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Inventory (Chapter 16) What is Inventory? (to p2) How Inventory works: two ways of ordering based on three elements Inventory models (to p3) (to p4) (to p5) 1 Inventory - to study methods to deal with “how much stock of items should be kept on hands that would meet customer demand” Objectives are to determine: a) how much to order, and b) when to order (to p1) 2 Two basic types of Inventory Systems 1) continuous (fixed-order quantity) • an order is placed for the same constant amount when inventory decreases to a specified level, ie. Re-order point 2) periodic (fixed-time) • an order is placed for a variable amount after a specified period of time • used in smaller retail stores, drugstores, grocery stores and offices (to p1) 3 3 basic inventory elements 1. Carrying cost, Cc • Include facility operating costs, record keeping, interest, etc. 2. Ordering cost, Co • Include purchase orders, shipping, handling, inspection, etc. 3. Shortage (stock out) cost, Cs • Sometimes peanlties involved; if customer is internal, work delays could result (to p1) 4 Inventory models Here, we only study the following three different models: 1. Basic model (to p6) 8th ed: 1, 15, 19, 9th ed: 1, 10, 13, 2. Model with “discount rate” (to p23) 8th: 24, 26; 9th: 17, 18 3. Model with “re-order points” (to p31) 8th: 36, 38; 9th: 26,28 5 1. Basic model The basic model is known as: “Economic Order Quantity” (EOQ) Models Objective is to determine the optimal order size that will minimize total inventory costs (to p7) How the objective is being achieved? 6 3 Basic EOQ models Three models to be discussed: 1. 2. 3. (to p8) Basic EOQ model EOQ model without instantaneous (to p15) receipt (to p18) EOQ model with shortages. (to p5) 7 The Basic EOQ Model • The optimal order size, Q, is to minimize the sum of carrying costs and ordering costs. • Assumptions and Restrictions: - Demand is known with certainty and is relatively constant over time. - No shortages are allowed. - Lead time for the receipt of orders is constant. (will consider later) - The order quantity is received all at once and instantaniously. How to determine the optimal value Q *? (to p10) 8 Determine of Q We try to • • • Find the total cost that need to spend for keeping inventory on hands = total ordering + stock on hands Determine its optimal solution by finding its first derivative with respect to Q How to get these values? 1. Find out the total carrying cost 2. Find out the total ordering cost 3. Total cost = 1 + 2 4. d (Total cost) /d Q = 0, and find Q* (to p10) 9 The Basic EOQ Model We assumed that, we will only keep half the inventory over a year then The total carry cost/yr = Cc x (Q/2). Then , Total cost = TC C D C Q Q 2 o c Total order cost = Co x (D/Q) Finding optimal Q* (to p11) 10 The Basic EOQ Model • EOQ occurs where total cost curve is at minimum value and carrying cost equals ordering cost: TC C D C Q Q 2 o min Then, c Q * 2C D C o * (How to obtain this?) c •Where is Q* located in our model? (to p12) 11 The Basic EOQ Model • Total annual inventory cost is sum of ordering and carrying cost: To order inventory To keep inventory TC C D C Q Q 2 o c Figure 16.5 The EOQ cost model Try to get this value Examples (to p13) 12 The Basic EOQ Model Example Consider the following: Model parameters : Cc $0.75, Co $150, D 10,000yd Optimal order size : Q** 2CoD 2(150)(10,000) 2,000 yd Cc (0.75) Total annual inventory cost : TC min Co D Cc Qopt (150) 10,000 (0.75) (2,000) $1,500 Q* 2 2,000 2 Number of orders per year : D 10,000 5 Q * 2,000 Order cycle time 311 days 311 62.2 store days 5 D / Q* No. of working days/yr Note: You should pay attention that all measurement units must be the same (to p14) Consider the same example, with yearly 13 The Basic EOQ Model EOQ Analysis with monthly time frame Model parameters : Cc $0.0625 per yd per month, Co $150 per order, D 833.3 yd per month Optimal order size : Q* 2CoD 2(150)(833.3) 2,000 yd Cc (0.0625) Total monthly inventory cost : TC min Co D Cc Q * (150) (833.3) (0.0625) (2,000) $125 per month 2 2,000 2 Q* Total annual inventory cost ($125)(12) $1,500 (unit be based on yearly) 12 months a year 14 (to p7) The EOQ Model with Noninstantaneous Receipt The order quantity is received gradually over time and inventory is drawn on at the same time it is being replenished. (to p16) Example: Let p = production, d = demand, then the total cost (TC) = Figure 16.6 The EOQ model with noninstantaneous order receipt Always greater than 0 why? 15 The EOQ Model with Noninstantaneous Receipt Model Formulation p daily rate at which the order is received over time ( or production rate) d daily rate at which inventory is demanded Q Total annual inventory cost :TC Co D Cc 1 d Q 2 p Optimal order size : Q* Assuming placing an order/yr 2CoD Cc(1 d / p) (to p17) Example 16 The EOQ Model with Noninstantaneous Receipt Example Let, Co $150, Cc $0.75 per unit, D 10,000 yd per year, 10,000/311 32.2 yd per day, p 150 yd per day Optimal order size : Q* 2CoD 2(150)(10,000) 2,256.8 yd 32.2 d 0.751 Cc 1 150 p Total minimum annual inventory cost : TC min Co Production run length D Q * d (10,000) (2,256.8) 32.2 Cc 1 (150) (.075) 1 $1,329 Q* 2 p (2,256.8) 2 150 Q * 2,256.8 15.05 days p 150 Number of orders per year (production runs) D 10,000 4.43 runs Q * 2,256.8 d 32.2 Maximum inventory level Q * 1 2,256.81 1,772 yd p 150 ( to p7) 17 The EOQ Model with Shortages Here, we allow Q being shortage, so that we could borrow or replenish the stocks later Max level of inventory • In the EOQ model wth shortages, the assumption that shortages cannot exist is relaxed. • Assumed that unmet demand can be backordered with all demand eventually satisfied. Shortage = S/Q Shortage What we needed On hand = (Q-S)/Q Total cost is t1 + t2 = S/Q + (Q-S)/Q = 1 18 (to p19) The EOQ Model with Shortages ½*base* height = ½ * (Q-S) * (Q-S)/Q = ½ * (Q-S)2 /Q Area = ½ * (S/Q) * S = ½ * S2 /Q Total cost = (to p20) • In the EOQ model wth shortages, the assumption that shortages cannot exist is relaxed. • Assumed that unmet demand can be backordered with all demand eventually satisfied. Shortage = S/Q Shortage What we needed On hand = (Q-S)/Q 19 The EOQ Model with Shortages Total cost Total shortage costs total carrying costs total ordering cost Total inventory cost : TC Cs Optimal order quantity : Q* S2 (Q S ) 2 D Cc Co 2Q 2Q Q 2CoD Cs Cc Cc Cs Cc Cc Cs Shortage level : S * Q * Example (to p21) 20 The EOQ Model with Shortages Example Let, Co $150, Cc $0.75 per yd, Cs $2 per yd, D 10,000 yd Optimal order quantity : Q* 2CoD Cs Cc 2(150)(10,000) 2 0.75 2,345.2 yd Cc Cs 0.75 2 Shortage level : S* Q * Cc 2,345.2 0.75 639.6 yd Cc Cs 2 0.75 2 2 2 2 Total inventory cost :TC Cs S Cc (Q * S*) Co D (2)(639.6) (0.75)(1,705.6) (150)(10,000) 2,345.2 2Q * 2Q * Q * 2(2,345.2) 2(2,345.2) $174.44 465.16 639.60 $1,279.20 (to p22) 21 The EOQ Model with Shortages Additional Parameters in Example Number of orders D 10,000 4.26 orders per year Q 2,345.2 Maximum inventory level Q S 2,345.2 639.6 1,705.6 yd Time between orders t = Q/D days per year 311 73.0 days between orders number of orders 4.26 Q S 2,345.2 - 639.6 0.171 or 53.2 days D 10,000 Time during which there is a shortage t2 S 639.6 0.064 year or 19.9 days D 10,000 Time during which inventory is on hand t1 (to p7) 22 2. Model with “discount rate” Price discounts are often offered if a predetermined number of units is ordered or when ordering materials in high volume. How do we decide if we should order more to take advantage of the discount being offered? (to p24) 23 Quantity Discounts with Constant Carrying Costs Analysis Approach Solution method: 1. We first determine the optimal order size, Q* 