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Project
Management
Dr. Ron Lembke
Operations Management
What’s a Project?
• Changing something from the way
it is to the desired state
• Never done one exactly like this
• Many related activities
• Focus on the outcome
• Regular teamwork focuses on the
work process
Examples of Projects
• Building construction
• New product introduction
• Software implementation
• Training seminar
• Research project
Why are projects hard?
• Resources-– People, materials
• Planning– What needs to be done?
– How long will it take?
– What sequence?
– Keeping track of who is supposedly doing what, and getting them to do it
IT Projects
• Half finish late and over budget
• Nearly a third are abandoned before
completion– The Standish Group, in Infoworld
• Get & keep users involved & informed
• Watch for scope creep / feature creep
Project Scheduling
• Establishing objectives
• Determining available resources
• Sequencing activities
• Identifying precedence relationships
• Determining activity times & costs
• Estimating material & worker requirements
• Determining critical activities
PERT & CPM
• Network techniques
• Developed in 1950’s
• CPM by DuPont for chemical plants
• PERT by U.S. Navy for Polaris
missile
• Consider precedence relationships
& interdependencies
• Each uses a different estimate of
activity times
• Completion date?
• On schedule? Within budget?
• Probability of completing by ...?
• Critical activities?
• Enough resources available?
• How can the project be finished early at the least cost?
Questions Answered by
PERT & CPM
PERT & CPM Steps
• Identify activities
• Determine sequence
• Create network
• Determine activity times
• Find critical path
• Earliest & latest start times
• Earliest & latest finish times
• Slack
Activity on Node (AoN)
2
4? Years
EnrollReceive
diploma
Project: Obtain a college degree (B.S.)
1 month
Attend class,
study etc.
1
1 day
3
Activity on Arc (AoA)
4,5 ?
Years
Enroll
Receive
diploma
Project: Obtain a college degree (B.S.)
1 month
Attend class, study,
etc.
11 day
2 3 4
AoA Nodes have
meaning
Graduating
SeniorApplicant
Project: Obtain a college degree (B.S.)
1
Aluma
Alumnus
2 3 4
Student
Precedence Relationships
– Scenario 1a
• Hierarchy of what needs to be done, in what order– Time required for each thing
• For me, the hardest part– I’ve never done this before. How do I know
what I’ll do when and how long it’ll take?
– I think in phases
– The farther ahead in time, the less detailed
– Figure out the tricky issues, the rest is details
– A lot will happen between now and then
– It works not badly with no deadline
Precedence Relationships
Mudroom
Mudroom Remodel
• Big-picture sequence easy:
– Demolition
– Framing
– Plumbing
– Electrical
– Drywall, tape & texture
– Slate flooring
– Cabinets, lights, paint
• Hard: can a sink fit?
D
W
DW
Before:
After:
Project Scheduling
Techniques
• Gantt chart
• Critical Path Method (CPM)
• Program Evaluation &
Review Technique (PERT)
Gantt Chart
J F M A M J J
Time PeriodActivity
Design
Build
Test
J F M A M J J
Time PeriodActivity
Design
Build
Test
Day before Thanksgiving, 2015, late afternoon, 27ºF
What do these 3 guys have in common?
Network Example
You’re a project manager for Bechtel.
Construct the network.Activity Predecessors
A --B AC AD BE BF CG DH E, F
Network Example -
AON
A
C
E
F
BD
G
H
Z
Network Example -
AON
A
B
E
D
CF
H
G
Z
Network Example -
AON
A
B
E
D
C
F
HG
Z
Network Example -
AOA
2
4
51
3 6 8
7 9
A
C F
EB
D
H
G
AOA Diagrams
2 31A
C
BD
A precedes B and C, B and C precede D
2 41A C
B
D
3
5
4
Add a phantom arc for clarity.
Scenario 1a
Task Description Days Pred
A Drawings 1 --
B Get & Dye fabric 3 --
C Build stage 4 A
D Sew fabric, frame back 2 B
E Rent, deliver, hang lights 1 A
F Paint stage 2 C
G Test assembly 1 D,F
H disassemble, deliver 1 G
I Assemble, touch-up 1 E,H
Scenario 1a Network
Diagram
3
START
A
1
B
3
E
1
C
4
D
2
F
2
G
1
H
1
I
1 Finish
Scenario 1a Longest
Path
3
START
A
1
B
3
E
1
C
4
D
2
F
2
G
1
H
1
I
1Finish
Paths: Days
AEI 3
ACFGHI 10
BDGHI 8
Critical Path Analysis
• Provides activity information
• Earliest (ES) & latest (LS) start
• Earliest (EF) & latest (LF) finish
• Slack (S): Allowable delay
• Identifies critical path
• Longest path in network
• Shortest time project can be
completed
• Any delay on critical activities delays
the whole project
• Activities have 0 slack
Critical Path
Analysis Example
Event
ID Pred. Description Time
(Wks)
A None Prepare Site 1
B A Pour fdn. & frame 6
C A Buy shrubs etc. 3
D B Roof 2
E D Do interior work 3
F C Landscape 4
G E,F Move In 1
Network Solution
A
EDB
C F
G
1
6 2 3
1
43
Earliest Start & Finish
Steps
• Begin at starting event & work forward
• ES = 0 for starting activities
• ES is earliest start
• EF = ES + Activity time
• EF is earliest finish
• ES = Maximum EF of all predecessors for
non-starting activities
Activity ES EF LS LF Slack
A 0 1
B
C
D
E
F
Activity A
Earliest Start Solution
For starting activities, ES = 0.
