INTRODUCTION

Network Analysis

Project

A project may be defined as a combination of interrelated activities which must be executed in a certain order before the entire task can be completed.

The aim of planning is to develop a sequence of activities of the project so that the project completion time and cost are properly balanced and that the successive demand of key resources is avoided.

Network Analysis

To meet the object of systematic planning, the managements have evolved a number of techniques applying network strategy. PERT and CPM are two of the many network techniques which have been widely used for planning , scheduling and controlling the large and complex projects.

DIFFERENCE BETWEEN PERT & CPM

The project which comprise of the variable type activities associated with probabilistic type time estimates, employ PERT version of the networks and the project comprising of the deterministic type of activities are handled by CPM version of networks.

PERT is event oriented and CPM is activity oriented.

 

Phases of Project management

Planning

Scheduling

Controlling

Planning

Setting the objective of the project and

assumption to be made.Development of WBS. Depending upon the

objective, the extent of control desire and the availability of the computational aids, the project is broken down in to clearly definable activities.

Estimating activity duration, their resources requirements and costs, and

Establishing the interdependence relationships between the activities.

Scheduling

In this phase the activities are kept according to precedence order and then determining-

Start and finish times for each activity.Critical path on which the activities

require special attention, andSlack and float for critical paths.

Controlling

The controlling phase is to follow up to the planning and scheduling phase and involves-

Making periodical progress reports.Reviewing the progress.Analyzing the status of the project, andManagement decisions regarding

updating, crashing and resource allocation, etc.

Network Diagram (Arrow Diagram)

Activity: Physically identifiable part of a project which consumes time and resources. It is represented by an arrow, tail represents Start and head represents the finish of an activity.

  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 generally represented by a numbered circle. The head event is always numbered higher than the tail event.

  Path: An unbroken chain of activity arrows

connecting the initial events to some other event is called a path.

Network

Network is a graphical representation of logically and sequentially connected arrows and nodes representing activities and events of a project. Networks are also called arrow diagrams.

  Network construction: Firstly the project is split into activities. Start and

end events, precedence order for all activities are decided. All activities are kept in a logical sequence by using notations. Then check:

Which activity must be completed before starting particular activity?

What activity/ies follow this? What activities must be performed concurrently with

this?

Types of Activities

Activity which must be completed before starting particular activity is called a predecessor activity and those which must follow a particular activity are called successor activity. E.g. B is successor to A, means A is predecessor of B, this can be written as B>A.

Dummy activity: An activity which only determines the dependency of one activity over the other, but does not consume any time and resources is called dummy activity. These activities are represented by dotted line arrows.

To remember

Looping: The arrow diagram must flow from left to

right and there should be no looping in between.

 Dangling: The situation where activities other than

the final activity do not have any successor activity called dangling and it must be avoided.

Draw network diagram for the following

For following precedence relationship draw network diagram.

B >A ; C >B ; D, E, H > C ; F > D ; G>E

Draw network diagram for the following

Precedence relationships are : A, C, D can start immediately.

E>B, C ; F, G > D ; H, I >E, F ; J > I, G ; K> H ; B>A

PERT Computations

FORWARD PASS COMPUTATION Start calculating from left to right that is

from initial event to end event. Initial event is assigned 0 time and calculate

the time at which that event is expected to occur at the earliest. This is called Earliest expected time(TE) .

For event j TE j =Maximum of all (T E i + t E

ij ) for all ij .

PERT Computations

BACKWARD PASS COMPUTATION  Start calculating from right to left that is from

last event to first event. For Last event time of occurrence is decided

i.e. same as TE . This is called Contractual Obligation Time denoted by Ts. The objective of backward pass computation is to calculate Latest allowable occurrence time(TL).

For event i TL i =Minimum of all (T L j - t E

ij ) for all ij .

Is the latest allowable occurrence time for event i.

For the following network diagram compute T E and T L

using forward and backward computations. Also compute slack and Find critical path

Network diagram

Calculation of TE, TL and slack

Q 2 Find the Earliest Expected time and latest occurrence time for all activities and also critical path.

Network Diagram

Network with computations

Calculation of TE, TL and slack

Time Estimates

PERT system is based on three time estimates of the performance time of an activity.

 1 Optimistic time estimate (to) :2 Pessimistic time estimate (tp):3 Most likely time estimate (tm):

Time Estimates

From these time estimates we can calculate mean time and standard deviation of time estimates for each activity.

The formulae For mean μ=( to+4tm + tp)/6 Standard deviation σ = (tp –to)/6 (Squaring both sides we get variance).

