Project ManagementProject ManagementProject ManagementProject Management

Project ManagementProject Management

Management of big projects that consist of large number of activitiesnumber of activities pose complex problems in planning,

scheduling and controli ll h th ti iti h t bespecially when the activities have to be

performed in a specified technological sequence

CPM (Critical Path Method) and PERT (Programme Evaluation & Review T h i ) f l id f ffi i t j tTechnique) are useful aids for efficient project management

h diff i h i h h bl d h They differ in their approach to the problem and the solution technique

Critical Path Method (CPM)Critical Path Method (CPM)

Assumes that activity times are proportional to the amount of resources allocated to them

By changing the level of resources the activity times and project completion ti b i dtime can be varied

It does not incorporate uncertainty in job p y jtimes and assumes prior experience with similar projects from which the relationship between resources and jobrelationship between resources and job times are available

Critical Path Method (CPM)

CPM then evaluates the trade_off

Critical Path Method (CPM)

_between project cost and project completion time

It is mostly used in construction/production projectswhere there is prior experience in handling similar projectshandling similar projects

Programme Evaluation and Review Technique (PERT)Technique (PERT)

PERT incorporates uncertainty in activity times in its analysis

It determines the probability of p ycompletion of various stages by specified deadlines

It also calculates the expected time to complete the project

Because of its ability to handle uncertainties in job times, PERT is mostly used in research and developmentused in research and development projects

Analysis by CPM/PERTAnalysis by CPM/PERT

Uses the network formulation to t th j t ti iti d th irepresent the project activities and their

ordering relations Network Formulation: Network Formulation:

Arcs/Lines in the network represent individual jobs in the project

Nodes represent specific points in time which mark the completion of one or more jobs in the projectj p j Nodes are generally called Events

Direction on the arc/line is used to /represent the job sequence

C

1 2 3A B

5

1 2 3

Start Completion43

• It is assumed that any job directed towards a node must be completedb f j b di d f hbefore any job directed away from that node can begin

Example 1Example 1

Consider 7 jobs A, B, C, D, E, F & G with j , , , , ,the following job sequence:

b d Job A precedes B & C Jobs C & D precede E

Job B precedes D Job B precedes D Jobs E & F precede G

Example 1 – Network FormulationExample 1 – Network Formulation

Job A is not preceded by any Job, thus, it is the starting Job

Job A precedes B & C

2 4Start

C

B

1 A

3

B

Example 1 – Network Formulation

J b B d D

Example 1 – Network Formulation

Job B precedes D

2 4Start

C

B D

1 A

3

B D

Example 1 – Network FormulationExample 1 – Network Formulation

2 4 5Start

C E

B D

1 A

3

B D

Jobs C and D precede E(Since Jobs C & D precede a job, it is expected that they will be completed together)

Example 1 – Network FormulationExample 1 – Network Formulation

F

2 4 5 6Start

Completion

C E G

B D

1 A

3

B D

Jobs E and F precede G

Job F is not preceded by any Job, thus, it can start right in the beginning

Example 2Example 2

Consider a project with five jobs A, B, C, p j j , , ,D & E with following job sequence:

Job A precedes C and D Job B precedes D Job C and D precede E

The completion time for A B C D & E are 3 The completion time for A, B, C, D & E are 3, 1, 4, 2, 5 days respectively.

1 2 4 5Start A C E

543

3

Completion543

B1

D2

0

Dummy jobDummy job

Dummy job Doesn’t really exist in the project

It i t id d bt i th j b

3 Dummy jobDummy job

It is necessary to avoid any doubt in the job sequence Completion time is always zero

In this network event 3 represent the completion of job In this network event 3 represent the completion of job B and dummy job

Since the dummy job is completed as soon as A is l t d t 3 ti ll fl t th l ti fcompleted, event 3 essentially reflects the completion of

jobs A and B

Rules (About dummy job)Rules (About dummy job)

Each activity is represented by one and lonly one arc

Each activity must be identified by two distinct end nodesend nodes

1 2A

BA

3B

21 OR

B 1 3A

2B

RulesRules

To maintain the correct precedence To maintain the correct precedence relationships, the following questions must be answered as each activity is added to the network:

What activities must immediately precedethe current activity?y

What activities must follow the current activity?

h i i i l What activities must occur concurrentlywith the current activity?

