Network Techniques for Project Managements

8/4/2019 Network Techniques for Project Managements

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Dr. Manoj Kumar TiwariProfessor

Department of Industrial Engineering &Management

I I T Kharagpur


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This is the path that has the longest lengththrough the project.

The shortest time that a project can conceivably

be finished is the critical path. The following computational procedure are

employed for determining the critical path.

1. Earliest Occurrence time (EOT)2. Latest Occurrence time (LOT)

3. Slack


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EOT refer to the time when the event can becompleted at the earliest.

EOT (i) =Max [EOT (k)+ d(k, i) ]

EOT (1)=0

EOT (2)= EOT (1)+ d(2, 1)= 0+13=13

EOT (3)= EOT (1)+ d(3, 1)

= 0+12=12 12

15

2

3

13

2

48

15

2

Duration ofactivity (k, i)

Earliestoccurrence

time of event k( k precedes i)

0

13

12


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12

1

0

5

13

2

28

12

3

13

2

48

20

15

2

=13+2 =15

EOT (4)= Max[(EOT (2)+ d(2, 4)), (EOT(3)+ d(3,4) )]

=

=20

EOT (5)= Max[(EOT (2)+ d(2, 5)), (EOT(4)+ d(4,5) )]

=

=28

12+8=20

13+15=28 20+2=22


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Earliest Star Time (EST) of an activity isobtained as;

EST (i, j) = EOT(i)

Earliest finish Time (EFT) of an activity isobtained as:

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


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LOT refer to the latest allowable time by which anevent can occur.

LOT (i) =Min [LOT (j) - d(i, j) ]

LOT (5)=28 LOT (4)= LOT (5) - d(4, 5)

= 28-2=26 LOT (3)= LOT (4)- d(3, 4)

= 26-8=18

Latest

occurrencetime for j ( jfollows i)

Duration ofactivity (i, i)

12

1

0

5

13

2

28 28

12 18

3

13

2

48

2620

15

2


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LOT (2)=Min[(LOT (5) - d(2, 5) ), (LOT (4) - d(2, 4))]

= Min[(28-15=13), (26-2=24)]

= 13

LOT (1)= Min[(LOT (2) - d(1, 2) ), (LOT (3) - d(1, 3))]

= Min[(13-13=0), (18-12=6)]

= 0

12

100

5

13

2

13

28 28

12 18

3

13

2

48

2620

15

2


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Latest finish Time (LFT) of an activity isobtained as:

LFT (i, j) = LOT(j)

Latest Star Time (LST) of an activity is obtainedas;

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


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Slack is a difference of LOT and EOT for eachevent.

The critical path is marked by events whichhave 0 slack.

It stars with the beginning events andterminates with the end event.

Here, path 1-2-5 is critical path

Event LOT EOT Slack(LOT-EOT)

5 28 28 0

4 26 20 6

3 18 12 6

2 13 13 0

1 0 0 0


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Network diagram for critical path (having 0slack of events)

12

1

00

5

13

2

13

28 28

12 18

3

13

2

48

2620

15

2

0

0

0

66


11/46

Float is the maximum amount of time that this activity can be

delay in its completion before it becomes a critical activity, i.e.,delays completion of the project

There are three measures of floats.1. Total float, 2. Free float, 3. Independent float

Total float (TF): It is the extra time available to completethe activity if it is started as early as possible, without delayingthe completion of the project.

TF (i,j)= LOT (j)- EOT (i)- d(i, j)

TF(2,4)= LOT (4)- EOT (2)- d(2, 4)=26-13-2=11

2 4

LOT(4)=26

LOT=13

d(2,4)=2

EOT=13

EOT=13


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Free float (FF): It is the float under most favorablecondition.

FF (i,j)= EOT (j)- EOT (i)- d(i, j)

FF(2,4)= EOT (4)- EOT (2)- d(2, 4)

=20-13-2=5

Independent float (TF): It is the float under mostadverse condition.

IF (i,j)= EOT (j)- LOT (i)- d(i, j)

IF(2,4)= EOT (4)- LOT (2)- d(2, 4)

=20-13-2=5


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The table summaries all types of time and floats . The critical activities (1-2) and (2-5) have zero (0) floats. This shows that

there is no flexibility associated with them.

