Chapter 11 Inventory Management

Types of Inventories Raw materials & purchased parts Work in progress Incoming students Work in progress Current students Finished-goods inventories (manufacturing firms) or merchandise (retail stores) Graduating students Replacement parts, tools, & supplies Goods-in-transit to warehouses or customers Students on leave

Functions of Inventory To meet anticipated demand To smooth production requirements To decouple components of the production-distribution To protect against stock-outs To take advantage of order cycles To help hedge against price increases or to take advantage of quantity discounts To permit operations

Inventory performance measures and levers Inventory level Low or high Customer service levels Can you deliver what customer wants? Right goods, right place, right time, right quantity Inventory turnover Cost of goods sold per year / average inventory investment Inventory costs, more will come Costs of ordering & carrying inventories Decisions: Order size and time

Inventory Counting Systems A physical count of items in inventory Periodic/Cycle Counting System: Physical count of items made at periodic intervals How much accuracy is needed? When should cycle counting be performed? Who should do it? Continuous Counting System System that keeps track of removals from inventory continuously, thus monitoring current levels of each item

Inventory Counting Systems (Cont’d) Two-Bin System - Two containers of inventory; reorder when the first is empty Universal Bar Code - Bar code printed on a label that has information about the item to which it is attached 214800 232087768 RFID: Radio frequency identification

Key Inventory Terms Lead time: time interval between ordering and receiving the order, denoted by LT Holding (carrying) costs: cost to carry an item in inventory for a length of time, usually a year, denoted by H Ordering costs: costs of ordering and receiving inventory, denoted by S Shortage costs: costs when demand exceeds supply

Effective Inventory Management A system to keep track of inventory A reliable forecast of demand Knowledge of lead times Reasonable estimates of Holding costs Ordering costs Shortage costs A classification system

ABC Classification System Classifying inventory according to some measure of importance and allocating control efforts accordingly. Importance measure= price*annual sales A - very important B - mod. important C - least important Annual $ volume of items A B C High Low Few Many Number of Items

Inventory Models Fixed Order Size - Variable Order Interval Models: 1. Economic Order Quantity, EOQ 2. Economic Production Quantity, EPQ 3. EOQ with quantity discounts All units quantity discount 3.1. Constant holding cost 3.2. Proportional holding cost 4. Reorder point, ROP Lead time service level Fill rate Fixed Order Interval - Variable Order Size Model 5. Fixed Order Interval model, FOI Single Order Model 6. Newsboy model

1. EOQ Model Assumptions: Only one product is involved Annual demand requirements known Demand is even throughout the year Lead time does not vary Each order is received in a single delivery Infinite production capacity There are no quantity discounts

Profile of Inventory Level Over Time The Inventory Cycle Profile of Inventory Level Over Time Q Usage rate Quantity on hand Reorder point Time Receive order Place order Receive order Place order Receive order Lead time

Average inventory held Length of an inventory cycle From one order to the next = Q/D Inventory held over entire inventory cycle Area under the inventory level = ½ Q (Q/D) Average inventory held = Inventory held over a cycle / cycle length = Q/2

Total Cost Annual carrying cost ordering Total cost = + Q 2 H D S TC =

Figure 11-4

Cost Minimization Goal Order Quantity (Q) The Total-Cost Curve is U-Shaped Ordering Costs QO Annual Cost (optimal order quantity)

Deriving the EOQ Using calculus, we take the derivative of the total cost function and set the derivative (slope) equal to zero and solve for Q. The total cost curve reaches its minimum where the carrying and ordering costs are equal.

EOQ example Demand, D = 12,000 computers per year. Holding cost, H = 100 per item per year. Fixed cost, S = $4,000/order. Find EOQ, Cycle Inventory, Optimal Reorder Interval and Optimal Ordering Frequency. EOQ = 979.79, say 980 computers Cycle inventory = EOQ/2 = 490 units Optimal Reorder interval, T = 0.0816 year = 0.98 month Optimal ordering frequency, n=12.24 orders per year. Notes:

Optimal Quantity is robust

Total Costs with Purchasing Cost Annual carrying cost Purchasing TC = + Q 2 H D S ordering PD Note that P is the price.

