network techniques for project management meaning of project & pm basic network techniques

Network techniques for Project Management

• Meaning of Project & PM• Basic Network techniques

Introduction to Project Management

• A temporary endeavour undertaken to create a unique

product or service (PMI, 2004, P.5)

• Variety and unique in nature

• Mainly used for large R& D activities

• Project management mainly developed by military

Two basic Network techniques

• PERT an acronym for Program Evaluation Review Technique• Originally developed to facilitate the planning and

scheduling the Polaris Fleet Ballistic Missile project of US• PERT is Probabilistic• CPM an acronym for Critical Path Method• developed in 1957 by the DuPont company in US to solve

scheduling problems in industrial setting.• CPM is concerned with the trade off between cost and time.

Definitions used in PERT CPM

• Activity: A time consuming task. An effort that require

resources and take a certain time for completion i.e.

designing a part, connecting a bridge girder

• Event: A milestone. Is a specific point in time indicating

beginning or ending of one or more activities. Does not

consume time or resources.

Rules for Network Construction

• Each activity must have a preceding and a succeeding

event

• Each event should have a distinct number

• There should be no loops in project network

• Each activity is represented by a uniquely numbered

arrow.

Example of PERTActivity Optimist

icMost likely

Pessimistic

Te=ta+4tm+tb/6

VarianceV=(b-a)/6

1-2 9 12 21 13 2

1-3 6 12 18 12 2

2-4 1 1.5 5 2 1

3-4 4 8.5 10 8 1

2-5 10 14 24 15 2.33

4-5 1 2 3 2 0.33i. Draw the network diagram ii. Determine critical pathiii. Calculate event slack and activity floativ. Find the standard deviation of CP durationv. Compute the probability of completing the project in 25

weeks

PERT Analysis

Uses 3 time estimates for each activityOptimistic time (a)Pessimistic time (b)Most likely time (m)

These estimates are used to calculate an expected value and variance for each activity (based on the Beta distribution)

PERT Analysis

Expected activity time (t)t = (a + 4m + b)

6Variance = [ (b – a) / 6 ]2

Standard deviation = SQRT(variance) = (b – a) 6

Project Variance and STD Deviation

• Project variance (σp2)

= ∑ (variances of all critical path activities) σp

2 = 0.11 + 0.11 + 1.0 + 1.78 + 0.11

= 3.11

• Project standard deviation (σp)

= SQRT (Project variance) σp = SQRT ( 3.11) = 1.76

PERT Analysis

Z = (Target time – expected time)σp

Z = (16 - 15) = 0.57 1.76

This means 16 weeks is 0.57 standard deviations above the mean of 15 weeks.

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CPM

• Cost -Time Relationship

• Expediting projects

• Expediting activities requires additional resources,

which means increasing the cost of the project.

• Slope gives us the cost increase associated with a

reduction of one unit of the activity duration

• Slope =Cc-Cn/Tn-Tc

Example 1

i. Find the all normal schedule and costii. Find the all-crash schedule and costiii.Find the least cost plan for all crash time scheduleiv.Find the least cost plan for an intermediate time schedule of 22

weeks.

Activity Activity Normal Crash Cost SlopeCc-Nc/Tn-Tc

week Cost Week

Cost

A 0-1 5 100 4 140 40

B 0-2 9 200 7 300 50

C 1-2 7 250 4 340 30

D 1-3 9 280 7 340 30

E 2-4 5 250 2 460 70

F 2-5 11 400 7 720 80

G 3-5 6 300 4 420 60

I 4-5 8 80 6 140 30

CPM Contd..all normal solution

• 0-1-2-4-5 is the critical path• Duration=5+7+5+8=25 weeks.• Cost=1860

00

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55

A5

E5

C7

B9

D9

F11

G6H0

I8

CPM Contd.. All crash solution

• 0-1-3-4-5 is the critical path • Duration is 17 days• Cost=2860

0

1

4

2

3

5

A4

E2C4

B7

D7

F7

G4H0

I6

Find the minimum cost for the Crash time

• 0-1-3-4-5 is the critical path • Duration is 17 days• Cost=2860

0

1

4

2

3

5

A4

E2C4

B7

D7

F7

G4H0

I6


• Step 1- Get the non critical activities with their slope• Step-2 Select the activity which has largest slope & expand as much as

possible to achieve as a large a cost reduction is possible• Step-3 Expand the next largest activity slope and continue until all path

become critical.

