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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
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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)
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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
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Project Characteristics
• Single unit• Many related activities• Difficult operations planning and inventory
control• General purpose equipment• High labor skills
Building Construction
Research Project
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Figure 3.1
Before Start of project Duringproject Timeline project
Planning the Project
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Figure 3.1
Before Start of project Duringproject Timeline project
Scheduling the Project
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Figure 3.1
Before Start of project Duringproject Timeline project
Controlling the Project
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Figure 3.1
Planning, Scheduling, Controlling
Before Start of project Duringproject Timeline project
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BudgetsDelayed activities reportSlack activities report
Time/cost estimatesBudgetsEngineering diagramsCash flow chartsMaterial availability details
CPM/PERTGantt chartsMilestone chartsCash flow schedules
Planning, Scheduling, Controlling
Before Start of project Duringproject Timeline project
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Project Management Techniques
• Gantt chart• Critical Path Method
(CPM)• Program Evaluation
and Review Technique (PERT)
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TimeJ F M A M J J A S
DesignPrototypeTestReviseProduction
A Simple Gantt Chart
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• 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
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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?
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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
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Perform a Critical Path Analysis
Determining the Project Schedule
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
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Perform a Critical Path Analysis
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
Determining the Project Schedule
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Perform a Critical Path Analysis
Figure 3.10
A
Activity Name or Symbol
Earliest Start ES
Earliest FinishEF
Latest Start
LS Latest Finish
LF
Activity Duration
2
Determining the Project Schedule
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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
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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
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0
ES
0
EF = ES + Activity time
ES/EF Network for Milwaukee Paper
0
Activity time
Start
Activity
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Start0
0
0
2
EF of A = ES of A + 2
0
ES of A
ES/EF Network for Milwaukee Paper
2
Activity time
Activity
A
A
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Start0
0
0
A
2
20
3
EF of B = ES of B + 3
0
ES of B
ES/EF Network for Milwaukee Paper
3Activity time
Activity
BB
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C
B
Start0
0
0
A
2
20
2
ES = 2
ES/EF Network for Milwaukee Paper
4
EF = ES + 2
2Activity time
Activity
C0 3
3
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C
B
3
0 3
Start0
0
0
A
2
20
3
ES = Max. (2,3)
ES/EF Network for Milwaukee Paper
7
EF = ES + 4
4Activity time
Activity
DD
2 4
2
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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
ES/EF Network for Milwaukee Paper
3 7
4
Completed Forward Pass
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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
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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
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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
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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
LS/LF Times for Milwaukee Paper
10
LS = LF – Activity time Next
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LS/LF Times for Milwaukee Paper
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
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LS/LF Times for Milwaukee Paper
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
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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
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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
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LS/LF Times for Milwaukee Paper
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
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• 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
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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
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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
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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
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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
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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 (
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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
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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? ?