Project Management Using Network Analysis By Prof. Nadpurohit.

Project Management Using Network Analysis

By

Prof. Nadpurohit

• Project Planning & Scheduling • CPM & PERT Techniques

• WBS •Network drawing• Network Analysis• Updating of Network

• Project Cost v/s Project Completion time • Time Cost Trade off• Crashing of Project Completion time

• Resource Management• Types of resources• Resource Smoothening & Resource Leveling

• Project Control•Earned Value Management System

Sept 2010 Slide 2 of 64 Project Management Using Network Analysis, By Prof. Nadpurohit

References:• PMBOK• Project Management by K Nagarajan, New age International Publications• Projects by Dr. Prasanna Chandra


“A Project”

Temporary endeavor undertaken to create a unique Product, Service or Result


Project Characteristics

Temporary:

means every Project has a finite duration (a definite beginning and a definite end). Projects are not ongoing efforts.

Temporary characteristics of a Project applies toOpportunityProject team working as a unit

Temporary characteristics of a Project does not apply to:Product, Service or result generated by the

Project.


Project Characteristics (continued)The driving forces that create stimuli for a Project are typically referred to as Problems, Opportunities or Business requirements.

End of Project is reached when

Projects objectives have been achieved.

When it becomes clear that the Project objectives will not or cannot be met and the Project is terminated.

The need for the Project no longer exists and the Project is terminated.


Project Characteristics (continued)

Uniqueness an important characteristic of Product, Service or Result

Project can create

A Product or artifact which is quantifiable, and which is either an end item itself or a component item.

A capability to perform a service, such as business functions supporting production or distribution.

A result such as outcomes or documents


Relation between Product and Project Life Cycle

Project Life Cycle

Initial

IDEA

Intermediate Final

Pro

du

ct

Op

erat

ion

s

ProductLife

CycleBusinessPlan

Upgrade

Div

estm

ent


Project v/s Operational work

Organization

Performs Work to achieve objectives

Projects. Temporary and Unique. Purpose to achieve objective And then terminate

Operations. Ongoing & repetitive. Objective is to sustainbusiness

Both Projects & Operations are performed by people, constrained by resources & are planned, executed and controlled


Some examples of Projects

Developing a new Product or ServiceEffecting a change in organization structure

Designing a new transportation vehicleDeveloping a new or modified Information system

Constructing a buildingBuilding a water system for a communityRunning a campaign for a Political Party

Implementing a new business ProcessResponding to a contract solicitation


Projects and Strategic Planning

Projects are a means of organizing activities that cannot be addressed with the organizational normal operational limits

Projects are often utilized as a means of achieving an organizations strategic plans


Projects and Strategic Planning

Strategic considerations that lead to Project authorization

. Market demand

. Organizational need

. Customer request

. Technological advance

. Legal Requirement


• Industrial / Commercial projects are complex in nature.

• Handling of these projects calls for a systematic approach which takes into account– Identification of numerous activities– Required time and other resources– Availability of these resources– Coordination and control


Program Evaluation and Review Technique - PERT

• Developed by US Department of Navy for their “Polaris missile” program in 1958.

• Is a form of network analysis technique consisting of– Translating the proposed job into a model by drawing

a network of activities involved.– Evaluating the network and manipulating it to the

extent possible so as to ensure that there will be a reasonable chance of achieving the objective (viz. completing the job) with in the constraints of time and other resources.

– Using the network to monitor and control the job.


PERT• Is of special relevance to

– One-time tasks such as construction of dams, factories, bridges etc.

– Manufacture of ships, missiles and such other activities which though repetitive are of huge proportion

– R&D projects– Major repairs and overhaul of plant /

machinery– Organization of large events / conferences.


Critical Path Method (CPM)

• CPM originated in Du Pont de Nemours company almost at the same time in 1958.

• CPM is also a Network analysis method and differs from PERT only in the approach to the network analysis.


Gantt Chart

• A planning and control tool used for scheduling jobs and depicting the progress made from time to time.

• In a Gantt chart, a division of space represents both an amount of time and an amount of work to be done in that time



Gantt Chart• Merits

– Remarkably compact for the amount of information it provides for planning and controlling jobs.

– Easy to draw and read– Its pictorial representation of time serves as

an aid to minimize idleness and loss of time.