RulesRules

Consider the following segment of a project: Consider the following segment of a project:

Activity C starts immediately after A and Bhave been completedhave been completed.

Activity E starts after B only has been completed.

A C

2

A C 3A C

2

EB2

B E

Incorrect Correct

In the Project Management Problem If the completion time of all the activities

and their technological sequence are known

It is called a Simplified Problem

In such a problem we need to In such a problem, we need to

Determine the minimum time in which the Determine the minimum time in which the project can be completed

Identify the critical jobs which can causeIdentify the critical jobs which can cause delay in the project completion

Simplified Project Management Problem can be formulated as a LPPcan be formulated as a LPP

1 2 4 5A C E

543

3

543

B1

D20

Let us consider Example 2f f

3

Let ti be time of occurrence of event i, where

i 1 2 5 i = 1, 2, …., 5 t1 = Start of the project t5 – t1 = Time of completion of project and t5 t1 = Time of completion of project and

objective is to minimize this duration

LP FormulationLP Formulation

Minimize Z = t5 - t1

Subject tot2 - t1 ≥ 3t - t ≥ 1 Constraints show that theConstraints show that the timetimet3 - t1 ≥ 1t3 – t2 ≥ 0t4 – t2≥ 4

Constraints show that the Constraints show that the time time available for completingavailable for completing a job a job should be ≥ should be ≥ time required to time required to completecomplete that jobthat job

t4 – t3 ≥ 2t5 – t4 ≥ 5ti≥ 0

completecomplete that jobthat job

i

• This can be solved by simplex method• Set t1 = 0 and chose the values of ti as small as

possible to satisfy the constraintspossible to satisfy the constraintst1 = 0, t2 = 3, t3 = 3, t4 = 7, t5 = 12;Z = tZ = t55 -- tt1 1 = 12 days= 12 days

We can also identify critical jobs in the process We need to identify those constraints that

will be satisfied as equations in the optimalwill be satisfied as equations in the optimal solution, and the jobs corresponding to those constraints are the critical jobs

Example

t2 – t1 ≥ 3 Satisfied as Equation; Criticalitical - Job At t ≥ 1 Not satisfiedt3 – t1 ≥ 1 Not satisfiedt4 – t2 ≥ 4 Satisfied as Equation; Criticalitical - Job Ct4 – t3 ≥ 2 Not satisfiedt – t ≥ 5 Satisfied as Equation; Criticalitical - Job Et5 – t4 ≥ 5 Satisfied as Equation; Criticalitical - Job E

Critical PathTh th i th j t t k ti th The path in the project network connecting the starting event and the ending event such that it passes through the critical jobs is called a critical path

Fi di iti l th i t k i

1 2 4 5

Finding a critical path in a network is equivalent to finding the longest pathlongest path

For a large project, the number of constraints may be too many for handling as LPP

A direct approach is available to solve such problems A direct approach is available to solve such problems using Network Analysis

Solution by Network AnalysisSolution by Network Analysis

Earliest timeEarliest time of node j, Uj, is the earliest time t hi h t jat which event j can occur

Event j can occur as soon as all the jobs directed towards node j are completed

1A towards node j are completed

e. g., j occurs as soon as A, B and C are completed

The earliest time of node j is given by2 jB

A

The earliest time of node j is given byUj = Max (U1 + t1j ; U2 + t2j; U3 + t3j) t1j, t2j and t3j are completion times of jobs A, B and C3

C

General form Uj = Max (Ui + tij)

i represents all arcs (i, j)

Earliest time of last node gives the earliest time of project completion

Example 2Example 2

1 2 4 5A C E

543

3

543

B1

D20

Uj can be calculated as follows U1 = 0

3

1 U2 = U1 + t12 = 3 U3 = Max(U1 + t13 ; U2 + t23 ) = Max (1,3) = 3 U4 = Max(U2 + t24 ; U3 + t34 ) = Max (7,5) = 7 U5 = U4 + t45 = 7+5 = 12