A
EDB
C F
G
1
6 2 3
1
43
Activity ES EF LS LF Slack
A 0 1
B 1 7
C 1 4
D 7 9
E 9 12
F 4 8
G 12 13
Earliest Start Solution
A
EDB
C F
G
1
6 2 3
1
43
Latest Start & Finish
Steps
• Begin at ending event & work backward
• LF = Maximum EF for ending activities
• LF is latest finish; EF is earliest finish
• LS = LF - Activity time
• LS is latest start
• LF = Minimum LS of all successors for
non-ending activities
Activity ES EF LS LF Slack
A 0 1
B 1 7
C 1 4
D 7 9
E 9 12
F 4 8
G 12 13 13
Earliest Start Solution
A
EDB
C F
G
1
6 2 3
1
43
Activity ES EF LS LF Slack
A 0 1 0 1 B 1 7 1 7 C 1 4 5 8 D 7 9 7 9 E 9 12 9 12 F 4 8 8 12 G 12 13 12 13
Latest Finish
Solution
A
EDB
C F
G
1
6 2 3
1
43
Activity ES EF LS LF Slack
A 0 1 0 1 0
B 1 7 1 7 0
C 1 4 5 8 4
D 7 9 7 9 0
E 9 12 9 12 0
F 4 8 8 12 4
G 12 13 12 13 0
Compute Slack
Critical Path
A
EDB
C F
G
1
6 2 3
1
43
Scenario 1b ES/EF
times
START
A
1
0 1
B
3
0 3
E
1
1 2
C
4
1 5
D
2
3 5
F
2
5 7
G
1
7 8H
1
8 9
I
1
910
Finish
Scenario 1c LF/LS
times
START
A
1
0 1
0 1
B
3
0 3
2 5
E
1
1 2
8 9
C
4
1 5
1 5
D
2
3 5
5 7
F
2
5 7
5 7
G
1
7 8
7 8
H
1
8 9
8 9
I
1
910
9 10 Finish
Scenario 1c
Finish
START
A
1
0 1
0 1
B
3
0 3
2 5
E
1
1 2
8 9
C
4
1 5
1 5
D
2
3 5
5 7
F
2
5 7
5 7
G
1
7 8
7 8
H
1
8 9
8 9
I
1
910
9 10
Other notation
• Compute ES, EF for each
activity, Left to Right
• Compute, LF, LS, Right to Left
C 7LS LF
ES EF
Example #2
A 21
E 5D 2B 4
C 7 F 7
G 2
21 28 28 35
35 37
28 3325 2721 25
0 21
Example #2
A 21
E 5D 2B 4
C 7 F 7
G 2
21 28 28 35
35 37
28 3325 2721 25
0 21
F cannot start until C and D are done.
G cannot start until both E and F are done.
Example #2
A 21
E 5D 2B 4
C 7 F 7
G 2
22 26
0 21
26 28 30 35
35 37
21 28 28 35
21 28 28 35
35 37
28 3325 2721 25
0 21
E just has to be done in time for G to start at 35, so it has slack.
D has to be done in time for F to go at 28, so it has no slack.
Example #2
A 21
E 5D 2B 4
C 7 F 7
G 2
22 26
0 21
26 28 30 35
35 37
21 28 28 35
21 28 28 35
35 37
28 3325 2721 25
0 21
E just has to be done in time for G to start at 35, so it has slack.
D has to be done in time for F to go at 28, so it has no slack.
Gantt Chart - ES
0 5 10 15 20 25 30 35 40
A
B
C
D
E
F
G
Gantt Chart - LS
0 5 10 15 20 25 30 35 40
A
B
C
D
E
F
G
Another Example
A 1
B 4
C 3
D 7
E 6
F 2
H 9
I 4
G 7
Solved Problem 1
A 10 1
0 1
B 41 5
1 5
C 36 9
1 4
D 72 9
1 8
E 65 11
5 11
F 29 11
8 10
H 99 18
8 17
I 418 22
18 22
G 711 18
11 18
Can We Go Faster?
