Network Analysis [Compatibility Mode]

7/31/2019 Network Analysis [Compatibility Mode]

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Operational Research Level 4Operational Research Level 4

Prepared byPrepared by T.M.J.A.CoorayT.M.J.A.CoorayDepartment ofDepartment of MathematicsMathematics

MA 4020MA 4020 --Network AnalysisNetwork Analysis 11


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Given:Given: Statement of workStatement of work written description of goalswritten description of goals

work & time frame of projectwork & time frame of project Work Breakdown StructureWork Breakdown Structure

Be able to: develo recedenceBe able to: develo recedencerelationship diagram which showsrelationship diagram which showsse uential relationshi of ro ect activitiesse uential relationshi of ro ect activities



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Popular tool for project schedulingPopular tool for project scheduling Gra h with bar re resentin time for eachGra h with bar re resentin time for each

tasktask

Also shows slack for activitiesAlso shows slack for activities amoun o me ac v y can e e ayeamoun o me ac v y can e e aye

without delaying project)without delaying project)



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20 4 106 8Month

Activity

Design house andobtain financing

Lay foundation

Order and receivematerials

Build house

Select paint

Select carpet

Finish work


31 5 7 9


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Critical Path Method (CPM)Critical Path Method (CPM) -- u on em ng onu on em ng on-- anan

-- deterministic task timesdeterministic task times -- -- --

Pro ect Evaluation & Review Techni uePro ect Evaluation & Review Techni ue(PERT)(PERT)

-- U.S. Navy, Booz, Allen & HamiltonU.S. Navy, Booz, Allen & Hamilton -- multiple task time estimates( probabilistic)multiple task time estimates( probabilistic)

-- activityactivity--onon--arrow network construction (AOA)arrow network construction (AOA)



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In AON, nodes represent activities &In AON, nodes represent activities &arrows show recedence relationshi sarrows show recedence relationshi s In AOA, arrows represent activities &In AOA, arrows represent activities &

nodes are events for oints in timenodes are events for oints in timeAn event is the completion or beginningAn event is the completion or beginning

of an activitof an activitA dummy shows precedence for twoA dummy shows precedence for two

activities with same start & end nodesactivities with same start & end nodes



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Network consists of branches & nodesNetwork consists of branches & nodes

Branch



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1 32Construct forms Pour concrete



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Consider the following table which describesConsider the following table which describesthe activities to be done to build a housethe activities to be done to build a house

ActivityActivity predecessors Durationpredecessors Duration

Design house and obtain financingDesign house and obtain financing --

BB Lay foundationLay foundation A 2A 2Order and receive materialsOrder and receive materials

DD Build houseBuild house B,C 3B,C 3

Select paintSelect paint ,,

FF Select carpetSelect carpet E 1E 1


n s wor n s wor ,,


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3

DummyLayLay foundation

422 3

Order materialr er ma er a

Incorrect Correctprecedence

relationshipprecedence

relationship



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3Lay foundation Dummy

1 2 4 6 73

2

1

3 1Desi n house Order and

Finish

work

Build

house

5

1

and obtain

financing

receive

materialsSelect

carpet

Select

paint



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A path is a sequence of connected activitiesA path is a sequence of connected activities

running from the start to the end node in arunning from the start to the end node in a

The critical path is the path with the longestThe critical path is the path with the longesturation in t e networuration in t e networ

time of the critical path (under normaltime of the critical path (under normalcircumstances)circumstances)



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path1:path1: 11--22--33--44--66--773 + 2 + 0 + 3 + 1 = 9 months; the critical path3 + 2 + 0 + 3 + 1 = 9 months; the critical path

path2:path2: 11--22--33--44--55--66--77

3 + 2 + 0 + 1 + 1 + 1 = 8 months3 + 2 + 0 + 1 + 1 + 1 = 8 monthspa :pa : -- -- -- --

3 + 1 + 3 + 1 = 83 + 1 + 3 + 1 = 8 monthsmonths

-- -- -- -- --

3 + 1 + 1 + 1 + 1 = 73 + 1 + 1 + 1 + 1 = 7 monthsmonths



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(Housebuilding example)(Housebuilding example)

