IOE 419 Mark S. Daskin Service Operations Management IOE Department Winter, 2017 University of Michigan Problem set 4 DUE: MONDAY - February 20, 2017 Points: 100 points total Problem 1: Babette has to determine how many snowplowings she should contract for every October. She estimates that the distribution of the number of plowable snowfalls is given by the values below: Snowfalls 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Prob 0.01 0.03 0.05 0.06 0.08 0.1 0.12 0.12 0.1 0.1 0.08 0.05 0.04 0.03 0.02 0.01 She can contract for each snowplowing at a cost of $70 per plowing. If the number of actual snowplowings needed is less than the number she contracts for, she loses the money she paid. In other words, if she pays $700 for 10 snowplowings and there are only 6 plowable snowfalls, she does not get the remaining $280 back; she loses it. If the number of plowable snowfalls exceeds the number she contracted for, she has to pay $100 per extra snowfall. In other words, if she contracts for 10 and there are actually 12, her cost for the season is $700 (for the contract) plus $200 (for the two extra snowfalls above the contracted number). a) Find the optimal number of snowplowings she should contract for. b) What is the expected annual cost of this service? c) What is the expected number of snowfalls in excess of the number she contracts for? d) What is the expected number of snowplowings that she loses or that she pays for but does not need? 1 IOE 419 Mark S. Daskin Service Operations Management IOE Department Winter, 2017 University of Michigan Problem 2: Each November (roughly speaking) you will have to determine how much of your salary to set aside in a Flexible Spending Account if you work at a company that offers this benefit. See section 5.6.2. You estimate that the distribution of qualified healthcare expenditures for the coming year is given by the following distribution: f X ( x ) = e x where = 0.0025 . Your marginal tax rate is 25%. a) What is the expected value of the qualified healthcare expenditures? b) How much should you set aside for the FSA? Give the value to the nearest dollar. c) If you can only set aside a multiple of $5 each month, how much should you set aside each year? d) What is the expected value of your healthcare expenditures over and above what you have set aside in the FSA in part (c)? e) What is the expected value of the amount that you have left in the FSA account at the end of the year if you use the value you found in part (c)? Problem 3: (This question relates to section 5.5 which we will NOT explicitly cover in class. You will need to read that section on your own.) Under some mild assumptions, the demand during a lead time approximately follows a Normal distribution. For example, if the true demand during a day follows a Poisson distribution with a mean of per day and the lead time is fixed at days and > 30 , the demand during a lead time is approximately Normal with a mean of and a variance of . You are the owner of a chain of brick and mortar camera stores, ShutterPix Photo. You have 100 stores nationwide. Each store sells an average of 3 Canon PowerShot SX 720 HS cameras (retail price of $325) per day. The lead time for replenishing your stock is 5 days. (Note that = 15 in this case and so the Normal approximation does not work well.) a) If you want the probability of stockout to be less than 0.025 at any store, how large should the safety stock be at each store? b) How many units of safety stock should you carry nationwide? c) If the cost of carrying a unit in inventory is $50 per year (due to obsolescence), what is your annual cost of safety stock, nationwide? d) The bad news is that you compete with a large online photo company that sells the same camera, but that ships cameras from a central warehouse nationally. The company (we will call them Gargantuan) also wants to be sure that it can satisfy 97.5% of its customers during a lead time (also of 5 DAYS). It sees a daily demand of 300 cameras (the same as your nationwide daily demand). How much safety stock does Gargantuan need and how much does this cost? 2 IOE 419 Mark S. Daskin Service Operations Management IOE Department Winter, 2017 University of Michigan e) Now, just imagine this issue played out over not just one SKU (stock keeping unit) but over 100 model types and manufacturers. Does this go part of the way toward explaining why many local photography shops are now out of business? Briefly explain your answer. f) What actions could you as the owner of ShutterPix take to mitigate the advantage that Gargantuan has and thereby reduce the required amount of safety stock that you carry? Problem 4: You want to minimize the total annual cost of ordering inventory and holding inventory. Monthly demand varies significantly as shown below. The fixed order cost is $800 per order and the inventory holding cost per item per month is $2. Assuming you have to place an order in January, a) What is the optimal ordering policy? When do you order and how much do you order at each of those times? (Assume you order at the beginning of a month for at least that month and possibly for additional months as well.) b) Now find the optimal total cost for every possible number of orders per year (from 1 through 12). This will involve adding a new constraint to your model. Plot the total cost versus the number of orders per year. c) For what values of the number of orders per year is the total cost within 10 percent of the optimal total cost? Now relax the assumption that you have to place an order in January. d) Now what is the optimal ordering policy (without any constraints on the number of orders per year)? How much better is this - in absolute dollars and in percentage terms - than is the policy you found in part (a) above? 3 5 INVENTORY DECISIONS IN SERVICES Brian waited patiently as the shoe sales clerk went into the back room to find the shoes he wanted. The sales clerk returned 5 minutes later with three boxes of shoes. "Good," Brian thought, "He found just what I wanted." "I'm afraid we do not have your size 9 in stock, but I brought out an 8j and a9\\- Maybe one of them will fit. Also, we have the black in your size. Are you sure you really need another pair of brown shoes?" the sales clerk said as he plopped the three boxes at Brian's feet. "Maybe the store at the other end of the mall has my size," Brian suggested as he stood up to leave. 5.1 WHY IS INVENTORY IN A SERVICE MODELING BOOK? Inventory is generally not considered a part of service operations. In fact, many authors assert that one of the key differentiators between the service and manufacturing sectors of the economy is that goods can be inventoried, and services cannot be carried in inventory. If a flight leaves from Chicago to New York on Service Science, by Mark S. Daskin Copyright 2010 by John Wiley & Sons, Inc 285 286 INVENTORY DECISIONS IN SERVICES Monday at 9:00 A.M. with 38 empty seats, the airline cannot put those empty seats into an inventory bank for later sales. The potential revenue from those seats is simply lost. If a lawyer in private practice elects to go on vacation for 2 weeks at the end of August, she cannot put that time into inventory for later billing to clients. The time is simply lost. However, a basic understanding of inventory theory and the modeling of inventory decisions is essential to understanding the service industries. To see why this is the case, let us first consider a local shoe store. If the owner of the store expects to sell shoes, she must maintain an inventory of shoes for sale. She must balance the cost of holding inventory against the cost of lost sales that result when a customer wants to buy a pair of shoes but the store does not have the style, color, or size needed by the customer. In addition, the owner does not want to have to pay for too many shipments from the various suppliers. Thus, she would like to have relatively few large shipments from a small number of suppliers. However, this means that she will have a relatively meager assortment of shoes because she is ordering from a small number of suppliers. Such