Transcript 311_Session27Ardavan2

Operations Management
Session 27: Project Management
Scheduling the Project
From Action Plan and WBS to Gantt chart and project
network.
 Gantt Chart
 Project Network


Activity-on-arrow
Activity-on-node
 CPM and PERT
 Risk analysis involves determining the likelihood that
a project can be completed on time


Statistics
Simulation
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Scheduling the Project
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History
Late 1950s


Critical Path Method (CPM)

Dupont De Nemours Inc. developed the method

Deterministic activity durations
Program Evaluation and Review Technique (PERT)
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
U.S. Navy, Booz-Allen Hamilton, and Lockeheed Aircraft

Probabilistic activity durations
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 Activity
The Language of
PERT/CPM

Task or set of tasks

Takes time and needs resources
 Precedence Relationships

The immediate predecessor activities
 Event


Completion of one or more activities (to allow the next activity or activities
to start)
Zero duration, zero resource
 Milestones

Significant events – showing completion of a significant portion of the
project
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The Language of PERT/CPM
 Network

Diagram of nodes connected by directed arcs

Shows technological relationships among activities
 Path

A set of connected activities such that each activity on both sides is
connected to one and only one other activity (with exception!) .
 Critical Path

A path where a delay in any of its activities will delay the project

The longest path on the network

The shortest time to complete the project
 Critical Time

The total time to complete all activities on the critical path
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Two Types of Network
Diagrams
 Activity-on-Arrow Network (Arrow Diagramming
Method)

Easier to show events and milestones

More compatible with network theory techniques

Sometimes requires dummy (artificial) activities
 Activity-on-Node Network (Precedence
Diagramming Method)

Easier representation

No dummy activity
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Activity on Arrow
Network
An Activity is an arc with two nodes at its beginning and its end
a
b
c
d
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AoA: Activity
Predecessors
A list of immediate predecessors is needed.
Task
Predecessor
a
--
b
a
c
b
Task
Predecessor
a
---
b
--
c
b
d
a
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a
b
c
a
d
b
c
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AoA: Activity
Predecessors
Task
Predecessor
a
---
b
a
c
a
d
a
Task
Predecessor
a
---
b
--
c
--
d
a,b,c
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b
a
c
d
a
d
b
c
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AoA May Need Dummy
Activity
 Two activities have the same starting and ending nodes
 A single activity connects to two or more nodes
Task
Predecessor
a
---
b
a
c
a
d
b,c
b
d
a
c
 Try this: a,b  c and a,d  e
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AoA: A Power Plant
Construction Project
Task
Description
Predecessor
a
Design & engineering
---
b
Select site
a
c
Select vendor
a
d
Select personnel
a
e
Prepare site
b
f
Manufacture generator
c
g
Prepare operation manual
c
h
Install generator
e,f
i
Train operators
d,g
j
Obtain license
h,i
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AoA: A Power Plant
Construction Project
b
a
e
c
d
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g
h
i
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a:b:a
c:a
d:a
e: b
f:c
g:c
h:e,f
i: d,g
j: h,i
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Draw AoN Network
a:b:a
c:a
d:a
e: b
f:c
g:c
h:e,f
i: d,g
j: h,i
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Draw AOA
Activity Predecessor Duration
a
6
b
2
c
5
d
a
4
e
a
3
f
c
8
g
b,e,f
9
h
c
7
i
b,e,f
4
j
h
9
k
d,i
6
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Transform into AON
2
b=2
1
3
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d=4
4
h=9
6
g=9
5
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a:b:c:d:a
e: a
f:c
g:b,e,f
h:c
i: b,e,f
j: h
k:d,i
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Draw AoN Network
a:b:c:d:a
e: a
f:c
g:b,e,f
h:c
i: b,e,f
j: h
k:d,i
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Critical Path and Critical
time
 The critical path is the shortest time in which a project can
be completed
 If a critical activity is delayed, the entire project will be
delayed.
 There may be more than one critical path.
 Brute force approach to finding critical path:
1.
identify all possible paths from start to finish
2.
sum up duration of activities on each path
3.
largest total indicates critical path
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Critical Path Method: The
Network
4
6
A1
A3
S
4
3
A4
A6
3
2
A2
A5
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E
Find the Critical Path.
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Critical Path Method:
Paths
4
6
A1
A3
S
3
A2
How many path?
E
4
3
A4
A6
2
10
11
8
A5
Critical Path is the longest path. It is the shortest time to complete the project
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Forward Path; Earliest
Starts
6
4
4
A1
00
4
A3
4
10
10
3
4
0
S
A4
4
0
4
3
0
A2
0
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8
5
A6
8
11
A5
3
3
2
E
11 11
3
5
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Forward Path
Max = 30
10
30
20
35
35
35
5
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Backward Path; Latest
Starts
0 4
4
A1
00
00
S
03
5
4
6 11
5
4
4
4
0
3
4
6 2
6
6
A2
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3
10
8
8
3
E
11
10
8
8
A5
3
0
4
A4
4
3
11
A3
4
11
8
8
5
3 1111
A6
8
11
8
5
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11
Backward Path
30
30
30
Min = 35
35
45
30
5
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Activity Slack
 Slack, or float: The amount of time a noncritical task can be
delayed without delaying the project
 Slack—LFT – EFT
or
LST – EST
 EST—Earliest Start Time

