Networks - CPM and PERT

8/7/2019 Networks - CPM and PERT

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Activity. It is a clearly defined project element, a job or atask which requires the consumption of resources includingtime. It is denoted by an arrow.

Event.An event describes the start or completion of anactivity. It is denoted by a numbered circle.

Path. A path is an unbroken chain of activities from theinitiating node to some other node, generally to the lastnode indicating the end or completion of the project.

Dummy Activity. A dummy activity is that activity which hasa logical function only and consumes no time or resources.

It is denoted by a dotted arrow. There are two types ofdummies:

Identity Dummy. It helps to keep the designation of eachactivity unique or different from another.

Dependency Dummy. It helps to keep the logic correct.

Definitions

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Activity arrows should be drawn from left to right

indicating progressive approach towards the

ultimate objective or the final event.

Crossing of activity arrows should be avoided.

Arrows should be drawn as straight or bent

lines but not curved lines.

Avoid use of unnecessary dummies.Activities are set in the order of their execution.

Events are set in the order of their occurrence.

Rules and Conventions

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Head event number should be greater than tail

event number. No event is numbered until the

tail event of each activity arrow ending into that

event has been numbered.

There should be no danglers or loops. Danglers

are activities which lead no where. All activities

must be connected to events and the finishing

activities must be connected to the finish event

of the project.

Rules and Conventions

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It is used where activity duration is known withcertainty.

Activities are identified.

Dependency of activities is determinedNetwork is drawn

Earliest start times and latest finish times arecalculated

Critical path and critical activities are identified

Critical Path Method

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Rasoi Appliances wants to launch a newly

designed microwave oven.

Activities required for the launch have beenidentified. Their relationship with each other

and the activity duration have also been

determined.

Example

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Activity

Symbol

Activity Description Predecessor Time

(days)

A Develop advertising plan - 6

B Develop promotion and training

materials plan

- 7

C Develop training plan - 8

D Schedule Radio,TV and print media

advertising

A 20

E Develop advertising copy A 18

F Prepare promotional material for in-

store introduction

B 9

G Prepare material for training of

stores representatives

B 8

H Conduct pre-introduction advertising

campaign

D, E 7

I Select store representatives for

training

C 2

J Conduct training G, I 14

K Final in-store launch of product F, H and J 10


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8

C

A

B

6

7

Drawing a Network. We can start withactivities A, B and C as they have no

predecessors.

Consider the activities which can follow only A or B or C asthese have been completed. D and E follow A, F and Gfollow B and I follows C. Add these activities to the net

work.

Activity H is dependent on the completion of Activities Dand E. Activity J is dependent on completion of Activities Gand I. Add these activities to the network.

A dummy activity L has been drawn. This is an identitydummy. Add activity K which can start only when activitiesH, F and J are completed.

Start from the left and number the nodes as we move to theright of the network in the order of their appearance. If two ormore nodes are on the same line, number from top to bottom.

If the dummy L had not been put, then node 5 would havebeen eliminated and activity E also would have ended atnode 7. In that case, a reference to activity between nodes

2-7 would refer to both D and E causing problems ofunique reference.

D

E

F

G

I

20

18

9

8

2

L

J

H

7

14

K

101

2

3

4

5

6

7

8 9


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The earliest that the project can start is attime zero.

The earliest finish time for an activity is the

earliest start time + activity duration.The earliest start time of an activity which is

dependent on two or more activities is thetime at which all the preceding activities arecompleted.

Calculating Earliest Start Time (EST)

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C

A

B

6

7

8

D

E

F

G

I

20

18

9

8

2

L

J

H

7

14

K

10

0

6

7

8

26

24

15

33 43

1

2

3

4

5

6

7

8 9

At Node 2, EST is = EST at Node 1 + Activity Duration of A

= 0 + 6 = 6

At Node 3, EST is = EST at Node 1 + Activity Duration of B

= 0 + 7 = 7

At Node 4, EST is = EST at Node 1 + Activity Duration of B

= 0 + 8 = 8

Node 7 is the starting event for activity J. Activity J can start only when

activities I and G are completed. Activity I can be completed earliest by 8 + 2

= 10, while G can end earliest on 7 + 8 = 15. Hence the earliest that J can

start is the end of 15th day.

