33 Project Management 03... · CPM/PERT Gantt charts Milestone charts Cash flow schedules Planning, Scheduling, Controlling Before Start of project During ... • Gantt chart •

© 2011 Pearson Education, Inc. publishing as Prentice Hall

3 Project Management3

SCM 352

Boeing 787 Dreamliner

“Delays are a natural part of the airplane supply business. They promise an unreasonable delivery date to lock up customers before a competitor can deliver an aircraft. The trick is to not have too long of a delay that it impacts your credibility.”

—Steve SwensonWall Street Journal

January 19, 2011

• Global Company Profile: Bechtel Group• Importance of Project Management• Project Planning

– Work Breakdown Structure• Project Scheduling• Project Management Techniques: PERT

and CPM

Outline


Bechtel Projects

• Building 26 massive distribution centers in just two years for the internet company Webvan Group ($1 billion)

• Constructing 30 high-security data centers worldwide for Equinix, Inc. ($1.2 billion)

• Building and running a rail line between London and the Channel Tunnel ($4.6 billion)

• Developing an oil pipeline from the Caspian Sea region to Russia ($850 million)

• Expanding the Dubai Airport in the UAE ($600 million), and the Miami Airport in Florida ($2 billion)

• Building a new subway for Athens, Greece ($2.6 billion)• Building a natural gas pipeline in Thailand ($700 million)• Building a highway to link the north and south of Croatia

($303 million)


Strategic Importance

Bechtel Project Management:Iraq war aftermathInternational workforce, construction professionals, cooks, medical personnel, securityMillions of tons of supplies

Boeing paid more than $5 billion in penalties and delay concessions to airlines that were forced to wait for their planes, resulting in older aircraft being kept in service and new routes designed specifically for the Dreamliner being added. (1/19/11, WSJ)Las Vegas Monorail

Cost overruns and penalties

http://online.wsj.com/article/SB10001424052748703954004576089703794210100.html?mod=djem_jiewr_OM_domainid#project%3DBOEING1001_shell%26articleTabs%3Dinteractive

Boeing 787 Dreamliner Timeline


Project Characteristics

• Single unit• Many related activities• Difficult operations planning and inventory

control• General purpose equipment• High labor skills

Building Construction

Research Project


Figure 3.1

Before Start of project Duringproject Timeline project

Planning the Project


Figure 3.1


Scheduling the Project


Figure 3.1


Controlling the Project


Figure 3.1

Planning, Scheduling, Controlling


© 2011 Pearson Education, Inc. publishing as Prentice Hall Figure 3.1

BudgetsDelayed activities reportSlack activities report

Time/cost estimatesBudgetsEngineering diagramsCash flow chartsMaterial availability details

CPM/PERTGantt chartsMilestone chartsCash flow schedules

Planning, Scheduling, Controlling



Project Management Techniques

• Gantt chart• Critical Path Method

(CPM)• Program Evaluation

and Review Technique (PERT)


TimeJ F M A M J J A S

DesignPrototypeTestReviseProduction

A Simple Gantt Chart


• Network techniques• Developed in 1950’s

• CPM by DuPont for chemical plants (1957)• PERT by Booz, Allen & Hamilton with the U.S.

Navy, for Polaris missile (1958)• Consider precedence relationships and

interdependencies• Each uses a different estimate of activity times

CPM and PERT


Questions PERT/CPM can answer

1. When will the entire project be completed?2. What are the critical activities or tasks in the project?3. Which are the non-critical activities?4. What is the probability the project will be completed

by a specific date?5. Is the project on target, behind, or ahead of schedule?6. Is the money spent equal to, less than, or greater than

the budget?7. Are there enough resources available to finish the

project on time?8. If the project must be finished in a shorter time, what

is the way to accomplish this at least cost?


