Chapter 13 - Inventory Management CHAPTER 13 INVENTORY MANAGEMENT Answers to Discussion and Review Questions 1. Inventories are held (1) to take advantage of price discounts, (2) to take advantage of economic lot sizes, (3) to provide a certain level of customer service, and (4) because production requires some in-process inventory. 2. Effective inventory management requires (1) cost information, information on demand and lead time (amounts and variabilities), an accounting system, and a priority system (e.g., A-B-C). 3. Carrying or holding costs include interest, security, warehousing, obsolescence, and so on. Procurement costs relate to determining how much is needed, vendor analysis, inspection of receipts and movement to temporary storage, and typing up invoices. Shortage costs refer to opportunity costs incurred through failure to make a sale due to lack of inventory. Excess costs refer to having too much inventory on hand. 13-1 Chapter 13 - Inventory Management 4. The RFID (Radio Frequency Identification) chip tags are beginning to be used with consumer products and they contain bits of data, such as product serial number. Scanners will automatically read the information on an RFID chip into a database, so the companies can keep track of sales and inventory. Keeping track of inventory will enable suppliers to keep track of trends and react to market changes. In addition, RFID chips will assist in increasing the speed of communication on a supply chain. The information between parties will travel faster, which will improve the responsiveness of buyers and ordering information on the supply chain. The risk of using RFID chip tags stems from privacy concerns. It is feared that computer pirates will figure out security controls and be able to scan shoppers’ merchandise and determine what they have bought. In order to avoid this risk, companies are considering turning of RFID tags once the items are purchased. 5. It may be inappropriate to compare the inventory turnover ratios of companies in different industries because the production process, requirements and the length of production run varies across different industries. The shorter the production time, the less the need for inventory. In addition, the material delivery lead times may vary between different industries. The higher the variability of lead time and the longer the lead time, the greater the need for inventory. As supplier reliability increases, the need for inventory decreases. The industries with higher forecast accuracies have less of a need for inventories. 6. a. Only one product is involved. b. Annual demand requirements are known. c. Demand is spread evenly throughout the year so that the demand rate is reasonably constant. d. Lead time does not vary. e. Each order is received in a single delivery. f. There are no quantity discounts 7. The total cost curve is relatively flat in the vicinity of the EOQ, so that there is a “zone” of values of order quantity for which the total cost is close to its minimum. The fact that the EOQ calculation involves taking a square root lessens the impact of estimation errors. Also, errors may cancel each other out. 8. As the carrying cost increases, holding inventory becomes more expensive. Therefore, in order to avoid higher inventory carrying costs, the company will order more frequently in smaller quantities because ordering smaller quantities will lead to carrying less inventory. 9. Safety stock is inventory held in excess of expected demand to reduce the risk of stockout presented by variability in either lead time or demand rates. 10. Safety stock is large when large variations in lead time and/or usage are present. Conversely, small variations in usage or lead time require small safety stock. Safety stock is zero when usage and lead time are constant, or when the service level is 50 percent (and hence, z = 0). 11. Service level can be defined in a number of ways. The text focuses mainly on “the probability that demand will not exceed the amount on hand.” Other definitions relate to the percentage of cycles per year without a stockout, or the percentage of annual demand satisfied from inventory. (This last definition often tends to confuse students in my experience.) Increasing the service level requires increasing the amount of safety stock. 13-2 Chapter 13 - Inventory Management 12. The A-B-C approach refers to the classification of items stocked according to some measure of importance (e.g., cost, cost-volume, criticalness, cost of stockout) and allocating control efforts on that basis. 13. In effect, this situation is a “quantity discount” case with a time dimension. Hence, buying larger quantities will result in lower annual purchase costs, lower ordering costs (fewer orders), but increased carrying costs. Since it is unlikely that the compressor supplier announces price increases far in advance, the purchasing agent will have to develop a forecast of future price increases to use in determining order size. Unlike the standard discount approach, the agent may opt to use trial and error to determine the best order size, taking into account the three costs (carrying, ordering and purchasing). In any event, it is reasonable to expect that a larger order size would be more appropriate; although obsolescence may also be a factor. 14. Annual carrying costs are determined by average inventory. Hence, a decrease in average inventory is desirable, if possible. Average inventory is Qo/2, and Qo decreases (run size model) if setup cost, S, decreases. 15. The Single-Period Model is used when inventory items have a limited useful life (i.e., items are not carried over from one period to the next). 16. Yes. When excess costs are high and shortage costs are low, the optimum stocking level is less than expected demand. 17. A company can reduce the need for inventories by: a. using standardized parts b. improved forecasting of demand c. using preventive maintenance on equipment and machines d. reducing supplier delivery lead times and delivery reliability e. utilizing reliable suppliers and improving the relationships in the supply chain f. restructuring the supply chain so that the supplier holds the inventory g. reducing production lead time by using more efficient manufacturing methods h. developing simpler product designs with fewer parts. Taking Stock 1. a. If we buy additional amounts of a particular good to take advantage of quantity discounts, then we will save money on a per unit purchasing cost of the item. We will also save on ordering cost because since we bought a larger quantity, we will not have to order this item as frequently. However, as a result of ordering larger quantity, we will have to carry larger inventory in stock, which in turn will result in an increase in inventory holding cost. b. If we treat holding cost as a percentage of the unit price, as the unit price increases, so will the holding cost, and as a result, if we are using the EOQ approach, we will have to place a smaller order, resulting in a lesser inventory. On the other hand, if we use a constant amount of a holding cost, the inventory decisions will not be affected by the changes in unit cost (price) of the item. 13-3 Chapter 13 - Inventory Management c. Conducting a cycle count once a quarter instead of once a year will result in more frequent counting, which will result in an increase in labor and overhead costs. However, the more frequent counting would also lead to less errors in inventory accuracy and more timely detection of errors, which in turn would lead to timely deliveries to customers, less work in process inventory, more efficient operations, improved customer service, and assurance of material availability. 