So, welcome to the review session for Module Five. The same format I have, this time I have chosen four types of problems for you, and I will show you the problem, explain to you the problem, and then with a little 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 a doctor, and the doctor is prescribing prescriptions. And those get filled by the pharmacist. And the pharmacist has a quality assurance person 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 that the doctor makes a certain percentage of errors, 2%. And the pharmacist makes some errors, though most of them actually get caught by the QA person, the quality assurance person. And so, basically, they get fixed, and you can think of that as a real work glue. And then finally, even the patient is able to recognize some of the defects. And so the question is, first, what is the likelihood that the patient is presented with the wrong medication? And then, finally, what is the question, what is the likelihood that 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. All right. The way that you want to think about this, this is really probability theory. And I 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 2% probability, or make things correctly, which happens with a 98% probability. Now if the doctor makes a mistake, really, then the pharmacist doesn't really matter, and its a fifty-fifty 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, it's left to the pharmacist now to mess up, if you wish. So in 99% of the cases, things will look good. All right? So the pharmacist will do the right thing. And then only 1% of the cases will the pharmacist make a mistake, a defect. Now, if you look at those defects then, in 97% of the cases this defect is going to be fixed by the pharmacy's QA person. And in 3% of the cases, it will be presented to the patient, and then the patient is not particularly good at recognizing, they just see the script and the bottle with the pills. So, if they have the wrong pills in the bottle the patient can't tell. And so only 10% then, of the patients are able to catch this as the patient. So, this is catch by patient. And then 90% here, are leading to the negative, get an undesirable outcome. So let's look at the question, once we have this logic, and you really see the Swiss cheese argument here, that in order for the medication, especially in the right branch here, to end up 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's really a big probability term. If you look at the first question, the first question is looking at the patient being presented with a wrong medication. So we're looking at this node here in the tree, so it was a 2% probability the doctor makes a mistake, and that leads to the wrong medication getting to the patient. And then, we're looking at this case independent of whether the patient catches it or not, was a probability of 0.98 times 0.01, the probability of pharmacists making a mistake times 0.03, that probability that the QA person doesn't catch it, we're going to get a total probability that corresponds to 2.0294%. And then in the second question, we really are interested in this kind of this final outcome here at the bottom. So those 2% here, 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 0.01 times 0.03 times now 0.9, which is the probability that the patient doesn't catch the mistake. And that gets us a total probability of 0.010265, I'm sorry I have 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 are going to look at how this impacts the capacity calculations. And so, specifically, you are looking here at the process that consists of four steps, step one, step two, step three, and step four. And these four steps are, 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 will be with you in just a moment to give you the answers. All right. So the first question here looks at how many times basically does a flow unit that is served here as demand, how 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 the demand here of D. For the previous step, station number three, we have 1.3 Ds, right? Because there's a 30% rework loop here, and please note that the assumption is that 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 into step two, you have to adjust for the scrap, and there's a 50% yield loss. That means you have to go, basically, through 2.6 times D units at station two. And, though there's no yield loss here at station one, 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. All right. 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 a 1 over 4, reflecting the 4 minute processing time. And 1 over 4 units per minute at the first step, then 1 over 3 at the second step. A 1 over 5 at the next step. And then 1 over 2 at the last step. And that allows me to compute the implied utilization as a ratio of the demand rate, 2.6D, divided by the capacity, which is 1 over 4. And that gives me here, 10.4D. In the same way, I'm going to make it a 7.8D here, right? So, 2.60 times 3, or equivalent D divided by 1 over 3. 6.5D here. And 2D here for station number 4. And the biggest implied utilization is a bottleneck, and that is here at station number one. Okay? So, station number one is the bottleneck. Now, what's the process capacity? Well, the process capacity, assuming there is enough demand, is driven by the capacity of station one. And actually, let me take back what I just said a moment ago. Even if there is not enough demand, the process capacity is driven by the bottleneck capacity here at station one. And so, I have to figure out, well, what can this guy, or girl, produce? And it's fifteen units per hour, of course. Fifteen units per hour is basically one unit every four minutes. But the problem, of course, is that 50% of that stuff will be scrap. 