Download Project acceleration

Question (Project acceleration): A project has been subdivided into five tasks. The structure of the
project, the activity durations, limits of acceleration of the activities, and acceleration costs are shown in
the table below.
Activity
A
B
C
D
E
Immediate
predecessor


A, B
A, C
B, C
Normal time
(days)
6
7
3
5
6
Crash time
(days)
3
5
2
4
5
Unit cost of
acceleration
10
20
40
30
60
(a) Draw the project network and, given normal times, determine the critical path. What is the project
duration?
(b) Accelerate the project as much as possible, one day at a time to its shortest possible duration.
(c) The firm has signed a contract that specifies that the project must be finished within 13 days. Should
the project not be finished at that time, it calls for a penalty of $100. What is your advice to the
manager?
Solution:
(a) Given normal times, the project network is as shown below:
In this small network, we can get away with enumerating that paths rather than go through the usual
forard and backward recursions. The paths are A  D, A  C  D, A  C  E, B  C  D, B  C  E, and B
 E. The critical (i.e., longest) path in the network is B  C  E, and the project duratin is 16 days.
(b) Accelerating the project by one day could be accomplished by either accelerating B, or C, or E by one
day. The respective costs are $20, $40, and $60, so that we decide to accelerate activity B by one day.
Given the new activity durations, the project network and the shortest paths are shown below.
There are now two critical paths whose lenghts are 15 each. Further acceleration would either have to
accelerate activities A and B simultanously at a cost of $10 + $20 = $30, or activity C at a cost of $40, or
activity E at a cost of $60. The cheapest option is to accelerate activities A and B simultaneously. The
resulting network and its shortest paths are shown below.
Note that the structure of the critical paths has not changed in the last step. The lengths of the critical
paths are now 14. Any further acceleration will again have to either simultaneously accelerate activities A
and B (which is no longer possible since the duration of activity B is at its crash time), or activity C at a
cost of $ 34, or activity E at a cost of $60. The least expensive option is to accelerate activity C. The
resulting network is shown below.
The project duration is now 13 and the structure of the critical paths has not changed. Hence any further
acceleration will have to either simultaneously accelerate A and B (no longer possible as B is at crash
time), accelerate C (which is also not possible as activity C is at crash time), or accelerate acitivity E. The
last is the only option left at this time, so we decide to accelerate acitivity E by one day. The resulting
network is shown below.
The project duration at this time is 12. The structure of the critical paths has changed, and any further
project acceleration must now either simultaneously accelerate activities A and B (not possible as B is at
crash time), accelerate activity C (not possible as C is at crash time), or simultaneously accelerate
activities D and E (which is not possible as activity E is at crash time). As none of the potential options
any longer exists, we have reached the end of the procedure and the present duration of 12 days is the
shortest possible project duration. The individual steps and their costs are summarized in the table below.
Project duration
Activity durations
Critical path(s)
16
15
14
6, 7, 3, 5, 6
6, 6, 3, 5, 6
5, 5, 3, 5, 6
BCE
ACE & BCE
ACE & BCE
Costs of
acceleration
0
20
50
13
12
5, 5, 2, 5, 6
5, 5, 2, 5, 5
ACE & BCE
ACD & ACE
BCD & CE
90
150
(c) Since it costs $90 to accelerate the project to the 13 day time limit specified in the contract and the
penalty costs for noncompliance are $100, our advice to the manager is to accelerate the project to a
duration of 13.