ASQ CQE Certified Quality Engineer – Section I – Management and Leadership (18 Questions) Part 2

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  • January 18, 2023
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33. 1B-2d CPM – Float and Critical Path

So in the previous slide, this is something which we plotted this activity network diagram where we had activity ABCD and E. And based on what is the predecessor of that, we plotted this activity network diagram. And now let’s talk about float and critical path. We did talk about that earlier as a layman to understand this. That the same activity activity set of activities. If I do A, B and D top one, it takes me eight days to reach to the end. If I go to A, C and E, it takes me seven plus one plus two is equal to ten days to complete. And if I go for A-B-E and end, this takes me seven plus 411 plus 213 days. There were three paths. This we talked earlier.

So now let’s look at the definition of float. Float or the slack. There’s a different name for float which is slack as well. It’s the amount of time that a task can be delayed without causing a delay. So now if you look at this one activity number D, we talked about that the highest amount of time it takes is in this middle path which is 13 days. So the top path also can be delayed up to 13 days because even if I complete the top line before 13 days, my project is not going to finish before 13 days. So what do I have is in D? If I look at that, if I look at this activity number D, this activity number D I can delay and how much I can delay 13 minus eight.

So there is a five days of float here. So I can delay five days here. So instead of two days, if this activity takes seven days also, then also it’s not a big deal. So activity number D has five days of float here. Now we are just looking at this as a layman. Next slide, I will go very systematically to draw that. So for the time being, just understand as a layman that okay. The top line is eight days, the middle is 30 in days. So there is five days of extra amount of cushion or the float which I can use on activity number D. Can I use that on activity number B or A? No, because if I use that on activity number A or B that will affect the middle line as well. The middle line will also increase. Instead of 13 days, this will increase. So that’s a float and the critical path is the path which has zero float or which is the longest period. So zero float here is on this activity which is abe. So in this case, abe is my critical path abe. So if you are managing the project and you know that abe is your critical path, then you don’t want to delay any of these activities.

So if you delay A, B or E, that will delay your whole project. You have some margin in delaying D. You have some margin in delaying C how much you can delay D. We talked about that you can delay C also how much you can delay C. Because if you look at the bottom path, this is ten days. So there is a still a question of three days. So you can delay C by three more days. So instead of one day, if it takes four days, not a big deal. Your project will still get completed in 13 days. So this was about float and the critical path. So the critical path is the path which has no float.

And the float is some extra amount or some extra cushion which you have which you can use to delay some activities. Not all activities can be delayed. If an activity cannot be delayed, it has zero floor, then you can say that that activity is on the critical path. So what we will do in the next slide is we will use the same thing but then we will go systematically. Instead of, let’s say five activities, if you have 100 activities that also you should be able to understand that how you calculate the critical path, how you calculate the flow for each of these activities. Let’s see that on the next slide you.

34. 1B-2d CPM – Forward and Backward Pass

Previously. Also we looked at this table where we had five activities ABCDE and we plotted a network diagram for that and based on our intuition, based on looking at that we could find out that the critical path in this is Abe. Because Abe was taking 13 days and rest other paths were taking less than that. And we also found out that there is some float in D and C but that was something easy for a smaller one. If you have a complex list of activities then you need to have some systematic way of finding out the critical path and float. Let’s see that here. How do we do that? So to do that we make one box for each activity. And let me draw that box here. So for each activity A-B-C-D-E we will make a box like this which will have four corners. Let me draw these four corners of that and in these four corners we will put early start. So this will be early start first one and then we’ll be early finish EF and then in the bottom we will have late start and late finish.

This we will do for each of these activities. Early start, early finish, late start, late finish. Top two things early start and early finish we will be doing in forward pass. So as we move forward we will find out early start and early finish and late start and late finish we will find out in backward pass. So once we move from last to first we will understand that once we actually do that and then float would be difference between these two. This is float. The float is the difference between the early start or the late start. Any activity which you can start at the earliest or you need to start the latest. The difference in that would be the float. Same way float could be calculated the difference between EF and LF as well. Early finish and late finish. The difference of that also would be float and you will see that both of these are same. Either you find out float by difference between late start and early start or late finish and early finish. So both of these differences you will see are equal.

