Ihab Saad – Solved example on Network Compression
AI: Summary ©
The speakers discuss the network compression process and how it can be compressible and effective by removing certain critical paths. They emphasize the importance of considering all critical paths and reducing the duration of activity until it reaches its maximum cost. The total cost curve is reduced by the duration of activity, and the total cost curve is a function of the number of critical paths. The speakers also explain how to calculate the total cost of the network and avoid unnecessary costs.
AI: Summary ©
Music.
Today we're going to talk about a network compression example. So
we're going to solve this problem and see what are the sequential
steps to solving a network compression problem. What we have
in front of us here is a table that shows activities, their
predecessors, the durations, normal and crash, and the cost
also normal and crash. So it's required to calculate the normal
lease cost and crash durations for the following project. Calculate
the cost associated with each duration. Indirect costs are $120
per day. So basically, for each day that you save from the
duration, you're going to save one $20
so the first step is looking at these activities. First of all,
we're going to try to draw this as a network, or as a Gantt chart to
determine, first we're going to draw it as a as a network to
determine the critical path. We're going to look at where are the
critical activities, which activities are compressible, and
so on and so forth.
The total direct cost for this project can be obtained by adding
the normal cost for all activities, which is equal to
4950,
so as you can see here, when you add these normal costs,
that's the sum of the direct cost of the project before compression.
The total indirect cost can be obtained by solving the network
and multiplying the total duration by the indirect cost per day. So
the total cost of the project is total direct plus total indirect.
The total direct was measured here by adding the normal cost before
compression, the total indirect is going to be equal to the duration
of the project that we calculate from the network times $120 per
day, which is the indirect cost per day when we add these two
numbers, the 4950, and the duration times 120 that would give
us the total cost of the project before compression. What we're
going to be doing gradually after each cycle of compression. We're
going to compare this cost to the initial total cost before
compression, and we're going to determine whether we are on the
right side of the curve, which means, as we compress the project
because total cost is going down, or on the left side of the curve,
which means, as we keep compressing the project, the total
cost goes up.
So here's the network based on the information that was given in the
table. We drew the network. We have the durations. So what we're
going to do next is basically calculate this network.
And here are the dates based on the network calculation, which
shows that the network the project is going to be expected to finish
on day 25
and has one critical path, path which is A, C, E, H. That's our
critical path for critical activities, the other activities
we notice here, for example, on activity B, we have only one day
of total float, so very close to being critical. Activity f, also
one day. G1, day, and activity d2, days of total float. So this is a
very tight network, which means after compression by one day
depending on the activity that we're going to compress, we might
create some new critical paths. So we should expect more critical
paths to appear as we compress this step.
The total indirect cost in this case is going to be 25 days times
one $20 which is 3000 so the total cost before compression is going
to be 7950,
now we're going to start looking based on the table at which
activities are compressible, effective. In this case, by the
way, as you notice, all the activities are finished to start,
which means all of them are effective.
Therefore we're going to be looking at which our activities
are critical, Ach, which are compressible, we're going to look
at the table which are effective. All of them are effective. And
then we're going to look at the critical compressible, effective,
least cost activity, which is going to cost the least to
compress. This is going to be our prime candidate for compression.
So here's our critical path,
looking at the other paths, ABH, 24 which means only one day of
total float. ACF, H, 23 two days of total float, one day, two days,
etc.
Now looking at the cost loops for the different activities. From the
initial table, we look at crash during crash cost minus normal
cost divided by normal duration minus crash duration, which is
delta c divided by delta t. So in this case, it's 100 divided by
one. It.
It is 150 divided by two. It is 120 divided by three. It is
200 divided by four, 100 divided by two, zero activity f is
incompressible, and then 150 and 120 looking at the critical
activities highlighted in red, we notice that the cheapest critical
activity, critical, compressible effective, is going to be activity
C,
by how many days can activity c be compressed? It can be compressed
by three days, but we know that we cannot compress three in one step
by three days, because, again, other paths have only one day of
total float, or maybe maximum of two. So if we compress C by three
days, we definitely are going to create a new critical path without
noticing it. So we have to keep an eye on the creation of new
critical paths.
So cycle one is going to be to compress activity c by one day, as
the total float on these only one day. Once C is compressed, B
becomes critical. The cost of compression, in this case, based
on the table, is going to be $40
so if I tell you right now that to compress the project by one day,
by compressing activity C, you're going to incur an extra cost of
$40 for the compression, which is the cost slope of C and you're
going to save $120
so the net is going to be savings of $80 that's definitely a good
situation. So right now, we are still moving on the right side of
the curve as we compress the total cost goes down. So now 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 on more than one critical path, if you
have more than one critical path, you have either to compare. You
have to compare both paths by exactly the same amount, and that
can be achieved either by compressing one activity on each
path or finding a common activity on both paths that's cheaper to
compress than the sum of the two activities on the different paths.
