Dr. Ardavan Asef-Vaziri:	Thank you very much for attending this session. Today, I'm going to talk about process flow analysis. The backbone of the process flow analysis is The Little's Law. Our whole session today will be around the Little's Law. Professor Little is a retired professor at UC Berkeley. But in the past 50 years, the name of the waiting line analysis and process flow analysis is tied with the name of Professor Little. He is retired, but he is still very active, and that is what we should also do. Work and work and appreciate life. Let me start with a simple problem. The problem is so simple that I'm sure all of you will get it. But the difficulty is this, as soon as we go to another example, not you, but many students have difficulties. Let's look at this situation. Suppose we have a container. That container has capacity of 100 gallons of liquid. Suppose a dense liquid comes into this container at the rate of 10 gallons per hour, and goes out at the rate of 10 gallons per hour. That is the basic concept in all stable systems. Inflow is equal to outflow. Let's suppose each molecule of this dense liquid, which comes into the system, cannot go out until all the molecules that before it was there. Have gone out. How long does it take for this molecule? We just came in to go out. How long does it take? Hundred gallons are there. We are emptying at a rate of 10, so it takes 10 hours for the other molecules to go out, and then this one goes out. That is simply 100/10=10. Flow time is 10 hours. Let's go through another similar example. Suppose here we have a container with capacity of nine and suppose nine gallons of that dense liquid are already there, and liquid, a rate of nine gallons per hour come in, and at rate of nine gallons per hour goes up. How long does it take for this molecule, which is in a fraction of second to come into the system, to go through the system and go out? How long does it take from this point to this point? For this molecule to arrive at this point, everything else, everything else before it should have already gone out. We need to empty this nine gallons, and we are emptying them at rate of 9/hour, so it takes 1 hour. For this molecule to go out, that is the flow time. What about this one? Twenty gallons are there. Flow rate of one gallon per hour comes in, and at rate of one gallon per hour goes out. How long does it take to empty this? Indeed, we empty this, but other flow come and fill it out. This container always has 20 gallons. This one always has nine gallons, and this one always has 100 gallons on average. It takes 10 hour for a flow, you need to pass through this process 1 hour to pass through this process and 20 hours to pass that process. Now, let me make the problem a little bit complicated. Not complicated, but a little bit not very easy. Suppose out of this 10, 9 goes here and out of this 10, one goes over there. Here, this is one, and this is nine. Ten goes out. Now we have a system. System is a set of components and the set of interrelationship between them. We have a system. Ten gallons per hour comes in, 10 gallons per hour goes out. All flow units pass this one. But some flow units go this way, and some flow units go this way. A mixed blue and green comes in. A mix blue and green goes out. Blue and green both go through the first process. Green goes through the bottom process. Blue goes through the top process, and then they go out. How long does it take for a molecule of this liquid from the time that it comes in until the time that it goes out. Let me answer this question by going through another problem. We have a coffee shop here close to the intersection of Plumber and reseda. The door of this coffee shop opens once every one minute on average. One customer comes in. In a stable system, if one customer comes in per minute, on average, one customer should go out per minute. Input should be equal to output in long run. For one hour, for half an hour, it is possible that input is greater than output or output is greater than input. But in long term, input is equal to output in all stable systems. If input is greater than output, people will be there and what will happen after 1 hour, 2 hours, 3 hours, 4 hours, 5 hours, there will be no room for anyone to enter that system and system collapses. What will happen if what goes out is greater than what comes in? It is impossible. For one minute it is possible, for half an hour, it is possible. For one hour it is possible, but in long term, it is not possible. How could number of people who go out be more than number of people who In long term, it is not possible. Therefore, for example, every 5 minutes, five customers come in, and five customers go out. Or at least the total number of people who come in per day is equal to the total number of people who go out. Therefore, when we divide it by the number of hours and the number of minutes on average, what comes in per minute is equal to what goes out per minute. What comes in per hour is equal to what goes out per hour. What is the throughput of the coffee? The report of this coffee shop is one customer per minute. One customer per minute comes in, one customer per minute goes out. We show it by R in the idles low. R in this problem is equal to one. Throughput is the flow rate in a stable system. Stable system input is equal to output, and throughput is average of this flow rate.Flow rate goes up and down, but throughput is the average of it. Throughput is 1/minute or 60/hour. If the system works for 10 hours, then it is 600 per day. Throughput always carries a time component with it. How at what rate people or products come in and go out? So if you go back to the original problem, the containers that I explained over there, throughput of the first container was ten, throughput of the container in bottom was nine, and throughput of the container on top was 1/hour. Throughput of the system was 10/hour. If a day is 24 hours, then throughput over there is 240 gallons per day. Inventory is what is inside