# Ihab Saad – Example on Schedule Compression

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## AI: Transcript ©

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.