# Ihab Saad – Example on Schedule Compression

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.
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Music. Hello again. Now we start talking about an example on

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network compression based on what we have learned so far about which

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activities to compress and how to achieve time reduction or time

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acceleration. So here we have an example calculate the normal least

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cost and crash durations for the following project, and calculate

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the cost associated with each duration, whereas indirect costs

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are \$120 per day. So the costs given here in this table are

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basically direct cost for each activity. It shows the activity,

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its immediate predecessor, normal duration and crash duration, which

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which shows the amount of compressibility that the activity

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has, in addition to the cost, normal cost and crash cost, how

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much would it cost to compress this activity by that amount of

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time? So for example, activity A can be compressed by one day,

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which is the difference between the normal and the crash duration

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at an extra cost of \$100 which is the difference between the crash

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cost and the normal cost. And we have that for all the activities.

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Notice here, for example, that activity f has a normal duration

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of four days and a crash duration of also four days, which makes

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delta T for activity f equal to zero. Therefore activity f is

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incompressible. If activity f ends up being a critical activity, it

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would be excluded from our efforts to compress that activity, because

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it is, it is incompressible in the first place. Same thing applies

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here to the delta C. It doesn't have any delta cost because,

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again, it cannot be compressed. So our first step is looking at this

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table, first of all, building the network, and then looking at the

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delta t and delta c and the cost slope for the different

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activities.

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The total direct cost for this project can be obtained by adding

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the normal cost for all activities, which is equal to

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\$4,950

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which is basically the sum of these costs here normal adding

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these costs, that's going to add up to 4950,

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the total indirect costs can be obtained only by solving the

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network and multiplying the total duration by the indirect cost per

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day, which again is \$120

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per day. Therefore we have to perform the forward pass, backward

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pass, and determine the expected early finish of that network.

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Multiply that time, the number of days by 120 and that will give us

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the total indirect cost for this project before compression.

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So here's our network that we have drawn based on the information

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with the IPAS and so on, and it shows basically

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several paths. Here's one path, here's a second path, here's a

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third path, and here's the fourth path. So we have four paths to

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this network, A, B, E, H, A, C, E, H, A, C, F, G, and ADG. These are

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the four different paths to this network.

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So the next step is to perform our calculations.

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And here's a very simple display of these calculations. It shows

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the early start for activity is zero, duration five days, so the

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early finish is zero, is five and so on. So it's a very simple

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calculation of the network time which shows that this project is

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going to end on day 25

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therefore this this network expected to finish on day 25 has

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one critical path which is going to be A, C, E, H, as it appears

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from The identical numbers on the early and late sites,

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and the total indirect cost is 25 times 120

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25 days times 125 \$20 per day. With a total of \$3,000

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we had already calculated the direct cost to be 4950 so the

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total cost for this project before compression is 79 \$50

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and here it shows the critical path that we talked about.

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So the next step is to draw the table to show the delta t, the

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delta c, the cost slope for the different activities.

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And here's the duration for each path, A, B, E, h. Now the duration

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for each path is assuming that the other paths do not exist. So if,

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for example, C did not exist, activity c did not exist, then the

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start date for activity e would have been 12. The start date for

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activity h would have been 18, and the completion date would have

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been 24 Same thing here, if the two top paths did not exist, or

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three top paths did not exist, and we had on the ADG, then activity G

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would start on day 16 and end on day 23 and the project would end

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on day 23 that's how we.

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Determine the length of each path.

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Here's the table. Based on the calculations of the crash and

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normal durations and crash and normal costs, we obtain the cost

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slope, which is the difference in cost divided by the difference in

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time. So for the first activity, 100 divided by one for B, 150

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divided by two and so on. And we find, again, we remind that

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activity f is incompressible. Now the ones in red are the critical

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activities. So these are our primary candidates for

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compression. They have to be critical compressible effective

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and among the critical, compressive and effective

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activities, we will select the one with the lowest cost low going

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back to this network here, we find that we do not have any start to

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start or any finish to finish relationships. It's a very simple

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network with all finish to start relationships. Therefore all the

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critical activities are effective in compressing the project

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duration, none of the critical activities are not none is

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incompressible. Therefore, again, all the critical activities are

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candidates. So the main criteria in determining which activity to

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So looking at this table here, we look at the critical activities

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and the cost slopes. Activity A has 100 day, \$100 per day.

