Parts Emporium Case Study
1. The short answer is that higher inventories do not provide an advantage in any of the nine competitive priority categories. The important point is that firms must have the “right amount” of inventory to meet their competitive priorities. The only relevant costs considered in this chapter are ordering costs, holding costs, and stockout costs. In the economic order quantity (EOQ) model, costs of placing replenishment orders tradeoff against the costs of holding inventory. Under the assumptions of the EOQ, average inventory is one-half of the order quantity. The number of orders placed per year varies inversely with order quantity.
When we consider stockout costs, an additional inventory (safety stock), is held to trade-off costs of poor customer service or costs for expediting shipments from unreliable suppliers. In the lean systems chapter, we see order quantities (lot sizes) that are much smaller than the “ideal” suggested by the EOQ model. As a result, lean systems average inventory is also much lower. Are there some other relevant costs of holding inventory that we have not considered in the EOQ model? If there are, a firm that ignores these costs will make the wrong inventory decisions.
Parts Emporium Case Study Essay Example
These wrong decisions will make the firm less competitive. Let’s examine the relationships between inventory and the nine competitive priorities discussed in the operations strategy chapter. We compare competitors H and L. They are similar in all respects except H maintains much higher inventory than does L. 1. Low-cost operations. Costs include materials, scrap, labor, and equipment capacity that are wasted when products are defective. When a process drifts out of control, competitor H’s large lot sizes tend to result in large quantities of defectives.
The EOQ does not consider the cost of defectives, and erroneously assumes that setup costs are constant. Small lots cause frequent setups, but the cost per setup decreases due to the learning curve. Competitor L will enjoy competitive advantages with lower setup, materials, labor, equipment, and inventory holding costs. 2. Top quality. Superior features, durability, safety, and convenience result from improved designs. High inventories force competitor H to choose between scrapping obsolete designs or delaying introduction of product improvements until the old inventory is consumed.
In either case, L gains a competitive advantage. 3. Consistent quality. Consistency in conforming to design specifications requires consistency in supplied materials, setups, and processes. Small lots made frequently tend to increase consistency. Again, advantage goes to L. 4. Delivery speed. Large lots take longer to produce than small lots. A customer will wait less time for competitor L to set up and produce orders made in small batches. 5. On-time delivery. Contrary to expectations, large inventories do not equate to on-time delivery. It’s more like, lots of inventory equals lots of chaos.
Big lots make big scheduling problems. Big lots get dropped, mishandled, and pilfered. Most lean companies experience dramatic improvement in on-time delivery. 6. Development speed. This response is similar to that given for top quality. Low inventories result in getting new designs to the market more quickly. 7. Customization. Lean companies usually don’t claim an advantage in customization. However, large inventories provide no advantage with regard to customization either. It remains unlikely that a customized product will be found in inventory, no matter how large. 8. Variety.
Mass customizers compete on service or product variety. They will keep products at raw material or component levels until a customer orders a specific configuration. Inventories are at as low a level as possible. 9. Volume flexibility. Lean (low inventory) companies tend to produce the same quantity of every product every day, but they claim considerable volume flexibility from month to month. On the other hand, a large finished goods inventory can be used to absorb volume fluctuations. In summary, a case can be made that several competitive priorities are not considered in the EOQ model.
It is sometimes difficult to place a dollar value on these competitive advantages, but the advantages invariably go to the low-inventory, small lot-size firm. So if the EOQ is too large, what is the “ideal” lot size? According to the lean philosophy, the “ideal” lot size is one. 2. The continuous review system requires the determination of two parameters: the order quantity and the reorder point. The ordering cost for each firm will decrease, which means that the economic order quantities will decrease. Because of this, there may be some implications for the logistics system.
Smaller, more frequent shipments could require more costly less-than-truckload shipments. In addition, while the order quantities will decrease, the reorder points will also decrease because the lead times will be smaller. The supply chain should experience smaller pipeline inventories as a consequence. If the new information system also reduces the variance of demand or lead times, there can be additional safety stock savings. However, all of these benefits will come at some additional expense for the incorporation of the new system. There will be capital costs for equipment and potential training costs involved.
3. Organizations will never get to the point where inventories are unneeded. Inventories provide many functions and should be managed, not eliminated. It is impossible to eliminate uncertainties in the provision of products or services. In addition, unless materials can be transported instantaneously, there will always be pipeline inventories. Cycle inventories will exist unless we universally get to the point where production of single units is feasible. PROBLEMS 1. Lockwood Industries First we rank the SKUs from top to bottom on the basis of their dollar usage.
