# Ihab Saad – Solved example on Network Compression

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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.