Thus, the earliest starting time is same as the latest starting time and theearliest finishing time is the same as the latest finishing time.

Activity(i,j)

Duration Earlieststarttime(i,j)

Earliestfinish

time(i,j)

Lateststarttime(i,j)

Latestfinishtime(i,j)

Totalfloat

Freefloat

Independentfloat(weeks)

a (1,2) 13 0 13 0 13 0 0 0

b (1,3) 12 0 12 6 18 6 0 0c (2,4) 2 13 15 24 26 11 5 0

d (3,4) 8 12 20 18 26 6 0 0

e (2,5) 15 13 28 13 28 0 0 0

f (4,5) 2 20 22 26 28 6 6 0


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1. Early start schedule: The schedule in which all activitystart as early as possible. All event occurs at their earliest because all activates start at their

earliest starting time and finish at their earliest finish tine. There may be time lags b/w the completion of certain activities and

the occurrence of events which these activities lead to.

All activities emanating from an event begin at the same time. it shows a continuous attitude toward the project and a desire to

minimize the possibility of delay.

2. Late start schedule: The schedule arrived at when allactivities are started late as possible.

1. All events occurs at their latest b/c all activities start at their latest

finishing time.2. Some activities may start after a time lag subsequent to the

occurrence of the preceding events.3. All activities leading to an event are completed at the same time. It shows a desire to commit resources late as late posible.


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.

0 5 1510 20 25 30

1

2

3 4

5

Activity(i,j)

Duration Earliest start time(i,j)

Earliest finish

time (i,j)

a (1,2) 13 0 13

b (1,3) 12 0 12

c (2,4) 2 13 15

d (3,4) 8 12 20

e (2,5)15 13 28

f (4,5) 2 20 22


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.

0 5 1510 20 25 30

Activity(i,j)

Duration Latest start time(i,j) Latest finish time (i,j)

a (1,2) 13 0 13

b (1,3) 12 6 18

c (2,4) 2 24 26

d (3,4) 8 18 26

e (2,5) 15 13 28

f (4,5) 2 26 28

1

2

3 4

5


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Project Evaluation and Review Technique(PERT) U S Navy (1958) for the POLARIS missile program

Multiple task time estimates (probabilistic nature) Activity-on-arrow network construction

Non-repetitive jobs (R & D work)

PERTis based on the assumption that an activitysduration follows a probability distribution instead ofbeing a single value


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Three time estimates are required to compute the

parameters of an activitys duration distribution: pessimistic time (tp ) - the time the activity would take if

things did not go well

most likely time (tm ) - the consensus best estimate of theactivitys duration

optimistic time (to ) - the time the activity would take ifthings did go well

Mean (expected time): te =t

p

+ 4 tm

+ to

6

Variance: Vt = (stander deviation ())2 =

tp - to6

tp - to6

2


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Variability in PERT analysis is measured by variance, standerdeviation.

Variance = sum of variances of activity duration on the criticalpath

The standard deviation of the project duration probabilitydistribution is computed by adding the variances of the criticalactivities (all of the activities that make up the critical path) andtaking the square root of that sum

Stander deviation = (sum of variances of activity durationon the critical path )1/2

Thus, the stander deviation of critical path duration

= (4+5.43)1/2

=3.07

activity tp to =(tpto)/6 Variance=2

(1-2) 21 9 2 4.00

(2-5) 24 10 2.33 5.43


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Draw the network.

Analyze the paths through the network andfind the critical path.

The length of the critical path is the mean of theproject duration probability distribution whichis assumed to be normal

Probability computations can now be madeusing the normal distribution table.