At Node 6, EST is = EST at Node 2 + Activity Duration of D =

6 + 20 = 26, and (from Node 5 to Node 7) 24 + 0 = 24. Earliest H can start

is 26.

At Node 5, EST is = EST at Node 2 + Activity Duration of E

= 6 + 18 = 24

At Node 8, we get

From 7 to 8, 26 + 7 = 33. From 3 to 8, 7 + 9 = 16

From 6 to 8, 15 + 14 = 29. Earliest K can start is 33.

At Node 9, EST = EST at Node 8 + duration of activity K

= 33 + 10 = 43


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The project will take 43 days.

Start form Node 9 as 43 days and work

backward to find out the latest time whenthe starting event of an activity must occur,

or the latest time by which all preceding

activities must finish so that the project is

not delayed

Latest Finish Time

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C

A

B

6

7

8

D

E

F

G

I

20

18

9

8

2

L

J

H

7

14

K

10

0

6

7

8

26

24

15

33 43

4333

26

19

26

17

0

11

6

At Node 8, LFT is = LFT at Node 9 Activity Duration of K

= 43 10 = 33

At Node 6, LFT is = LFT at Node 8 Activity Duration of H

= 33 7 = 26At Node 7, LFT is = LFT at Node 8 Activity Duration of J

= 33 14 = 19

At Node 5, LFT is = LFT at Node 7 Activity Duration of L

= 26 0 = 26

At Node 3, LFT is = LFT at Node 6 Activity Duration of G or

LFT at Node 8 Activity Duration of F

= 19 8 =11 or 33 9 = 24

If event 3 occurs at 24 then the project will get delayed. LFT is = 11

At Node 4, LFT is = LFT at Node 6 Activity Duration of I

= 19 2 = 17At Node 2, LFT at Node 5 Activity Duration of E or

LFT at Node 7 Activity Duration of D.

Take the smallest value. At Node 2, LFT is = 6

At Node 1, LST = 0

1

2

3

4

5

6

7

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Spare time in an activity is called float. It isused to economise on resources withoutaffecting the overall duration of the project.

Types of floats Total float

Free float

Interference float

Independent float

Floats

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Total Float. It is the spare time available on any given

activity if the tail event occurred at its earliest timeand the head event at its latest time.

Total Float = Time Latest at Head Time Earliest at

Tail Activity Duration.

= 30 4 3 = 23

14

4 26

308

G

3

Total Float. = 30 4 3 = 23

Free Float. The spare time available on an activity if both the tail and the

head events occur at their earliest time. If this spare time is used up

during the execution of this activity, it will have no effect on subsequent

activities.

Free Float = Time Earliest Head Time Earliest Tail

Activity Duration

= 26 4 3 = 19

Total Float. = 30 4 3 = 23

Free Float. = 26 4 3 = 19

Interference Float. It is equal to total float less free float. If this float is usedup in an activity, it will interfere with the availability of floats available for

subsequent activities.

Interference Float = Total Float Free Float

= 23 19 = 4

Total Float. = 30 4 3 = 23

Free Float. = 26 4 3 = 19

Interference Float. = 23 19 = 4Independent Float. The spare time available in an activity which is

neither affected by the use of float by preceding activities nor

does it affect the float available in subsequent activities.

Independent Float = Time Earliest Head Time

Latest Tail Duration.= 26 8 3 = 15


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Activity Total Float

(TL Head TE

Tail Duration)

Free Float

(TE Head TE

Tail Duration )

Interference

Float

(Total Float Free Float)

Independent

Float

(TE Head TLTail Duration)

A 6 0 6 =0

B 11-0-7=4 7-0-7=0 4-0=4 7-0-7=0

C 17-0-8=9 8-0-8=0 9-0=9 8-0-8=0

D 26-6-20=0

E 26-6-18=2 24-6-18=0 2-0=2 24-6-18=0

F 33-7-9=17 33-7-9=17 17-17=0 33-11-9=13

G 19-7-8=4 15-7-8=0 4-0=4 15-11-8 (0)

H 33-26-7=0

I 19-8-2=9 15-8-2=5 9-5=4 15-17-2 (0)

J 33-15-14=4 33-15-14=4 4-4=0 33-19-14=0

K 43-33-10=0

Calculation of floats


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All activities whose floats are zero arecalled critical activities. They are critical asthey have no spare time available for their

execution. Critical activities have the sameEFT and LST at their start and finishnodes and have no float.

