M. Sundaram Tenn. Tech1 MANAGING PROJECTS USING NETWORK TECHNIQUES.

M. Sundaram Tenn. Tech 1

MANAGING PROJECTS USING NETWORK TECHNIQUES


Project Management Project Management is one of the world’s most in-demand skill sets and is one of the fastest growing professional disciplines in North America. Project Management is used by large corporations, governments, and smaller organizations to standardize and reduce the tasks necessary to complete a project in the most effective and efficient manner.

Engineers that master in project management skills may lead improvement initiatives that result in measurable growth in return on investment, economic value added, sales growth, customer satisfaction and retention, market share, time to market, employee satisfaction, and employee motivation.

PMI provides certification in project management


Network based Techniques-Outline

Project Planning – An Introduction Development of Project Network Identifying Critical paths Probabilistic Analysis in PERT Networks How to use the Normal table Project Cost Control Resource Allocation EVA for monitoring progress of Projects


PERT and CPM

• PERT (Program Evaluation Review Technique) was developed by a joint team set up by the U.S. Navy Special Projects Office that included representatives from Lockheed Aircraft Corporation (Prime contractor of the POLARIS program) and from the consulting company Booze, Allen, and Hamilton.

• The objective of this team was to develop an integrated planning and control system for the Polaris missile submarine program which would help avoid the time and cost overruns that had plagued other such development programs.

• An important feature of the PERT approach is its statistical treatment of the

uncertainty in activity time estimate which involved the collection of three separate time estimates and the calculation of probability estimates of meeting specified schedule dates.

• The three estimates used are: Optimistic time, Pessimistic time, and Most likely time.

• PERT networks are usually to manage projects that have several uncertain activities


PERT and CPM- Contd.

• CPM (Critical Path Method) evolved from a parallel joint effort initiated originally at DuPont and later expanded to include Remington Rand Univac and Mauchly Associates.

• The two key differences of this approach from PERT: (1) the use of only one time estimate for each activity (and thus no statistical treatment of uncertainty) and (2) the inclusion, as an integral part of the overall scheme, of a procedure for time/cost tradeoff to minimize the sum of direct and indirect project costs.

• An important common feature of both PERT and CPM is the use of a network diagram for project representation in which arrows represent activities ("activity-on arrow").

• A modification of this approach involves the representation of activities by circles, with arrows indicating precedence ("activity-on-node").


Applications of PERT and CPM

• Construction projects (e.g.) buildings, highways, houses, and bridges.)

• Preparation of bids and proposals for large projects.

• Maintenance planning of oil refineries, ships, chemical plants and other large scale operations.

• Planning for relocating a facility

• Manufacture and assembly of large and complex products such as airplanes, ships, and mainframe computers.

• Simple projects such as home remodeling, moving to a new house, and home cleaning and painting.

• Design and development of new products

• Facilities planning and implementation of new layouts in manufacturing.

• Development of computer Software packages


What is a design project?Design Project – Unique sequence of activities (work tasks) required to be performed in developing a product.

costcost

Design projectDesign project

performanceperformance

timetime

Changing the length of any side of the project triangle affects the other sides!

Changing the length of any side of the project triangle affects the other sides!


Project Planning

A Simple Project Network

Design Fabricate1 32


Managing a design project?

Design problem –FUNCTION(customer & company requirements)

Solution - FORM(manufacturing specifications)

Activities

(decision making processes)Develop a project plan

then execute the plan


Why should we plan a design project?

WHAT ? ……...scope of work tasksWHEN ? ……...scheduleHOW MUCH?..budgetWHO?………...organization chart,

responsibilities table

Without a clear roadmap, how will you get where you need to go?

In planning a design project we make decisions which answer the following questions

In planning a design project we make decisions which answer the following questions


s ite vis itQ F D /H o QE ng. C ha ra c te ris tic sC o ns tra intsS a is fa c tio n c urve sS e le c t s tra te gyD e ve lo p p la nD e s ign re vie w m e e ting

P ro b le m F o rm ula tio n

G e ne ra te a lte rna tive c o nc e p tsA na lyze a lte rna tive sE va lua te a lte rna tiveR e fineD e s ign re vie w m e e ting