Then we partition them into classes. The analysis was done using OM Explorer Tutor12. 1—ABC Analysis. Cumulative % Cumulative % SKU # Description Qty Used/Year Value Dollar Usage Pct of Total of Dollar Value of SKUs Class 4 44,000 $1. 00 $44,000 60. 0% 60. 0% 12. 5% A 7 70,000 $0. 30 $21,000 28. 6% 88. 7% 25. 0% A 5 900 $4. 50 $4,050 5. 5% 94. 2% 37. 5% B 2 120,000 $0. 03 $3,600 4. 9% 99. 1% 50. 0% B 6 350 $0. 90 $315 0. 4% 99. 5% 62. 5% C 8 200 $1. 50 $300 0. 4% 99. 9% 75. 0% C 3 100 $0. 45 $45 0. 1% 100. 0% 87. 5% C 1 1,200 $0. 01 $12 0. 0% 100. 0% 100. 0% C Total $73,322
The dollar usage percentages don’t exactly match the predictions of ABC analysis. For example, Class A SKUs account for 88. 7% of the total, rather than 80%. Nonetheless, the important finding is that ABC analysis did find the “significant few. ” For the items sampled, particularly close control is needed for SKUs 4 and 7. 2. Stock-Rite Inc. Computing the annual usage value for each SKU and rank ordering them highest to lowest, we get: SKU Annual Value ($) Cumulative Value ($) D205 9,690 9,690 U404 6,075 15,765 A: 55% A104 3,220 18,985 L205 3,035 22,020 B: 22% L104 2,005 24,025 S104 1,604 25,629 X205
1,603 27,232 C: 23% X104 1,500 28,732 One classification might be to group the top two items (i. e. , 25% of the SKUs) in A class accounting for 55% of the total value. The next two SKUs would be classified as B and the last four as C. The dollar usage percentages don’t exactly match the predictions of ABC analysis. For example, Class A SKUs account for only 55% of the total, rather than 80%. Nonetheless, the important finding is that ABC analysis did find the “significant few. ” For the items sampled, particularly close inventory management is needed for SKUs D205 and U404. 3. Yellow Press, Inc. a.
Economic order quantity b. Time between orders 4. Babble Inc. a. D =( 400 tapes/month)(12 months/yr) = 4,800 tapes/year b. Time between orders years or 2. 5 months 5. Dot Com a. b. Optimal number of orders/year = (32,000)/400 = 80 orders c. Optimal interval between orders = 300/80 = 3. 75 days d. Demand during lead time = L = (5 days)(32,000/300) = 533 books e. Reorder point = L + safety stock = 533 + 0 = 533 books f. Inventory position = OH + SR – BO = 533 + 400 – 0 = 933 books 6. Leaky Pipe Inc. a. b. Optimal number of orders = (30,000)/(775) = 38. 7 or 39 c. Optimal interval between orders = (300)/(39) = 7.
69 days d. Demand during lead time = L = (4 days)(30,000/300) = 400 units e. Reorder point = L + safety stock = 400 + 0 = 400 units f. Inventory position = OH + SR – BO = 400 +775 – 0 = 1175 units 7. Sam’s Cat Hotel a. Economic order quantity = 90/week D = (90 bags/week)(52 weeks/yr) = 4,680 S = $54 Price = $11. 70 H = (27%)($11. 70) = $3. 16 = 399. 93, or 400 bags. Time between orders, in weeks b. Reorder point, R R = demand during protection interval + safety stock Demand during protection interval = L = 90 * 3 = 270 bags Safety stock = z? dLT When the desired cycle-service level is 80%, .
= 15 = 25. 98 or 26 Safety stock = 0. 84 * 26 = 21. 82, or 22 bags c. Initial inventory position = OH + SR – BO = 320 + 0 – 0 320 – 10 = 310. Because inventory position remains above 292, it is not yet time to place an order. d. Annual holding cost Annual ordering cost When the EOQ is used these two costs are equal. When , the annual holding cost is larger than the ordering cost, therefore Q is too large. Total costs are $789. 75 + $505. 44 = $1,295. 19. e. Annual holding cost Annual ordering cost Total cost using EOQ is $1,263. 60, which is $31. 59 less than when the order quantity is 500 bags.
