>> Today I'm going to teach when to order. By when to order, I do not mean time in terms of quantity. At which level of inventory should I place an order to be able to satisfy the demand with a good probability? This is called perpetual inventory system, when time of order is defined in terms of quantity. We also have periodical inventory system when at the beginning of the week, at the beginning of the month, at the beginning of the season, you place an order. It is defined in terms of time but quantity is valuable. In this situation, quantity is fixed. Quantity that we ordered is fixed, and quantity at which we place an order is fixed. We first need to learn or refresh our mind about normal distribution. This is how I refer to a normal distribution. Variable X has a normal distribution with the mean of mu and the standard deviation of sigma, and we all know that sigma is a coefficient to represent variability, but a better form of it is when we divide sigma by mu and define something called coefficient of variations. Mu can be positive, negative, but sigma is always positive. You may watch this lecture on YouTube for an introduction to normal distribution, but here I assume you know almost nothing about normal distribution and very briefly I explain almost everything that you need for the order. The first thing we need to pay attention is normal distribution is frequently seen in nature and in industry, weight of people, test scores, life of a light bulb, number of votes in an election, are examples of normal distribution. Let me explain three basic distributions, uniform, normal, and exponential. This is uniform distribution, normal distribution, and exponential distribution. All three distributions are continuous probability distribution. That means probability of a random variable to be equal to a specific value is zero. Probability of a random variable to be equal to a specific value, anything, integer, decimal point, or whatever, probability of being equal to that is equal to zero. Probability is not defined for a single value in continuous probability distribution. It is always defined for the range, for the range. Look, the width of the red part is constant in all three distributions, but depending on when that distance falls, the area of the red part of the shape differs. In uniform distribution, it really doesn't matter. Wherever it goes, the area is the same. The probability is uniformly distributed everywhere. Uniform probability distribution usually has an upper bound and a lower bound. It start from a point and ends at the point. Along these two points, if the width of the red shape is constant, then the area is constant, but down here in normal distribution, when we get closer to average, the area of the red shape is larger compared to when we go to the two tails. Normal distribution starts from negative infinity and goes to positive infinity. It can be any number. Exponential distribution, the third one, start from zero and goes to infinity. The area of the red shape gets larger when we get closer to small values, and when we go to larger values, it is quite small again. Probabilities in uniform are similar everywhere. In normal, are higher when we get closer to average. In exponential, is higher when we get closer to small values. In all three distribution, probability of X being equal to a specific value is zero. Why? Because these are continuous random variables. Any distance that you imagine, no matter how small it is, for example from 1 to 2, between 1 to 2, we have millions and millions and billions and billions of numbers. The whole probability is 1. 1 divided by an extremely, extremely, extremely large number, all divided by infinity is 0. So we never talk about probability of a random variable to be equal to a specific value in continuous probability distribution, we always talk about the probability of a random variable to be between two values, and that is defined by the area, but that area is computed using different formulas in these three different distributions. We first discuss normal distribution. We need to learn two Excel functions. Given X, if I have the value of the random variable, for example I have this value and I want to find it out, what is the probability of my random variable to be less than or equal to this value? We said to be equal is 0 but we always define it as less than or equal. If I have X, and I am going to find the probability of my random variable to be less than or equal to X, I use this Excel formula and I can compute probability. I use the function NORM.DIST and then here I enter X. Here I enter average of the normal distribution. Here I enter standard deviation, and here I enter 1. I enter 1 for probability and I enter 0 for height. I only need height if I want to draw to the curve to get a neat, normal curve. Nine-nine percent of time, we place 1 over there because we are talking about probabilities. The second situation is when I do have the probability. I do have this probability and I want to find this X value. What is the X value for which the probability of being less than or equal to X is equal to this? In that case, this is what am I looking for and this is the Excel formula, X is equal to NORM dot inverse, probability, average, and standard deviation. Read this problem. Don't move forward. Stop my lecture and try to answer this question. So I have the value of 9. What is this probability? We need NORM distribution function. X is 9. Mu is 10. Standard deviation is 1.5, and 1. We place them over there. What about this one? Here is at least, so it is 11 and more. Unfortunately, Excel cannot give us 11 and more, but it can give us 11 and less, but we know that probability of everything is equal to 1, therefore, if I find 11 and less, which could be this probability, then if I subtract it from 1, that would be 11 or more. I am looking for the red part. I can get the blue part from Excel, which is this one. Tells me, give me 11 and give me 10 which is the average and give me 1.5 which is standard deviation and 1. 