a small assortment might not meet the needs of her clientele and, as a result, she may lose customers. Also, since she wants to receive relatively few shipments per year to avoid excess shipping costs, she will necessarily have to order in larger quantities, resulting in larger average inventory-carrying costs. In short, how should she balance the desire for low inventories with the need to satisfy her customers and to avoid large shipping costs? Alternatively, consider again the case of an airline. Once an aircraft departs with an empty seat, the potential revenue from that seat is lost. The seat cannot be sold again. Thus, the airline has a desire to fly with as large a load factor as possible. (The load factor on a flight is the ratio of the number of seats that are sold to the number of passenger seats on the aircraft. Typically, airlines operate with a fleet-wide load factor of roughly 80 percent.) One way to do this would be to discount the fares to attract additional customers. But doing so only increases the breakeven load factor, or the load factor the airline must achieve to cover its costs. Also, if the load factor is very high, it is unlikely that new passengers will be able to make a reservation on the flight on which they want to fly as that flight is likely to be full. Like shoe customers, many of these flyers may elect to travel using other carriers. To manage the inventory of seats available on each flight, airlines have instituted complex and sophisticated yield and revenue management systems that set prices for the seats and the number of seats that can be sold at each fare level. The problem is further complicated by the fact that some of the seats on a Chicago to Los Angeles flight leaving at noon may be for passengers flying directly from Chicago to Los Angeles. Other passengers on the flight might be continuing to destinations in the Far East, while still others may have originated on the East Coast of the United States earlier in the day. Passengers for whom the Chicago to Los Angeles flight leg is only one of several on which they will be flying typically do not pay for that flight leg separately; rather, they buy a single ticket for all of the flight legs (including return flights) in their itinerary. EOQA BASIC INVENTORY MODEL 287 Finally, consider the problem faced by an individual employee who needs to determine how much money to set aside in a flexible spending account (FSA). These accounts operate as follows. An employee can place money in such an account on a monthly basis for qualifying health care expenditures. The money that is set aside is tax free. Thus, if an employee is in a 30 percent tax bracket, one dollar spent from post-tax dollars is the equivalent of about $1.43 of FSA money. Thus, the employee would prefer to spend money on health care expenditures from the FSA account rather than using post-tax dollars on health care. There are two problems that the employee faces. First, the amount of money to set aside on a monthly basis for a particular year must typically be determined during the open enrollment period, which is generally sometime in the fall of the year preceding the year in which the money can be spent. At the time the employee must decide how much to set aside, he does not typically know what his expenditures during the coming year will be. The second complication is that the money in an FSA account is there on a "use-it-or-lose-it" basis. If the employee does not incur qualifying expenditures of at least the amount in the FSA account, the funds placed in the account are lost. In essence, the amount of money in the FSA account is an inventory of money that must be managed by the employee. Similar problems arise in the purchase of seasonal goods. There is likely to be a single point in time at which the purchases can be made. If an inventory remains at the end of the season, the goods must be sold at a significant loss. On the other hand, if too few are purchased, the retailer will incur lost sales (and disgruntled customers). Most of the examples we have discussed so far entail uncertain demand: In the case of the shoe store, the demand for shoes is uncertain. In the airline case, the demand for seats at any given fare is uncertain. In the FSA example, the need for money for health care expenditures is uncertain. To understand how to deal with problems with uncertain demand, it is useful to begin by examining a very simple model in which the demand is known with certainty. 5.2 EOQA BASIC INVENTORY MODEL The economic order quantity (EOQ) model is, perhaps, the most basic of all inventory models. This model deals explicitly with the tradeoff between ordering very infrequently to avoid multiple shipping charges and ordering in small quantities (but very often) to avoid large inventory-carrying costs. Figure 5.1 illustrates the key relationships and tradeoffs associated with this simple inventory model. The model assumes that inventory is used at a constant rate of R units per unit time as shown by the slope of the on-hand inventory. Note that all of the downward sloping lineswhether they are dashed lines or solid lineshave the same slope. The solid lines represent the case in which orders occur relatively infrequently. An order of size Q arrives and is depleted at rate R. Exactly QIR time periods later, the inventory is completely expended and another order of size Q arrives. The process repeats. The total ordering cost 288 INVENTORY DECISIONS IN SERVICES Figure 5.1. Basic economic order quantity relationships per unit time (e.g., per year) is relatively low because the order size is large and relatively few orders per unit time are needed to satisfy the demand. On the other hand, the inventory-carrying cost is large because the average inventory on hand is QI2. The dashed lines represent a case in which we place orders twice as often. Therefore, the ordering costs per unit time are twice as large as they are in the solid line case, but the inventory holding cost is half what it is in the solid line case. To make this more concrete, the EOQ model makes the following assumptions: There is no lead time for shipments from the supplier. Backorders are not permitted. There is only a single SKU (stock keeping unit) of interest. There are no quantity discounts associated with large orders. The demand is deterministic. The demand is static. We will use the following notation in discussing the EOQ model: Inputs R C h H S Annual usage rate Unit cost of an item Holding cost per year per dollar of inventory carried Annual holding cost per year per item of inventory carried (equal to hC) Fixed cost per order including all fixed order placement costs and shipping costs. 289 EOQA BASIC INVENTORY MODEL Decision variable Q Order quantity With this notation, we can write the total annual cost of buying, ordering, and holding inventory of the item in question as a function of the order quantity, Q, as follows: TC{Q) = CR + [^S + [^hC. (5.1) The first term of equation (5.1) gives the total cost of purchasing R items every year, excluding the fixed ordering costs. Each item costs C dollars and we must buy R every year to satisfy demand. Each year there will be R/Q orders placed and the fixed cost of placing an order is S. Thus, the second term represents the fixed ordering costs. If each order is for Q units, on average there will be Q/2 items on hand. Thus, the annual holding cost is given by the final term of (5.1). It is worth noting that the first term of (5.1) does not depend at all on the order quantity, Q. We simply need to pay the supplier CR every year to buy R units. We can buy them one at a time or all at once, or even buy them once every decade. The annual purchase cost of the R units will be the same: CR. Therefore, in finding the optimal order quantity, we can ignore this term. All future references to the total cost will omit this term for this reason. Figure 5.2 plots the ordering cost, the second term of (5.1), and the inventorycarrying cost, the third term of (5.1) as a function of the order quantity, Q. In this case, the unit cost was C = 100, the