Transcript Project Management Chapter 3 part 3

Project Management
Chapter 10 part 3
Crashing the Project (Time-Cost)
Project Crashing (Time-Cost)
(Reducing the length of time of a project)
 Estimated time is fixed... no more.
 Can reduce through additional
resources
 Manpower,
equipment
 Direct cost of activity is always increased
“I am here to put you back on schedule”
-Darth Vader
 Need to reduce time of project because:
 Requirement to complete in specified time
frame
 Economic advantage
 Three kinds of costs
 Crash costs (activity direct costs)
 Can be given directly, or need to determine
through linear interpolation
 Administration costs (or project indirect costs)
 Penalty costs
 Incur crash costs to avoid administration and
penalty costs
Procedure
 Always crash one period at a time!
 1. Identify options – critical activities
 2. Select least expensive
 3. Savings? Implement if so. Update all paths
 4. Repeat 1-3 until no cost savings are possible
Note: the critical path will change – can have two
5
5
3
3
B
C
D
A
3
5
2
4
3
2
Admin Cost =
$500
X
2
Y
2
3
Z
1
A -$250
B-$100
C-$200
DX-$350
Y-$125
ABCD 16
AXYZ 15
Cost
Save
Net
Cumul
Crash Costs
Z-$325
What activity do we crash
first?
5
5
3
3
B
C
D
A
3
5
2
4
3
2
Admin Cost =
$500
ABCD 16
AXYZ 15
Cost
Save
Net
Cumul
X
2
Y
2
3
Z
1
Crash Costs
A -$250
B-$100
C-$200
DX-$350
Y-$125
Z-$325
5
5
3
3
B
C
D
A
3
5
2
4
3
2
Admin Cost =
$500
ABCD 16
AXYZ 15
Cost
Save
Net
Cumul
X
2
Y
2
3
Z
1
Crash Costs
A -$250
B-$100
C-$200
DX-$350
Y-$125
Z-$325
5
5
3
3
B
C
D
A
3
5
2
4
3
2
Admin Cost =
$500
ABCD 16
AXYZ 15
Cost
Save
Net
Cumul
X
2
Y
2
3
Z
1
Crash Costs
A -$250
B-$100
C-$200
DX-$350
Y-$125
Z-$325
5
5
3
3
B
C
D
A
3
5
2
4
3
2
Admin Cost =
$500
ABCD 16
AXYZ 15
Cost
Save
Net
Cumul
X
2
Y
2
3
Z
1
Crash Costs
A -$250
B-$100
C-$200
DX-$350
Y-$125
Choose B because it
is the cheapest of the
three alternatives (A,
B, or C)
Z-$325
4
5
3
3
B
C
D
A
3
5
2
4
3
2
X
2
Admin Cost =
$500
B
ABCD 16
AXYZ 15
15
15
Cost
Save
Net
Cumul
100
500
400
400
Y
2
3
Z
1
Crash Costs
A -$250
B-$100
C-$200
DX-$350
Y-$125
Z-$325
4
5
3
3
B
C
D
A
3
5
2
4
3
2
X
2
Admin Cost =
$500
ABCD 16
AXYZ 15
Cost
Save
Net
Cumul
Y
22
3
Z
11
Crash Costs
A -$250
B-$100
C-$200
DX-$350
Y-$125
15
15
100
500
400
400
Z-$325
What activity(ies) do we
crash next?
