Transcript Planning Fundamentals

Henry Yuliando
LOGO
Oct 2012
Project Time Planning and
Networks
Project Management
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Context
A uniqueness implies that every
project must be defined a new and a
scheme created telling everyone
involved what to do.
Deciding and specifying what they
have to do is the function of project
definition , the output of which is a
project plan.
Making sure they do it right is the
function of project control.
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Learning from Past Projects
Learning from Past Projects
 While developing a project plan, the project
manager refer to earlier, similar projects
(plans, procedures, successes, and failures).
 Ideally the project manager is provided with
planning assistance in the form of lessons
learned, best practices, suggested
methodologies and templates, and even
consulting advice derived from experience in
past projects.
Project Management Activities
 Planning
 Objectives
 Resources
 Work break-down
schedule
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 Organization
 Controlling
 Monitor, compare, revise, action
 Scheduling
 Project activities
 Start & end times
 Network
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Project Planning, Scheduling,
and Controlling
Figure 3.1
Before
project
Start of project
Timeline
During
project
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Project Planning, Scheduling,
and Controlling
Figure 3.1
Before
project
Start of project
Timeline
During
project
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Project Planning, Scheduling,
and Controlling
Figure 3.1
Before
project
Start of project
Timeline
During
project
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Project Planning, Scheduling,
and Controlling
Figure 3.1
Before
project
Start of project
Timeline
During
project
Time/cost estimates
Project Planning,
Scheduling,
Budgets
and Controlling
Engineering diagrams
Cash flow charts
Material availability details
Budgets
Delayed activities report
Slack activities report
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CPM/PERT
Gantt charts
Milestone charts
Cash flow schedules
Figure 3.1
Before
project
Start of project
Timeline
During
project
Project Planning
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 Establishing
objectives
 Defining project
 Creating work
breakdown
structure
 Determining
resources
 Forming
organization
Project Organization
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 Often temporary structure
 Uses specialists from entire
company
 Headed by project manager
 Coordinates activities
 Monitors schedule
and costs
 Permanent
structure called
‘matrix organization’
A Sample Project Organization
President
Human
Resources
Marketing
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Project 1
Project 2
Figure 3.2
Finance
Design
Quality
Mgt
Production
Mechanical
Engineer
Test
Engineer
Technician
Electrical
Engineer
Computer
Engineer
Technician
Project
Manager
Project
Manager
Project Management Techniques
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 Gantt chart
 Critical Path Method
(CPM)
 Program Evaluation
and Review
Technique (PERT)
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Six Steps PERT & CPM
1. Define the project and prepare the
work breakdown structure
2. Develop relationships among the
activities - decide which activities
must precede and which must
follow others
3. Draw the network connecting all of
the activities
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Six Steps PERT & CPM
4. Assign time and/or cost estimates
to each activity
5. Compute the longest time path
through the network – this is
called the critical path
6. Use the network to help plan,
schedule, monitor, and control the
project
A Comparison of AON and AOA
Network Conventions
Activity on
Node (AON)
(a)
A
C
B
Activity
Meaning
A comes before
B, which comes
before C
A
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(b)
(c)
Figure 3.5
C
B
A and B must both
be completed
before C can start
Activity on
Arrow (AOA)
A
B
C
A
B
C
B
A
C
B and C cannot
begin until A is
completed
B
A
C
A Comparison of AON and AOA
Network Conventions
Activity on
Node (AON)
A
C
B
D
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(d)
A
C
(e)
B
Figure 3.5
D
Activity
Meaning
C and D cannot
begin until A and
B have both
been completed
C cannot begin until
both A and B are
completed; D
cannot begin until B
is completed. A
dummy activity is
introduced in AOA
Activity on
Arrow (AOA)
A
C
B
D
A
C
Dummy activity
B
D
A Comparison of AON and AOA
Network Conventions
Activity on
Node (AON)
A
B
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(f)
Figure 3.5
C
D
Activity
Meaning
B and C cannot
begin until A is
completed. D
cannot begin until
both B and C are
completed. A
dummy activity is
again introduced
in AOA.
