Speaker 1:	Thank you very much for attending this session. Today, we will talk about Reorder Point ROP. If your brain is not fresh regarding normal distribution, it will not be a bad idea if you watch this 8 minutes lecture on normal distribution and see different problems in normal distribution. We have already talked about the characteristics of forecasting techniques. All forecasting techniques have four characteristics. Long-range forecasts are less accurate than short-range forecasts. Aggregate forecasts are more accurate than individual forecasts. Forecasts are usually and always incorrect. Therefore, forecasts should be accompanied by a measure of forecast error. Usually, as forecasts, we talk about the average of the forecast and the measure of forecast error, we usually talk about standard deviation or standard deviation. Divided by the average, which is called the coefficient of variation. To summarize, forecasts are always incorrect, so they are not constant. We cannot say it's this much and remains at this value. They have an average and a standard deviation. In our inventory model, the assumption was that inventory is known and it is consumed at a constant rate. When I have this much inventory, it's consumed in a linear fashion like this. therefore, if from the time that I can place an order until the time that I receive it, if it is this much, therefore, whenever inventory on hand reaches the volume, which is equal to demand during lead time, we place an order, and we expect to get it at this point, and then we have inventory again, and we consume it again. But we may not consume the inventory like that. It may not be in a linear fashion. It may be something like this. However, at some point, we reach the point when the inventory on hand is equal to the average demand during lead time. If we place an order at that point, at that volume of inventory, then we expect to consume it like this. But we may consume it like this, faster or slower, much faster. Slower, faster, slower, demand will not be consumed like this. Indeed, demand during lead time has a normal distribution where the average is average demand during lead time, and then it also has a standard deviation. If we place an order when inventory and hand are equal to average demand during lead time, it is possible we encounter a shortage here or a shortage here or a shortage here. we may have extra inventory at this point. Therefore, we place an order when inventory on hand is equal to average demand during lead time plus something. That something we refer to it as safety stock, I safety. Therefore, the order point is a point when inventory on hand, how much inventory we do have in our storage is equal to average demand during lead time plus an extra cushion, which we call safety stock. Let's start with problem A. Suppose in the later field technology game 2, the average demand during lead time from the time that we place an order until the time that we receive it is 4 units. The standard deviation of demand during lead time is 6 units. Compute reorder point when we want to have a service level of 95% and risk of 5%. What do we mean? If I have this much, we want this probability to be 95%. this probability of being out of stock to be 5%. This is average demand during lead time, probability of no stockout if we have this much inventory. that is also called service level. This is safety stock, which we will put on top of the average demand during lead time, and this is the reorder point. And that probability is risk or probability of stockout. Demand during lead time is 40, the standard deviation of demand during lead time is 6, and we want to have a service level of 95%. If we place an order when the inventory on hand is equal to 40, there is a 50% probability that we can satisfy the demand, and a 50% probability that we will have a stockout. We have a shortage. We are out of stock. over here, the service level is 50 and the risk is 50. But we want a service level to be 95 and probability of stock out to be 5%. Therefore, we want this area to be 95% and this area to be 5%. That is quite simple. We have a function, norm inverse, where we give probability to that function, and it will give us this value, the stock that you should have. Reorder point is equal to norm inverse, probability is 95%. Average is 40, and standard deviation is 6. It will give us 49.87. Reorder point is equal to average demand during lead time plus safety stock. These are notations that we may use. For safety stock, we may use Isafety, or just IS. Safety stock is the gap between the er point and average lead time demand. That is 9.87. By adding 9.87 to 40, we were able to reduce the risk from around 50% to say 5%. We can round it to 50 and for 50, we reach a service level slightly more than 95%. If you put it into the normal function, the normal distribution, 50, average 40, standard deviation 6, and this one, we always put it over there if you want to find probability, and then it will give us this probability, which is 95.22. It is 22.22% larger than 95% because we increase this 49.87 to 50. If you are comfortable with these two normal distribution functions, we are fine. Otherwise, it is better if you look at that 8-minute lecture, which I have on normal distribution. If you are fine with this first problem, then I will discuss 3 other problems which differ from this problem, but we transform them into this problem, and then just follow the solution procedure that we have. We refer them to problem B, problem C, and problem D, and we compare them with problem A. Look at problem B. In problem B, we do have demand, but not during the lead time per day. The standard deviation of demand is not during the lead time, but per day. Instead of having a normal distribution with its average