^M00:10:00
Then I will give this area, but the whole area is equal to 1. Therefore if you subtract it from 1, you get your answer. Now let us solve this problem. What is the probability that the battery life is between 11 and 13 hours? So that is 11 and that is 13. Both of them are on the right-hand side of the average. It really doesn't matter. We just compute the probabilities. In order to compute this probability, we first compute the probability of less than or equal to 13. That is just a simple formula in Excel. X is equal to 13. Average and standard deviation are 10 and 1.5, and 1 is for probability. I compute that value, then I go and compute less than or equal to 11, which is over there. Instead of 13, I put 11 in the formula and I compute it. Now I have less than or equal to 11 and less than or equal to 13. All I need to do is to take this part out of this part and the rest will be this part, and that is the rest. What about this? Exactly the same thing, just numbers are changed. I go and compute this part, then I compute this part, and then I subtract them from each others and this part, as simple as that. The same formula, X, mu, sigma, and 1, 11.5, and the rest are the same. That is the probability. 9.5 and the rest are the same. Two probabilities, subtract, that is the answer. Exactly similar problem. Just go ahead and solve it in 30 second. Open an Excel sheet. 10 average, 1.5 standard deviation, 8, 6, equal to NORM. I really don't need to remember everything. NORM distribution, that is this one. What is your X? My X is this. What is your mean? That is my mean, and lock it. In Mac, for lock it you need to push command-T. That will lock it. Standard deviation, F4 and lock it, and 1, and I get the probability. Then just copy it down. 8 will become 6 while the other two, which are absolute, referencing remain the same and then you go here and say 8 minus 6 probability, 8%, 8.7%, and that is what we have. What about this one? A little bit different. Here we don't have X. We have probability. We have this probability which is 15% and we want to find this point here. We will use a function NORM inverse and then enter this probability here and the rest is the same, and it doesn't need 1 or 0, so we enter probability, then mean, then the standard deviation, then it will tell us this point is 8.45. The bottom 15% of the battery lives is at most 8.45. Now look at this one, top 10%. Can you compute it? I know this is 10% and I want to know where is this point? The good news is if on the right-hand side I have 10%, on the left-hand side I have 90% Therefore I can go to Excel. I can enter 90% and I can find this point, because 90% on the left, 10% on the right. Therefore I go to Excel. Instead of 10%, I enter 1 minus 10%, average, standard deviation, and I get this number. Honestly, this is all you need to know about normal distribution, as long as our course is concerned. Now let's go to the order point. Look, when I had a lecture on forecasting, I talked about some features of all forecasting techniques. Forecasts are usually or always inaccurate. You can never think that what you forecast is exactly what will happen and that is why whenever we forecast, we say this is our average forecast. We don't know what the actual demand would be or what the actual that variable will come out. This is our average forecast and we need to always accompany our average forecast with the standard deviational forecast. The degree of variability which is associated with that average. The third fact about forecasting is aggregate forecast are more accurate than individual forecasts. For example, forecasts for Ford Motor cars is more accurate than forecast for Mustang. The variability or coefficient of variation for all Ford products is smaller than for Mustang, and then for all cars of all manufacturers, variability is less than that of Ford Motor alone, and forecast for next year economy is more accurate, less variability, less coefficient of variation, compared to forecast for car industry, and forecast for summation of 10 days' demand is more accurate than forecast for one day demand. The variability for 10 days is less than variability in one day. Up and downs cross each others out in people's language. In mathematical language, it is the variance of those 10 random variable which are add up, so if I have 10 random variables, each one has a standard deviation. When I add them up, it is not their standard deviations which is added up. It is their variances, so it'd be 10 times the variance of one of them, and then when we translate variance into standard deviation, we should put it under square root and then 10 goes under the square root. Therefore, standard deviation of summation of 10 random variables is not 10 times the standard deviation of one of them, but 10 under square root which is much less, and the long-range forecasts are less accurate that short-term forecasts. The more we move into the future, our knowledge about the future will get less accurate and then variability goes up. Indeed, if we look at the time from now until when we need that forecast as a lead time, standard deviation of forecast for L time units away from now, that doesn't go under square root. Why