So, welcome to the review of session for Module five. The same format, I have, this time I've chosen four types of problems for you and I will show you the problem, explain to you the problem and then with a little bit of delay, I'll walk through the solutions. Here is the first problem. The first problem is about a pharmacy and the way that the prescription process works, is you basically you have you have a doctor and the doctor is prescribing you know, prescriptions and, and those get filled by the pharmacist. And, the pharmacist has a quality assurance person you know, working with him or her, And that person catches a good number of any potential problems and then the medication ends up with the patient. Now the problem though, as you can see in the question, The problem is the doctor makes a certain percentage of errors, Two%. And the pharmacists makes some, some errors. Most of them, most of them actually get caught by a QA person, the quality assurance. And so basically they get fixed and you can think of that as a, as a [INAUDIBLE] and then finally, even the patient, even the patient is able to recognize some of the defects. And so, the question here is, first what is the likelihood that the patient is presented with the wrong medication? And then finally, what is the question that the, what's the likelihood that phe Patient ends up actually taking a medication that is not right for him or her. Take some time, put me on pause, and then give it a shot yourself and then, look at my solution. Alright, the way that you want to think about this, this is really probability theory, and it, it starts, I think it's helpful to just start with the doctor here. So the doctor, basically can make a mistake, which happens with a two percent probability or make things correctly, which happens with a 98% probability. Now, if the doctor makes a mistake, really the pharmacist doesn't really matter, and it's a 50-50 shot between the adverse outcome, That the patient gets the wrong medication, And the outcome that basically the problem is fixed by the patient. Now, for the 98% cases that the pharmacist, the doctor makes the right decision is left to the pharmacist now to mess up, if you wish. So in 99% of the cases, things will go look good. Alright? So the pharmacist will do the right thing. And then only one percent of the cases will the pharmacist make a mistake, a defect. If you look at those defect, then, In 97% of the cases, This defect is going to be fixed by the pharmacy's QA person. And in three percent of the cases, it will be presented to the patient and then, you know, the patient is not particularly good at, at, at, kind of recognizing they just see the scrip and the bottle with the pills. So if they have the wrong pills in the bottle, the patient can't tell. And so only ten%, then, of the patients are able to catch this as the patient. You know, this is catch by patient and then, 90% here are leading to the negative and undesirable outcome. So let's look at the question in front of once you have this logic, and you really see the, the Swiss Cheese argument here, That in order for the medication, especially in the in, in, in, in the right branch, if you have to end up in the, in, in, in, in the body of the patient, There are lots of things that have to go wrong, right? The pharmacist has to make a mistake. The inspection people has to miss it and the patient has to miss it and so, it is really a big probability term. If you look at the first question, the first question is, is looking at the patient being presented with the wrong medication. So, we are looking at this note here in the tree. So, there is a two percent possibility that the doctor makes a mistake and that leads to the wrong medication getting to the patient and, then, you know, we are looking at this case independent of whether the patient catches it or not with a probability of 0.98 times 0.01, the probability of the pharmacist making a mistake, then 0.03, the probability that the QA person doesn't catch it. We going to get a total probability that corresponds to 2.2, 2.0. 294%. And then in the second question, we really are interested in this kind of, this final outcome here at the bottom. So those two percent here, those, those 0.02, we're only interested in half of those cases because the other half is caught by the patient, plus 0.98 times this, same as the [INAUDIBLE] 0.01 times 0.03 times, now, 0.9 which is the probability that the patient doesn't catch the mistake. And that gives us a total probability of 0.010265. Sorry, I had to look at the computer for this. This is a little bit more than what I can do in my head, But, that gives you the answer to the questions. Now, in the previous problem, you just had to compute the defect probabilities, basically at the end of the process. Now in this question here, we're going to look at how this impacts the capacity calculations. And so, specifically, you are looking here at a process that consists out of four steps, Step one, step two, step three, and step four. And these four steps, have quality problems. In particular, the second step here has a defect probability of, 50%. 