8. Sam’s Cat Hotel, revisited a. If the demand is only 60 bags per week, the correct EOQ is: D = (60 units/wk)(52 wk/yr) = 3,120 bags = 326. 54, or 327 bags If the demand is incorrectly estimated at 90 bags, the EOQ would be incorrectly calculated (from problem 7) as 400 bags. The total cost, working with the actual demand, is: We can see clearly now that the cost penalty of Sam’s difficulty in foreseeing demand for kitty litter is $21. 31 ($1,053. 20 – $1,031. 89). b. If S = $6, and , the correct EOQ is: = 108. 85, or 109 bags The total cost, working with the actual ordering cost, is
If the reduced ordering cost continues to be unseen, the cost penalty for not updating the EOQ is (573. 91 – 343. 96) = $229. 95. 9. A Q system (also known as a reorder point system) = 300 pints/week = 15 pints a. Standard deviation of demand during the protection interval: = 15 = 45 pints b. Average demand during the protection interval: Demand during protection interval = L = 300 * 9 = 2700 pints c. Reorder point R = average demand during protection interval + safety stock Safety stock = z? dLT When the desired cycle-service level is 99%, z = 2. 33. Safety stock = 2. 33 * 45 = 104. 85 or 105 pints
R = 2,700 + 105 – 0 = 2,805 pints 10. Petromax Enterprises a. b. Safety stock = z? dLT = = (1. 28)(125) = 277. 13 or 277 units Reorder point= average lead time demand + safety stock = (3)(50,000/50) + 277 = 3,277 units 11. A continuous review system for door knobs. Find the safety stock reduction when lead time is reduced from five weeks to one week. Standard deviation of demand during the (five-week) protection interval is = 85 door knobs. Desired cycle service level is 99% (therefore z = 2. 33). Safety stock required for five-week protection interval: Safety stock = = 2. 33(85) = 198. 05, or 198 door knobs
Safety stock required for one-week protection interval ?dLT = = = 85 door knobs = 85/ = 38. 01 door knobs. Safety stock = = 2. 33(38. 01) = 88. 57 or 89 door knobs Safety stock reduction Reduction = 198 – 89 = 109 door knobs. 12. A two-bin system. “The two-bin system is really a Q system, with the normal level in the second bin being the reorder point R. ” Find cycle-service level, given: L = 2 weeks = 5 bolts = 53 bolts/weekR = 120 bolts Safety stock = R – L = 120 – (53*2) = 14 bolts Safety stock = z? dLT = 14 bolts ?dLT = = 5 = 7. 07 bolts z(7. 07) = 14 z = 1. 98 When z = 1. 98, the cycle-service level is 97.
67%. 13. Nationwide Auto Parts a. Protection interval (PI)= P + L = 6 +3 = 9 weeks Average demand during PI= 9 (100) = 900 units Standard deviation during PI= = 60 units b. Target inventory= (P+L) + z? P+L = 900 + (1. 96)(60) = 1,018 c. Order quantity= Target inventory – IP = 1,018 – 350 = 668 units presuming no SR or BO 14. A P system (also known as a periodic review system). Find cycle-service level, given: L = 2 weeks P = 1 week (P + L) = 218 boxes = 40 boxes T = 300 boxes T = Average demand during protection interval + Safety stock T = 218 + z(40) = 300 boxes z = (300 – 218)/40 = 2. 05 When z = 2.
05, cycle-service level is 97. 98 or 98%. 15. A Successful Product Annual Demand, D = (200)(50) = 10,000 units, H = ((0. 20)(12. 50)) = 2. 50 a. Optimal ordering quantity b. Safety stock = = (2. 33)(16) = 74. 56 or 75 units c. Safety stock will now be: (2. 33)(16) = 52. 72 or 53 units % reduction in safety stock= (75 – 53)/75 = 29. 33% d. Safety stock will be= (2. 33)(8) = 37. 28 or 38 units % reduction in safety stock= (75 – 38)/75 = 49. 33% 16. Sam’s Cat Hotel with a P system a. Referring to Problem 7, the EOQ is 400 bags. When the demand rate is 15 per day, the average time between orders is (400/15) = 26.
