4

Table 4 Data of Example 2

Acti

vity

Imm

ediate

predece

ssor(s)

Duration

(weeks) Mean

duration(weeks)

Variance(weeks)

ot

ot

ot

A __ 4 4 1

0

5 1.00

B __ 1 2 9 3 1.78

C __ 2 5 1

4

6 4.00

D A 1 4 7 4 1.00

E A 1 2 3 2 0.11

47

F A 1 5 9 5 1.78

G B,C 1 2 9 3 1.78

H C 4 4 4 4 0.00

1 D 2 2 8 3 1.00

J E, G 6 7 8 7 0.11

(c) The critical path calculations are shown in Figure 4.

From Figure 4, the critical path is C—D1—G—J and the corresponding expected project

completion time is 16 weeks.

Fig .4 Project network with critical path calculations.

(e)The probability of corresponding the project on or before 35 weeks is computed as shown

below.

The sum of the variance of the critical activities = 4+1.78+0.11= 5.89weeks

Therefore, the standard deviation, 43.289.5 ==σ weeks.

Also ( ) =≤ 35xP ( ) .9999.082.7zP43.21635xP =≤=

−

≤σµ−

48

EXERCISES

1.Define 'project', and give some application areas of project management. Explain different phases of

project management.

2. Distinguish between CPM and PERT.

3. Discuss the guidelines for constructing a project network.

4. Define the following: (a) total float, (b) free float, and (c) critical path.

5. What are the time estimates used in PERT?

6. Discuss the cost trade-off in project crashing.

7. Distingush between resource levelling and resource allocation.

8. A construction company has listed down various activities that are involved in constructing a

building. These are summarized along with predecessor(s) details in the table 9.##.

Table 9.##

Activity Immediate predecessor(s)

A __

B __

C A

D B

E A, B

F C, D

G F, B

H E, G

I H, G

J I, F

K J, L

L A

M K

Draw a project network for the above project.

9. Consider the details of a project as shown in the table 9.##:

TABLE 9.##

Activity Immediate predecessor(s) Duration (months)

49

A __ 4

B __ 8

C __ 5

D A 4

E A 5

F B 7

G B 4

H C 8

I C 3

J D 6

K E 5

L F 4

M G 12

N H 7

O 1 10

P J, K, L 5

Q M, N, 0 8

(a) Construct the CPM network.

(b) Determine the critical path.

(c) Compute total floats and free floats for non-critical activities.

10.The following table gives data on normal time and Cost and crash time and cost jor, project:

Activity Normal Crash

Time(days) Cost(Rs.) Time(days) Cost(Rs.)

1-2 6 60 4 100

1-3 4 60 2 200

2-4 5 50 3 150

2-5 3 45 1 65

3-4 6 90 4 200

4-6 8 80 4 300

5-6 4 40 2 100

6-7 3 45 2 80

50

Total cost(Rs.) 470

The indirect cost per day is Rs. 10.

(i) Draw the network for the project.

(ii) Find the critical path.

(iii) Determine minimum total time and corresponding cost.

11.The following table gives data on normal time and cost and crash time and cost for a project.

(a) Draw the network and identify the critical path.

(b) What is the normal project duration and associated cost?

(c) Find out total float for each activity.

(d) Crash the relevant activities systematically and determine the optimum project time and cost.


Time(weeks) Cost(Rs.) Time(weeks) Cost(Rs.)

1-2 3 300 2 400

2-3 3 30 3 30

2-4 7 420 5 580

2-5 9 720 7 810

3-5 5 250 4 300

4-5 0 0 0 0

5-6 6 320 4 410

6-7 4 400 3 470

6-8 13 780 10 900

7-8 10 1,000 9 1,200

Total cost(Rs.) 4,220

Indirect costs are Rs. 50 per week.

12. A project schedule has the following characteristics.

Activity 1-2 1-3 2-4 3-4 3-5 4-9 5-6 5-7 6-8 7-8 8-10 9-10

Time(days) 4 1 1 1 6 5 4 8 1 2 5 7

From the above information, you are required to

(i) Construct a network diagram.

(ii) Compute the earliest event time and latest event time

51

(iii) Determine the critical path and total project duration

(iv) Compute total, free float for each activity.

13.The following Table shows the job of a project with their duration in days. Draw the network and

determine the critical path. Also calculate all the free floats of each activity.