50% of the units here are going to, go to scrap. And, then the next step has a rework loop, has a 30% probability that you're going to have to rework the unit, and then you're going to sell it over here. Again, as usual, take some time and I'll be with you in just a moment to give you the answers. Alright. So the first question here looks at, how many times, basically, does a flow unit that is served here has demand. How do, often does it have to be handled at each of the resources? And so, the way to think about this is really, well, For the last step here, step number four, every unit of demand has to be processed, and so that gives us a demand here of D. For the previous step, station number three, we have one point 3D's, right, Because there's a 30% rework loop here. And please note that the assumption is that a rework always works. So you don't have to think about a rework of a rework of a rework. So that means really every unit has to go through there, and then 30% of the units have to go through there twice. Now, as you then go to step two, you have to adjust for the scrap and there's a 50% year loss, that means you have to go basically through 2.6 times D units at station two. And though there is no year loss here at station one while you need to do one unit at station one for every unit at station two, and so that's also 2.6D. But to answer the question here for the third step here, it's basically 1.3 times the amount of demand. Alright, the next question is the bottleneck and how to find the bottleneck? Well, let's go back to the basics here and just compute the capacity level of each of the resources. Basically, we have one over four, reflecting the four minute processing time, And one over four units per minute at the first step, Then one over three at the second step. A one over five at the third step, And then, a one over two at the last step. And, that allows me to compute the implied utilization as a ratio of the demand trait, 2.6D divided by the capacity, which is one over four, And that gives me here 10.4D. The same way I'm going to get a 7.8d here, right? So, 2.60 times three, or equivalent to divided by one over three. 6.5d here and 2d here for station number four. And, the biggest implied utilization is the bottleneck and that is here, station number one, okay? So station number one is the bottleneck. Now, what's the process capacity? Well, the process capacity, assuming there's enough demand, is driven by the by the capacity of station one, and actually let me take what I just said a moment ago, if there is not enough demand, suppose this capacity is driven by bottleneck capacity of station one, And so, I have to figure out what, what can of this guy overall can produce, and it's fifteen units per hour of course. So, at fifteen units per hour is, is basically one unit every four minutes. But the problem of course, the problem is 50% of that stuff will be scrapped, And so really only 7.5 units per hour will make it to the end. Notice that the rework loop is not impacting this number, because the rework does not involve the bottleneck, it really has no impact on the overall process capacity. Just one final note or observation, if you let me do this quickly, Is let me quickly again comment on the rework loop. Instead of putting the capacity here as one over five or that suppose the same time is five minutes, can take also more probabilistic approach, and I think I did this in class earlier on, you can see that with, with the probability of 70% I'm going to spend five minutes, and then with the probability of 30% I'm gonna spend ten minutes. And so, you get a 6.5 minutes per unit as your processing time and that's one over 6.5 as the capacity. And so alternatively, you can just ignore the rework loop here for the, the, the sake of demand and just put a D here, And just ask really, how many unique flow units have to show up at station number three. And then instead of then dividing by one over five, you divide by one over 6.5, And that gives us the same 6.5 here. So you can choose either one of these approaches, Whatever is easier for you. All right. The next question is about chicken eggs and I went on Google here to inform myself about the average weight of this chicken eggs. I hope I got this right, I believe it's 47 gram, And, I have this eco-friendly farmer here who is concerned about the output of his, well, it is not really his output, but the output of his the chicken. And he takes a sample and finds them that, they have basically, a mean of 47 and a weight of two grams. However you can only make money of them if it fall into this specification interval between 44 and 50. And so, you want to basically do some of this sigma calculations that we talked about in class. Try it out. Alright. The first one is relatively easy, I would argue. Right? The first one is a capability score. And remember this capability score looks at the upper specification of a limit minus the lower specification limit, Which is 44 divided by six times the standard deviation. And so in this case here, that is simply six divided by six times two, which is 0.5. Now that is a relatively low capability score. If you just go back through the slide that we discussed in class, you notice that you know, you are somewhere between a one sigma and a two sigma process, and so this corresponds now to a bunch of defects. And that's what the second question is about. So what percentage of the eggs fall within the specification limits