2. In making inventory decisions involving holding costs, setting inventory levels and deciding on quantity discount purchases, the materials manager, plant manager, production planning and control manager, the purchasing manager, and in some cases the planners who work in production planning and control or purchasing departments should be involved. The level and the nature of involvement will depend on the organizational structure of the company and the type of product being manufactured or purchased. 3. The technology has had a tremendous impact on inventory management. The utilization of bar coding has not only reduced the cost of taking physical inventory but also enabled real time updating of inventory records. The satellite control systems available in trucks and automobiles has enabled companies to determine and track the location of in-transit inventory. Critical Thinking Exercise 1. 2. 3. Including a wider range of foods provide fast food companies with a competitive edge in terms of improving customer satisfaction and service. However, it has also complicated the operational function of the company. Expansion of menu offerings can create problems for inventory management because there are more ingredients and inventory items to order and to control the levels of inventory. This means higher labor costs in terms of placing orders, increased storage facility needs and increased need for coordinating the shipments from the supplier so that deliveries can be at low cost and efficient. Increasing the variety of items on the menu will also cause problems with forecasting. Since we have more items on the menu, it is likely that the demand for current menu items will decrease. The forecasts for all items will need to be revised. If we are not able to estimate this possible decrease, then the forecasting problem will result in excess or insufficient inventory levels. a. How important is the item? For example, does it relate to a holiday or other important event, such as graduation cards? b. Are comparable substitutes readily available? c. What competitor alternatives are available to customers? d. Is this an occasional occurrence, or indicative of a larger, perhaps ongoing, problem? Among considerations are: How many stamps does he now have? Does he know how many he has? If so, how many? What is his usage rate or current need for stamps? What else does he need the cash for today? Can he get more money at a bank or ATM? How long will it be before he will return to the post office? Will the post office be closed for a holiday or a Sunday? Can he buy stamps elsewhere in case he runs low? How convenient is it for him to visit the post office? 13-4 Chapter 13 - Inventory Management Can he purchase “forever stamps” and temporarily avoid a price increase? 4. Student answers will vary. Memo Writing Exercises 1. Cost of carrying inventory must be weighted against the following costs: a. cost of shortages (finished goods inventory) b. cost of failures (work-in-process inventory) c. cost of supplier reliability d. cost of ordering e. cost of setups f. cost of quantity discounts g. cost of smoothing production for seasonal products. 2. The possible advantages of using a single supplier include: a. obtaining a discount due to additional volume purchased from the supplier b. building trust and working with the supplier so that the material will be delivered in a timely fashion to avoid stockouts and excess inventory. The possible advantages of using multiple suppliers include: a. the adverse effect of tardiness will be felt much less when there are multiple sources for the materials. The main advantage of using the fixed order interval model is the reduction in ordering cost because orders for different parts are aggregated during the order interval. The main disadvantage of using the fixed order interval model is that the company faces the risk of experiencing shortages during the fixed interval. In this particular situation, the advantages of using a single supplier may be offset by using the fixed order interval model. 13-5 Chapter 13 - Inventory Management Solutions 1. a. Item 4021 Usage 90 Unit Cost $1,400 9402 300 12 3,600 C 4066 30 700 21,000 B 6500 9280 150 10 20 1,020 3,000 10,200 C C 4050 80 140 11,200 C 6850 2,000 10 20,000 B 3010 4400 400 5,000 20 5 8,000 25,000 C B In descending order: Item Usage x Cost 4021 $126,000 1. b. Usage x Unit Cost $126,000 Category A 4400 25,000 B 4066 6850 21,000 20,000 B B 4050 9280 11,200 10,200 C C 3010 8,000 C 9402 3,600 C 6500 3,000 228,000 Category A B C Percent of Items 11.1% 33.3% 55.6% Category A Percent of Total Cost 55.3% 28.9% 15.8% 13-6 Chapter 13 - Inventory Management 2. The following table contains figures on the monthly volume and unit costs for a random sample of 16 items for a list of 2,000 inventory items. Item K34 K35 K36 M10 M20 Z45 F14 F95 F99 D45 D48 D52 D57 N08 P05 P09 Unit Cost Usage 10 200 25 600 36 150 16 25 20 80 80 250 20 300 30 800 20 60 10 550 12 90 15 110 40 120 30 40 16 500 10 30 Dollar Usage 2,000 15,000 5,400 400 1,600 16,000 6,000 24,000 1,200 5,500 1,080 1,650 4,800 1,200 8,000 300 Category C A B C C A B A C B C C B C B C a. Develop an A-B-C classification for these items. [See table.] b. How could the manager use this information? To allocate control efforts. c. It might be important for some reason other than dollar usage, such as cost of a stockout, usage highly correlated to an A item, etc. 3. D = 1,215 bags/yr. S = $10 H = $75 a. Q 2 DS 2(1,215)10 18 bags H 75 b. Q/2 = 18/2 = 9 bags c. D 1,215 bags 67.5 orders Q 18 bags / orders 13-7 Chapter 13 - Inventory Management TC Q / 2H d. D S Q 18 1,215 (75) (10) 675 675 $1,350 2 18 e. Assuming that holding cost per bag increases by $9/bag/year Q= 2(1,215)(10) 17 bags 84 TC 17 1,215 (84) (10) 714 714.71 $1,428.71 2 17 Increase by [$1,428.71 – $1,350] = $78.71 4. D = 40/day x 260 days/yr. = 10,400 packages S = $60 H = $30 a. Q0 2DS H b. TC Q D H S 2 Q 2(10,400 )60 203 .96 204 boxes 30 204 10,400 (30 ) (60 ) 3,060 3,058 .82 $6,118 .82 2 204 c. Yes d. TC 200 200 10,400 (30 ) (60 ) 2 200 TC200 = 3,000 + 3,120 = $6,120 6,120 – 6,118.82 (only $1.18 higher than with EOQ, so 200 is acceptable.) 