• Demerits– Does not depict the interdependencies of jobs– Does not aid getting information on the

probability of completion of a job.


Stages in application of PERT/CPM

• Drawing the network– The purpose of this is to identify all the events /

activities that are essential for completion of a Project and to bring out their interrelationship to satisfy the technological / logical sequencing requirements.

• Network Analysis– By incorporating the time required for completing

each of the activities in the network, the project duration as well as the criticality of the activity can be found out. At this stage it is also possible to compute the probability of completing the Project or part of the project in a given specified time



• Resource Allocation and Scheduling– Based on the network calculation and assessment of

resources required for each of the activities, the plan is translated into a time schedule. If it is possible to expedite the activities by incurring additional cost, the economics of doing so are also examined before finalizing the schedule.



• Project control– This stage calls for periodic updating of the network

amongst other things to monitor the progress of the Project, and making necessary changes in the schedules to ensure completion of the Project as close to the target time as possible.


Concepts / Definitions

• ProjectTemporary endeavor undertaken to create a unique

Product, Service or Result.

• Work breakdown structure (WBS)A work breakdown structure is a deliverable oriented

grouping of project elements that organizes and defines the total scope of the project.

Work not in the WBS is outside the scope of the project. As with the scope statement, the WBS is often used to

develop or confirm a common understanding of project scope. Each descending level represents an increasingly detailed description of the project elements.



• ActivityIs a task or item of work to be done, that consumes time, effort, money or other resources.


• Event– Events do not consume resources.– Events are represented by circle– An Event is realized only when all activities

terminating at it are completed.



• Network diagram – Shows how the tasks will flow from beginning

to end



Steps involved in drawing a Network

• Identify the End objective and define the events which are precedent to the End objective

• List out the activities required to accomplish these events

• Draw the events and activities and connect them according to the defined logic.

• Now for every activity answer the following questions– What should precede it– What needs to follow it– What can be done concurrently

• Modify redraw the diagram to suit answers for above.• Check for errors• Number the events as per convention.


Activity sequencing: tools and techniques

• Network diagram – Shows how the tasks will flow from beginning

to end– Types

• Precedence diagramming method (PDM) • Arrow diagramming method (ADM)


• Precedence diagramming method (PDM) : method of constructing a project network diagram to represent the activities using nodes and to represent dependencies using arrows

• Also called activity on node

D

C

B

A

start finish



• PDM could have four types of dependencies– Finish-to-start – Finish-to-finish– Start-to-start– Start-to-finish



• Arrow diagramming method (ADM) : method of constructing project network diagram using arrows to represent activities and connecting them at nodes to show their dependencies. The nodes are Events




• ActivityIs a task or item of work to be done, that consumes time, effort, money or other resources.It is represented by an arrow with its head indicating direction of the progress in the project. i.e. The tail of the arrow marks the commencement of the activity and the Head of the arrow marks the completion of the activity.

5

The number 5 over the arrow indicates the Time required for completion of the activity


• Event– An activity is bound by 2 events referred to as Tail

event and Head event.– Events do not consume resources.– Events are represented by circle– An Event is realized only when all activities

terminating at it are completed.


Tail Event

HeadEvent

Activity

1 2 A

Activity A or 1-2 requiring 5 units of time, Event 1 Event 2

5


• Network– A network is a graphic representation of a

Project’s operation and is composed of activities and events


Event 1 is Start of Project and does not have any activity before it

Event 5 is End of Project and does not have any activity after it

4

3

2

1 5

A

B

C

D

E

GF

F is a Dummy activityIt does not consume any resourceIt is introduce in Network to preserve logic


Rules for drawing a Network

• Only 1 activity may connect any 2 Events• Except for the Start and the End event,

every event must have at least 1 activity entering it and at least 1 activity leaving it.

• Every activity must have a Tail event and a Head event.

• There must be no loops in the Network, i.e. no continuous path must go through one Event 2 times.


NetworkErrors in drawing Network -

Looping

1 2

3

How to determine Looping in a NetworkHead event must always have number higher than Tail event


NetworkErrors in drawing Network -

Dangling

1

4

2 3

Event 1 is Start of Project Event 3 is end of Project

Ways to avoid Danglinga) All Activities must start and finish with an

Eventb) Except for the Start and End Event, each event must

have at least 1 activity entering and 1 activity leaving it.


