Inventory Class Notes Why pay attention to Inventory May be the largest current asset A shortage may cause a stockout and an unhappy customer A shortage may result in a production stoppage Too much inventory can mean higher carrying costs and negatively affect profits. Inventory Functions Decoupling - When a manufacturing or other process consists of two or more separate activities and where the product of one activity may be a component of another activity, it is important to have an inventory of these products so that the manufacturing process can operate efficiently and without delay. Storing resources - Farmers store hay Smoothing out irregularities in supply – Demand for many products is seasonal. Manufactures must often produce more products in the off-season and then store them until the demand increases. Buying or producing in lots or batches – to reduce costs by getting quantity discounts. Allows organizations to cope with perishable materials – perishable goods. Storing Labour – oil producers Avoiding stockouts and shortages. 1 Two Inventory Decisions 1. How much to order 2. When to order A major objective is to minimize total inventory costs. Some of the most significant costs are: 1. 2. 3. 4. The actual cost of the items. The cost of Ordering. The cost of Carrying (holding ) Inventory. The cost of Stockouts. Economic Order Quantity (EOQ) Model Assumptions 1. Demand D is known and occurs at a constant rate. 2. Lead Time is constant. Lead time is the time between the placement of the order and the receipt of the order. 3. The order quantity Q is the same for each order. The inventory level increases by Q units every time an order is received. 4. The cost per order Co, is constant and does not depend on quantity ordered. 5. The purchase cost per unit C, is constant and does not depend on quantity ordered. 6. The inventory holding cost per unit per time period, Ch, is constant. The total inventory holding cost depends on both Ch and the size of the inventory. 7. Shortages such as stock-outs or backorders are not permitted. 8. The inventory position is reviewed continuously. As a result, an order is placed as soon as an inventory position reaches the reorder point. 2 See figure 6.2 on page 189 a description of inventory usage over time. Economic Order Quantity Model – EOQ EOQ is a model designed to determine that particular quantity to order that will minimize total inventory costs. Types of Inventory Costs Ordering Costs include all of the costs incurred into getting a unit into the firm’s inventory and are expressed as a dollar cost per order. See table 6.1 on page 188. Holding costs (carrying costs) are the costs incurred because a firm owns or maintains inventories. Some of the items included are: interest expense, obsolescence, storage space, storage administration, taxes, insurance, and theft. Holding costs can be expressed as a percentage of average inventory value or as a cost per unit per time period. See table 6.1 on page 188. How Many to Order 2DCo Q OPTIMAL UNITS PER ORDER C h Q = number of pieces per order (cases, truckloads etc.) EOQ = Q* = optimal # of pieces to order D = annual demand in units Co = ordering cost of each order Ch = holding cost per unit per year 3 Graphically in figure 6.3 it shows that when EOQ assumptions are met, total cost is minimized when