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Managing Flow Variability : Safety Inventory

Managing Flow Variability : Safety Inventory. Forecasts Depend on: (a) Historical Data and (b) Market Intelligence. Demand Forecasts and Forecast Errors Safety Inventory and Service Level Optimal Service Level – The Newsvendor Problem Demand and Lead Time Variability

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Managing Flow Variability : Safety Inventory

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  1. Managing Flow Variability: Safety Inventory Forecasts Depend on: (a) Historical Data and (b) Market Intelligence. Demand Forecasts and Forecast Errors Safety Inventory and Service Level Optimal Service Level – The Newsvendor Problem Demand and Lead Time Variability Pooling Efficiency through Centralization and Aggregation Shortening the Forecast Horizon Levers for Reducing Safety Inventory

  2. Four Characteristics of Forecasts • Forecasts are usually (always) inaccurate (wrong).Because of random noise. • Forecasts should be accompanied by a measure of forecast error.A measure of forecast error (standard deviation) quantifies the manager’s degree of confidence in the forecast. • Aggregate forecasts are more accurate than individual forecasts.Aggregate forecasts reduce the amount of variability – relative to the aggregate mean demand. StdDev of sum of two variables is less than sum of StdDev of the two variables. • Long-range forecasts are less accurate than short-range forecasts.Forecasts further into the future tends to be less accurate than those of more imminent events. As time passes, we get better information, and make better prediction.

  3. Service Level and Fill Rate Within 200 time intervals, stockouts occur in 20. Probability of Stockout = # of stockoutintervals/Total # of intervals = 20/200 = 0.1 Risk = Probability of stockout= 0.1 = 10% Service Level = 1-Risk = 1=0.1 = 0.9 = 90%. Suppose that cumulative demand during the 200 time intervals was 25,000 units and the total number of units short in the 20 intervals with stockouts was 4,000 units. Fill rate = (25,000-4,000)/25,000 = 21,000/25,000 = 84%. Fill Rate = Expected Sales / Expected Demand Fill Rate = (1- Expected Stockout )/ Expected Demand

  4. μand σ of Demand During Lead Time Demand during lead time has an average of 50 tons. Standard deviation of demand during lead time is 5 tons. Acceptable risk is no more than 5%. Find the re-order point. Service level = 1-risk of stockout = 1-0.05 = 0.95. Find the z value such that the probability of a standard normal variable being less than or equal to z is 0.95.

  5. Given a 95% SL 95% Probability The table will give you z Probability z Value using Table Go to normal table, look inside the table. Find a probability close to 0.95. Read its z from the corresponding row and column. Normal table 0.05 z Second digit after decimal Up to the first digit after decimal 1.6 Z = 1.65

  6. F(z) z 0 The standard Normal Distribution F(z) F(z) = Prob( N(0,1) <z) Risk Service level z value 0.10.9 1.28 0.050.95 1.65 0.010.99 2.33

  7. Excel: Given Probability, Compute z

  8. Relationship between z and Normal Variable x z = (x-Average x)/(Standard Deviation of x) x = Average x +z (Standard Deviation of x) LTD = Average lead time demand σLTD = Standard deviation of lead time demand ROP = LTD + zσLTD ROP = LTD + Isafety

  9. Demand During Lead Time is Variable N(μ,σ) Demand of sand during lead time has an average of 50 tons. Standard deviation of demand during lead time is 5 tons Assuming that the management is willing to accept a risk no more that 5%. Compute safety stock. LTD = 50, σLTD = 5 Risk = 5%, SL = 95%  z = 1.65 Isafety= zσLTD Isafety = 1.64 (5) = 8.2 ROP= LTD + Isafety ROP = 50 + 1.64(5) = 58.2 Risk Service level z value 0.10.9 1.28 0.050.95 1.65 0.010.99 2.33 • When Service level increases • Risk decreases • z increases • Isafetyincreases

  10. Example 2; total demand during lead time is variable Average Demand of sand during lead time is 75 units. Standard deviation of demand during lead time is 10 units. Under a risk of no more that 10%, compute SL, Isafety, ROP. What is the Service Level? Service level = 1-risk of stockout = 1-0.1 = 0.9 What is the corresponding z value? SL (90%)  Probability of 90%  z = 1.28 Compute the safety stock? Isafety = 1.28(10) = 12.8 ROP = LTD + Isafety ROP = 75 + 12.8 = 87.8

  11. Service Level for a given ROP Example Compute the service level at GE Lighting’s warehouse, LTD= 20,000, sLTD= 5,000, and ROP= 24,000 ROP = LTD + I safety 24000 = 2000 + I safety I safety = 4,000 I safety = z sLTD 4000 = z(5000) z = 4,000 / 5,000 = 0.8 SL= Prob (Z ≤ 0.8) from Normal Table