Activity on
Arrow (AOA)
A
Dummy
activity
B
D
C
AON Example
Milwaukee Paper Manufacturing's
Activities and Predecessors
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Activity
A
Description
Build internal components
Immediate
Predecessors
—
B
Modify roof and floor
—
C
Construct collection stack
A
D
Pour concrete and install frame
A, B
E
Build high-temperature burner
C
F
Install pollution control system
C
G
Install air pollution device
D, E
H
Inspect and test
F, G
Table 3.1
AON Network for Milwaukee
Paper
A
Activity A
(Build Internal Components)
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Start
Start Activity
B
Activity B
(Modify Roof and Floor)
Figure 3.6
AON Network for Milwaukee
Paper
Activity A Precedes Activity C
A
C
B
D
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Start
Activities A and B Precede
Activity D
Figure 3.7
AON Network for Milwaukee
Paper
F
A
C
E
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Start
H
B
D
G
Arrows Show Precedence
Relationships
Figure 3.8
AOA Network for Milwaukee
Paper
2
4
Dummy
Activity
1
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C
(Construct
Stack)
3
D
5
(Pour
Concrete/
Install Frame)
6
H
(Inspect/
Test)
7
Figure 3.9
Determining the Project
Schedule
Perform a Critical Path Analysis
 The critical path is the longest path
through the network
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 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
Determining the Project
Schedule
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Perform a Critical Path Analysis
Activity
A
B
C
D
E
F
G
H
Description
Build internal components
Modify roof and floor
Construct collection stack
Pour concrete and install frame
Build high-temperature burner
Install pollution control system
Install air pollution device
Inspect and test
Total Time (weeks)
Time (weeks)
2
3
2
4
4
3
5
2
25
Table 3.2
Determining the Project Schedule
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Perform a Critical Path Analysis
Earliest start (ES) = earliest time at which an activity can start,
Activity
Description
Time
(weeks)
assuming all predecessors have
been
A
Build internal
components
2
completed
Modify
roof andtime
floor
EarliestBfinish (EF)
= earliest
at which an activity can3be
finished
C
Construct
collection stack
2
D start (LS)
Pour=concrete
andatinstall
4
Latest
latest time
whichframe
an activity can start
so as to not delayburner
the completion time
E
Build high-temperature
4 of
the entirecontrol
projectsystem
F
Install pollution
3
Latest
= air
latest
time by
which an activity has to
G finish (LF)
Install
pollution
device
5 be
finished so as to not delay the completion
H
Inspect and test
2
time of the entire project
Total Time (weeks)
25
Determining the Project Schedule
Perform a Critical Path Analysis
Activity Name or
Symbol
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Earliest
Start
Latest
Start
A
ES
EF
LS
LF
2
Earliest
Finish
Latest
Finish
Activity Duration
AON Example
Milwaukee Paper Manufacturing's
Activities and Predecessors
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Activity
A
Description
Build internal components
Immediate
Predecessors
—
B
Modify roof and floor
—
C
Construct collection stack
A
D
Pour concrete and install frame
A, B
E
Build high-temperature burner
C
F
Install pollution control system
C
G
Install air pollution device
D, E
H
Inspect and test
F, G
Forward Pass
Begin at starting event and work forward
Earliest Start Time Rule:

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
If an activity has only one immediate predecessor, its ES
equals the EF of the predecessor
If 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 forward
Earliest Finish Time Rule:
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
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
ES/EF Network for Milwaukee Paper
ES
EF = ES + Activity time
Start
0
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0
0
...
EF of A =
ES of A + 2
ES
of A
0
Start
0
A
0
2
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0
2
...
0
A
2
0
Start
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0
0
2
EF of B =
ES of B + 3
ES
of B
B
0
3
3
...
0
A
2
2
0
Start
2
0
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0
0
B
3
2
C
3
4
...
0
A
2
2
0
Start
2
C
4
2
0
= Max (2, 3)
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0
D
3
0
B
3
7
3
4
...
0
A
2
2
2
0
Start
C
4
2
0
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0
0
B
3
3
3
D
4
7
...
0
A
2
2
2
0
Start
C
2
7
3
0
4
0
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4
4
F
E
8
13
4
0
B
3
3
3
D
4
7
H
15
2
G
8
13
5
Figure 3.11
Backward Pass
Begin with the last event and work backwards
Latest Finish Time Rule:

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
If an activity is an immediate predecessor for just a single
activity, its LF equals the LS of the activity that immediately
follows it
If 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)
...
Begin with the last event and work backwards
Latest Start Time Rule:
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
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
...