lead time demand and its standard deviation, standard deviation of lead time demand, we have a different curve. Average is daily demand and its standard deviation is the standard deviation of daily demand. Then we also have lead time, which is four days here. We want to transform this problem into this problem, and we already know how to solve this problem. Let's go through the transformation process. Lead time is fixed, and it is L. Demand per day is a random variable, and it has an average of R and standard deviation of Sigma R. Here, lead time is fixed. Demand per day is variable, and we do have its average and its standard deviation. Demand during lead time has an average of lead time demand and the standard deviation of sigma lead time demand. We want to translate the summation of these two into this. In this situation, demand during lead time is equal to lead time times average daily demand. This part is quite simple to compute. Daily demand is 10, and lead time is 4. We multiply 10 by 4, and that is 40. Therefore, lead time demand is 40. This part was translated into the previous problem. What about this one? This one is also easy. Standard deviation of lead time demand is equal to square root of lead time times standard deviation of daily demand, as simple as that. This is what we have, and we can mathematically prove it. But you don't need to be worried about the proof. Just remember that when lead time is fixed, and demand is variable, then the standard deviation of lead time demand is equal to the square root of lead time. Do not forget square root times standard deviation of daily demand. Therefore, that is also easy to compute. I have a square root of 4, and then the standard deviation of daily demand is 3 times 3, and that would be 2 times 3, which is 6. If demand is variable and lead time is fixed, L is lead time. In this situation is 4 days. Demand per day, we show it in bold R. Average demand per day is 10. The standard deviation of daily demand is 3. Lead time demand, average demand during lead time is equal to 4 times 10, which is 40, and the standard deviation of lead time demand is equal to the square root of lead time times standard deviation of daily demand, and it is 6. We had a problem like this, we translate it into a problem like this, and this is the first problem that we have already solved. we just follow the procedure we used to solve that problem. But just in order to not repeat the same problem because these data were coming from the same problem instead of lead time of four and service level of 95%. We try service level of 90% and lead time of nine days just to have another practice problem. If you want a solution to this problem, we already have it on page 8. As I explained, to have more practice, we solve it for these assumptions. Assume lead time is nine days, average daily demand, 10, standard deviation of daily demand, three. Lead time demand is nine multiplied by ten, and nine. Standard deviation of lead time demand is square root of lead time, which is square root of nine, which is three multiplied by standard deviation of daily demand, which is three. So now we are working with a standard normal distribution like this. Average is 90 and standard deviation is nine, we are looking for such a point where this area is 10%, and this area is 90%. That is straightforward implementation of normal distribution. 101.53 is our reorder point. The difference with average demand during lead time is safety stock. We usually round up ROP or safety stock, therefore, we may round it up to 12. ROP will be 102. When it is 102, it leads to a service level slightly greater than 90%. Yes, service level is 90.88. I did a bit more than 90% because we rounded this ROP from 101.50 something to 102. Now let's look into problem C. We have the same problem, but this time, demand per day is 10 units, and lead time is nine days and a standard deviation of two days. So here, demand is fixed and lead time is variable. How do we find which one is fixed? Which one is variable? The one which has standard deviation is variable. Here, it says, and lead time is nine days, and the standard deviation of lead time is two days. That nine days is average for lead time. Then standard deviation of two days. Formula for this problem is a little bit different than the other problem. In this problem, there is no square root. Again, we are trying to translate this problem into problem A, and as soon as we translate it into problem A, then we already know how to solve it. Demand per day is fixed. Lead time is variable. It has an average and a standard deviation. We need to take this one and this one and use this to to translate the problem into this. Average demand during lead time, as usual, is average lead time times demand or average demand time lead time. The standard deviation of lead time demand does not have any square root. It is simply standard deviation of lead time times daily demand. Standard deviation of lead time demand. In the previous situation, where lead time was fixed was square root of lead time multiplied by standard deviation of daily demand. Standard deviation of lead time demand. The current situation doesn't have any square root. Because R is fixed and L is variable, here I have standard deviation of L, and here I do have R. This is the difference between these two formula. As soon as we write these two formulas on a piece of paper, the rest is straightforward. Lead time is variable and demand is fixed. Bold L is used to show lead time. Regular L, capital L is used as average lead time. In this case, it is nine days. Demand is fixed and it is ten per day. Lead time demand is multiplication of these two numbers, which is nine days. Standard deviation of lead time demand is equal to daily demand multiplied