Hence, the minimum duration of project is 12 daysdays However, in this process, critical jobs are not

identified We need to calculate Latest time of an event

Latest time of node i, Vi, is the latest time , i,at which event i can occur without delaying the project beyond its earliest time7

F

The project will not be delayed if the three jobs F, G, & H are completed by V7, V8 & V9

8i G

F

H This is possible if

Vi = min [V7 – ti7 ; V8 – ti8 ; V9 – ti9 ]9

H

General form Vi = min [Vj – tij] Vi = min [Vj tij]

Where j represent all nodes for which arc (i, j) exists

Example 2

1 2 4 5A C E

543

3

543

B1

D20

To calculate the latest time of the events, We set the latest time of the last event equal to its

3

We set the latest time of the last event equal to its earliest time and work backwards

V5 = U5 = 12 V4 = V5 – t45 = 7 V3 = V4 – t34 = 5 V2 = min [( V4 – t24) ; (V3 – t23 )] = min [3,5] = 3 V1 = min [( V2 – t12) ; (V3 – t13 )] = min [0,4] = 0

Example 2

The EARLIEST and LATEST Times are represented pin the network as follows:

(0 0) (3 3) (7 7) (12 12)

1 2 4 5A C E

543D

(0, 0) (3, 3) (7, 7) (12, 12)

3

B1

20

(3, 5)

The difference between the latest time and earliest ti f t i ll d th l k ti f th ttime of an event is called the slack time of that event Slack time denotes how much delay can be tolerated in Slack time denotes how much delay can be tolerated in

reaching that event without delaying the project completion date

For the problem slack time of various events is For the problem, slack time of various events is 1 0 2 0 3 2

Those events that have zero slack time are critical events

3 2 4 0 5 0

Critical Path is a path such that the jobs on this path have zero slack time

11 2 4 5AA CC EE

543

Summary of the Results of Network AnalysisNetwork Analysis

Event Earliest time

Latest time

Slack time

Remark

1 0 0 0 Critical

2 3 3 0 Critical

3 3 5 2 Non-critical

4 7 7 0 Critical

5 12 12 0 Critical

Project ScheduleProject Schedule

i jJ

i jtijUi

VjEarliest occurrence time of i

Latest occurrence time of j

Completion time of J

For J Earliest starting time = Ui

Latest starting time = Vj – tijj j

Earliest finish time = Ui + tij Latest finish time = Vj

(0 0) (3, 3) (7, 7) (12, 12)1 2 4 5

A C E543

BD2

(0, 0) ( , ) ( , ) (12, 12)

3

B1

2

(3, 5)

Project Schedule of Example 2Project Schedule of Example 2

11 22 33 44 55 66 77 88

Job Expected duration

Earliest start

Latest Start

Earliest finish

Latest finish

Slack time

Remarks

(days) U (6)-(2) (3)+(2) V (6)-(5)(days) Ui (6) (2) (3)+(2) Vj (6) (5)

A 3 0 0 3 3 0 Critical

B 1 0 4 1 5 4 NoncriticalB 1 0 4 1 5 4 Noncritical

C 4 3 3 7 7 0 Critical

D 2 3 5 5 7 2 Noncritical

E 5 7 7 12 12 0 Critical

Exercise ProblemExercise Problem Consider a project consisting of nine jobs with

the following precedence relationship and time g p pestimates

Job Predecessor Time (days)A - 15A 15B - 10C A, B 10D A, B 10E B 5E B 5F D, E 5G C, F 20H D, E 10H D, E 10I G, H 15

Draw the project network Determine the earliest completion time of the project Determine the earliest completion time of the project Prepare the project schedule Identify the critical path

Project networkProject network

1 3A 5C 6G 7I(0,0) (15,15) (30,30) (50,50) (65,65)

1 315 510 620 715

B10

D10

H10

0 F 5

2 4E5

10 10 10

(25 25)(10,15) (25,25)

Slack time1 = 0 ; 2 = 5 ; 3 = 0 ; 4 = 0 ; 5 = 0 ; 6 = 0; 7 = 0

Critical Path

1 = 0 ; 2 = 5 ; 3 = 0 ; 4 = 0 ; 5 = 0 ; 6 = 0; 7 = 0

1 3A15 4D

10 5F5 6G

20 7I15