ACTIVITY Predecessor

A -

B -

C B

D B

E A

F A

G F

H C.E,G

I F

BC

BD A

E

AF

FG

FI

H

C

E

G

1

3

2

4

56

A

BC

D

E

H

F

IG


• A Project consists of a series of activities labeled A, B, C, D, E, F, G, H, and I. The interdependency of activities is as followsA<D,E; B,D<F; C<G; B<H; F,G<I and G<H.

AD

E

B

D

F C G B H

G

F

IG H


1

3

47

2

8

AE

D

B F

ICG

HD1

D2

6

95

D3 D4

Earliest Event Time Latest Event Time

0

01

E

L4

E

L6

E

L7

E

L3

E

L2

E

L5

A(23)

B(8)

C(20)

G(19)

H(4)

D(16)E(24)

F(18) I(10)

D1D2


0

01

E

L4

E

L6

E

L7

E

L3

E

L2

E

L5

A(23)

B(8)

C(20)

G(19)

H(4)

D(16)E(24)

F(18) I(10)

D1D2

Determination of Earliest Event Time of EventStart from the 1st Event i.e. Start of the Project. The Earliest Event time for it = 0Consider the next event 2.Consider all activities terminating at that event i.e. C. C starts at event 1 and requires 20 days. For event 2 occur activity C has to be completed. Hence event 2 can occur only after the time event 1 has occurred and after C is completedi.e after 0 + 20 days = 20 daysSimilarly event 3 will occur after event 1 has occurred and activity A is completedi.e. after 0 +23 days = 23 daysFor event 4 to occur we need to complete activities B and D. The earliest that event 4 can occur is after the completion of activity B and D. i.e. Maximum of (0 + 8), (23 + 16) = 39 days


Hence earliest completion of Project time is 67 days

0

01

39

L4

57

L6

67

L7

23

L3

20

L2

39

L5

A(23)

B(8)

C(20)

G(19)

H(4)

D(16)

E(24)

F(18) I(10)

D1D28,39

20

23

39, 5743, 67, 47

39, 39


0

01

39

L4

57

L6

67

L7

23

L3

20

L2

39

L5

A(23)

B(8)

C(20)

G(19)

H(4)

D(16)E(24)

F(18) I(10)

D1D2

Determination of Latest Event Time of EventStart from the last Event i.e. End of the Project. The Latest Event time for it = Project completion time = 67 daysConsider the next earlier event 6.Consider all activities starting from that event i.e. I. I requires 10 days. Latest time that event 6 can occurWith out delaying the Project completion time is 67 – 10 = 57 days Similarly Latest Time for event 5 = Minimum of (57 – 0) or 67 – 4 = 57 days.For event 4 Latest time is Minimum of (57 – 0) or (57 – 18) = 39 daysFor event 3 the latest event time = Minimum of (39 – 16) or (67 – 24) = 23 daysFor event 2 the latest event time = 57 -19 = 38 days

For event 1 Latest time is Minimum of (38 – 20), (39 – 8), (23 – 23) = 0


0

01

39

394

57

576

67

677

23

233

20

382

39

575

A(23)

B(8)

C(20)

G(19)

H(4)

D(16)E(24)

F(18) I(10)

D1D2

57, 63

5739, 57

23, 43

38

0, 31, 18


0

01

39

394

57

576

67

677

23

233

20

382

39

575

A(23)

B(8)

C(20)

G(19)

H(4)

D(16)E(24)

F(18) I(10)

D1D2

Activity I,j Normal Time

Earliest Time Latest Time Total Float or Slack

availableStart Finish Start Finish

C (1-2) 20 0 20 18 38 18

A (1-3) 23 0 23 0 23 0

B (1-4) 8 0 8 31 39 31

G (2-5) 19 20 39 38 57 18

D (3-4) 16 23 39 23 39 0

E (3-7) 24 23 47 43 67 20

D1 (4-5) 0 39 39 57 57 18

F (4-6) 18 39 57 39 57 0

D2 (5-6) 0 39 39 57 57 18

H (5-7) 4 39 43 63 67 24

I (6-7) 10 57 67 57 67 0

Activities that have Total float = 0 or No Slack available are the activities on the critical path.

