project planning and budgeting recall the four stages project definition and conceptualization...

Project Planning and Budgeting

Recall the four stages• Project Definition and

Conceptualization• Project Planning and Budgeting• Project Execution and Control• Project Termination and

Closeout

Ch 17 - 2

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

Ch 17 - 3

Project Planning

• Statement of work– written description of goals, work & time

frame of project• Activities require labor, resources & time• Precedence relationship shows sequential

relationship of project activities

WORK BREAKDOWN 1

WORK BREAKDOWN 2

GANTT CHART

Ch 17 - 4© 2000 by Prentice-Hall IncRussell/Taylor Oper Mgt 3/e

Simplified Project Network

1 32Construct forms Pour concrete

NETWORK CHART 1

NETWORK CHART 2

Ch 17 - 5

Elements Of Project Planning• Define project objective(s)• Identify activities• Establish precedence relationships• Make time estimates• Determine project completion time• Compare project schedule objectives• Determine resource requirements to meet

objective


Work breakdown structure (WBS)

– determine subcomponents, activities & tasks


Gantt Chart

• Popular tool for project scheduling• Graph with bar for representing the time for

each task• Provides visual display of project schedule• Also shows slack for activities

– (amount of time activity can be delayed without delaying project)

Ch 17 - 8

20 4 106 8

31 5 7 9Month

Month

Activity

Design house andobtain financing

Lay foundation

Order and receivematerials

Build house

Select paint

Select carpet

Finish work

A Gantt Chart

Ch 17 - 9

CPM/PERT• 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

Ch 17 - 10

The Project Network

Network consists of branches & nodes

1 32

Branch

Node

Ch 17 - 11

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

1 2 4 6 7

3

5

32

0

1

31

1

1

Lay foundation

Design house and obtain financing

Order and receive materials

Dummy

Finish work

Select carpet

Select paint

Build house

Ch 17 - 13

Critical Path

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

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

• Project cannot be completed in less than the time of the critical path

Ch 17 - 14

All Possible Paths

A: 1-2-3-4-6-73 + 2 + 0 + 3 + 1 = 9 months; the critical path

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

C: 1-2-4-6-7

3 + 1 + 3 + 1 = 8 monthsD: 1-2-4-5-6-7

3 + 1 + 1 + 1 + 1 = 7 months


Concurrent Activities

4

3

2

DummyLay foundation

2 3

Lay foundation

Order materialOrder material

Incorrect precedence relationship

Correct precedence relationship


Early Times(House-building example)

• 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


Computing Early Times

– ES23 = max (EF2) = 3 months

– ES46 = max (EF4) = max (5,4) = 5 months

– EF46 = ES46 + t46 = 5 + 3 = 8 months

– EF67 =9 months, the project duration


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 time– LSij = 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


Early And Late Times

1 2 4 6 7

3

5

32

0

1

31

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=5, EF=8

LS=5, LF=8

( )ES=6, EF=7

LS=7, LF=8

( )ES=8, EF=9

LS=8, LF=9

( )ES=3, EF=4

LS=4, LF=5

( )ES=5, EF=6

LS=6, LF=7


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

ActivityLS ES LF EF Slack (S)1-2* 0 0 3 3 02-3 3 3 5 5 02-4 4 3 5 4 13-4* 5 5 5 5 04-5 6 5 7 6 14-6* 5 5 8 8 05-6 7 6 8 7 16-7* 8 8 9 9 0

* Critical path


Probabilistic Time Estimates

• Reflect uncertainty of activity times• Beta distribution is used in PERT

b - a6

( )Variance: 2 =

a = optimistic estimatem = most likely time estimateb = pessimistic time estimate

Where,

2

Mean (expected time): a + 4m + b6

t =


Example Beta Distributions

m = t ba

P (

time)

ba

P (

time)

tm

b

mta

P (

time)


PERT Example

1

2

4

6

73 5 9

8

Manual Testing

Dummy

System Training

Dummy

System Testing

Orientation

Position recruiting

System development

Equipment installation

Equipment testing and modification

Final debugging

System changeover

Job training


Activity Information

1 - 2 6 8 10 8 .441 - 3 3 6 9 6 1.001 - 4 1 3 5 3 .442 - 5 0 0 0 0 .002 - 6 2 4 12 5 2.783 - 5 2 3 4 3 .114 - 5 3 4 5 4 .114 - 8 2 2 2 2 .005 - 7 3 7 11 7 1.785 - 8 2 4 6 4 .447 - 8 0 0 0 0 .006 - 9 1 4 7 4 1.007 - 9 1 10 13 9 4.00

Time estimates (wks) Mean Time VarianceActivity a b c t 2


Early And Late Times1 - 2 8 0.44 0 8 1 9 11 - 3 6 1.00 0 6 0 6 01 - 4 3 0.44 0 3 2 5 22 - 5 0 0.00 8 8 9 9 12 - 6 5 2.78 8 13 16 21 83 - 5 3 0.11 6 9 6 9 04 - 5 4 0.11 3 7 5 9 2

