Project Management for m

8/3/2019 Project Management for m

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PROJECT MANAGEMENT

Project Management is the act of organizing, planning, scheduling, controlling and

directing a scheme or programme of activities with the purpose of ensuring that all

activities related to the scheme are completed on schedule. The word ³project´ can be defined as a set of activities or progammes which is being undertaken in order

to achieve some predetermine goals. It could be a small set of activities divided

into tasks which take time and resources to get finished. The execution of projects

requires that we know in advance the likely amount of time and resources it will

consume. There are different techniques of managing projects. One of the most

common methods is the PERT ; also there is CPM .

Both techniques are aspects of Network Analysis. There are also other techniques

such as Resource Scheduling; PERT/COST, e.t.c

The last two mentioned deal with how to minimize cost, time, space, personnel,

machinery, money and other resources.

Another important technique is called Crashing. In Crashing, especially in PERT

the project analyst is interested in working under pressure to meet the targets. It

should be noted that the PERT analysis is designed to compute the probability of

having completion time of a project being behind schedule.

The project Manager therefore tries to trade off/or strike a balance between cost

saving and meeting set targets by time reduction below a given schedule. The

analyst must ensure that the benefit of crashing far outweighs the attendant cost.

It is important that the organization should have a sound knowledge of techniques

of managing projects and that is what these are designed to achieve.

PERT is an acronym for Project Evaluation and Review Technique while CPM

denotes Critical Path Method. In CPM, project management is deterministically

analyzed. The sister concept, PERT is moored on stochasticity. The latter statement finds relevance in Murphy¶s Law as follows:

1. It is not as easy as it looks

2. It will cost more than you estimated



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3. It will take more time than you think

4. If anything will go wrong, it will.

The laws apply to every situation. Much earlier CPM was considered to be

restricted or to be concerned with only cost and time trade-off. Based on the

premise, the technique was grossly involved in float computations, resource

leveling and other collateral issues. By resources we mean the 5Ms - Men,

Material, Money, Method, Machines.

On the other hand the PERT technique is concerned with probabilistic

determination of activity times particularly the earliest start time of a project.

Therefore the decision problem of the project manager would be as follows:

1. To consider carefully the length of time the project will take.

2. The determination of the Earliest Start Time (EST), Latest Start Time

(LST), Earliest Finished Time (EFT), Latest Finished Time (LFT) of

each activity of the scheme of work.

3. To determine the most critical task to be done if the project is to be

completed on schedule.

Drawing from the experience of Murphy¶s Law, the project manager may wish to

have 3 distinct estimates of time for each activity namely:

1. Optimistic time (a)

2. Pessimistic time (b)

3. The most likely time (m)

Today, the dividing line between PERT and CPM has vanished and both of themare now done concurrently and that is the reason why in some texts it is written as

PERT/CPM. In other words, the two techniques have fused together. One

remarkable comment about PERT is that it is built on the philosophy that the

earliest start time of a project is normally distributed. In point of fact, it is not

exactly a normal distributed function. Rather it is exactly a beta () distribution.



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But for the purpose of mathematical tractability, analysts often take the beta

distribution as a normal distribution.

CPM

We shall define some basic methods of network analysis. These include:

1. Events: It is a particular undertaking at a point in time. It is represented by

circle.

2. Activity: This denotes an aspect or a component of a job. It is represented by

an arrow.

3. Dumm y activity: It is represented by a dotted arrow. It is one that does notinvolve cost and time expenditure.

To make the concept clearer, we shall take a hypothetical building project. We will

start by enumerating the activities associated with building a house; determine their

various durations as well as EST, EFT, LST, and LFT. We shall also examine the

predecessors in order to ascertain the precedence diagram.

Table 1: Precedence Data for Building a House

Activity Preceded By

A Building Plan

B Procure Sand and Gravel

C Clear Site

D Process C of O A

E Lay Foundation C

F Erect Building Walls D, B, E

G Make Frames D, B, EH Fix Frame G

I Roofing F, H

J Plastering I

K Electrical/Plumbing J

L Ceiling J



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M Inspection K

N Painting M, L

O Commissioning N

Usually a precedence of the events has to be specified so that a Network diagram

can be constructed as follows:

Figure 1: Network diagram for building construction

Fig (1) depicts the network diagram for the data given in table 1. It is a chain of

events and activities linked together.

Notations



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Figure 2: Events Window

Each event is annotated on a window to explain the EST, LST, EFT, LFT in

window 1 and 2 respectively. The arc or branch specifies the activity duration and

has been accordingly annotated in the network. The first step is to inspect the precedence diagram and sketch the network. Then we start with the forward pass,

beginning with event 0. Then we go from event to event according to the sequence

of numbering and in the process we start inserting the EST for the first window of

each event. This process is called the forward pass. When we get to event 11, we

will reverse and start filling the LST for the second window of each event. Please

observe that the EST of a head event is equal to the EFT of a tail event; and LFT of

a tail event is equal to LST of a head event. These statements are evident from

figure 2. Observe also that inspection is a dummy activity because it does not

consume resources so to say.

Computation of Network Time Under CPM

At this stage we shall take it that the activity times are known. Accordingly, we

reproduce a modified version of Table 1 which is needful for subsequent

computations.

Table 2: Precedence data incorporating activity ti mes

Activity Preceded By Time (Week)

A Building Plan 3

B Procure Sand and Gravel 1

C Clear Site 1

D Process C of O A 1

E Lay Foundation C 1



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F Erect Building Walls D, B, E 4

G Make Frames D, B, E 3

H Fix Frame G 1.5

I Roofing F, H 1

J Plastering I 2K Elect/Plumbing J 2.5

L Ceiling J 2

M Inspection K

N Painting M, L 2

O Commissioning N 1

1

0

0 0

3 3

2

1 1

3

4 4

4

7 7

5

8.5 8.5

6

9.5 9.5

8

14 14

11

17 17

Commissioning

1 week

(O

Erect buildg wall

4 wks

Roofing

1 wkPlastering

2 wks

Makeframes3 wks

Fix frames1.5 wks

Buildingplan3 wks

Process C of O1 wk

Sand/ gravel 1 wk

Clear land1 wk

Foundation1 wk

El2.

Figure 3: Network Diagram



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As can be seen in the Figure, the activity duration is stated on each arc or branch of

the network. The circles, as explained in figure 2, indicate the events.

The first step after sketching the network diagram is the use of the forward pass to

compute the network EST and then fix them in the first side window of eventcircles. When this has been completed, a backward pass is used to compute the

network LST and we use it to fill the second side window of the event circles.