Hence Critical path is A-D-F-I


Free Float, Interference Float, Independent Float

10

0

27

13

48

15 617

18

728

28

38

85

12

12

A (7)

B (6)

C (8) D

E (3) I (10)

F (5)

K (3)

G (4)

H (5)

J (16)

Activity I:Total Float = 28 – 17 - 10 = 1 dayActivity F:Total Float = 18-7-5 =6 daysIf F absorbs all its float then I has no float, but if F absorbs only 5 days of its float then I still has its 1 day floatThe float in an activity which can be used with out affecting float of any succeeding activity is called as Free Float

Interference Float = Difference between Total float and Free float

Consider activity K: If the tail event of this K activity is reached as late as possible and the Head event of the activity starts as early as possible then the float for K is 17-8-3 = 6 days. This float of 6 days whose absorption does not affect neither preceding activity nor the subsequent activity is called Independent Float

Total Float = Free float + Interference FloatTotal Float = Lj – Ei – t, Free Float = Ej – Ei – t,. Interference Float = Lj – Ej,, Independent Float = Ej – Li - t


UPDATING PROJECT

• Can be done in 2 ways– Use the revised time estimate of incomplete activities

and calculate from initial event the earliest completion time and the latest completion time of each event in the usual manner to know the Project completion time.

– Change the complete work to zero duration and represent all the activities already finished by an arrow called the Elapsed time arrow. Events in the revised network are renumbered.


• The network for a project is shown below. A review of the project after 15 days reveals that – Activities 1-2, 1-3, 2-3, 2-4 and 3-4 are completed.– Activities 3-5 and 4-6 are in progress and need 2 and 4 days more resp.– The revised estimate shows that activity 8-9 will take only 8 days but

activity 7-9 will need 10 days.• Draw the new network after updating the Project and determine

the Critical path.


10

0

23

3

37

7

413

13

514

15

616

16

719

26

824

249

34

34

2

3

46

7

7

5

9

3

12

810

8

Network at the start of the Project


10

0

23

3

37

7

413

13

514

15

616

16

719

26

824

249

34

34

2

3

46

7

7

5

9

3

12

810

8


10

0

23

3

37

7

413

13

1517

18

1619

19

1722

25

1827

2719

35

35

2

3

46

7

2

5

9

4

128

8

10

Updated Network after day 15 of the Project

1415

15

1315

15

(Finished 4-6) 2

(Finished 3-5) 8

Review at the end of 15th day

Activity Time reqd. (Days)

Activity Status

1-2 0 completed

1-3 0 completed

2-3 0 completed

2-4 0 completed

3-4 0 completed

3-5(13-15) 2 In Progress


5-7(15-17) 5 Not started






New critical path1-2-3-4-14-16-18-19

Project duration increased by 1 day to 35 days

Review at the end of 15th day

Activity Time reqd. (Days)

Activity Status

1-2 0 completed

1-3 0 completed

2-3 0 completed

2-4 0 completed

3-4 0 completed









New critical path1-10-16-18-19

Project duration increased by 1 day to 35 days


2nd Method

10

0

23

3

37

7

413

13

514

15

616

16

719

26

824

249

34

34

2

3

46

7

7

5

9

3

12

810

8


10

010

15

15

1619

19

1517

1817

22

25

1827

2719

35

3515

4

12

8

9

2

5

10

8

Updated Network after day 15 of the Project

Probability Assessment in PERT Network

• 3 time estimates are used for every activity• Most optimistic time (to): it is the duration of the

activity when everything goes well. It is assumed that such performance can be improved only in about 1% of the cases

• Most likely time (tm): Normal time the activity would take. It is the time that would occur most often if the activity was to be repeated several times under the same conditions.

• Most Pessimistic time (tp): It is the longest duration expected under the assumption that every thing goes wrong. There is only 1 % chance that the activity will extend beyond this value.

• Expected Time (te) is given by (to + 4 tm + tp)/6• Standard Deviation ( σ ) is given by (tp - to)/6• Variance is given by ( σ )2



• The range specified by optimistic time and pessimistic time is assumed to enclose every possible estimate of the duration of the activity.

• The most likely time estimate may not coincide with the midpoint (to+ tp)/2.

