Schedule Compression and Time-Cost Trade-Off - Construction Project Scheduling and Control - page 223

3. We accelerate path 4 by 2 days (we may have to do so in two steps—one day at
a time—or we may be able to do so in one step; this point is explained later).
Now, the duration of the project is 25 days.
4. If the shortened activity falls on more than one path, all of the paths with the
activity will also be shortened.
5. Assuming that the shortened activity is not on path 5, we now have two paths
with 25 days' duration: paths 4 and path 5. To accelerate the project further, we
must shorten both paths. If the two paths share one or more activities, cutting
the duration of a shared activity will shorten both paths. Otherwise, we must
cut the duration of an activity on each path. We cut 2 days from each path.
Now, the duration of the project is 23 days.
6. Three paths are now tied as the longest: paths 3, 4, and 5. We repeat step 5 for
these paths in order to cut 1 day from each path and subsequently cut 1 day
from the duration of the project. The duration is now 22 days.
7. Note again that when we cut the duration of an activity, several paths may be
affected. For example, if we cut an activity that is on both paths 4 and 2, path 2
is now cut unintentionally from 17 days to 16 days.
8. Path 6 has now “joined the club” and become critical. We repeat step 6.
9. Clearly, the further we accelerate the project, the more paths must be short-
ened and the more activities must be involved. Consequently, we may need to
compress six or seven activities to cut only 1 day from the project duration.
This fact is why the direct cost of accelerating is not linear (although it may be
linear for short intervals).

It is interesting to note that an increase in the duration of the critical path by x
days will result in an increase in the duration of the project by x days as well. However,
the reverse is not necessarily true. A decrease in the duration of the critical path by x
days will result in a decrease in the duration of the project by y days, where

>

0 y ≤ x

So, if we have a project with the critical (longest) path = 150 days and we cut this
path to 144 days ( x = 6 days) but we find that the project duration = 146 days (then
y = 4 days), the explanation for this is that there is another path with duration = 146
days, and it has become the new critical path. We need to:

a. Keep what we did to the original critical path but also cut the new critical path
by two days, or
b. Reverse what we did and cut the original critical path by only 4 days and then
deal with the new situation (two critical paths) on the basis of the steps dis-
cussed earlier. We may find an activity that both paths share, and it may be
easier and less costly to crash this activity.