Job 1-2 1-3 1-4 2-5 3-7 4-6 5-7 5-8 6-7 6-9 7-10 8-10

Duration(days) 10 8 9 8 16 7 7 7 8 5 12 10

9-10 10-11 11-12

15 8 5

14. The activities involved in Alpha Garment Manufacturing Company are listed with their time

estimates as in the following table:

Activity Description Immediate

predecessor(s)

Duration(days)

A Forecast sales volume __ 10

B Study competitive market __ 7

C Design item and facilities A 5

D Prepare production plan C 3

E Estimate cost of production D 2

F Set sales price B, E 1

G Prepare budget F 14

Draw the network for the given activities and carry out the critical path calculations.

15. Consider the following data of a project.


a m b

52

A __ 3 5 8

B __ 6 7 9

C A 4 5 9

D C A 4 5

D B 3 5 8

E A 4 6 9

F C, D 5 8 11

G C, D, E 3 6 9

H F 1 2 9





(e) If the probability of completing the project is 0.9, find the expected project completion time.

16. Consider the following table summarizing the details of a project:


a m b

A __ 4 4 10

B __ 1 2 9

C __ 2 5 14

D A 1 4 7

E A 1 2 3

F A 1 5 9

G B, C 1 2 9

H C 4 4 4

1 D 2 2 8

J E, G 6 7 8

K F, H 2 2 8

L F, H 5 5 5

M 1, J, K 1 2 9

N L 6 7 8

53




(d) What is the probability of completing the project on or before 35 weeks'?

(e) If the probability of completing the project is 0.85, find the expected project completion time.

17. Consider the following data of project.


Time (weeks) Cost (Rs.) Time (weeks) Cost (Rs.)

1-2 7 600 4 840

1-3 11 200 9 First week: Rs. 70

Second week: Rs. 80

2-3 10 800 8 1000

2-4 6 500 4 760

2-5 16 100 9 380

3-4 6 200 4 360

3-5 9 500 4 960

4-5 8 300 5 500

If the indirect cost per week is Rs. 300, find the optimal crashed project completion time.

18. Consider the data of a project as shown in the following table:



1-2 8 800 5 950

1-3 5 500 3 700

1-4 9 600 6 1050

2-5 10 900 8 1300

3-5 5 700 3 1100

3-6 6 1200 5 1500

4-6 7 1300 5 1400

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5-7 2 400 1 500

6-7 4 500 2 900

If the indirect cost per week is Rs. 300, find the optimal crashed project completion time.

19. Consider the problem of project scheduling as shown in the following table:

Activity Duration (weeks) Manpower requirement

1-2 4 9

1-3 8 5

2-3 10 7

2-4 6 6

3-4 9 8

3-5 5 12

4-5 7 7

Obtain a schedule which will minimize the peak manpower requirement and also smooth out period-to-

period variation of manpower requirement.

20. Consider the problem of project scheduling as shown in the following table:


1-2 5 8

1-3 4 10

1-4 6 8

2-4 10 10

2-5 4 7

3-5 4 4

4-6 8 12

5-6 9 6

Obtain a schedule which will minimize the peak manpower requirement and also smooth out period-to-

period variation of manpower requirement.

21. Consider the network scheduling problem as shown in the following table:


1-2 6 10

1-3 4 6

55

1-4 9 8

2-6 5 4

3-6 8 7

4-6 2 5

6-7 7 4

Schedule the activities of the project with a maximum limit on the manpower requirement as 15.

22. The table below provides cost and time estimates of seven activities of a project.

Activity Time estimates Direct cost estimates

Normal Crash Normal Crash

1-2 2 1 10 15

1-3 8 5 15 21

2-4 4 3 20 24

3-4 1 1 7 7

3-5 2 1 8 15

4-6 5 3 10 16

5-6 6 2 12 36

(i) Draw the project network corresponding to normal time.

(ii) Determine the critical path and the normal duration and normal cost of the project.

(iii) Crash the activities so that the project completion time reduces to 9 weeks, with minimum

additional cost.

23. Table below shows jobs, their normal time and cost and crash time and cost estimates for a project.

Job Normal time(days) Cost(Rs.) Crash time(days) Cost(Rs.)

1-2 6 1,400 4 1,900

1-3 8 2,000 5 2,800

2-3 4 1,100 2 1,500

2-4 3 800 2 1,400

3-4 Dummy __ __ __

3-5 6 900 3 1,600

4-6 10 2,500 6 3,500

5-6 3 500 2 800

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9,200

Indirect cost for the project is Rs. 300 per day.

(i) Draw the network of the project.

(it) What is the normal duration and cost of the project ?

(iii) If all activities are crashed, what will be the minimum project duration and corresponding cost ?

(iv) Find the optimal duration & minimum project cost.

(Ans. (ii) 20 days, Rs. 15,200 (iii) 12 days, Rs. 16,534.34

(iv) 17 days ; Rs. 15,000.)

24. The following time-cost table (time in weeks, cost in rupees) applies to a project. Use it to arrive at

the network associated with completing the project in minimum time at minimum cost.