4
5
3
3
B
C
D
A
3
5
2
4
3
2
X
2
Admin Cost
= $500
B
Y
2
3
Z
1
Crash Costs
A -$250
B-$100
C-$200
DX-$350
Y-$125
ABCD 16
AXYZ 15
15
15
Cost
Save
Net
Cumul
100
500
400
400
Z-$325
4
5
3
3
B
C
D
A
3
5
2
4
3
2
X
2
Admin Cost
= $500
BB
ABCD 16
AXYZ 15
15
15
Cost
Save
Net
Cumul
100
500
400
400
Y
2
3
Z
1
Crash Costs
A -$250
B-$100
C-$200
DX-$350
Y-$125
Z-$325
4
5
3
3
B
C
D
A
3
5
2
4
3
2
Admin Cost
= $500
X
2
B
Y
2
3
Z
1
Crash Costs
A -$250
B-$100
C-$200
DX-$350
Y-$125
ABCD 16
AXYZ 15
15
15
Cost
Save
Net
Cumul
100
500
400
400
Z-$325
4
5
3
3
B
C
D
A
3
5
2
4
3
2
Admin Cost
= $500
X
2
B
Y
2
3
Z
1
Crash Costs
A -$250
B-$100
C-$200
DX-$350
Y-$125
ABCD 16
AXYZ 15
15
15
Cost
Save
Net
Cumul
100
500
400
400
Z-$325
4
5
3
3
B
C
D
A
3
5
2
4
3
2
Admin Cost =
$500
X
2
B
Y
2
3
Z
11
Crash Costs
A -$250
B-$100
C-$200
DX-$350
Y-$125
ABCD 16
AXYZ 15
15
15
Cost
Save
Net
Cumul
100
500
400
400
Z-$325
4
5
3
3
B
C
D
A
3
5
2
4
3
2
X
2
Admin Cost =
$500
B
ABCD 16
AXYZ 15
15
15
Cost
Save
Net
Cumul
100
500
400
400
Y
2
3
Z
1
Crash Costs
A -$250
B-$100
C-$200
DX-$350
Y-$125
Z-$325
3
5
3
3
B
C
D
A
3
5
2
3
3
2
Admin Cost
= $500
X
Y
2
B
2
BY
3
Z
1
Crash Costs
A -$250
B-$100
C-$200
DX-$350
Y-$125
ABCD 16
AXYZ 15
15
15
14
14
Cost
Save
Net
Cumul
100
500
400
400
225
500
275
675
Z-$325
3
5
3
3
B
C
D
A
3
5
2
3
3
2
X
Y
2
2
Admin Cost =
$500
B
ABCD 16
AXYZ 15
15
15
14
14
Cost
Save
Net
Cumul
100
500
400
400
225
500
275
675
BY
3
Z
1
Crash Costs
A -$250
B-$100
C-$200
DX-$350
Y-$125
Z-$325
What activity(ies) do we
crash next?
3
5
3
3
B
C
D
A
3
5
2
3
3
2
Admin Cost =
$500
X
Y
2
B
2
BY
3
Z
1
Crash Costs
A -$250
B-$100
C-$200
DX-$350
Y-$125
ABCD 16
AXYZ 15
15
15
14
14
Cost
Save
Net
Cumul
100
500
400
400
225
500
275
675
Z-$325
3
5
3
3
B
C
D
A
3
5
2
3
3
2
Admin Cost =
$500
X
Y
2
B
2
BY
3
Z
1
Crash Costs
A -$250
B-$100
C-$200
DX-$350
Y-$125
ABCD 16
AXYZ 15
15
15
14
14
Cost
Save
Net
Cumul
100
500
400
400
225
500
275
675
Z-$325
3
5
3
3
B
C
D
A
3
5
2
3
3
2
Admin Cost =
$500
X
Y
2
B
22
BY
3
Z
11
Crash Costs
A -$250
B-$100
C-$200
DX-$350
Y-$125
ABCD 16
AXYZ 15
15
15
14
14
Cost
Save
Net
Cumul
100
500
400
400
225
500
275
675
Z-$325
3
5
3
2
B
C
D
A
3
5
2
3
3
2
Admin Cost =
$500
X
Y
2
B
2
BY
A
3
Z
1
Crash Costs
A -$250
B-$100
C-$200
DX-$350
Y-$125
ABCD 16
AXYZ 15
15
15
14
14
13
13
Cost
Save
Net
Cumul
100
500
400
400
225
500
275
675
250
500
250
925
Z-$325
3
5
3
2
B
C