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Activity C, \$40 per day. E, \$50 per day and age, \$120

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per day. Therefore, looking at this, we found out, find out that

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activity c is the least expensive to compress, so this is going to

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be our primary candidate for compression.

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And that is compression cycle number one, we're going to conduct

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these compressions in different cycles, as I mentioned several

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times before, the time compression or network compression problem is

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not a hard problem at all. The concept is very simple. It's just

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a repetitive calculation of the forward and the backward paths

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after each cycle of compression, compressing each activity by one

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day at a time, and then recalculating the dates for the

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network. So in cycle one compress activity c by one day. Now we

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notice that looking at the network again,

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look at the total float on activity B,

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13 and 12. So the total float for this activity is only one day,

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which means if I reduce the duration of this activity by one

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day, if I end this activity on day 12,

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going in the backward pass, we're going to find out that the late

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finish for activity B is going to become 12, which means activity B

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is going to lose its first day, or its only day of total float, which

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is going to make it a critical activity. So we're going to create

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a new critical path going through A, B, E, H.

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So this is the first cycle activity c is compressed, and B

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will become critical with the cost of compression is going to be \$40

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therefore, to proceed with the second cycle, we have to check the

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cost slopes of activities to be compressed. Remember that in order

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to achieve an effective compression of more than one

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critical path, we must compress all critical paths by the same

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amount. So if we have multiple critical paths that need to be

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compressed, it's

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not enough to compress only one of them, because if you compress one

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and leave the others incompressed or uncompressed, that would make

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them longer, and they would remain critical, whereas the one that has

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been compressed is not critical anymore. So in order to achieve

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that compression, we have to compress all the critical paths by

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exactly the same amount, and this can be achieved either by

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compressing one activity on each one of these paths by the same

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amount, or looking for a common activity that is shared by all of

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these critical paths, and compressing this activity, which

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is going to result in a complete In a compression for all the

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existing critical paths,

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the net cost of compression here is savings of \$80 why? Because we

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have spent an extra 80 \$40 to compress activity C, but at the

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same time, we reduced the total project duration by one day, which

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saved us \$120 in indirect costs.

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And here's the calculation after the first cycle of compression.

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Notice that each cycle is going to be done in a different color, so

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that we can identify which activities have been affected and

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which dates have changed.

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So here, when we compress activity C from eight to seven, perform the

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calculations in the backward pass here I have activity B with zero

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total float.

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Therefore now we have a new critical path in addition to the

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old one. So AC.

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Eh is still critical and now a, b, eh has become another critical

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path.

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Now let's look at the other paths. We notice here that activity G has

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a total float of only one day, so does activity f and also so does

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activity D. Therefore,

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any additional compression to this path more likely is going to

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result in these activities becoming critical, except if we

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compress activity A, because compressing activity A is going to

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affect all the paths so the non critical activities are going to

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keep their total float. So now let's look again at the cost of

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compression. In order to compress this path, these two paths at the

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same time by the same amount, we can either compress activity A by

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itself because it's joined between the two, or we can compress

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activity e by itself, or we can compress activity h by itself,

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therefore A or B and C together, or H by itself, or or E by itself,

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or H by itself. Let's look at the cost for this compression.

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So we find that activity c is still the least expensive,

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followed by E, followed by a, followed by h. Now if I want to

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compress C, I have also to compress B at the same time.

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Therefore the cost of compressing one extra day is going to be 40

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plus 75 that would be 115 to achieve that compression. Or

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another option is to compress activity e by itself, or to

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compress activity A by itself, or activity h by itself. So

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based on the information from cycle one, the least expensive

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activity to compress is activity e, again, from here, from this

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table, this activity, this expensive activity to compress, is

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activity e, with a cost of \$50 this result in converting all the

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activities on the network into critical activities, because,

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again, all their meaning activities had only one day of

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total float. Once we compressed activity e, all of these non

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critical activities will lose that total float, thus becoming

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critical. And the net cost of this compression is a saving of \$70

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because again, we spent an extra \$50 we saved 120 so the net is

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savings of \$70

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and now we can see the new numbers, and it shows clearly that

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all the other activities have become critical.