1-2 2 800 1 1,400

1-3 5 1,000 2 2,000

1-4 5 1,000 3 1,800

2-4 1 500 1 500

2-5 5 1,500 3 2,100

3-4 4 2,000 3 3,000

3-5 6 1,200 4 1,600

4-5 3 900 2 1,600

(Ans. Crash activity 1-3 by 3 weeks, 4-5 by 1 week and 3-4 by 1 week, giving an optimal 7-weeks

duration project at a cost of Rs. 11,800.)

25. The utility data for a network is given below. Crash the network to minimum project duration and

determine the project cost for that duration.



0-1 1 5,000 1 5,000

1-2 3 5,000 2 12,000

1-3 7 11,000 4 17,000

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2-3 5 10,000 3 12,000

2-4 8 8500 6 12,500

3-4 4 8,500 2 16,500

4-5 1 5,000 1 5,000

(Ans. 10 days, Rs. 74,000.) S. Consider a network problem for which the utility data are given in the

table. Crash systematically the

activities and determine the optimum project duration and cost.

Activity Normal Crash T∆ C∆ CT

∆∆

Ttime

(days)

Cost

(Rs.)

Time

(days)

Cost

(Rs.)

1-2 8 100 6 200 2 100 50

1-3 4 150 2 350 2 200 100

2-4 2 50 1 90 1 40 40

2-5 10 100 5 400 5 300 60

3-4 5 100 1 200 4 100 25

4-5 3 80 1 100 2 20 10

Total 580

Indirect cost Rs. 70 per day. (Ans. 11 days; Rs. 1,760.)

26. The utility data for a network is given in the table. Crash the network to minimum possible

duration, and find the corresponding optimal project cost. The total normal cost of the project is Rs.

9,00,000 only.

Activity Duration (weeks) Cost slope

(Thousands of Rs.) Normal Crash

1-2 4 2 7

1-3 7 4 4

58

2-3 7 4 3

2-4 5 3 12

3.-5 6 3 6

4-5 4 1 8

4-6 10 6 10

5-6 6 6 __

5-7 8 6 10

6-8 6 4 12

7-8 5 2 7

10. The activity data for a project is given in the table. Find the

(a) normal duration and the normal cost.

(b) optimal duration and the minimum total cost.

(c) minimum duration and the associated project cost.

Activity Normal Crash T∆ C∆ CT

∆∆

Time

(weeks)

cost

(Rs.)

Time

(weeks)

cost

(Rs.)

1-2 8 7,000 3 10,000 5 3,000 600

1-3 4 6,000 2 8,000 2 2,000 1,000

2-3 0 0 0 0 0 0 0

2-5 6 9,000 1 11,500 5 2,500 500

3-4 7 2,500 5 3,000 2 500 250

4-6 12 10,000 8 16,000 4 6,000 1,500

5-6 15 12,000 10 16,000 5 4,000 800

5-7 7 12,000 6 14,000 1 2,000 2,000

6-8 5 10,000 5 10,000 0 0 __

7-8 14 6,000 7 7,400 7 1,400 200

7-9 8 6,000 5 12,000 3 600 2,000

8-9 6 6,000 4 7,800 2 1,800 900

Total 86,500 Total 1,15,700

Indirect cost Rs. 1,000 per week.

(Ans. (a) 41 days; Rs. 1,27,500 (b) 30 days; Rs. 1,25,700

(c) 25 days; Rs. 1,32,700.)

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27. (a) The following data pertains to a CPM network. It is desired to compress the project to the least

possible duration day by day and estimate the extra cost.

Activity Normal time Crash time Cost slope

1-2 3 2 700

1-3 7 4 200

2-3 5 3 100

2-4 8 6 200

3-4 4 2 400

(b) If there is an indirect cost of Rs. 800 per day, what will be the least project duration and the

associated cost ? (Ans. (a) 8 days, Rs. 2,100;

(b) 9 days, Rs. 8,400.)

28. A small maintenance project consists of jobs in the table below. With each job is listed its normal

time and a minimum or crash time in days. The cost in Rs. per day of each job is also given.

Activity Normal time Crash time Cost slope

1-2 9 6 20

1-3 9 5 25

1-4 15 10 30

2-4 5 3 10

3-4 10 6 15

4-5 2 1 40

(a) What is the normal project length and the minimum project length ?

(b) Determine the minimum crashing cost of schedules ranging from normal length down to, and

including, the minimum length schedule. That is, if L = length of the schedule, find the costs of

schedules which are L, L - 1, L - 2 and so on.

(c) Overhead costs total Rs. 60 per day. What is the optimum length schedule in terms of both

crashing and overhead cost? List the scheduled duration of each job for your solution.