Total Costs with PD Cost Adding Purchasing cost doesn’t change EOQ TC with PD TC without PD PD Quantity Adding Purchasing cost doesn’t change EOQ

2. Economic Production Quantity Production done in batches or lots Capacity to produce a part exceeds the part’s usage or demand rate Assumptions of EPQ are similar to EOQ except orders are received incrementally during production This corresponds to producing for an order with finite production capacity

Economic Production Quantity Assumptions Only one item is involved Annual demand is known Usage rate is constant Usage occurs continually Production rate p is constant Lead time does not vary No quantity discounts

Economic Production Quantity Usage Production & Usage Production & Usage Usage Inventory Level

Average inventory held (Q/p)(p-D) p-D D Q/p Time Q/D Average inventory held=(1/2)(Q/p)(p-D) Total cost=(1/2)(Q/p)(p-D)H+(D/Q)S

EPQ example Demand, D = 12,000 computers per year. p=20,000 per year. Holding cost, H = 100 per item per year. Fixed cost, S = $4,000/order. Find EPQ. EPQ = EOQ*sqrt(p/(p-D)) =979.79*sqrt(20/8)=1549 computers Notes:

3. All unit quantity discount Cost/Unit Order Quantity 5,000 10,000 Two versions Constant H Proportional H $3 $2.96 $2.92

3.1.Total Cost with Constant Carrying Costs TCb Total Cost Decreasing Price TCc Annual demand*discount EOQ Quantity

3.1.Total Cost with Constant Carrying Costs TCb Total Cost Decreasing Price TCc Annual demand*discount EOQ Quantity

Price a > Price b > Price c Example Scenario 1 Price a > Price b > Price c a b c Total Cost TCa TCb TCc Q*=EOQ Quantity

Price a > Price b > Price c Example Scenario 2 Price a > Price b > Price c a b c Total Cost TCa TCb TCc EOQ Q* Quantity

Price a > Price b > Price c Example Scenario 3 Price a > Price b > Price c a b c Total Cost TCa TCb TCc EOQ Q* Quantity

Price a > Price b > Price c Example Scenario 4 Price a > Price b > Price c a b c Total Cost TCc TCa TCb Q*=EOQ Quantity

3.1. Finding Q with all units discount with constant holding cost Note all the price ranges have the same EOQ. Stop if EOQ=Q1 is in the lowest cost range (highest quantity range), otherwise continue towards quantity break points which give lower costs Quantity Total Cost 1 2

All-units quantity discounts Constant holding cost A popular shoe store sells 8000 pairs per year. The fixed cost of ordering shoes from the distribution center is $15 and holding costs are taken as $12.5 per shoe per year. The per unit purchase costs from the distribution center is given as C3=60, if 0 < Q < 50 C2=55, if 50 <= Q < 150 C1=50, if 150 <= Q where Q is the order size. Determine the optimal order quantity.

Solution There are three ranges for lot sizes in this problem: (0, q2=50), (q2=50, q1=150) (q1=150,infinite). Holding costs in all there ranges of shoe prices are given as H=12.5, EOQ is not feasible in the lowest price range because 138.6 < 150. The order quantity q1=150 is a candidate with cost TC(150)=8000(50)+8000(15)/150+(12.5)(150)/2 =401,900 Let us go to a higher cost level of (q2=50, q1=150). EOQ=138.6 is in the appropriate range, so it is another candidate with cost TC(138.6)=8000(55)+8000(15)/138.6+(12.5)(138.6)/2 =441,732 Since TC(150) < TC(132.1), Q=150 is the optimal solution. Remark: In these computations, we do not need to compute TC(50), why? Because TC(50) >= TC(132.1).