Determining the Project Schedule

Perform a Critical Path Analysis• The critical path is the longest path through the

network• The critical path is the shortest time in which

the project can be completed• Any delay in critical path activities delays the

project• Critical path activities have no slack time


Perform a Critical Path Analysis


Time ImmediateActivity Description (weeks) Predecessors

A Build internal components 2 –B Modify roof and floor 3 –C Construct collection stack 2 AD Pour concrete and install frame 4 A, BE Build high-temperature burner 4 CF Install pollution control system 3 CG Install air pollution device 5 D, EH Inspect and test 2 F, G

Total Time (weeks) 25



Activity Description Time (weeks)A Build internal components 2B Modify roof and floor 3C Construct collection stack 2D Pour concrete and install frame 4E Build high-temperature burner 4F Install pollution control system 3G Install air pollution device 5H Inspect and test 2

Total Time (weeks) 25

Earliest start (ES) = earliest time at which an activity can start, assuming all predecessors have been completed

Earliest finish (EF) = earliest time at which an activity can be finished

Latest start (LS) = latest time at which an activity can start so as to not delay the completion time of the entire project

Latest finish (LF) = latest time by which an activity has to be finished so as to not delay the completion time of the entire project




Figure 3.10

A

Activity Name or Symbol

Earliest Start ES

Earliest FinishEF

Latest Start

LS Latest Finish

LF

Activity Duration

2



Begin at starting event and work forwardEarliest Start Time Rule:

If an activity has only a single immediate predecessor, its ES equals the EF of the predecessorIf an activity has multiple immediate predecessors, its ES is the maximum of all the EF values of its predecessors

ES = Max. {EF of all immediate predecessors}

Forward Pass


Begin at starting event and work forwardEarliest Finish Time Rule:

The earliest finish time (EF) of an activity is the sum of its earliest start time (ES) and its activity time

EF = ES + Activity time

Forward Pass


0

ES

0

EF = ES + Activity time

ES/EF Network for Milwaukee Paper

0

Activity time

Start

Activity


Start0

0

0

2

EF of A = ES of A + 2

0

ES of A


2

Activity time

Activity

A

A


Start0

0

0

A

2

20

3

EF of B = ES of B + 3

0

ES of B


3Activity time

Activity

BB


C

B

Start0

0

0

A

2

20

2

ES = 2


4

EF = ES + 2

2Activity time

Activity

C0 3

3


C

B

3

0 3

Start0

0

0

A

2

20

3

ES = Max. (2,3)


7

EF = ES + 4

4Activity time

Activity

DD

2 4

2


H

2

13 15

G

5

8 13

E

4

4 8

F

3

4 7

D

C

2

2 4

B

3

0 3

Start0

0

0

A

2

20

Figure 3.11


3 7

4

Completed Forward Pass


Begin with the last event and work backwardsLatest Finish Time Rule:

If an activity is an immediate predecessor for just a single activity, its LF equals the LS of the activity that immediately follows itIf an activity is an immediate predecessor to more than one activity, its LF is the minimum of all LS values of all activities that immediately follow it

LF = Min. {LS of all immediate following activities}

Backward Pass


Begin with the last event and work backwardsLatest Start Time Rule:

The latest start time (LS) of an activity is the difference of its latest finish time (LF) and its activity time

LS = LF – Activity time

Backward Pass


E

4

F

3

G

5

H

2

4 8 13 15

4

8 13

7

D

4

3 7

C

2

2 4

B

3

0 3

Start0

0

0

A

2

20

LF = EF of Project

1513

LS = LF – Activity time

LS/LF Times for Milwaukee Paper

BackwardPass


E

4

F

3

G

5

H

2

4 8 13 15

4

8 13

7

D

4

3 7

C

2

2 4

B

3

0 3

Start0

0

0

A

2

20

1513LF = Min. (LS of

following activity)

13


10

LS = LF – Activity time Next



E

4

F

3

G

5

H

2

4 8 13 15

4

8 13

7

D

4

3 7

C

2

2 4

B

3

0 3

Start0

0

0

A

2

20

15134 8

8 13

10 13

LF = Min. (4, 10)