• Hence it is justified to assume that the duration of each activity follows Beta (β) distribution with its unimodal point occurring at tm and its end points at to & tp



In a PERT Network the critical path comprises of 6 activities whose estimated duration in days is given below If the Project is scheduled for completion with in 42 days, what is the probability of achieving the schedule

Activity Optimistic Time Most likely Time Pessimistic Time

1 3 6 12

2 8 9 10

3 6 8 10

4 2 5 8

5 3 9 12

6 3 4 5



1 3 6 12

2 8 9 10

3 6 8 10

4 2 5 8

5 3 9 12

6 3 4 5

Activity Time Estimate Expected Time T

Std. Dev. σ Variance (σ)2

t1 t2 t3

1 3 6 12 6.5 1.5 2.25

2 8 9 10 9.0 0.33 0.11

3 6 8 10 8.0 0.67 0.45

4 2 5 8 5.0 1.00 1.00

5 3 9 12 8.5 1.5 2.25

6 3 4 5 4.0 0.33 0.11

TE = 41 6.17

Expected Time for completion of Project, TE = 41 daysVariance of Project Time = 6.17Hence σ for project = √6.17 = 2.48 daysScheduled time of the Project TS = 42 daysStandard normal variate = Z = (TS – TE)/σ = (42 -41)/2.48 = 0.4From table for area under the normal curve for value Z = 0.4 The area = 0.5 + 0.1544 = 0.6554Hence the probability of completing the project by schedule time is 65.54%


Example

The following table gives the data for the activities of a small Project. Draw the Network and determine the Expected completion of the Project. What is the probability that the Project would take 5 more days for completion from the Expected completion time. Find the Project completion time that will have 95% probability of success (2010, MU)


1 - 2 1 4 7

1 - 3 5 10 17

2 - 4 3 3 3

2 - 6 1 4 7

3 - 4 8 15 26

3 - 5 2 4 8

4 - 5 5 5 5

5 - 6 2 5 8


Sept 2010 Project Management Using Network Analysis, By Prof. Nadpurohit Slide 56 of 64

PERT CPMThe result of analysis are expressed in terms of events. PERT is hence said to be Event oriented

The results of calculations are considered in terms of activities of the Project. CPM is activity oriented

The duration of activity is computed from multiple time estimates. This is done to take into account time uncertainties. These estimates are used to arrive at the probable achievement of any given scheduled date of Project completion.

CPM uses only single time estimates and hence does not consider time uncertainties

Does not usually consider costs Attempts to bring out relationship between Project duration and cost on the premise that it would be possible to manipulate project duration with in certain limits by incurring certain costs. Thus a shorter duration means deployment of additional resources and hence incidence of additional costs.


Time-Cost Trade-Off

• Project costs– Direct costs

• Those expenses that can be directly charged to each and every activity of the Project. E.g. Manpower costs, material consumed, equipment cost, etc.

– Indirect costs• These are related to overall Project duration.

E.g. service and administration costs.• Any reduction in the Project duration leads to

reduction of these costs.


• Normal duration of an activity– That duration of the activity for which the cost is

minimum

• Normal cost of the activity– The minimum cost of the activity at its normal

duration

Time-Cost Trade-Off


• Cost Slope– Cost Slope for an activity =

(Crash cost – Normal cost) / (Normal time – Crash time)

Time-Cost Trade-Off

Minimum Project Time

Op

tim

um

Pro

ject

Co

st

Pro

jec

t T

ota

l C

os

t

Project Duration

Project Total Cost Curve

Project Indirect Cost Curve

Project Direct Cost Curve

Total Project Cost Time Relationship


• Optimal Project Duration– Is the one which results in minimum overall cost

of the Project.– To determine this optimal duration we should

know• Normal duration of each activity• Normal cost of each activity• Cost of crashing the activity and hence the Cost slope of

the activity• Extend to which the activity can be crashed and • Indirect cost of Project per unit time of Project duration

Time-Cost Trade-Off


Example for determination of Optimal Project duration and Minimum Project cost

• The following table shows details of a Project.The indirect cost is Rs. 400/- per day of Project duration.