C o nc e p t D e s ign

D e ve lo p a rc hite c tureG e ne ra te c o nfigura tio ns

D F AD F M

A na lyze

E va lua teR e fineD e s in re vie w m e e ting

C o nfigura tio n D e s ign

P a ra m e tric p ro b le m fo rm ula tio nG e ne ra te a lte rna tive s

F M E AF a ult tre e sF ishb o ne d ia gra m s

A na lyze a lte rna tive s

E va lua te a lte rna tive sO p tim iza tio nM ulti- a ttrib ute o p t.D e s ign re vie w m e e ting

P a ra m e tric D e s ign

D e ta il d ra w ingsA sse m b ly d ra w ingsIllus tra tio nsP ro je c t R e p o rtP ro to typ e te s t re p o rtsO ra l p re se nta tio nsD e s ign re vie w m e e ting

D e ta il D e s ign

W id ge tD e s ign

Work breakdown structure


1.0 Design Problem Formulation1.1 Visit Site,

Meet with customers, determine desired attributes and parameters1.2 Complete QFD/HOQ

Determine requirements, engineering characteristics1.3 Satisfaction Curves,

Determine the satisfaction curves for each engineering characteristic.1.4 Create EDS

List in-use purposes for the productList product performance requirements

1.5 Conduct BenchmarkingResearch existing products that are currently availableContact manufacturers and request brochuresAnalyze the competition for functionality and performance

Scope of work (partial)


Project Name Date

Task Smith Johnson Tully Hughs Person n Hours1.1 6 R 1 1 2 2 121.2 3 3 R 2 3 3 141.3 1 2 3 6 6 R 181.4 2 1 2 R 2 4 111.5 4 1 1 3 R 5 141.6 3 2 2 R 2 2 111.7 2 1 2 5 R 3 13

m-th task

Total hours 21 11 13 23 25 93

R - Resposible engineer, in-charge

Responsibilities table

Who does whatassist

responsible


Project schedule


Project Budget


Elements of Project Management

Project team Individuals from different departments within

company

Matrix organization Team structure with members from different

functional areas depending on skills needed

Project manager Leader of project team


Project Planning

Statement of work Written description of goals, work & time

frame of projectActivities require labor, resources &

timePrecedence relationship shows

sequential relationship of project activities


Project Control

All activities identified and included Completed in proper sequence Resource needs identified Schedule adjusted Maintain schedule and

budget Complete on time


A Gantt Chart

Popular tool for project schedulingPopular tool for project schedulingGraph with bar for representing the time Graph with bar for representing the time

for each taskfor each taskProvides visual display of project scheduleProvides visual display of project scheduleAlso shows slack for activitiesAlso shows slack for activities

Amount of time activity can be delayed Amount of time activity can be delayed without delaying projectwithout delaying project


A Gantt Chart

| | | | |Activity

Design house and obtain financing

Lay foundation

Order and receive materials

Build house

Select paint

Select carpet

Finish work

00 22 44 66 88 1010MonthMonth

MonthMonth11 33 55 77 99

Figure 6.2Figure 6.2


CPM/PERT- A Little History

Critical Path Method (CPM) DuPont & Remington-Rand (1956) Deterministic task times Activity-on-node network construction

Project Eval. & Review Technique (PERT) US Navy, Booz, Allen & Hamilton Multiple task time estimates Activity-on-arrow network

construction


The Project Network

Network consists of arcs & nodes

1 32

ArcArc

NodeNode



Network Construction

In AON, nodes represent activities & arrows show precedence relationships

In AOA, arrows represent activities & nodes are events for points in time

An event is the completion or beginning of an activity A dummy shows precedence for two activities with

same start & end nodes


Project Network for a House

3322 00

11

33

11 1111

1 2 4 6 7

3

5

Lay Lay foundationfoundation

Design house Design house and obtain and obtain financingfinancing

Order and Order and receive receive materialsmaterials

DummyDummy

Finish Finish workwork

Select Select carpetcarpet

Select Select paintpaint

Build Build househouse



Concurrent Activities

2 3

Lay foundationLay foundation

Order materialOrder material

(a)(a) Incorrect precedence Incorrect precedence relationshiprelationship

(b)(b) Correct precedence Correct precedence relationshiprelationship

3

42

DummyDummyLay Lay foundationfoundation

Order materialOrder material

11

22 00


Example Problem -1

Develop an activity-on-arrow (AOA) type network for the precedence relationship shown below.