13-8 Chapter 13 - Inventory Management 5. D = 750 pots/mo. x 12 mo./yr. = 9,000 pots/yr. Price = $2/pot S = $20 H = ($2)(.30) = $.60/unit/year a. Q0 TC 2DS H 2(9,000 )20 774 .60 775 .60 774 .6 9,000 (.60 ) (20 ) 2 774 .6 TC = 232.35 + 232.36 = 464.71 If Q = 1500 TC 1,500 9,000 (.6) (20) 2 1,500 TC = 450 + 120 = $570 Therefore the additional cost of staying with the order size of 1,500 is: $570 – $464.71 = $105.29 b. 6. Only about one half of the storage space would be needed. u = 800/month, so D = 12(800) = 9,600 crates/yr. H = .35P = .35($10) = $3.50/crate per yr. S = $28 Present TC : a. Q0 800 9,600 (3.50 ) (28) $1,736 2 800 2DS H TC at EOQ: 2(9,600 )$28 391 .93 [round to 392 ] $3.50 392 9,600 (3.50 ) (28) $1,371 .71 . Savings approx. $364.28 per year. 2 392 13-9 Chapter 13 - Inventory Management 7. H = $2/month S = $55 D1 = 100/month (months 1–6) D2 = 150/month (months 7–12) a. Q0 2DS H D1 : Q0 2(100 )55 74 .16 2 D 2 : Q0 2(150 )55 90 .83 2 b. The EOQ model requires this. c. Discount of $10/order is equivalent to S – 10 = $45 (revised ordering cost) 1–6 TC74 = $148.32 TC 50 50 100 ( 2) ( 45 ) $140 * 2 50 TC 100 100 100 ( 2) ( 45 ) $145 2 100 TC 150 150 100 ( 2) ( 45 ) $180 2 150 7–12 TC91 = $181.66 TC 50 50 150 ( 2) (45 ) $185 2 50 TC100 100 150 ( 2) ( 45 ) $167 .5 * 2 100 TC150 150 150 ( 2) ( 45 ) $195 2 150 13-10 Chapter 13 - Inventory Management 8. D = 27,000 jars/month H = $.18/month S = $60 a. 2DS 2(27 ,000 )60 4,242 .64 4,243 . H .18 Q TC= Q D H S 2 Q TC4,000 = $765.00 TC4,243 = $736 .67 $1.32 Difference 27,000 4,000 (60) 765 (.18) 2 4,000 TC4000 = 27 ,000 4,243 TC4243 = 60 763 .68 (.18) 2 4,243 b. Current: For D 27 ,000 6.75 Q 4,000 D to equal 10, Q must be 2,700 Q Q 2DS So 2,700 H 2(27,000)S .18 Solving, S = $24.30 c. the carrying cost happened to increase rather dramatically from $.18 to approximately $.3705. 2DS 2(27,000)5 0 Q 2,700 H H Solving, H = $.3705 13-11 Chapter 13 - Inventory Management 9. p = 5,000 hotdogs/day u = 250 hotdogs/day D= 250/day x 300 days/yr. = 75,000 hotdogs/yr. 300 days per year S = $66 H = $.45/hotdog per yr. a. 2DS p H pu Q0 2(75,000 )66 5,000 4,812 .27 [round to 4,812] .45 4,750 b. D/Qo = 75,000/4,812 = 15.59, or about 16 runs/yr. c. run length: Qo/p = 4,812/5,000 = .96 days, or approximately 1 day 10. p = 50/ton/day u = 20 tons/day D= 20 tons/day x 200 days/yr. = 4,000 tons/yr. 200 days/yr. S = $100 H = $5/ton per yr. 2DS p 2(4,000 )100 50 516 .40 tons [10,328 bags] H pu 5 50 20 a. Q0 b. I max Q 516 .4 (p u ) (30 ) 309 .84 tons [approx . 6,196 .8 bags ] P 50 Average is I max 309 .48 : 154 .92 tons [approx. 3,098 bags] 2 2 c. Run length = Q 516 .4 10 .33 days P 50 d. Runs per year: D 4,000 7.75 [approx. 8] Q 516 .4 e. Q = 258.2 TC = I max D H S 2 Q TCorig. = $1,549.00 TCrev. = $ 774.50 Savings would be $774.50 13-12 Chapter 13 - Inventory Management 11. S = $300 D = 20,000 (250 x 80 = 20,000) H = $10.00 p = 200/day u = 80/day a. Q0 2DS p 2(20,000 )300 H pu 10 200 200 80 Q0 = (1,095.451) (1.2910) = 1,414 units b. Run length = Q 1,414 7.07 days P 200 c. 200 – 80 = 120 units per day d. I max Q 1,414 (p u ) (200 80 ) 848 .0 units P 200 848 ÷ 80/day = 10.6 days - 1.0 setup 9.6 days No, because present demand could not be met. e. 1) Try to shorten setup time by .40 days. 2) Increase the run quantity of the new product to allow a longer time between runs. 3) Reduce the run size of the other job.] f. I max In order to be able to accommodate a job of 10 days, plus one day for setup, there would need to be an11 day supply at Imax, which would be 880 units on hand. Solving the following for Q, we find: Q Q ( p u) (200 80) 880 units P 200 Q = 1,467. Using formula 13-4 for total cost, we have TC @ 1,467 units = $8,489.98 TC @ 1,414 units = $8,483.28 Additional cost = $6.70 13-13 Chapter 13 - Inventory Management 12. p = 800 units per day d = 300 units per day Q0 = 2000 units per day a. Number of batches of heating elements per year = 75,000 37 .5 batches per year 2,000 b. The number of units produced in two days = (2 days)(800 units/day) = 1600 units The number of units used in two days = (2 days) (300 units per day) = 600 units Current inventory of the heating unit = 0 Inventory build up after the first two days of production = 1,600 – 600 = 1,000 units Total inventory after the first two days of production = 0 + 1,000 = 1,000 units. c. Maximum inventory or Imax can be found using the following equation: pd 800 300 2,000 I max Q0 (2,000)(.6 25) 1,250 units 800 p Average inventory I max 1,250 625 units 2 2 d. Production time per batch = Q 2,000 2.5 days P 800 Setup time per batch = ½ day Total time per batch = 2.5 + 0.5 = 3 days Since the time of production for the second component is 4 days, total time required for both components is 7 days (3 + 4). Since we have to make 37.5 batches of the heating element per year, we need (37.5 batches) x (7 days) = 262.5 days per year. 262.5 days exceed the number of working days of 250, therefore we can conclude that there is not sufficient time to do the new component (job) between production of batches of heating elements. An alternative approach for part d is: The max inventory of 1,250 will last 1250/300 = 4.17 days 4.17 – .50 day for setup = 3.67 days. Since 3.67 is less than 4 days, there is not enough time. 13-14 Chapter 13 - Inventory Management 13. D = 18,000 boxes/yr. S = $96 H = $.60/box per yr. a. Qo = 2DS H 2(18,000 )96 2,400 boxes .60 Since this quantity is feasible in the range 2000 to 4,999, its total cost and the total cost of all lower price breaks (i.e., 5,000 and 10,000) must be compared to see which is lowest. TC2,400 = 2,400 18,000 (.60) ($96) $1.20(18,000 ) $23,040 2 2,400 TC5,000 = 5,000 18,000 (.60) ($96) $1.15(18,000 ) $22,545 .6 [lowest ] 2 5,000 10,000 18,000 (.60) ($96) $1.10(18,000 ) $22,972 .80 2 10,000 Hence, the best order quantity would be 5,000 boxes. Lowest TC TC10,000 = TC 2,400 b. 5,000 10,000 Quantity D 18,000 3.6 orders per year Q 5,000 13-15 Chapter 13 - Inventory Management 14. a. S = $48 D = 25 stones/day x 200 days/yr. = 5,000 stones/yr. Quantity 1 – 399 400 – 599 600 + TC490 = Unit Price $10 9 8 a. H = $2 Q 2DS 2(5,000 )48 489 .90 H 2 490 5,000 2+ 48 + 9 (5,000) = $45,980 2 490 600 5,000 2+ 48 + 8 (5,000) = $41,000 2 600 600 is optimum. TC600 = b. H = .30P EOQ $8 2(5,000 )48 447 NF .30 (8) TC ( Not feasible at $8/stone) EOQ $9 2(5,000 )48 422 .30 (9) 422 447 600 Quantity (Feasible) Compare total costs of the EOQ at $9 and lower curve’s price break: Q D TC = (.30P) + (S) +PD 2 Q TC422 = 422 [.30($9)] + 2 5,000 ($48) + $9(5,000) = $46,139 422 600 5,000 [.30($8)] + ($48) + $8(5,000) = $41,120 2 600 Since an order quantity of 600 would have a lower cost than 422, 600 stones is