Largest EFT of all predecessors
 EFT—Earliest Finish Time

EST + duration for this task
 LFT—Latest Finish Time

Smallest LST of following tasks
 LST—Latest Start Time

LFT – duration for this task
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Computing Slack Times
EST
EFT
Task = duration
slack = xxxx
LST
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LFT
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Critical Path, Slacks
0
4
4
5
11
11
A3
A1
0
6
4
4
4
S
4
E
10
8
11
8
A4
3
3
6
4
6
A2
0
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3 11
A6
8
8
8
11
A5
3
3
5
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Slack Times Example
Task Pred. Dur.
Task Pred. Dur.
a
--
4
g
c,d
1
b
--
3
h
e
4
c
a
3
i
f
5
d
a
2
j
e,g
6
e
b
6
k
h,i
1
f
b
4
For each task, compute ES, EF, LF, LS, slack
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Slack Times Example
c=3
slack=
LST
Task=dur
slack=xxx
EST
g=1
slack=
a=4
slack=
EFT
j=6
slack=
d=2
slack=
Finish
Start
e=6
slack=
h=4
slack=
b=3
slack=
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LFT
k=1
slack=
f=4
i=5
slack=
slack=
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Activity Times in PERT
 Optimistic (a)

Activity duration to be ≤ a has 1% probability.  ≥ a has 99%
probability
 Pessimistic (b)

Activity duration to be ≥ b has 1% probability  ≤ b has
99% probability
 Most likely (m)

The mode of the distribution
 All possible task durations (or task costs) can be
represented by statistical distributions
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Beta Distribution: The Probability
Distribution of Activity Times
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Activity Expected Time and
Variance
 Mean, “expected time”
TE = (a + 4m + b)/6
 Standard deviation, 
 = (b-a)/6
 Variance
2 = [(b-a)/6]2
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95% & 90% Levels
 If we replace 99% with 95% or 90% levels

Activity duration to be ≤ a has 5% probability

Activity duration to be ≥ b has 5% probability
(b  a )

3.3

Activity duration to be ≤ a has 10% probability

Activity duration to be ≥ b has 10% probability
(b  a )

2.6
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Probability of Completing
the Critical Path on Time
 We assume the various activities are statistically
independent of each other
 Individual variances (and mean) of the activities on a path
can then be summed to find the variance (mean) of the
path
 Determine the mean and standard deviation of the critical
path
 Compute the probability of critical path being ≤ a
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The Probability of Completing the
Critical Path on Time
Z
DCP  m CP

2
CP
DCP = the desired completion date of the critical path
mCP= the sum of the TE for the activities on the critical path
2CP = the sum of the variances of the activities on the critical path
Given Z, the probability of having the standard normal variable
being ≤ Z is the probability of completing the project in a time ≤
D
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Selecting Risk and
Finding D
Select the probability of meeting the completion date and solve
for the desired date, D
DCP  mCP  Z CP
Using the probability, you can compute Z and then solve for D
3/31/2016
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Probability of Completing a
Project on Time
 Find all paths in the network
 Compute mean and standard deviation of each path
 Compute the probability of completing each path in ≤ the
given time
 Calculate the probability that the entire project is completed
within the specified time by multiplying these probabilities
together
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Critical Path Method: Paths
Suppose all activities have beta distribution
4,1
A1
6,2
A3
E
3,1
4,1
S
A4
3,0.5
A2
A6
2,0.5
A5
10
11
8
The first number is the mean; the second is standard deviation.
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Probability of Completing
CP in 12 days
What is the probability of competing the critical path in a maximum
of 12 days?
DCP = the desired completion date of the critical path
mCP= 4+4+3 = 11
2CP = 12+12+12 = 3
CP = 1.73
Z
DCP  mCP
 CP
12  11

 0.58
1.73
Z= 0.58  P(z≤0.58) = 0.72
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Selecting Risk and
Finding CP Time
With a probability of 90%, in how many days will the CP be
completed?
From Standard Normal Table  Z 90% = 1.28
DCP  mCP  Z CP
mCP  11,  CP  1.73, Z %90 1.28
DCP  11  1.28 1.73  11  2.21  13.21
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Probability of Completing
The Project in 12 days
The probability of completing the critical path in not more than 12
days was 0.72. We need to compute this probability for blue path
and green path too, and then multiply these probabilities
mCP= 6+4 = 10
2CP = 12+22 = 5 Z= (12-10)/2.24 = 0.89  P(z≤0.89) = 0.81
CP = 2.24
mCP= 3+2+3 = 8
2CP = 0.52+0.52+12 = 1.22 Z= (12-8)/1.1 = 3.63  P(z≤3.63) ≈ 1
CP = 1.1
The probability of competing the Project in not more than 12 days is
0.72×0.81×1 = 0.58
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