by standard deviation of lead time. That is 10 multiplied by two, which is 20. Therefore, we just implement the normal inverse function and we will get the order point has 116 rounded up. If we subtract the order point from average demand during lead time, we get safety stock. Let me show you what is the difference between problem B in which demand daily demand was variable or weekly demand was variable or hourly demand was variable. Demand is variable, but lead time is fixed, and problem C when lead time was variable and demand was fixed. This is the difference. Click on this. Look, these are different situations for problem B and problem C. Problem B is the blue one. Lead time is fixed. For example, here is five days, we always have five columns, but their heights are different. Problem C is the red one. We may have 1, 2, 3, 4, 5, different reds, but their heights are always equal. The reason for having two different formulas is this. This situation is summation of summation of L random variables. L random variables, in this case, five random variables are added to each other, and then they form the demand during lead time. This one is not L random variables. It is just one random variable. Then random variable is lead time, which is multiplied by a constant, and that constant is constant demand per day or per period. A random variable, L is multiplied by a constant, which is R, and that forms the demand during lead time. In this situation, L random variables are added to each other's summation of L random variables. A random variable multiplied by a constant. That is why the two formulas are different. Now, let's go to case D where both the demand and lead time are variable. Specifically in this case, average demand per day is 300. Standard deviation of demand per day is 100. Lead time is ten days and standard deviation of lead time is three days, so both demand per day and lead time or random variables. We remember the case when lead time was fixed and demand per day was a random variable, and this was the formula for average and for standard deviation. If we square the standard deviation, we get variance, and this is the value for the situation when demand was fixed and lead time was random variable. The demand during the lead time was the same for both situation, computation was the same, standard deviation was computed in a different way. If we squared it and get variance, we have the formula here. Now, for the case, D, when both situations happen together, still, the average demand during lead time is computed the same. It is average lead time multiplied by average demand. Variance is summation of these two variances. We take this one, put it over there, put it plus sign, and put it over there, and then we compute variance of the demand during lead time. This one plus this. Now we can compute square root of the variance and get the standard deviation, and that's it. Now we have a formula to compute average demand during lead time and a formula to compute standard deviation of demand during lead time, and therefore, we can transform problem D into problem A, and we already know how to solve problem A. Demand during lead time, lead time is 10 days, demand per day, 300. If we multiply these two numbers, we get 3,000 units. Then we get lead time and multiplied by standard deviation of daily demand to the power of two. That is ten multiplied by 100^2 plus daily demand, which is 300^2, multiplied by a standard deviation of lead time square, which is 3^2. This is 10,000 multiplied by 10 is 100,000. This is 300 times three, which is 900. 900^2 is equal to 81 and four zeros in front of it. You can compute it in any way you like. If you add these two numbers together, we have 910,000, but it's under square root. Its square root is 954. Now we have the average, which is 3,000 and we have standard deviation, which is 954, 3000. And we are looking for this point to warranty 95% service level. Implement the normal distribution function, normal inverse, and we get 4,569.2, we round it up. We get 4570. This point is four, five, 70. That is reorder point. If I subtract demand during lead time, it will give me the safety stock, and safety stock is equal to 1,570. We covered four types of problems in reorder point where we had demand and standard deviation of demand during lead time. Then we had three different other problems. In one of them, demand was variable, lead time was fixed. In another, lead time was variable, demand was fixed, and in the third one, both of them were variable. We were able to translate all these three problems into problem A, and we know how to solve problem A. A little bit more practice, let's see if we can solve this problem. Please read it and see how do you feel. Suppose we are in Game-II, where we need to handle EOQ and ROP. The average lead time demand is 140 units. For example, you have used formulas that we discussed, and you have come out with this average demand during lead time. Standard deviation of demand during lead time is 40, so we are in problem A, or if you are in any of the three problems, we have already transformed them into problem A. Each time inventory reaches this value, we order this much. Suppose holding cost, H, is equal to $60 per year, and that is what the holding costs in the game is. We have these parameters, we have lead time demand, 140, standard deviation of lead time demand, reorder point, annual ordering costs per unit of product, demand per day. Because it says 300 units is the demand or supply of 15 days, demand per day or supply per day is 20. Each time we order 15 days of supply or 15 days of demand, which is 300. We have all this information. Compute the service level. I have ROP, I have average lead time demand, and I have a standard deviation of lead time demand. A