holding cost rate was h = 0.15, and the annual holding cost per item was, therefore, H =15. The fixed ordering cost per order was S = 5000 and the annual demand was R = 150,000. The fixed cost per year, shown as the downward sloping dashed line in Figure 5.2, decreases rapidly with the order quantity as fewer and fewer orders need to be placed each year as the order size increases. On the other hand, the inventory-carrying cost, shown as the linear dashed line, increases with the order quantity. For small order sizes, the total cost, shown as the heavy solid line, is dominated by the fixed order costs. The total cost, therefore, initially decreases with the order quantity. At some point, however, the inventory-carrying costs dominate and the total cost begins to increase with the order quantity. To minimize (5.1), we will treat Q, the order size as a continuous variable. We can then take the derivative of the total annual cost and equate it to 0 to obtain the order size that minimizes the total annual cost. [Note that because the total cost function is convex, or U-shaped, we can be sure that the value that we obtain in this way actually does minimize (5.1).] The resulting optimal order quantity is given by Q> = M. (5.2) 290 INVENTORY DECISIONS IN SERVICES Figure 5.2. EOQ cost components This equation makes sense. As the annual demand, R, increases, the optimal order size also increases. Similarly, as the fixed cost per order, S, increases, the optimal order size goes up so that we place fewer orders per year. Also, as the cost of the item, C, or the carrying cost rate, h, increases, the optimal order size decreases. With the values used to generate Figure 5.2, the optimal order size is 10,000 units and we order 15 times per year (R/Q* = 150,000/10,0000 = 15). The optimal total cost, if we order Q* units, is given by TC*(Q*) = y/2hCRS. (5.3) For the case illustrated in Figure 5.2, the optimal total cost is $150,000 per year. Again, this excludes the cost of purchasing the items, which is fixed at $15,000,000, and is independent of the order quantity. In deriving the optimal total cost, we can show that at the optimal order quantity, the fixed ordering cost given by the second term of equation (5.1) is exactly equal to the inventory holding cost, given by the third term of equation (5.1). This is also evident from Figure 5.2. The minimum of the solid total cost curve occurs at the point at which the fixed cost and holding cost curves intersect. Figure 5.2 suggests that the total cost curve is relatively flat at the optimal point. This means that small changes in the order quantity should not have a big impact on the total cost. To verify this, suppose the actual order quantity is Q = aQ*. In other words, instead of ordering exactly the optimal quantity, Q*, we order a multiple a of that quantity. In this case, the total ordering and inventory-carrying cost (excluding again the purchase cost of the items) is given by 291 EOQA BASIC INVENTORY MODEL Figure 5.3. Percent error in total cost as a function of deviations in the order quantity TC(aQ*) = - ^ + ^ h C y ' aQ* 2 = y/2hCRs( + - \\ \\2a 2) (5.4) Equation (5.4) shows that the ratio of the total cost when the order quantity is Q = aQ* rather than the optimal order quantity Q* to the optimal total cost is TC(aQ*)J TC{Q*) {2a ay 2) V } Figure 5.3 plots the ratio of the total fixed and inventory-carrying cost as a function of a, the fraction of the optimal order quantity that is actually ordered. As expected, the error is 0 percent when a - 1 or we order exactly 100 percent of the optimal order quantity. More importantly, the difference in total cost is less than 1 percent as long as the order quantity is between 87 percent and 115 percent of the optimal order quantity. If the order quantity is between 73 percent and 137 percent of the optimal quantity, the total cost is still less than 5 percent more than the optimal total cost. When we order half the optimal order quantity or twice the optimal amount, the total cost is only 25 percent more than the optimal total cost. 292 5.3 INVENTORY DECISIONS IN SERVICES EXTENSIONS OF THE EOQ MODEL As indicated in section 5.2, the economic order quantity model makes a number of very restrictive assumptions. In this section, we begin to relax some of these assumptions. 5.3.1 Positive Travel Time Perhaps the most obvious assumption made by the EOQ model is that the orders show up instantly, as soon as the inventory from the last order is exhausted. This assumption is easy to relax as long as the lead timethe time between when an order is placed and when it is receivedis deterministic or perfectly predictable. If the lead time is time units (measured in the same time units as the annual demand and all of the other parameters of the model), then we simply place an order when there are TR units on hand. In the example that led to Figure 5.2, if the lead time is exactly 3 days, then we should reorder when there are 1233 items 3 in inventory because 1233 = 150,0001365 5.3.2 Backorders The standard EOQ model assumes that we do not allow stockouts or backorders. In fact, even when demand is entirely deterministic, allowing backorders can be beneficial. Figure 5.4 modifies Figure 5.1 to allow for backorders in each inventory cycle. B items are backordered each cycle and the demand for these items is filled when the new order of size Q arrives, resulting in a maximum inventory during each cycle of / = Q - B items. Note that the average inventory is no longer III items because there is a period of time during each cycle of duration BIR during which there is no inventory on hand. Time Figure 5.4. EOQ model with backorders 293 EXTENSIONS OF THE EOQ MODEL In this case, the objective is to minimize the sum of three cost components: (a) the fixed ordering costs, (b) the inventory-carrying costs as in the standard EOQ model, and (c) the backorder costs. We must now optimize over two decision variables: the order quantity, Q, and the maximum number of backordered items per cycle, B. If we let pC be the cost per item per year that is backordered (analogous to the holding cost per item per year of hC), then we can show that the optimal order quantity is O* =f3l\\PE = 0* \\R (5 6) The optimal maximum number of backorders is given by B =Q * *ip(P+hy Finally, the optimal total cost is given by TC{Q*, B*) = y/2hCRSj-2. \\ p +h (5.7) Equations (5.6) and (5.7) emphasize the relationship between this model and the standard EOQ model. Equation (5.6) shows that the optimal order quantity in the presence of backorders is larger than when backorders are not permitted. In fact, the two are equal only when the penalty cost per dollar of backordered items, p, is infinite, or at least extremely large relative to the holding cost per unit per year, h. The total cost, on the other hand, is always less than the total cost of the EOQ model that does not allow for backorders. Again, the two are equivalent, only when the penalty cost per dollar of backorders is extraordinarily large. For the example of Figure 5.2, if the penalty cost per dollar of backorders is 0.12, the optimal order size is 15,000, as opposed to 10,000 for the E O Q model. Only 10 orders are placed each year, as opposed to 15, when backorders are not permitted. The optimal maximum backorder quantity is 8333, a large quantity relative to the order size. The total cost including the penalty costs for backorders, however, drops from $150,000 in the EOQ case to $100,000 in this case; a savings of 33.3 percent. If the penalty per dollar of backordered items is 0.341, slightly more than twice the holding cost per dollar of 0.15, the optimal order size drops to 12,000, the maximum number of backordered items goes down to 3667, and the total cost increases to $125,000, including the backorder penalty costs. Figure 5.5 plots the percentage savings in total cost as a result of allowing backorders versus the ratio of the backorder penalty cost to the holding cost (plh). The savings exceeds 10 percent of the E O Q cost as long the backorder cost is less than four times the holding cost. 294 INVENTORY DECISIONS IN SERVICES Figure 5.5. Percent savings in total cost as a function of the backorder/holding cost ratio Time Figure 5.6. EOQ model with production at the beginning of each phase 5.3.3 Production versus Order Delivery Another extension to the basic EOQ model is to allow for a production phase at the beginning of each cycle, as shown in Figure 5.6. When inventory is depleted, a production phase begins, as opposed to inventory being delivered from a supplier. Production occurs at an annual rate of P units and so inventory is accumulated during the production phase at a rate of P - R units. The fixed cost, S, can now be thought of as a fixed setup cost for production. In this case, we can show that the optimal maximum inventory on hand is a* Production : 2RS P-R _n* hd P UEOQ IP-R (5.8) 295 EXTENSIONS OF THE EOQ MODEL The optimal total cost is JlhCRsJ^-. 