the optimum order size. c. ROP = 25 stones/day (6 days) = 150 stones. TC600 = 13-16 Chapter 13 - Inventory Management 15. D = 4,900 seats/yr. H = .4P S = $50 Range 0–999 1,000–3,999 4,000–5,999 6,000+ P $5.00 4.95 4.90 4.85 H $2.00 1.98 1.96 1.94 Q 495 497 NF 500 NF 503 NF Compare TC495 with TC for all lower price breaks: 495 4,900 TC495 = ($2) + ($50) + $5.00(4,900) = $25,490 2 495 1,000 4,900 TC1,000 = ($1.98) + ($50) + $4.95(4,900) = $25,490 2 1,000 4,000 4,900 TC4,000 = ($1.96) + ($50) + $4.90(4,900) = $27,991 2 4,000 6,000 4,900 TC6,000 = ($1.94) + ($50) + $4.85(4,900) = $29,626 2 6,000 Hence, one would be indifferent between 495 or 1,000 units TC 495 497 500 503 1,000 4,000 Quantity 13-17 6,000 Chapter 13 - Inventory Management 16. D = (800) x (12) = 9600 units S = $40 H = (25%) x P For Supplier A: 2( 9 ,600 )( 40 ) Q13.6 475.27 ( not feasible ) (.25 )( 13.6 ) Q13.8 2( 9 ,600 )( 40 ) 471.81 (.25 )( 13.8 ) 9 ,600 471.81 TC 471.81 ( 40 ) [( 13.8 )( 9 ,600 )] 3.45 471.81 2 TC 471.81 813.88 813.87 132 ,480 TC 471.81 $134 ,107.75 9 ,600 500 TC 500 ( 40 ) [( 13.6 )( 9 ,600 )] (.25 )( 13.6 ) 500 2 TC 500 768 850 130 ,560 TC 500 $132 ,178 * For Supplier B: Q13.7 2( 9 ,600 )( 40 ) 473.53 (.25 )( 13.7 ) 9 ,600 473.53 TC 471.81 ( 40 ) [( 13.7 )( 9 ,600 )] (.25 )( 13.7 ) 473.53 2 TC 471.81 810.93 810.92 131,520 TC 471.81 $133 ,141.85 Since $132,178 < $133,141.85, choose supplier A. The optimal order quantity is 500 units. 13-18 Chapter 13 - Inventory Management 17. D = 3600 boxes per year Q = 800 boxes (recommended) S = $80 /order H = $10 /order If the firm decides to order 800, the total cost is computed as follows: D Q TC H S (P * D) 2 Q 800 3,600 TC Q 800 $10 $80 (3,600 x 1.1) 2 800 TC Q 800 4,000 360 3,960 8,320 Even though the inventory total cost curve is fairly flat around its minimum, when there are quantity discounts, there are multiple U shaped total inventory cost curves for each unit price depending on the unit price. Therefore when the quantity changes from 800 to 801, we shift to a different total cost curve. If we take advantage of the quantity discount and order 801 units, the total cost is computed as follows: D Q TC H S (P * D) 2 Q 801 3,600 TC Q 801 $10 $80 (3,600 x 1.0) 2 801 TC Q 801 4,005 359 .55 3,600 7,964 .55 The order quantity of 801 is preferred to order quantity of 800 because TCQ=801 < TCQ=800 or 7964.55 < 8320. EOQ 2DS H 2(3,600 )(80 ) 240 boxes 10 D Q TC EOQ H S (P * D) 2 Q 240 3,600 TC EOQ $10 $80 (3,600 x 1.1) 2 240 TC EOQ 1,200 1,200 3,960 6,360 The order quantity of 800 is not around the flat portion of the curve because the optimal order quantity (EOQ) is much lower than the suggested order quantity of 800. Since the EOQ of 240 boxes provides the lowest total cost, it is the recommended order size. 13-19 Chapter 13 - Inventory Management 18. Daily usage = 800 ft./day Lead time = 6 days Service level desired: 95 percent. Hence, risk should be 1.00 – .95 = .05 This requires a safety stock of 1,800 feet. ROP = expected usage + safety stock = 800 ft./day x 6 days + 1,800 ft. = 6,600 ft. 19. Expected demand during LT = 300 units dLT = 30 units a. Z = 2.33, ROP = exp. demand + Zd LT 300 + 2.33 (30) = 369.9 = 370 units b. 70 units (from a.) c. smaller Z = less safety stock ROP smaller: 20. LT demand = 600 lb. d LT = 52 lb. risk = 4% Z = 1.75 a. ss = Zd LT = 1.75 (52 lbs.) = 91 lbs. b. ROP = Average demand during lead time + safety stock ROP = 600 + 91 = 691 lbs. c. With no safety stock risk is 50%. 21. d d LT SL a. b. = 21 gal./wk. = 3.5 gal./wk. = 2 days = 90 percent requires z = +1.28 90% 0 1.28 z-scale 6 8.39 gallons ROP d(LT) z LT (d ) 21(2/7) 1.28 (2/7) (3.5) 8.39 gallons 10 2 Q d(OI LT ) Z d OI LT A 21 1.28(3.5) 12 / 7 8 33 .87 7 7 or approx. 34 gal./wk. Average demand per day = 21 gallons / 7days per week = 3 gallons = Average demand during lead time = (3 gallons) (2 days) = 6 gallons 2 L (3.5) 1.871 7 ROP 8 6 1.069 L 1.871 Z is approximately 1.07. From Appendix B, Table B, the lead time service level is .8577. Risk of stockout is 1 - .8577 = .1423 Z 13-20 Chapter 13 - Inventory Management c. 1 day after From a, ROP = 8.39 2 more days on hand = ROP – 2 gal. = 6.39 6.39 = 21 (2/7) + Z P (stockout) = ? solving , Z d = 21 gal./wk. d = 3.5 gal./wk. 22. 23. 24. d ROP LT ss Risk 2 / 7 (3.5) 6.39 6 .208 .21 1.871 From Appendix B, Table B, Z=.21 gives a risk of 1 – .5832 = .4168 or about 42% = 30 gal./day = 170 gal. = 4 days ss = Zd LT = 50 = 50 gal. = 9% Z = 1.34 Solving, d LT = 37.31 3% Z = 1.88 x 37.31 = 70.14 gal. D ROP LT LT = 85 boards/day = 625 boards = 6 days = 1.1 day SL 96% Z = 1.75 ROP = d x LT + Z d LT 625 = 85 x 6 + Z (85) 1.1 Z = 1.23 10.93% .1093 approx. 11% 2 ROP = dLT + Z LT d d LT 2 4(4) 144(1) d = 12 units/day LT = 4 days = 12 (4) + 1.75 d = 2 units/day LT = 1 day = 48 + 1.75 (12.65) = 48 + 22.14 = 70.14 13-21 2 Chapter 13 - Inventory Management 25. LT = 3 days D = 4,500 gal 360 days/yr. d S = $30 H = $3 4 ,500 12.5 / day 360 d = 2 gal. Risk = 1.5% Z = 2.17 a. Qty. Unit Price Qo = 2DS 300 H 1 – 399 $2.00 400 – 799 1.70 800+ 1.62 TC = Q/2 H + D/Q S + PD TC300 = 150 (3) + 15 (30) + 2(4,500) = $9,900 TC400 = 200(3) + 11.25(30) + 1.70(4,500) = $8,587.50* TC800 = 400(3) + 5.625(30) + 1.62(4,500) = $8,658.75 b. ROP = d LT + Z LT d = 12.5 (3) + 2.17 = 37.5 + 7.517 = 45.02 gal. 26. d d LT S H a. 3 (2) = 5 boxes/wk. = .5 boxes/wk. = 2 wk. = $2 = $.20/box D = .5 boxes/wk. x 52 wk./yr. = 26 boxes/yr. 2DS 2(260 )2 Q0 72 .11 round to 72 H .20 b. ROP = d (LT) + z LT (d) ROP d (LT ) 12 5(2) z 2.83 LT ( d ) 2 (.5) Area under curve to left is .9977, so risk = 1.0000 – .9977 = .0023 c. Q 0 d(OI LT) z d OI LT A Thus, 36 = 5(7 + 2) + z(.5) 7 2 – 12 Solving for z yields z = +2.00 which implies a risk of 1.000 – .9772 = .0228. 13-22 Chapter 13 - Inventory Management 27. d d LT LT SL = 80 lb. = 10 lb. = 8 days = 1 day = 90 percent, so z = +1.28 a. ROP = d (LT )+z 2 LT 2 d d 2 LT 8(10) 2 80 2 (1) 2 = 640 + 1.28 (84.85) = 748.61 [round to 749] b. E(n) = E(z) dLT = .048(84.85) = 4.073 units = 80 (8) + 1.28 28. D = 10 rolls/day x 360 days/yr. = 3,600 rolls/yr. d = 10 rolls/day LT = 3 days H = $.40/roll per yr. d = 2 rolls/day S = $1 a. 2DS 2(3,600)1 Q0 134.16 [round to 134] H .40 b. SL of 96 percent requires z = +1.75 ROP = d (LT) + z LT (d) = 10(3) + 1.75 (3) (2) = 36.06 [round to 36] c. E(n) = E(z) d LT = .016( LT )(d) = .016 3 (2) = .055/cycle D 3600 E(N) = E(n) = .0554 = 1.488 or about 1.5 rolls Q 134 d. 1 SLannual E ( z ) dLT 3.464 (.016 ) .000413 Q 134 .16 SLannual 1 .000413 .9996 13-23 Chapter 13 - Inventory Management 29. (Partial Solution) SLannual = 99% a. SS = ? b. Qo = 179 cases E(z) d LT Q E(z) d LT .99 = 1 – 179 Solving, E(z) = 0.358 From Table 12–3, Z = 0.08 Hence, the probability of a stockout is 1-.8577=.1423. SLannual = 1 – risk = ? dLT = 80, d LT = 5 a. ss = Zd LT = .08 (5) = .40 cases b. 1 – .5319 = .4681 30. d = 250 gal./wk. H = $1/month d = 14 gal./wk. S = $20 LT = 10 wk. .5 Yes SLannual = 98% a. Q0 2DS 2(13,000 )20 208 .16 208 H 12 SLannual = 1 – (z) dLT Q dLT = LT d .02 = (z) 9.90 208 E(z) = .42 z = –.04 SL = .4840 = 5 (14) = 9.90 b. E(n) = E(z) dLT 5 = E(z)9.90 E(z) = .505 z = –.20 SL = .4207 31. D = 250 gal./wk. x 52 wk./yr. = 13,000 gal./yr. SS = –.04(9.90) = –.40 SS = zdLT SS = –.20(9.90) = –1.98 units FOI Q = d (OI + LT) + zd OT LT A SL = .98 = d (16) + 2.05d 16 – A Cycle 1 640 + 2.05(3) 16 – 42 = 622.6 → 623 units OI = 14 days LT = 2 days 2 640 + 2.05(3) 16 – 8 = 656.6 → 657 units D = 40/day 3 640 + 2.05(3) 16 – 103 = 561.6 → 562 units d = 40/day 2 = 3/day 13-24 Chapter 13 - Inventory Management 32. 