doesn't go? Because in the previous situation it was L random variables which were added to each other, but here there is one random variable which is multiplied by something. Statistically, we can prove that when we add several random variables the variances get added to each others and therefore the summation or the number, which is in this example was 10, will go under the square root, but when we talk about time from now to hundred days from now, an average hundred days which it could be 50 days or 150 days, the standard deviation of this lead time will be multiplied by demand per day to come out with the total standard deviation of that time period. Now let's look at reorder point problem. We do have two types of ordering, periodic inventory system or periodic ordering, and perpetual ordering. In periodic ordering, we order at the beginning of the period, beginning of the week, beginning of the month, beginning of the season, so reorder point, the point at which we order, is defined in terms of time. But we also have perpetual inventory system in which reorder point, the point at which we order, is defined not in terms of time but in terms of how much inventory do we have on hand? 
^M00:20:05
Here we discuss perpetual inventory system. So we have an inventory and it gradually goes down. We have this much inventory and each day we consume some of it, down, down, down and at each of these point, this is what we have on hand. This is what we have on hand. This is what we have on hand. Goes down, down, down, down, down, down, down and we have consumed all of our inventory and we have nothing here. So suppose this is our lead time. For example, this is two days, three days, four days, whatever. Our y axis was quantity or inventory, and our x axis is time. Suppose if I place an order here, I will get the order over there, and because this is how my inventory is consumed, if I need the order over there and I need to place an order here, therefore I should have this much inventory on hand and that is what we call it reorder point. That is lead time from the time that we place an order until the time that we get that line shows our consumption, therefore, my material here, I should place an order here and when I am here, during this time, I will consume this much. Therefore, when inventory on hand is this much, I will place an order, average demand during lead time. But the reality is that inventory is not consumed in a linear fashion. It will go up and down. Demand, every day demand is not equal to the other days. When and where inventory on hand reaches the average of what I need during lead time, I place an order. I have assumed that on average the inventory will go down like this based on the average demand. However, inventory may go down like this because the demand is low and therefore at the time when I get the order, I already have this much inventory on hand. On the other hand, inventory may go down quite fast, faster than the average. In that case, for a period, I will not have inventory. This is what I have on hand. This is time. This is the inventory. It will go like this down, but I estimate that it is going down in a linear fashion. This is my lead time, therefore I'm expecting it to go like that and when the order is received, I have nothing on hand, and this is my reorder point. This is average demand during lead time. And over there in this situation, reorder point is equal to average demand during lead time, but if I place an order over there, I may consume the inventory quite fast. Therefore for a period, I may be out of material. That is a risk, and usually we want to reduce that risk because if I place an order, when inventory on hand is equal to the average demand during lead time, there is 50% probability that the demand during lead time exceeds what I have and there is 50% probability that it is less than what I have, 50% probability of overstock, 50% probability of understock. Cost of understock is usually more than overstock. Therefore, I usually like to reduce the probability of understock. When I place an order, inventory on hand is not equal to the average demand during the lead time. It is equal to the average demand during the lead time plus safety stock. I order at the higher level. If I order at the higher level, probability of overstock is more than 50% and probability of understock is less than 50%. The probability of understock is referred to as risk. The risk of the probability is referred to as service level. When I place an order, I do know the demand during lead time based on its average and its standard deviation, but I don't know exactly what it is. It will have a normal distribution. If I place an order when inventory on hand is equal to average demand, 50/50, 50% probability of overstock, 50% probability of risk, understock, 50% service level, 50% risk, but if I go ahead and add something to it as safety stock and place an order when inventory on hand is this much, not this much, in that case, service level is more than 50% and risk is less than 50%. There's a probability that during lead time, demand is quite high or quite low, high or low, high or low. Safety stock makes it possible that probability of understock to be less