21/46

X +

Z

Probability

x X+2 X+3X-X-2X-3

Range Probability

0.682

0.954

0.998

x

X +

X + 2

X + 3


22/46

The probability of completion is completed withinspecified time as follows:

1. First find,

2. Then, obtain cumulative probability up to Z bylooking probability distribution of standerdeviation.

Z =

D - T

Where, T = project mean time

= project standard mean time

D = (proposed ) specified time


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Z Cumulative Probability-3.0 0.001

-2.8 0.003

-2.6 0.005

-2.4 0.008

-2.2 0.014

-2.0 0.023

-1.8 0.036

-1.6 0.055

-1.4 0.081

-1.2 0.115

-1.0 0.159

-0.8 0.212

-0.6 0.274

-0.4 0.345

-0.2 0.421

0.0 0.500

0.2 0.579

0.4 0.655

0.6 0.726

0.8 0.788

1.0 0.841

1.2 0.885

1.4 0.919

1.6 0.945

1.8 0.964

2.0 0.977

2.2 0.986

2.4 0.992

2.6 0.995

2.8 0.997

3.0 0.999


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Given that:

project mean time (T)=28

project standard mean time ()=3.07

proposed (specified)

time (D)

Z Cumulative

probability20 (20-28)/3.07 = -2.6 0.005

25 (25-28)/3.07= -1.0 0.159

30 (30-28)/3.07= -0.6 0.726

Fined thevalue from

previoustable.


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Critical Path Method (CPM) E I Du Pont de Nemours & Co. (1957) for construction of new

chemical plant and maintenance shut-down

Deterministic task times

Activity-on-node network construction

Repetitive nature of jobs In contrast of PERT model, CPM model is developed for projects

which are relatively risk-free

PERT approach is probabilistic while CPM approach isdeterministic

In CPM network analysis we work with single time estimates. In CPM network basically we analyze variation in activity time as

a result of change in recourse assignment.

The main thrust of CPM analysis is on time cost relationship.


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1. The cost associated with a project has twocomponents :

o Direct cost: It is incurred on direct material and direct labor.

o Indirect cost: It consist of overhead items like indirect supplies, rent,insurance, managerial services etc.

2. Activity of the project can be expedited by crashingwhich involves employing more recourses:

3. Crashing reduces time but enhances direct costs b/c offactors like overtime payment, extra payment, and

wastage.The relationship b/w time and direct activity

cost cab be reasonably approximated by a downwardstraight line.


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Crashing:

reducing project time by expending additionalresources

Crash time:

an amount of time an activity is reduced

Crash cost:

cost of reducing activity time

Goal reduce project duration at minimum cost


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.Crashing activity

Crash time

Crash cost

Normal Activity

Normal

time

Normal

cost

Slope = crash cost per unit time

Activity duration


29/46

4. Indirect costs associated with the projectincreases linearly with project duration.

project duration

Linear

relation


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CPM analysis seeks to examine the consequences of crashing ontotal cost.

Indirect cost shows linear relation with project duration thus, therelationship b/w direct cost and project duration is important andobeys following procedure.

The procedure used in this respect is generally as follows:1. Obtained the critical path in network. Determine the project

duration and direct cost.

2. Examine the cost-time slope (crash time per minute) of activitieson critical path obtained and crash the activity which has the least

slope.3. Construct new critical path after crashing as per previous step.

And further determine project duration and cost.

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


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The project network depict the activities, duration and directactivity cost.

The indirect cost is Rs 2,000 per week.

1

8

42

53

7

69 3

10

5

7

9

5 6


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Normal and crash time and cost is listed in the table

Activity Time (week) Cost (Rs) Cost to expedite per week

Normal Crash Normal Crash

1-2 8 4 3,000 6,000 750

1-3 5 3 4,000 8,000 2,000

2-4 9 6 4,000 5,500 500

3-5 7 5 2,000 3,200 600

2-5 5 1 8,000 12,000 1,000

4-6 3 2 1/2 10,000 11,200 2,400

5-6 6 2 4,000 6,800 7006-7 10 7 6,000 8,700 900

5-7 9 5 4,200 9,000 1,200

45,200 70,400


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Find out critical path of the network.

Earliest Occurrence time (EOT)

1

8

42

53

7

6

9 3

10

5

7

9

5 6

8

5

17

1312

1920

30

22

Projectduration

is 30 week


34/46

.

