Ihab Saad – Example on Schedule Compression
AI: Summary ©
The speakers discuss the maximum compressible time for a network, using a table and calculations based on various variables. They explain the process of compressing a network and identifying critical and effective activities, emphasizing the importance of starting with the lowest cost activity and compressing all critical paths by the same amount to achieve maximum savings. The speakers also discuss reducing duration by two days while still saving some money, compressing all paths by the same amount, and being still critical. The projects are designed to compress all paths by the same amount, and the projects are still critical.
AI: Summary ©
Music. Hello again. Now we start talking about an example on
network compression based on what we have learned so far about which
activities to compress and how to achieve time reduction or time
acceleration. So here we have an example calculate the normal least
cost and crash durations for the following project, and calculate
the cost associated with each duration, whereas indirect costs
are $120 per day. So the costs given here in this table are
basically direct cost for each activity. It shows the activity,
its immediate predecessor, normal duration and crash duration, which
which shows the amount of compressibility that the activity
has, in addition to the cost, normal cost and crash cost, how
much would it cost to compress this activity by that amount of
time? So for example, activity A can be compressed by one day,
which is the difference between the normal and the crash duration
at an extra cost of $100 which is the difference between the crash
cost and the normal cost. And we have that for all the activities.
Notice here, for example, that activity f has a normal duration
of four days and a crash duration of also four days, which makes
delta T for activity f equal to zero. Therefore activity f is
incompressible. If activity f ends up being a critical activity, it
would be excluded from our efforts to compress that activity, because
it is, it is incompressible in the first place. Same thing applies
here to the delta C. It doesn't have any delta cost because,
again, it cannot be compressed. So our first step is looking at this
table, first of all, building the network, and then looking at the
delta t and delta c and the cost slope for the different
activities.
The total direct cost for this project can be obtained by adding
the normal cost for all activities, which is equal to
$4,950
which is basically the sum of these costs here normal adding
these costs, that's going to add up to 4950,
the total indirect costs can be obtained only by solving the
network and multiplying the total duration by the indirect cost per
day, which again is $120
per day. Therefore we have to perform the forward pass, backward
pass, and determine the expected early finish of that network.
Multiply that time, the number of days by 120 and that will give us
the total indirect cost for this project before compression.
So here's our network that we have drawn based on the information
with the IPAS and so on, and it shows basically
several paths. Here's one path, here's a second path, here's a
third path, and here's the fourth path. So we have four paths to
this network, A, B, E, H, A, C, E, H, A, C, F, G, and ADG. These are
the four different paths to this network.
So the next step is to perform our calculations.
And here's a very simple display of these calculations. It shows
the early start for activity is zero, duration five days, so the
early finish is zero, is five and so on. So it's a very simple
calculation of the network time which shows that this project is
going to end on day 25
therefore this this network expected to finish on day 25 has
one critical path which is going to be A, C, E, H, as it appears
from The identical numbers on the early and late sites,
and the total indirect cost is 25 times 120
25 days times 125 $20 per day. With a total of $3,000
we had already calculated the direct cost to be 4950 so the
total cost for this project before compression is 79 $50
and here it shows the critical path that we talked about.
So the next step is to draw the table to show the delta t, the
delta c, the cost slope for the different activities.
And here's the duration for each path, A, B, E, h. Now the duration
for each path is assuming that the other paths do not exist. So if,
for example, C did not exist, activity c did not exist, then the
start date for activity e would have been 12. The start date for
activity h would have been 18, and the completion date would have
been 24 Same thing here, if the two top paths did not exist, or
three top paths did not exist, and we had on the ADG, then activity G
would start on day 16 and end on day 23 and the project would end
on day 23 that's how we.
Determine the length of each path.
Here's the table. Based on the calculations of the crash and
normal durations and crash and normal costs, we obtain the cost
slope, which is the difference in cost divided by the difference in
time. So for the first activity, 100 divided by one for B, 150
divided by two and so on. And we find, again, we remind that
activity f is incompressible. Now the ones in red are the critical
activities. So these are our primary candidates for
compression. They have to be critical compressible effective
and among the critical, compressive and effective
activities, we will select the one with the lowest cost low going
back to this network here, we find that we do not have any start to
start or any finish to finish relationships. It's a very simple
network with all finish to start relationships. Therefore all the
critical activities are effective in compressing the project
duration, none of the critical activities are not none is
incompressible. Therefore, again, all the critical activities are
candidates. So the main criteria in determining which activity to
start with is going to be the cost slope.