than probability of overstock, risk to be much less than service level, usually, with 90% probability, we have the material, with 10% we don't have. Now let's try to solve this problem based on the knowledge we have about normal distribution. We know that from the time that we place an order until the time that we get it, the demand has an average of 200 and standard deviation of 25. Now if we place an order when we have 200, there is 50% risk and 50% service level, but we like to increase this green area and we want to reduce the red part such that service level is much higher than risk. How do we do that? By adding safety stock to the average demand during lead time such that this probability is something that we like, and this probability is something that we like, but don't forget summation of red and green is always 1. We want to reduce red in favor of green. Average demand during lead time, service level we want it more than 50%. Therefore, we add safety stock to average demand during lead time, and that is R, reorder point, and this small probability is risk. We want to have 90% service level, 10% risk, average demand 200, standard deviation 25. If you place an order when inventory on hand is 200, 50% red, 50% blue. We don't want it. We want this area, service level, to be larger, this area, risk, to be smaller. Service level in this case is 50%. We want a situation like this. Here is risk. Here is service level. We go to Excel. We type NORM inverse. 90% is service level, this probability. 200 is average, and 25 is the standard deviation. 
^M00:30:02
We get 232 and that is our reorder point. 232 is about 32 units more than 200 and that is what we refer to it as Isafety, safety stock. If we place an order when inventory on hand is 232, 233, service level goes up from 50% to 90% and risk, probability of stockout comes down from 50% to 10%. By adding 32-point-something to our average demand during lead time and place an order when inventory on hand is that much, we increase our service level from 50% to 90% and risk is reduced from 50% to 10%. Now let's see if you can solve this problem. Here we do not have the distribution of demand during lead time. We do have the distribution of demand per day or per period, and we have the lead time in terms of days or periods, days, weeks, month. We need to be able to translate this problem into the previous problem as soon as we can transform it, then we can use the previous problem to solve it. Previous problem, you had demand during lead time. Average and standard deviation here we don't have. Here we have lead time and average and standard deviation of demand per day. If we can translate this problem into the previous problem, we are done. Lead time is fixed. We refer to it as L. Demand per day is a random variable, normally distributed random variable, with average of R and standard deviation of sigma R. The difference is this. In the previous problem we had lead time demand which was normally distributed with average of lead time demand. This one we write it a little bit bold to show that it is a random variable, and this one is average of that random variable and this is standard deviation of that random variable. So if lead time is L and demand per day is R, therefore it is a common sense that average lead time demand is L times R. This is quite simple, common sense, and we can use this formula that standard deviation of demand during lead time is square root of lead time multiplied by standard deviation of demand per day. So we have this formula and we have this formula, and therefore this problem is already translated into the previous problem. If demand is a variable and lead time is fixed, if L is lead time, which in this case is 16 days, if R is demand per day or demand per period, which here is 20 per day, if sigma of R is standard deviation of demand per day or per period, which is 5, so note that here lead time is defined in terms of days. It could have been defined in terms of period, weeks, months. This is because that is this, this is demand per day. If that was week, here I needed demand per week. Because that is days here, I have standard deviation of demand per day. I have all this information. Lead time demand is averaged demand during lead time, it is lead time multiplied by average demand and that is 320. Standard deviation of lead time demand is square root of lead time, square root of 16 multiplied by a standard deviation of demand per day. This 16 is over there. 