Example: Q1 feasible stop 3.2. Summary of finding Q with all units discount with proportional holding cost Note each price range has a different EOQ. Stop if Q1=“EOQ of the lowest price” feasible Otherwise continue towards higher costs until an EOQ becomes feasible. In each price range, evaluate the lowest cost. Lowest cost is either at an EOQ or price break quantity Pick the minimum cost among all evaluated Total Cost Example: Q1 feasible stop 1 Quantity

3.2. Finding Q with all units discount with proportional holding cost Total Cost 2 Example: Q1 infeasible, Q2 feasible, Break point 1 is selected since TC1 < TC2 1’ Quantity

3.2. Finding Q with all units discount with proportional holding cost Stop if 1 is feasible, otherwise continue towards higher costs until a EOQ becomes feasible. Evaluate cost at all alternatives Total Cost 1’ 2 Quantity

All-units quantity discounts Proportional holding cost A popular shoe store sells 8000 pairs per year. The fixed cost of ordering shoes from the distribution center is $15 and holding costs are taken as 25% of the shoe costs. The per unit purchase costs from the distribution center is given as C3=60, if 0 < Q < 50 C2=55, if 50 <= Q < 150 C1=50, if 150 <= Q where Q is the order size. Determine the optimal order quantity.

Solution There are three ranges for lot sizes in this problem: (0, q2=50), (q2=50, q1=150) (q1=150,infinite). Holding costs in there ranges of shoe prices are given as H3=(0.25)60=15, H2 =(0.25)55=13.75 H1 =(0.25)50=12.5. EOQ1 is not feasible because 138.6 < 150. The order quantity q1=150 is a candidate with cost TC(150)=8000(50)+8000(15)/150+(0.25)(50)(150)/2 =401,900 Let us go to a higher cost level of (q2=50, q1=150). EOQ2=132.1 is in the appropriate range, so it is another candidate with cost TC(132.1)=8000(55)+8000(15)/132.1+(0.25)(55)(132.1)/2 =441,800 Since TC(150) < TC(132.1), Q=150 is the optimal solution. We do not need to compute TC(50) or EOQ3, why?

Types of inventories (stocks) by function Deterministic demand case Anticipation stock For known future demand Cycle stock For convenience, some operations are performed occasionally and stock is used at other times Why to buy eggs in boxes of 12? Pipeline stock or Work in Process Stock in transfer, transformation. Necessary for operations. Students in the class Stochastic demand case Safety stock Stock against demand variations

4. When to Reorder with EOQ Ordering Reorder Point - When the quantity on hand of an item drops to this amount, the item is reordered. We call it ROP. Safety Stock - Stock that is held in excess of expected demand due to variable demand rate and/or lead time. We call it ss. (lead time) Service Level - Probability that demand will not exceed supply during lead time. We call this cycle service level, CSL.

Optimal Safety Inventory Levels An inventory cycle Q ROP time Lead Times Shortage

Safety Stock Quantity Maximum probable demand Expected demand LT Time Expected demand during lead time Maximum probable demand ROP Quantity Safety stock

Inventory and Demand during Lead Time ROP Inventory= ROP-DLT ROP Upside down Inventory DLT: Demand During LT LT Demand During LT

Shortage and Demand during Lead Time Demand During LT LT Shortage DLT: Demand During LT ROP Shortage= DLT-ROP Upside down ROP

Cycle Service Level Cycle service level: percentage of cycles with shortage

DLT : Demand during lead time LT and demand may be uncertain.

Reorder Point ROP = E(DLT) + z σDLT ROP Risk of a stockout Service level Probability of no stockout Expected demand Safety stock z Quantity z-scale ROP = E(DLT) + z σDLT

Finding safety stock from cycle service level CSL Note we use normal density for the demand during lead time The excel function normsinv has default values of 0 and 1 for the mean and standard deviation. Defaults are used unless these values are specified.

Example: Safety inventory vs. Lead time variability D = 2,500/day; D = 500 L = 7 days; CSL = 0.90 Normsinv(0.9)=1.3, either from table or Excel If LT=0, DLT=sqrt(7)*500=1323 ss=1323*normsinv(0.9)=1719.8 ROP=(D)(L)+ss=17500+1719.8 If LT=1, DLT=sqrt(7*500*500+2500*2500*1)=2828 ss=2828*normsinv(0.9)=3676 ROP=(D)(L)+ss=17500+3676

Expected shortage per cycle First let us study shortage during the lead time Ex:

