Ihab Saad – Solved example on Network Compression

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.
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Music.

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Today we're going to talk about a network compression example. So

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we're going to solve this problem and see what are the sequential

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steps to solving a network compression problem. What we have

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in front of us here is a table that shows activities, their

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predecessors, the durations, normal and crash, and the cost

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also normal and crash. So it's required to calculate the normal

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lease cost and crash durations for the following project. Calculate

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the cost associated with each duration. Indirect costs are \$120

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per day. So basically, for each day that you save from the

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duration, you're going to save one \$20

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so the first step is looking at these activities. First of all,

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we're going to try to draw this as a network, or as a Gantt chart to

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determine, first we're going to draw it as a as a network to

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determine the critical path. We're going to look at where are the

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critical activities, which activities are compressible, and

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so on and so forth.

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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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4950,

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so as you can see here, when you add these normal costs,

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that's the sum of the direct cost of the project before compression.

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The total indirect cost can be obtained by solving the network

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and multiplying the total duration by the indirect cost per day. So

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the total cost of the project is total direct plus total indirect.

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The total direct was measured here by adding the normal cost before

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compression, the total indirect is going to be equal to the duration

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of the project that we calculate from the network times \$120 per

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day, which is the indirect cost per day when we add these two

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numbers, the 4950, and the duration times 120 that would give

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us the total cost of the project before compression. What we're

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going to be doing gradually after each cycle of compression. We're

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going to compare this cost to the initial total cost before

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compression, and we're going to determine whether we are on the

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right side of the curve, which means, as we compress the project

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because total cost is going down, or on the left side of the curve,

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which means, as we keep compressing the project, the total

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cost goes up.

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So here's the network based on the information that was given in the

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table. We drew the network. We have the durations. So what we're

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going to do next is basically calculate this network.

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And here are the dates based on the network calculation, which

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shows that the network the project is going to be expected to finish

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on day 25

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and has one critical path, path which is A, C, E, H. That's our

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critical path for critical activities, the other activities

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we notice here, for example, on activity B, we have only one day

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of total float, so very close to being critical. Activity f, also

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one day. G1, day, and activity d2, days of total float. So this is a

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very tight network, which means after compression by one day

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depending on the activity that we're going to compress, we might

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create some new critical paths. So we should expect more critical

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paths to appear as we compress this step.

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The total indirect cost in this case is going to be 25 days times

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one \$20 which is 3000 so the total cost before compression is going

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to be 7950,

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now we're going to start looking based on the table at which

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activities are compressible, effective. In this case, by the

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way, as you notice, all the activities are finished to start,

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which means all of them are effective.

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Therefore we're going to be looking at which our activities

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are critical, Ach, which are compressible, we're going to look

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at the table which are effective. All of them are effective. And

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then we're going to look at the critical compressible, effective,

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least cost activity, which is going to cost the least to

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compress. This is going to be our prime candidate for compression.

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So here's our critical path,

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looking at the other paths, ABH, 24 which means only one day of

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total float. ACF, H, 23 two days of total float, one day, two days,

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etc.

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Now looking at the cost loops for the different activities. From the

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initial table, we look at crash during crash cost minus normal

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cost divided by normal duration minus crash duration, which is

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delta c divided by delta t. So in this case, it's 100 divided by

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one. It.

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It is 150 divided by two. It is 120 divided by three. It is

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200 divided by four, 100 divided by two, zero activity f is

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incompressible, and then 150 and 120 looking at the critical

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activities highlighted in red, we notice that the cheapest critical

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activity, critical, compressible effective, is going to be activity

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C,

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by how many days can activity c be compressed? It can be compressed

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by three days, but we know that we cannot compress three in one step

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by three days, because, again, other paths have only one day of

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total float, or maybe maximum of two. So if we compress C by three

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days, we definitely are going to create a new critical path without

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noticing it. So we have to keep an eye on the creation of new

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critical paths.

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So cycle one is going to be to compress activity c by one day, as

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the total float on these only one day. Once C is compressed, B

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becomes critical. The cost of compression, in this case, based

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on the table, is going to be \$40

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so if I tell you right now that to compress the project by one day,

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by compressing activity C, you're going to incur an extra cost of

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\$40 for the compression, which is the cost slope of C and you're

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going to save \$120

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so the net is going to be savings of \$80 that's definitely a good

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situation. So right now, we are still moving on the right side of

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the curve as we compress the total cost goes down. So now to proceed

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with the second cycle, we have to check the cost slopes of

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activities to be compressed. Remember that in order to achieve

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an effective compression on more than one critical path, if you

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have more than one critical path, you have either to compare. You

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have to compare both paths by exactly the same amount, and that

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can be achieved either by compressing one activity on each

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path or finding a common activity on both paths that's cheaper to

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compress than the sum of the two activities on the different paths.

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So in this case, for after the compression cycle. Number one, we

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have savings of \$80

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and that shows now the numbers based on the compression cycle of

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activity. One notice here that the duration of C has been reduced to

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seven days, which resulted in the new dates in green. And now on the

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backward pass, we notice that activity B has become critical. So

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now we have

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two critical paths, A, B, E, H and A, C, E, H, A, C, E, H,

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activity D still has one day. Activity f has one day and

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activity G, still as one day

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now the options to compress this network.