5 is here. 5 is not under parentheses, so it will become 4 times 5, which is 20, and that's it. Now the problem has been translated into the previous problem. Problem was originally like this. We had demand per day and the standard deviation of demand per day, now we have translated into demand during lead time and the standard deviation of demand during lead time and this is exactly the same as the previous problem and we know how to solve it. Lead time demand is 320. Standard deviation of lead time demand is 20. Service level is 90%. Plug it in into NORM inverse because we do have probability, we do have mean, and we do have standard deviation, and we get reorder point. We need to place an order when the inventory on hand is this much or 346, and the gap between that one and this one is our safety stock. Safety stock is 26. By adding 26 to 320, we go from a 50% risk to 10% risk, so this is the answer to the problem. This is reorder point. This is average demand during the lead time, and this is our safety stock. Now let's go to another problem. Please read this problem and see if you can find the difference between this problem and previous problem. So we had a problem a) in which we had average and the standard deviation of demand during lead time. We had problem b) in which lead time was fixed but we had the average and standard deviation of demand per period, per day, per week and so on, in whatever the lead time is defined in terms of that unit. Now let's see if you can find the difference between problem c) and problem b). In problem b) lead time was fixed, demand per period was variable. In this problem, demand per period is fixed, lead time is variable. In the previous problem, we had several random variables. L are random variables if the lead time is defined is defined as L. We had L random variables which we should add them together. Here we only have one random variable, and that random variable is lead time, and that random variable is multiplied by demand which is fixed. We need to know what is the average demand during lead time and we need to know what is the standard deviation of demand during lead time. If you can answer these questions, then problem c) also is translated into problem a) and we are done. Demand per day is fixed. Lead time is a random variable. It has an average and standard deviation and we want to use this data to compute average and standard deviation of demand during lead time. Average demand during lead time in this case, also the same as previous case, is average lead time multiplied by demand per day. This is our average lead time and this is demand per day and that is average demand during lead time, but standard deviation of demand during lead time, which we again, show it by sigma of LTD, does not have a square root over there. It's simply R multiplied by standard deviation of lead time. Why? Because this problem is one random variable which is multiplied by a constant. Problem c) has one random variable which is lead time and that lead time is multiplied by a constant which is demand per day. The other problem, problem b), had L random variables, which are added together to define the demand during lead time. That is why the previous one has square root, this one does not have the square root, but that is an statistics issue. 
^M00:40:04
We just need to remember these and I will give you these formulas. What we need to know is which formula should we use when demand is a variable and lead time is fixed compared to when lead time is variable and demand is fixed. If lead time is variable and demand is fixed, if L is lead time as a random variable and L, not bold L, is average lead time, which in this case was 16 days, and R is demand per period or per day, which is 20 here, lead time demand, average lead time demand is R times L. R is 20. L is 16. Average lead time demand is 320. Standard deviation of lead time is 4. Standard deviation of lead time demand is R times standard deviation of lead time. Standard deviation of lead time was 4. R was equal to 20 per day. Therefore, lead time demand is 20 times 4, which is 80. Now we do have all the information and problem c) also is translated into problem a). Average lead time demand 320, standard deviation of lead time demand 80, service level 90, plug it into the normal distribution, NORM inverse because we have probability. We have average. We have standard deviation. We get X, which is our ROP, reorder point, 422 or 423. If we subtract it from 320, it is 103, which is our safety stock. ROP, lead time demand, safety stock. Now let me once again a little bit explain about the three problems. Part a) or problem a), in that problem we had average demand during the lead time no matter how long that lead time is. We had the average demand during that period, and we also had standard deviation of demand during that period, and we just used our knowledge in normal distribution and we solved the problem. Then we talked about part b) and part c) or problem b) and problem c). In problem b), lead time was fixed. It was for example, 5 days, so lead time is fixed. What is variable? It is demand, demand per period, demand per day. How do we know what is fixed, what is variable by a standard deviation? If in a problem I don't give you standard deviation about something, that means it is fixed. In this specific situation, we give standard deviation with respect to daily demand, and lead time doesn't have standard deviation, so it is fixed. In problem 2, it is lead time which has standard deviation and demand does not have standard deviation. Therefore, in this case, demand per day is fixed and it is lead time which is variable. In both cases, we showed how to transfer the problem into the original problem and then use the knowledge in normal distribution and solved the original problem. In the first problem, or indeed in problem b) here, demand per day is multiplied by lead time and that is average demand during lead time. In