Management must exercise strict control

to ensure that critical activities areexecuted as per schedule.

The path through these activities is calledcritical path.

Identifying Critical Activities and Critical

Path

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All activities whose floats are zero arecalled critical activities. They are critical asthey have no spare time available for their

execution. Critical activities have the sameEFT and LST at their start and finishnodes and have no float.

Management must exercise strict control

to ensure that critical activities areexecuted as per schedule.

The path through these activities is calledcritical path.

Identifying Critical Activities and Critical

Path

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C

A

B

6

7

8

D

E

F

G

I

20

18

9

8

2

L

J

H

7

14

K

10

0

6

7

8

26

24

15

33 43

4333

26

19

26

17

0

11

6

A,D,H and K are critical activities. Critical Path is 1 -2 7- 8 - 9

1

2

3

4

5

6

7

8 9


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Computer packages draw activity on node

networks

Nodes represent the activities.

Arrows merely show their logical relationship.

Nodes representing activities which start the

project may be connected to a milestone

start if desired.

Dummy activities are not used.

Arrows may cross each other.

Activity on Node Network

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For our purpose the node shall be drawn as

follows


20

Earlieststart time Identity Earliestfinish

time

Total

Float

Description

Latest

start time

Duration Latest

finishtime


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A, B and C start the project


21

A

6

C

8

B

7


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Activities D and E follow A; F and G follow B;

and I follows C. Let us add these to the network.

22

A

6

C

8

B

7

D

20

E

18

F

9

I

2

G

8


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Activity H is dependent on completion of D and E; and J is

dependent on completion of G and I. Let us add these to the

network.

23

A

6

C

8

B

7

D

20

E

18

F

9

I

2

G

8

H

7

J

14


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Activity K is dependent on completion of H and J.

24

A

6

C

8

B

7

D

20

E

18

F

9

I

2

G

8

H

7

J

14

K

10


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Let us compute EST by a forward pass. EFT = EST +

Duration.

25

0 A 6

6

0 C 8

8

0 B 7

7

6 D 26

20

6 E 24

18

7 F 16

9

8 I 10

2

7 G 15

8

26 H 33

7

15 J 29

14

33 K 43

10


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Let us compute LFT by a backward pass. LST = LFT

Duration.

26

0 A 6

0 6 6

0 C 8

9 8 17

0 B 7

4 7 11

6 D 26

6 20 26

6 E 24

8 18 26

7 F 16

24 9 33

8 I 10

17 2 19

7 G 15

11 8 19

26 H 33

26 7 33

15 J 29

19 14 33

33 K 43

33 10 43


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Let us compute total float.

Total Float = LFT EFT = LST EST

27

0 A 6

0

0 6 6

0 C 8

9

9 8 17

0 B 7

4

4 7 11

6 D 26

0

6 20 26

6 E 24

8 18 26

7 F 16

17

24 9 33

8 I 10

9

17 2 19

7 G 15

4

11 8 19

26 H 33

0

26 7 33

15 J 29

4

19 14 33

33 K 43

0

33 10 43


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Activities A, D, H and K have zero float and are critical.

The critical path is A D H K, and the project will

take 43 days.