ESES -- earliest time activity can startearliest time activity can start

network to determine ES timesnetwork to determine ES times= ac v y me= ac v y me

ESESijij = maximum (EF= maximum (EFii)) i

EFEFijij = ES= ESijij + t+ tijij ESES1212 = 0= 0


EFEF1212 = ES= ES1212 + t+ t1212 = 0 + 3 = 3 months= 0 + 3 = 3 months


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--ESES

2323

= max (EF= max (EF22

) = 3 months) = 3 months

-- ESES4646 = max (EF= max (EF44) = max (5,4) = 5 months) = max (5,4) = 5 months

-- EFEF4646 = ES= ES4646 + t+ t4646 = 5 + 3 = 8 months= 5 + 3 = 8 months

-- EFEF6767 =9 months, the project duration=9 months, the project duration



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LSLS -- latest time activity can be startedlatest time activity can be started

without delaying the projectwithout delaying the project

Backward pass starts at end of network toBackward pass starts at end of network to

determine LS timesdetermine LS times LFLF -- latest time activity can be completedlatest time activity can be completed

without delaying the projectwithout delaying the project LSLSijij = LF= LFijij -- ttijij LFLFijij = minimum (LS= minimum (LSjj))



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If a deadline is not given take LF of theIf a deadline is not given take LF of the

project to be EF of the last activityproject to be EF of the last activity 6767 ==

LSLS6767 = LF= LF6767 -- tt6767 = 9= 9 -- 1 = 8 months1 = 8 months 5656 66

LSLS5656 = LF= LF5656 -- tt5656 = 8= 8 -- 1 = 7 months1 = 7 months

2424 44 ,,monthsmonths

LSLS = LF= LF -- tt = 5= 5 -- 1 = 4 months1 = 4 months



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

= , =

LS=5, LF=5

= = ES=8, EF=9

1 2 4 6 73

2

1

3 1ES=0, EF=3

LS=5, LF=8 LS=8, LF=9

ES=3, EF=4

5

1LS=0, LF=3 ES=6, EF=7

LS=7, LF=8

LS=4, LF=5

ES=5, EF=6

= =,



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Activit SlackActivit Slack

-- --

Activities on critical path have ES = LS &Activities on critical path have ES = LS &==

SSijij = LS= LSijij -- ESESijij

ijij == ijij -- ijij

SS = LS= LS -- ESES = 4= 4 -- 3 = 1 month3 = 1 month



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Total slack/float or Slack of an activityTotal slack/float or Slack of an activity

TotalTotal slack/slack/ floatfloat meansmeans thethe amountamount ofof

withoutwithout affectingaffecting thethe entireentire projectprojectcomp e oncomp e on meme..

TheThe activitactivit onon aa iveniven athath shareshare thethemaximummaximum possiblepossible slackslack of of thethe activityactivity

.. SumSum of of thethe possiblepossible slacksslacks of of thethe


ac v esac v es cancan nono exceeexcee ee max mummax mumslackslack alongalong thatthat pathpath..


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an activity which does not affect itsan activity which does not affect itsimme iate successors.imme iate successors.

FSFSijij = ES= ESjj EFEFijijFF34=ES4FF34=ES4--EF34 =5EF34 =5--5=05=0FF24=ES4FF24=ES4--ES24=5ES24=5--4=14=1



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ActivityActivity ESES LSLS EFEF LFLF Slack (S) Free slackSlack (S) Free slack

11--2*2* 00 00 33 33 0 00 0

22--33 33 33 55 55 0 00 0

--

33--4*4* 55 55 55 55 0 00 0

44--55 55 66 66 77 1 01 0

44--6*6* 55 55 88 88 0 00 0

55--66 66 77 77 88 1 11 1

--

* Critical path* Critical path



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0 2 4 6 8 10

Activity

Design house and

Lay foundation

Order and receivema er a s

Build house

Select aint

Select carpet

Finish work


1 3 5 7 9


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Reflect uncertainty of activity timesReflect uncertainty of activity times Beta distribution is used in PERTBeta distribution is used in PERT

a + 4m + bt =

b - a

6

(Variance: 2 =

where

2

a = optimistic estimate

m = most likely time estimate


b = pessimistic time estimate


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(tim

e)

(tim

e)

ba m btam

ime)

t

P(


m = t ba


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Equipment testingand modification

DummyEquipmentinstallation

Finaldebugging

1 83 5 9

System

Training

TestingSystem

Testing

development

changeover

Job

Orientation

oston

recruitingtraining



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Activit InformationActivit InformationTime estimates (wks) Mean Time Variance

Activity a m b t 2

1 - 2 6 8 10 8 .44

1 - 3 3 6 9 6 1.001 - 4 1 3 5 3 .44

2 - 5 0 0 0 0 .00

2 - 6 2 4 12 5 2.78

3 - 5 2 3 4 3 .11

4 - 5 3 4 5 4 .11

4 - 7 2 2 2 2 .00

5 - 8 3 7 11 7 1.785 - 7 2 4 6 4 .44

7 - 8 0 0 0 0 .00

-


.