the second case also, average of lead time is multiplied by demand per day and it forms the average demand during lead time, but for standard deviation, they are different, where the demand per day is variable, lead time goes under the square root, and then is multiplied by standard deviation of demand. Lead time goes under square root. In part c) or problem c), there is no square root. Demand per day is here. It is simply multiplied by standard deviation of lead time. You may ask why one of them has square root, the other one doesn't have? Because these two problems are very different and I will show it to you. Let me click on these two graphs. Look at the blue graph and red graph, and let me know what do you see? Blue graph is problem b). Red graph is problem c). What is the difference? In problem b) or blue graph, I always have five bars and that is where each of them goes for one of the days, because lead time is five days. Lead time is fixed and I have five random variables. Each time they are different. In this problem, demand during lead time is computed by adding five random variables, or in general, L random variables. The red graph is entirely different. Sometimes we have three bars, sometimes five, sometimes six, sometimes seven, four, five, but the height of the bars are the same. In this case, it is one single random variable, which is lead time, and it is multiplied by demand per day. The blue graph is summation of L random variable. The red graph is one random variable multiplied by a number. That is the difference and statistically, we can prove that the standard deviation are computed differently. Now let's solve this problem. Please read the problem, understand it, clarify what is what, and then we move forward together. Average demand during lead time, lead time demand, is 20,000 units. Standard deviation of lead time demand is 5,000. Whenever inventory reaches 24,000, we place an order, so this 24,000 is our ROP, and we order 14 days of supply. The quantity that we order is equal to 28,000 which is equivalent to 14 days, so if we divide it by 14, we get demand per day to be equal to 28,000 divided by 14. Inventory holding cost, which we show it by H, is 2. Lead time demand 20,000, standard deviation of lead time demand 5,000, reorder point 24,000, carrying cost $2 per unit per year, demand per day 2,000, Q or EOQ 28,000. Compute the service level, so this is like problem a) or part a). We shouldn't be worried about translating the problem to the original problem. It is already the original problem. What is the service level? If this is demand during lead time, I should find what is my average demand during lead time and I should find at which level do I place an order? Then I can go to normal table and compute this area which is the risk or this area which is the service level. Average demand during lead time is 20,000. Standard deviation of demand during lead time is 5,000. Reorder point is when inventory reaches 24,000. Therefore, the problem becomes quite simple. I have this X. I have the average. I have the standard deviation. 
^M00:49:59
I can go ahead and compute this probability and that is my service level. Normal distribution, reorder point, average demand during lead time, and standard deviation, and we want to compute probability. We don't want to throw out the curve, so we put 1, and the probability or the service level is about 79%, and risk is the difference between 100% and this one which is 21%. Therefore, in this situation, probability of serving customer is almost four times of probability of not being able to serve the customer. This is 21%. This is 79%, about 79%. Compute the cycle inventory. Cycle inventory is always defined as half of what we ordered. So in this case, we ordered 28,000 and cycle inventory is half of it, which is 14, and this is the logic. We ordered Q units at the beginning of the period. By the end of the period, it goes to 0 or by the end of the cycle. Therefore, during the cycle, our average inventory is Q plus 0 divided by 2, or simply Q divided by 2. So cycle inventory always half of what we ordered, 28,000 divided by 2, 14,000. Compute the average inventory. Average inventory is defined as cycle inventory plus safety stock. Safety stock is not divided by 2. What we ordered is divided by 2, but safety stock is not, and I will explain. We start consuming what we have until we reach to a point that we realize that what we have on hand is equal to the average demand during lead time. During this period, which is our lead time, we will consume this much product. Gradually, goes to zero, if we consume it based on the average. Because we don't consume it based on average, we place an order when we have something more than average demand during lead time and that is what we call it safety stock. Therefore, our reorder point is demand during lead time plus safety stock. When the inventory on hand is equal to average demand during lead time plus safety stock, we place an order. If we consume it based on the average, it will go to zero here and therefore by the end of the period, still we will have the safety stock, but the whole story of the safety stock is around this fact, that we know the demand is not