50 wk./yr. P34 D = 3,000 units d d LT unit cost H S Risk P35 D = 3,500 units = 60 units/wk. = 4 units/wk. = 2 wk. = $15 = (.30)(15) = $4.50 = $70 = 2.5% d d LT unit cost H S Risk = 70 units/wk. = 5 units/wk. = 2 wk. = $20 = (.30)(20) = 6.00 = $30 = 2.5% Q = (OI + LT) d + z Q P 34 2(3,000 )70 305 .5 306 units 4.50 LT d – A QP35 = 70 (4 + 2) + 1.96 4 2 (5) – 110 QP35 = 420 + 24 – 110 ROPP34 = d x LT+ z LT d QP35 = 334 units ROPP34 = 60(2) + 1.96 2 (4) = 131.1 33. a. Item H4-010 H5–201 P6-400 P6-401 P7-100 P9-103 TS-300 TS-400 TS-041 V1-001 Annual $ volume 50,000 240,800 279,300 174,000 56,250 165,000 945,000 1,800,000 16,000 132,400 Classification C B B B C C A A C C Item H4-010 H5-201 P6-400 P6-401 P7-100 P9-103 TS-300 TS-400 TS-041 V1-001 Estimated annual demand 20,000 60,200 9,800 14,500 6,250 7,500 21,000 45,000 800 33,100 Ordering cost 50 60 80 50 50 50 40 40 40 25 b. 13-25 Unit holding cost ($) .50 .80 8.55 3.60 2.70 8.80 11.25 10.00 5.00 1.40 EOQ 2,000 3,005 428 635 481 292 386 600 113 1,087 Chapter 13 - Inventory Management 34. Cs = Rev – Cost = $4.80 – $3.20 = $1.60 Ce = Cost – Salvage = $3.20 – $2.40 = $.80 Cs $1.60 1.6 SL = = = = .67 Cs + Ce $1.60 + $.80 2.4 Since this falls between the cumulative probabilities of .63(x = 24) and .73(x = 25), this means Don should stock 25 dozen doughnuts. 35. Cs = $88,000 Ce = $100 + 1.45($100) = $245 Cs $88,000 a. SL = = = .9972 Cs + Ce $88,000 + $245 Using the Poisson probabilities, the minimum level stocking level that will provide the desired service is nine spares (cumulative probability = .998). 13-26 .4545 –.11 0 78.9 80 x Demand 19 20 21 22 23 24 25 26 27 . . . z-scale doz. doughnuts P(x) .01 .05 .12 .18 .13 .14 .10 .11 .10 . . . Cum. P(x) .01 .06 .18 .36 .49 .63 .73 .84 .94 . . . [From Poisson Table with = 3.2] x Cum. Prob. 0 .041 1 .171 2 .380 3 .603 4 .781 5 .895 6 .955 7 .983 8 .994 9 .998 . . . . . . Chapter 13 - Inventory Management b. SL Cs Cs Ce .041 Cs C s 245 .041(C s 245 ) C s .041C s 10 .045 C s .959 C s 10 .045 C s $10 .47 Carrying no spare parts is the best strategy if the shortage cost is less than or equal to $10.47 ( C s 10.47 ). 36. Cs = Rev – Cost = $5.70 – $4.20 = $1.50/unit Ce = Cost – Salvage = $4.20 – $2.40 = $1.80/unit $1.50 Cs $1.50 SL = = = = .4545 $3.30 Cs + Ce $1.50 + $1.80 The corresponding z = –.11 So = d – z d = 80 – .11(10) = 78.9 lb. 37. d = 80 lb./day d = 10 lb./day d = 40 qt./day A stocking level of 49 quarts translates into a z of + 1.5: d = 6 qt./day S– d Ce = $.35/qt z= = 49 – 40 = 1.5 Cs = ? d 6 S = 49 qt. This implies a service level of .9332: .9332 0 1.5 z-scale 40 49 quarts Cs Cs Thus, .9332 = Cs + Ce Cs + $.35 Solving for Cs we find: .9332(Cs + .35) = Cs; Cs = $4.89/qt. SL = Customers may buy other items along with the strawberries (ice cream, whipped cream, etc.) that they wouldn’t buy without the berries. 13-27 Chapter 13 - Inventory Management 38. Cs = Rev – Cost = $12 – $9 = $3.00/cake Ce = Cost – Salvage = $9 – ½ ($9.00) = $4.50/cake Demand is Poisson with mean of 6 Cs $3.00 SL = = = .40 Cs + Ce $3.00 + $4.50 Since .40 falls between the cumulative probability for demand of 4 and 5, the optimum stocking level is 5 cakes. 39. 40. [From Poisson Table with = 6.0] Demand Cum. Prob. 0 .003 1 .017 2 .062 3 .151 4 .285 .40 5 .446 6 .606 . . . . . . Cs = $.10/burger x 4 burgers/lb. = $.40/lb. Ce = Cost – Salvage = $1.00 – $.80 = $.20/lb. Cs $.40 SL = = = .6667. Cs + Ce $.40 + $.20 The appropriate z is +.43. So = + z = 400 + .43(50) = 421.5 lb. Cs = $10/machine Ce = ? S = 4 machines .6667 Demand Freq. 0 .30 1 .20 2 .20 3 .15 4 .10 5 .05 1.00 0 .43 z-scale 400 421.5 lb. Cum. Freq. .30 .50 .70 .85 .95 1.00 $10 .95. $10 + Ce Setting the ratio equal to .85 and solving for Ce yields $1.76, which is the upper end of the range. Setting the ratio equal to .95 and again solving for Ce , we find Ce = $.53, which is the lower end of the range. b. The number of machines should be decreased: the higher excess costs are, the lower SL becomes, and hence, the lower the optimum stocking level. c. For four machines to be optimal, the SL ratio must be between .85 and .95, as in part a. Setting the ratio equal to .85 yields the lower limit: Cs .85 = Solving for Cs we find Cs = $56.67. Cs + $10 Setting the ratio equal to .95 yields the upper end of the range: Cs .95 = Solving for Cs we find Cs = $190.00 Cs + $10 a. For four machines to be optimal, the SL ratio must be .85 13-28 Chapter 13 - Inventory Management 41. a. Ratio Method # of spares Probability of Demand 0 0.10 1 0.50 2 0.25 3 0.15 Cumulative Probability 0.10 0.60 0.85 1.00 Cs = Cost of stockout = ($500 per day) (2 days) = $1000 Ce = Cost of excess inventory = Unit cost – Salvage Value = $200 – $50 = $150 SL Cs 1,000 .869 C s C e 1,000 150 Since 86.9% is between cumulative probabilities of 85% and 100%, we need to order 3 spares. b. Stocking Level 0 1 2 3 Tabular Method Demand = 0 Prob. = 0.10 $0 .10(1)($150)=$15 .10(2)($150)=$30 .10(3)($150)=$45 Demand = 1 Prob. = 0.50 .50(1)($1000)=$500 $0 .50(1)($150)=$75 .50(2)($150)=$150 Demand = 2 Prob. = 0.25 .25(2)($1000)=$500 .25(1)($1000)=$250 $0 .25(1)($150)=$37.50 We should order 3 spares. 13-29 Demand = 3 Prob. = 0.15 .15(3)($1000)=$450 .15(2)($1000)=$300 .15(1)($1000)=$150 $0 Expected Cost $1,450 $565 $255 $232.50 Chapter 13 - Inventory Management 42. a. Ratio Method: Demand and the probabilities for the cases of wedding cakes are given in the following table. Demand 0 1 2 3 Probability of Demand 0.15 0.35 0.30 0.20 Cumulative Probability 0.15 0.50 0.80 1.00 Cs = Cost of stockout = Selling Price – Unit Cost = $60 – $33 = $27 Ce = Cost of excess inventory = Unit Cost – Salvage Value = $33 – $10 = $23 SL Cs 27 .54 Cs Ce 27 23 Since the service level of 54% falls between cumulative probabilities of 50% and 80%, the supermarket should stock 2 cases of wedding cakes. b. Tabular Method Stocking Level 0 1 2 3 Demand = 0 Prob. = 0.15 $0 .15(1)($23)=$3.45 .15(2)($23)=$6.90 .15(3)($23)=$10.35 Demand = 1 Prob. = 0.35 .35(1)($27)=$9.45 $0 .35(1)($23)=$8.05 .35(2)($23)=$16.10 Demand = 2 Prob. = 0.30 .30(2)($27)=$16.20 .30(1)($27)=$8.10 $0 .30(1)($23)=$6.90 The supermarket should stock 2 cases of wedding cakes. 