Expected shortage per cycle If we assume that DLT is normal,

Example: Finding expected shortage per cycle Suppose that the demand during lead time has expected value 100 and stdev 30, find the expected shortage if ROP=120. z=(120-100)/30=0.66. E(z)=0.153 from Table 11-3. Expected shortage = 30*0.153=4.59

Fill rate Fill rate is the percentage of demand filled from the stock In a cycle Fill rate = 1-(Expected shortage during LT) / Q For normally distributed demand

Example: computing the fill rate Suppose that the demand during lead time has expected value 100 and stdev 30, find the expected shortage if ROP=120. Compute the fill rate if order sizes are 200. Compute the annual expected shortages if there are 12 order cycles per year. Expected shortage per cycle=4.59 from the last example. Fill rate = 1-4.59/20=0.7705 Annual expected shortage=12*4.59=55.08.

Determinants of the Reorder Point The rate of demand The lead time Demand and/or lead time variability Stockout risk (safety stock)

5. Fixed-Order-Interval Model Orders are placed at fixed time intervals Order quantity for next interval? Suppliers might encourage fixed intervals May require only periodic checks of inventory levels Items from same supplier may yield savings in: Ordering Packing Shipping costs May be practical when inventories cannot be closely monitored

FOI compared against variable order interval models A single order must cover the demand until the next order arrives. Exposure to random demand during not only lead time but also before. Requires higher safety stock than variable order interval models. May provide savings in set up / ordering costs.

6. How many to order for the winter? Parkas at L.L. Bean Notes: Mention that L.L. Grain is a mail order company deciding on the number of units of a Fall jacket to order. An estimate of demand using past information and expertise of buyers is given here. What should the appropriate order quantity be? In general distribution may not be known. Discuss methodology used in the Matching supply and demand article as a possibility in deciding on demand uncertainty (distribution).

Parkas at L.L. Bean Cost per parka = c = $45 Sale price per parka = p = $100 Discount price per parka = $50 Holding and transportation cost = $10 Salvage value per parka = s = $40 Profit from selling parka = p-c = 100-45 = $55 Underage cost=$55 Cost of overstocking = c-s = 45-40 = $5 Overage cost=$5 Notes: What information is required to make the ordering decision? Stress cost of understocking and overstocking. How to evaluate these costs for this example?

6. Single Period Model Single period model: model for ordering of perishables and other items with limited useful lives Shortage cost: generally the unrealized profits per unit, $55 for L.L.Bean. We call this underage. Excess cost: difference between purchase cost and salvage value of items left over at the end of a period, $5 for L.L.Bean. We call this overage.

Parkas at L.L. Bean Expected demand = 10 (‘00) parkas Expected profit from ordering 10 (‘00) parkas = $499 The underage and overage probabilities after ordering 1100 parkas P(D>1100):Probability of underage P(D<1100):Probability of overage Notes: Show detailed calculation of expected profit.

Optimal level of product availability p = sale price; s = outlet or salvage price; c = purchase price CSL = Probability that demand will be at or below reorder point At optimal order size, Expected Marginal Benefit from raising order size = =P(Demand is above stock)*(Profit from sales)=(1-CSL*)(p - c) Expected Marginal Cost = =P(Demand is below stock)*(Loss from discounting)=CSL*(c - s). Let Co= c-s; Cu=p-c, then the optimality condition is (1-CSL*)Cu = CSL* Co, CSL* = Cu / (Cu + Co) Notes:

Parkas at L.L. Bean Notes: Discuss marginal benefit and marginal cost of each jacket. We keep increasing order size as long as expected benefit exceeds expected cost.

Order Quantity for a Single Order Co = Cost of overstocking = $5 Cu = Cost of understocking = $55 Q* = Optimal order size Notes: Explain how formula is derived using decision tree.

Optimal Order Quantity without Normal Demands 0.917 Notes: Optimal Order Quantity = 13(‘00)

Operations Strategy Too much inventory Wise strategy Tends to hide problems Easier to live with problems than to eliminate them Costly to maintain Wise strategy Reduce lot sizes Reduce set ups Reduce safety stock Aggregate negatively correlated demands Remember component commonality Delayed postponement