5.3.4 (5.9) Multiple Items One of the other major limitations of the standard EOQ model is that it assumes that there is a single SKU. In fact, it is often the case that multiple items or SKUs are ordered at a single time. In this section, we outline three different ways of dealing with inventory problems with multiple SKUs. The first approach is the nave approach, in which each SKU is treated independently. To illustrate this approach, consider the problem illustrated in Table 5.1. Three items are considered: a low-quality item, a medium-quality item, and a high-quality item. The demand is dominated by the medium-quality item. The cost of purchasing the high-quality item from the supplier is twice that of the low-quality item. The holding cost rate is 20 percent per item per year for all three of the SKUs. Finally, the fixed cost is divided into two parts: a fixed cost per order and an incremental cost per line item on the overall order. Thus, if all three items are ordered jointly, the fixed cost is composed of a single payment of $125, which is the fixed cost per order, plus a line-item cost of $75 for each of the three items, resulting in a total fixed cost of $350. In the nave approach, each item is treated independently. In other words, we simply apply equation (5.2) to find the optimal order quantity for each SKU. Table 5.2 illustrates the results. The total annual cost is $7000. By dividing each item's demand by the order size, we find that the items are ordered 2.5,10, and 5 times per year for the low-, medium-, and high-quality items, respectively. See Multi-item EOQ (individual orders).xls in the online appendix. In fact, since the fixed cost of placing an order is divided into two partsa cost per order and a line-item or SKU costwe might want to try to improve on the cost by placing a single order for all three items and simply determining the TABLE 5.1. Example inputs for multiple SKU examples Cost Fixed Order Cost Incremental Order Cost Total Fixed Cost Annual Demand Holding Cost Rate Holding Cost Low Medium High 125 125 75 200 100 0.2 25 200 125 75 200 1000 0.2 40 250 125 75 200 200 0.2 50 296 INVENTORY DECISIONS IN SERVICES TABLE 5.2. Naive model results Fixed Cost Per Year Order Size Avg Inventory Inventory Cost Total Cost Overall Cost Low Medium High 500.00 40.00 20.00 500.00 1000.00 7000.00 2000.00 100.00 50.00 2000.00 4000.00 1000.00 40.00 20.00 1000.00 2000.00 TABLE 5.3. Optimal total cost with a single joint order Line-Item (SKU) Order Cost Fixed Cost Per Order Order Size Avg Inventory Inventory Cost Total Cost Overall Cost Low Medium High 649.52 1082.53 11.55 5.77 144.34 1876.39 6062.18 649.52 649.52 115.47 57.74 2309.40 2958.92 23.09 11.55 577.35 1226.87 joint order frequency. Let S0 be the fixed cost per order (the $125 value above). Furthermore, let / be the set of items to be ordered jointly, S be the line-item or SKU cost per item for SKU j e J, and h, R, and c be the cost per dollar per year of inventory, the demand per year, and the purchase cost per item of SKU / e J. Finally, we let X be the unknown number of orders per year. This is only the decision variable. The total annual cost can be written as a function of the number of orders per year, X, as follows: M TC(X)= So + 5 , X + V yw X J (5.10) Taking the derivative of (5.10) with respect to the number of orders per year, X, and setting the derivative to 0, we obtain X* = 5 So+ ; V yw Y (5.11) J Applying this formula to the data shown in Table 5.1, we find that the optimal number of orders per year, X, is 8.66. The total cost, as shown in Table 5.3, is $6062, or a reduction of 13.4 percent relative to the naive approach. EXTENSIONS OF THE EOQ MODEL 297 See Multi-item EOQ (single order).xls in the online appendix. In comparing the single joint order with the naive approach, we find that we are ordering much too often for the low-quality products. Therefore, we might consider a third approach, in which we determine how often each year to place orders and then, how often to include each SKU in the grand order. In other words, we may elect to place an order every month, but to include the low demand items only every quarter. While there are a number of approaches to solving this problem (Chopra and Meindl, 2004), perhaps the easiest way to do so is to let the Excel Solver attack the problem. Note that this problem is inherently non-linear. The Excel Solver is better for such problems than is What's Best, which focuses on linear optimization problems. 298 INVENTORY DECISIONS IN SERVICES See Multi-item EOQ (suborders).xls in the online appendix. 5.3.5 Limited Space or Value of the Goods in Inventory A final limitation of the EOQ model that can be relaxed relatively easily is that there are no capacity restrictions on the size or value of the inventory on hand at any time. To relax this assumption or limitation, we introduce the following notation in line with our previous notation: Inputs and sets / set of SKUs to be analyzed Sj fixed cost per order of purchasing item y e / Rj annual demand of SKU j & J 299 EXTENSIONS OF THE EOQ MODEL hj Cj j y annual holding cost per dollar of inventory of item j e J cost per item of SKU j & J value or size of a single item of SKU ; e 7 maximum allowable value or size of all items in inventory Decision variables Qj order quantity for item j e J With this notation, we can find the optimal order quantities by solving the following optimization problem. ye/ s.t. Sj L jj /?,.,<7 (5.13) y>0 (5.14) ye/ y/e/ The objective function (5.12) minimizes the sum of the ordering costs and the inventory holding costs. As before, it ignores the fixed annual cost of purchasing the items, which would be given by ]c--. Constraint (5.13) limits the total ye/ value or size of the inventory. Clearly, if , refers to the physical size of the items, then y must also refer to size and they must be in compatible units (e.g., cubic feet). Finally, constraint (5.14) simply requires all order quantities to be non-negative. If the solution to (5.12) without considering constraint (5.13) happens to satisfy the constraint, then we are done. In that case, each of the order quantities would simply be the corresponding EOQ value as given by (5.2). If these values violate constraint (5.13), the optimal order quantities can be computed using the following equation: Qf = J, ' ' \\hjcj + aj V/e/ (5.15) In this equation, a is an unknown multiplier. As a increases, the order quantity will decrease. Eventually, for a sufficiently large value of a, the order quantities will be just small enough so that they satisfy constraint (5.13). As shown below, the value of a can be found using a number of spreadsheet-based techniques. If the constraint coefficients, p}, are all proportional to the unit costs, c, then (5.15) can be further simplified to Q* = XQfOQ VjeJ (5.16) 300 INVENTORY DECISIONS IN SERVICES where = 0' jeJ (5 17) In other words, the optimal order quantities are each proportional to the standard EOQ values given by (5.2), in this case. EXTENSIONS OF THE EOQ MODEL 301 302 INVENTORY DECISIONS IN SERVICES EXTENSIONS OF THE EOQ MODEL 303 304 5.4 INVENTORY DECISIONS IN SERVICES TIME-VARYING DEMAND The standard economic order quantity model assumes that the demand rate is constant