So this is something which we need to do. So let’s start with start activity. So we start this project on day number zero and then we move to activity number A. And just like this we will make a big box for activity number A. And my activity number A is of duration of two days. So let me put that here. The first thing I put is early start. So when we can start earliest earliest we can start at it on day zero. Because that’s something which you can start straight away on day zero and if you start it on day zero it takes two days. So when you will have this done you will have this done in day number two. So it takes two days. So this is early start, early finish. Then what about bottom these two things? Late start, late finish. We will not talk about that now.

We will talk about that in the backward pass. Let’s move forward in the forward pass. So we have A and then the next activity is activity number B. Let’s draw that in B. When we can have B started, we can start B on day number two because A has to finish. So A finished on day number two. So we start B on day number two. What’s the duration of activity B? Four days. So when we will finish this, four plus two is equal to six days. On 6th day I will finish activity number B. Coming to the next one, which is activity number D. Just keep an eye here because that’s where I’m taking all these numbers from. So activity number D. D is of the duration of two days. Early start, early finish. Early start is six because B finishes on 6th day. D will start on the 6th day. Takes two days.

Make it eight. That finishes the top line. Now coming to the bottom one, after A, we have activity number C. So let’s draw here. This is my activity number C. When we can have activity number C started, we can have activity number C started on day number two because A has to finish that. So the early finish of A will be the early start of activity number C. C has number of days as one. So this will complete on day number three. From C we go to activity number E. Now here there is slight complication, but it’s not complication also. Now. When does E start? For E to start, both B and C needs to be completed. C gets completed on day three. Can we start activity number E on day three? No, we cannot. Because we need to complete activity number B as well. So both B and C will get completed on day number six. So day number six is the day when we can have E started, not the three, which is the early finish of C. But we are looking at the early finish of B. When B finishes, then only you will be able to start E.

E is of seven days. So this will make it 13 once this is done and then everything is done, and then we go to the end point of this project. So this is the diamond shape end point of the project. The project is ending on day number 13. So this takes 13 days. This was the forward pass. We moved from start to the end. Now let’s start moving from end to the start. Backward pass. So in backward pass we are looking at two things, late start and late finish. Let’s look at E first. When is the day when we can have by latest finish activity number E. Activity number E needs to be finished the latest by 13th day, not to delay the project. So 13th day will be the late finish for activity number E. And if I take seven days out of that the latest start, this has to be on the day number six. So 13 minus seven is equal to six. That’s easy. Same thing let’s do with D, activity number D.

When is the latest finish of activity number D? When we went forward then we saw that on day number eight this activity got finished. But when is the latest we can finish D? Latest we can finish D is by day 13th because that’s the day when project gets finished. So day 13 is the latest you need to finish D. And from 13 if you take out to two, which is the duration of this activity, then the latest start of this activity should be on the day number eleven. All right, so let’s move back, let’s move back to C. So activity number C, since E is starting on day number six, so the finishing of C should be on day number six. On day number six, activity number C should finish and six minus one is equal to five. So let us start would be day number five.

Similarly if I look at B, if I look at B and I’m looking at the latest finish when my B should be finished, my B should be finished on day number eleven or day number six. It should finish by day number six because if it doesn’t finish by day number six, activity E starts suffering. So out of eleven and out of these six I have to choose six here. On day number six my B needs to be finished. If B finishing on day number six, then B needs to start on day number two because six minus four is equal to two. Similarly if I go back to A when my A should finish, my A should finish either on day number two here or day number five when B and C has to start. But then I have to select number two second day because any delay after that will delay my activity number B.

So the latest finish for activity number A is day number two and the latest start would be two minus two is equal to zero. This is something which we have drawn. Now it might look a little bit complicated initially, but once you do once or twice, do some practice here because I’m sure there is something or other will be coming related to whatever you are learning here related to activity network Diagram finding out float finding out the critical path making table something will come in CQE exam. So once you have made this diagram now it’s much easier. And as I earlier said, that float is the difference between early start and late start. That means late start minus early start late finish minus early finish. So let me put it here. Float is equal to late finish minus early finish or is equal to late start minus early start. These two things 13 let me put another color here. 13 -13 is zero. So there is a zero. Float here. Here is zero float here.