Probability of Meeting the Scheduled Dates:

tE ij= Average time /duration of activity

= (to+4tm + tp)/6

Standard deviation σ for network = Square root of Sum of the variances along the Critical Path

______ = √∑ σij 2

We calculate probability using normal distribution (Standard normal variate Z)

Z= (Ts -TE )/σ

What is the probability that the project will be completed in20days? The diagram is given below.

Where P(0<Z<0.48)=0.1846

Solution

Expected time duration TE =19 daysContractual obligation time Ts= 20days

Standard Deviation for project =σ (for activity durations of all activities on critical path

= √( ∑ σij 2 )

= 2.08Normal Variate Z= (Ts-Te)/ σ = o.48From the probability table, probability= 0.6846

CPM computations

Earliest Start time: Earliest Start time of an activity is the earliest occurrence time of an event from which the activity emanates. Earliest start time for activity ij is denoted by Esi & is same as Earliest occurance time of event i.

Esi = TEi

  Latest Completion Time:

This is the Latest Occurrence time of the event at which the activity terminates. Latest Completion time of an activity ij is denoted by Lcj

Lcj = TLj

Latest Start Time:This is the Latest completion time of the activity minus the activity duration. For activity ij, the latest start time is,

Lsij = Lcj - tE ij

  Earliest Completion Time:

This is the earliest start time of the activity plus the duration of the activity. For activity ij, the earliest completion time is

  Ec ij = Esi+ tE ij = TE

i + tE ij

Float

Float of an activity has the same significance as the slack of the events. Slack corresponds to events, hence to PERT, while float corresponds to activities & hence to CPM.

There are four types of floatsTotalFreeIndependentInterfering

Total Float

The total float is equal to the difference between the maximum time available time to perform the activity and the activity duration time. The maximum time available time for any activity is from the earliest start time to the latest completion or finish time.

Thus, for an activity (i-j), the total float is given by:  TF

ij =Lcj - Esi - tE ij = Lsij - Esi

=Col 5-Col 3

In other words, it is the difference between the maximum time available for the activity and the actual time it takes to complete. Thus, total float indicates the amount of time by which the actual completion of an activity can exceed its earliest expected completion time without causing any delay in the project duration.

Free Float

It is defined as that portion of the total float within which an activity can be manipulated without affecting the float of the succeeding activities. It can be determined by subtracting the head event slack from the total float of an activity.

  i.e. FF

ij = T F ij – (slack of head event j)

= (Esj - Esi) - tE ij

= Esj - Ec ij

or The free float indicates the value by which an activity

in question can be delayed beyond the earliest starting point without affecting the earliest start, and therefore the total float of the activities following it.

Independent float

It is defined as that portion of the total float within which an activity can be delayed for start without affecting float of the preceding activities. It is computed by subtracting the tail event slack from the free float.

i.e. IF ij = FF

ij – {slack of tail event i}

IF ij = Esj - Lci - tE

ij  The independent float is always either equal to a

less than the free float of an activity. If a negative value is obtained, the independent float is taken to be 0.

Interfering float

Utilization of the float of an activity can affect the float of subsequent activities in the network. Thus, interfering float can be defined as that part of the total float which causes a reduction in the float of the successor activities. In other words, it can be defined as the difference between the latest finish time of the activity under consideration and the earliest start time of the following activity, or 0, whichever is larger. Thus, interfering float refers to that portion of the activity float which cannot be consumed without affecting adversely the float of the subsequent activity or activities.

EXAMPLE

A small assembly plant assembles PCS through 9 interlinked stages according to following precedence process:-

i)Draw network & find critical path. ii) Tabulate total, free & independent float

Activity (1)

Duratn (2)

1-2 41-3 121-4 102-4 82-5 63-6 84-6 105-7 106-7 06-8 87-8 108-9 6

Solution

Cost analysis, Contracting and Updating

Project Cost Each activity of a project consumes some

resources and hence has cost associated with it. In most cases cost of activity depends on the amount of time consumed by it. Thus by increasing the cost, the project duration can be reduced. The aim is always strike the balance between the costs and time and to obtain the optimum project schedule. The optimum project schedule implies lowest possible cost & the associated optimum time for the project. The total cost of any project consists of the direct & indirect cost involved in its execution.

Direct Cost

The Direct cost is directly dependent on the amount of resources involved in the execution of the individual activities. This is the cost of the materials equipments and labor required to perform the activity. As the activity is compressed, the direct cost goes on increasing. If it is compressed beyond a point C, the cost increases rapidly for significant change in activity duration time. This point is called as Crash point. Crash time is thus the minimum activity duration to which an activity can be compressed by increasing the resources and hence increasing the direct cost.   Crash cost – Normal Cost Cost Slope = Normal Time - Crash Time

Indirect Cost

It can be divided into two parts: Fixed indirect cost is due to general admin

expenses, license fee, insurance cost and taxes and does not depend on progress of project.

