1. Chapter 8 - Project Management
Lecture 2
Today’s lecture covers the followings:
To study “project crashing” concept
LP formulation for project
management problem
The use of QM (Try it yourself!)
Tutorial: Chapter 8: 8th Ed:Q26 and Q30
9tj Ed: Q21 and Q23
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Chapter 8 - Project Management

2. 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?
If so, how!
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Chapter 8 - Project Management

3. 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?
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Chapter 8 - Project Management

4. Chapter 8 - Project Management
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
How to do that:
Question: What criteria should it be based on when deciding to crashing critical times?
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Chapter 8 - Project Management

5. Chapter 8 - Project Management
Example – crashing (1)
Max weeks can be crashed
Normal weeks
The critical path is 1-2-3, the completion time =11
How? Path: = 5+6=11 weeks
Path: 1-3 = 5 weeks
Now, how many days can we “crash” it? 2
6(3)
5 (1) 3 1
5(0)
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Chapter 8 - Project Management

6. Chapter 8 - Project Management
Example – crashing (1) 2
6(3)
5 (1) 3 1
5(0)
The maximum time that can be crashed for:
Path = = 4
Path 1-3 = 0
Total weeks can be crashed = = 4
Are we to use up all these 4 weeks?
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Chapter 8 - Project Management

7. Chapter 8 - Project Management
Example – crashing (1)
3(0)
4(0) 2
6(3)
5 (1) 3 1
5(0)
If we used all 4 days, then path 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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Chapter 8 - Project Management

8. Chapter 8 - Project Management
Example – crashing (1)
Such as 2
6(3)
5 (1) 3 1
8(0)
Now, we cannot use all 4 days (Why?)
Because path 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 completion time to the same time
As path (HOW?)
Chapter 8 - Project Management
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9. Chapter 8 - Project Management
Example – crashing (1)
Solution: 2
6(3)
5 (1) 3 1
8(0)
We can only reduce total time for path = 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
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Chapter 8 - Project Management

10. Chapter 8 - Project Management
Example – crashing (1)
Solution:
4(0)
4(1) 2
6(3)
5 (1) 3 1
8(0)
3(0) OR
5 (1)
6(3) 2 1 3
8(0)
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Now, paths and 1-3 are both critical paths
Chapter 8 - Project Management

11. Chapter 8 - Project Management
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?
Procedural step.
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Chapter 8 - Project Management

12. Project Crashing and Time-Cost Trade-Off Example Problem (1 of 3)
F= (D-C)/(A-B)
E=(A-B) A B D F C
Table 8.5
Normal Activity and Crash Data for the Network in Figure 8.16
Note: A,B,C,D are given Note: we will use F values to decide
We need to compute E and F which path to crash!
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13. Chapter 8 - Project Management
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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Chapter 8 - Project Management

14. Chapter 8 - Project Management
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
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More example!
Chapter 8 - Project Management
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15. Chapter 8 - Project Management
Step 1
Using CPM, the critical path is
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Chapter 8 - Project Management

16. Chapter 8 - Project Management
Step 2
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Chapter 8 - Project Management

17. Section1 Section 2 Section 3 Section 4
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
Section Section Section Section 4
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18. Section1 Section 2 Section 3 Section 4
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) $$200
Section Section Section Section 4
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19. Section1 Section 2 Section 3 Section 4
Step 3:
We now crashed them one section at a time as follows:
5(0)
3(0)
8(3) $500
9(0)
7(0)
12(3) $$7000
12(5) $400
4(1) $7000
4(1) $3000
4(3) $$200
4(3) $$200
Section Section Section Section 4
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20. =(5*$400)+(3*$500)+(3*$7000)+(1*$7000)= 31,000
Step 4:
We now crashed them one section at a time as follows:
5(0)
3(0)
8(3) $500
9(0)
7(0)
12(3) $$7000
12(5) $400
4(1) $7000
4(1) $3000
4(3) $$200
4(3) $$200
Total crash cost
=(5*$400)+(3*$500)+(3*$7000)+(1*$7000)= 31,000
Total crashed weeks= =12
Note: critical path is
Completion time = = 24
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21. Chapter 8 - Project Management
Crashed cost
10(5)
5(4) 1 3 4
4(1)
4(2) 2
How to solve this problem?
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Chapter 8 - Project Management

22. Chapter 8 - Project Management
Further detail steps
Determine the critical path
Crash the critical path to the level where other non-critical paths become a critical one
Consider for further crashing until all possible crashing resources were consumed!
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Chapter 8 - Project Management

23. Chapter 8 - Project Management
Critical path
10(5)
5(4) 1 3 4
4(1)
4(2) 2
The critical path is 1-3-4, completion time is 10+5 = 15
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Chapter 8 - Project Management

24. Crash to a level to which other non-critical path is introduced
5(0)
3(2)
10(5)
5(4) 1 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
So, we try to reduce the critical path to this level !
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Chapter 8 - Project Management

25. Crash all resources until no further can be reduced!
2(0)
5(0)
3(1)
10(5)
5(3) 1 3 4
4(3)
3(2)
4(2) 2
Both critical
Paths = 7
Stop, since no more resources can be reduced in path 1-3-4
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Chapter 8 - Project Management

26. 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
crashing model
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Chapter 8 - Project Management

27. Chapter 8 - Project Management
LP formulation
General linear programming model is:
minimize Z = cixi
subject to
xj - xi  tij for all activities i  j
xi, xj  0
where xi = earliest event time of node i
xj = earliest event time of node j
tij = time of activity i  j
LP formulation for the project management
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Chapter 8 - Project Management

28. Chapter 8 - Project Management
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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Chapter 8 - Project Management

29. Chapter 8 - Project Management
Objective is Minimize Z = x1 + x2 + x3 + x4 + x5 + x6 + x7
Subject to 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
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Do you know how to read the results from the LP output?
Chapter 8 - Project Management

30. Chapter 8 - Project Management
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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Chapter 8 - Project Management

31. 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
We now y to represent these
How?
Chapter 8 - Project Management
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32. Chapter 8 - Project Management
Min 400y y y24 + 0y y y y y67
And all yij <= their total allowance crash time
A complete model is shown in next slide
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Chapter 8 - Project Management

33. Chapter 8 - Project Management
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 = $400y y y y y y y67
subject to 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 x x7 - x6  4
x7  xj, yij  0
Max crashing time
for critical path
i.e. total allowable
crashed time
New set of equations
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CPM value
Chapter 8 - Project Management