If you look at that, there is a float of three days which will be same as late start and early start. So three days float here. If you look here, there is a float of five days. There is a float of five days. Float of five days means activity number D can get delayed by five days. Float in B is zero zero a is also float is zero. So now if you look at this, look at all those activities which have zero float. So what are those activities? Zero float activities are A, B and E. So that means my critical path is critical path in this particular set of activity is A then B and E.

That means you really cannot afford to delay any of these three activities. A, B and E. These needs to be completed on time for the project to be completed on time. Float is in D and E. D has a float of float is equal to five and C has a float of three days. So C can get delayed by three days no problem. Activity D gets delayed by five days, no problem. But any other activity, if that gets delayed, the project gets delayed. So this was active activity Network Diagram how to find critical path how to find flow once again very important from CQE exam point of view. Expect one, two or even more question from this particular knowledge of drawing. Activity network diagram finding out critical path finding out floor.

35. 1B-2d PM Tools – PERT (Program Evaluation and Review Technique)

Previously when we talked about CPM, the critical path method, there we plotted a network diagram and in that particular example which we took, there were a number of activities A, B, Cde and for example activity A took two days to complete, activity B took four days to complete and so on. And based on that we plotted the network diagram and there in addition to number of days for each activities, we had predecessor and successor relationship also, which activity happens after what B happens once A has been completed. So based on that we plotted a network diagram and we found out the critical path and we found out the float. That’s what we did previously. Now, coming to Pert pert is program evaluation and review technique.

The only major change which you need to understand here between Pert and CPM is that in Pert, pert network acknowledges that there will be a time variation. So for example, in CPM we said that activity A takes two days to complete. But in part we understand that activity A might take slightly longer time or slightly shorter time depending on number of situations. So activity time is not fixed. So there is some range in which this activity gets completed rather than giving a fixed number that activity A gets completed in two days, b, in four days we give a range here range which consists of the most optimum time, most pessimistic time and most likely time.

So we have optimistic time, pessimistic time and the most likely time. So we give three numbers here instead of giving one number for the time period. So what we use here is we use the probabilistic approach that what’s the probability of this particular activity happening in best case scenario, in worst case scenario and in most likely case scenario. And based on these three numbers we calculate a number which is the expected time. Expected time is calculated by adding optimistic time four times the most likely time, the pessimistic time, and then dividing this number by six. On the next slide we will look at an example of this as well.

So in addition to expected time being a probabilistic approach, we would like to have a standard deviation as well, just to know that what’s the spread of this the standard deviation is calculated by Pessimistic minus optimistic divided by six. So these two formulas you need to remember and there’s a good chance that you might get something related to this in your CQE exam as well.

So on the next slide let’s look at a very simple example and probably that’s what you might get something like that in your exam as well. Let’s take the same example of activity A. Let’s say what we did in CPM, activity A was for two days. So instead of two days, probably that was a very optimistic value, that two days was the optimistic value. So now instead of giving two days as the time for the activity. Now, if I give you three numbers, one is the most optimistic time. Let’s say the optimistic time is two days. So let’s put it here. Most likely this will get done in three days. So let’s put that as the most likely time. Most likely is equal to three days. And the pessimistic.

Pessimistic means if everything goes wrong and in worst case scenario, this might take even up to seven days, supplier might not be able to supply you something, your workers might go on strike. Whatever pessimistic things might happen based on that, the most pessimistic time to get this activity done is seven days. Now, based on this, if I ask you to find out the expected time for this activity is equal to and if you remember the formula earlier in formula we said that optimistic time, let’s put it O plus four times the most likely time plus the pessimistic time divided by six. So this was the formula.

So here if we put the value optimistic is two days plus four into most likely is three days plus pessimistic is seven divided by six. And if we add these, this comes out to be 21 by six. So this is four multiplied by three is equal to 1212, plus two is equal to 1414, plus seven is equal to 21 divided by six is equal to 3. 5 days. So the expected time for this activity will be 3. 5 days. That’s how you calculate that. And if I ask you to calculate the standard deviation, standard deviation will be, as I earlier said, pessimistic minus optimistic divided by six. Pessimistic value here is seven. Optimistic value is two days divided by six. So this gives you five divided by six or 0. 83 days. So that’s how you calculate the expected time and standard deviation in case of pulp.

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