Time-Cost Models
1. Identify the critical path
2. Find cost per day to expedite each node on
critical path.
3. For cheapest node to expedite, reduce it as
much as possible, or until critical path
changes.
4. Repeat 1-3 until no feasible savings exist.
Time-Cost Example
• ABC is critical path=30
Crash cost Crash
per week wks avail
A 500 2
B 800 3
C 5,000 2
D 1,100 2
C 10B 10A 10
D 8
Cheapest way to gain 1
Week is to cut A
Time-Cost Example
• ABC is critical path=29
Crash cost Crash
per week wks avail
A 500 1
B 800 3
C 5,000 2
D 1,100 2
C 10B 10A 9
D 8
Cheapest way to gain 1 wk
Still is to cut A
Wks Incremental Total
Gained Crash $ Crash $
1 500 500
Time-Cost Example
• ABC is critical path=28
Crash cost Crash
per week wks avail
A 500 0
B 800 3
C 5,000 2
D 1,100 2
C 10B 10A 8
D 8
Cheapest way to gain 1 wk
is to cut B
Wks Incremental Total
Gained Crash $ Crash $
1 500 500
2 500 1,000
Time-Cost Example
• ABC is critical path=27
Crash cost Crash
per week wks avail
A 500 0
B 800 2
C 5,000 2
D 1,100 2
C 10B 9A 8
D 8
Cheapest way to gain 1 wk
Still is to cut B
Wks Incremental Total
Gained Crash $ Crash $
1 500 500
2 500 1,000
3 800 1,800
Time-Cost Example
• Critical paths=26 ADC & ABC
Crash cost Crash
per week wks avail
A 500 0
B 800 1
C 5,000 2
D 1,100 2
C 10B 8A 8
D 8
To gain 1 wk, cut B and D,
Or cut C
Cut B&D = $1,900
Cut C = $5,000
So cut B&D
Wks Incremental Total
Gained Crash $ Crash $
1 500 500
2 500 1,000
3 800 1,800
4 800 2,600
Time-Cost Example
• Critical paths=25 ADC & ABC
Crash cost Crash
per week wks avail
A 500 0
B 800 0
C 5,000 2
D 1,100 1
C 10B 7A 8
D 7
Can’t cut B any more.
Only way is to cut C
Wks Incremental Total
Gained Crash $ Crash $
1 500 500
2 500 1,000
3 800 1,800
4 800 2,600
5 1,900 4,500
Time-Cost Example
• Critical paths=24 ADC & ABC
Crash cost Crash
per week wks avail
A 500 0
B 800 0
C 5,000 1
D 1,100 1
C 9B 7A 8
D 7
Only way is to cut C
Wks Incremental Total
Gained Crash $ Crash $
1 500 500
2 500 1,000
3 800 1,800
4 800 2,600
5 1,900 4,500
6 5,000 9,500
Time-Cost Example
• Critical paths=23 ADC & ABC
Crash cost Crash
per week wks avail
A 500 0
B 800 0
C 5,000 0
D 1,100 1
C 8B 7A 8
D 7
No remaining possibilities to
reduce project length
Wks Incremental Total
Gained Crash $ Crash $
1 500 500
2 500 1,000
3 800 1,800
4 800 2,600
5 1,900 4,500
6 5,000 9,500
7 5,000 14,500
Time-Cost Example
C 8B 7A 8
D 7
No remaining possibilities to
reduce project length
Wks Incremental Total
Gained Crash $ Crash $
1 500 500
2 500 1,000
3 800 1,800
4 800 2,600
5 1,900 4,500
6 5,000 9,500
7 5,000 14,500
• Now we know how much it costs us to save any number of weeks
• Customer says he will pay $2,000 per week saved.
• Only reduce 5 weeks.
• We get $10,000 from customer, but pay $4,500 in expediting costs
• Increased profits = $5,500
Ex.
AOA
AON
• Paths:
• ABF: 18
• CDEF: 20
C5
B 10A 6
D 4 E 9
F2
Activity Avail Cost
A 0 ----
B 2 500
C 1 300
D 3 700
E 2 600
F 1 800
Gantt charts to
keep track
• Original
A 6 B 10
C 5 D 4 E 9
2
F 2
Activity Avail Cost
A 0 ----
B 2 500
C 1 300
D 3 700
E 2 600
F 1 800
Consider CDEF
C is cheapest, crash 1 day
A 6 B 10
C 4 D 4 E 9
1
F 2
C5
B10A 6
D 4 E 9
F2
C4
B10A 6
D 4 E 9
F2
Revised Pictures: Activity Avail Cost
A 0 ----
B 2 500
C 0
D 3 700
E 2 600
F 1 800
Consider
DEF
E cheapest
Crash E 1
day
Keep Track with Gantt Charts
A 6 B 10
C 4 D 4 E 8
F 2
C4
B10A 6
D 4 E8
F2
New Pictures:
Activity Avail Cost
A 0 ----
B 2 500
C 0
D 3 700
E 1 600
F 1 800
All Activities
Critical
ABF: 18
CDEF: 18
Options: Crash:
• F $800 – last one
• B, AND D or E
• E is cheaper than D
• $500+$600 =
$1,100 > 1,000
• Cost > Benefit
A 6 B 10
C 4 D 4 E 8
F 1
C4
B10A 6
D 4 E8
F1
Scenario 2
Now, need to do it in 7 days!