8 - 9 1 10 13 9 4.00


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Early And Late TimesEarly And Late Times

Activity t ES EF LS LF S FS?

11 -- 22 88 0.440.44 00 88 11 99 11

11 -- 33 66 1.001.00 00 66 00 66 00

--

22 -- 55 00 0.000.00 88 88 99 99 11

22 -- 66 55 2.782.78 88 1313 1616 2121 88-- ..

44 -- 55 44 0.110.11 33 77 55 99 22

44 -- 77 22 0.000.00 33 55 1414 1616 1111

-- ..55 -- 77 44 0.440.44 99 1313 1212 1616 33

77 -- 88 00 0.000.00 1313 1313 1616 1616 33


-- ..

88 -- 99 99 4.004.00 1616 2525 1616 2525 00


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( )ES=8, EF=13LS=16 LF=21

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

( )

E = , EF=

LS=1, LF=98

0

5

4( )

ES=13, EF=17

LS=21 LF=25

1 83 5 9= =

( )ES=0, EF=6

LS=0, LF=6 93

( )ES=9, EF=16

LS=9, LF=16= =

( ),

LS=6, LF=9

ES=0, EF=3 04 ( )ES=9, EF=13

LS=12, LF=16

( ),

LS=16 LF=25

LS=2, LF=53 4

2

( )= , =

LS=5, LF=9( )

,

LS=16 LF=16


( )ES=3, EF=5LS=14, LF=16


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Project variance is theProject variance is the sum of the variancessum of the variancesalong the critical pathalong the critical path

22 22 22 22 221313 3535 5858 8989

= 1.00 +0.11 + 1.78 + 4.00= 1.00 +0.11 + 1.78 + 4.00

= 6.89 weeks= 6.89 weeks



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completed (project completion time is )completed (project completion time is )within a s ecified eriod of timewithin a s ecified eriod of time

wherewhere

== ==

Z =x -

pp = project standard deviation= project standard deviation

Z = number of standard deviations of x fromZ = number of standard deviations of x from



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Probability),(~ X

=z

Z

= t Timex



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Standard Normal Distribution OfStandard Normal Distribution Of

transformed Project Timetransformed Project TimeProbability

)1,0(~ NZ

Z

=0 Timez



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What is the probability that the project isWhat is the probability that the project is

completed within 30 weeks?completed within 30 weeks?

P(XP(X 30) = ?30) = ?22 = 6.89 weeks= 6.89 weeks

== 6.896.89 = 2.62 weeks= 2.62 weeks

Z = xZ = x -- =30=30 -- 25 = 1.9125 = 1.91

P ZP Z 1.91 = ?1.91 = ?

2.62



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Determining Probability

From Z Value

Z 0.00 0.01 .. 04 0.09. . .

+0.4719

. . . .

1.9 0.4713 0.4767

.. .. ..

P( x < 30 weeks) = 0.50+ 04719

= 0.9719


= 25 Time (weeks)x = 30

What is the probability that the project willWhat is the probability that the project will


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What is the probability that the project willWhat is the probability that the project will

22 - 25 -32.62 2.62=Z = = -1.14

P(Z< -1.14) = 0.1271

P( x< 22 weeks) = 0.1271



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Useful at many stages of projectUseful at many stages of projectmana ementmana ement Mathematically simpleMathematically simple

Gives critical path & slack timeGives critical path & slack time

Provides project documentationProvides project documentation Useful in monitorin costsUseful in monitorin costs



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Networks generated provide valuableNetworks generated provide valuableproject documentation and graphicallyproject documentation and graphicallypoint out who is responsible for variouspoint out who is responsible for various

project activitiesproject activitiesApplicable to a wide variety of projectsApplicable to a wide variety of projects andand

industriesindustriesUseful in monitoring not only schedules, butUseful in monitoring not only schedules, but



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Assumes clearly defined, independent,Assumes clearly defined, independent,& stable activities& stable activities Specified precedence relationshipsSpecified precedence relationships

beta distributionbeta distribution

u v m mu v m m

OverOver--emphasis on critical pathemphasis on critical path

MA 4020MA 4020 N t k A l iN t k A l i 3939

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Network Analysis [Compatibility Mode]