67 or about 27 days. The lead time is 3 weeks ? 6 days per week = 18 days. If the review period is set equal to the EOQ’s average time between orders (27 days), then the protection interval (P + L) = (27 + 18) = 45 days. For an 80% cycle-service level z = 0. 84 = 41. 08 Safety stock = = 0. 84(41. 08) = 34. 51 or 35 bags T = Average demand during the protection interval + Safety stock T = (15*45) + 35 = 710 b. In Problem 7, the Q system required a safety stock of 22 bags to achieve an 80% cycle-service level. Therefore, the P system requires a safety stock that is larger by (35 – 22) = 13 bags.
c. From Problem 7, inventory position, IP = 320. The amount to reorder is T – IP = 710 – 320 = 390. 17. Continuous review system. a. Economic order quantity. or 894 units Time between orders (TBO) = Q/D = 894/20,000 = 0. 0447 years = 2. 32 weeks b. Weekly demand = 20,000/52 = 385 units For a 95% cycle-service level, z = 1. 65 Safety stock: = (1. 65)(100) = 233. 34, or 233 units Now solve for R, as R = L + Safety stock = 385(2) + 233 = 1,003 units c. i. Annual holding cost of cycle inventory ii. Annual ordering cost d. With the 15-unit withdrawal, IP drops from 1,040 to 1,025 units.
Because this level is above the reorder point (1,025 > 1,003), a new order is not placed. 18. Periodic review system a. From Problem 17, or 894 units Number of orders per year = = 20,000/894 = 22. 4 orders per year. weeks P is rounded to 2 weeks. b. For a 95% cycle-service level, z = 1. 65. Therefore Safety stock = 200 units Safety stock = 1. 65(200) = 330 units, T = Average demand during the protection interval + Safety stock T = (385 * 4) + 330 = 1,870 units c. In Problem 17, with a Q system the safety stock is 233 units. Therefore, (330 – 233) = 97 more units of safety stock are needed. 19.
Continuous review system a. Economic order quantity b. Safety stock. When cycle-service level is 88%, z = 1. 18. Safety stock = = (1. 18)(12) = 20. 03, or 20 units c. Reorder point R = L + Safety stock = 64(2) + 20 = 148 units. d. If Q = 200 and R = 180, average inventory investment is higher than necessary to achieve an 88% cycle-service level. The larger order quantity increases average cycle stock by 20 units, and the higher reorder point increases safety stock by 32 units. 20. Periodic review system a. From problem 19, EOQ = 160 weeks P is rounded to 3 weeks. b. For an 88% cycle-service level, z = 1.
18. Therefore Safety stock = 26. 83 units. Safety stock = 1. 18(26. 83) = 31. 66, or 32 units T = average demand during the protection interval + Safety stock T = (64 * 5) + 32 = 352 units 21. Wood County Hospital a. D = (1000 boxes/wk)(52 wk/yr) = 52,000 boxes H = (0. l5)($35/box)=$5. 25/box The savings would be $3,229. 16 – $2,861. 82 = $367. 34. b. When the cycle-service level is 97%, z = 1. 88. Therefore, Safety stock = = (1. 88)(100) = 1. 88(141. 42) = 265. 87, or 266 boxes R = L + Safety stock = 1000(2) + 266 = 2,266 boxes c. In a periodic review system, find target inventory T, given: P = 2 weeks
L = 2 weeks Safety stock = = 200 units. Safety stock = 1. 88(200) = 376 units T = Average demand during the protection interval + Safety stock T = 1000(2 + 2) + 376 T = 4,376 units The table below is derived from OM Explorer Solver—Inventory Systems. Notice that the total cost for the Q system is much less than that of the P system. The reason is that the optimal value of P was not used here. The optimal value is weeks. Continuous Review (Q) system Periodic Review (P) System z = 1. 88 Time Between Reviews (P) 2. 00 Weeks ? Enter manually Safety Stock 266 Standard Deviation of Demand d During Protection Interval
200 Reorder Point 2266 Safety Stock 376 Annual Cost $4,258. 32 Average Demand During Protection Interval 4000 Target Inventory Level (T) 4376 Annual Cost $7,614. 00 22. Golf specialty wholesaler a. Periodic Review System or 179 1-irons or 4. 0 weeks When cycle-service level is 90%, z = 1. 28. Weekly demand is (2,000 units/yr)/(50 wk/yr) = 40 units/wk L = 4 weeks Safety stock: z = (1. 28) = 10. 86, or 11 irons T = (P+L) + Safety stock = 40(4+4) + 11 = 331 irons. b. Continuous review system Safety stock = = (1. 28)(3) = 1. 28(3)(2) = 7. 68, or 8 irons R = L + Safety stock = 40(4) + 8 =168 irons
23. Osprey Sports. a. The economic order quantity is = 289. 83, or 290 lures. b. The safety stock and reorder point are = 12. 41 lures The z value for a 97 percent cycle-service level = 1. 88. The safety stock = 1. 88 (12. 41) = 23. 33, or 23 lures The reorder point = + Safety stock = (4)(10) + 23 = 63 lures. c. The total annual cost for this continuous review system is + (H)(Safety stock) = = $312. 83 24. Farmer’s Wife a. The continuous review system is specified by the fixed order quantity and the reorder point. We will use the EOQ for the order quantity. The order quantity is: = 244.