0
A
2
2
2
0
Start
C
2
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7
3
0
4
0
Figure 3.12
4
4
F
E
8
13
13
4
0
B
3
3
H
2
15
15
LS = LF
D – Activity timeG
3
7
4
8
13
5
LF = EF
of Project
...
0
A
2
2
2
0
Start
C
10
2
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3
7
13
E
LF = Min(LS
of 8following
4
activity)
0
0
Figure 3.12
4
4
F
13
13
4
0
B
3
3
3
D
4
7
G
8
13
5
H
2
15
15
...
LF = Min(4, 10)
0
A
2
2
2
0
Start
2
C
2
4
4
4
10
0
4
4
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0
Figure 3.12
0
B
3
3
3
D
4
7
E
4
F
3
7
13
8
13
8
13
G
8
13
8
13
5
H
2
15
15
...
0
0
0
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0
Start
0
Figure 3.12
A
2
2
2
2
2
C
2
4
4
4
10
0
4
0
4
0
1
B
3
3
3
4
4
D
4
E
4
F
3
7
13
8
13
8
13
G
7
8
13
8
8
13
5
H
2
15
15
Computing Slack time
After computing the ES, EF, LS, and LF times for
all activities, compute the slack or free time for
each activity
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
Slack is the length of time an activity can be delayed without
delaying the entire project
Slack = LS – ES
or
Slack = LF – EF
...
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Earliest
Start
Activity
ES
A
B
C
D
E
F
G
H
0
0
2
3
4
4
8
13
Earliest
Finish
EF
Latest
Start
LS
Latest
Finish
LF
Slack
LS – ES
2
3
4
7
8
7
13
15
0
1
2
4
4
10
8
13
2
4
4
8
8
13
13
15
0
1
0
1
0
6
0
0
On
Critical
Path
Yes
No
Yes
No
Yes
No
Yes
Yes
Table 3.3
...
0
0
0
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0
Start
0
A
2
2
2
2
2
C
2
4
4
4
10
0
4
0
4
0
1
B
3
3
3
4
4
D
4
E
4
F
3
7
13
8
13
8
13
H
2
15
15
G
7
8
13
8
8
13
5
Figure 3.13
Activity on Arrow
2
C
4
A
F
E
dummy
1
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B
3
D
5
G
6
H
7
ES – EF Gantt Chart
for Milwaukee Paper
1
A Build internal
components
B Modify roof and floor
C Construct collection
stack
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D Pour concrete and
install frame
E Build high-temperature
burner
F Install pollution control
system
G Install air pollution
device
H Inspect and test
2
3
4
5
6
7
8
9
10 11 12 13 14 15 16
LS – LF Gantt Chart
for Milwaukee Paper
1
A Build internal
components
B Modify roof and floor
C Construct collection
stack
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D Pour concrete and
install frame
E Build high-temperature
burner
F Install pollution control
system
G Install air pollution
device
H Inspect and test
2
3
4
5
6
7
8
9
10 11 12 13 14 15 16
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Variability in Activity Times
 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
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...
 Three time estimates are required
 Optimistic time (a) – if everything goes
according to plan
 Most–likely time (m) – most realistic
estimate
 Pessimistic time (b) – assuming very
unfavorable conditions
...
Estimate follows beta distribution
Expected time:
t = (a + 4m + b)/6
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Variance of times:
v = [(b – a)/6]2
...
Estimate follows beta distribution
Probability
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t = (a + 4m + b)/6
Probability of
1 in 100 of
< a occurring
v = [(b − a)/6]2
Probability of
1 in 100 of > b
occurring
Activity
Time
Optimistic
Time (a)
Most Likely Time
(m)
Pessimistic Time
(b)
Computing Variance
Activity
a
A
B
C
D
E
F
G
H
1
2
1
2
1
1
3
1
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Optimistic
Most
Likely
Pessimistic
Expected
Time
Variance
m
b
t = (a + 4m + b)/6
[(b – a)/6]2
2
3
2
4
4
2
4
2
3
4
3
6
7
9
11
3
2
3
2
4
4
3
5
2
.11
.11
.11
.44
1.00
1.78
1.78
.11
Table 3.4
Probability of Project Completion
Project variance is computed by
summing the variances of critical
activities
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sp2 = Project variance
= (variances of activities
on critical path)
...