provided by the local distributor? So for that, I have to leave my PowerPoint quickly and, and jump in to Excel and so, what I want to find out, is really I want to find out, from the normal distribution, the specification limit, the upper specification limit is a 50. All right? And so I have a distribution, a normal distribution was 47 mean and a two standard deviation, and I'm looking at the cumulative normal here, and so I have basically 93.3% of zx below 50 grams. And so one minus that, probability here gives me the probability that this egg is going to be too, too heavy. So six percent of the cases. As far as, the probability that this egg is too light is concerned, I can again look at the normal distribution and I can look at the scenario that I have a 44, and that normal distribution was 47 mean and two as a standard deviation. And that is, surprise, surprise, also 6.6%, because the mean of 47 is just right in the middle of the confidence interval. So if I add up those probabilities, since air can't be too heavy and too light at the same time, I'm going to get a, a probability of 13.36, Percent that the egg is outside the confidence interval or the specification interval. And then, one minus that probability, equals to 86%, is the probability that I had been fishing for in the question. Alright. So that's you know, that's the kind of the number two here. I'm just going to write, see Excel, And I'm going to put the Excel spreadsheet up there on the Wiki, And then the third question looks at basically how much does the standard deviation have to be improved to get to a CP score of two thirds. Okay? And so we just take the same equation as above, which was 50 minus 44 divided by six times, And now, we leave the sigma as a variable, as an unknown to be solved for and that we want to be equals to two third. And so that is equivalent to, you know, that's basically this is six. Six divided by six cancels out and then we have a one over a sigma equals to two thirds, and that means that sigma would have to be reduced from the current state, which was two grams would have to be reduced to 1.5 grams. I have to disclose here that in the questions that we're doing, we'll always assume that the current mean is actually in the middle of the specification interval. It gets a little tricky otherwise, but, I'm sure you can figure it out in Excel, And I promise to not test you on the exam with that special nasty type of question. Alright. My last question is a very creative one. It's a, a word matching problem in the context of the Toyota production system. And the way this works is I've, provided you here with, seven descriptions of, of, of managerial practices in operations and I have seven Japanese terms here, below. And what I want you to do is I want you to go ahead and, read these statements and match them to the Japanese words. Go ahead. All right, let's tackle the first one. let's just go A through G through each of these ones and just think about what Japanese words comes to mind. So examples of this includes working to working, medical workers making unnecessary movements, working on defects, idle time, all of the stuff that shouts out waste, waste, waste and the Japanese term for that is muja. So we have A and right here, Second, a system that enables a line worker to signal that she or he needs assistant, assistance from his supervisor and that's used to implement the Jidoka principle, So that is the Andon Cord. Now the Andon Cord, this cord that goes adjacent to the line, and that helps people to alert the supervisor. And notice at, at that time the light starts blinking, It's not that all of the factory, the entire production line stops, but just the line segment and even that is not necessarily always stop the course typically even at Toyota. You're just going to have some buffers in there. So that's the Andon Cord. C, A brainstorming technique that helps you, find root causes, Of usually undesirable outcomes. That is the Fishbone diagram, right, also known as the Ishikawa diagram, And please don't think that Ishikawa is a fish in Japanese, but I think it's just named by the inventor, Though, I really have to do close-set, I speak absolutely no Japanese, I'm sorry for that. Then Part D, workers at Toyota make suggestions to process improvement. It's not just management that comes up with these suggestions and that is the classic Kaizen process. Kaizen the process of continuous improvement. How do you control the amount of work and process inventory? That is an easy one, Especially since we're running out of options here and that looks like Kanban, Number F or letter F. If a plant uses this technique, the adjacent cars on the line would be mixed models, Different colors, some with no sun roof. Things like that. You would be producing to demand. There was a mixed model production principle that we talked about in the variety module and that's the idea of Heijunka.. And then finally making production problems visible, and stopping the prod production upon detection of defects, that's the logic of detect, stop, alert, and that's the idea behind Jidoka. Okay, if I'd been a little more creative, maybe, these letters here, if I would read them now, would spell out something beautiful. Sorry they are just what they are here, but that at least concludes this review