95, or 245 cows. The safety stock is: = 61. 64 cows. The z value for a 90 percent cycle-service level = 1. 28. The safety stock = 1. 28 (61. 64) = 78. 90, or 79 cows. The reorder point = + Safety stock = (30)(8) + 79 = 319 cows b. The system would operate as follows: Whenever the stock of cows drops to 319, order 245 more cows. c. The total annual cost for this continuous review system is + (H)(Safety stock) = = $243. 06 25. Muscle Bound To find the cycle-service level, we must determine the standard deviation of demand during lead time and then use the equation for total annual cost to solve for z.
We will use the EOQ for the ordering quantity. The standard deviation of demand during lead time is = 5,078. 14 barbells The economic order quantity is = 3,538. 36, or 3,538 barbells The total annual cost (with z as a variable) is + (H)(Safety stock) = = $16,000 We now solve for z z = = 0. 8785, or 0. 88. This value of z corresponds to a cycle-service level of 81 percent. 26. Georgia Lighting Center. Using the demand data given in the problem statement, we extended text Table 12. 2 below the dashed line in the following way. The beginning inventory for day 7 is the ending inventory for day 6, which is 27 units.
The demand for day 7 is 7 units, which leaves 20 units in inventory at the end of day 7. No orders are open to the supplier; consequently the inventory position is 20 units. Because 20 units exceeds the reorder point of 15 units, no new order is placed. Continuing in this manner, the inventory position at the end of day 9 drops below the reorder point; consequently a new order for 40 units is placed. That order will be received three business days later, or day 12. The complete simulation results with Q = 40 and R = 15 are: Open Beginning Orders Daily Ending Inventory Amount Day Inventory Received Demand
Inventory Position Ordered 1 19 — 5 14 14 40 2 14 — 3 11 51 — 3 11 — 4 7 47 — 4 7 40 1 46 46 — 5 46 — 10 36 36 — Sat 6 36 — 9 27 27 — Mon7 27 — 7 20 20 — 8 20 — 4 16 16 — 9 16 — 2 14 14 40 10 14 — 7 7 47 — 11 7 — 3 4 44 — 12 4 40 6 38 38 — 13 38 — 10 28 28 — 14 28 — 0 28 28 — 15 28 — 5 23 23 — 16 23 — 10 13 13 40 17 13 — 4 9 49 — 18 9 — 7 2 42 — TOTAL 343 a. The average ending inventory is: or 19 units b. No stockouts occurred during any of the three cycles. myomlab Advanced PROBLEMS 1. Office Supply Shop The screen shot below is taken from OM Explorer Solver – Demand During Protection Interval Simulator.
It shows the results of 500 trials. a. Given the simulation, the value of R must yield a service level that meets or exceeds the desired value of 95%. That value of R is 71 pens, which will yield a cycle service level of 96. 4%. b. The average demand during the protection interval is 35 pens. Since the reorder point is 71, the safety stock must be 71 – 35 = 36 pens. The high level of safety stock is necessary because of the high variance in the demand during protection interval distribution and the high variance in lead time. 2. Grocery store. a.