Project variance is computed by
summing the variances of critical
Project variance
activities
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s2p = .11 + .11 + 1.00 + 1.78 + .11 = 3.11
Project standard deviation
sp =
=
Project variance
3.11 = 1.76 weeks
...
PERT makes two more assumptions:
 Total project completion times follow a
normal probability distribution
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 Activity times are statistically independent
...
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Standard deviation = 1.76 weeks
15 Weeks
Figure 3.15
(Expected Completion Time)
...
What is the probability this project can be
completed on or before the 16 week
deadline?
Z=
due
date
–
expected date
of completion
/sp
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= (16 wks – 15 wks)/1.76
= 0.57
Where Z is the number of standard
deviations the due date lies from the
mean
...
From Appendix I
What is the probability
be
.00
.01this project
.07 can.08
completed
on or before the 16.52790
week .53188
.1 .50000 .50399
deadline?.2 .53983 .54380
.56749 .57142
due
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.5
.6
expected date
Z.69146
= date .69497
− of completion
.71566 /s
.71904
p
.72575
.72907
.74857
.75175
= (16 wks − 15 wks)/1.76
= 0.57
Where Z is the number of standard
deviations the due date lies from the
mean
...
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Probability
(T ≤ 16 weeks)
is 71.57%
0.57 Standard deviations
15
Weeks
Figure 3.16
16
Weeks
Time
...
Probability of
0.99
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Probability of
0.01
2.33 Standard
deviations
From Appendix I
0
Figure 3.17
2.33
Z
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...
 Variability of times for activities
on noncritical paths must be
considered when finding the
probability of finishing in a
specified time
 Variation in noncritical activity
may cause change in critical path
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What Project Management Has Provided So Far
 The project’s expected completion
time is 15 weeks
 There is a 71.57% chance the
equipment will be in place by the
16 week deadline
 Five activities (A, C, E, G, and H)
are on the critical path
 Three activities (B, D, F) have
slack time and are not on the
critical path
 A detailed schedule is available
Trade-Offs And Project Crashing
It is not uncommon to face the
following situations:
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 The project is behind schedule
 The completion time has been
moved forward
Shortening the duration of the
project is called project crashing
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Factors to Consider When Crashing A Project
 The amount by which an activity is
crashed is, in fact, permissible
 Taken together, the shortened
activity durations will enable us to
finish the project by the due date
 The total cost of crashing is as
small as possible
Steps in Project Crashing
1. Compute the crash cost per time period. If
crash costs are linear over time:
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(Crash cost – Normal cost)
Crash cost
per period = (Normal time – Crash time)
2. Using current activity times, find the critical
path and identify the critical activities
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...
3. If there is only one critical path, then
select the activity on this critical path
that (a) can still be crashed, and (b) has
the smallest crash cost per period. If
there is more than one critical path,
then select one activity from each
critical path such that (a) each selected
activity can still be crashed, and (b) the
total crash cost of all selected activities
is the smallest. Note that a single
activity may be common to more than
one critical path.
...
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4. Update all activity times. If the
desired due date has been
reached, stop. If not, return to
Step 2.
Crashing The Project
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Time (Wks)
Activity Normal Crash
A
B
C
D
E
F
G
H
2
3
2
4
4
3
5
2
1
1
1
2
2
2
2
1
Cost ($)
Normal
Crash
22,000
30,000
26,000
48,000
56,000
30,000
80,000
16,000
22,750
34,000
27,000
49,000
58,000
30,500
84,500
19,000
Crash Cost Critical
Per Wk ($) Path?
750
2,000
1,000
1,000
1,000
500
1,500
3,000
Yes
No
Yes
No
Yes
No
Yes
Yes
Table 3.5
Crash and Normal Times and Costs for Activity B
Activity
Cost
Crash
$34,000 —
Crash Cost/Wk =
Crash $33,000 —
Cost
$34,000 – $30,000
3–1
$4,000
=
= $2,000/Wk
2 Wks
=
$32,000 —
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$31,000 —
$30,000 —
Normal
Cost
Figure 3.18
Crash Cost – Normal Cost
Normal Time – Crash Time
Normal
—
|
1
Crash Time
|
2
|
3
Normal Time
Time (Weeks)
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Example :
Desired completion time from 32 days to 29 day
Picture 11.