28

0 A 6

0

0 6 6

0 C 8

9

9 8 17

0 B 7

4

4 7 11

6 D 26

0

6 20 26

6 E 24

8 18 26

7 F 16

17

24 9 33

8 I 10

9

17 2 19

7 G 15

4

11 8 19

26 H 33

0

26 7 33

15 J 29

4

19 14 33

33 K 43

0

33 10 43


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Lags show the minimum amount of time a

dependent activity must be delayed to begin

or end. Lags can also be used to constrain

the start and finish of an activity. Lagrelationships can be classified as:

Start to Start

Start to Finish

Finish to Start

Finish to Finish

Lags

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Start to Start

30

Digging trench

for cable 5 kms

Filling of trench

5 kms

Laying of cable

5 kms

Lag 1 day

Lag 1 day


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Start to Finish

31

Testing

Documentation

Lag 2 days


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Finish to Start

32

Pouring

concrete

Removing

formwork

Erecting

formwork

Lag 2

weeks


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Finish to Finish

33

Testing

Assembling prototype Lag 2 weeks


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Lags

34

Testing

Assembling prototype Lag 2 weeks

Lag 3 weeks

A combination of lags may also be used


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Some activities in a project can be completed in shortertime by employing extra resources. But the duration of allactivities cannot be reduced by increasing resourcesbecause of their nature or because of the restrictions onemployment due to space constraints and so on.

If an activity can be completed earlier, extra cost onextra resources will have to be incurred, but if thisreduces the overall duration of the project this will resultin reduction of the overhead costs.

Completing an activity in a shorter time than normal is

referred to as activity crashing and the additional cost iscalled crash cost.

Cost Crashing

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Example

36

Activity Dependency NormalTime

CrashTime

NormalCost

Crash Cost(Increase per day)

A START 4 4 4000 -

B START 8 6 8000 1500

C F, D, FINISH 3 3 600 -

D B 6 5 900 150

E START 7 5 350 100

F A 15 12 9000 900

G B 12 10 1200 200

H G, FINISH 10 8 1000 150J L, FINISH 5 4 1000 300

K E 9 7 900 150

L G & K 11 8 2200 350

Fixed cost is Rs 500 per day.


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Draw the network.

Calculate EST and LFT using normal times of activityduration.

Identify critical activities and path.

The duration of the project will reduce only if activitieson critical path are crashed.

Crash the activity which is cheapest to crash.

After crashing any activity, recompute the timings andidentify critical path. The path may change causingthe critical activities to change.

Continue crashing activities in this manner till theobjective has been achieved, i.e. lowest cost of project or minimum time.

Solution

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10

31

13

B

E

D

G

K JL

H

I

8

6

7

12

5119

20

20

7

80

C

2 5F

9

A

4

3

15

36

194

20B, G, L and J are critical activities and should be considered for crashing

Total cost Rs 47150

Direct cost Rs 29150

Indirect cost Rs 18000

3111

36

20

80

18

33

6

4 7 8

38


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Activity Crash by

Days

Cost of crashing per

day

B 2 1500

G 2 200

L 3 350

J 1 300

Activity G is the cheapest to crash. Crash G by 2 days.Recompute the timings and identify the critical path.


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10

29

13

B

E

D

G

K JL

H

I

8

6

7

10

5119

18

18

7

80

C

2 5F

9

A

4

3

15

34

194

18 29

9

34

18

80

16

31

6

4 7 8

Total cost Rs 46550



Cost reduced by Rs 600

Time reduced by 2 days

B, G, L and J are still critical activities. B, L and J should be considered for

crashing

40


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41

Activity Crash by

Days


day

B 2 1500

L 3 350

J 1 300

Activity J is the cheapest to crash. Crash J by 1 days.Recompute the timings and identify the critical path.


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42

10

29

13

B

E

D

G

K JL

H

I

8

6

7

10

4119

18

18

7

80

C

2 5F

9

A

4

3

15

33

194

18 29

9

33

18

80

15

30

6

4 7 8

Total cost Rs 46350



Cost reduced by Rs 800


B, G, L and J are still critical activities. B, and L should be considered for

crashing


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43

Activity Crash by

Days


day

B 2 1500

L 3 350

Activity L is the cheapest to crash. Crash L by 3 days.Recompute the timings and identify the critical path.


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45

Activity Crash by

Days


day

B 2 1500

Crash activity B by 2 days

Recompute the timings and identify the critical path.


