use of pert and cpm model in projects

8/4/2019 Use of Pert and Cpm Model in Projects

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USE OF PERT AND CPM MODEL IN

PROJECTS

Presented by,

Sathya.S (10UTB18), Sheeba Shankari.M.C (10UTB19)


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CONTENT OF THE PRESENTATION Introduction to Network techniques (PERT and CPM Model in

Projects)

Projects and its characteristics that make them amenable to be analyzed

by PERT and CPM

Development of Project Network

Time estimation

Determination of Critical Path

Scheduling of the activities when resources are limited

Crashing of the Project

Network Cost controlling


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INTRODUCTION TO NETWORK TECHNIQUES

Focus to Implement the project

Completion of numerous activities - Resources

Purpose of network technique

Two network techniques:

PERT Model

- Planning and Scheduling

- Risky and Uncertain Projects

- Probabilistic

CPM Model

- Scheduling and Cost controlling in industrial projects

- Risk free Projects

- Deterministic


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PROJECTS THAT ARE AMENABLE TO

ANALYSIS BY PERT AND CPM

Launching a spaceship

Research and Development Program

Construction of a plant

Building of a river valley project

Training of a manpower

Starting a new venture

Characteristics of the above projects that make them amenable to

analysis by PERT and CPM:

- Project broke down into set of jobs

- Jobs must be performed into certain sequence

- Jobs can be start or stop in an independent manner


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TERMINOLOGY USED IN NETWORKS

Activity

Event

Network

Path

Critical

Forward method

Backward method

2

43

1 ReceiveGuests

TakeDinner


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RULES FOR NETWORK CONNECTION

Each activity should have a preceding and

a succeeding event.

Denoted numerically by a pair of

preceding and succeeding event.

Each event should have a unique number.

No loops in the project network

Not more than one activity can have the

same preceding and succeeding event.

1

2 3

1 2


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DETERMINATION OF CRITICAL PATH

Calculate the Earliest occurrence time (EOT) for each event- Earliest Starting time

EST(i, j) = EOT(i)

- Earliest Finishing time

EFT(i, j) = EOT(i) + d(i, j)

Calculate the Latest occurrence time (LOT) for each event

- Latest Starting time

LST(i, j) = LOT(j) - d(i, j)

- Latest Finishing time

LFT(i, j) = LOT(j)


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CONT.

Calculate the Slack for each event

Slack = LOT - EOT

Obtain the Critical and Slack paths

Compute the Activity floats

Total float (i, j) = LOT(j) - EOT(i) - d(i, j)

Free Float (i, j) = EOT(j) - EOT(i) - d(i, j)

Independent Float (i, j) = EOT(j) - LOT(i) - d(i, j)


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SCHEDULING WHEN RESOURCES ARE

LIMITED

Critical activities Vs Non critical activitiesTwo Schedules:

Early Start Schedule

All Events occur at their earliest time (EST & EFT)

May have Time lags in completion of certain activities

All activities emanating from an event begin at the same time

Late Start Schedule

All Events occur at their Latest time (LST & LFT)

Certain activities may start after a time lag

All activities leading to an event complete at the same time


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Problem1

Scheduling to match availability of Manpower

Resource Constraint - only 12 men available for the project

1

4

53

2

2 days (b) 1 day (e)

2 days (c)

2

8

6


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Event EOT LOT Activity Duration EST EFT

1 0 0 a(1-2) 10 0 1

2 1 3 b(1-3) 2 0 2

3 2 2 c (1-5) 8 0 2

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

5 5 5 e(3-5) 6 2 3

f(3-4) 5 2 5


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Problem- 2

Scheduling to match the release of funds

Total Cost = 15,600,000

Resource Constraint:

Release of funds 1st year - 6,900,000

2nd year - 6,800,000

3rd year - 1,900,000

Activity Duration in

months

Cost per Month Cost Rs.

(1-2) 13 200,000 2,600,000

(1-3) 12 500,000 6,000,000

(2-4) 2 1,000,000 2,000,000

(3-4) 8 250,000 2,000,000

(2-5) 15 100,000 1,500.000

(4-5) 2 750,000 1,500,000


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Release of funds:

1st year - 6,900,000 2nd year - 6,800,000 3rd year - 1,900,000

Activity( i, j )

Duration(Months)

Early StartTime

EarlyFinishTime

LatestStart Time

LatestFinishTime

(1 - 2) 13 0 13 0 13

(1 - 3) 12 0 12 6 18

(2 - 4) 2 13 15 24 26

(3 - 4) 8 12 20 18 26

(2 - 5) 15 13 28 13 28

(4 - 5) 2 20 22 26 28


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CONT..

