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Chapter 11

Chapter 11. Inventory Management. Types of Inventories. Raw materials & purchased parts Incoming students Work in progress Current students Finished-goods inventories ( manufacturing firms ) or merchandise ( retail stores ) Graduating students Replacement parts, tools, & supplies

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Chapter 11

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  1. Chapter 11 Inventory Management

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

  3. 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

  4. 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

  5. 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

  6. 0 214800 232087768 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 hasinformation about the item to which it is attached • RFID: Radio frequency identification

  7. 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

  8. 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

  9. High A Annual $ volume of items B C Low Few Many Number of Items 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

  10. 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

  11. 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

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

  13. 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

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

  15. Figure 11-4

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

  17. 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.

  18. 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.

  19. Optimal Quantity is robust

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

  21. Cost Adding Purchasing costdoesn’t change EOQ TC with PD TC without PD PD 0 Quantity EOQ Total Costs with PD

  22. 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

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

  24. Economic Production Quantity Usage Production & Usage Production & Usage Usage Inventory Level

  25. Average inventory held (Q/p)(p-D) D p-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

  26. 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

  27. 10,000 5,000 Order Quantity 3. All unit quantity discount Cost/Unit Two versions Constant H Proportional H $3 $2.96 $2.92

  28. 3.1.Total Cost with Constant Carrying Costs TCa TCb Total Cost Decreasing Price TCc Annual demand*discount EOQ Quantity

  29. 3.1.Total Cost with Constant Carrying Costs TCa TCb Total Cost Decreasing Price TCc Annual demand*discount EOQ Quantity

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

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

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

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

  34. Total Cost 1 2 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

  35. All-units quantity discountsConstant 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.

  36. 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).

  37. 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

  38. 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

  39. 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

  40. All-units quantity discountsProportional 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.

  41. 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?

  42. 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

  43. 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.

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

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

  46. Inventory and Demand during Lead Time ROP 0 Inventory= ROP-DLT ROP Upside down Inventory 0 DLT: Demand During LT LT 0 Demand During LT

  47. 0 Shortage= DLT-ROP ROP Upside down DLT: Demand During LT 0 Shortage LT 0 Demand During LT Shortage and Demand during Lead Time ROP

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

  49. DLT : Demand during lead timeLT and demand may be uncertain.

  50. Service level Risk of a stockout Probability of no stockout Quantity ROP Expected demand Safety stock 0 z z-scale Reorder Point ROP = E(DLT) + z σDLT

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