D
A
3
5
2
3
3
2
Admin Cost =
$500
X
Y
2
B
2
BY
A
3
Z
1
Crash Costs
A -$250
B-$100
C-$200
DX-$350
Y-$125
ABCD 16
AXYZ 15
15
15
14
14
13
13
Cost
Save
Net
Cumul
100
500
400
400
225
500
275
675
250
500
250
925
Z-$325
3
4
3
2
B
C
D
A
3
5
2
2
3
2
Admin Cost =
$500
X
Y
2
B
3
Z
2
BY
A
1
CY
Crash Costs
A -$250
B-$100
C-$200
DX-$350
Y-$125
ABCD 16
AXYZ 15
15
15
14
14
13
13
12
12
Cost
Save
Net
Cumul
100
500
400
400
225
500
275
675
250
500
250
925
325
500
175
1100
Z-$325
3
4
3
2
B
C
D
A
3
5
2
2
3
2
Admin Cost
= $500
X
Y
2
B
3
Z
2
BY
A
1
CY
Crash Costs
A -$250
B-$100
C-$200
DX-$350
Y-$125
ABCD 16
AXYZ 15
15
15
14
14
13
13
12
12
Cost
Save
Net
Cumul
100
500
400
400
225
500
275
675
250
500
250
925
325
500
175
1100
Z-$325
3
4
3
2
B
C
D
A
3
5
2
2
3
3
X
Y
Z
2
2
1
2
Admin Cost
= $500
B
BY
A
CY
CZ
ABCD 16
AXYZ 15
15
15
14
14
13
13
12
12
11
11
Cost
Save
Net
Cumul
100
500
400
400
225
500
275
675
250
500
250
925
325
500
175
1100
525
500
-25
1075
We
lose
$$, so
do not
crash
CZ
3
4
3
2
B
C
D
A
3
5
2
2
3
3
X
Y
Z
2
2
1
2
B
BY
A
CY
ABCD 16
AXYZ 15
15
15
14
14
13
13
12
12
Cost
Save
Net
Cumul
100
500
400
400
225
500
275
675
250
500
250
925
325
500
175
1100
Conclusions
 The most economical duration of the project
is 12 weeks.
 Achieved by crashing B twice, A once, C
once and Y twice.
 Spent $900 in increased direct costs.
 Avoided $2000 in administrative costs.
 Net savings $1100.
Interpolation
Crash cost =
$1,500
Crash Time
Total Cost = $9,000
Normal Time
Total Cost = $3,000
Slope =
(CC – NC)/(NT-CT)
CC=Crash Cost
NC=Normal Cost
NT=Normal Time
CT =Crash Time
CT=3
Time
NT=7
Example 2 Network
A
F
C
E
B
D
H
G
Activity
Normal Time
Crash Time
Normal Cost
Crash Cost
A
2
1
22000
22750
B
3
1
30000
34000
C
2
1
26000
27000
D
4
3
48000
49000
E
4
2
56000
58000
F
3
2
30000
30500
G
5
2
80000
84500
H
2
1
16000
19000
Critical Path
 Four paths in the network:




Path 1: A – C – F – H: 9 weeks
Path 2: A – C – E – G – H: 15 weeks
Path 3: A – D – G – H: 13 weeks
Path 4: B – D – G – H: 14 weeks
 Path 2 is critical
Activity
Normal Time
Crash Time
Normal Cost
Crash Cost
A
2
1
22000
22750
B
3
1
30000
34000
C
2
1
26000
27000
D
4
3
48000
49000
E
4
2
56000
58000
F
3
2
30000
30500
G
5
2
80000
84500
H
2
1
16000
19000
Crashing Example 2
 Recall the project duration was 15 weeks
 Suppose that project deadline is now 13
weeks
 How should the project be crashed?