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Department of Engineering Management & Technology

EMT 6904 Integrated Studies in Engineering Management

Introduction to Network Analysis

There are three common types of project management methods: Gantt Chart, CriticalPath Method- CPM, andProject Evaluation and Review Techniques - PERT. CPM andPERT are network analysis method.

1 Gantt Chart

A Gantt chart is a simple graphical representation showing the time needed for all projectactivities, as well as the earliest times when activities can begin.

The chart consists of a timescales across the bottom, activities are listed down the left-

hand side, and times when activities should be done are blocked off in the body of thechart.

Slackis the amount of time an activity can be delayed without delaying the project.

We can use arrow to denote the actual activity time during project execution.

Network Analysis -- Page 1 of 17 -- Mar/2007

Activities

A

B

C

D

E

F

Month 0 2 4 6 8 10

Duration ofActivity

Slack Duration ofCritical Activity

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Department of EMT EMT6904 Integrated Studies in Engineering Management

Gantt chart technique is used primarily in the planning of resources, and to monitorprogress through the execution phase of a project.

Advantages

- The chart shows clearly the state of each activity at any point in the project.- Any delays, with necessary expediting, rescheduling and preparation, can be

identified easily.- The method is suitable for using in small projects with relatively few activities and

simple straight linear relationships.

Disadvantages

- The chart does not show clearly the inter-relationships of the project activities.

- It is difficult to determine how long the start of a project can be delayed withoutadversely affecting the schedule.- As the project proceeds, adjustment and re-scheduling are difficult using a Gantt

Chart due to deviations from the original plan if there are too many activitiesinvolved.

- Decisions based on the Gantt Chart tend to be arbitrary.

2 Network Analysis

Critical Path Method (CPM) and Project Evaluation and Review Technique (PERT)are very similar and uses a network to show precedence inter-relationships betweenproject activities.

Both methods allow the management to determine the most critical activities in thechain of events leading to project completion.

The only real difference between the CPM and the PERT is in the timing.- CPM assumes that each activity has a fixed duration which is known exactly.- PERT assumes the duration can vary according to a known distribution.

The choice of the method used depends on the size and complexity of project, as wellas the resources available (e.g. computer software). On small projects the Gantt Chartmay be preferred, but on the large projects, critical path systems will almost certainly

be required for cautious analysis.


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2.1 Project Networks

It consists of two basic components:Branches andNodes.

A project network consists of a series of nodes connected by arrows or arcs and an eventis the completion or beginning of an activity in a project.

There are two different approaches for the constructing a network.

(1) Activity-on-Arrow(AOA)

Arrows represent activities and nodes represent events for points in time at whichactivities begin and end.

This approach will be adopted during the lecture.

(2) Activity-on-Node (AON)

Nodes represent activities and arrows show precedence relationships.

Sometimes, a dummy activity is required in the AOA approach to clarify the precedencerelationships between two activities. A dummy activity has an activity time zero andrequires no resources.


BranchNode

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The figure below shows the AOA and AON Approaches for several activity relationshipscommonly encountered.

AOA AON Activity Relationships

S T US T U S precedes T, which precedesU.

S

T

US

T

U

S and T must be completedbefore U can be started.

T

U

ST

U

S

T and U cannot begin until Shas been completed.

S

T

U

V

S

T

U

V

U and V cannot begin until both S and T have beencompleted.

T

S U

V

Dummy

S

T

U

V

U cannot begin until both S andT have been completed; Vcannot begin until T has been

completed.

S T V

UDummy

S T V

U

T and U cannot begin until Shas been completed; V cannot

begin until both T and U havebeen completed.


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2.2 The Critical Path

A network path is a sequence of connected activities that runs from the start node to theend node in the network. The critical path is the longest path through a networkindicating the minimum time in which a project can be completed.

From the network given above, four possible completion paths can be observed:

Path #1 : A-B-Dummy-D-G

Total activity time = 3 + 2 + 0 + 3 + 1 = 9Path #2 : A-B-Dummy-E-F-G

Total activity time = 3 + 2 + 0 + 1 + 1 + 1 = 8Path #3 : A-C-D-G

Total activity time = 3 + 1 + 3 + 1 = 8Path #4 : A-C-E-F-G

Total activity time = 3 + 1 + 1 + 1 + 1 = 7

Path #1, A-B-Dummy-D-G, is the critical path since it is the longest path with theminimum completion time for the project - 9 months.

Detailed analysis

i. Activities B and C cannot start until activity A has been completed. Therefore, thesoonest start time for all activities starting after A is 3 months.

ii Dummy activity is completed after 5 months but activity C is completed at the ends of4 months. Then, the soonest finishing B & C is 5 months.

.