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So here's the original critical path,

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here's the second critical path,

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and here's the third critical path, each one shown in a

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different color that shows when did these activities become

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critical? That's very important, again, as we mentioned before, to

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use different colors, so that we do not lose track of these changes

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and which activity was changed at which cycle and so on and so

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forth.

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Now the project ends on day 23 so we have managed to reduce the

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duration by two days while still saving some money, which is good.

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Our next step,

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again, all the activities now are critical,

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so we need to look at which activities need to be compressed.

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In order to compress this network, again, the options are, I have to

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compress

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all the paths by the same amount. So it's either compressing, for

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example, H with G or E with G, or B and C together with D and G,

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because D is on the separate path from G. So remember that we have

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three paths. So it's either B, C, G and D together, or

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E, G and D together, or H, G and D together, because f is

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incompressible. Or maybe looking at activity A, which is the common

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activity in all paths, if we compress activity A, that might

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end up being less expensive.

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So looking at the cost A by itself, is 100

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B, C, with any other activity, that activity that already exceeds

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the 100. So that's not our choice.

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Therefore, we find out that the cheapest activity to be compressed

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is going to be activity A.

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Since we now have four critical paths, compressing the network

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requires compressing all the paths by the same amount. This can be

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achieved by compressing at least one activity on each path. And

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again, here it shows the analysis of the costs together. So by

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compressing activity a one day, the net cost would.

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Be a savings of \$20 because the extra cost is \$100 and the savings

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of \$120

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now look at these numbers, 165, 202 70 and so on. That is more

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than the 120 that I'm going to be ending up saving. Therefore, if I

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compress, after compressing activity A, if I try to compress

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any other activities, that's going to be on the more expensive side,

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because that's going to end up costing me more money. So I'm not

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saving any money by reducing the duration. I'm actually spending

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more money.

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And here, once we have compressed activity A, all the activities are

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still critical,

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and now we end the project on day 22

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so here it shows the different compression cycles that We have

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done and which activities became critical when

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now the question is, where to stop? Are we going to keep

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compressing this network? Or should we stop here? And the

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answer to this question depends on what's the objective we need to

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achieve if we need to compress the network to the maximum. It is

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still compressible, because again, we can compress B and C and D and

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G, for example, at the same time, and that's going to result in

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another day of compression.

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So we can do that, but that's going to be, as we have already

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noticed, it's going to be on the more expensive side, because we

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are not saving any money by doing that, we are spending more money

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if we need to compress the network until it's not economically wise,

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so compress it any further, then we have to stop here, because

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against compressing further is going to result in more expenses,

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and that's not economically meaning. If we need to compress

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the network by a certain number of days. Let's say we all we wanted

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to do is compress the network by three days. We have already

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achieved that. So here we can stop. So it depends on how the

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question is being asked. So the question can be asked as compress

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the network to the maximum regardless of the cost. In this

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case, we would keep compressing or compressing it. Compress the

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network until it's not economically feasible or

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meaningful to compress it any further, which means we're going

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to stop as soon as the savings are not achieved, which means the cost

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of compression exceeds the savings resulted from reduction in

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indirect costs or compressed by a certain number of days. As soon as

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we reach that number of days we are going to start, we're going to

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stop our conversion.

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And this is basically a graphical representation of that concept.

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Here we have the direct cost curve, which shows that the

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shorter the duration, the higher the cost of direct costs. Here's

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the indirect cost, which shows the opposite, the shorter the

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duration, the lower the indirect cost. And then here's the sum of

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the two curves together. It's adding the numbers from the two

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curves together. It shows that for a certain portion,

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as you compress,

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the total cost goes down, which is what we have been doing so far,

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and then you reach a certain point, and the cost starts rising

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up again.

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So this is where we had stopped originally, right now in this

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problem. And this is going to be called the right side of the total

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cost curve. And this is the left side of the total cost curve. On

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the right side of the total cost curve, the indirect cost is more

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effective. So with the reduction, the total curve trend follows the

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same trend as the indirect cost, which is reduction as you

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compress. On the left hand side, the curve, the total cost follows

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the curve or the trend of the direct cost. As we compress, the

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total cost goes up so again, until we reach a certain point where we

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cannot compress anymore. So this is basically in a nutshell how we

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can achieve network compression or time acceleration.

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I hope this helps you understand the concept and you can practice

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on some additional problems. Thank you and see you later. You.

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