Notice that in the case of forward pass for event 0, the EST is zero. For event 2,

under forward pass EST is 3 weeks, i.e. the activity duration. For event 3, using the

forward pass, there are three possibilities:

i. The sum of network activities of building plan and processing of C of O (3 +

1 = 4) could be entered in the first side window of event 3.

ii. Also for the activity C we could enter 1 in the first side window of event 3.

iii. The sum of activity of clear site and lay foundation (1+1 = 2) could be

entered in the first side window of event 3.

The sum of activity of building plan and processing of C of O, i.e 4, is the highest

and accordingly entered into the first side window of event 3. The computations of

the rest EST follow the same procedure.

In the case of backward pass, we normally start at the last event (11) and walk backward. For instance, the LST of commissioning event is 17 1 = 16

To obtain the LFT for ceiling, we subtract the 2 weeks for painting from 16 and we

have 14. However, for the LFT of plastering, there are two possibilities but we

shall take the right decision by passing through the critical path. In order to

determine the path, (0 + 2.5) which represents the sum of activity times for

inspection and electrical/plumbing, 14 ± 2.5 = 11.5. However, passing through the

path of ceiling activity gives (14 ± 2 = 12). The path with higher value is selected

and the outcome registered in the right side window of event 7.

From what we have done so far on backward pass, it is evident that the optimal

step is to do the subtraction along the critical path. The rest computations are

similarly obtained and the comprehensive network diagram is drawn as depicted in

figure 3 above.



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Float Calculation for CPM

Having established the network for the building project, we need to compute the

float under critical path method. And to do that, we need the following formulae:

Slacks, S = LST ± EST

Total Float TF = LFT EST D

Free Float FF = EFT EST D

Independent Float IF = EFT LST D

Where:-

D = Activities Duration

EST = Earliest Start Time

LST = Latest Start Time

EFT = Earliest Finish Time

LST = Latest Start Time

TF = Total Float

FF = Free Float

IF = Independent Float

Table 3:

Activities EST LST EFT LFT AD(D)

SLACK (S)

TF FF IF

A 0 0 3 3 3 0 0 0 0

B 0 0 4 4 1 0 3 3 3

C 0 0 1 3 1 0 2 0 0

D 3 3 4 4 1 0 0 0 0



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E 1 3 4 4 1 2 2 2 0

F 4 4 8.5 8.5 4 0 0.5 0.5 0.5

G 4 4 7 7 3 0 0 0 0

H 7 7 8.5 8.5 1.5 0 0 0 0

I 8.5 8.5 9.5 9.5 1 0 0 0 0J 9.5 9.5 11.5 11.5 2 0 -4 0 0

K 11.5 11.5 14 14 2.5 0 0 0 0

L 11.5 11.5 14 14 2 0 0.5 0.5 0

M 14 14 14 14 0 0 0 0 0

N 14 14 16 16 2 0 0 0 0

O 16 16 17 17 1 0 0 0 0

The CPM has been used to compute all the floats/slack in the Network. It may be

instructive to ascertain the practical relevance of the floats/slack calculated. In this

regard they represent a kind of buffer for adjustment for delays incurred in some

tardy task or activities. Thus they enable the project Manager to make up for loss

times so that the network critical path is not delayed.

Programme Evaluation Review Technique (PERT)

The CPM approach deals with deterministic way of analyzing project network. The

PERT version is concerned with stochastic (probabilistic) method of estimating the

network Earliest Start Time. With this approach, we are expected to have 3different probabilistic estimates of time namely:-

1) Optimistic time (a)

2) Pessimistic time (b)

3) Most likely time (m)

The estimate of the expected time is given by:

=

The variance is given by: =



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=

This network earliest start time (EST) is assumed to be normally distributed. In

other words, if we have a sequence of expected time for all activities in the

network, it is possible to average it in form of class intervals and then obtain the

frequency of the scores found in each interval.

The mid-class mark is plotted against the frequency to obtain a normal distribution

as follows:

Fig.4: Normal Distribution of (Network Earliest Start Time)

In reality, the actual graph is indeed a ballpark or approximation of normal

distribution in the sense that it is actually -distribution; and -distribution is a

class of Wiebull distribution with appropriate values of scale and shape parameters.

The approximation makes it possible or feasible for us to use a Normal

Distribution table for such computations. This is because the objective of using a

Normal Distribution is to enable us compute the probability of the network time

0



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being behind a given schedule. Accordingly, in practice, the project manager

works conscentiously to meet deadline given by the clients.

The points sketch above is the basic philosophical foundations of PERT.

Estimation of Probability Distribution

Recall

E(x) =

If = 0, then is called an unbiased estimator.

However if 0), then is a biased estimator.

Now

E(x) = . f(x)



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= . p(x)

where f(x) = p(x) = probability density function

Variance of X, = E

= E ( 2x + )

= E( ) 2E(x) + E( )

= E( ) 2. + ( )

= E( ) 2 + ( )

= E( )

Note:

(1) In this application the variance has been given as:

=

(2) The Central Limit Theory (CLT) can be evoked and used in conjunction

with the fact that a and b represent pessimistic and optimistic times

respectively to compute the variance for the various activities of the network

under review or consideration.

Table 4:

Activities Schedule

time ( )

Expected

time ( )

Variance

( )



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A 4 4.2 0.25

B 1.6 1.8 0.11

C 2 1.8 0.06

D 1.6 1.7 0.11

E 2 1.9 0.11F 5.1 5 0.25

G 3 3.6 0.08

H 2 2.5 0.17

I 2 1.8 0.11

J 2 2.7 0.11

K 3 3.1 0.06

L 3 2.4 0.03

M 0 0 0

N 3 2.7 0.06

O 2 1.4 0.01

The values of a, b, and m can be estimated for every activity of the network.

The usual way to obtain this data is to gather information pertaining to the

various activity times of similar building projects earlier executed. If the

number of such similar projects is 10 for example, the least time taken for every

activity of the project can be searched from the tabulation and be used as an

estimate of optimistic time, a. correspondingly, the pessimistic time, b willrepresent the highest time for the activity in the tabulation. And the most likely

time, m will be the average of the listings in the appropriate row.

For the C of O example, the activity is represented by D. Along row D the least

entry there is 1 and that represents the optimistic time. Also the maximum

number there is 3 and that represents the pessimistic time. Therefore to compute

the most likely time we sum all the entries in row D and divide by 10 and this

gives the value of m. we have used the concept of mini max and maxi max as

our decision rule in the foregoing analysis.