A3

B 2

Dummy 0

C1

D3

E, 1F, 1

G1

months

start at 3 months

A3

B2

Dummy 0

C1

D3

E, 1F, 1

G1

months

event at 5 months

start at 3 months

A3

B2

Dummy 0

C1

D3

E 1F 1

G1

months

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iii. Activity G cannot start until after 8 months (follow the pathe A-B-D) or after 7months (follow the path A-B-E-F). Since all activities have to be completed beforeactivity G can start, then the soonest activity G realised is 8 months.

iv. Obviously, the project time is 9 months by adding 1 month for activity G. Then, thecritical path is determined, A-B-D-G, which is the longest path in the network.

2.3 Activity Scheduling

Step 1 - Determination of theEarliest Activity Times

The earliest activity time is the maximum time in which all preceding activities have beencompleted.

The network analysis starts by finding the earliest possible time for each activity. Aforward pass, i.e. calculation starts from the first node of the network, is performedthrough the network to determine theEarliest Activity Times for each activity.

Assuming

ESi = Earliest Start Time for an activity i

EFi = Earliest Finish Time for an activity i

then,

ESi = maximum(EFprevious)EFi = ESi + ti

where ti is the duration of activity i


start at 8 months

event at 5 months

start at 3 months

A3

B 2

Dummy 0

C1

D3

E 1F 1

G1

months

start at 8 months

event at 5 months

start at 3 months

A3

B2

Dummy 0

C1

D3

E 1F 1

G1

months finish in 9 months

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The same example is used to demonstrate how the earliest activity times can bedetermined.

Activity A

ESA = 0EFA = ESA + tA = 0 + 3 = 3 months

Activity B

ESB = max(EFprevious) = 3 monthsEFB = ESB + tB = 3 + 2 = 5 months

Activity Dummy

ESdummy = max(EFprevious) = 5 monthsEFdummy = ESdummy + tdummy = 5 + 0 = 5 months

Activity C

ESC = max(EFprevious) = 3 monthsEFC = ESC + tC = 3 + 1 = 4 months

Activity D

ESD = max(EFprevious) = 5 monthsEFD = ESD + tD = 5 + 1 = 6 months

etc. ...

Step 2 - Determination of theLatest Activity Times

A latest activity time is determined to check how long each activity can be delayedwithout extending the project time.

Having gone through the network and found the earliest time for each activity, abackward pass is made through the network to determine theLatest Activity Times. The

procedure for this is almost the reverse of the forward pass used to find the earliestactivity times. The calculation starts at the last node of the network and works backwardto the first node.

Assuming

LSi = Latest Start Time for an activity i


32 0

1 3

1 1

1

(ES=0,EF=3)

(ES=3,EF=5) (ES=5,EF=5)

(ES=3,EF=4)

(ES=5,EF=8)

(ES=5,EF=6)(ES=6,EF=7)

(ES=8,EF=9)

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LFi = Latest Finish Time for an activity i

then,

LSi = LFi ti

LFi = minimum(LSfollows)

where ti is the duration of activity i

Activity G

LFG = 9 monthsLSG = LFG tG = 9 1 = 8 months

Activity F

LFF = min(LSfollows) = 8 monthsLSF = LFF tF = 8 1 = 7 months

Activity D

LFD = min(LSfollows) = 8 monthsLSD = LFD tD = 8 3 = 5 months

Activity E

LFE = min(LSfollows) = 7 monthsLSE = LFE tE = 7 1 = 6 months

Activity C

LFC = min(LSfollows) = 5 monthsLSC = LFC tC = 5 1 = 4 months

etc. ...


(LS=0,LF=3)

(LS=3,LF=5) (LS=5,LF=5)

(LS=4,LF=5)

(LS=5,LF=8)

(LS=6,LF=7) (LS=7,LF=8)

(LS=8,LF=9)

3

2 0

1 3

1 1

1

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2.4 Activity Slack

Considering the network example used before,

i. Activities on the critical path

All activities must be start and finish exactly on time and cannot bedelayed for whatever reasons.

The earliest start times and the latest start time of all activities on thecritical path are equal, i.e. the path A-B-D-G in the example.

All activities on the path have to be done at a fixed time and if they arelate the whole project is delayed.

The length of the path determines the overall project duration, 9 monthsin this case.

ii. Non-Critical Activities

The earliest start times and the latest start time of all activities not on thecritical path are not equal, i.e. the activities C, E and F in the example.

Activities have some flexibility in timing which may be delayed orextended without necessarily affecting the overall project duration.

Slack time S, the amount of flexibility in the timing of the activity,exists.

Si = LSi ESiorSi = LFi EFi

e.g.