Task Normal Crash $/day
B 3 2 $1,000
C 4 1 2,000
D 2 1 500
F 2 1 800
START
A
1
0 1
0 1
B
3
0 3
2 5
E
1
1 2
8 9
C
4
1 5
1 5
D
2
3 5
5 7
F
2
5 7
5 7
G
1
7 8
7 8
H
1
8 9
8 9
I
1
910
9 10
Not critical
Not critical
Cheapest
$800 gain 1 day
Finish
Scenario 2
Now, need to do it in 7 days!
Task Normal Crash $/day
B 3 2 $1,000
C 4 1 2,000
D 2 1 500
F 2 1 800
START
A
1
0 1
0 1
B
3
0 3
1 4
E
1
1 2
8 8
C
4
1 5
1 5
D
2
3 5
4 6
F
1
5 6
5 6
G
1
6 7
6 7
H
1
7 8
7 8
I
1
8 9
8 9
Not critical
Not critical
Used up
Only optionDay 1 $800
Day 2 $2,000
Finish
Scenario 2
Now, need to do it in 7 days!
Task Normal Crash $/day
B 3 2 $1,000
C 3 1 2,000
D 2 1 500
F 2 1 800
START
A
1
0 1
0 1
B
3
0 3
0 3
E
1
1 2
8 8
C
3
1 4
1 4
D
2
3 5
3 5
F
1
4 5
4 5
G
1
5 6
5 6
H
1
6 7
6 7
I
1
7 8
7 8
Used up
Day 1 $800
Day 2 $2,000
Something on both paths
C, and B or D, D cheaper
Day 3 $2,500
Total $5,300
Finish
Scenario 2
Now, Done in 7 days!
Task Normal Crash $/day
B 3 2 $1,000
C 2 1 2,000
D 1 1 500
F 2 1 800
START
A
1
0 1
0 1
B
3
0 3
0 3
E
1
1 2
5 6
C
2
1 3
1 3
D
1
3 4
3 4
F
1
3 4
3 4
G
1
4 5
4 5
H
1
5 6
5 6
I
1
6 7
6 7
Used up
Day 1 $800
Day 2 $2,000Day 3 $2,500
Total $5,300
Used up
Finish
What about
Uncertainty?
PERT Activity Times
• 3 time estimates
• Optimistic times (a)
• Most-likely time (m)
• Pessimistic time (b)
• Follow beta distribution
• Expected time: t = (a + 4m + b)/6
• Variance of times: v = (b - a)2/36
Example Activity
• a = 2, m = 4, b = 6
• E[T] = (2 + 4*4 + 6)/6 = 24/6 = 4.0
• σ2 = (6 – 2)2 / 36 = 16/36 = 0.444
Example
Activity a m b E[T] variance
A 2 4 8 4.33 1
B 3 6.1 11.5 6.48 2
C 4 8 10 7.67 1
Project 18.5 4
Complete in 18.5 days, with a variance
of 4.
CBA
4.33 6.48 7.67
Sum of 3 Normal
Random Numbers
15
20
2
X
35
30
2
X
10
10
2
X
10 20 30 40 50 60
60
60
2
X
Average value of the sum is
equal to the sum of the averages
Variance of the sum is equal to
the sum of the variances
Notice curve of sum is more spread
out because it has large variance
Back to the Example:
Probability of <= 21 wks
18.5 21
Average time = 18.5, st. dev = 2
21 is how many standard deviations
above the mean?
21-18.5 = 2.5.
St. Dev = 2,
so 21 is 2.5/2 = 1.25 standard
deviations above the mean
Book Table says area between
Z=1.25 and –infinity is 0.8944
Probability <= 21 wks
= 0.8944 = 89.44%
Benefits
of PERT/CPM
• Useful at many stages of project
management
• Mathematically simple
• Use graphical displays
• Give critical path & slack time
• Provide project documentation
• Useful in monitoring costs
Limitations
of PERT/CPM
• Clearly defined, independent, &
stable activities
• Specified precedence
relationships
• Activity times (PERT) follow
beta distribution
• Subjective time estimates
• Over emphasis on critical path
Conclusion
• Explained what a project is
• Drew project networks
• Compared PERT & CPM
• Determined slack & critical path
• Found profit-maximizing crash decision
• Computed project probabilities