Ordering costs are equal to Holding costs. Mathematically this is shown as: 𝑄 𝐷 𝐶ℎ = 𝐶𝑜 2 𝑄 Solving for Q we get: 2DCo Q OPTIMAL UNITS PER ORDER C h Sumco Pump Company Example EOQ = Q* = ? D = 1000 units Co = $10/order Ch = $0.50 per unit per year Q 2(1000)(10) = √40,000 = 200 units 0.50 TC D Co Q C IS TOTAL COST 2 h Q 1000 (10) 200 (0.5) 50 + 50 = 100 2 200 Since Annual Demand is 1000 units and there are 200 units per order, then there must be 5 orders per year. TC Average Inventory is 𝑸 𝟐 = 𝟐𝟎𝟎 𝟐 = 𝟏𝟎𝟎 Do problem 6-20 4 6-20. D = 100,000; Co = $10; Ch = $0.005 a. Q* 2 100, 000 10 20, 000 number 6 screws 0.005 b. Number of orders per year D 100, 000 5 Q 20, 000 Total ordering cost = 5($10) = $50 per year c. Average inventory Q 20, 000 10, 000 units 2 2 Total holding cost = 10,000(0.005) = $50 per year When to Order Reorder Point (ROP) ROP = (demand per day) X (lead time in days) Example D = Annual Demand = 8000 units d = Daily demand = 40 units Q = Order Quantity = 400 units L = Lead Time = 3 days ROP = (demand per day) X (lead time in days) = D x L ROP = (40) X (3) = 120 Do 6-27 5 Production Run Model Eliminating the Constant Supply Rate Assumption in EOQ Assuming Production rate exceeds demand rate During the production run excess inventory is built up. When the production run ends, inventory is gradually reduced until the next production run is started. Where: Q = number of pieces per order, or production run Ch= holding cost per unit of inventory per year Cs = setup cost (replaces ordering cost) D = Annual Demand in units d= daily demand rate p = daily production rate t = length of production run in days 𝑸∗ = √ 𝑪 𝟐𝑫𝑪𝒔 𝒅 (𝟏− ) 𝒉 𝒑 = Optimal Production Quantity 6 Brown Manufacturing Annual Demand = D = 10,000 units Setup Costs = Cs = $100 Carrying Cost = Ch = $0.50 per unit per year Daily Production Rate = p = 80 units daily Daily Demand rate = d = 60 units daily 𝑸∗ ∗ = √ 𝑪 𝑸 = √ 𝟐𝑫𝑪𝒔 𝒅 (𝟏− ) 𝒉 𝒑 = Optimal Production Quantity 𝟐(𝟏𝟎,𝟎𝟎𝟎)(𝟏𝟎𝟎) 𝟎.𝟓(𝟏− 𝟔𝟎 ) 𝟖𝟎 = 4000 units Length of the production cycle = 𝐷 𝑄 𝑝 Production runs per year = 𝑄 = = 4000 80 10,000 4000 𝑑 = 50 𝑑𝑎𝑦𝑠 = 2.5 Maximum Inventory Level = 𝑄(1 − 𝑝) Students Do 6-28 7 6-27. a. EOQ 2 2,500 18.75 2 DCo 250 units . Ch 1.5 b. Average inventory Annual holding cost Q 250 125 2 2 Q Ch 125 1.5 $187.50 2 c. Number of orders per year Annual ordering cost D 2,500 10 Q 250 D Co 10 18.75 187.5 Q d. Total cost = $187.5 + $187.5 + 2500(15) = $37,875 e. With 250 days per year, and 10 orders per year, the number of days between orders = 250/10 = 25 days. f. ROP = d × L = (2500 units per year/250 days per year) × 2 = 20 units 6-28. a. daily demand = 2500/250 = 10 units per day b. Q* 2 2,500 25 2 DCs 324.92 d 10 1.48 1 Ch 1 p 50 c. 324.92/50 = 6.5 days. Inventory sold = (10 units/day)(6.5 days) = 65 units. d. Maximum inventory level = Q(1 – d/p) = 324.92(1 – 10/ 50) = 259.94 Average inventory = 0.5(Maximum inventory level) = 0.5 (259.94) = 129.97 Annual holding cost = (average inventory) Ch = 129.97 (1.48) = $192.35 e. Number of production runs = D/Q =2500/324.92 = 7.694 Annual setup cost = (D/Q)Cs = 7.694(25) = $192.35 f. Including the cost of production, the annual cost is $192.35 + $192.35 + 2,500(14.80) = $37,384.71 g. ROP = d × L = 10 × 0.5 = 5 units 8 Quantity Discount Models 1. For each discount price(C), compute EOQ = √ 2𝐷𝐶𝑜 𝐼𝐶 2. If EOQ < minimum for discount, adjust the quantity to Q = minimum for discount. 