over time. For many products and services this is simply not the case. In fact, for many retailers in the United States, the Christmas holiday season generates a large percentage of their annual sales. For example, more than one in every six dollars spent in the retail jewelry business is spent in December. Other big months in this industry are February (with nearly 9 percent of all sales due in part to Valentine's Day) and May (with 9.3 percent of all annual jewelry sales due partially to Mother's Day). Most department stores exhibit similar trends with December accounting for a disproportionately large percentage of the sales revenue. The ratio between gasoline sales in July to those in December is over two to one. Conversely, the dollar value of fuel dealerships sales (e.g., home heating fuel) in January is over twice that in the summer months of June, July, and August. Other retail outlets experience significantly less seasonal fluctuation. The monthly dollar value of grocery sales ranges from a low of 7.77 percent of the annual sales in February to a high of 8.72 percent in May (U.S. Census Bureau, 2009). This low range of fluctuation reflects the fact that people have to eat year-round. In this section, we outline two different approaches to dealing with seasonal demand, depending on whether or not the planning horizon is finite or (effectively) infinite. A finite horizon would occur if we need to plan for a single year, and we do not care about what happens after the end of that year. For many typical demand profiles, this may be perfectly fine as we are likely to want to place an order in January in any case. More generally, however, it may be that we need to plan for a long time horizon. What occurred this past January, in terms of sales, is likely to occur again next January. Any plan that artificially truncates the planning horizon at the end of a year (or any other month) may miss opportunities to save money that would accrue due to the use of a longer (effectively infinite) planning period. We will begin, however, with a finite planning horizon model. Specifically, we will consider the problem of planning for time-varying monthly demands. Figure 5.15 illustrates the monthly percentage of the annual sales for paint and wallpaper stores. Relatively little home remodeling apparently occurs during the holiday season (November and December) and the deep winter (January and February). The peak is in July (exactly the month we decided to repaint our bedroom and bathroom). As before, we will assume that there are two relevant costs associated with the management of inventory at a store selling paint and wallpaper: the fixed cost associated with ordering any inventory, S, and the holding cost per item per year, H = hC. We will assume that orders can be placed only for an entire month's worth of inventory. In other words, if we place an order at the beginning of January, it can cover January's inventory needs, or January and February's needs, and so on. It cannot cover January and part of February's needs, however, with a new order to arrive in mid-February. TIME-VARYING DEMAND 305 Figure 5.15. Percent of annual dollar sales by month for paint and wallpaper stores Figure 5.16. Inventory on hand for inventory computation To illustrate the computation of the amount of inventory on hand, consider the three cases illustrated in Figure 5.16, each of which corresponds to receiving inventory at time 1, the beginning of January. If we had only ordered enough inventory for one month, the inventory begins at 200 and declines steadily to 0 at the end of the month. The average inventory on hand during this time is 100 306 INVENTORY DECISIONS IN SERVICES units. If we had ordered for two months, in addition to the 100 average units for January, during January we would hold all 1400 units of inventory to be used in February. During February, the average inventory would be 700. Thus, ordering for two months would result in 100 + 1400 + 700, or 2200 item-months of inventory. Finally, if we order for three months, during January and February we would have an additional 300 units of inventory to be used in March. In March, the average inventory would be only 150 units. Thus, ordering for three months would result in 2950 item-months of inventory. We can now let c be the cost of (1) placing an order and (2) holding inventory that arrives at the beginning of month / and that lasts for / months. In the example above, if the cost of holding an item in inventory for one month is $2, and the fixed cost of placing an order is $1000, then we would have cn = 1000 + 2 100 = 1200, c12 = 1000 + 2 2200 = 5400 and c13 = 1000 + 2 2950 = 6900. This definition of costs allows us to represent the inventory planning problem with monthly demands as a shortest path problem. Figure 5.17 shows the structure of such a shortest path problem if we need to plan for the first six months of a year. Darker lines represent larger inventory quantities. Thus, the darkest, topmost line represents placing a single order at the beginning of the year for six months of inventory. At the other extreme, the links connecting successive nodes Figure 5.17. Inventory planning as a shortest path problem 307 TIME-VARYING DEMAND represent six separate orders, each for a single month's inventory. Notice that the number of nodes is one more than the number of periods for which we are planning. In the example shown in Figure 5.17, we are planning for six months, and so there are seven nodes. The link between the June and July nodes, for example, represents the inventory purchased at the beginning of June for use during that month. Now consider the demand data shown in Table 5.4. These values correspond to a paint store with annual sales of 10,000 units. The projected sales in each month are proportional to the national average values shown in Figure 5.15. If the fixed cost of placing an order, shipping the items, and receiving them is $10,000 and the monthly carrying cost per item is $2, the optimal solution, as shown in Figure 5.18, is to order every three months, in January, April, July, and October. The total cost is $69,467, as shown in the computations below. cyeB_i4pf = 10,000 + 2J + 1.5706 + 2.5795[ = 16,809 ^Apr-Jul = 10,000+ 2 { + 1.5-945+ 2.5-9391 = 18,406 = 10,000 + 2 { ^ U l . 5 - 9 5 7 + 2.5-897J = 18,370 = 10,000 + 2 J + 1.5662 + 2.560} = 15,882 Total Cost = 69,467 See Time Dependent Demand as Shortest Path.xls in the online appendix. TABLE 5.4. Example monthly demand data Month January February March April May June July August September October November December Demand Rate 716 706 795 876 945 939 1014 957 897 891 662 601 308 INVENTORY DECISIONS IN SERVICES (ug) (sep) (octj (NOV) (Dec) Uan) Figure 5.18. Optimal solution to the inventory problem of Table 5.4 with a fixed order cost of $10,000 and a per unit monthly carrying cost of $2 For finite horizon planning models of the sort outlined above, the first order must be placed so that inventory arrives at the beginning of the first period (e.g., for January). In some cases, this may not be the optimal strategy. We now present a model in which time wraps around itself in the model. In other words, there is technically no end of the planning horizon. An order can be placed to arrive at the beginning of November for four months of inventory. This order will then satisfy demands in November, December, January, and February. We simply assume that the demand in next year's January will be identical to the demand this year in January. To model such an infinite planning horizon problem, we define the following notation. Inputs and sets P set of planning periods (e.g., the set of months in a year, or the set of days in a week or in a month) c fixed order cost plus inventory-carrying cost of an order that arrives at the beginning of period i and is to be used for inventory during the next j periods. This is calculated as before but with the appropriate wrap-around notation, so that an order that arrives in November (e.g., i = 11) for four periods would serve demand in November, December, January, and February