Transcript Quality and reliability management in projects

‘Crashing’ – reducing task
durations by increased costs
(lecture)
Definition of crashing
• Obtaining reduction in time at an increased cost
(increasing the employed resources).
• Cost-slope:
the cost of reducing duration time by unit time.
• Let’s see the following example:
0
6
0
0
0
4
0
4
4
6
b
4
4
2
8
8
c
7
a
0
0
1
1
2
1
10
0
e
9
9
2
11
11
0
14
f
0
6
6
0
11
d
6
5
11
11
3
14
Procedure for crashing
1. Crash one time unit at a time
2. Only crashing critical path activities has any effect
on TPT
3. Crash the that activity first that is the cheapest to
reduce in time
4. Be aware of multiple critical paths
5. Stop crashing when:
•
the crash-time is reached at every activity,
•
benefits of possible crashing are lower than crashing
costs.
Crashing table
• If the costs to reduce times are known, then a
table can be set up showing the relative costs
for the reduction in time of each activity by a
constant amount.
• Crash-time is the minimum duration of an
activity. It is given by technical factors.
Activity
Duration
Float
Crash
time
Cost-slope
(label)
(day)
(day)
a
4
0
2
100
b
2
0
2
150
c
2
1
1
110
d
5
0
3
e
2
1
1
f
3
0
2
200 Benefit of
160 reducing TPT
by one day:
500
400 €/day
(€/day)
Solution method
1. step: identify the critical activities
2. step: find the critical activity with cheapest crash
cost, and if its cost slope is lower than the daily
benefit from crashing, reduce its duration with
one day. If there is no activity to crash, or it is
too costly, stop crashing and go to step 4.
3. step: reidentify the critical path, and go back to
step two.
4. step: identify the newest critical path, TPT and
the total net benefit of crashing.
Solution
Path durations
normal
step 1
step 2
step 3
step 4
step 5
Path / activity
crached
crashed
–
a
a
d
d, c
none
a-b-c-e-f
13
12
11
11
10
–
a-b-d-f
14
13
12
11
10
–
Cost:
–
100
100
200
310
–
Cumulated net
benefit:
–
300
600
800
890
–
After crashing:
– there are two critical paths
– TPT is 10 days
– total benefit of crashing is €890
Example 2 (for individual work)
b
d
e
2
2
5
0
a
0
7
g
3
3
c
f
3
3
• Identify the critical path and the TPT.
Example 2 (for individual work)
0
3
0
0
5
5
b
0
0
0
3
2
3
7
7
d
5
5
2
0
12
e
7
7
5
0
12
3
12
a
0
0
0
7
3
1
6
3
6
c
4
3
15
g
0
3
0
3
12
3
15
9
f
7
9
3
12
Critcal: a-b-d-e-g
TPT: 15
Using tbe table on the next slide, calculate the
optimal TPT with crashing.
Activity
(label)
Normal
duration (day)
Float
(day)
Crash
time
Costslope
(€/day)
a
3
0
1
500
b
2
0
1
550
c
2
1
1
150
d
5
0
3
900
e
5
0
4
400
f
3
3
2
100
g
3
0
3
200
• What is the new TPT?
10 days
• What is the total profit on
crashing?
€3000
Benefit of
reducing TPT
by one day:
1200 €/day
Reading
• Textbook chapter 8. pp. 61-63.
Thank you for listening