1

8

42

53

7

6

9 3

10

5

7

9

5 6

8

7

17

1421

20

30

9

0


35/46

The critical path of the network is (1-2-4-6-7)

The project duration is 30 week

The total direct cost is Rs 45,200

1

8

42

53

7

6

9 3

10

5

7

9

5 6

0

2

0

1

0

00


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

Time cost slop of activity = (crash cost-normal cost)/ (normaltime crash time)

Time cost slop of activity 1-2=(6000-3000)/(8-4)=750


Time cost slop of activity 4-6=(11200-1000)/(3-21/2)=2400


Minimum


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Activity 2-4 has lowest slop on critical path.

Now, the new critical path is 1-2-5-6-7 The project duration is 29 week

The total direct cost is Rs 46,700

1

8

42

53

7

66 3

10

5

7

9

5 6

Henceactivity 2-4is crashed to

6 (as pergiven table)

(45,200-4,000+5,500)


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Activity 5-6 has the lowest slop on the critical path.

Now the new critical path is (1-2-4-6-7)

The project duration of this path is 27 week

The total direct cost = Rs 49,500

1

8

42

53

7

66 3

10

5

7

9

5 2

Hence ,activity 5-

6 iscrashed to 2

(as pergiven datain the table)

(46,700-4,000+6,800)


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Activity1-2 has lowest slop on the new path

Now the new critical path is (1-2-4-6-7) The project duration of this path is 24 week


1

4

42

53

7

66 3

10

5

7

9

5 2

Hence ,activity 1-2is crashed to

4 (as per

given data inthe table)

(49,500 -3,000+6,000)


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Activity6-7 has lowest slop on the new path

Now there are two new critical path (1-3-5-6-7) & (1-3-5-7)

The project duration of both path is 21 week


1

4

42

53

7

66 3

7

5

5

9

5 2


7 (as per


(52,500 -6,000+8,700)


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For both the critical path a common activity 3-5 has lowest slop

Now , new critical path (1-2-4-6-7) The project duration of the path is 20 week


1

4

42

53

7

66 2

1/2

10

5

5

9

5 2


5 (as per


(55,200 -2,000+3,200)


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For both the critical path a common activity 4-6 has lowest slop

Now the critical path is again (1-2-4-6-7)

The project duration of both path is 19 1/2 week


1

4

42

53

7

66 2

1/2

10

5

5

9

5 2

Hence ,activity 4-6is crashed to21/2 (as per


(56,400 -10,000+11,200)


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With crashing the possible activities direct cost increases and

indirect cost decreases.

Crashed activities Project Duration Direct cost Indirect cost Total cost

None 30 45,000 60,000 105,200

(2-4) 29 46,700 58,000 104,700(2-4) and (5-6) 27 49,500 54,000 103,500

(1-2), (2-4), and (5-6) 24 52,500 48,000 100,500

(1-2), (2-4), (5-6) and (6-7) 21 55,200 42,000 97,200

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

(6-7)

20

56,400 40,000 96,400(1-2), (2-4), (3-5)(,5-6), (4-6)and (6-7)

191/2 57,600 39,000 96,600

Direct cost+ Indirect

cost

(Projectduration) *

(Indirect cost perweek=2000)


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Crashing costs increase as project duration decreases

Indirect costs increase as project duration increases

Reduce project length as long as crashing costs are less than indirect costs

Direct cost

Indirect cost

Total project cost

time

Time-Cost Tradeoff


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Useful at many stages of project management

Mathematically simple

Give critical path and slack time

Provide project documentation

Useful in monitoring costs

How long will the entire project take to be completed? What are

the risks involved?

Which are the critical activities or tasks in the project which could

delay the entire project if they were not completed on time?

Is the project on schedule, behind schedule or ahead of schedule?

If the project has to be finished earlier than planned, what is the

best way to do this at the least cost?

PERT/CPM can answer the following important questions:


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Clearly defined, independent and stable activities Specified precedence relationships Over emphasis on critical paths Deterministic CPM model Activity time estimates are subjective and depend

on judgment PERT assumes a beta distribution for these time

estimates, but the actual distribution may be

different PERT consistently underestimates the expectedproject completion time due to alternate pathsbecoming critical

Documents

Network Techniques for Project Managements