2. We then compare with any lower total cost with a discount price and accept the one has the minimum total cost Example (to p25) 24 Quantity Discounts with Constant Carrying Costs Example (1 of 2) Consider the following example: The following discount schedule is offered, which size of order should we subscript? Co $2,500, Cc $190 per unit, D 200 Quantity 1- 49 50 – 89 90 + Then, we have Q* Price $1,400 1,100 900 Falls in this section, Now we compare the TC of this, to The next discount class, 90+ 2CoD 2(2,500)(200) 72.5 Cc 190 (to p26) 25 Quantity Discounts with Constant Carrying Costs Example (2 of 2) - Compute total cost at eligible discount price ($1,100): CoD Q* Cc PD Q* 2 (2,500)(200) (72.5) (190) (1,100)(200) $233,784 (72.5) 2 TC* - Compare with total cost of with order size of 90 and price of $900: CoD Q* TC* Cc PD Q* 2 (2,500)(200) (190)(90) (900)(200) $194,105 (90) 2 - Because $194,105 < $233,784, maximum discount price should be taken and 90 units ordered. Do we always take the offer? (to p27) 26 No • It depends on the carrying cost • Example: (to p28) • Carrying Costs as a Percentage of Price 27 Quantity Discounts with Carrying Costs as a Percentage of Price Example (1 of 2) - Consider the same example, but with 15% of TC as carrying cost - Then, the carrying cost for each category is as follows: - Data: Co = $2,500, D = 200 computers per year Quantity Price (TC) Carrying Cost 0 - 49 $1,400 1,400(.15) = $210 50 - 89 1,100 90 + 900 1,100(.15) = 165 900(.15) = 135 (to p29) Chapter 16 - Inventory Management 28 Quantity Discounts with Carrying Costs as a Percentage of Price Example (2 of 2) - Compute optimum order size for purchase price without discount and C c = $210: Q* 2CoD 2(2,500)(200) 69 Cc 210 - Compute new order size: Q* 2(2,500)(200) 77.8 165 - Compute minimum total cost: TC (note: 69 falls onto 50-89 category) CoD Q* (2,500)(200) (77.8) Cc PD 165 (1,100)(200) $232,845 Q* 2 77.8 2 - Compare with cost, discount price of $900, order quantity of 90: TC (2,500)(200) (135)(90) (900)(200) $191,630 90 2 - Optimal order size computed as follows: Q* 2(2,500)(200) 86.1 135 (less than 90 as needed) - Since this order size is less than 90 units , it is not feasible,thus optimal order size is 77.8 units. (to p5) 29 Quantity Discount Model Solution with QM for Windows Exhibit 16.4 30 3. Model with “re-order points” • The reorder point is the inventory level at which a new order is placed. • Order must be made while there is enough stock in place to cover demand during lead time. • Formulation: R = dL, where d = demand rate per time period, L = lead time Then R = dL = (10,000/311)(10) = 321.54 Working days/yr What would happen? (to p32) 31 Reorder Point • Inventory level might be depleted at slower or faster rate during lead time. • When demand is uncertain, safety stock is added as a hedge against stockout. Two possible scenarios No Safety stocks! Safety stock! We should then ensure Safety stock is secured! How? (to p33) 32 Determining Safety Stocks Using Service Levels • We apply the Z test to secure its safety level, R d L (Zd L ) Safety stock Reorder point Average sample demand How these values are represented in the diagram of normal distribution? 33 (to p34) Reorder Point with Variable Demand R d L Zd L where R reorder point d average daily demand L lead time d the standard deviation of daily demand Z number of standard deviations corresponding to service level probabilit y Zd L safety stock Example (to p35) 34 Reorder Point with Variable Demand Example Example: determine reorder point and safety stock for service level of 95%. d 30 yd per day, L 10 days, 5 yd per day d For 95% service level, Z 1.65 (Appendix A) R d L Z L 30(10) (1.65)(5)( 10 ) 300 26.1 326.1 yd d Safety stock is second term in reorder point formula : 26.1. (to p5) 35 Determining the Reorder Point with Excel Exhibit 16.5 36