SE = LSE ESE= 6 5

= 1 month


3

2 0

1 3

1 1

1

(ES=0,EF=3)(LS=0,LF=3)

(ES=3,EF=5)(LS=3,LF=5)

(ES=5,EF=5)

(LS=5,LF=5)

(ES=3,EF=4)(LS=4,LF=5)

(ES=5,EF=8)(LS=5,LF=8)

(ES=5,EF=6)(LS=6,LF=7)

(ES=6,EF=7)(LS=7,LF=8)

(ES=8,EF=9)(LS=8,LF=9)

3

2 0

1 3

1 1

1

S=0

S=0 S=1

S=0S=0 S=0

S=1 S=1

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There is no slack for the activities on the critical path and all otheractivities not on the critical path do have slack.

The slack on activities E and F is called shared slack, which means the

sequence of activities E-F can be delayed 1 month jointly without delaying theproject.

Information on slack is because decisions can be made regardingreallocations of resources.

Resources can be taken temporarily from activities with slack and givento other activities that are behind schedule until the slack is used up.


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Working Example - CPM (Textbook, Problem 17-3, Page 862)

Question:

Activity Immediate precedence Time (Weeks)

A - 4B - 7C A 8D A 3E B 9F C 5G C 2H D, E, F 6I B 5

a. Construct the project network, determine the critical path and indicate the project

completion time and slack for each activity.

b. Construct a Gantt chart.

Suggested Solution:

a.

Activity Time ES EF LS LF Slack

A 4 0 4 0 4 0

B 7 0 7 1 8 1

C 8 4 12 4 12 0

D 3 4 7 14 17 10

E 9 7 16 8 17 1

F 5 12 17 12 17 0

G 2 12 14 21 23 9

H 6 17 23 17 23 0

I 5 7 12 18 23 11


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a. Network

Project completion time = _________ weeks

Critical Path

Critical path : __________________

b. Gantt Chart


Activities

A

B

C

D

E

F

2

G

H

I

0 4 6 8 10 12 14 16 18 20 22 24

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2.5 Probabilistic Time Estimates

In the CPM-type approach, assumption is made that activity times are known withcertainty. However, this is not the case in the real life situation.

Probabilistic time estimates are used in the PERT-type approach to deal with uncertaintycaused by various reasons like labour shortage, weather, supply delays, or accidents.

Experience suggests that activity times can usually be described by a Beta distributionsince no prior knowledge of the shapes of the distributions of activity times have beenknown in a unique project network.

Activity times are stated in terms of three reasonable time estimates:

i. the optimistic time (a) is the shortest time in which the activity can be completed,if all goes exceptionally well.

ii. the most likely time (m) is the probable time required to perform the activity.iii. the pessimistic time (b) is the longest estimated time required to perform an

activity.

Two assumptions are required for the estimation of the mean and variance of a beta

distribution.

i. three times estimates, a, m, and b, can be estimated accurately.ii. the standard deviation, , of the activity time is one-sixth of the range b-a.

Then,

mean (expected time):6

4 bmat

++=

variance :2

2

6

=

ab


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2.6 Probabilistic Network Analysis

Probabilistic analysis of a project network is the determination of the probability ofmeeting the project completion deadline given the mean and variance of a normallydistributed completion time.

An assumption is made that the duration time of one activity does not depend on that ofany other activity, which implies that an expected project time and variance can bedetermine by summing up the individual expected activity times and variances.

Making use of the central limit theorem, the sum of a group of independent, identicallydistributed random variables approaches a normal distribution as the number of randomvariables increases.

In the order words, the mean of the normal distribution is the sum of the expected activitytimes on the path. In the case of the critical path, it is the earliest expected finish time for

the project, tp.

Then,

=x

Z

where = tp = project mean timex = the proposed project timeZ= no. ofx is from the mean

This value ofZ is then used to find the corresponding probability in the given NormalDistribution Table.


Z

= tp

x

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Working Example - Probabilistic Network Analysis

Question:

Activity Precedence Optimistic time

a

Most likely time

m

Pessimistic time

bA - 4 5 12

B - 2 3 10

C - 8 10 12

D A 6 6 6

E A 2 5 8

F A 3 6 15

G B, D 4 4 4

H C, E 5 7 15

I F, G 6 8 10

a. Draw the project network

b. Determine the critical path and the expected project completion time and variance

c. Determine the probability that the project will be completed in 26 days or less.

Suggested Solution:a.

b.

For the activity A: t = expected time = 6

4 bma ++

=6

)12()5(4)4( ++= ( 6 )

2 = variance =2)

6(

ab

=2)

6

)4()12((

= ( 1.78 )

...


6

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b.