So in this case, for after the compression cycle. Number one, we
have savings of $80
and that shows now the numbers based on the compression cycle of
activity. One notice here that the duration of C has been reduced to
seven days, which resulted in the new dates in green. And now on the
backward pass, we notice that activity B has become critical. So
now we have
two critical paths, A, B, E, H and A, C, E, H, A, C, E, H,
activity D still has one day. Activity f has one day and
activity G, still as one day
now the options to compress this network.
We can look at activity B and C together. We can look at activity
e, we can look at activity h, or we can look at activity A,
whichever is going to be cheaper to compress.
So now let's look at the numbers again. B by itself, $75
C, $40 so if you want to compress B and C together, then it's going
to cost 115,
e costs 50, which is cheaper, H costs 120 which is more expensive,
a costs 100 which is cheaper than B and C combined. So probably the
cheapest one to compress right now would be activity e, which can be
compressed by up to two days. But again, going back here, we notice
that the total float in this case is only one day on the other
paths. So we cannot compress e in one step by two days.
Therefore the next step is to compress e by one day. This
results in converting all the activities on the network into
critical activities, a very interesting situation, but before
we forget that resulted in an extra cost of $50
and resulted in savings of $120
therefore the net is going to be savings of $70
so now we have reduced The duration of activity, E from six
to five, and you notice here the purple numbers now all the
activities have zero total float, which means, right now we have how
many critical paths. We have four separate critical paths, A, B, E,
H, A, C, E.
Maximum, we have to keep compressing until we exhaust all
the compressibility on at least one path. Once one critical path
becomes incompressible, then the whole network becomes
incompressible, because
if you try to shorten the other paths, that critical path is going
to stay longer, and that defeats the purpose of compression. So
once one critical path at least, becomes exhausted with its
compression, then the problem is over. We cannot compress any
further if we need to compress the network until it's not
economically wise to compress it any further, we stop as soon as
the cost of compression cannot be offset by the reduction to the
indirect cost. So the increase in the cost slopes of the activities
being compressed cannot be offset by the reduction or the savings
from the indirect cost. If we need to compress the network by a
certain number of days, then we need to stop as soon as we reach
that time duration equivalent to that number of days. For example,
if my project is four days behind schedule, why should I compress
more than four days? It's not going to help me, because I'm
probably going to pay more to compress beyond four days. So I
should stop at the four days and avoid paying the liquidated
damages.
And this is basically the the idea that we're talking about, that
total cost curve. This is going to be the right side, and this is
going to be the left side right now, we have already reached this
point, which is the lowest
total cost, which is the optimum duration. So that's after three
days of compression. Compressing any further is going to result in
an increase to the cost. If you want to try, you can try on your
own to keep compressing this network to the maximum. But this
is basically where we're going to stop assuming that the question
was to compress the network until it was not economically feasible
to compress any further, which is at this point and we are already
there.
So basically this is our solved example on network compression.
Remember, again, a few things to remember when solving this
problem. First of all, you have to draw the network correctly. You
have to calculate the normal durations in the network. To get
the expected completion date, you calculate the total indirect cost
by multiplying that date times the cost per day, the indirect cost
per day, the total normal cost, or the total direct cost, can be
calculated by adding all the normal costs. Adding the total
indirect plus the total direct gives you the total cost of the
project before compression. Always keep an eye on the near critical
activities which have very low amount of total float, because
once you start compressing, you may cause these activities to
become critical, and therefore creating additional critical
paths, which is going to require parallel compression of these
paths. Remember that If you have two critical paths, and you
compress one activity on each path,
that results in only one day of shortening of the project, because
these paths are in parallel. So it's not you compressed one day on
activity B, for example, and one day on activity C, but this
resulted, since there are apparent paths, resulted in only one day of
compression to the whole project. If you have three critical paths,
and you compress one activity on each again, you compress three
activities by one day each. But since they were running in
parallel, that's that resulted in only one day of compression for
the whole project. If you have a total float, the minimum total
float on the network, let's say, is five days. Then in this case,
if an activity can be compressed by three days, you can do that in
one step to save yourself some time, because you know that by
compressing this activity by three days, you are still gonna have a
minimum of two days of total float on the other paths, so you're not
gonna create any new critical paths. One last thing, and this is
very tricky part here, you may have some activities that are
ineffective on one path,
yet they might be effective on another path. So for example, if I
have a network, the initial critical path had a start to start
relationship and a finish to finish relationship.
If you remember, per the definition of the effective
activities, the start to start would make the first activity
ineffective. The finish to finish would make the second activity
ineffective. So on that particular path, these activities are
ineffective. But let's say we have created a new critical path and
that contain these same activities. Let's say the first
and the last activities in a finish to start relationship with
the other activities. Then the first and the last activities are
still ineffective on the first path, because the start to start
and finish to finish relationship, but they are effective activities
on the second path.
Because they are finished to start activities so they can be used on
one path and not on another. Remember that? Because, again,
that might be a question on a test.
Good luck, and
I'll see you on the test. You.