piece of cake, just implement the SL formula. If you put this one here and this one here, and this one here, and as usual, we put one over there to compute probabilities. We only put zero over there if you want to draw a normal curve, which looks neat, not like what I draw by hand here. Other than that, this last parameter is always one, and then we got service level 88%, and risk is 12%. Compute the cycle inventory. Cycle inventory is always half of what we order. We order throughout the cycle, it goes to zero. If we don't have safety stock, average inventory and cycle inventory are equal. If we have safety stock, we should add it to the cycle inventory to compute the average inventory. Each time you order 300, the cycle inventory is half of it, which is 150. Compute the average inventory. Average inventory is cycle inventory plus safety stock. Let's see how much safety stock is here. The order point is 187, the average demand during lead time is 140; therefore, safety stock is 47. Unlike cycle inventory, which is half of what we order, for safety stock, we should put it over there as it is. We shouldn't divide it by two. I try to explain it here using this graph. Look, suppose this is how we have consumed the product, and we have reached to this point. This height is demand during lead time. If demand is over there and this is lead time, if we consume at a constant rate, that's how it is consumed. But we also need safety stock to make sure that if demand is higher than what we have thought, we do not see a shortage here. Here we see extra inventory, but that is fine. Usually, the cost of understock is higher than the cost of overstock. That is why the service level of 50% is not good. We want to increase that service level to above 50 to close to 90, 95, 99 and reduce the risk to 10%, 5%, 1%, and so on. In this case, at the end of the period, safety stock is not there. In this green case, we have safety stock, and we have even more than safety stock. In this case, we have more than safety stock. In this case, consume safety stock, more than safety stock, consumed safety stock, fully consumed safety stock, more than safety stock, consume safety stock. Sometimes, safety stock is fully there, and we have something even more than safety stock, and sometimes we have consumed safety stock fully or partially. Therefore, we have a normal distribution here. Sometimes we have less than average, sometimes we have average. On average, we have average, and average is that safety stock. Therefore, average inventory is cycle inventory, which is Q/2 plus the entire safety stock, not a portion of it. At the order point, we place an order for 300 units. Cycle inventory or Icycle is 150, Isafety is 47. Therefore, the average inventory is Icycle + Isafety, which is 197. On average, we have this much inventory. Carrying cost of one unit per year, in the game, is $60. Therefore, 197, 60 is total inventory holding cost per year. Compute the average flow time, average inventory here, and throughput is 300/15, which is 20 per day, so throughput is 20 per day, average inventory is 197, and therefore, flow time is 9.85 days. Let's go through this last problem and complete our discussion on a reorder point. We order four weeks of supply when the stock, the inventory on hand, reaches 525. Demand per week has an average of 400 and a standard deviation of 125. Compute the risk of stockout. We have a situation like this: average 400, standard deviation 125, and then whenever inventory reaches 525, we place an order. This is service level, and this is risk. To compute the service level, we just put these numbers into a normal distribution function. Norm inverse, 525, 400, 125, 1, will get us 84%. If we subtract it, the remainder, which is around 16% is risk. Compute Isafety for 80%, 90%, 95%, and 99% service levels. The purpose of this part is to show that as the service level gets better, the cost also goes higher because safety stock goes higher. Therefore, the purpose of this part is to create an efficient frontier for the trade-off between service level and cost. Companies compete with each other in four dimensions. Cost, quality, time, and variety. Service is a part of quality. Here, we want to recognize the relationship between cost and quality. Lead time demand is 400, standard deviation of lead time demand is 125. If the service level is 0.8, we use this formula, and we compute ROP. If it is 0.9, we put 0.9 in this formula and compute ROP. 0.95, we put it into this formula here and compute 606. We put 0.99 here, we put average here, which is 400, and we put standard deviation here, and then we compute ROP. That's quite straightforward. If we subtract this column from this column, we get safety stock, Isafety. These are safety stocks, and these are service levels. Here we are talking about service levels is 80%, 90%, 95%, 99%. If you assume this is 100, the service level is 100%. Going from 80%, we go to 90%, that is going from 100% to 113%. When we go from 80% to 99%, if 80% is considered 100%, 99% is considered, 124%. We can show these percentages' improvement in service level, which is a measure of quality here. But look at the other dimension. This is a cost dimension, and we measure it in terms of safety stock. At 80% service level, we need 105 safety stock. Let's compute it as 100% as a base. When the service level goes to 90%, safety stock required is 160; 160 compared to 100 is 152%. When the service level goes to 99, the required safety stock for that is 291 because ROP goes up to 691. This is 291 compared to 105 is 276%. By looking at this column, I can identify quality, and by looking at this column, I can identify cost. This is the efficient frontier to trade off cost and quality. Thank you very much for attending this session.