0

1

4

2

3

5

A 4

E 2

C 4

B 7

D 7

F 7

G 4H 0

I 6


• Step 1- Get the non critical activities with their slope =B-50 C-30,E-70,G-60,F-80• Step-2 Select the activity which has largest slope & expand as much as possible to

achieve as a large a cost reduction is possible= Select F for 2 days expansion cost 160

• Step-3 Expand the next largest activity slope and continue until all path become critical.=E one day is possible at$70 in similar manner G by 2 days cost 120 finally activity B can be expanded by one day “ $50 saving.

• So total cost savings are :160+70+120+50=400.Thus the revised cost for 17 days are 2860-400=$2460

00

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A4

E2C4

B7

D7

F7

G4H0

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Determine the least cost plan for any desired number of days ie Least cost plan

for 22 days• First step is to list all critical activities and their slope

• The activity with the least slope will be compressed first, because decreasing the

project time by one day will result in smallest increase in cost.in this case activity C

& I can be selected, arbitrarily C is selected but how many days -the most is up to

is crash time 4 days here. However such a reduction may create one or more

additional critical path. The minute an additional CP is created, the compression

should be stopped. Thus here C is reduced from 7 days to 4 days at a cost of 90.So

the least cost plan for 22 days is 1860+90=1950

Critical activities Activities by events Slope

A 0-1 40

C 1-2 30

E 2-4 70

I 4-5 30

Example2• The following data were obtained from a study of the time

required to overhaul a small power plant.

i. Find the all normal schedule and costii. Find the all-crash schedule and costiii. Find the least cost plan for all crash time scheduleiv. Find the least cost plan for an intermediate time schedule of 17

weeks.

Activity Crash Schedule Normal Schedule

Week cost Week cost

1-2 3 6 5 4

1-3 1 5 5 3

2-4 5 7 10 4

3-4 2 6 7 4

2-6 2 5 6 3

4-6 5 9 11 6

4-5 4 6 6 3

6-7 1 4 5 2

5-7 1 5 4 2



i. Find the all normal schedule and cost 1-2-6-7=16 ,1-2-4-6-7=31, 1-2-4-5-7=25,1-3-4-6-7=28,1-3-4-7=22 and cost= 4+4+6+2=16

ii. Find the least cost plan for all crash time scheduleiii.Find the least cost plan for an intermediate time schedule of 17

weeks.

Activity Crash Schedule Normal

1-2 3 6 5 4

1-3 1 5 5 3

2-4 5 7 10 4

3-4 2 6 7 4

2-6 2 5 6 3

4-6 5 9 11 6

4-5 4 6 6 3

6-7 1 4 5 2

5-7 1 5 4 2

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i. Find the all crash schedule and cost 1-2-6-7=6 ,1-2-4-6-7=14, 1-2-4-5-7=13,1-3-4-6-7=9,1-3-4-5-7=8 and cost= 6+7+9+4= 26

ii. Find the least cost plan for all crash time scheduleiii.Find the least cost plan for an intermediate time schedule of 17

weeks.

Activity Crash Schedule Normal

1-2 3 6 5 4

1-3 1 5 5 3

2-4 5 7 10 4

3-4 2 6 7 4

2-6 2 5 6 3

4-6 5 9 11 6

4-5 4 6 6 3

6-7 1 4 5 2

5-7 1 5 4 2

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Example2The following data were obtained from a study of the time


i. Find the least cost plan for all crash time schedule.ii. Find the least cost plan for an intermediate time schedule of

17 weeks.

Activity Crash Schedule

Normal Schedule

SlopeCc-nc/nt-ct

1-2 3 6 5 4 1

1-3 1 5 5 3 0.5

2-4 5 7 10 4 0.6

3-4 2 6 7 4 0.4

2-6 2 5 6 3 0.5

4-6 5 9 11 6 0.5

4-5 4 6 6 3 1.5

6-7 1 4 5 2 0.5

5-7 1 5 4 2 1

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Project Management Framework

network techniques for project management meaning of project & pm basic network techniques

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