each gallon or what is inside all those gallons. Inventory of the first gallon was 100. Inventory of the gallon in the bottom was nine gallons. Throughput, throughput, throughput, throughput, throughput, throughput, has a time unit associated with it, 10/hour. 1/hour, 9/hour. 1/hour, 9/hour,10/hour. Throughput has a time unit associated. Inventory is a number, 100 gallons, 20 gallons, 9 gallons. There is no time associated. Inventory is in numbers. Indeed, throughput translates inventory from numbers to time. That is what the Little's law does to change the dimension of inventory from unit to time, and we will see. Inventory in the coffee shop is the average number of people who are needing. The data of the problem tells us there are always four people waiting and one person at service. Five people on average are always inside the system. In real life, these five people sometimes is nothing, sometimes it is ten, on average, they are five. You may imagine that we think we are in an airplane far above, and then we look at the general patterns in a stable system. Throughput is 1/minute, and five people on average are there. As soon as one person leaves, these people come forward and another person will come in and on average, always five people are there. What is inventory? Inventory is five. On average, five customers are in the coffee shop. Inventory, we show it by I. R is equal to one flow unit. Flow unit is the general term that we use it a lot. In this coffee shop, a flow unit is a customer. In that example which we were discussing before, a flow unit is a molecule of that dense liquid. How long does it take on average for a customer to go out? Very simple, the same as what we discussed before. It's a container with five molecules, empty at a rate of one molecule per minute. Therefore, flow time is equal to inventory divided by throughpu. That is 5/1=5. Five, what? Five minutes. Why? Because this throughput is in terms of units per minute? Let me provide you with another explanation of the Litter's Law. We have 1, 2, 3, 4 customers waiting here. One customer is with the server. This is the server, the process. Another customer is about to arrive. This customer is st fraction of second to leave. This customer is almost here. It has a fraction of second to go out. This red goes over there, this red goes over there, this red goes over there. Therefore, this is a red, this is a red, this is a red, and this green comes in here. In the fraction of second that this customer is going to leave the system, looks back over its shoulder. How many people he will see, or she will see here? Five, five people are there. That is inventory when this customer is leaving the system. These people came in at what rate at rate of 1/minute. When this customer came in, when he was in this situation, when he looked over his shoulder, he saw nothing. Now, when he looks over his shoulder, he sees five customers. Five people are there. They came in in what rate 1/minute? How long does it take for five customers to go behind him? Because at the first time when he was going to come into the system, he looked over his shoulder and he saw nothing. Now, when he's leaving the system, he looks over his shoulder and he sees five people. Those people came in at rate of 1/minute. It takes five minutes for five people to come in at a rate of 1/minute. Therefore, from the time that he had no one behind him up to the time that he sees five people behind him, it took him five minutes to go from this point to this point, and that is what we call it the flow time. Flow time is equal to inventory divided by throughput. Flow time was equal to inventory divided by throughput, and therefore, throughput times flow time is equal to inventory, and that is the Little's Law. Again, throughput here was 1/minute or we can say 60/hour, or if the shop works for eight hours, you may say 480/day. In all these cases, we will get the same results. For example, suppose R, instead of 1/minute, R = 60/hour. Throughput times flow time is equal to inventory. Throughput is 60, flow time we don't know, inventory is five. Therefore, flow time is equal to 5/60 or 1/12. Here we got 1/12 before we had five. Are they the same thing? Yes, because five is five minutes and 1/12 is 1/12 hours and 1/12 hours is five minutes. Here we have shown that if I have 100 of Item A and 1,000 of Item B, but if throughput of A is 4/day and throughput of B is 200/day, here, 100/4, that is 25 days of inventory. But 1,000/200, that is five days of inventory. This 100 is more than the 1,000. I encourage you to read these slides, Slide 10 and Slide 11. Inventory is measured. It has several dimensions, space, units, time, and value. Now let's make the problem a little complicated. Suppose 40% of customers who come into this coffee shop at Plumbers and [inaudible] 40% of them go into the second line when they get exotic coffee or whatever and percent of them, as soon as they are done here, they leave the coffee shop. 40% go there, and on average, four people are in the second one. How long does it take a person who gets a black coffee and leave? How long does it take a person who get exotic coffee and leave? And how long does it take a person who comes into this coffee shop? He's a prototype customer. He's a mixture of these two. How long does it take him or her to come into this coffee shop and go out? What is the average flow time? Don't tell me that four is too much because we don't have any control on that customers in the second line. It depends on how fast the person who is in charge of exotic coffee is doing his job. It's just possible that we have four over there or 20 over there or two over there or 20 over there. It depends on the speed of the person who serve that part of the system. Again, one person per minute comes in, 60% get black coffee and leave, 40%. Exotic coffee. In the first line, five people in the second line, four people on average. What is the flow time of a person who get a black coffee, exotic coffee, or in general, what is the flow time of a customer? Here is the line for everyone. Throughput is equal to 1/minute, inventory is equal to five, flow time is equal to inventory divided by throughput and that is 5/1, which is equal to five. Five what? Five minutes? Because throughput was in terms of minutes. T_1 = 5. Throughput is equal to 1/minute. But 40% of it goes this way. Therefore, 40% of one is 0.4. 