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We can look at activity B and C together. We can look at activity

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e, we can look at activity h, or we can look at activity A,

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whichever is going to be cheaper to compress.

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So now let's look at the numbers again. B by itself, \$75

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C, \$40 so if you want to compress B and C together, then it's going

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to cost 115,

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e costs 50, which is cheaper, H costs 120 which is more expensive,

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a costs 100 which is cheaper than B and C combined. So probably the

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cheapest one to compress right now would be activity e, which can be

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compressed by up to two days. But again, going back here, we notice

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that the total float in this case is only one day on the other

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paths. So we cannot compress e in one step by two days.

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Therefore the next step is to compress e by one day. This

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results in converting all the activities on the network into

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critical activities, a very interesting situation, but before

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we forget that resulted in an extra cost of \$50

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and resulted in savings of \$120

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therefore the net is going to be savings of \$70

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so now we have reduced The duration of activity, E from six

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to five, and you notice here the purple numbers now all the

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activities have zero total float, which means, right now we have how

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many critical paths. We have four separate critical paths, A, B, E,

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H, A, C, E.

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Maximum, we have to keep compressing until we exhaust all

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the compressibility on at least one path. Once one critical path

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becomes incompressible, then the whole network becomes

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incompressible, because

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if you try to shorten the other paths, that critical path is going

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to stay longer, and that defeats the purpose of compression. So

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once one critical path at least, becomes exhausted with its

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compression, then the problem is over. We cannot compress any

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further if we need to compress the network until it's not

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economically wise to compress it any further, we stop as soon as

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the cost of compression cannot be offset by the reduction to the

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indirect cost. So the increase in the cost slopes of the activities

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being compressed cannot be offset by the reduction or the savings

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from the indirect cost. If we need to compress the network by a

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certain number of days, then we need to stop as soon as we reach

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that time duration equivalent to that number of days. For example,

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if my project is four days behind schedule, why should I compress

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more than four days? It's not going to help me, because I'm

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probably going to pay more to compress beyond four days. So I

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should stop at the four days and avoid paying the liquidated

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damages.

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And this is basically the the idea that we're talking about, that

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total cost curve. This is going to be the right side, and this is

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going to be the left side right now, we have already reached this

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point, which is the lowest

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total cost, which is the optimum duration. So that's after three

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days of compression. Compressing any further is going to result in

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an increase to the cost. If you want to try, you can try on your

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own to keep compressing this network to the maximum. But this

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is basically where we're going to stop assuming that the question

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was to compress the network until it was not economically feasible

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to compress any further, which is at this point and we are already

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there.

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So basically this is our solved example on network compression.

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Remember, again, a few things to remember when solving this

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problem. First of all, you have to draw the network correctly. You

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have to calculate the normal durations in the network. To get

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the expected completion date, you calculate the total indirect cost

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by multiplying that date times the cost per day, the indirect cost

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per day, the total normal cost, or the total direct cost, can be

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indirect plus the total direct gives you the total cost of the

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project before compression. Always keep an eye on the near critical

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activities which have very low amount of total float, because

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once you start compressing, you may cause these activities to

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become critical, and therefore creating additional critical

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paths, which is going to require parallel compression of these

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paths. Remember that If you have two critical paths, and you

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compress one activity on each path,

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that results in only one day of shortening of the project, because

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these paths are in parallel. So it's not you compressed one day on

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activity B, for example, and one day on activity C, but this

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resulted, since there are apparent paths, resulted in only one day of

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compression to the whole project. If you have three critical paths,

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and you compress one activity on each again, you compress three

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activities by one day each. But since they were running in

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parallel, that's that resulted in only one day of compression for

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the whole project. If you have a total float, the minimum total

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float on the network, let's say, is five days. Then in this case,

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if an activity can be compressed by three days, you can do that in

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one step to save yourself some time, because you know that by

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compressing this activity by three days, you are still gonna have a

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minimum of two days of total float on the other paths, so you're not

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gonna create any new critical paths. One last thing, and this is

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very tricky part here, you may have some activities that are

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ineffective on one path,

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yet they might be effective on another path. So for example, if I

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have a network, the initial critical path had a start to start

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relationship and a finish to finish relationship.

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If you remember, per the definition of the effective

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activities, the start to start would make the first activity

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ineffective. The finish to finish would make the second activity

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ineffective. So on that particular path, these activities are

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ineffective. But let's say we have created a new critical path and

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that contain these same activities. Let's say the first

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and the last activities in a finish to start relationship with

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the other activities. Then the first and the last activities are

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still ineffective on the first path, because the start to start

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and finish to finish relationship, but they are effective activities

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on the second path.

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Because they are finished to start activities so they can be used on

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one path and not on another. Remember that? Because, again,

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that might be a question on a test.

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Good luck, and

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I'll see you on the test. You.

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