43. Cs = $99, Ce = $200 SL = 99___ 99 + 200 = 0.3311. z = −0.44. Overbook: 18 – 0.44(4.55) = 15.998, or 16 tickets. 44. Mean usage = 4.6 units/day Standard dev. = 1.265 units/day LT = 3 days ROP = 18 Using equation 12-13: 18 = 4.6(3) + z(1.265)√3 Solving, z = 1.92 which gives a service level of 97.26%. 13-30 Demand = 3 Prob. = 0.20 .20(3)($27)=$16.20 .20(2)($27)=$10.80 .20(1)($27)=$5.40 $0 Expected Cost $41.85 $22.35 $20.35 $33.35 Chapter 13 - Inventory Management Case: UPD Manufacturing 1 Students must recognize that without demand variability, the fixed order interval order quantity equation reduces to: Q = d(LT + OI) – Available (because there is no safety stock) Since Available = d x LT, the fixed order interval order quantity equation Q further reduces to the following: Q = d x OI = (6) (89) = 534 units Therefore, ordering at six-week intervals requires an order quantity of 534 units. On the other hand, the optimal order quantity is determined by using the basic EOQ equation. Q 2(89 )(32 ) 2dS 267 h .08 The weekly total cost based on optimal order quantity EOQ is given below: d Q 89 267 TC EOQ S H 32 .08 2 267 2 Q TC EOQ 10 .67 10 .68 21 .35 / week The weekly total cost based on six-week fixed order interval (FOI) order quantity is given below: d Q 89 534 TC FOI S H 32 .08 2 534 2 Q TC FOI 5.33 21 .36 26 .69 / week Weekly savings of using EOQ rather than 6-week FOI is 26.69 – 21.35 = $5.34 The annual savings = (52 weeks) ($5.34 /week) = $277.68 2. The total annual savings as a result of switching from six-week FOI to EOQ are relatively small and switching to the optimal order quantity may not be warranted. However, even though the absolute value of the savings is relatively small, the percentage of savings is approximately 25% (5.34 / 21.35). Therefore if FOI approach is used with other parts or components as well, the total potential loss may be significant. 13-31 Chapter 13 - Inventory Management Case: Harvey Industries In order to improve the current inventory control system, the new president may want to consider the following: 1. Computerize the inventory control system. There are too many parts for the current manual system. 2. Currently, no paper work is used when items are withdrawn from the stockroom when they are needed on the shop floor. Harvey Industries may either want to establish a procedure for recording the transactions in the stockroom or invest in a bar coding system. If a bar coding system is purchased, it has to be coordinated with the new computerized inventory control system. Establishing a cycle counting procedure may be very helpful also. As a result of these actions, the inventory accuracy should substantially improve. 3. It appears that utilization of ABC inventory classification system is needed. The company should never experience stockouts in their basic “C” items. ABC analysis will allow Harvey Industries to establish an appropriate degree of control over items in terms of order quantity and ordering frequency. Case: Grill Rite The president’s stance on steady output conflicts with seasonal demand. However, it is unlikely that this will change. One alternative might be to identify a complementary product that would offset seasonal demand for electric grills. The main problem is inventory management. One advantage of having a single, centralized warehouse is the lower need for safety stock due to the canceling effect of random variability in orders from the various regions. Conversely, with separate warehouses, each warehouse needs a relatively larger safety stock to guard against variations in demand. What is needed is overall control of the system that would take into account seasonal variations in demand and achieve a better match between regional demand and supply. This might involve making or improving regional forecasts. In any case, improved system visibility is essential: direct access to regional warehouse data by the main warehouse is needed in order to be able to coordinate and set priorities on inventory shipments to regional warehouses. That should take care of most of the problem. It may also pay to examine the feasibility of shipping from one warehouse to another when a shortage occurs. Relevant costs would include transaction costs, transportation costs, versus the potential increase in profit by making up a shortage. 13-32 Chapter 13 - Inventory Management Case: Farmers Restaurant 1. Inventories are crucial not only to Farmers Restaurant, but to businesses in general. Customer satisfaction and customer return is contingent upon proper inventory management. If customers visit the Farmers Restaurant and are unable to receive the food they desire due to stockout, the customer may be dissatisfied and may not return to Farmers Restaurant. In addition to customer satisfaction, total food costs are also important to businesses. In the restaurant industry if too much of a product is ordered and not used; it could result in product wastage due to the items expiring. This would result in an increase in total food costs when the goal is to keep food costs low! Overstocking products can also negatively affect Farmers Restaurant and other businesses. By having more products on-hand than needed, Farmers Restaurant is tying up funds that might be more productive elsewhere. Overall, it is imperative that Farmers Restaurant try to successfully manage inventory levels due to these potential/existing problems. 2. A fixed-interval ordering system is appropriate. 3. Q = ˉd (OI + LT) + zσd √(OI + LT) – A OI = Time between orders LT = Lead Time It is expected demand between orders and lead time + safety stock – inventory on hand Q = 5 (3+2) + 1.64 (3.5) √(3 + 2) – 3 = 25 + 13 – 3 = 35 (Therefore, 18 2-packs.) 4. 12 = 5 x 2 + z (3.5) (1.41) 12 = 10 + z (4.935) 2 / 4.935 = z (4.935) / 4.935 Z = 0.41 Service Level = 0.6554 1 – 0.6554 = 0.3446 Therefore, the risk for stockout is 34.46% which is high 13-33 Chapter 13 - Inventory Management Operations Tour: Bruegger’s Bagel Bakery 1. If too little inventory is maintained, there is a risk of stockout and potential lost sales. In addition, if there is not sufficient work-in-process inventory, the production process may become too inefficient, raising the cost of production. On the other hand, if too much inventory is maintained, the carrying cost may become excessively high. 2. a. Customers judge the quality of bagels by their appearance (size, shape and shine), taste and consistency. Customers are also very interested in receiving high service quality. b. At each stage of the production process, workers check bagel quantity. c. Steps for Bruegger’s Bagel Bakery Operations: a. Purchase ingredients from suppliers b. Mix the dough c. Shape the dough d. Ship to stores e. Store the bagels f. Boil and malt g. Bake h. Sell to customers The company can improve quality at each step by monitoring output more carefully and training and education of the employees. 