(Ans. (a) 20 days, 12 days;

(b) 12 days, Cc = Rs. 335, C7, = Rs. 1,055

(c) 15 days.)

29. The utility data fora network is given below. The activity durations are in days and the cost in

60

rupees. The indirect cost per day is Rs. 250. Determine the optimum project schedule.


Time (weeks) cost (Rs.) Time (weeks) cost (Rs.)

1-2 4 600 2 800

1-3 2 500 1 900

2-4 6 1,000 3 1,750

2-5 4 1,200 4 1,200

3-5 5 1,000 3 1,200

3-7 10 2,500 5 3,500

4-5 5 1,300 5 1,300

5-6 8 2,000 6 2,100

5-7 0 0 0 0

6-8 7 2,000 7 2,000

7-8 8 1,600 5 1,780

(Ans. Crash 1-2 by 2 days, 2-4 by 1 day, 5-6 by 2 days;

Rs. 13,000, 25 days.)

30. Following table gives the activities in a construction project and other relevant information:

Activity Predecessor Time (days) Direct cost (Rs.)


A __ 4 3 60 90

B __ 6 4 150 250

C __ 2 1 38 60

D A 5 3 150 250

E C 2 2 100 100

F A 7 5 115 175

G D, B, E 4 2 100 240

Indirect costs vary as follows

Days 15 14 13 12 11 10 9 8 7 6

Cost (Rs.) 600 500 400 250 175 100 75 50 35 25

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(a) Draw an arrow diagram for the project.

(b) Determine the project duration which will return in minimum total project cost.

31. A Project has 7 activities. The relevant data is given below.

Activity Predecessor Time (days) Direct cost (Rs.)


A __ 7 5 500 900

B A 4 2 400 600

C A 5 5 500 500

D A 6 4 800 1,000

E B, C 7 4 700 1,000

F C, D 5 2 800 1,400

G E, F 6 4 800 1,600

The project manager wishes to complete the project in the minimum possible time. However, he is not

authorised to spend more than Rs. 5,000 on crashing. Suggest the least cost schedule.

Assume that there is no indirect or utility cost. (Ans. 14 weeks, cost: Rs. 4,600.)

32. The time and cost estimates of different activities of a project and their precedence relationships are

given below.

Activity Predecessor Time (weeks) Direct cost (Rs.)


A __ 6 4 10,000 14,000

B __ 4 3 5,000 8,000

C A 3 2 4,000 5,000

D B 8 3 1,000 6,000

E B 14 6 9,000 13,000

F C, D 8 4 7,000 8,000

Overhead costs amount to Rs. 1,000 per week.

(i) Draw the network and show the critical path.

(ii) Crash the network to the minimum possible duration. What will be the critical activities after such

62

crashing? (Ans. (i) B-D-F (ii) 10 weeks; A-C-F, B-D-F.)

33. The basic cost-time data for jobs in a project are giver. below.

Job Normal Crash Crashing cost

per day(Rs./day) Time (days) Cost(Rs.) Time(days) Cost(Rs.)

A 3 140 2 210 70

B 6 215 5 275 60

C 2 160 1 240 80

D 4 130 3 180 50

E 2 170 1 250 80

F 7 165 4 285 40

G 4 210 3 290 80

H 3 110 2 160 50

The activities dependencies are as below:

(i) A, B, C are starting activities (ii) Activities D, E and F can start when A is completed, (iii) C can

start after B and D are completed, (iv) H can start after C and E are completed, (v) G, H and F are final

activities. Draw the network and indicate the critical path and the total time of completing the project.

If the project is to be completed in 8 days, what is the minimum cost to be incurred? Indicate this

cheapest schedule.

34. A list of activities along with their precedence requirement, normal time and cost and crash time

and cost are given in the table below.

Activity Predecessor Normal Crash

Time (days) Cost(Rs.) Time (days) Cost(Rs.)

A __ 4 300 2 450

B A 9 600 5 960

C A 6 620 4 780

D B 4 320 3 395

E B, D 6 1440 3 1,980

F C, D 4 350 2 470

G E, F 3 270 2 335

(a) Draw the network.

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(b) What are the normal and crash costs and durations of the project ? How many days will be saved

and what will be the project cost if all the activities are crashed to the maximum possible extent ? It is

expected that earlier completion of the project will result in additional profit of Rs. 50 per day. The

indirect cost is Rs. 80 per day.

(c) Analyse the project for optimum duration cost.

Documents

Chap 9 Project Managementdocshare02.docshare.tips/files/20953/209533229.pdfCPM and PERT are special applications. Network It is the graphic representation of logically and sequentially