Activity Predecessor

A -

B A

C A

D B,C

E C

F D,E


Solution


Example Problem -2Develop an A-O-A type network from the description below.

Activity Predecessor

A -

B -

C A

D B

E A

F C,D

G C,D,E

H F


Solution


Critical Path

A path is a sequence of connected A path is a sequence of connected activities running from start to end activities running from start to end node in networknode in network

The critical path is the The critical path is the path with the longest path with the longest duration in the networkduration in the network

Project cannot be Project cannot be completed in less than completed in less than the time of the critical the time of the critical pathpath


The Critical Path

A: 1-2-3-4-6-73 + 2 + 0 + 3 + 1 = 9 months

B: 1-2-3-4-5-6-73 + 2 + 0 + 1 + 1 + 1 = 8 months

C: 1-2-4-6-73 + 1 + 3 + 1 = 8 months

D: 1-2-4-5-6-73 + 1 + 1 + 1 + 1 = 7 months

33

22 00

11

33

11 11

111 2 4 6 7

3

5




DummyDummy






The Critical Path

A: 1-2-3-4-6-73 + 2 + 0 + 3 + 1 = 9 months

B: 1-2-3-4-5-6-73 + 2 + 0 + 1 + 1 + 1 = 8 months

C: 1-2-4-6-73 + 1 + 3 + 1 = 8 months

D: 1-2-4-5-6-73 + 1 + 1 + 1 + 1 = 7 months

33

22 00

11

33

11 11

111 2 4 6 7

3

5




DummyDummy





The Critical Path

The Critical Path


The Critical Path 33

22 00

11

33

11 1111

1 2 4 6 7

3

5




DummyDummy





1 2 4 6 7

3

5

3

2 0

1

3

1 1

Start at 3 months

Start at 5 months

1

Finish at 9 months

Start at 8 months

Activity Start Times


Early Times

ES - earliest time activity can start Forward pass starts at beginning of

CPM/PERT network to determine ES times EF = ES + activity time

ESij = maximum (EFi) EFij = ESij - tij

ES12 = 0 EF12 = ES12 + t12

= 0 + 3 = 3 months

2003

33

22 00

11

33

11 11

111 2 4 6 7

3

5




DummyDummy






Computing Early Times

ES23 = max EF2 = 3 monthsES46 = max EF4 = max 5,4 = 5 monthsEF46 = ES46 + t46 = 5 + 3 = 8 monthsEF67 = 9 months,

the project duration

33

22 00

11

33

11 11

111 2 4 6 7

3

5




DummyDummy






Computing Early Times 33

22 00

11

33

11 11

111 2 4 6 7

3

5




DummyDummy





1 2 4 6 7

3

5

3

2 0

1

3

1 1

1

(ES = 0, EF = 3)

(ES = 5, EF = 8)

(ES = 3, EF = 5)

(ES = 3, EF = 4)

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

(ES = 8, EF = 9)

(ES = 5, EF = 5)

Early Start and Finish Times


Late Times

LS - latest time activity can start & not delay project

Backward pass starts at end of CPM/PERT network to determine LS times

LF = LS + activity timeLSij = LFij - tij

LFij = minimum (LSj)


Computing Late Times

LF67 = 9 months LS67 = LF67 - t67 = 9 - 1 = 8 months LF56 = minimum (LS6) = 8 months LS56 = LF56 - t56 = 8 - 1 = 7 months LF24 = minimum (LS4) = min(5, 6)

= 5 months LS24 = LF24 - t24 = 5 - 1 = 4 months

33

22 00

11

33

11 11

111 2 4 6 7

3

5




DummyDummy






Computing Late Times

LF67 = 9 monthsLS67 = LF67 - t67 = 9 - 1 = 8 monthsLF56 = minimum (LS6) = 8 monthsLS56 = LF56 - t56 = 8 - 1 = 7 monthsLF24 = minimum (LS4) = min(5, 6)

= 5 monthsLS24 = LF24 - t24 = 5 - 1 = 4 months

33

22 00

11

33

11 11

111 2 4 6 7

3

5




DummyDummy






1 2 4 6 7

3

5

3

2 0

1

3

1 1

1

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

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

ES = 3, EF = 4LS = 4, LF = 5( )ES = 0, EF = 3

LS = 0, LF = 3( )

