Chapter 8 - Project Management Chapter topics

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Lecture 2
Today’s lecture covers the followings:
1. To study “project crashing” concept
2. LP formulation for project
management problem
3. The use of QM (Try it yourself!)
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Tutorial: Chapter 8: 8th Ed:Q26 and Q30
9tj Ed: Q21 and Q23
Chapter 8 - Project Management
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Project Crashing
Basic Concept
In last lecture, we studied on how to use CPM to
determine solution for a project problem
There, we determine its critical path and completion
time.
Question:
Can we cut short its project completion
time?
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If so, how!
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Project Crashing
Solution!
Yes, the project duration can be reduced by
assigning more resources to project
activities
But, doing this would somehow increase
our project cost!
How do we strike a balance?
Chapter 8 - Project Management
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Trade-off concept
Here, we adopt the “Trade-off” concept
• ie, we attempt to “crash” some “critical”
events by allocating more sources to them,
and also to maintain a balance that the
shortening time is not less than the normal
activities
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• How to do that:
• Question: What criteria should it be based
on when deciding to crashing critical times?
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Example – crashing (1)
Max weeks can be crashed
Normal weeks
5 (1)
1
2
6(3)
3
5(0)
The critical path is 1-2-3, the completion
time =11
How? Path: 1-2-3 = 5+6=11 weeks
Path: 1-3 = 5 weeks
Now, how many days can we “crash” it?
Chapter 8 - Project Management
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Example – crashing (1)
5 (1)
1
2
6(3)
3
5(0)
The maximum time that can be crashed for:
Path 1-2-3 = 1 + 3 = 4
Path 1-3 = 0
Total weeks can be crashed = 4 + 0 = 4
Are we to use up all these 4 weeks?
Chapter 8 - Project Management
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Example – crashing (1)
4(0)
5 (1)
1
3(0)
2
6(3)
3
5(0)
If we used all 4 days, then path 1-2-3 has
(5-1) + (6-3) = 7 completion weeks
Now, we need to check if the completion time for path 1-3 has lesser than 7
weeks (why?)
Now, path 1-3 has (5-0) = 5 weeks
Since path 1-3 still shorter than 7 weeks, we used up all 4 crashed weeks
Question: What if path 1-2 has, say 8 week completion time?
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Example – crashing (1)
Such as
5 (1)
2
6(3)
3
1
8(0)
Now, we cannot use all 4 days (Why?)
Because path 1-2-3 will not be critical path anymore as
path 1-3 would now has longest hour to finish
Rule: When a path is a critical path, it will stay as a critical path
So, we can only reduce the path 1-2-3 completion time to the same time
As path 1-2. (HOW?)
Chapter 8 - Project Management
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Example – crashing (1)
Solution:
5 (1)
2
6(3)
3
1
8(0)
We can only reduce total time for path 1-2-3 = path 1-2,
that is 8 weeks
If the cost for path 1-2 and path 2-3 is the same then
We can random pick them to crash so that its completion
Time is 8 weeks
Chapter 8 - Project Management
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Example – crashing (1)
Solution:
4(0)
5 (1)
3
1
OR
8(0)
5 (1)
1
2
4(1)
6(3)
2
3(0)
6(3)
3
8(0)
Now, paths 1-2-3 and 1-3
are both critical paths
Chapter 8 - Project Management
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Time-cost Trade-off
In this subject, the decision for “crashing” the
project is based on the trade-off between
“time and cost”
The method is called “Time-cost Trade-off”
How it works?
– We determine an average crash cost for each event
• How to do that?
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– Procedural step.
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Chapter 8 - Project Management
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Project Crashing and Time-Cost Trade-Off
Example Problem (1 of 3)
A
B
C
D
E=(A-B)
F= (D-C)/(A-B)
F
Table 8.5
Normal Activity and Crash Data for the Network in Figure 8.16
Note: A,B,C,D are given
We need to compute E and F
Note: we will use F values to decide(to p11)
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which path to crash!
Time-Cost Trade-Off
Steps:
1. use “normal cost” to determine the critical path
2. for each event, compute their average crash cost
3. for each section of critical path, crash their
maximum time by retaining this section be part
of the “critical” path.
4. compute total crashing costs and completion
time
Example
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Example: trade-off
•
•
•
•
•
Consider the same example as show in below
Step 1 determine it critical path
Step 2 determine all average unit crash cost
Step 3 crashing events with minimum costs
Step 4 compute crashed weeks and costs
More example!