ak 1 if an order that arrives at the beginning of period i, to be used for the next j periods satisfies the inventory needs of period k; 0 if not In other words, aijk = \\ if i < k < i + j - 1, and ajk = 0 otherwise, where we allow the appropriate wrap-around of the indices. That is, if we are planning monthly, and i - 11, meaning that an order arrives at the beginning of November, and j = 4 meaning that the order is for four periods, then NOVANOV = NovADec = Q-NOVAMI = MM,* = 1 whilefljvov,4,z= 0 for all other months Z. Decision variables X 1 if an order arrives at the beginning of period i and is to be used for inventory during the next ;' periods; 0 if not With this notation, we can formulate the problem as follows: \\p\\ Min (5-18) 309 TIME-VARYING DEMAND TABLE 5.5. Comparison of infinite horizon and finite horizon annual costs for the data of Table 5.4 with a per unit per month holding cost of $10 Finite Horizon Model Fixed Cost Total Cost First Order Order/Yr Total Cost % Over Infinite Horizon Cost 1,000 2,500 5,000 6,500 8,000 8,700 9,000 9,500 10,000 25,000 50,000 80,000 100,000 250,000 500,000 61,995 79,995 109,995 127,505 143,065 150,025 152,695 156,585 159,695 245,155 347,335 437,795 494,755 794,755 1,062,145 January January January January January February February February January January January January January January April 12 12 12 11 10 9 8 7 6 5 4 3 2 2 1 61,995 79,995 109,995 127,505 143,065 150,065 152,735 156,625 159,695 245,155 347,335 437,795 494,755 794,755 1,096,075 0.00% 0.00% 0.00% 0.00% 0.00% 0.03% 0.03% 0.03% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 3.19% s.t. \\p\\ ^ < 1 VIEP (5.19) 7=1 \\P\\ = 1 *kGP (520) X:je{0,l} (5.21) ieP 7=1 VieP;j = l,...,\\P\\ The objective function (5.18) minimizes the total cost. If an order arrives at the beginning of period i for the next j periods (i.e., for periods i,i + 1,... ,i + j-l), then the cost including the fixed costs of placing an order as well as the inventorycarrying costs are c and Xt). = 1. This cost then contributes to the objective function. Constraint (5.19) states that in each period, we can place at most one order. Constraints (5.20) ensure that the demand in each period is served or covered by exactly one order. Constraints (5.21) are standard integrality constraints. Table 5.5 compares the results of using this model with the finite horizon (shortest path) model discussed earlier in this section for the demand data shown in Table 5.4. The holding cost per unit of inventory per month was fixed at $10, while the fixed cost per order was varied from $1000 to $500,000. Note that what is important is not the absolute value of the fixed cost, but rather the ratio of the fixed cost per order to the holding cost per item per month. To understand why this is so, imagine what would happen if we changed the units from dollars to 310 INVENTORY DECISIONS IN SERVICES Euros or to the Israeli Shekel. As long as the ratio remained the same, the order policy would remain unchanged. In most cases, the two models agree completely. Only in those cases in which the infinite horizon model of (5.18)-(5.21) begins with an order in a month other than January are the costs different. In only one casethe last one shown in Table 5.5is the percentage difference between the costs of the two models significantly different. Note that in both models, as the fixed cost increases relative to the holding cost, we order less frequently and in larger quantities, holding more inventory each period. This is as expected. See Time dependent demand with wrap around.xls in the online appendix. 5.5 UNCERTAIN DEMAND AND LEAD TIMES The models discussed so far ignore uncertainty in either the demand or the lead time, which is the time between when an order is placed and when the order is received. In fact, both are likely to be uncertain or stochastic. Thus, there is a need to plan for this uncertainty. Throughout the discussion below, we treat only the case in which there is a continuous review of the inventory on hand. An alternative approach, which leads to the need for more inventory to ensure a given customer service level, is to make inventory replenishment decisions periodically. The reader interested in such systems can consult any one of a number of operations management and supply chain texts, including Chopra and Meindl (2004). To motivate the need to account for variations in both the demand and the lead time, consider an inventory manager who knows that the average daily demand is 3 units and that the average lead time is 4 days. If both the lead time and the demand are deterministic or not subject to random variations, there will never be any stockouts if the manager places an order when the inventory gets to 12 units (or 3 units per day times 4 days for the lead time). Just as the inventory runs out, a new shipment will appear and a new inventory cycle will begin. Now, however, suppose the lead time is again deterministic, but the demand is subject to variability. Specifically, suppose the demand follows a Poisson distribution with a mean of 3 units per day. If the inventory manager again orders when the inventory on hand gets to 12 units, the probability of not being able to meet all of the demand before the shipment arrives is simply the probability of 13 or more demands. This is 0.424. (The reader should verify that he or she knows how to compute this from the Poisson distribution.) If the manager decides to order when the inventory hits 20 (as opposed to 12), the probability of stocking out drops to 0.012. However, on average, there will be more inventory left in the system when the new order arrives. The average inventory left when a shipment arrives is the safety stock. Alternatively, we can think of the safety stock as the difference between the reorder point (the inventory at which we place an order for more of the product) and the expected demand during a lead time. UNCERTAIN DEMAND A N D LEAD TIMES 311 To illustrate these concepts further, consider the simulation results of a stochastic inventory system shown in Figure 5.19. The simulation runs until 400 demands are realized. The reorder point (shown with the heavy black line) is 12 and the lead time is fixed at 4 days. After each demand is realized, the inventory is updated. When the inventory reaches the reorder point, a new order for 55 units (the order size) is placed and the time until the arrival of the new order is computed. If an order has arrived between the time of the previous customer's demand and the current customer's demand, the order quantity is added to the inventory on hand. The mean daily demand is 3 units. Seven orders are placed in the example shown. The demand during each of the seven lead times does not exceed the available inventory and so no stockouts occur. The fact that no stockouts are shown in Figure 5.19 is a coincidence. In fact, the same conditions are likely to result in stockouts as shown in Figure 5.20. In Figure 5.19. Stochastic inventory simulation (no stockouts) Figure 5.20. Stochastic inventory simulation (with stockouts) 312 INVENTORY DECISIONS IN SERVICES Figure 5.21. Stochastic inventory simulation with exponentially distributed lead times this case, three of the seven orders result in stockouts, as shown by the diamonds, which indicate negative or backordered inventory. Finally, if the lead time itself is subject to variability, there is an even greater likelihood of a stockout during a lead time for a fixed reorder point. Figure 5.21 shows the results of a simulation under identical conditions, except that the lead time is now exponentially distributed with a mean of 4 days. Again, seven orders are placed over the course of roughly 130 days. A small number of backorders occur during the first lead time. The second lead time is very short and no backorders occur. The third lead time is longer and results in about 10 backordered items. Over 30 items are backordered during the fourth lead time. The fifth and sixth lead times are very short. In fact, the sixth lead time is nearly instantaneous. Finally, the seventh lead time is long enough