42

LS = LF – Activity timeNext



E

4

F

3

G

5

H

2

4 8 13 15

4

8 13

7

D

4

3 7

C

2

2 4

B

3

0 3

Start0

0

0

A

2

20

15134 8

8 13

10 13

0 0

1 4 4 8

2 40 2

Complete Network


After computing the ES, EF, LS, and LF times for all activities, compute the slack or free time for each activity

Slack is the length of time an activity can be delayed without delaying the entire project

Slack = LS – ES or Slack = LF – EF

Computing Slack Time


Earliest Earliest Latest Latest OnStart Finish Start Finish Slack Critical

Activity ES EF LS LF LS – ES Path

A 0 2 0 2 0 YesB 0 3 1 4 1 NoC 2 4 2 4 0 YesD 3 7 4 8 1 NoE 4 8 4 8 0 YesF 4 7 10 13 6 NoG 8 13 8 13 0 YesH 13 15 13 15 0 Yes

Table 3.3

Computing Slack Time



E

4

F

3

G

5

H

2

4 8 13 15

4

8 13

7

D

4

3 7

C

2

2 4

B

3

0 3

Start0

0

0

A

2

20

15134 8

8 13

10 13

0 0

1 4 4 8

2 40 2

Critical Path


• CPM assumes we know a fixed time estimate for each activity and there is no variability in activity times

• PERT uses a probability distribution for activity times to allow for variability

• 3 time estimates– Optimistic time (a) – if everything goes according to plan– Most-likely time (m) – most realistic estimate– Pessimistic time (b) – assuming very unfavorable conditions

• Time estimates follow beta distribution• Expected time: t = (a + 4m + b)/6• Variance of times: v = [(b – a)/6]2

PERT Activity Times


TaskImmediate

Predecesors Optimistic Most Likely PessimisticA None 3 6 15B None 2 4 12C A 6 12 30D A 2 5 8E C 5 11 17F D 3 6 15G B 3 9 27H E,F 1 4 7I G,H 4 19 28

PERT with 3 Activity Time Estimates


Task Optimistic Most Likely Pessimistic ExpectedA 3 6 15 7B 2 4 12 5C 6 12 30 14D 2 5 8 5E 5 11 17 11F 3 6 15 7G 3 9 27 11H 1 4 7 4K 4 19 28 18

Expected Time =Optimistic + 4(Most Likely) + Pessimistic

6

Expected Time


H

4

32 36

32 36

E

21 32

G5 16

25 36

F

7

12 19

25 32

C7 21

217

D

5

7 12

2520

A

7

70

70

B

5

0 5

2520

Start0

0

0

00

K36 54

36 54

14 11

1811

21 32

Critical Path

Duration = 54 weeks


What is the probability of finishing this project in less than 53 weeks?

Z = D - TE

cp2σ∑

Example

tTE = 54

p(t < D)

D = 53


Task Optimistic Most Likely Pessimistic VarianceA 3 6 15 4B 2 4 12C 6 12 30 16D 2 5 8E 5 11 17 4F 3 6 15G 3 9 27H 1 4 7 1K 4 19 28 16

Variance Along Critical Path

(Sum of variance along critical path) =∑σ 2 = 41

6Activity variance, = Pessimistic - Optimistic )2 2σ (


0.156- =4154-53=T - D = Z

2cp

E

∑σ

p(Z<-0.156) = 0.5-0.0636 = 0.436, or 43.6% (Appendix 1, A3)

tTE = 54

p(t < D)

D = 53


0.156- =4154-53=T - D = Z

2cp

E

∑σ

There is a 43.6% probability that this project will be completed in less than 53 weeks.

p(Z<-0.156) = 0.5-0.0636 = 0.436, or 43.6% (Appendix 1, A3)

tTE = 54

p(t < D)

D = 53

Thank You

Questions? ?

Documents

33 Project Management 03... · CPM/PERT Gantt charts Milestone charts Cash flow schedules Planning, Scheduling, Controlling Before Start of project During ... • Gantt chart •