So looking at this table here, we look at the critical activities
and the cost slopes. Activity A has 100 day, $100 per day.
Activity C, $40 per day. E, $50 per day and age, $120
per day. Therefore, looking at this, we found out, find out that
activity c is the least expensive to compress, so this is going to
be our primary candidate for compression.
And that is compression cycle number one, we're going to conduct
these compressions in different cycles, as I mentioned several
times before, the time compression or network compression problem is
not a hard problem at all. The concept is very simple. It's just
a repetitive calculation of the forward and the backward paths
after each cycle of compression, compressing each activity by one
day at a time, and then recalculating the dates for the
network. So in cycle one compress activity c by one day. Now we
notice that looking at the network again,
look at the total float on activity B,
13 and 12. So the total float for this activity is only one day,
which means if I reduce the duration of this activity by one
day, if I end this activity on day 12,
going in the backward pass, we're going to find out that the late
finish for activity B is going to become 12, which means activity B
is going to lose its first day, or its only day of total float, which
is going to make it a critical activity. So we're going to create
a new critical path going through A, B, E, H.
So this is the first cycle activity c is compressed, and B
will become critical with the cost of compression is going to be $40
therefore, to proceed with the second cycle, we have to check the
cost slopes of activities to be compressed. Remember that in order
to achieve an effective compression of more than one
critical path, we must compress all critical paths by the same
amount. So if we have multiple critical paths that need to be
compressed, it's
not enough to compress only one of them, because if you compress one
and leave the others incompressed or uncompressed, that would make
them longer, and they would remain critical, whereas the one that has
been compressed is not critical anymore. So in order to achieve
that compression, we have to compress all the critical paths by
exactly the same amount, and this can be achieved either by
compressing one activity on each one of these paths by the same
amount, or looking for a common activity that is shared by all of
these critical paths, and compressing this activity, which
is going to result in a complete In a compression for all the
existing critical paths,
the net cost of compression here is savings of $80 why? Because we
have spent an extra 80 $40 to compress activity C, but at the
same time, we reduced the total project duration by one day, which
saved us $120 in indirect costs.
And here's the calculation after the first cycle of compression.
Notice that each cycle is going to be done in a different color, so
that we can identify which activities have been affected and
which dates have changed.
So here, when we compress activity C from eight to seven, perform the
calculations in the backward pass here I have activity B with zero
total float.
Therefore now we have a new critical path in addition to the
old one. So AC.
Eh is still critical and now a, b, eh has become another critical
path.
Now let's look at the other paths. We notice here that activity G has
a total float of only one day, so does activity f and also so does
activity D. Therefore,
any additional compression to this path more likely is going to
result in these activities becoming critical, except if we
compress activity A, because compressing activity A is going to
affect all the paths so the non critical activities are going to
keep their total float. So now let's look again at the cost of
compression. In order to compress this path, these two paths at the
same time by the same amount, we can either compress activity A by
itself because it's joined between the two, or we can compress
activity e by itself, or we can compress activity h by itself,
therefore A or B and C together, or H by itself, or or E by itself,
or H by itself. Let's look at the cost for this compression.
So we find that activity c is still the least expensive,
followed by E, followed by a, followed by h. Now if I want to
compress C, I have also to compress B at the same time.
Therefore the cost of compressing one extra day is going to be 40
plus 75 that would be 115 to achieve that compression. Or
another option is to compress activity e by itself, or to
compress activity A by itself, or activity h by itself. So
based on the information from cycle one, the least expensive
activity to compress is activity e, again, from here, from this
table, this activity, this expensive activity to compress, is
activity e, with a cost of $50 this result in converting all the
activities on the network into critical activities, because,
again, all their meaning activities had only one day of
total float. Once we compressed activity e, all of these non
critical activities will lose that total float, thus becoming
critical. And the net cost of this compression is a saving of $70
because again, we spent an extra $50 we saved 120 so the net is
savings of $70
and now we can see the new numbers, and it shows clearly that
all the other activities have become critical.