to result in nearly 20 backorders. In the case of stochastic demand or lead times, there is a need to determine not only how much inventory to order each cycle, but also when to order during each cycle, or the reorder point. The reorder point determines the magnitude of the safety stock or average inventory left at the end of a cycle. Although there are many different approaches to solving this problem, perhaps the easiest way to attack the problem is to determine the order size using the EOQ model. Then, all we need to do is to determine when to place the order as a function of the demand and lead time uncertainty and the goals of the inventory system or manager. Two performance metrics are typically employed. The fill rate is the fraction of the total demand that is filled from inventory. The cycle service level is the probability that an inventory cycle will not entail a stockout. If the demand during any period (e.g., any day) is relatively large, then it can often be approximated quite well by a normal distribution. We let be the demand during a single (e.g., daily) period, \\ be the variance of the demand during any period, and L be a random variable denoting the duration (in periods) 313 UNCERTAIN DEMAND AND LEAD TIMES of the lead time with mean and variance L and \\ respectively. The mean demand during a lead time is given by the product of the per period demand and the average duration of the lead time. In other words, VLT = dLIf the demands in each period are independent, and the lead time is constant or deterministic, the variance of the lead time demand is given by LT = ^Lad- In other words, the variance of the demand during a lead time is simply equal to the variance of the per period demand, \\ , multiplied by the number of periods in a lead time, L, If there is variability in the lead time as well, the variance of the demand during a lead time is given by = _2 + 2, (5.22) where the second term accounts for the variability of the lead time. We can now use the mean and variance of the demand during a lead time to determine the appropriate reorder point for a given level of service. We will begin with the cycle service level. If the inventory manager wants to ensure that the probability of a stockout during an inventory cycle is less than or equal to a, then the probability that the demand is less than or equal to the reorder point during a lead time must be at least 1 - a. In other words, we require fROP-LT^ >1-, where () is the cumulative normal distribution evaluated at x, and ROP is the reorder point. Solving for the reorder point, we find ROP = LT + z1.aoLT, (5.23) where the probability of a standard normal random variable being less than or equal to Z\\-a is 1 - a. To compute the fill rate, we need to compute the expected number of demands during a cycle that are backordered or not filled from inventory, E(B). Clearly, all such demands will occur during a lead time or after an order has been placed. This is given by E(B)= J (x-ROP)f(x)dx, 314 INVENTORY DECISIONS IN SERVICES where f(x) is the distribution of demand during a lead time. With normally distributed demands with a mean LT and a variance of a2LT during a lead time, we can show that this is equal to E(B) = -ss 1- ( ss V \\^LT +(^LT^ ' SS ^ (5.24) y^LT J Jl In (5.24), ss is the expected safety stock given by ROP - LT = ROP - and B is a random variable denoting the number of backordered items per cycle. The fill rate is then given by fill rate = 1 - ^ 9 , Q (5.25) where Q is the order quantity, or average inventory used during a cycle. Solving for the reorder point that ensures that the fill rate is at least is difficult, even if the order quantity, Q, is known. We would need to solve i-^l-. Q Solving for the expected number of backorders and substituting (5.24) for E(B), we obtain -(ROP-LT) 1- ({ROP-^ + aLT<> \\ROP-LT) -Q(l-). (5.26) This is clearly a rather intimidating equality to solve for the reorder point, ROP, because the reorder point appears in the arguments of both the cumulative standard normal function as well as the standard normal density function. UNCERTAIN DEMAND AND LEAD TIMES 315 316 5.6 INVENTORY DECISIONS IN SERVICES NEWSVENDOR PROBLEM AND APPLICATIONS The analysis of section 5.5 assumed, in effect, that the inventory system would continue indefinitely. In other words, we were interested in minimizing the long run average cost per unit time. In many contexts, however, a single decision must be made about inventory levels before the demand is realized. If there is excess inventory on hand after the realization of the demands, the additional inventory is sold at a salvage price that is typically below the original purchase cost. Thus, the vendor incurs a loss on any excess inventory on hand at the end of the period. On the other hand, if the manager orders too little, there is a large likelihood of a stockout that incurs a loss of goodwill, a loss of potential sales and, possibly, additional penalty costs. Consider, for example, the problem faced by a small boutique dress shop. Each year it has to determine how many high fashion dresses of each size, color, and style to order. Each dress costs $100 to purchase and sells for $300. Thus, the store makes $200 on each sale. However, unsold dresses at the end of the season must be sold below cost to a salvage shop for only $70. Thus, the boutique loses $30 on any dress that is not sold at the end of the season. Prior to the season, the manager has estimated the probability of selling anywhere between zero and six dresses of a particular size, color, and style. These values are shown in Table 5.6. NEWSVENDOR PROBLEM AND APPLICATIONS 317 TABLE 5.6. Example probabilities of dress sales Sales Probability Buy 3 0 1 2 3 4 5 6 Total 0.05 0.10 0.15 0.20 0.25 0.15 0.10 1.00 -90 140 370 600 600 600 600 485 TABLE 5.7. Net profit as a function of number of dresses purchased Purchase 2 3 4 -30 200 200 200 200 200 200 -60 170 400 400 400 400 400 -90 140 370 600 600 600 600 -120 110 340 570 800 800 800 188.5 354 485 570 Sales Probability 0 1 0 1 2 3 4 5 6 Total 0.05 0.10 0.15 0.20 0.25 0.15 0.10 1.00 Net Profit 0 0 0 0 0 0 0 0 5 -150 80 310 540 770 1000 1000 597.5 MAX 6 -180 50 280 510 740 970 1200 590.5 Suppose the boutique buys three dresses. If no dresses are purchased, the boutique loses $90 or three dresses times a loss of $30 per dress. If one dress is sold, the boutique makes $200 on the sale, but loses $60 on the two unsold dresses for a net profit of $140. If two dresses are purchased, the stores nets $370, or two sales at $200 each minus a loss of $30 on the single unsold dress. Finally, if the demand is three or more, the boutique makes $600. These values are also shown in Table 5.6. Multiplying the probabilities by the net profits and summing the resulting values gives an expected profit if the store buys three dresses of $485. Table 5.7 summarizes the expected net profit as a function of the number of dresses purchased. In this case, it is best to purchase five dresses, resulting in an expected profit of $597.50. 