3. The basic ingredients can be purchased using either fixed order interval or fixed order quantity models. EOQ production lot size model is most appropriate for deciding the size of production quantity. 4. If there is a bagel-making machine at each store, the company would have to invest in more machinery, more space for production and storage, and more worker training for the production of bagels. However, the lead time to make the bagels will be shortened. The shorter lead time will provide faster, more flexible response to customer demands and fresher bagels. Enrichment Module: Advanced EOQ Problem This enrichment module consists of an additional EOQ problem to further solidify the concepts associated with the Economic Production Model. The question forces the students to work backwards through the EOQ production equation. Problem A company produces plastic powder in lots of 2000 pounds, at the rate of 250 pounds per hour. The company uses powder in an injection molding process at the steady rate of 50 pounds per hour for an eight-hour day, five days a week. The manager has indicated that the setup cost is $100 for this product, but “We really have not determined what the holding cost is.” 13-34 Chapter 13 - Inventory Management a. b. What weekly holding cost per pound does the lot size imply, assuming the lot size is optimal? Suppose the figure you compute for holding cost has been shown to the manager, and the manager says that it is not that high. Would that mean the lot size is too large or too small? Explain. Solution to Enrichment Module Problem a. Q 2,000 lbs. p 250 lbs. / hr. u d 50 lbs. / hr. S $100 D (50 lbs. / hr.) x (8 hrs. / day) x (5 days / week ) 2,000 lbs. / week Q 2 DS H p x pd 2,000 2(2,000 )(100 ) 250 x H 250 50 2,000 400 ,000 x(1.25) H 2,000 500 ,000 H (2,000 ) 2 500 ,000 H 500 ,000 $. 125 / week 4,000 ,000 Decreasing the value of carrying cost (H) will result in an increase in the lot size. Since holding inventory is not as expensive, the firm can afford to carry more inventory and therefore produce a larger batch. H b. 13-35 Chapter 13 - Inventory Management Enrichment Module 2: Inventory Model with Planned Shortages In most cases, shortages are undesirable and should be avoided. However, in certain circumstances, it may be desirable to plan and allow for shortages. Planned shortages are implemented for high dollar volume items where the inventory carrying cost is very high. The model discussed in this section refers to the specific type of shortages called backorders. When a customer attempts to purchase an out-of-stock item, the firm does not lose the sale. The customer waits until the purchased order arrives from the supplier. If there were no additional cost associated with backordering, there would be no incentive for the firm to maintain any inventory. However, there are costs associated with backordering. The tangible part of the backorder cost involves the cost of expediting the delivery (special delivery) and production of the backordered item. The intangible part of the backordering cost involves the loss of goodwill due to the fact that the customers are forced to wait for their orders. The longer the waiting period, the higher the backorder cost due to loss of goodwill. There is a direct trade-off between the inventory carrying cost and the cost of a planned shortage in the form of backorders. In many cases the cost of backorders can be easily offset by the reduction in carrying costs. The model discussed in this section will not be valid if a customer decides not to wait for the backorder. The fixed order quantity inventory model with planned shortages (backorders) is very similar to the basic EOQ model. When the reorder point is reached, a new economic order quantity (Q) is placed. Figure 1 shows the schematic representation of this model. The size of the backorder is B units and the maximum inventory is Q–B units. The average size of the backorder is B/2 for each order cycle. T is defined as the amount of time between two successive orders (a complete order cycle). t1 is the part of the order cycle where the customer orders are met from stock. In other words, during t1 there is positive inventory level. On the other hand, t2 is the period of time in the order cycle where the inventory is depleted and all the customer orders are placed on backorder (stockout period). Symbol definitions used to explain various concepts are listed below. H = carrying cost per unit per year S = setup cost per batch (lot) D = annual demand Q* = optimal order quantity B = size of the backorder CB = backorder cost per unit per year B* = optimal planned backorder quantity T = Q/D (length of the complete order cycle in years) or T = Q/d (length of the complete order cycle in days) t1 = (Q – B)/D or (Q – B)/d (time period during which inventory is positive) t2 = B/D or B/d (time period during which there is no inventory) In this model, the average inventory is not Q/2 or not even (Q – B)/2 because during the shortage period there are no units in inventory. The average inventory calculation for this model can be explained with the following example: 13-36 Chapter 13 - Inventory Management A large local car dealership orders a certain brand of automobiles from a car manufacturer located in Detroit. Order quantity (Q) is 500 units, annual demand (D) is 7500 and the firm operates 300 working days per year. Due to the high holding costs, the company plans to backorder (B) 200 cars per order cycle. Determine the average inventory. d = (D/number of operational days) = 7,500/300 = 25 units (daily consumption) T = Q/d = 500/25 = 20 days. (time between orders is 20 days) t1 = (Q – B)/d = (500 – 200)/25 = 300/25 = 12 days (time period during which there is no shortage) t2 = B/d = 200/25 = 8 days (time period during which there is no inventory) The dealership will carry an average of (Q – B)/2 units during t1 and no units during t2. Therefore, total number of unit days during the inventory cycle can be computed by multiplying t1 with (Q – B)/2 Number of unit days of inventory/cycle t1 * Number of unit days of inventory/cycle QB 2 (Q B) (Q B) d 2 (Q B) 2 2d In other words, an average of 150 units are carried in inventory for 12 days and zero units are carried for 8 days (shortage period). Therefore, total number of unit days of inventory during the complete order cycle is (150)(12) = 1800. Since there are a total of 20 days in the complete order cycle, the average inventory can be computed by dividing the total number of unit days of inventory by the number of days in the inventory cycle. In this example, the average inventory is equal to 1,800/20 or 90 units. Therefore, the average inventory can be computed by using the following formula: (Q B ) 2 2d Average inventory Q d Average Inventory (Q - B) 2 2Q Using