 Variable indirect cost depends on the time

consumed by the project & consists of overhead expenditure, supervision, interest on capital and depreciation, etc. It is assumed that indirect cost increases linearly with time.

Crashing the NetworkExample 1

For the following information about a project, draw network. What is normal duration cost of project? Which one should be selected for crashing? What will be project duration and total cost after first crashing?

Find the optimum duration and cost. (Indirect cost is Rs.1000 per week)

Cost Table

Draw the Network Diagram

ii) Normal duration cost of project is = direct cost of all activities + indirect cost

for 41 weeks = 86500 + 41*1000 =1,27,50  

Calculation of Cost Slope

First find all possible paths and their duration, among which select critical path. On this path identify the activities which can be crashed, with their cost of crashing. Now select the one which is costing minimum.

Paths Normal duration Crash duration 1-3-4-6-8-9 34 24 1-2-3-4-6-8-9 38 25critical 1-2 -5 -6-8-9 40 24 1-2-5-7-8-9 41 critical 21 1-2-5-7-9 29 15

Crashing

On the critical path, Activities Cost of crashing/unit

time 1 -2 600 2-5 500 5-7 2000 7-8 200 (can be crashed by 1

weeks & will

8-9 900 add cost of 200)

Now the critical path is 1-2 -5 -6-8-9 with max duration.   Paths Normal duration Crash

duration 1-3-4-6-8-9 34 24 1-2-3-4-6-8-9 38 25critical 1-2 -5 -6-8-9 40 24 1-2-5-7-8-9 41 critical (crashed by 1) 21 1-2-5-7-9 29 15

Select the activity to be crashed-

Activities Cost of crashing/unit time 1-2 600 2-5 500 to be crashed by 2 weeks 5-6 800 6-8 - can’t be crashed 8-9 900

Paths Normal duration Crash duration

1-3-4-6-8-9 34 24

1-2-3-4-6-8-9 38 25critical

1-2 -5 -6-8-9 40 38 24

1-2-5-7-8-9 41 40 38 21 1-2-5-7-9 29 27

15

Paths Normal duraton Crash duration 1-3-4-6-8-9 34 32 24 1-2-3-4-6-8-9 38 (34) 32 25critical 1-2 -5 -6-8-9 40 38 (34) 32 24 1-2-5-7-8-9 41 40 38 (34) 32 21 1-2-5-7-9 29 27 (23) 15 (18)

To reduce duration to 34 from 38 select activities with min cost slop- Activities Cost of crashing/unit time 1-2 600 (crash 4 times will reduce the project duration to 34)2-5 500 5-6 8006-8 - 8-9 900 (crash 2times will reduce duration to 32)7-8 2002-3 na3-4 2504-6 15005-7 20001-3 1000

Select the activity to be crashed & reduce the duration for the project

Crashing

New critical path is 1-2-3-4-6-8-9 duration 38 on which activity 3-4 is costing least for crashing, hence duration 7 is reduced to 5 ie by 2 weeks. And keep on doing the same method till the project duration becomes 25(Crash critical).No further crashing is possible.

Final table

Thus optimum duration is 30 weeks and optimum project cost is Rs. 123400.

Example 2

The time & cost estimates & precedence relationship of the different activities constituting a project are given below:

i)Draw project network diagram & find critical path. ii) If a dead line of 17 weeks is imposed for the completion of

the project , what activities will be crashed? What would be the additional cost and the critical activities

after crashing the project?

Cost Slop Calculation

Network

Path Normal Duration Crash duration 1-2-4-5-6 15 11 1-2-5-6 22critical 16 critical 1-2-4-6 9 7 1-3-5-6 20 14 1-3-6 13 8 But deadline is to complete project in 17 weeks, hence

Activities to be crashed and cost involved

Activities to be crashed and cost involved Critical path Alternative cost(ΔC/ΔT) Crash activity

duration Activity 1-2-5-6 A(1-2) 1000 H(twice) E(2-5) 2000 (5-6)

21 H(5-6) 500  1-2-5-6 A(1-2) 1000 A E(2-5) 2000 (1-2) 19   1-2-5-6 E(2-5) 2000 E(2-5) 18 1-3-5-6 E(2-5) 2000 E(2-5), 17 B(1-3) 200 B(1-3)  1-3-5-6 E(2-5) 2000 not required C(1-3) 3000

Answer

Total cost= normal cost + crashing cost = 36000 +(H twice)500*2+

(A)1000+ (E twice) 2000*2+(B)200 = 36000 +6200 = 42200