Table 5:

Activities

A 3 3.5 5 4 4.5 6 3.8 4 3 3.2



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B 1 1.5 2 2.5 1 3 1 2 2.5 1

C 2 2 2.5 1 1.5 2.5 1 1.5 2 1.5

D 1 1 1.5 2 1.5 1.5 1.3 2 2.1 3

E 2 1.5 2 1 1.5 3 2.5 1 1.5 2

F 5 6 4 4.5 5 5.5 7 4 4 4G 3.5 3.5 3 4 3 4.5 4.7 3 3 2

H 2 2 1.5 3 3.5 1.5 3 2 4 1

I 1.5 1.5 2 3 2.5 1 1 1 2 1

J 3 3 2 2 2 2.5 3 2 2.5 4

K 3 3 3 2.5 2.5 2.5 3 4 4 2

L 2 2.5 3 2.5 3 2 2 2.5 2 2

M 0 0 0 0 0 0 0 0 0 0

N 2.5 2.5 3 3.5 3 2 2 2 3 3

O 1.5 2 1.5 1.5 1.5 1 1 1 1.5 1.6

With these data, the values of a, b and m can be computed for each activity of

the project.

The BLU E which means Best Linear Unbiased Estimator could be expressed

as:

=

Table 6: Project Expected Ti mes

Activities A b m =

A 3 6 4.04 4.19 4B 1 3 1.75 1.83 1.6

C 1 2.5 1.75 1.75 2

D 1 3 1.69 1.79 1.6

E 1 3 1.81 1.87 2

F 4 7 4.9 5.13 5

G 3 4.7 3.47 3.60 3



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H 1.5 4 2.4 2.52 2

I 1 3 1.7 1.80 2

J 2 4 2.5 2.66 2

K 2.5 4 3.0 3.08 3

L 2 3 2.35 2.40 3M 0 0 0 0 0

N 2 3.5 2.65 2.68 3

O 1 1.6 1.41 1.37 2

Table 7:

Events Expected

time ( )

Corresponding

variance for

EST = sum of

activities

variances

Standard

deviation

()

Schedule

time ( )

given

Z = P(z)

0 0 0 0 0 0 0 01 3 0.25 0.500 4 1 0.2 0.4207

2 1 0.06 0.245 2 1 4.082 0.00003

3 4 0.36 0.600 6 2 3.33 0.0012

4 7 0.44 0.660 9 2 3.03 0.0012

5 8.5 0.61 0.780 10.5 2 2.56 0.0052

6 9.5 0.72 0.849 9 -0.5 -0.589 0.2810

7 11.5 0.83 0.911 15 3.5 3.842 0.0010

8 14 0.89 0.943 18 4 4.242 0.00003

9 14 0.89 0.943 18.5 4.5 4.772 0.0000310 16 0.95 0.975 20 4 4.103 0.00003

11 17 0.96 0.980 21 4 4.082 0.00003

Some Notes on Normal Distribution



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Normal distribution arises when a given data set is arranged into class intervals

of predetermined class width and frequency of each class interval is determined

from the data set.

When the frequency is plotted against the mid-class mark, gives a frequencydistribution of the sort

Fig. 5: Normal Distributions with various scale and shape parameters

Most times the distribution obtained may be biased if it is skewed to the left or

to the right. Notice that in Fig. 5 above, distribution 1 is skewed to the left hand

and distribution 2 is unbiased, distribution 3 has peakedness and therefore

biased.

Distribution 4 is skewed to the right, but all are normal distributions. The only

difference is that there is biasedness in 1, 3, and 4 respectively. Therefore

peakedness and skewness are all attributes of biasedness.

f

Mid Mark



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The standard deviation, gives a measure of variability from central tendency

which is the mean (expectation). In the analysis under consideration, focus shall be

sharpened by the search for standard normal variable expressed as:

Z =

Ultimately, we seek for p(z > )

Figure 6:

We need to compute the Z variables and then access the normal probability

table and read off the associated probability for being behind .

Observe that under the normal probability table, the vertical column or the left

contain the Z variables while the top columns give the decimal associated with

the Z variable.

E.g. suppose we wish to compute

P(Z > ) = P(Z > 1.550)

1 (Z)



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Access the table and read off the column 1.5 and get the remaining decimal

figures from the top headings just like it is with normal logarithm table.

Notice that normal distribution values are presented in different ways. Its

presentation structures have to be properly understood i.e.

Figure 7:

OR

0



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Figure 8:

Figure 9:

To compute the probability for activity A being behind schedule (see Table Y)

P(0 < z < 0.2) = 0.5000 ± 0.0793

Z = 0.4

0

0



20

= 0.4207

Observe further that in some statistical table the probability can be read off

directly without subtraction. Students should therefore be careful in studying

the table to be used so as to know what to do.

Assignment

Question:

(a) Define the followings

(i) Free float

(ii) Independent Float

(iii) Network analysis

(b) Given the precedence data below.

Table 8: Hou se Constr uction Data

S/ No Activities Must be

preceded by

Optimistic

time (a)

Pessimistic

time (b)

Most likely

time (m)



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activity

1 Lay foundation 3.0 6.0 4.0

2 Excavate for drain 1.5 2.0 1.7

3 Lay drains 2 1.5 2.1 2.0

4 Brick work 1, 2, 3 13.0 17.0 15.05 Plastering 4 4.0 6.0 4.8

6 Carpentry 4 7.2 9.3 8.4

7 Roofing 6 9.0 11.0 10.0

(c) Given the schedule time is as follows:

Table 9: Events Sched ule

Events Schedule Time

1 0

2 2

3 5

4 18

5 26

6 37

Required :-

(i) Sketch the network and indicate the critical path.

(ii) Estimate the probability of the network events being behind schedule

Solution:

1(a)

i. F ree F loat :-



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This is the amount of time an activity can be delayed without affecting the

commencement of a subsequent activity at its earliest start time, but may

affect float of a previous activity.

ii. I

ndependent F

loat :-

This is the amount of time an activity can be delayed, when all the preceding

activities are completed as late as possible and all succeeding activities

completed as early as possible. Independent float therefore does not affect

the float of either preceding or subsequent activities.

iii. Network Analysis:-

Network Analysis is a generic term for a family of related techniques

developed to aid management to plan and control projects.