Activity Expectedtime, t

Variance2

ES EF LS LF Slack

A 6 1.78 0 6 0 6 0

B 4 1.78 0 4 8 12 8C 10 0.67 0 10 6 16 6

D 6 0 6 12 6 12 0

E 5 1 6 11 11 16 5

F 7 2 6 13 9 16 3

G 4 0 12 16 12 16 0

H 8 1.67 11 19 16 24 5

I 8 0.67 16 24 16 24 0

Critical path : ( ADGI )

The expected completion time = ( 6+6+4+8 ) = ( 24 ) days

Variance = ( 1.78+0+0+0.67 ) = ( 2.45 )

c. The following normal probability distribution describes the probability analysis.

Z =x -

=

2.45)(

)24(-)26(= ( 1.28 )

From the given normal distribution table, the probability is ( 0.8997 )

Therefore, P(x < 26) = ( 89.97% )


P(x

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Normal Distribution Table

The function tabulated is (z) where (z) is the cumulative distribution function of astandardised Normal variablez.

Thus (z) =

zx

dxe2/

2

2

1

is the probability that a standardised Normal variable

selected at random will be greater than a value of z=

x.

x.00 .01 .02 .03 .04 .05

.06 .07 .08 .09

0.0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.53590.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.57530.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.61410.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.65170.4 0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.6879

0.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.7224

0.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.75490.7 0.7580 0.7611 0.7642 0.7673 0.7704 0.7734 0.7764 0.7794 0.7823 0.78520.8 0.7881 0.7910 0.7939 0.7967 0.7995 0.8023 0.8051 0.8078 0.8106 0.81330.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315 0.8340 0.8365 0.8389

1.0 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.86211.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790 0.8810 0.88301.2 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.8980 0.8997 0.90151.3 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.91771.4 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.9319

1.5 0.9332 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 0.9429 0.94411.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.95451.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 0.9608 0.9616 0.9625 0.96331.8 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9699 0.9706

1.9 0.9713 0.9719 0.9726 0.9732 0.9738 0.9744 0.9750 0.9756 0.9761 0.9767

2.0 0.97725 0.97778 0.97831 0.97882 0.97932 0.97982 0.98030 0.98077 0.98124 0.981692.1 0.98214 0.98257 0.98300 0.98341 0.98382 0.98422 0.98461 0.98500 0.98537 0.985742.2 0.98610 0.98645 0.98679 0.98713 0.98745 0.98778 0.98809 0.98840 0.98870 0.988992.3 0.98928 0.98956 0.98983 0.99010 0.99036 0.99061 0.99086 0.99111 0.99134 0.991582.4 0.99180 0.99202 0.99224 0.99245 0.99266 0.99286 0.99305 0.99324 0.99343 0.99361

2.5 0.99379 0.99396 0.99413 0.99430 0.99446 0.99461 0.99477 0.99492 0.99506 0.995202.6 0.99534 0.99547 0.99560 0.99573 0.99585 0.99598 0.99609 0.99621 0.99632 0.996432.7 0.99653 0.99664 0.99674 0.99683 0.99693 0.99702 0.99711 0.99720 0.99728 0.997362.8 0.99744 0.99752 0.99760 0.99767 0.99774 0.99781 0.99788 0.99795 0.99801 0.998072.9 0.99813 0.99819 0.99825 0.99831 0.99836 0.99841 0.99846 0.99851 0.99856 0.99861

3.0 0.99865 0.99869 0.99874 0.99878 0.99882 0.99886 0.99889 0.99893 0.99896 0.99900

3.1 0.99903 0.99906 0.99910 0.99913 0.99916 0.99918 0.99921 0.99924 0.99926 0.999293.2 0.99931 0.99934 0.99936 0.99938 0.99940 0.99942 0.99944 0.99946 0.99948 0.999503.3 0.99952 0.99953 0.99955 0.99957 0.99958 0.99960 0.99961 0.99962 0.99964 0.999653.4 0.99966 0.99968 0.99969 0.99970 0.99971 0.99972 0.99973 0.99974 0.99975 0.99976

3.5 0.99977 0.99978 0.99978 0.99979 0.99980 0.99981 0.99981 0.99982 0.99983 0.999833.6 0.99984 0.99985 0.99985 0.99986 0.99986 0.99987 0.99987 0.99988 0.99988 0.999893.7 0.99989 0.99990 0.99990 0.99990 0.99991 0.99991 0.99992 0.99992 0.99992 0.999923.8 0.99993 0.99993 0.99993 0.99994 0.99994 0.99994 0.99994 0.99995 0.99995 0.999953.9 0.99995 0.99995 0.99996 0.99996 0.99996 0.99996 0.99996 0.99996 0.99997 0.99997


z

(z)

0

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