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

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

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

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

Early and Late Start and Finish Times


Activity Slack

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

Activities not on critical path have slack Sij = LSij - ESij

Sij = LFij - EFij

S24 = LS24 - ES24 = 4 - 3 = 1 month


Activity Slack Data

33

22 00

11

33

11 11

111 2 4 6 7

3

5


Design Design house and house and obtain obtain financingfinancing

Order Order and and receive receive materialsmaterials

DummyDummy





Activity Activity LSLS ESES LFLF EFEFSlackSlack*1-2*1-2 00 00 33 33

00*2-3*2-3 33 33 55 55

002-42-4 44 33 55 44

11*3-4*3-4 55 55 55 55

004-54-5 66 55 77 66

11*4-6*4-6 55 55 88 88

005-65-6 77 66 88 77

11*6-7*6-7 88 88 99 99

00* Critical path* Critical path


Activity Slack Data

33

22 00

11

33

11 11

111 2 4 6 7

3

5




DummyDummy



SelectSelect paintpaint


ActivityActivity LSLS ESES LFLF EFEF SlacksSlacks

*1-2*1-2 00 00 33 33 00*2-3*2-3 33 33 55 55 002-42-4 44 33 55 44 11

*3-4*3-4 55 55 55 55 004-54-5 66 55 77 66 11

*4-6*4-6 55 55 88 88 005-65-6 77 66 88 77 11

*6-7*6-7 88 88 99 99 00* Critical path* Critical path

1 2 4 6 7

3

5

3

2 0

1

3

1 1

1

Activity Slack

S = 0

S = 1S = 1

S = 1

S = 0 S = 0

S = 0

S = 0


Other Methods of Determining the Critical Path

• Simple Method

• Tabular Method

• Enumeration of all paths


Simple Method


Enumeration of all paths


Probabilistic Time Estimates

Reflect uncertainty of activity timesBeta distribution is used in PERT


Probabilistic Time Estimates

Reflect uncertainty of activity timesBeta distribution is used in PERT

aa = optimistic estimate = optimistic estimatemm = most likely time estimate = most likely time estimatebb = pessimistic time estimate= pessimistic time estimate

wherewhere

Mean (expected time):Mean (expected time): tt = =aa + 4 + 4mm + + bb

66

Variance:Variance: 22 = =bb - - aa

66

22


Example Beta Distributions

PP(t

ime)

(tim

e)

PP(t

ime)

(tim

e)

PP(t

ime)

(tim

e)

TimeTimeaa mmtt bbaa mm tt bb

m m = = tt

TimeTime

TimeTimebbaa


Kat Tech Company

System System changeover changeover

2

4

6

1 73 5 9

8

Manual Manual TestingTesting

DummyDummy

System System TrainingTraining

DummyDummySystem System TestingTesting

OrientationOrientation

Position Position recruitingrecruiting

System System developmentdevelopment

Equipment Equipment installationinstallation

Equipment testing Equipment testing and modificationand modification

Final Final debuggingdebugging

Job Job trainingtraining

aa

bb

cc

dd

ee

ff

gg

hh

ii

jj kk

ll

mm


Activity Estimates

1 - 21 - 2 66 88 1010 88 0.440.441 - 31 - 3 33 66 99 66 1.001.001 - 41 - 4 11 33 55 33 0.440.442 - 52 - 5 00 00 00 00 0.000.002 - 6 2 - 6 22 44 1212 55 2.782.783 - 5 3 - 5 22 33 44 33 0.110.114 - 54 - 5 33 44 55 44 0.110.114 - 84 - 8 22 22 22 22 0.000.005 - 75 - 7 33 77 1111 77 1.781.785 - 85 - 8 22 44 66 44 0.440.447 - 87 - 8 00 00 00 00 0.000.006 - 96 - 9 11 44 77 44 1.001.007 - 97 - 9 11 1010 1313 99 4.004.00