1(b)



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(i)

(ii) Estimate the probability of the network event being behind schedule

Calculation of expected times are as depicted here under:

, Lay foundation =

=

= 4.17

, Excavation Drains =

=

= 1.72



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, Lay Drains =

=

= 1.93

Brick work =

=

= 15

, Plastering =

=

= 4.87

, Carpentry =

=

= 8.35

, Roofing =

== 10.0

For Variance, the calculations are as shown below:



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Lay foundation, = = = 0.25

Excavate for drains, = = = 0.007 0.01

Lay drains, = = = 0.01

Brick work, = = = 0.44

Plastering, = = = 0.11

Carpentry, = = = 0.12

Roofing, = = = 0.11

For Standard deviation:

Lay foundation, = = = 0.5

Excavate drains, = = = 0.1

Lay drain, = = = 0.1

Brick work, = = = 0.67

Plastering, = = = 0.33



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Carpentry, = = = 0.35

Roofing, = = = 0.33

Summarily, the expected times and the variances are computed and depicted in

Table 10.

Table 10: S ummary of Com putations

S/No Activity Expected

duration ( )

Variance ( ) Standard

deviation ()

1 Lay foundation 4.17 0.25 0.5

2 Excavate drains 1.72 0.01 0.1

3 Lay drains 1.93 0.01 0.1

4 Brick work 15.00 0.44 0.67

5 Plastering 4.97 0.11 0.33

6 Carpentry 8.36 0.12 0.35

7 Roofing 10.00 0.11 0.33

Now add the appropriate expected duration and variance to obtain the table below.

Table 11:

Events Expected Earliest time

from network

Corresponding variance

of earliest time = sum of

activity variance

1 0.00 0.002 1.72 0.01

3 4.17 0.25

4 19.17 0.69

5 23.53 0.81

6 37.53 0.92



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Now using the above table, we can then compute the probability of each event

being behind a given schedule.

= = = 0

= = = 2.80

= = = 1.06

= = = 1.41

= = = 1.70

= = = 0.55

Table 12:

Event Schedule

time (S)

Network

Earliest time

(t)

Normal z-

value (z =

)

Probability of earliest time

being greater than schedule

time for normal table

1 0 0.00 0.0000

2 2 1.72 2.80 0.0026

3 5 4.17 1.66 0.0485

4 18 19.17 -1.41 0.9207

5 26 27.53 -1.69 0.9543

6 37 37.53 -0.54 0.7054

Further Consideration on CPM

Basic Ti me Analysis

Some definition:



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We shall use a simple network to define the following terms;

y Earliest Start time (Head event): EST ( = Earliest Finish Time: EFT)

y Latest Start time (Head event): EST ( = Latest Finish Time: LFT)

y Earliest Start time (Tail event): EST ( = Earliest Start Time: EST)

y Latest Start time (Tail event): LST ( = Latest Start Time: LST)

Figure 10 shows the development of a network.

Fig. 10

Critical Path ( F ig ure 11 refers)



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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 duration times. There

may be more than one critical path in a network and it¶s possible for the critical

path to run through a dummy.

Fig. 11: Critical path

y EST of a Head Event ( F ig ure 12 refers)

Fig. 12: Forward Pass



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² The earliest start time (EST) of a head event is obtained by adding

onto the EST of the tail event, the linking activity duration starting

from event 0, time 0 and working through the network.

² Where two or more routes arrive at an event, the longest route time

must taken as, for example, at activity F, D, and E but route B, D

takes the longer route.

² The EST in the finish event (event 5) is the project duration

y L ST ± Latest Start Ti me ( F ig ure 13 refers)

Fig. 13: Backward pass

To compute the above we need to need to work backwards and that is why it has

been termed Backward Pass.

² Starting at the finish event No 5, insert the LST (i.e day 9) and work

backwards through the network deducting each activity duration

from the previously calculated LST.



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² Where the tails of activities B and C join event No 1, the LST for C

is day 3 and the LST for B is day 1. The lowest number is taken as

the LST for event No 1 because if event No 1 occurred at day 3 then

activities B and C could not be completed by day 7 as required and

the project would be delayed.

F loat com putations ( F ig ure 14 refers)

Figure 14(a)



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Figure 14(b)

(i)

(ii) Total Float = Latest Headtime (LFT) Earliest Tailtime (EST)

Activity Duration(D)

= 50 10 10

= 30 days

(iii) Free Float = Earliest Headtime (LFT) Earliest Tailtime (EST)

Activity Duration

= 40 10 10

= 20 days

(iv) Independent Float:



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= Earliest Head time (EFT) Latest Tailtime (LST)

Activity Duration

= 40 20 10

= 10 days

Please note further as follows:

(a) The most important type of float is Total F loat . This is because it is

involved with the overall project duration. On some occasion the term F loat

is used without qualification. In such cases assume that Total F loat is

implied.

(b) The total float can be calculated separately for each activity but it is often

useful to find the total float over chains of non-critical activities between

critical events. For example in Figure 11 the only non-critical chain of

activities are C, E for which the following calculations can be made.

Non-critical Time Time Total Float

Chain Required Available over chain

C, E 3 + 1 = 4 days 7 1 = 6 days = 2 days

i.e. =

Total float = 6 4

= 2 days

Worked Example on Float calculations

Consider the activities involved in the building of a boat as sketched out below:

Activity Preceding

Activity

Activity

Description

Activity Duration

A Design Hull 9

B Prepare Boat Shed 3



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C A Design mast and

mast mount

8

D A Obtain Hull 2

E A Design Sails 3

F A Obtain mast mount 2G C Obtain mast 6

H C Design Rigging 1

J B, D Prepare Hull 4

K F, J Fit mast mount to

hull

1

L E, H, G, K Step mast 2

M E, H Obtain sails and

rigging

3

N L, M Fit sails and

rigging

4

Sol ution:

The network is sketched below. The critical path is shown with two transverse

lines (ll): A, C, G, L, N with a duration of 20 days.



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Activity EST LST EFT LFT D TF FF IF

0 0 9 9 9

B 0 0 11 18 3 15 8 8

9 9 17 17 8

D 9 9 11 18 2 7

E 9 9 18 22 3 10 6 6

F 17 17 19 22 2 3

17 17 23 23 6

H 17 17 18 22 1 4

I

J 11 18 19 22 4 7 4

K 19 22 23 23 1 3 1

23 23 25 25 2

M 18 22 25 25 3 4 4

25 25 29 29

* = Critical Activities

Computation of floats along Non-critical paths

Non-critical chain Time required Time available Total float over

chain

B, J, K 8 23 15

D, J, K 7 14 7

F, K 3 6 3

E, M 6 16 10

H, M 4 8 4

E, DUMMY 3 14 9

H, DUMMY 1 6 5

Slack

This is the difference between the EST and LST for each event. Strictly it does not

apply to activities but on occasion the terms are confused and unless expressly

stated event slack is always implied. Events on the critical path have zero slack.