4 - 8 2 0.00 3 5 14 16 115 - 7 7 1.78 9 16 9 16 05 - 8 4 0.44 9 13 12 16 37 - 8 0 0.00 13 13 16 16 36 - 9 4 1.00 13 17 21 25 87 - 9 9 4.00 16 25 16 25 0

Activity t 2 ES EF LS LF S


Network With Times

1

2

4

6

73 5 9

8

( )ES=8, EF=8

LS=9, LF=9

( )ES=6, EF=9

LS=6, LF=9

( )ES=3, EF=5

LS=14, LF=16

( )ES=0, EF=3

LS=2, LF=5

( )ES=0, EF=6

LS=0, LF=6

( )ES=0, EF=8

LS=1, LF=9

3

80

5

4

4

7

0

2

93

6

( )ES=3, EF=7

LS=5, LF=9

4 ( )ES=9, EF=13

LS=12, LF=16

( )ES=9, EF=13

LS=9, LF=16

( )ES=13, EF=13

LS=16 LF=16

( )ES=13, EF=25

LS=16 LF=25

( )ES=13, EF=25

LS=16 LF=25

( )ES=8, EF=13

LS=16 LF=21


Project Variance

• Project variance is the sum of variances on the critical path

2132

352

572

792

100 011 178 4 00

6 89

. . . .

. weeks


Probabilistic Network AnalysisDetermine probability that project is completed within

specified time

where = tp = project mean time

= project standard deviationx = proposed project timeZ = number of standard deviations x is from mean

Z = x -


Normal Distribution Of Project Time

= tp Timex

Z

Probability


Probabilistic Analysis Example

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

2 6 89

6 89 2 62

30 252 62

191

191 0 9719

.

. .

..

( . ) .

weeks

weeks

Zx

P Z


Determining Probability From Z Value

Z 0.00 0.01 ...0.09

1.9 0.4713 0.4719 …0.4767

......

......

= 25 Time (weeks)

x = 30

P( x<= 30 weeks) = 0.9719


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

= 25 Time (weeks)

x = 22

P( x<= 22 weeks) =0.1271

Z

P Z

22 252 62

32 62

114

114 01271. .

.

( . ) .


Project Crashing• Crashing is reducing project time by

expending additional resources• Crash time is an amount of time an activity

is reduced• Crash cost is the cost of reducing the activity

time• Goal is to reduce project duration at

minimum cost


House-building Network

Activity times in weeks

1 2 4 6 7

3

5

12

80

4

124

4

4


Normal Activity And Crash Data

1-2 12 7 $3,000 $5,000 5 $4002-3 8 5 2,000 3,500 3 5002-4 4 3 4,000 7,000 1 3,0003-4 0 0 0 0 0 04-5 4 1 500 1,100 3 2004-6 12 9 50,000 71,000 3 7,0005-6 4 1 500 1,100 3 2006-7 4 3 15,000 22,000 1 7,000

$75,000 $110,700

TotalNormal Crash Allowable CrashTime Time Normal Crash Crash Time Cost per

Activity (wks) (wks) Cost Cost (wks) Week


Network With Crashing Costs

1 2 4 6 7

3

5

128

0

4

124

4

4

$7,000$7,000

$500

$3,000$400

$200 $200

Activity 1-2 can be crashed a total of 5 weeks for $2000Crash cost per week = Total crash cost/Total crash time

= $2,000/5 = $400 per week


Normal And Crash Relationships

1242 6 8 10 140

1,000

3,000

4,000

5,000

7,000

2,000

6,000

$

Weeks

Crashed activity

Normal activity

Crash cost

Normal cost

Crash time Normal time

Slope = crash cost per week


Crashing Solution

1-2 12 7 5 $400 $2,0002-3 8 5 3 500 1,5002-4 4 3 0 3,000 03-4 0 0 0 0 04-5 4 1 0 200 04-6 12 9 3 7,000 21,0005-6 4 1 0 200 06-7 4 3 1 7,000 7,000

12 $31,500

Normal Crash Crash Crash CrashingTime Time Time Cost per Cost

Activity (wks) (wks) Used Week Incurred


Crashed Project

1 2 4 6 7

3

5

12 7

8 5

0

4 3

12 9

44

4 3

Original time Crashed times


Time-Cost Relationship • 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


Time-Cost TradeoffC

ost

($

)

Project Duration

Crashing

Total cost

Indirect cost

Direct cost

Time

Minimum cost = optimal project time

project planning and budgeting recall the four stages project definition and conceptualization...

Documents

project slide

network project

project network network

project objectives

e project network

project completion time

budgeting slide

critical path slide