TIME ESTIMATES (WKS)TIME ESTIMATES (WKS) MEAN TIMEMEAN TIME VARIANCEVARIANCE

ACTIVITYACTIVITY aa mm bb tt σσ22

2

4

6

1 73 5 9

8


2

4

6

1 73 5 9

8

Early and Late Times

For Activity 1-2For Activity 1-2

aa = 6,= 6, m m = 8, = 8, bb = 10 = 10

t t = = = 8 weeks= = = 8 weeksaa + 4 + 4mm + + bb

666 + 4(8) + 106 + 4(8) + 10

66

22 = = = week = = = weekbb - - aa

66

2210 - 610 - 6

66

224499


2

4

6

1 73 5 9

8

Early and Late Times

ACTIVITYACTIVITY tt σσ22 ESES EFEF LSLS LFLF SS

1 - 21 - 2 88 0.440.44 00 88 11 99 111 - 31 - 3 66 1.001.00 00 66 00 66 001 - 41 - 4 33 0.440.44 00 33 22 55 222 - 52 - 5 00 0.000.00 88 88 99 99 112 - 6 2 - 6 55 2.782.78 88 1313 1616 2121 883 - 5 3 - 5 33 0.110.11 66 99 66 99 004 - 54 - 5 44 0.110.11 33 77 55 99 224 - 84 - 8 22 0.000.00 33 55 1414 1616 11115 - 75 - 7 77 1.781.78 99 1616 99 1616 005 - 85 - 8 44 0.440.44 99 1313 1212 1616 337 - 87 - 8 00 0.000.00 1313 1313 1616 1616 336 - 96 - 9 44 1.001.00 1313 1717 2121 2525 887 - 97 - 9 99 4.004.00 1616 2525 1616 2525 00


Kal Tech Company

2

4

6

1 73 5 9

8

ES = 9, EF = 16ES = 9, EF = 16LS = 9, LF = 16LS = 9, LF = 16













88

55

44

66 33 77

99

33

22

44 00

00


Kal Tech Company

2

4

6

1 73 5 9

8














88

55

44

66 33 77

99

33

22

44 00

00

2 = 2 + 2 + 2 + 2

= 1.00 + 0.11 + 1.78 + 4.00

= 6.89 weeks

13 35 57 79

Total project variance


Probabilistic Network Analysis

Determine probability that project is Determine probability that project is completed within specified timecompleted within specified time

wherewhere == ttpp = project mean time = project mean time

== project standard deviationproject standard deviationx x == proposed project timeproposed project timeZZ = = number of standard deviations number of standard deviations xx

is from meanis from mean

ZZ = =xx - -


Normal Distribution Of Project Time

= = ttpp TimeTimexx

Z

ProbabilityProbability


Kal Tech Company

What is the probability that the project What is the probability that the project is completed within 30 weeks?is completed within 30 weeks?

= 25= 25 Time (weeks)Time (weeks)xx = 30 = 30

PP((xx 30 weeks) 30 weeks)


Kal Tech Company


Example 6.2Example 6.2



22 = 6.89 weeks= 6.89 weeks

= 6.89= 6.89

= 2.62 weeks= 2.62 weeks

ZZ ==

==

= 1.91= 1.91

xx - -

30 - 2530 - 252.622.62


Kal Tech Company




22 = 6.89 weeks= 6.89 weeks

= 6.89= 6.89


ZZ ==

==

= 1.91= 1.91

xx - -

30 - 2530 - 252.622.62

From Table A.1, a From Table A.1, a ZZ score of 1.91 score of 1.91 corresponds to a probability of 0.4719.corresponds to a probability of 0.4719.Thus Thus PP(30) = 0.4719 + 0.5000 = 0.9719(30) = 0.4719 + 0.5000 = 0.9719


Kal Tech Company




0.3729


Kal Tech Company


22 = 6.89 weeks= 6.89 weeks

= 6.89= 6.89


ZZ ==

==

= -1.14= -1.14

xx - -

22 - 2522 - 252.622.62



0.3729


What is the probability that the What is the probability that the project is completed within 22 project is completed within 22 weeks?weeks?

22 = 6.89 = 6.89

= 6.89= 6.89

= 2.62 = 2.62

ZZ ==

==

= -1.14= -1.14

xx - -

22 - 2522 - 252.622.62

From Table A.1, a From Table A.1, a ZZ score of -1.14 corresponds score of -1.14 corresponds to a probability of 0.3729.to a probability of 0.3729.Thus Thus PP(22) = 0.5000 - 0.3729 = 0.1271(22) = 0.5000 - 0.3729 = 0.1271

= 25= 25 Time Time (weeks)(weeks)

xx = 22 = 22


0.3729

Kal Tech Company


Another Example problemFor the PERT network given below, Determine the critical path and the expected length of the critical path.

i. Compute the probability of completing the project in 20 days. ii. What is the likely project duration that the project manager can be confident with 95%

certainty?