Original Project Network – Crashing Example
Critical path 1367 (B F I) shown in boldface, 32 days length
…
Table 4. Normal and crash data
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Activity
A
B
C
D
E
F
G
H
I
Total
Time (days)
Cost ($)
Normal
Crash
Normal
Crash
Crash Cost
per Day
7
8
9
11
8
10
12
13
14
6
6
7
8
5
7
10
11
10
600
750
900
1100
850
1000
1300
1400
1500
750
900
1100
1400
1200
1300
1500
1500
2000
150
75
100
100
116.66
100
100
50
125
$9400
…
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 Based on the critical path, the activity that can
be shortened in the cheapest manner is activity
B, at an incremental cost $75 per day.
(see table 4).
 Crashing activity to B to the maximum extent
possible (i.e. normal time – crash time = 8 – 6 =
2 days) would reduce the completion time for
activities B F I to 32 – 2 = 30 days, at a total
project cost of $9400 + $150 = $9550.
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…
Picture 12.
Project Network – Activity B crashed 2 days
The new critical path 1257 (A D H) shown in boldface,
31 days length
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…
Picture 13.
Project Network – Activity B crashed 2 days; activity H crashed 2 days
The new critical path 1367 (A D H) shown in boldface,
30 days length
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…
Project Network – Activity B crashed 2 days; activity H crashed 2 days;
Picture 14. activity F crashed 1 day.
The new critical path 1257 and 1367 shown in boldface,
29 days length (desired completion time)
…
Table 5. Summary of time-cost tradeoffs
Step
0
1
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2
3
Action
No crashing in network
Activity B crashed by 2
days
Activity H crashed by 2
days
Activity F crashed by 1
day
Critical Path
1367
1257
Total Project
Completion
Time
Total Cost
($)
32
9400
31
9550
30
9650
29
9750
1367
1257
1367
Making crashing decision using LP
 Based on the previous case:
xi  timeof occurenceof eventi; i  1,2,...,7
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yi  amountof timecrashedfor activityj; j  A, B,...,I
 Minimize (crash cost) Z
= $150yA + $75yB + $100yC + $100yD + $116.66yE +
$100yF + $100yG + $50yH + $125yI
 Project completion-date constraint
x7  29
 Activity crash-time constraint
yA  1 ; yB  2 ; yC  2 ; yD  3 ; yE  3 ; yF  3 ; yG  2
;
yH  2 ; yI  4
…
 Constraint describing the network :
1.
2.
3.
The occurrence time for an event must be equal to, or greater than, the activity completion time for
all activities leading into the node that represents that event.
The start time for an activity is equal to the occurrence time of its preceding event.
The time required to complete an activity is equal to its normal time minus the length of time it is
crashed.
 Begin with the event occurrence time for event 1 = 0 or x1 = 0
For event 2
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Occurrence time  time required to
for event 2
complete activity A
+ start time for
activity A
(x1 = 0)
Normal time – Crash time
for activity A for activity A
x2  7 – yA + 0
or x2 + yA  7
And for following events are
…
 Event 3 : x3  8 - yB + 0 = x3 + yB  8
 Event 4 : x4  9 - yC + x2 = x4 - x2 + yC  9 (note that activity C begins with event 2,
x2)
 Event 5 : x5  11 - yD + x2 = x5 - x2 + yD  11 (for the path from activity D)
: x5  8 - yE + x3 = x5 - x3 + yE  8 (for the path from activity E)
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 Event 6 : x6  10 - yF + x3 = x6 - x3 + yF  10
 Event 7 : x7  12 - yG + x4 = x7 - x4 + yG  12 (for the path from activity G)
: x7  13 - yH + x5 = x7 - x5 + yH  13 (for the path from activity H)
: x7  14 - yI + x6 = x7 - x6 + yI  14 (for the path from activity I)
 With xi  0 for i = 1,2,..,7
yj  0 for j = A,B,..,I
…
 The optimal solution
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x1 = 10
x2 = 7
x3 = 6
x4 = 17 (the slack for activity C = 1 day)
x5= 18
x6 = 15
x7 = 15
yA = 0
yB = 2
yC = 0
yD = 0
yE = 0
yF = 1
yG = 0
yH = 2
yI = 0
 The solution values of yB = 2, yF = 1, and yH = 2, indicate that activity B, F
and H must be crashed by 2, 1, 2 days, respectively.