Chapter 8 - Project Management
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Step 1
• Using CPM, the critical path is
1-2-3-4-6-7
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Step 2
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Step 3:
First, we cluster each segment of critical path into sections that can
be crashed and to consider to crash them one section at a time
Section1
Section 2
Section 3
Section 4
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Step 3:
We now add the normal and crashed time and cost to each segment
8(3) $500
12(3) $$7000
12(5) $400
4(1) $7000
4(1) $3000
4(3) $$200
4(3) $$200
Section1
Section 2
Section 3
Section 4
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Step 3:
We now crashed them one section at a time as follows:
5(0)
8(3) $500
9(0)
7(0)
12(3) $$7000
12(5) $400
3(0)
4(1) $7000
4(1) $3000
4(3) $$200
4(3) $$200
Section1
Section 2
Section 3
Section 4
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Step 4:
We now crashed them one section at a time as follows:
5(0)
8(3) $500
9(0)
7(0)
12(3) $$7000
12(5) $400
3(0)
4(1) $7000
4(1) $3000
4(3) $$200
4(3) $$200
Total crash cost
=(5*$400)+(3*$500)+(3*$7000)+(1*$7000)= 31,000
Total crashed weeks= 5+3+3+1=12
Note: critical path is 1-2-3-4-6-7
Completion time = 7+5+0+9+3 = 24
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Crashed cost
10(5)
1
5(4)
3
4
4(1)
4(2)
2
How to solve this problem?
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Further detail steps
1. Determine the critical path
2. Crash the critical path to the level where
other non-critical paths become a critical
one
3. Consider for further crashing until all
possible crashing resources were
consumed!
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Critical path
10(5)
1
5(4)
3
4
4(1)
4(2)
2
The critical path is 1-3-4, completion time is 10+5 = 15
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Crash to a level to which other noncritical path is introduced
5(0)
10(5)
1
3(2)
5(4)
3
4
4(3)
4(2)
2
Both critical
Paths = 8
The non-critical path is 1-2-4, has the processing time = 4+4 = 8
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So, we try to reduce the critical path to this level !
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Crash all resources until no further
can be reduced!
2(0)
3(1)
5(3)
5(0)
10(5)
1
3
4(3)
3(2)
4
4(2)
2
Both critical
Paths = 7
Stop, since no more resources can be reduced in path 1-3-4
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Formulating the CPM/PERT Network as a Linear
Programming Model
- The objective is to determine the earliest time the project can be completed
(i.e., the critical path time).
• normal CPM
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• crashing model
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LP formulation
General linear programming model is:
minimize Z = cixi
subject to
xj - xi  tij for all activities i  j
x i, x j  0
where xi = earliest event time of node i
xj = earliest event time of node j
tij = time of activity i  j
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• LP formulation for the project management
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LP for the CPM
• Let consider a simple problem as outlined as
follows:
Let xi be denote as each node i
And segment of say path 1-2 as x2-x1
Then
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Objective is Minimize Z = x1 + x2 + x3 + x4 + x5 + x6 + x7
x2 - x1  12
(for path 1-2)
x3 - x2  8
(for path 2-3)
x4 - x2  4
(for path 2-4)
x4 - x3  0
(for path 3-4)
x5 - x4  4
(for path 4-5)
x6 - x4  12
(for path 4-6)
x6 - x5  4
(for path 5-6)
x7 - x6  4
(for path 6-7)
xi, xj  0
Do you know how to read the results from the LP output?
Subject to
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General concept
• All formulation of CPM is used, except we
need one more variable to represent the
crashed cost per unit of each path
• Example!
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Consider again the following crashed cost as an example
- Objective is to reduce the project duration from 36 to 30 weeks at the minimum possible crash cost.
Our objective is to min these
How?
We now y to represent these
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Min 400y12 + 500y23 + 3000y24 + 0y34 + 200y45 + 7000y46 + 200y56 + 7000y67
And all yij <= their total allowance crash time
A complete model is shown in next slide
Chapter 8 - Project Management
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The CPM/PERT Network as a Linear Programming Model
Example Problem Project Crashing - Model Formulation
xi = earliest event time of node i
xj = earliest event time of node j
yij = amount of time by which activity i  j is crashed (i.e., reduced)
minimize Z = $400y12 + 500y23 + 3000y24 + 200y45 + 7000y46 + 200y56 + 7000y67
subject to
Max crashing time
for critical path
i.e. total allowable
crashed time
New set of equations
CPM value
y12  5
y12 + x2 - x1  12
y23  3
y23 + x3 - x2  8
y24  1
y24 + x4 - x2  4
y34  0
y34 + x4 - x3  0
y45  3
y45 + x5 - x4  4
y46  3
y46 + x6 - x4  12
y56  3
y56 + x6 - x5  4
y67  1
x67 + x7 - x6  4
x7  30
xj, yij  0
Chapter 8 - Project Management
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