So here's the original critical path,
here's the second critical path,
and here's the third critical path, each one shown in a
different color that shows when did these activities become
critical? That's very important, again, as we mentioned before, to
use different colors, so that we do not lose track of these changes
and which activity was changed at which cycle and so on and so
forth.
Now the project ends on day 23 so we have managed to reduce the
duration by two days while still saving some money, which is good.
Our next step,
again, all the activities now are critical,
so we need to look at which activities need to be compressed.
In order to compress this network, again, the options are, I have to
compress
all the paths by the same amount. So it's either compressing, for
example, H with G or E with G, or B and C together with D and G,
because D is on the separate path from G. So remember that we have
three paths. So it's either B, C, G and D together, or
E, G and D together, or H, G and D together, because f is
incompressible. Or maybe looking at activity A, which is the common
activity in all paths, if we compress activity A, that might
end up being less expensive.
So looking at the cost A by itself, is 100
B, C, with any other activity, that activity that already exceeds
the 100. So that's not our choice.
Therefore, we find out that the cheapest activity to be compressed
is going to be activity A.
Since we now have four critical paths, compressing the network
requires compressing all the paths by the same amount. This can be
achieved by compressing at least one activity on each path. And
again, here it shows the analysis of the costs together. So by
compressing activity a one day, the net cost would.
Be a savings of $20 because the extra cost is $100 and the savings
of $120
now look at these numbers, 165, 202 70 and so on. That is more
than the 120 that I'm going to be ending up saving. Therefore, if I
compress, after compressing activity A, if I try to compress
any other activities, that's going to be on the more expensive side,
because that's going to end up costing me more money. So I'm not
saving any money by reducing the duration. I'm actually spending
more money.
And here, once we have compressed activity A, all the activities are
still critical,
and now we end the project on day 22
so here it shows the different compression cycles that We have
done and which activities became critical when
now the question is, where to stop? Are we going to keep
compressing this network? Or should we stop here? And the
answer to this question depends on what's the objective we need to
achieve if we need to compress the network to the maximum. It is
still compressible, because again, we can compress B and C and D and
G, for example, at the same time, and that's going to result in
another day of compression.
So we can do that, but that's going to be, as we have already
noticed, it's going to be on the more expensive side, because we
are not saving any money by doing that, we are spending more money
if we need to compress the network until it's not economically wise,
so compress it any further, then we have to stop here, because
against compressing further is going to result in more expenses,
and that's not economically meaning. If we need to compress
the network by a certain number of days. Let's say we all we wanted
to do is compress the network by three days. We have already
achieved that. So here we can stop. So it depends on how the
question is being asked. So the question can be asked as compress
the network to the maximum regardless of the cost. In this
case, we would keep compressing or compressing it. Compress the
network until it's not economically feasible or
meaningful to compress it any further, which means we're going
to stop as soon as the savings are not achieved, which means the cost
of compression exceeds the savings resulted from reduction in
indirect costs or compressed by a certain number of days. As soon as
we reach that number of days we are going to start, we're going to
stop our conversion.
And this is basically a graphical representation of that concept.
Here we have the direct cost curve, which shows that the
shorter the duration, the higher the cost of direct costs. Here's
the indirect cost, which shows the opposite, the shorter the
duration, the lower the indirect cost. And then here's the sum of
the two curves together. It's adding the numbers from the two
curves together. It shows that for a certain portion,
as you compress,
the total cost goes down, which is what we have been doing so far,
and then you reach a certain point, and the cost starts rising
up again.
So this is where we had stopped originally, right now in this
problem. And this is going to be called the right side of the total
cost curve. And this is the left side of the total cost curve. On
the right side of the total cost curve, the indirect cost is more
effective. So with the reduction, the total curve trend follows the
same trend as the indirect cost, which is reduction as you
compress. On the left hand side, the curve, the total cost follows
the curve or the trend of the direct cost. As we compress, the
total cost goes up so again, until we reach a certain point where we
cannot compress anymore. So this is basically in a nutshell how we
can achieve network compression or time acceleration.
I hope this helps you understand the concept and you can practice
on some additional problems. Thank you and see you later. You.