3. For each EOQ or adjusted Q, compute total cost 𝑇𝐶 = 𝐷 𝑄 𝐶𝑜 + 𝐶ℎ + 𝐷𝐶 𝑄 2 4. Choose the lowest cost quantity. Quantity Discounts D = 5000 Category 1 2 3 Co = 49 Order Size 0 to 999 1000 to 1999 2000 and over Ch = 20% of unit cost Unit Cost $5.00 $4.80 $4.75 Q 1 2DCo 2(5000)49 700 IC (0.20)(5.00) Q 2 2DCo 2(5000)49 714 (0.20)(4.80) IC Q 3 2DCo 2(5000)49 718 IC (0.20)(4.75) 9 Since Q2 and Q3 are not high enough to qualify for the discount they must be adjusted upward to the nearest quantity that would allow the product to be purchased at the discount price. Therefore Q2 = 1000 and Q3 = 2000 In the original EOQ model the annual purchase cost was not included because it was constant and never affected by the inventory order policy. However, with discounts, the annual purchase cost (DC) is included in the equation for total cost. TC D Co Q C DC 2 h Q 700 ( 5000 ) (1.00) (5000)(5.00) 25,700 TC (49) 1 700 2 1000 (0.96) (5000)(4.80) 24725 TC (5000) (49) 2 1000 2 2000 (0.95) (5000)(4.75) 24,822.50 TC (5000) (49) 3 2000 2 10 6-38. D = 1,000; unit cost = $50; Co = $40; Ch = 0.25 × unit cost Q 2 1, 000 40 80 0.25 50 With discount, unit cost = (1 – 0.03) × $50 = $48.50 Qd* 2 1, 000 40 81.22 0.25 48.50 which should be adjusted to minimum orderable quantity (i.e., 200). Original total cost 1, 000 50 1, 000 80 40 0.25 50 80 2 = $51,000 Discount cost 1, 000 48.50 1, 000 200 40 0.25 48.50 $49,912.50 200 2 Therefore, North Manufacturing should take the discount. Safety Stock Safety Stock is additional inventory carried by some businesses in order to avoid stockout situations during times when either demand or lead time is unusually high. Two Important Factors are: Stockout Cost Holding Cost Stockout cost usually involves lost sales and lost goodwill. How to determine the level of safety Stock – Service Level. Service level is the percentage of time the customers’ demands are met. When demand during lead time is normally distributed, we can determine the safety stock level using the normal distribution. 11 Where ROP = (Avg. demand during lead time) + ZσdLT Where Z is the number of standard deviations for a given service level. σdLT is the standard deviation of demand during lead time. Hinsdale Company – SKU A3378 Demand during lead time is normally distributed The mean µ = 350 The standard Deviation σ = 10 95% service level is desired. Therefore Z = 1.65 Since ROP = (Avg. demand during lead time) + ZσdLT, then ROP = 350 + 1.65(10) = 366.5 or about 367 units. Since demand during lead time is 350 then the safety sock level is 17 units. Do Problem 39 6-39. = 60; = 7 Safety stock for 90% service level = Z (at 0.90) = 7 × 1.28 = 8.96 9 6-40. Ch = $0.50; = 600; = 7 Safety stock for 90% service level 9 Carrying cost = 9 × 0.5 = $4.50 Safety stock for 95% service level = 7 × 1.65 12 Carrying cost = $6.00 Safety stock for 98% service level = 7 × 2.05 15 Carrying cost = $7.50 12 Omit 206 -209 Single Period Inventory Models Some products are either of no value or a much reduced value if they are not sold during the period for which they were produced or manufactured. For example: Newspapers, Magazines and Perishable foods Marginal Analysis With Discrete Distributions Marginal Profit (MP) the additional profit from one additional unit Marginal