The target level (T) should be 150 tubes of Happy Breath Toothpaste. This result comes from OM Explorer Solver – Demand During Protection Interval Simulator. b. Using OM Explorer once again, the cycle-service level for T = 150 would be 97. 8%. Eliminating the variance in supply lead times will significantly increase the cycle service level of the inventory. 3. Floral shop a. The EOQ for the continuous review system would be as follows. The demand during protection interval distribution is shown below. To attain at least a 90% cycle service level, the florist needs to set the reorder point at 166 baskets.
b. As the output from OM Explorer Solver – Q-System Simulator shows, the average cost per day is $274. 74. EXPERIENTIAL LEARNING: SWIFT ELECTRONIC SUPPLY, INC. This in-class exercise allows students to test an inventory system of their design against a new demand set. On the day of the simulation, students should come with sufficient copies of Table 1. Table Table1 12. 6 | Simulation Evaluation Sheet Day 1 2 3 4 5 6 7 8 9 10 Beginning inventory position Number ordered Daily demand Day-ending inventory Ordering costs ($200 per order) Holding costs ($0. 05 per piece per day) Shortage costs ($2 per piece)
Total cost for day Cumulative cost from last day Cumulative costs to date It is best to precede the simulation with a brief overview of the simulation process and the calculation of costs. The instructor may decide to require students to bring a computer to class and use a spreadsheet of their design to accomplish the tasks embodied in Table 1. Once everyone understands the simulation procedure, the instructor uses the “actual” demands in TN1, one at a time, and proceeding at a pace such that students have a chance to decide whether or not to order that period, how much to order, and calculate relevant costs.
The instructor can stop at any point, using TN2 to benchmark students’ results against any of the four provided systems in this manual. A good idea is to stop at the halfway point in the simulation and ask students what their total costs are. The variance is often quite high. The same benchmarking comparisons can be done at the end of the simulation. The instructor can use the students’ results to discuss differences in the systems tried, the importance of using safety stocks, and the value of perfect information. One of the provided systems in this manual utilizes the Wagner-Whitin (WW) approach, which is optimal for perfect forecasts.
The variance in student results will be greater if this exercise is used as a prelude to a discussion of formal inventory systems (such as the Q-system or P-system). Alternatively, the exercise can be used after a presentation of the formal systems to give students a practicum for the theory. TN3 shows the cost structure and system parameters for the EOQ-system, Q-system and P-system. All the relevant case information and derived data are on the left side of the sheet, and key computed parameters for three systems are presented on the right side of the sheet.
There are some other points that need to be addressed about TN3 through TN7: “Average Demand/day” and “Standard Deviation” come from a statistical analysis of the historical demand data in Table 12. 3. All the ordering quantities are rounded up as integers. Consequently, the associated costs might differ a little from what they actually are. The review time in the EOQ-system is actually up to the student. In TN4 we have used the EOQ divided by average daily demand. TN4 through TN6 show the application of the provided systems for the demand data in TN1.
TN7 shows the results from WW system. In all of our reported results, inventory levels at the start of the day are used to make inventory decisions. This is consistent with the daily purchasing routine at Swift. Economic Order Quantity (EOQ) System Under this system, students order the EOQ each and every review period, which using the case data would be 3 days, without any forecasts of future demand or consideration of demand variability. TN4 shows the performance of this system. Students may elect to use varying review periods. If so, their results will differ from TN4.
Q-system This system assumes that inventory levels are checked on a daily basis and compared to a “Reorder Point (RP). ” If actual inventory level goes below the RP, an order of EOQ is placed; if above, no order will be placed. In the provided results, the RP is calculated by adding safety stock to average demand during the two-day lead time. The safety stock is designed to meet the 95 percent cycle service level. TN5 shows the results of the Q-system. P-system The inventory level is reviewed every three days, which is determined by dividing EOQ by average demand.
The target inventory level is composed of two parts: “average demand during the protection interval,” which is the review period plus the lead time, and the “safety stock. ” Every review period (three days in the provided results), an order is placed to bring the inventory position up to the target inventory level. TN6 shows the performance of the P-system. Wager-Whitin (WW) System The WW system is based on dynamic programming and assumes all demands are known with certainty. Consequently, it provides an absolute lower bound on the solution found by the students.
The WW system assumes that stockouts are to be avoided. It is interesting to show the difference in total costs between the WW solution and another system because it demonstrates the cost of uncertainty. The solution using the WW system is shown in TN7. Also note that the lot sizes are shown in the day in which they must arrive. Actual release dates would be two days earlier. This implies that the first order for 1733 would have been placed in day 0, one day before the actual start of the simulation. TN 1. Actual Demand Data for Simulation CASE: PARTS EMPORIUM * A. Synopsis
This case describes the problems facing Sue McCaskey, the new materials manager of a wholesale distributor of auto parts. She seeks ways to cut the bloated inventories while improving customer service. Back orders with excessive lost sales are all too frequent. Inventories were much higher than expected when the new facility was built, even though sales have not increased. Summary data on inventory statistics, such as inventory turns, are not available. McCaskey decides to begin with a sample of two products to uncover the nature of the problems—the EG151 exhaust gasket and the DB032 drive belt.