36

TIME-COST TR ADE ±OFF

Oftentimes in project management we are confronted with the decision problem of

considering the advantage or disadvantage of reducing project duration and pay

extra cost for such reduction. If there appears to be a perceived advantage, then we

make the decision to crash. Crashing is a term applied to the method or technique

of reducing project or activity schedule time by employing more resources as, for

example, deploying more personnel, use of overtime or perhaps contracting aspects

of the work at higher cost in order to meet set target time. Sometimes this decision

becomes quite expedient.

To illustrate the concept sketched-out in the foregoing presentation. Let¶s consider

the data relating to two-machine maintenance project. It is assumed that the

management of this project has leaned or depended on past experience to give

perhaps what appears to be a dependable estimate of time so that there seems to be

no need to provide three estimates of time as we had done in the past. By three

estimates of the same activity we mean the following:

² Optimistic time, (a)

² Pessimistic time, (b) and

² Most likely time, (m)

The data are presented in what follows:

Table 16: Activity List for a Two-machine Maintenance Project

Activity Description Immediate Predecessors Expected Time (days)

A Overhaul machine 1 7

B Adjust machine 1 A 3

C Overhaul machine 11 6

D Adjust machine 11 C 3



37

E Test system B, D 2

Let¶s sketch the network

Fig. 17: Network of a two-machine Maintenance Project

Next we compute the critical path for this project. The detailed computation is

laid out below in a table.

Table 17: Activity Schedule for the Maintenance Project Earliest Latest Earliest Latest Slack Critical

Activity Start Start Finish Finish (LST EST) Path

Time (EST) Time (LST) Time (EFT) Time (LFT) for each activity

A 0 0 7 7 0 yes

B7

7

10 10 0 yes

C 0 0 6 7 1 No

D 6 7 10 10 1 No

E 10 10 12 12 0 yes

A 7

C

6

B

3

D3

E

2

1

2

3

4 5

0 0

6 7

10 12

7 7

10 12



38

The penultimate column indicates which path that is critical. Obviously it is the

path that has zero entries i.e. A ± B ± E or nodes 1 ± 2 ± 4 ± 5. This suggests that

the project completion time is:

7 + 3 + 2 = 12 days

Crashing Activity Times

Suppose we are confronted by supervening circumstances demanding that we cut

short the project duration to 10 days instead of the normal 12 days. Obviously this

would entail use of either overtime or more manpower with the concomitant

additional cost. To do this, we shall need some information such as those presented

below:

1. Estimated activity cost under the normal or expected activity time

2. Activity completion time under maximum crashing (that is, shortest possible

activity time)

3. Estimated activity cost under maximum crashing

Then define the following:

= normal activity time

= crashed activity time (at maximum crashing)

Cn = normal activity cost

Cc = crashed activity cost (at maximum crashing)

Set M = -

Where M = maximum possible activity time reduction due to crashing.

Also set

K =



39

K is called the crashing cost per activity.

Suppose, for example, that activity A scheduled to be finished in 7 days as is seen

in table 2 at a normal cost of N5,000. If the maximum crash activity time of 4 days

is permitted at a cost of N8,000, then we can carry out the following analysis for activity A.

= 7

= 4

Cc = 8,000

Cn = 5,000

MA = 7-4 = 3days

at crashing cost of

= N1,000 per day

We can sketch a graph to illustrate the analysis.

Activity Cost (N)

5000

6000

7000

8000



40

Activity time (days)

Fig. 18: Time-cost Relationship for Activity A

The next questions are

- Which activities to crash

- How much should the activities be crashed

In order to meet the 10days project completion deadline at minimum cost?

To answer these questions we need to do the following analysis:

Table 18: Crashing Details

Normal Crash Total normal Total crash Max. Crash Crash cost

Activity time time cost cost days per day

A 7 4 5000 8000 3 1000

B 3 2 2000 3500 1 1500

C 6 4 5000 9000 2 2000

D 3 1 2000 5000 2 1500

E 2 1 3000 5500 1 2500

N 17,000 N31, 000

4 5 6 7



41

Examination of the last column reveals that A has the least crash cost per day. We

can thus reduce activity A by two-days so that we can complete the Job in 10daysat the minimum additional cost of N2, 000.

Exercise

(i) Construct a PERT/CPM network for a project having the following

activities.

Table 19: Activity List

Activity Immediate predecessor

A

B

C A

D A

E C,B

F C,B

G D,E

The project is completed when both activities F and G are complete.

(ii) Suppose the project has the following activity times

Table 20: Activity List

Activity Time (months)

A 4

B 6

C 2

D 6E 3

F 3

G 5

(a) Find the critical path for the project network



42

(b) The project must be completed in yrs; do you anticipate difficulty in

meeting the deadline? Explain.

(iii) Suppose the following data are available for the above project.

Table 21: Crash data

Activity Normal Time Crash Time Total Normal

Crash (N

x ).

Total Crash Cost

(N x )

A 4 2 50 70

B 6 3 40 55

C 2 1 20 24

D 6 4 100 130E 3 2 50 60

F 3 3 25 25

G 5 3 60 75

Required

If there is compelling need to complete the project in 12 months,

(a) What activities should be crashed?

(b) What is the crash cost?

(c) What are the critical activities?

A Linear Programming Method for Crashing Decisions

In this approach we first define our decision variables

time of occurrence of event

= 1, 2, 3, 4, 5

Amount of crush time used for activity j

J = A, B, C, D, E

We re-sketch the network for ease of reference



43

Fig. 17 (Repeated): Network diagram

Earlier we had determined the total normal time project cost to be fixed at N17,

000 (see Table 18)

We can minimize the total project cost (normal cost plus crash cost) simply by

minimizing the crash cost. Thus our linear programming objective function

becomes:

Minimize ..............................(1)i j

j Z K y! 7

1000 1500 2000 1500 2500 A B c D E Z y y y y y ! ). ± (see Table 18)

Where K j is the crash cost for activity j, j = A, B, C, D, E on per-unit time basis.