Activity Node a m b

A 1 2 1 1 7

B 1 3 1 4 7

C 1 4 2 2 8

D 2 5 1 1 1

E 3 5 2 5 14

F 4 6 2 5 8

G 56 3 6 15


Solution


Project Crashing

Crashing is reducing project time Crashing is reducing project time by expending additional resourcesby expending additional resources

Crash time is an amount of time an Crash time is an amount of time an activity is reducedactivity is reduced

Crash cost is the cost of reducing Crash cost is the cost of reducing the activity timethe activity time

Goal is to reduce project duration Goal is to reduce project duration at minimum costat minimum cost


Housebuilding Network

121288 00

44 1212

44 44

441 2 4 6 7

3

5


121288 00

44 1212

44 44

441 2 4 6 7

3

5


$7,000 –

$6,000 –

$5,000 –

$4,000 –

$3,000 –

$2,000 –

$1,000 –

–| | | | | | |

0 2 4 6 8 10 12 14 Weeks

Normal activity

Normal time

Normal cost


121288 00

44 1212

44 44

441 2 4 6 7

3

5


$7,000 –

$6,000 –

$5,000 –

$4,000 –

$3,000 –

$2,000 –

$1,000 –

–| | | | | | |

0 2 4 6 8 10 12 14 Weeks

Crash cost

Crashed activity

Normal activity

Normal timeCrash time

Normal cost



$7,000 –

$6,000 –

$5,000 –

$4,000 –

$3,000 –

$2,000 –

$1,000 –

–| | | | | | |

0 2 4 6 8 10 12 14 Weeks

Crash cost

Crashed activity

Normal activity

Normal timeCrash time

Normal cost

Slope = crash cost per week

Total crash cost $2,000Total crash time 5

= = $400 per week


Normal Activity and Crash Data 1 2 4 6 7

3

5

TOTALTOTALNORMALNORMAL CRASHCRASH ALLOWABLEALLOWABLE CRASHCRASH

TIMETIME TIMETIME NORMALNORMAL CRASHCRASH CRASH TIMECRASH TIME COST PERCOST PERACTIVITYACTIVITY (WEEKS)(WEEKS) (WEEKS)(WEEKS) COSTCOST COSTCOST (WEEKS)(WEEKS) WEEKWEEK