B. Purpose The purpose of this case is to allow the student to put together a plan, using either a continuous review system (Q system) or a periodic review system (P system), for two inventory SKUs. Enough information is available to determine the EOQ and R for a continuous review system (or P and T for a periodic review system). Because stockouts are costly relative to inventory holding costs, a 95% cycle-service level is recommended. Inventory holding costs are 21% of the value of each item (expressed at cost).
The ordering costs ($20 for exhaust gaskets and $10 for drive belts) should not be increased to include charges for making customer deliveries. These charges are independent of the inventory replenishment at the warehouse and are reflected in the pricing policy. C. Analysis We now find appropriate policies for a Q system, beginning with the exhaust gasket. Shown here are the calculations of the EOQ and R, followed by a cost comparison between this continuous review system and the one now being used. The difference is what can be realized by a better inventory control system.
Reducing lost sales due to back orders is surely the biggest benefit. 1. EG151 Exhaust Gasket a. New plan Begin by estimating annual demand and the variability in the demand during the lead time for this first item. Working with the weekly demands for the first 21 weeks of this year and assuming 52 business weeks per year, we find the EOQ as follows: Weekly demand average = 102 gaskets/week Annual demand (D) = 102(52) = 5304 gaskets Holding cost = $1. 85 per gasket per year (or 0. 21 ? 0. 68 ? $12. 99) Ordering cost = $20 per order gaskets
Turning to R, the Normal Distribution appendix shows that a 95% cycle-service level corresponds to a z = 1. 65. We then use the EG151 data to find the standard deviation of demand. Standard deviation in weekly demand () = 2. 86 gaskets Standard deviation in demand during lead time R= Average demand during the lead time + Safety stock = 2(102) + 1. 65(4. 04) = 210. 66, or 211 gaskets b. Cost comparison After developing their plan, students can compare its annual cost with what would be experienced with current policies. Cost Category Current Plan Proposed Plan
Ordering cost $707 $313 Holding cost (cycle inventory) 139 314 TOTAL $846 $627 The total of these two costs for the gasket is reduced by 26 percent (from $846 to $627) per year. The safety stock with the proposed plan may be higher than the current plan, if the reason for the excess back orders is that no safety stock is now being held (inaccurate inventory records or a faulty replenishment system are other explanations). We cannot determine the safety stock level (if any) in the current system. The extra cost of safety stock for the proposed system is minimal, however.
Only seven gaskets are being proposed as safety stock, and their annual holding cost is just another $1. 85(7) = $12. 95. Surely the lost sales due to back orders are substantial with the current plan and will be much less with the proposed plan. One symptom of such losses is that 11 units are on back order in week 21. A lost sale costs a minimum of $4. 16 per gasket (0. 32. ? $12. 99). If 10 percent of annual sales were lost with the current policy, this cost would be $4. 16(0. 10)(5,304) = $2,206 per year. Such a loss would be much reduced with the 95% cycle-service level implemented with the proposed plan.
2. DB032 Drive Belt a. New plan The following demand estimates are based on weeks 13 through 21. Weeks 11 and 12 are excluded from the analysis because the new product’s start-up makes them unrepresentative. We find the EOQ as follows: Weekly demand average = 52 belts/week Annual demand (D) = 52(52) = 2704 belts Holding cost $0. 97 per belt per year (or 0. 21 ? 0. 52 ? $8. 89) Ordering cost $10 per order gaskets Turning now to R, where z remains at 1. 65, we use the data in the DB032 table to find: Standard deviation in weekly demand () = 1.
76 belts Standard deviation in demand during lead time belts R= Average demand during the lead time + Safety stock = 3(52) + 1. 65(3. 05) = 161. 03, or 161 belts b. Cost comparison After developing their plan, students again can compare the cost for the belts with what would be experienced with current policies. Cost Category Current Plan Proposed Plan Ordering cost $ 27 $115 Holding cost (cycle inventory) 485 114 TOTAL $512 $229 With the belt, the total of these two costs is reduced by 55 percent. The safety stock with the proposed plan