• Determine the optimum Project Duration and minimum Project cost

Activity

Immediate Predecessor

Normal

Time (Weeks)

Crash

Time (Weeks)

Normal

Cost (Rs.’000)

Crash

Cost (Rs.’000)

Cost Slop

e (Rs.’000)

A - 10 7 20 30 3.33

B - 8 6 15 20 2.5

C B 5 4 8 14 6

D B 6 4 11 15 2

E B 8 5 9 15 2

F E 5 4 5 8 3

G A, D, C

12 8 3 4 0.25


1

0

0

2

8

8

3

13

14

414

14

A(10)

5

16

21

6

26

26

B(8)

C(5)

D(6)

D1(0)

E(8)

F(5)

G(12)

10, 13, 14

26, 21

14

21

1614

13

8

9, 8, 13

4, 0


Activity


Normal

Time (Weeks)

Crash

Time (Weeks)

Normal

Cost (Rs.’000)

Crash

Cost (Rs.’000)

Cost Slop

e (Rs.’000)

A - 10 7 20 30 3.33

B - 8 6 15 20 2.5

C B 5 4 8 14 6

D B 6 4 11 15 2

E B 8 5 9 15 2

F E 5 4 5 8 3

G A, D, C

12 8 3 4 0.25


1

0

0

2

8

8

3

13

14

414

14

A(10)

5

16

21

6

26

26

B(8)

C(5)

D(6)

D1(0)

E(8)

F(5)

G(12)

10, 13, 14

26, 21

14

21

1614

13

8

9, 8, 13

4, 0

Normal Project Time = 26 daysNormal direct cost = Rs. 71,000/-Indirect cost = 400*26 = 10,400Total Project cost = Normal direct cost + Indirect cost = 71,000 + 10,400 = Rs. 81,400/-Critical path B-D-G (1-2, 2-4, 4-6)Crashing critical activity with minimum cost slope i.e. activity G by 4 days, we get

10

0

28

8

313

14

414

14

516

21

622

22

G(8)

F(5)

E(8)

D(6)

A(10)

B(8)

C(5)

D1(0)


Activity


Normal

Time (Weeks)

Crash

Time (Weeks)

Normal

Cost (Rs.’000)

Crash

Cost (Rs.’000)

Cost Slop

e (Rs.’000)

A - 10 7 20 30 3.33

B - 8 6 15 20 2.5

C B 5 4 8 14 6

D B 6 4 11 15 2

E B 8 5 9 15 2

F E 5 4 5 8 3

G A, D, C

12 8 3 4 0.25


Project Time = 22 daysDirect cost = Rs. 71,000/- + 4*250 = 72,000/-Indirect cost = 400*22 = 8,800Total Project cost = Direct cost + Indirect cost = 72,000 + 8,800 = Rs. 80,800/-Critical path B-D-G (1-2, 2-4, 4-6)Critical activity G has reached its crash limit and hence cannot be crashed furtherCrashing other critical activity with minimum cost slope i.e. activity D by 1 day, we get

10

0

28

8

313

14

414

14

516

21

622

22

G(8)

F(5)

E(8)

D(6)

A(10)

B(8)

C(5)

D1(0)

0

01

13

133

8

82

13

134

16

165

21

216

A(10)

B(8)

C(5)

D1(0)G(8)

D(5)

E(8)

F(5)


Activity


Normal

Time (Weeks)

Crash

Time (Weeks)

Normal

Cost (Rs.’000)

Crash

Cost (Rs.’000)

Cost Slop

e (Rs.’000)

A - 10 7 20 30 3.33

B - 8 6 15 20 2.5

C B 5 4 8 14 6

D B 6 4 11 15 2

E B 8 5 9 15 2

F E 5 4 5 8 3

G A, D, C

12 8 3 4 0.25


Project Time = 21 daysDirect cost = Rs. 72,000/- + 2000/- = 74,000/-Indirect cost = 400*21 = 8,400Total Project cost = Direct cost + Indirect cost = 74,000 + 8,400 = Rs. 82,400/-This Total Project cost (Rs. 82,400/- ) is > the Total Project cost for 22 days viz. Rs. 80,800/-Hence Optimal Project Duration is 22 days and Minimum Total Project cost is Rs. 80,800/-

0

01

13

133

8

82

13

134

16

165

21

216

A(10)

B(8)

C(5)

D1(0)G(8)

D(5)

E(8)

F(5)

Example for determination of Optimal Project duration and Minimum Project cost

• The network given below gives the normal activity durations and the table given below gives the data on the various costs and crash times. The indirect cost is Rs. 120/- per day of Project duration.