PERT/COST
 The first step in PERT/COST procedure is to
subdivide the project into components that can
be used to plan and schedule the cost
associated the project (budgeting process).
 Major steps :
1.
2.
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3.
4.
For each activity in the project, determine the aggregate cost
associated with the activities. This will be the budget for that
activity.
Given the expected activity time for each activity, convert the
budgeted cost for each activity into a cost per unit time period.
(assumed at uniform rate over time)
Using the expected activity times, perform the critical path
calculations to determine the critical path for the project.
Using the earliest and latest start times from the critical path
calculations, determine the amount of money that should be
spent during each time period in order to complete the project
by a desired date.
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…
Picture 15. Project network – PERT/COST example
…
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Table 6. Expected activity times and cost estimates
Activity
Expected activity
time (months)
Estimated cost
(Budget $)
Budgeted cost
per month
A
B
C
D
E
F
G
H
4
2
3
8
3
2
6
7
20,000
20,000
12,000
24,000
21,000
18,000
36,000
14,000
$ 5,000
10,000
4,000
3,000
7,000
9,000
6,000
2,000
Total budgeted cost =
165,000
…
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Table 7. Activity schedule and slack time
Activity
Earliest
Start
Time
(ES)
Latest
Start
Time
(LS)
Earliest
Finish
Time
(EF)
Latest
Finish
Time
(LF)
Slack
(S)
On Critical
Path?
A
B
C
D
E
F
G
H
0
0
4
4
7
7
12
9
0
7
6
4
9
9
12
11
4
2
7
12
12
11
18
16
4
9
9
12
12
11
18
18
0
7
2
0
2
2
0
2
Yes
No
No
Yes
No
No
Yes
No
…
Table 8. Budgeted Costs ($000), using earliest start
time
Month
Activity
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
A
5
5
B
10 10
5
5
20
20
C
4
4
4
D
3
3
3
12
3
3
3
E
7
7
7
F
9
9
3
3
18
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2
15 15 5
5
7
7
24
21
G
H
Total
Cost/
Month
Total
Cost to
Date
Totals
6
6
6
6
2
2
2
2
2
2
7 19 19 12 5
5
8
8
8
8
6
6
36
14
6
6
15 30 35 40 47 54 61 80 99 111 116 121 129 137 145 153 159 165
165
…
Table 9. Budgeted Costs ($000), using latest start time
Activitiy
A
Month
1
2
3
4
5
5
5
5
5
6
7
9 10 11 12 13 14 15 16 17 18
C
3
3
10 10
20
4
4
4
12
3
3
3
3
3
3
24
E
7
7
7
21
F
9
9
18
G
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Totals
20
B
D
8
H
Total
Cost/
Month
Total
Cost to
Date
6
6
6
6
6
6
36
2
2
2
2
2
2
14
7 17 17 19 19 12 8
8
8
8
8
8
165
2
5
5
5
5
3
3
5 10 15 20 23 26 33 50 67 86 105 117 125 133 141 149 157 165
…
180
160
140
Budgeted cost
earliest start
times
120
100
80
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60
40
Budgeted cost
latest start times
20
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
Monitoring and Controlling Project Costs
 Value of work completed =
(percent of completion ) x (total budgeted cost)
 Activity cost difference =
total actual cost – value of work completed
 Ex :
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Table 10. Activity cost and completion : end of month 9
Activity
Total
Budgeted
Cost ($)
A
B
C
D
E
F
G
H
Total
20,000
20,000
12,000
24,000
21,000
18,000
36,000
14,000
165,000
Percent of
Completion
Value of
Work
Completed
($)
Total Actual
Cost ($)
Activity Cost
Difference ($)
100
100
100
50
25
0
0
0
20,000
20,000
12,000
12,000
5250
0
0
0
18,000
22,000
15,000
13,000
5000
0
0
0
-2000
2000
3000
1000
-250
0
0
0
3750
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Problem in the application of PERT/CPM
 It may be difficult to divide a project into a set
of independent activities.
 It may be difficult to firmly establish the
precedence relationship among various
activities, and not all precedence relationship
can be anticipated before a project begins. (for
R&D projects)
 The PERT procedure is highly dependent on
being able to accurately make activity time
estimates.
 The theoretical foundation of the PERT
statistical procedure is subject to question
regarding to assumptions in expected activity
time and its variance. (not nearly as important
as the practical problems associated with
making accurate time estimates)