Today

1
Lecture 2

• Todays 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 EdQ26 and Q30
• 9tj Ed Q21 and
  Q23

(to p2)
(to p26)
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!

(to p3)
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?

(to p4)
4
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?

(to p5)
(to p11)
5
Example crashing (1)
Max weeks can be crashed
Normal weeks

• The critical path is 1-2-3, the completion time
  11
• How? Path 1-2-3 5611 weeks
• Path 1-3 5 weeks
• Now, how many days can we crash it?

2
6(3)
5 (1)
3
1
5(0)
(to p6)
6
Example crashing (1)

2
6(3)
5 (1)
3
1
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?
(to p7)
7
Example crashing (1)

3(0)
4(0)
2
6(3)
5 (1)
3
1
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?
(to p8)
8
Example crashing (1)

Such as
2
6(3)
5 (1)
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?)
(to p9)
9
Example crashing (1)

Solution
2
6(3)
5 (1)
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
(to p10)
10
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)
(to p4)
Now, paths 1-2-3 and 1-3 are both critical paths
11
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.

(to p12)
(to p13)
12
Project Crashing and Time-Cost Trade-OffExample
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
(to p11)
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!
13
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

(to p14)
14
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

(to p15)
(to p16)
(to p17)
(to p20)
More example!
(to p21)
15
Step 1

• Using CPM, the critical path is
• 1-2-3-4-6-7

(to p14)
16
Step 2
(to p14)
17
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
(to p18)
18
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
(to p19)
19
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
Section1 Section 2 Section 3
Section 4
(to p14)
20
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 (5400)(3500)(37000)(1
7000) 31,000 Total crashed weeks
533112 Note critical path is
1-2-3-4-6-7 Completion time 75093 24
(to p14)
21
Crashed cost
10(5)
5(4)
1
3
4
4(1)
4(2)
2
How to solve this problem?
(to p22)
22
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!

(to p23)
23
Critical path
10(5)
5(4)
1
3
4
4(1)
4(2)
2
The critical path is 1-3-4, completion time is
105 15
(to p24)
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 44 8 So, we try to reduce
the critical path to this level !
(to p25)
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)
4(2)
3(2)
2
Both critical Paths 7
Stop, since no more resources can be reduced in
path 1-3-4
(to p1)
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

(to p27)
(to p30)
(to p1)
27
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

(to p28)
28
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
(to p29)
29
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
(to p26)
Do you know how to read the results from the LP
output?
30
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!

(to p31)
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?
(to p32)
32
Min 400y12 500y23 3000y24 0y34 200y45
7000y46 200y56 7000y67
And all yij lt their total allowance crash time
A complete model is shown in next slide
(to p33)
33
The CPM/PERT Network as a Linear Programming
ModelExample 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 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


Max crashing time for critical path i.e. total
allowable crashed time
New set of equations
(to p26)
CPM value