Speaker 1:	Let's solve a simple problem about capacity, throughput, cycle time, and tack time. Two stations, it takes 6 minutes in the first cessation to complete the task, and it takes 5 minutes to complete the task in the second station. These are sequential tasks. A product goes to the first cessation, then to the second station. There are three machines, three resource units in Station 1, or we may generally call it resource pool one, and there are two machines, two resource units of a different type than the first three resource units. They are in Station 2. Two resource units in resource Pool 2, pre resource units in resource Pool 1. These are generic names that I like to use. Compute the capacity of this process. One product every 6 minutes. In station what? How many product? In 1 minute. 1*1/6. That is what we usually call it R_P1. That is 1/T_P. In this case, for Station 1, it is 1/6. For Station 2, 5 minutes, therefore, R_P1 in the Station 2 is equal to 1/5. 1/5, 1/6. If we had one machine in each station, obviously, the bottleneck was here. Because this one in each minute it can produce 1/6, and that is in 1/6*60 = 10, and the other one, 1/5 per minute, if I multiply by 60, that would be 12. Therefore, the bottle neck would have been here. But we don't have one machine in each station. We have more than one machine and we need to now consider what is the or capacity of one machine in the Station 1 is 1/6 per minute. But we have three of those machines. That is 1/6+1/6+1/6, and that is 3/6 per minute. In general, it is C/T_P. Therefore, capacity of the first station, is equal to 3/6 per minute. If you multiply it by 60, that would be 30/hour, capacity of the second one, 1/5 per minute, but we have two of them, and an hour is 60 minutes. This would be 24, capacity of 24, capacity of 30, the weakest is the bottleneck and the bottleneck is here. I repeat what I said, but before I repeat, there is also an alternative way to look at it. In the Station 1, we need 6 minutes. In an hour we have 60 minutes. Therefore, if we divide 60/6, capacity of one machine is ten, but we have three machines, therefore, the capacity is a little bit different than what I said. Following what I have already said, again, the capacity story. We can repeat the same logic for Station 2. First from the first perspective, then from the second perspective. It takes 5 minutes and an hour is 60. Therefore, 60/5 = 12, products per hour, and we do have two machines. 2*12 = 24. Or as I have already explained, R_P = c/T_P, 2/5 per minute. If you multiply by 60, that would be 24 and a chain is as strong as its weakest link. Station 2 is the boder line. Compute the cycle time in this process. Cycle time is all capability. How often can we produce a product? What is the time interval between two consecutive product that we can produce. Capacity is 24/hour. Therefore, it takes us 1/24 hour to send the next product or capability to send the next product out everyon over 24 hours. Not over 24 hours, which is a day, 24. That is, if I multiply by 60, that will give me 2.5 minutes. I could have also said as I have said here, an hour is 60 minutes and I can produce 24. Therefore, if you divide that s and the same logic is over there, I can produce 24. In 1 hour, how long does it take to produce one product? That is 1*1/24. These are just simple computations. We said our capacity is 24/hour. We could have said it is 24*60 minutes. What is the capacity in 1 minute? That would be 24/60 = 4/10 = 0.4. If capacity is 0.4/minute, cycle time is 1/0 0.4, and that is 2.5 minutes. I can do this. All of them just different look at the same problem with very simple manipulation. Suppose the utilization of this process is 75%, compute the throughput. A simple formula. Utilization is equal to throughput divided by capacity. Utilization is 0.75. Throughput we don't know, but we know capacity is 24. Therefore, if you do the multiplication, tripled is equal to 18/hour, or it is 18/60 = 3/10 = 0.3/minute. Our capacity is to produce 0.4/minute or 24/hour. But the demand is 0.3 or if I multiply by 60, 18 units per hour. It is 18, capacity 24, utilization, 75%. Compute the utilization of the least utilized station. Utilization of the most utilized station is 0.75 or 75%. Least utilized station, we have two stations. One, capacity of 30, the other one, capacity of 24, and 18/hour pass these stations. Utilization here is equal to 0.75. What is utilization? Here, this is the least utilized station. Least utilized station is a station which its utilization is not greater than the utilization of any other station. Bottleneck is the station or resource full where utilization is not smaller than the utilization of any other station. Here, if I want to compute utilization of Station 1, that would be R, which is 18, divided by the capacity of that station which is 30 and that would be 0.6 or 60%. Compute the tack time. Cycle time is our capability to produce products. The time interval that we need between producing one product and producing the next product. That is cycle time. Tag time is the same thing based on the demand. What demand wants from us? How often demand wants a product? The difference between two consecutive products that the market demands is called tag time. Market1 throughput. Market 1 18/hour. Our capability is 24/hour. That is with respect to throughput and capacity. But with respect to cycle time and tag time, cycle time we just realized. Cycle time is 2.5 minutes. What about tag time? For tag time we follow the same logic. Market 1 is 18/hour. Therefore, in 1/18 hours Market 1 product. If I multiply by 60, that would be every 3.33 minutes. I could have also said, demand is 18/hour. Therefore, it is 0.3/minute in 1 minute, demand is 0.3. In how much time demand become one, and that is 1*1/0.3, and that is 3.3333. Every 2.5 minutes, we can produce one product, we can send out the next product, but demand does not ask for that. It is enough if you can send out one product, every 3.333 minutes. Throughput 18, tag time, 3.33. Capacity 30 cycle time, 2.5. Capacity is always greater than throughput. Cycle time is always less than tact. Because cycle time is one over capacity and tact is one over through. Let me just add one point. We said utilization relatively measures throughput compared to capacity that is throughput divided by capacity, in this case, it is 18/24 or 75%. We also have safety capacity. That is the absolute difference between capacity and throughput R_P minus R. Here I divided them. Here I just subtract them. That is 24-18 = 6. Safety capacity is 6/hour. Utilization is 75%. Thank you again for attending this short session. 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