0.4/minute goes through this line, 0.6/minute goes out, 0.41/minute we leave. Therefore, if we look at this system, throughput is 0.4/minute, and inventory is 1,2,3,4. Throughput. 0.4/minute, inventory is equal to four, therefore, flow time is equal to four, divided by 0.4, and that is equal to 10. Therefore, it takes us 10 minutes to pass this line, and five minutes to pass this. People who get black coffee, they will be there for five minutes. People who get exotic coffee five minutes, and they will be over there for 10 minutes, that would be 15 minutes. Therefore, we do have two class of people. One group get black coffee or something close to that and leave in five minutes. The other the staying five minutes in the first line, they go to the second line, and they will also be there for 10 minutes, and therefore it takes them 5+10, 15 minutes to leave. How long does it take a customer to come into this system, get what he or she wants and leave? There are three ways to answer this question. The first 40% are there for 15 minutes and 60% are there for five minutes? That is equal to 6+3 = 9 Therefore, an average customer will be in this store in this coffee shop for nine minutes. It is a mixture of those people who get black coffee and those who get exotic. The second procedure can claim everyone will be there for five minutes. Everyone, no matter he or she gets black coffee or exotic coffee, that person will be there for five minutes. Five is there for everyone. Only 40% of people only 40% of people will be there for 10 more minutes. That is 5+(0.4*10), which is equal to 5+4, and that is nine. There is a third procedure, which does not need to go through who gets black coffee, who gets exotic coffee or so on and so forth. Therefore, if the system is really complicated, we really do not need to go through details. We draw a line around the todo system. Throughput is equal to 1/minute. Any objection? No. Inventory is equal to 5+4, which is nine. Throughput times flow time is equal to inventory. That means one times flow time is equal to nine, and therefore flow time is equal to nine. Three different ways to compute the flow time of a flow unit, a prototype flow unit inside this system. Now, let's go back and look at the three containers that we have at the beginning of this discussion. Ten comes in and 10 goes out. If I ask you what is the flow time in the system, how long does it take a molecule from this point to this point? Your answer is quite simple. Throughput is equal to 10/hour. Inventory is 100, 120 and 129. Inventory is equal to 129. Therefore, flow time is equal to inventory divided by throughput, and that is 12.9 hours. You can go through detailed computations and find it out how long does it take to go through this pass and how long does it take to go to the other pass and then compute the average. Weighted average, of course. I encourage you to go through all these slides and read them besides watching the lecture. Of course, if just by reading them, you understand the material, then you really don't need to watch the lecture. But I have explained everything in detail in these slides. Now we have a different system. We have this system. Customers coming at the rate of 1.5/minute. We have six people here, and we have three people here. This is the original line, this is the exotic line. One-third of customers go this way, 2/3 go this way. What is the flow time of the customers who get black coffee and leaf? What is the flow time of customers who get exotic coffee and leaf? And what is the flow time of a customer? Flow time here, T_1 is equal to inventory, which is six, divided by 1.5, and that is four minutes. It takes four minutes to go from this point to this point. Those people who get only black coffee, they will be there for four minutes and leave. Four minutes is the flow time of the customers who go only through the first process. For the second process, 1/3 of 1.5 will go through there. One-third of 1.5 is 0.5/minute. Flow comes in at rate of 1.5/minute, then it goes through two branches. One-third of it, which is 0.5/minute, goes through this branch, 2/3 of it, which is 1/minute, goes through the other branch. Here, throughput is 0.5/minute and inventory is equal to three. Flow time is equal to 3/0.5, and that is equal to six. It takes six minutes to go through this line. Now, how long does it take to pass through this system? Tutored will be there for four minutes and 1/3 will be there for 6+4, which is 10. That is 8/3+10/3, which is six. On average, a prototype customer will be there for six minutes. Those people who get black coffee and leave will be there for four minutes. Those who get exotic coffee will be there 6+4 minutes, which is 10 minutes. A different way to get the same results. Everyone should go through the first line, so it takes four minutes. One-third of people go through the second line. That is 1/3 and second line takes six minutes. The other 2/3, we'll go through this line, which takes zero minutes. That is why we even don't need to write it. 6*1/3 is 2+4 is six. Finally, we may also say, this is the total system. Nine people are there. That is inventory, and throughput is equal to 1.5 throughput times flow time is equal to inventory, flow time is equal to nine, divided by 1.5, which is six. I have put very detailed explanations of what I did on these pages. You may go through them. Thank you very much for paying attention to this talk. I will put a set of examples, a set of simple examples to help you to feel the nature of the Little's Law. Please watch them too, and then I will go through a little bit other problems.