5.6.1 Basic Model The model for this sort of problem is called the newsvendor problem, as its original application is to a newsvendor, like a local convenience store, that sells daily 318 INVENTORY DECISIONS IN SERVICES newspapers. As before, the manager of the store must determine how many papers to purchase from the publisher. The purchase price is c per paper and the sale price is p per paper. If, at the end of a day, any papers are not sold, they can be salvaged at a price of s per paper. Clearly, for the problem to make sense we require p> c> s. The probability that the demand will be exactly y papers is given by /(y) and D is the maximum demand. If the manager purchases X papers, the expected profit is given by X D E(profit\\purchaseX) = ^{p-c)yf(y) + y=0 (p-c)(X)f(y) y=X+l -L(x-y)f(y)(c-s). (5.27) y=0 The first term in (5.27) gives the expected profit due to the sale of X or fewer papers. The second term gives the expected profit when the demand exceeds X, the number of papers purchased. In this case, the sales are limited to X. Finally, the last term gives the expected salvage costs when the sales are X or less. Equation (5.27) can be written as E(profit\\purchase x X) = ^(p-s)yf(y)-(p-s)XF(X) + (p-c)X, (5.28) where F(y) is the probability that the demand is y or less. Now, suppose we purchase X + 1 items. The incremental profit of the last item is E{additional profit purchase X +1) = \\Z(p-s)yf(y)-(p-s)(x+i)F(x+V+(p-c)(x+i)\\ -t(p-s)yf(y)-(p-s)XF(x)+(p-c)x\\ = -(p-s)F(X) + (p-c). (5.29) If this incremental profit is positive, it is advantageous to increase the number of purchased papers from X to X + 1. This condition is given by F{X)BZ1= P-c p-s (p-c) + (c-s) (5.30) In (5.30), the term in the numerator,/? - c, can be thought of as the cost of underage or cu. This is the marginal net profit that is lost if we do not have enough inventory to satisfy all of the demand. Similarly, the second term in the denominator, c - s, can be thought of as the cost of overage, c0. This is the marginal net cost if we have inventory left over at the end of the season. Thus, the critical fraction in (5.30) can be written as p-c _ cost of underage _ cu (p-c) + (c-s) cost of underage + cost of overage cu + c0 To illustrate the use of condition (5.30), consider, again, the example shown in Table 5.6. In this case, we have cu= p - c = 300 - 100 = 200 and c0 = c - s = 100 - 70 = 30. The critical fractile in (5.30) is therefore = 0.87. Condition v ' 200 + 30 (5.30) says that we should order five items since F(4) = 0.75 and F(5) - 0.9. In other words, 5 is the smallest value of the number of items we can order such that F(Z) > 0.87. When the demand is normally distributed with mean and variance 2 (or can be well-approximated by the normal distribution), condition (5.30) states that we want to purchase a quantity Q such that JQZJL] = E^. = S^, ^ ; p-s cu+c0 (5.31) where () is the cumulative standard normal distribution evaluated at JC. Solving for Q, we get Q = + Z(p-C)i(p-S)0, where za is the value of the cumulative standard normal distribution such that the probability of obtaining a value less than or equal to za is . The use of this formula is illustrated in the next subsection. 5.6.2 Determining FSA Allocations At many firms, employees can set aside pre-tax dollars to be used for qualifying health care expenditures. Employees must typically decide how much to set aside for the coming year (e.g., 2011), sometime near the end of the preceding year (e.g., November of 2010). Any funds that are not used by the end of the year for which they are set aside are lost completely. Unfortunately, next year's expenditures are not known at the time the decision about how much to set aside needs to be made. 320 INVENTORY DECISIONS IN SERVICES The objective of a rational employee should be to maximize his or her expected post-tax, post-health-care income. Let us define the following inputs and decision variables: Inputs / a x gross income the marginal tax rate the health care expenditures, a random variable, which we will assume is normally distributed with a mean and variance 2 Decision variable Y by the amount of money to set aside in the FSA account for health care expenditures With this notation, the expected post-tax, post-health care income is given Gross income - taxes - FSA allocation - expected additional health care costs = I-a(I-Y)-Y-j(x-Y)f{x)dx (5.32) Y If we equate the derivative of this expression with respect to Y, the amount of money to set aside for health care expenditures, to zero, we find that the optimal value satisfies the condition F(Y) = a (5.33) or or = + . In the United States, the peak marginal tax rate is 35 percent, so the largest possible value of za is -0.385. Thus, under the current tax laws in the United States, the amount of money that you set aside in an FSA account should be less than the amount you expect to spend on health care costs in the coming year. To evaluate the actual expected post-tax, post-health-care funds available, we need to evaluate the final integral in (5.32). To do so, it is useful to know that 321 NEWSVENDOR PROBLEM AND APPLICATIONS Figure 5.24. Example post-tax post-health care funds fx exp I f\\ dx= \\ \\ = \\dx ^)-^~ where () and () are the standard normal density and cumulative distributions evaluated at x respectively. After some algebra, and using the expression above, we find that expression (5.32) becomes (/-)(1-) + (-) - - - (5.34) Figure 5.24 shows the available post-tax and post-health-care expenditure funds available to an individual in the 35-percent tax bracket, earning $100,000 as a function of the FSA allocation. (Note that this computation assumes that all income is taxed at the marginal rate, which is an overestimation. This should not significantly affect the optimal allocation, but the total available funds may be larger than those computed in the figure.) The individual's health care expenditures are normally distributed with a mean of $3000 and a standard deviation of $1000. The optimal allocation in this case is $2615, resulting in $62,680 in posttax, post-health-care available funds. While the error associated with not using 322 INVENTORY DECISIONS IN SERVICES an FSA account at all is only slightly more than 1 percent, the difference is $680, which is not money to sneer at, by any standards. For any allocation between $2000 and $3000, the difference is less than 0.1 percent, or about $63enough for a nice dinner for two. A natural question at this point in time is how does this result in (5.33) compare with the more general theoretical result shown in (5.31)? In this case, if we under-allocate funds to the FSA account, we have to spend a post-tax dollar on health care. Instead of this costing us 1 - a, it costs us $1, and so the cost of underage is a. Similarly, every dollar allocated to the FSA that is not spent is lost, but we did not have to pay taxes on that dollar and so it really only cost us 1 - a dollars. Thus c = a and c0 - 1 - a, resulting in a critical ratio in (5.31) of a. 5.6.3 Contracting for Services As a final example of the newsvendor problem, we consider the issue of contracting for services. For example, a homeowner may want her driveway plowed during the winter storm season. She must contract with a company for a given number of plowings during the coming winter. For example, she may elect to contract for five plowings at $50 each or a total of $250 for the winter. As in the FSA allocation example, if the number of snowfalls is less than the number she contracts for, she still has to pay the full $250. If it snows more than the contracted number of times, the service will still plow her driveway, but at a premium rate of $80 per additional snowfall. If the owner can estimate the probability mass function of the number of snowfalls during the coming winter, the key question is how many plowings she should contract for to minimize her expected cost. To formalize this, suppose she contracts for K plowings at a cost of c per snowfall. If the number of snowfalls exceeds K, she must pay e for each additional snowfall. Clearly, we require e > c. Otherwise, she should not contract for any plowings; she should simply pay e per plowing. Finally, let f(x) be the probability that there will be exactly x snowfalls during the coming winter. The expected cost to the homeowner is then Cost(K) = cK + e%U-K)fU)j>K (5-35) The first term of (5.35) captures the cost of the plowings for which she has contracted. The second term represents the expected number of additional plowings multiplied by the cost per additional plowing. Similarly, if she contracts for K + 1 plowings, the expected cost is given by Cost(K + l) = c(K + ) + e ^ (j-K-l)f(j). (5.36) j>K+l Subtracting (5.35) from (5.36), we find that the incremental cost of the (K + l)sl plowing is Figure 1.1 Gross domestic product over 60 years Based on the U.S. Department of Commerce, Bureau of Economic Analysis, National Income and Product Accounts Table 1.1.5 Gross Domestic Product (Seasonally adjusted at annual rates) Figure 3.28 Sample time dependent queueing analyzer output Figure 3.29 Sample probability of waiting results Figure 3.30 Steady state (re