Loss (ML) is the loss that occurs when a unit is stocked but not sold. The decision to stock is based on probability. We will decide to stock an additional unit only if the expected marginal profit from that unit exceeds the expected marginal loss. Let P = the probability of selling one additional unit. Then (1-p) is the probability the unit will not sell. Therefore, Expected Marginal Profit = P(MP) and Expected Marginal Loss = (1-P)ML. The decision to stock then is only if P(MP) ≥ (1-P)ML. Solving for P, we get 𝑃 ≥ 𝑀𝐿 𝑀𝐿 + 𝑀𝑃 Steps of Marginal Analysis with Discrete Distribution 13 𝑀𝐿 Determine the value of 𝑃 ≥ . 𝑀𝐿+𝑀𝑃 Construct a probability table with a cumulative probability column. Keep ordering inventory as long as the probability of selling 𝑀𝐿 at least one additional unit is greater than . 𝑀𝐿+𝑀𝑃 Café du Donut MP = 2 ML = 4 Daily Sales 4 5 6 7 8 9 10 Step 1. Prob. Of Demand Cumulative Prob. 0.05 P(x ≥ 4) = 1.00 0.15 P(x ≥ 5) = 0.95 0.15 P(x ≥ 6) = 0.80 0.20 P(x ≥ 7) = 0.65 0.25 P(x ≥ 8) = 0.45 0.10 P(x ≥ 9) = 0.20 0.10 P(x ≥ 10) = 0.10 1.00 𝑃 ≥ 𝑀𝐿 𝑀𝐿+𝑀𝑃 𝑃 ≥ 4 4+2 = 4 6 = 0.67 Therefore keep stocking units as long as there is at least a .67 chance of selling the unit. Step 2. Add a cumulative probability column to the table. Step 3. Order 6 cartons 14 Do problem 6-53 6-53. Demand Probability P(Demand _____) 50 0.05 1.00 75 0.10 0.95 100 0.20 0.85 125 0.30 0.65 150 0.20 0.35 175 0.10 0.15 200 0.05 0.05 a. ML = 20; MP = 80 – 20 = 60 ML/(ML + MP) = 20/(20 + 60) = 0.25 Stock 150 b. ML = 35; MP = 80 – 35 = 45 ML/(ML + MP) = 35/(35 + 45) = 0.4375 Stock 125 c. Demand Probability P(Demand _____) 50 0.25 1.00 75 0.25 0.75 100 0.25 0.50 125 0.25 0.25 ML = 20; MP = 100 – 20 = 80 ML/(ML + MP) = 20/(20 + 80) = 0.20 Stock 125 15 Marginal Analysis with the normal distribution Need to know µ, σ, ML, MP Newspaper Example – Joe’s Newsstand µ = 60 σ = 10 Step 1. 𝑃 ≥ 𝑀𝐿 𝑀𝐿+𝑀𝑃 ML = 0.20 𝑃 ≥ .20 .20+.30 MP = 0.30 = .20 .50 = 0.40 Step 2. Use the normal curve to highlight the area of interest and determine the z value. Step 3. Solve for X using 𝑍 = 𝑋−𝜇 𝜎 where 𝑋 = 𝜇 + 𝑍𝜎 𝑋 = 60 + 0.25(10) = 62.5 𝑛𝑒𝑤𝑠𝑝𝑎𝑝𝑒𝑟𝑠 Do problem 6-55 6-55. ML = 35; MP = 15 ML/(ML + MP) = 35/(35 + 15) = 0.70 Using the normal distribution table, the z-value associated with the upper 70% of the normal distribution is between –0.52 and –0.53. We will choose –0.53 since P(Z – 0.53) is slightly more than 0.70. The quantity to stock is X* = + z = 45,000 – 0.53(4,450) = 42,641.5. 16 Victorian Bakery specializes in making cheesecakes that are then sold to restaurants. The cheesecakes can be produced for $3 each and sold for $7 each. Any cheesecakes left at the end of the day are sold to a retirement home for $1 each. The demand for the cheesecakes is normally distributed with a mean of 62 and a standard deviation of 15. Using marginal analysis, recommend the level of demand the bakery should plan to meet in order to maximize the expected value of profit. 17 µ = 62 σ = 15 Step 1. 𝑃 ≥ 𝑀𝐿 𝑀𝐿+𝑀𝑃 ML = 2 𝑃 ≥ 2 4+2 MP = 4 = 2 6 = 0.3333 Step 2. Use the normal curve to highlight the area of interest and determine the z value. Step 3. Solve for X using 𝑍 = 𝑋−𝜇 𝜎 where 𝑋 = 𝜇 + 𝑍𝜎 𝑋 = 62 + 0.44(15) = 68.6 𝑐ℎ𝑒𝑒𝑠𝑒𝑐𝑎𝑘𝑒𝑠 18 ABC Analysis Inventory Group Dollar Usage (%) Inventory Items (%) A B C 70 20 10 10 20 70 Use Quantitative Control? YES SOMETIMES NO 19