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46

10

24

13

B

E

D

G

K JL

H

I

6

6

7

10

489

16

16

7

60

C

2 5F

9

A

4

3

15

28

194

16 24

7

28

16

60

12

27

6

4 7 8

B, G, L, J, E and K are now critical activities. Even if we crash E or K the

duration on path B G L J cannot be reduced any further. No further

crashing is necessary

Total cost Rs 47900



Cost increased by Rs 750



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47

Activity

Crashed

Crashed by

(Days)

Duration of

Project

Direct

Costs

Indirect

Costs

Total

Cost

Normal - 36 29150 18000 47150

G 2 34 29550 17000 46550

J 1 33 29850 16500 46350

L 3 30 30900 15000 45900

B 2 28 33900 14000 47900

The project can be completed in 30 days at a least cost of Rs 45900.

The minimum time required to complete the project is 28 days at a cost of

Rs 47900


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PERT Programme Evaluation andReview Technique is used inprobabilistic situations when the duration

of activities is not known with certainty.Three time estimates are used for activityduration pessimistic time, optimistic timeand most likely time.

Mean time for each activity is calculated.Network is drawn as for CPM with mean

times as activity durations

PERT Network

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Critical activities are identified.

Standard deviation on the critical path iscalculated.

Using the standard deviation and the meanand assuming normal distribution, projectduration with different levels of confidence orvice versa can be computed.

PERT Network

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Standard deviation cannot be added. Variance can be

added. Standard deviation on the critical path is

PERT - Network

50

4Mean Time

6

o m pt t t

!

St r i tip ot t!

2 2 ...A BSD SD


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PERT - Example

51

Activity Predecessor Optimistic Time(to)

Most likelyTime (tm)

PessimisticTime (tp)

A - 1 2 3

B - 1 2 3

C - 1 2 3

D A 1 2 9

E A 2 3 10

F B 3 6 15

G B 2 5 14

H D, E 1 4 7

J C 4 9 20

K J, G 1 2 9

L H, F, K,

Finish

4 4 4


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Solution

52

Activity OptimisticTime (to)

Mostlikely

Time (tm)

PessimisticTime (tp)

MeanTime

StandardDeviation

A 1 2 3 2 0.33

B 1 2 3 2 0.33

C 1 2 3 2 0.33

D 1 2 9 3 1.33

E 2 3 10 4 1.33

F 3 6 15 7 2.00

G 2 5 14 6 2.00

H 1 4 7 4 1.00

J 4 9 20 10 2.67

K 1 2 9 3 1.33

L 4 4 4 4 0


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Draw the network and find critical activities

53

5

1

1

1 98

7

6

5

4

3

2

B

HE

D

A

LF

C

G

J K

2

3

4 4

2

6

7

2 10 3

4

1915

122

6

0

2

2

1

91

5

1

2

2

0

1

1

7

6

C, J, K and L are critical activities. Critical path is 1 4 7 8 9

Project will be completed in 19 days or less with 50% confidence.


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54

Activity Optimistic

Time (to)

Most

likely

Time (tm

)

Pessimistic

Time (tp)

Mean

Time

Standard

Deviation

C 1 2 3 2 0.33

J 4 9 20 10 2.67

K 1 2 9 3 1.33

L 4 4 4 4 0

2 2 2 2

2 2 2 2

S. .

. 2. .

C J K LSD SD SD SD!

!

!


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55

With 84% level of confidence how much time would theproject take?

PERT assumes that the distribution of the total project

completion time is normal. 84% represents Mean + 1standard deviation. We can say with 84% level ofconfidence that the project will finish in 22 weeks.

What are the chances that the project will finish in 20weeks?

20 190.33

3

x

z

Q

W

!

! !

19 20

0.33 SD

From normal distribution tables the probability when z =0.33 is 0.6290.

There is a 63% chance that the project will finish in 20 weeks.


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PERT is a probabilistic model and is based on three timeestimates. It is used mainly for projects where the activitydurations are uncertain, like research and developmentprojects. Levels of confidence and probabilities can be

associated with the completion date of a project. CPM is based on certainty of the activity durations. It is

used for projects where there is certainty about the timethat each activity would take. The project completionduration is not probabilistic but is certain.

Differences PERT and CPM

56

Networks - CPM and PERT

Documents

Transcript of Networks - CPM and PERT