a similar logic, we can also develop the average backlog formula. The dealership will experience shortage (backorders) for 8 days during the order cycle. The average amount of backorder on a given shortage day is B/2. Based on this information, the total number of backorder unit days can be computed using the following equation: (t2) (B/2) = (B/D)(B/2) = B2 /2D. In our example, there are 8 days of a planned shortage period. During this period, an average of 200/2 = 100 units of backorders are realized. Therefore, the total number of backorder unit days during the order cycle is (8)(100) = 800 units. Since there are a total of 20 days in the order cycle, the average backorder quantity for the complete order cycle can be determined by dividing the total number of backorder unit days by the number of days in the complete inventory cycle. In this example, using the above equation, we obtain an average backorder quantity of 800/20 = 40 units. The general equation for the average backorder quantity is: 13-37 Chapter 13 - Inventory Management B2 Average backorder 2d Q d Average backorder B2 2Q Annual inventory carrying cost is still calculated by multiplying the average inventory with the inventory carrying cost per unit per year. The formula for the annual ordering (setup) cost is the same as it was for the basic EOQ model. The annual backorder cost is determined by multiplying the average backorder quantity with the backorder cost per unit per year. The annual inventory carrying cost is given by: (Q B) 2 H 2Q The annual ordering and backordering costs are given by the following respective formulas: D S Q B2 CB 2Q Therefore, the total annual inventory cost (TC) can be expressed by summing the annual inventory carrying cost, annual ordering cost and the annual backordering cost as shown in the following formula: TC (Q B) 2 D B2 H S CB 2Q Q 2Q Taking the first total derivative of the above total cost formula with respect to Q and setting the resulting equation to zero and solving for Q will result in the following optimal quantity (Q*) and optimal backorders (planned shortages) (B*) formulas: Q* 2 DS H C B H C B H B* Q * H B 13-38 Chapter 13 - Inventory Management Figure 1 An inventory situation with planned shortages Inventory Maximum Inventory Level Q–B Q Stockout B t1 Time QB d T = Q/d t2 Example: XYZ Company distributes a major part for the F–15 fighter jets. Due to the very high holding cost, the company wants to implement a model with planned shortages. The annual demand is 78,000 and the company operates 300 days per year. The annual carrying rate is 10% of the item cost and the unit cost of this item is $1,000. The setup cost per batch is estimated at $500. a. Determine the optimal order quantity and total annual inventory cost (setup cost + carrying cost) using the basic EOQ model with no backorders. b. If each unit backordered costs the company $200 per unit per year, what would be the optimal order quantity and the optimal size of the planned backorder? c. Determine the annual carrying cost, the annual setup cost, the annual backordering cost and the annual total inventory cost for the planned shortage model used in part b. d. Determine the values of t1, t2 and T in days. e. Should the company adopt the planned backorder model of part b or the basic EOQ model of part a which does not allow backorders? D = 81,000 units S = $500 d = 81,000/300 days = 27 units per day. H = ($1,000) (.10) = $100 CB = $200 a. Q* 2DS H Q* 2(81,000 )(500 ) 900 100 D 81,000 S (500 ) 45,000 900 Q annual setup cost = Q 900 annual carrying cost = (H) (100 ) 45,000 2 2 13-39 Chapter 13 - Inventory Management Q* 2DS ( H C B ) H C B Q* 2(81,000 )(500 ) 100 200 H 200 3 Q* 81,000 2 Q * 1,215 ,000 1,102 .3 H B Q * H CB 100 B (1,102 .3) 367 .433 100 200 c. Annual carrying cost (Q B) 2 H 2Q (1,102 .3 367 .43) 2 (100 ) 2(1,102 .3) 24,495 .77 D 81,000 Annual setup cost (S) (500 ) 1,102 .3 Q 36,741 .35 B2 C B Annual backordercost 2Q 367 .433 2 (200 ) 12,247 .55 2(1,102 .3) Let TC = Total annual inventory cost HC = annual inventory holding cost SC = annual setup cost BC = annual backordering cost TC = HC + SC + BC = 24,495.77 + 36,741.35 + 12,247.55 = 73,485.07 13-40 Chapter 13 - Inventory Management d. e. d 81,000 27 300 T Q 1,102 .3 40 .83 days d 27 t1 Q B 1,102 .3 367 .43 27 .22 days d 27 t2 B * 367 .43 13 .61 days d 27 The model with planned backorders is preferred because the total annual inventory cost of the basic inventory model is substantially higher than the total annual inventory cost of the planned backorder model. TCbasic EOQ = 90,000 TCbackorder = 73,485.07 90,000 – 73,485.07 = 16,514.93 The total cost savings equal 18.4% Problems The manager of an inventory system believes that inventory models are important decision-making aids. Although the manager often uses an EOQ policy, he has never considered a backorder model because of his assumption that backorders are “bad” and should be avoided. However, with upper management’s continued pressure for cost reduction, you have been asked to analyze the economics of a backordering policy for some products that can possibly be backordered. For a specific product with D = 800 units per year, C0 =$150, H = $10, and Cb = $20, what is the economic difference in the EOQ and the planned shortage or backorder model? If the manager adds constraints that no more than 35% of the units may be backordered and that no customer will have to wait more than 20 days for an order, should the backorder inventory policy be adopted? Assume 250 working days per year. Solution to Problem D = 800 units/year C0 = $150 H = $10/unit/year Cb = $20/unit/year Planned shortage model: 13-41 Chapter 13 - Inventory Management 2DS ( H C B ) H C B Q* 2( 800 )( 150 ) ( 10 ) 20 10 20 Q* 92 ,000 189 .737 units H 10 ( 189 .737 ) 63 .24 units B* Q * 30 H CB EOQ model: 2DS H Q* 2(800 )(150 ) 154 .92 units 10 Total cost planned shortage model: TC = Total annual inventory cost HC = Annual inventory holding cost SC = Annual setup cost BC = Annual backordering cost HC = (Q B) 2 (189 .737 63 .24 ) 2 H (10 ) $421 .68 2Q 2(189 .737 ) D 800 SC = S (150 ) $632 .45 Q 189 .737 B2 (63 .24 ) 2 C B 20 $210 .78 BC = 2Q 2(189 .737 ) TC = HC + SC + BC = 421.68 + 632.45 + 210.78 TC = $1,264.91 Total cost regular EOQ model: Q 154 .92 HC = H (10 ) $774 .60 2 2 D 800 S (150 ) $774 .60 Q 154 .92 SC = TC = HC + SC = 774.60 + 774.60 =$1,549.20 TCDifference = 1,549.20 – 1,264.91 = $284.29 Using the planned shortage model will result in savings of $284.29. Number of orders = D 800 4.216 orders Q 189 .737 13-42 Chapter 13 - Inventory Management D Q Expected annual number of units short = (B) Expected annual number of units short = (63.24)(4.216) = 266.44 d= D 800 3.2 units/day 250 250 t2 = 63 .24 19 .763 days 3.2 Since 19.763 < 20 and 266 .44 < .35, the backorder inventory policy should be adopted. 800 13-43