constant. It goes up and down. For example, after I place an order in this point, I may consume what I have like this. Therefore, when I receive the next order by the end of the lead time, I still have something more than safety stock. I have the safety stock and I have something more than that. However, my demand may be like this, so even before the end of the cycle, I am out of everything. I have fully consumed my safety stock. I have zero inventory. Here I have more than safety stock inventory. Here I have less than safety stock inventory or zero inventory. I may consume it like this. At this point, I receive my order and I have a lot. Inventory on hand by the end of the cycle is much more than safety stock, or it may be consumed like this. By the end of the cycle, I have consumed a part of my safety stock, or it may be l like this. I have more than safety stock. Like this, I have consumed a large portion of my safety stock, or like this or like this. Therefore, by the end of the cycle, sometimes I have the safety stock and something more than safety stock, and sometimes I have consumed safety stock entirely or partially. Therefore, as we discussed before, demand during lead time has normal distribution with its average equal to average demand during the lead time. Therefore, by the end of the cycle, because sometimes I have more than safety stock. Sometimes I consume parts of safety stock, it is quite logical to assume on average at the end of the cycle, safety stock is always there, and that is why when we compute average inventory, we divide what we have ordered by 2 and we call it cycle inventory but we do not divide safety stock. Cycle inventory is equal to what we ordered divided by 2. Average inventory is cycle inventory plus safety stock. At reorder point we ordered 28,000. Cycle inventory is 14,000. Safety stock is 4,000 because we placed the order when we have 24,000, while demand during lead time is 20,000. Therefore, 24,000 minus 20 is 4,000 safety stock, and therefore average inventory is 14,000 plus 4,000, 18,000. Compute the total holding cost per year. Average inventory is 18,000. Holding cost or carrying cost is 2. We multiply them by each other. Average holding cost per year is 36,000. Compute the average flow time. We go back to our Little's law. Average inventory is 18,000. 14 days of supply is 28,000, so 28,000 divided by 14 is equal to 2,000. This is R. This is I. Throughput times flow time is equal to inventory. R is 2,000 per day. I is 18,000. Flow time is 9 days, days because this 2,000 is stated in terms of days. Let's solve this problem. We place an order when inventory on hand is 525. Each week we consume 400. Lead time is 1 week and standard deviation of demand is 125. We have the reorder point. We have average demand during lead time, and we have standard deviation of demand. Put 1, and we get 84%. At the bottom part, I have tried to show an efficient frontier. Efficient frontier is between service and cost. When service level goes up, our service is better, but when service level goes up, safety stock should also go up, and safety stock has a cost because we should keep it throughout the year, and that is safety stock multiplied by holding cost. Therefore, that's a tradeoff, tradeoff between service and cost. Here in this table, I have tried to discuss this issue that when we try to have better service, we have no other choice to have more cost. Click on it, open object. This Excel sheet opens. This is service level. I come here and compute the reorder point, NORM inverse, 80% service level. I have average lead time demand and I do also have standard deviation of lead time demand, so I have those. Therefore, I compute reorder point. I go here, average demand during lead time, standard deviation of demand during lead time, and this is service level which is 90%. The previous one was 80%. And then I copy it down. I have it for everything. If we assume 80% service level is our 100%, then when we go from 80% to 90%, 90% divided by 80% multiply by 100, and that would be 113. 
^M01:00:03
We are going up by 13%. If I copy this down, here we go up by 19%. Here we go up by 24%. So this was 0 is equal to this one minus this one and lock it. 13% increase, 19% increase, and 24% increase, that is how my service increases, but safety stock is the reorder point minus average demand during lead time is equal to reorder point, which is here, minus average demand during lead time. Copy it down. Again here, if you assume this 100, then this would be 100 multiplied by what I'm here. Go up, divided by the original value and lock it, because you want to measure everything compared to the original value. 152, 195, 274, equal to 0, equal to this one minus this one, and lock it, and copy it down. Look at here. I increase my service by 13%. My carrying cost, my safety stock cost went up by 52%. Increased it by 19%, my safety stock cost went up by 95%. I increased it by 24%, cost went up by 176%. This is our efficient frontier. If we want to improve our service level like this from 100 to 105, 10, 15, 20 and 125, our cost will go up by this. 125% -- I mean 25% compared to 100, would leap to about 300%, and indeed, 300 minus 100 which is 200% increase in costs.