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 a rework does not involve the bottleneck, it really has no impact on the overall process capacity. Just one final not, or observation, if you'll let me do this quickly is, let me quickly, again, comment on the rework loop. Instead of putting the capacity here as 1 over 5 or the suppose the same time as five minutes. You can take, also, a more probabilistic approach, and I think I did the same class earlier on. You can say that with a probability of 70%, I'm going to spend five minutes. And then with a probability of 30%, I'm going to spend ten minutes. And so, you get a 6.5 minutes per unit as your processing time. And that's a 1 over 6.5 as a capacity. And so, alternatively, you can just ignore the rework loop here for the sake of demand, and just put a D here, and just ask really how many unique flow units will show up at station number three. And then, instead of then dividing by 1 over 5, you divide by 1 over 6.5, and that gives you 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 chicken eggs. I hope I got this right. I believe it's 47 grams and I have this eco-friendly farmer here who is concerned about the output of his, well, it's not really his output, but the output of his chicken. And he takes a sample, and finds then, that they have basically, a minimum 47 and a weight of 2 grams. However, he can only make money on them if they falls into the specification interval of between 44 and 50. And so you want to basically do some of the six sigma calculations that we talked about in class. Try it out. All right. The first one is relatively easy, I would argue. Right? The first one is a capability score. And remember, the capability score looks at the upper specification of a limit minus a lower specification limit, which is forty-four. Divided by six times the standard deviation. And so, in this case here, it is simply 6 divided by 6 times 2, or just 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're somewhere between 1 sigma and 2 sigma process, and so this corresponds not 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 jump into Excel. And so, what I want to find out is really, I want to find out, from the normal distribution the upper specification limit is at 50. All right? And so, I have a normal distribution with 47 mean, and a 2 standard deviation. And I'm looking at the cumulative normal here. And so, I have basically 93.3% of the eggs are below 50 grams. And so, 1 minus that probability here gives me the probability that this egg is going to be too heavy, so 6% 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 in that normal distribution was 47 mean and choose 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 the egg can't be too heavy and too light at the same time, I'm going to get a probability of 13.36% that the egg is outside the confidence interval, or the specification interval. And then 1 minus that probability equals to 86%, is the probability that I had been fishing for in the question. All right. So that's, you know, that's 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, by how much does the standard deviation have to be improved to get to a CP score of two-thirds? And so it just takes the same equation as above, which was 50 minus 44 divided by 6 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-thirds. And so, that is equivalent to, that's basically, this is 6. 6 divided by 6 cancels out, and then we have a 1 over sigma equals two-thirds, and that means that sigma would have to be reduced from the current state, which was 2 grams, would have to be reduced to 1.5 grams. I have to disclose here that in the questions that we are doing, we'll always assume that the current mean is actually in the middle of the specification interval. It gets a little trickier otherwise, but I'm sure you can figure it out in Excel, and I promise not to test you on the exam with that special nasty type of question. All right. My last question is a very creative one. It's 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 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 to each of these ones, and just think about what Japanese word comes to mind. So examples of this includes workers making unnecessary movements, working on defects, idle time, all of this stuff that shouts out waste, waste, waste. And the Japanese term for that is muda. So we have A entered here. Second, a system that enables a line worker to signal that she or he needs assistance from a 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 that at that time comes the light starts blinking. It's not that all of the factory, the entire production line stops. But just a line segment. And even that is not necessarily always stop, because 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 a fishbone diagram, also known as the Ishikawa diagram. And please don't think that Ishikawa stands for fish in Japanese, but I think that's just named by the inventor. Though I really have to disclose that 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 a classic Kaizan process. Kaizan, the process of continuous improvement. How do you control the amount of work and process inventory? That is an easy one, especially since we are running out of options here, and that looks like Kanban. Number F, or letter F. If a plant uses this technique, the adjacent course on the line would be mixed models, so different colors, sunroof no sunroof, things like that. You would be producing to demand. That 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 production upon detection of defects. That's the logic of the 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 session