Note that the x variables indicating event occurrence do not result in costs; thus

they have zero coefficients in the objectives function

The constraints for the model involves describing the network, limiting theactivity crash times, and meeting the project completion date, the constraints used

to describe the network are perhaps the most difficult. These constraints are based

on the following conditions.

1

0 0

2

7 7

3

6 7

4

10 10

5

12 12

A 7 B

3

C

6

D 3

E2



44

1. The time of occurrence of event must be greater than or equal to the

activity completion time for all activities leading into the node or event

2. An activity start time is equal to the occurrence time of its preceding node or

event

3. An activity time is equal to its normal time less the length of time it is

crashed.

The general expression for the three constraints stated above is given by:

[ í ] +

Time of occurrence event Actual time for activity j Start time for activity j 3 of 9

Starting with Event 2:

x = j = A, i = 1 = 0 (see network)

[ í ] + 0

Occurrence time for event 2 Actual time for activity A Start time for activity A ( =

0)

Note that at node 1, event occurrence time is zero (x1 = 0)

7 í

Or + 7 constraint 1

Event 3:

í +

í + 0

6 í

Or + 6 constraint 2



45

Event 4

Here observe that two activities namely B and D enter event (node) 4, we

therefore obtain the expression for our constraints as:

Branch B:

Branch D:

Re-arrange equation (3) and (4):

í + + 3

í + + 3 constraint 3 and 4

Event 5

constraint 5

Above constraints namely

(1), (2), (5), and (6), i.e. altogether, five (5) constraints

are necessary but not sufficient

to describe our CPM network.

The maximum allowable crash time constraints are:



46

See Table 18, column 6.

The desired project completion date is 10days as against the normal completion

date of 12 days. This new constraint can be expressed as

Negativity constraints:

Also

22)

Taken together, there are nine variables namely:



47

And eleven constraints namely:

Thus taking the non-negativity constraints into consideration, we can solve the

nine variable, eleven constraints problem to obtain the following solutions:

Recall that the objective function is:

Min. Z = 1000yA + 1500yB + 2000yC + 1500yD + 2500yE

Substitute for y j values, j = A, B, C, D, E.

Then we obtain:

Z = 1000(2) + 1500(0) + 2000(0) + 1500(1) + 2500(0)

= 2000 + 1500

= 3, 500

The solution values of yA = 2 and yD =1 tell us activity D must be crashed 1day

( 1,500) in order to meet the 10-day project completion deadline. Because of this

crashing, the time for activity A will be reduced to 7 ± 2 = 5 days, while the time

Eleven

constraints.



48

for activity D will be reduced to 3 -1 =2 days. The total project cost (normal cost

plus crashing cost) will be:

1700 + 2,000 = 3,700.

To generate the new activity schedule under crashing, we use the crashed

activity times and repeat the critical path calculations for the network. Doing this

provides the activity schedule shown in the table below. Note that in our final

solution, all activities are critical.

Resolving the linear programming model with alternative project completion

dates, days, will show the project manager the cost associated with

crashing the project to meet alternative dead lines.

Resource Scheduling

So far we had concerned ourselves with allocation of time and cost to projects. We

now want to focus on control of projects with due regard to resource limitations,

for example:

² Material shortage

² Limited number of skilled craftsmen

To be able to schedule the resource requirements for a project, the followingdetails are required:

1. The customary activity times, description and sequences as usual;

2. The resource requirements for each activity showing the classification of the

resource and the quantity required.

3. The resource in each classification that is available to the project. If variation

in availability is likely during the project life, those must also be specified.

4. Any management restrictions that need to be considered, e.g. which

activities may or may not be split or any limitations on labour mobility.

Exam ples



49

1. A simple project has the following time and resource data. For simplicity

only one resource of labour is considered but similar principles would apply

to other types of interchangeable resources.

Project D

ata

Resource constraint:

2 men only are available.

Required

i. Draw the activity times on a Gantt Chart based on their ESTs

ii. Based on ESTs, sketch the resource aggregation profile.

iii. Prepare the resource aggregation profile if there are only 2 men are

available.

iv. Do (iii) above with a 5-day constraint.

Sol ution:

Use forward and backward pass to estimate the EST and LST

i.

C

1

Activity Preceding Activity Duration (Days) Labour Requirements

A 1 2 men

B 2 1 man

C A 1 1 man

D 5 1 man

E B 1 1 man

F C 1 1 man

3 2

2

4



50

A 1 1 F

2 E

Figure 19: Network diagram

For EST

Slack reduces the duration LF ± slack

Critical path = activity D

i. Gantt chart

D

Activity Resource Duration ES LS EF LF Slack A 2 1 0 2 1 4 2

B 1 2 0

C 1 1 1 3 2 4 2

D 1 0

E 1 1 2 4 5 5 2

F 1 1 2 4 5 5 2

0

0

0

4

5 5

3

2 4

B

1

D

5



51

A C F

B E

Time

Time Scale (days)

Fig. 20: Gantt Chart

ii.

4

5

No of men

Required

1 2 3 4 50



52

Resource Availability

Level

Fig. 21: Resource Allocation based on ESTs

iii.

D

A B C E F

Fig. 22: Resource allocation with 2-man constraint

Total Float = LST ± EST ± D

A: 3 ± 0 ± 1 = 2

C

0 1

1

2

3

4

5

32 4 5

Extra time of

1 day is

likely to

result

Prepared with due regard to

f loats in order to smooth out

resource requirements

N o o f m

e n r e q u i r e d



53

B: 4 ± 0 ± 2 = 2

C: 4 ± 1 ± 1 = 2

D: 5 ± 0 ± 5 = 0

E: 5 ± 2 ± 1 = 2

F: 5 ± 2 ± 1 = 2

3 men are likely to be needed f or 2days

i.e. day 2 to day 4

Extra resources

days

Fig. 23: Resource Allocation prof ile based on 5-day constraint

Example 2:

0 1 2 3 4 5

1

2

3

4

5

D

B

A

C

E F



54

A project has the following activity durations and resource requirements.

Activity Preceding Activity Duration (days) Resource requirements (units)

(a) Assume no restrictions, show the network, critical path and resource

requirements on a day-by-day basis, assuming that starts are made on the

EST of each activity.

(b) Assume that there are only 6 units of resources, what would the plan be?