Variance = 2

Variance = Sum of variances of activity durations on critical

path.

Simple Example:

Variance for X is 2 = ((5-3)/6)2 = 0.11 Variance for X is 2 = ((7-1)/6)2 = 1

Y is less certain than X, so accept the contract with X

Optimistictime(weeks)

Most likelytime

Pessimistictime

Duration

Accountant X 3 4 5 4

Accountant Y 1 4 7 4


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CONT Probability of completion by a specified date:

Z Probability distribution of completing a project onthe specified date job

D Specified date to complete a job

T Mean Standard deviation for critical path duration

Simple Problem:

If the project has a critical duration of 28 weeks and its standarddeviation is 3.07, then what is the probability of completing a project in 20days.

Probability of completing project in 20 days is Z=(20-28)/3.07

Z = -2.6

From cumulative probability distribution table Z(-2.6) = 0.005


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CPM MODEL

Focus of CPM analysis is on variations in activity times as a result of changes in

resource assignments.

Main thrust of CPM is on time-cost relationships

CPM find the project schedule which have minimum total cost.

Assumptions in CPM analysis:

Cost associated with the project is of two components, Direct costs and

Indirect costs.

Crashing Project can be expedited which involves employing more

resources.

Time and Direct Cost of the project are inversely proportional

Indirect cost and time are directly proportional


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COST - TIME LINE

Activity time

Crashing activity

Crashtime

Crash

cost

Normal Activity

Normal

time

Normal

cost

Slope = crash cost per unit time


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INDIRECT COSTS

Ind

irectCostof

Project

Project Duration


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PROCEDURE

Obtain the critical path in the normal network. Determine the

project duration and direct cost.

Examine the cost-time slope of activities on the critical path

obtained and crash the activity which has the least slope.

Identify the new critical path after crashing, now determine the

project duration and cost.

Repeat steps 2 and 3 till activities on the critical path are

crashed.


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PROBLEM ON CRASHING

Indirect cost Rs 2000 per week

ActivityNormal

timeCrash time

NormalCost

CrashCost

Cost toexpediteper week

1-2 8 4 3000 6000 750

1-3 5 3 4000 8000 2000

2-4 9 6 4000 5500 500

3-5 7 5 2000 3200 600

2-5 5 1 8000 12000 1000

4-6 3 2 1/2 10000 11200 24005-6 6 2 4000 6800 700

6-7 10 7 6000 8700 900

5-7 9 5 4200 9000 1200

45200 70400


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PROJECT DURATION AND TOTAL COST

Exhibit Activities CrashedProject

durationin weeks

Totaldirectcosts

TotalIndirectcosts

Total Cost

A None 30 45,200 60,000 1,05,200

B (2-4) 29 46,700 58,000 1,04,700

C (2-4) and (5-6) 27 49,500 54,000 1,03,500

D (1-2),(2-4) and (5-6) 24 52,500 48,000 1,00,500

E

(1-2),(2-4),(5-6) and

(6-7) 21 55,200 42,000 97,200

F(1-2),(2-4),(3-5),(5-6)and (6-7)

20 56,400 40,000 96,400

G(1-2),(2-4),(3-5),(5-6), (4-6) and (6-7)

19 1/2 57,600 39,000 96,600


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COST CONTROLLING

Basic Principle of NCS:

Costs are planned, measured, analyzed and controlled in terms of

project activities.

Network Cost System

Cost Projection

Analysis and Control of costs

Monitoring and Control of cost:

Costs incurred to date

Budgeted Costs to date

Value of work done to date

Cost over-run(under-run) to date

Time over-run(under-run) to date


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EXAMPLES

TIME & COST, BUDGET & ACTUAL

EG 1:

If a budgeted cost for the particular activity is Rs.1,000

and it should be completed in 1day. The actual cost incurred is

Rs.800 for completing the 2/3rd of work in 1day. What is the

cost variance?


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Thank You

use of pert and cpm model in projects

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