1-21-2 1212 77 $3,000$3,000 $5,000$5,000 55 $400$400

2-32-3 88 55 2,0002,000 3,5003,500 33 500500

2-42-4 44 33 4,0004,000 7,0007,000 11 3,0003,000

3-43-4 00 00 00 00 00 00

4-54-5 44 11 500500 1,1001,100 33 200200

4-64-6 1212 99 50,00050,000 71,00071,000 33 7,0007,000

5-65-6 44 11 500500 1,1001,100 33 200200

6-76-7 44 33 15,00015,000 22,00022,000 11 7,0007,000

$75,000$75,000 $110,700$110,700



3

5



1-21-2 1212 77 $3,000$3,000 $5,000$5,000 55 $400$400

2-32-3 88 55 2,0002,000 3,5003,500 33 500500

2-42-4 44 33 4,0004,000 7,0007,000 11 3,0003,000

3-43-4 00 00 00 00 00 00

4-54-5 44 11 500500 1,1001,100 33 200200

4-64-6 1212 99 50,00050,000 71,00071,000 33 7,0007,000

5-65-6 44 11 500500 1,1001,100 33 200200

6-76-7 44 33 15,00015,000 22,00022,000 11 7,0007,000

$75,000$75,000 $110,700$110,700

12

8 0

4 12

4 4

41 2 4 6 7

3

5

$400

$500

$3,000 $7,000

$200 $200

$7,000



3

5



1-21-2 1212 77 $3,000$3,000 $5,000$5,000 55 $400$400

2-32-3 88 55 2,0002,000 3,5003,500 33 500500

2-42-4 44 33 4,0004,000 7,0007,000 11 3,0003,000

3-43-4 00 00 00 00 00 00

4-54-5 44 11 500500 1,1001,100 33 200200

4-64-6 1212 99 50,00050,000 71,00071,000 33 7,0007,000

5-65-6 44 11 500500 1,1001,100 33 200200

6-76-7 44 33 15,00015,000 22,00022,000 11 7,0007,000

$75,000$75,000 $110,700$110,700

7

8 0

4 12

4 4

41 2 4 6 7

3

5

$500

$3,000 $7,000

$200 $200

$7,000

Crash cost = $2,000



3

5



1-21-2 1212 77 $3,000$3,000 $5,000$5,000 55 $400$400

2-32-3 88 55 2,0002,000 3,5003,500 33 500500

2-42-4 44 33 4,0004,000 7,0007,000 11 3,0003,000

3-43-4 00 00 00 00 00 00

4-54-5 44 11 500500 1,1001,100 33 200200

4-64-6 1212 99 50,00050,000 71,00071,000 33 7,0007,000

5-65-6 44 11 500500 1,1001,100 33 200200

6-76-7 44 33 15,00015,000 22,00022,000 11 7,0007,000

$75,000$75,000 $110,700$110,700

7

7 0

4 12

4 4

41 2 4 6 7

3

5

$500

$3,000 $7,000

$200 $200

$7,000

Crash cost = $2,000 + $500 = $2,500


Crashing costs increase as project Crashing costs increase as project duration decreasesduration decreases

Indirect costs increase as project Indirect costs increase as project duration increasesduration increases

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

Time-Cost Relationship


Time-Cost TradeoffC

ost

($)

Co

st (

$)

Project durationProject duration

CrashingCrashing TimeTime

MMinimuinimum cost = optimal project timem cost = optimal project timeTotal project costTotal project cost

Indirect costIndirect cost

Direct costDirect cost


Earned -Value Analysis (EVA)•EVA is a widely used project control method that compares actual Vs budgeted expenses on a period-by-period basis.•Project control is comparing project progress to the plan so that corrective action can be taken when deviation from planned performance occurs.•The first step in performing an earned -value analysis is to calculate the budget cost of work scheduled(BCWS) for each time period.

- For example if weekly periods are chosen, the total budgeted amount

for a task will be distributed evenly during the scheduled period. The Table next page shows a project with three tasks. Task “A” is budgeted for $3,000. If it is scheduled for three weeks, the budget cost/week is $1,000.


Earned -Value Analysis (EVA)- Contd.• The next step is to determine the actual cost of work

performed (ACWP).• This may be entered weekly into the weekly column

labeled ACWP.• The next step is to calculate the budgeted cost of work

performed (BCWP) by estimating the percent completion for each task and multiplying the total budget amount.

• Next the schedule variance and the cost variance for each week will have to be calculated as below.- Schedule variance = BCWP – BCWS- Cost Variance = BCWP – ACWP


Example Problem

Project team Alpha has accumulated $5,000 of expenses as of the end of week#10. The budgeted cost of work scheduled for Week #10 is $6,500. The project has one task, which is about 45% complete at the end of week #10. Calculate the schedule and cost variances. Is the project ahead (or behind), and under (over) budget?


Solution• ACWP=$5000• BCWS=$6500• BCWP (for end of wk #10) = 45%(6500) = $2925• Schedule Variance = BCWP-BCWS

= 2925-6500= $-3575

• Cost Variance = BCWP-ACWP = 2925-5000 = $-2075

• The project is behind by $3575 of work. Also, for the work performed, it is over budget by $2075.


Another Problem

Team Delta is working on a project that has two work tasks, each worth $5,000, which are 45% and 35% complete as of Week #5. The actual expenses are $2,500 as of the end of Week #5. The budgeted cost of work scheduled for Week #5 is $6,500. Calculate the schedule and cost variances. Is the project ahead (or behind), and under (over) budget?


Solution• ACWP=$2500

• BCWS=$6500

• BCWP (end of wk #5)=45%(5000)+35%(5000) =2250+1750

= $4000

• Schedule Variance = BCWP-BCWS

= 4000-6500

= $-2500

• Cost Variance = BCWP-ACWP

= 4000-2500

= $ 1500


Resources

Manpower Money Machines Materials Method


Resource Allocation

What to do if resources are limited?

Allocate the limited resources so that the project may be completed on time using the resources optimally.

Two or more activities may require the same resource simultaneously resulting in increased project length.

What to do if resources are unlimited?

Very rarely a project may have the luxury of having unlimited resources.

If this pleasant situation were to arise, the responsibility of the project team lies in leveling the resources so that the same amount of the resources are used every period.

M. Sundaram Tenn. Tech1 MANAGING PROJECTS USING NETWORK TECHNIQUES.

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