• Determine the optimum Project Duration and minimum Project cost

1

0

0

2

5

5

3

11

11

417

17

96

6

5

10

Activity Normal Time (days)

Crash Time

(Days)

Normal Cost (Rs.)

Crash Cost (Rs.)

Cost Slope (Rs.)

1-2 5 4 170 240 70

1-3 9 6 310 550 80

2-3 6 4 80 200 60

2-4 10 8 130 230 50

3-4 6 4 110 290 90


1

0

0

2

5

5

3

11

11

417

17

96

6

5

10

1

0

0

2

5

5

3

11

11

417

17

96

6

5

10

Normal Project Time = 17 daysNormal direct cost = Rs. 800/-Total Project cost = Normal direct cost + Indirect cost / day x Duration of the Project = 800 + 120 x 17 = Rs. 2840Critical path 1-2, 2-3, 3-4Activity 1-3 and 2-4, each have a float of 2 daysCrashing activity critical with minimum cost slope i.e. activity 2-3 we get


Crash Time

(Days)

Normal Cost (Rs.)

Crash Cost (Rs.)

Cost Slope (Rs.)

1-2 5 4 170 240 70

1-3 9 6 310 550 80

2-3 6 4 80 200 60

2-4 10 8 130 230 50

3-4 6 4 110 290 90

Critical Path

1

0

0

2

5

5

3

10

10

416

16

96

5

5

10



Crash Time

(Days)

Normal Cost (Rs.)

Crash Cost (Rs.)

Cost Slope (Rs.)

1-2 5 4 170 240 70

1-3 9 6 310 550 80

2-3 6 4 80 200 60

2-4 10 8 130 230 50

3-4 6 4 110 290 90

1

0

0

2

5

5

3

10

10

416

16

96

5

5

10

Project Time = 16 daysDirect cost = Rs. 800/- + 60 = Rs. 860Total Project cost = Direct cost + Indirect cost / day x Duration of the Project = 860 + 120 x 16 = Rs. 2780Critical path 1-2, 2-3, 3-4Activity 1-3 and 2-4, each have a float of 1 daysCrashing critical activity with minimum cost slope i.e. activity 2-3 we get

1

0

0

2

5

5

3

9

9

415

15

96

4

5

10



Crash Time

(Days)

Normal Cost (Rs.)

Crash Cost (Rs.)

Cost Slope (Rs.)

1-2 5 4 170 240 70

1-3 9 6 310 550 80

2-3 6 4 80 200 60

2-4 10 8 130 230 50

3-4 6 4 110 290 90

1

0

0

2

5

5

3

9

9

415

15

96

4

5

10Project Time = 15 daysDirect cost = Rs. 860/- + 60 = Rs. 920Total Project cost = Direct cost + Indirect cost / day x Duration of the Project = 920 + 120 x 15 = Rs. 2720Critical path: All paths are critical viz. 1-2-3-4; 1-2-4 & 1-3-4Now Activity 1-3 and 2-4, do not have any float Activity 2-3 has reached its crash duration and cannot be reduced furtherCrashing optionsa) Crash 1-2 & 1-3 at a crash cost of 70+80=Rs.150b) Crash 2-4 & 3-4 at a crash cost of 50+90=Rs. 140

1

0

0

2

4

4

3

8

8

414

14

86

4

4

10

1

0

0

2

5

5

3

9

9

414

14

95

4

5

9



Crash Time

(Days)

Normal Cost (Rs.)

Crash Cost (Rs.)

Cost Slope (Rs.)

1-2 5 4 170 240 70

1-3 9 6 310 550 80

2-3 6 4 80 200 60

2-4 10 8 130 230 50

3-4 6 4 110 290 90

Crashing optionsa) Crash 1-2 & 1-3 at a crash cost of 70+80=Rs.150b) Crash 2-4 & 3-4 at a crash cost of 50+90=Rs. 140

1

0

0

2

4

4

3

8

8

414

14

86

4

4

10

1

0

0

2

5

5

3

9

9

414

14

95

4

5

9

a b

Option B looks to be the most beneficial

Project Time 14 daysDirect cost = 920+140=1060Indirect cost = 14x120=1680Hence Total Project cost = 1680+1060=2740



Crash Time

(Days)

Normal Cost (Rs.)

Crash Cost (Rs.)

Cost Slope (Rs.)