D

C 2

Fig. 24: Network diagram

A 6 3

B 3 2

C 2 2

D C 2 1

E B 1 2

F D 1 1

0

0 0

2 3

3

4 5

1

3 5

4

6 6

2

F

1

B

3

E

1

A

6

2

A

C



55

Time (days)

Fig.25: Gantt chart

B

E

C

D F

A

Time (days)

Fig. 26: Resource Allocation based on no Restriction

Floats:

F = LST ± EST ± D

A: 6 ± 0 ± 6 = 0

B: 5 ± 0 ± 3 = 2

C: 3 ± 0 ± 2 = 1

D: 5 ± 2 ± 2 = 1

D F

2 3 4 5 6

67

5

4

3

2

1

1 2 3 4 5

Resource

(units)



56

E: 6 ± 3 ± 1 = 2

F: 6 ± 4 ± 1 = 1

From the above analysis activities B and E have a float of 2 days. We can therefore

start the chain two days late. Having regard to this, we can now sketch the resource

allocation profile like so:

A : 6 ± 0 ± 6 = 0

B : 5 ± 0 ± 3 = 2

C : 3 ± 0 ± 2 = 1

D : 5 ± 2 ± 2 = 1

E : 6 ± 3 ± 1 = 2

F : 6 ± 4 ± 1 = 1

Resource

B E

Fig. 27:

Activity on Nodes

Points to note:

C

6

5

4

3

2

1

1 2 3 4 5 6

DF

A

Time (days)



57

1. This type of network, sometimes known as precedence diagram , is an

alternative form of presentation to the conventional type we had been using

since the commencement of this series of lectures.

Activities on node network are characterized by:

y Activities are shown on boxes, not as arrows

y No events appear

y Boxes are linked by lines to indicate their sequence or precedence

y A box appears for the start and another for the finish of a project

y No dummies are necessary

The activity on node approach is best illustrated by comparison with the procedure

for drawing conventional networks.

Example Conventional Network Activity on Node network

X Y

X

Y

B D

A

C

B

F

1

1 3

C

1

Z

(c). Activity C depends

upon activity A. Activity

D depends upon activity

A and B.

(b). Activity Z depends

upon activity X and Y

(a). Activity X depends

upon activity Y

(d). Activity B can start

immediately activity A

is complete. Durations

are 8 days and 6 days

respectively

Y

X

Y

Z

B D

CA

A

6 days 8 days6 days



58

A

1

2 B E

1

(i) conventional Network

Note further as follows:

i. Events as such do not appear in activity on node (A/N) networks

ii. The lines shown in A/N network indicate precedence and are not activities.

The activities are represented by the square boxes.

iii. From example (c), it will be seen that dummies are not necessary in A/N

networks.

iv. The time (6 days) shows on the A/N network (see (d) above) as known as

the depending time.

4

2

2 4



59

Note

(a) The lines joining the boxes are not activities; they are precedence as earlier

explained.

(b) The ESTs/LSTs in the boxes are calculated by the same process as

previously described but note that they are not the same values as shown in

the events of the conventional network.

(c) The EFT and LFT are calculated by adding the activity time on the EST and

LST respectively.

A 1

0 1

2 3

C 1

1 2

3 4

F 1

2 3

4 5

D 5

0 5

0 5

E 1

2 3

4 5

B 2

0 2

2 4

5

Finish

5

0

Start

0

. .

EST EFT

LST LFT

Duration

Activity

Code



60

(d) The critical path is found by the usual method of comparing the EST and the

LST (or EFT and LFT)

Exam ple

(a) Draw an activity or node diagram for the following project.

PP.72 Activity Preceding activity Duration

1 4

2 1 7

3 1 5

4 1 6

5 2 2

6 3 3

7 5 5

8 2,6 11

9 7,8 7

10 3 4

11 3

12 9,10,11 4

(b) Calculate the EST/LST/and LFT values for each box10 4

9 13

26 30



61

Exam ple

The following activity-on-node network shows the relationship between the eleven

activities which make up a particular stage of a large construction project. Within

each activity box, the letter denotes the µname¶ of the activity, and the number of

its duration in weeks.

START

B6 D12 G2

F6



62

Required:

a)

i. How long will it take to complete this stage of the project?

ii. What is the critical path?

iii. By how much would the duration of activity J have to increase before it

becomes critical?

iv. What would be effect, if any, on both the duration of the project and the

critical path if activity D does not have to be preceded by B?

b) The cash required to complete each activity is as follows:

Activity A B C D E F G H J K L

Cash (N1000) 30 18 12 12 8 12 6 15 16 12 6



63

What is the maximum amount of cash required in any one week and when

will this occur?

You should assume that all activities start at their earliest time and that the

necessary cash is used at a uniform rate throughout the duration of each

activity.

Sol ution:

The first stage is to calculate the EST/LST values for the network

A10

0 0

C3

13 15

E4

13 15

H5

17 19

K6

22 24

L2

30 30

B6

10 10

D12

16 16

G2

28 28

F6

16 22

Start

0

END

32 32

J4

17 20

0



64

(a)

(i) Completion time = 32 weeks

(ii) Critical path A, B, D, G, L

(iii) 3 weeks

(iv) The removal of the link from B to D would alter the critical path to A-

C-E-H-K-L, duration 30 weeks.

(b) Tables of EST and cash required per week are as shown below

Activity Duration Start Finish Cash per week (N000)

A 10 0 10 3

B 6 10 16 3

C 3 10 13 4

D 12 16 28 1

E 4 13 17 2

F 6 16 22 2

G 3 28 30 3

H 5 17 22 3

I 4 17 21 4

K 6 22 28 2

L 2 30 32 3

A: = 3 per week

B = = 3 per week



65

C = = 4 per week

From the above table the concurrent activities and cash requirement can be

established

110 A 3

1113 B, C

1416 B, E

17 D, E, F

1821 D, F, H, J

22 D, F, H

2328 D, K

2930 G 3

3132 L 3

Therefore the maximum cash requirement is N10,000 for the period 18 ± 21 weeks

where the activities D, F, H and J are in progress.

Pert/Cost

We have learnt that PERT and CPM are useful tools for project planning and

scheduling with due regards to time. We shall now concentrate on PERT/COST

that focus on scheduling and controlling projects using the criteria of cost

considerations. This type (version) enables project managers to maintain costs

associated with project executions within budget limits.

Generally, PERT/CPM contains detailed activities that may not be convenient for

analyzing associated cost of the project. In this regard, activities that share similar

Weeks

Activities in

progressCash

Required



66

cost classification may be grouped and called work packages. With this type of

grouping, control on project by sections, departments or even sub-contractors

undertaking such projects may easily control cost of projects by virtue of such

groupings (work-packages). The steps involved in the use of PERT/COST control

system are as follows:

1. There is need for the specification of network expected times (i.e duration).

This enables us to estimate the associated floats, i.e. total floats.

2. The budget or estimated cost needs to be specified.

3. From (1) and (2) above, the cost gradients (cost rates) can be estimated or

rather calculated.

4. Next, compute the total float based on our network estimation of the

following:

² EST

² LST

² EFT, and

² LFT

From the above analysis, the slack and critical paths are estimated

5. Next, we establish budgeted cost for each activity based on the knowledge of

EST earlier calculated in (4) above.

6. Next, we also establish similar budgeted cost for each activity on the basis of

LST.

7. From (5) and (6), the columns for monthly (weekly) costs and the associated

cumulative monthly cost as applicable, need to be computed by column

addition (month by month).

8. Another crucial step is to plot total budgeted cost (cumulative monthly cost)

against time. This will yield a looped graph whose enclose portion depicts



67

the feasible region of budget estimate. All budget planning and control are to

be done within this feasible region.

9. The final step involves the specification of actual cost and percentage

completion. Based on this information, we compute the following for eachactivity in the network.

Estimated budget, = ( /100) (1)

Where I = activity number

= percentage completion of activity i

= budget for activity i

Then í A = (2)

Where A = the actual budget that needs to given for each activity i

= the deviation or discrepancy between the actual A and estimated

or budget.

When > 0, we call such discrepancy or deviation over-run, if however, < 0,

we refer to such cost as under-run. Obviously, situations of over-run are never

desirable. Total project cost over-run to date is the sum of column over the

specified activity i.

On the other hand,

= ( /100)

And = = percentage cost overrun

Perhaps an example will help to drive the above concepts home.

Consider the following information



68

F 2

11 D 3

B

G 1

Activity

Expected time

(month)

Budgeted or

Estimated cost

Budgeted cost per

month(cost gradient)

A 2 10,000 5,000

B 3 30,000 10,000

C 1 3,000 3,000

D 3 6,000 2,000

E 2 20,000 10,000

F 2 10,000 5,000

G 1 8,000 8,000

Let¶s calculate the floats by first determining the EST and LST (forward pass and

backward pass) respectively.

2 F

11 3

B

3 G 1

C

A

E

2

1

3

4

5

6

2

3

2

1

0

2 6

3

C

A

E

D

2

1

3

4

5

6

2

3

2

0

5 6

5 7

8 8

1



69

LST =5 f or E because it will take 7 days to complete.

Note the following:

The number of months = 8, derived from the critical path duration: B ± D ± F = 3 +

3 +2 = 8months.

EFT = sum of activity durations along the relevant path(s) concerned.

LST:

For A, recall that the activity duration is 2 and latest finish time is 5 (seecircle 2), therefore if we start latest in the third day we shall still finish on the fifth

day. For E, the same as in A except that we 2days to do the work and the work

have to be finished in the 7th

month. Therefore we can start latest the 6th

month in

order to finish in the 7th

month.

We now move to step (5)

Activity EST LST EFT LFT LST-EST Slack

LFT-EST

Critical path?

A

0 2 5 3B 0 0 3 3 0 Yes

C 2 5 6 3

D 3 3 6 6 0 Yes

E 3 5 7 2

F 6 6 8 8 0 yes

G 5 7 6 8 2



70

Budgeted Cost Based on EST Schedule (N x )

Months

Activity 1 2 3 4 5 6 7 8

A

B

CD

E

F

G

5

10

5

10 10

32

10

2

10

2

8

5 5

Monthly

cost

Cumulative

(Total

project

cost)

15

15

15

30

13

43

12

55

12

67

10

77

5

82

5

87

Budgeted Cost Based on LST Schedule (N x )

Months

Activity 1 2 3 4 5 6 7 8

A

B

C

D

E

F

G

10 10 10

5

2

5

2

3

2

10 10

5 5

8



71

Monthly

cost

Cumulative

10

10

10

20

10

30

7

37

7

44

15

59

15

74

13

87

Region of feasible budgetfor total project cost

Earliest start time (EST)cost schedule

Latest start time (LST)cost schedule

100

90

80

70

60

50

40

30

20

10

1 2 3 4 5 6 7 8

B

u d g e t e d

t o t a l

p

r o j e c t e d

c o s t

( c u m

u l a t i v e )

N (000)

Graph of f easible budgets f or total project costs

At this juncture we require to know the activity cost estimate. Suppose we are

given the following:

Activity Budgeted cost Activity Budgeted cost

A N 10,000 E N 20,000

B 30,000 F 10,000

C 3000 G 8000

D 6000



72

Finally we use

= A í



73

Discrepancy equation to ascertain areas of over run and under run.

To do this, we need the actual cost of the relevant activities and the corresponding

percentage completion. Suppose, for instance, we are given such information as

below.

Activity Actual cost ( ) percentage completion

A 12,000 100

B 30,000 100

C 1000 50

D 2000 33

E 10,000 25F 0 0

G 0 0

Obviously,

= x (10,000) = N 10,000

= x (30,000) = 1500

= A í

= 12,000 ± 10,000

= 2,000

Also,

= A í

= 1000 í

But = 1,500

= 1000 ± 1500



74

= 500, a cost under run i.e. saving

A cost status report can now be prepared as follows:

Activity Actual Cost (AC) Estimated budgeted Difference

( ) Budget Cost ( ) D

= ( ) A 12,000 10,000 2000

B 30,000 30,000 0

C 1,000 1,500 -500

D 2,000 2,000 0

E 10,000 2,500 7,500

F 0 0 0G 0 0 0

N 55,000 N 46,000 N 4000

Note:

= x 30,000

= 30,000

= x 6,000

= 1980 § -N-2,000

= x 10,000

= 2,500

= x 0

= 0

= x 0



75

= 0

= 4,000

= 46,000

Over run = x 100 9%

Assignment

Given:

Activity Expected time Expected cost Budgeted cost per month

A 6 90 15

B 2 16 8

C 3 3 1

D 5 100 20

E 3 6 2

F 2 2 1

G 3 60 20

H 4 20 5

I 2 4 2

J 2 2 1

303



Routing

6 5 D 2 Cost Estimate

E 3

2

B

1

2 5

7

3 6

8 4

Product

desi nA

Prototype

C 3

F

Testing

G 3

Final Report

J 2

Pricing and

Forecast

21

Completion

4Market

Survey

H

Market

Research

plan

Market

Brochure

Documents

Project Management for m