1-2 5 4 170 240 70

1-3 9 6 310 550 80

2-3 6 4 80 200 60

2-4 10 8 130 230 50

3-4 6 4 110 290 90

Crashing optionsa) Crash 1-2 & 1-3 at a crash cost of 70+80=Rs.150b) Crash 2-4 & 3-4 at a crash cost of 50+90=Rs. 140

1

0

0

2

4

4

3

8

8

414

14

86

4

4

10

1

0

0

2

5

5

3

9

9

414

14

95

4

5

9

a b

Option c)Activity 2-3 which was at crash time has been relaxed by 1 day to 5 days. By doing this it would be still at a time < its normal time.Activity 1-2 has been crashed by 1 day to 4 days while activity 2-4 has been relaxed to original 10 daysThe crash cost from the original network would be 70+60+90=Rs.220Therefore the Direct cost for this option of 14 days Project Time = 800+220=1020.Indirect cost = 14x120=1680Hence Total Project cost = 1680+1020=2700 Hence this c) option would be the best option

c 1

0

0

2

4

4

3

9

9

414

14

95

5

4

10



Crash Time

(Days)

Normal Cost (Rs.)

Crash Cost (Rs.)

Cost Slope (Rs.)

1-2 5 4 170 240 70

1-3 9 6 310 550 80

2-3 6 4 80 200 60

2-4 10 8 130 230 50

3-4 6 4 110 290 90

Project Time = 14 days, Total Project cost = Direct cost + Indirect cost / day x Duration of the Project = 1020 + 120 x 14 = Rs. 2700

All activities are criticalActivity 1-2 is at its crash time & hence cannot be further crashedFor a Project duration of 13 days the crashing options are a) Crash 2-4 and 3-4 at a crash cost of 50+90=140b) Crash 1-3, 2-3 and 2-4 at a crash cost of 80+60+50=190Choosing a option we get

1

0

0

2

4

4

3

9

9

414

14

95

5

4

10

d

1

0

0

2

4

4

3

9

9

413

13

94

5

4

9

Total Project Cost:= Direct costs + Indirect cost / day x Duration of the project= 1020+140 + 120x13 = Rs. 2720/-

This total Project cost is > Rs. 2700/Hence Optimal Project duration at a minimum Total cost of Rs. 2700/- is 14 days


End of Chapter


Planning is essential for successful accomplishment of any task

The details and depth of planning is related to the complexity of the task to be achieved.

Major elements of planning• Objective

– Defining what is supposed to be achieved.

• Analysis– Gathering facts on what is required to reach the objective.

• Evaluation– Examining the time & effort involved in adapting various

ways to reach the end objective.

• Risk appraisal– Determining the allowance to be made for uncertainty.

• Decision– Specifying the method for implementation

• Schedule– Establishing the time schedule for implementation of the

segment of plan.


NetworkRestraints and Ladders

1 2

3 4

5 6

7 8

A

B

C

DE

Activity 1-3, Waiting to receive 1st lot for starting activity B

Activity 2-4, Time taken to process last lot to complete activity B


• Conditional diagramming methods – Allow for non-sequential activities such as

loops or conditional branches – Neither PDM nor ADM allows loops or

conditional branches

• Network templates – Standardized networks can be used for entire

project or portions (subnets or fragnets) of it

Activity sequencing: tools & techniques


• Project network diagrams– Schematic displays of projects activities and

logical relationships among them – May include full project details or have one or

more summary activities (hammocks)– E.g. PERT

Activity sequencing: outputs


• Project Planning & Scheduling (3 sessions)• CPM & PERT Techniques

• WBS •Network drawing• Network Analysis• Updating of Network

• Project Cost v/s Project Completion time (2 sessions)• Time Cost Trade off• Crashing of Project Completion time

• Resources allocation (1 session)• Resource Smoothening and Resource Leveling

• Project Control (1 session)• Earned Value Management


• Project: Characteristics & Types• Project Identification & Formulation• Project Management & Appraisal

• Project Cost Estimation & DPR• Project Financing• Project Appraisal

• Technical• Financial &• Market


• Arrow diagramming method (ADM) : method of constructing project network diagram using arrows to represent activities and connecting them at nodes to show their dependencies – (called activity-on-arrow AOA)

start finish

A

C

DE

F

B

GH

What type of activity is this?



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Project Management Using Network Analysis By Prof. Nadpurohit.

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