  12. Table returns probability Given z Probability Given z, Find the Probability 0.00 z Second digit after decimal Up to the first digit after decimal 0.8 z = 0.8 Probability = 0.7881 Service Level (SL) = 0.7881

  13. Service Level for a given ROP • SL= Prob (LTD ≤ ROP) • LTD is normally distributed • LTD = N(LTD, sLTD) • ROP = LTD + zσLTD • ROP = LTD + Isafety • Isafety = z sLTD • z = Isafety/sLTD • Then we go to table and find the probability

  14. Excel: Given z, Compute Probability

  15. μ and σ of Demand Per Period and Fixed Lead Time If Lead Time is fixed and Demand is variable L: Lead Time R: Demand per Period R:AverageDemand per Period Average Demand During Lead Time LTD = L×R R : Standard Deviation of Demand per Period Standard Deviation of Demand During Lead Time = LTD LTD =L R LTD = LR

  16. μ and σ of demand per period and fixed Lead Time Average demand of a product is 50 tons per week. Standard deviation of the weekly demand is 3 tons. Lead time is 2 weeks. Assume that the management is willing to accept a risk no more that 10%. z = 1.28 L= 2 weeks, R= 50 tons per week, R = 3 tons per week LTD = 2(50) = 100 LTD = LR LTD =L R LTD =2 3 = 4.24 ROP = LTD + Isafety = LTD + zLTD ROP = 100 + 1.28 × 4.24 ROP = 100 + 5.43

  17. μ and σ of Lead Time and Fixed Demand per Period If Demandis fixed and Lead Time is variable R: Demand per Period L: Lead Time L: Average Lead Time Average Demand During Lead Time LTD = L×R L : Standard Deviation of Lead Time Standard Deviation of Demand During Lead Time = LTD LTD = R L LTD = LR

  18. Lead Time Variable, Demand fixed Demand of sand is fixed and is 50 tons per week. The average lead time is 2 weeks. Standard deviation of lead time is 0.5 week. Under a risk of no more that 10%, compute ROP and Isafety. Acceptable risk; 10%  z = 1.28 R: 50 tons, L = 2 weeks, L = 0.5 week LTD = 2(50) = 100 LTD = LR LTD = R L LTD = 50 ×0.5 = 25 ROP = LTD + Isafety = LTD + zLTD ROP = 100 + 1.28 × 25 ROP = 100 + 32

  19. Both Demand and Lead Time are Variable R: demand rate per period R: Average demand rate σR:Standard deviation of demand L: lead time L: Average lead time σL:Standard deviation of the lead time LTD: demand during the lead time (a random variable) LTD: Average demand during the lead time σLTD:Standard deviation of the demand during lead time =

  20. Both Demand and Lead Time are Variable Lead time has mean of 10 days and a stddevof 2 days. Demand per day has a mean of 2000 and stddev of 1581. How much safety inventory is needed in order to provide a 95% service level? R: Average demand rate= 2000 units σR:Standard deviation of demand = 1581 L: Average lead time = 10 days σL:Standard deviation of the lead time = 2 days = 10(2000) = 20000 = =6402.78 =1.65 =1.65(6402.78) = 10565

  21. Optimal Service Level: The Newsvendor Problem An electronics superstore is carrying a 60” LEDTV for the upcoming Christmas holiday sales. Each TV can be sold at $2,500. The store can purchase each unit for $1,800. Any unsold TVs can be salvaged, through end of year sales, for $1,700. The retailer estimates that the demand for this TV will be Normally distributed with mean of 150 and standard deviation of 15. How many units should they order? Note: If they order 150, they will be out of stock 50% of the time. Which service level is optimal? 80%, 90%, 95%, 99%?? Cost =1800, Sales Price = 2500, Salvage Value = 1700 Underage Cost = Marginal Benefit = p-c = 2500-1800 = 700 Overage Cost = Marginal Cost = c-v = 1800-1700 = 100 Optimal Service Level = P(R≤ROP) = MB/(MB+MC)

  22. Optimal Service Level: The Newsvendor Problem Underage Cost =Cu = 2500-1800 = 700 Overage Cost = Co = 1800-1700 = 100 Optimal Service Level = P(R≤ROP) = MB/(MB+MC) SL = 700/800 = 0.875 Probability of excess inventory Probability of shortage R =N(150,15) ROP = LTD + Isafety = LTD + zσLTD = 150+1.15(15) Isafety = 17.25 = 18 ROP = 168 Risk = 12.5% 0.875 0.125 1.15

  23. Optimal Service Level: The Newsvendor Problem Demand for a product in the upcoming period is normally distributed with mean of 4000 and standard deviation of 1000. Unit Revenue = Sales Price = p = 30. Unit purchase cost = c = 10. Salvage value = v = 6. Goodwill cost = g = 1 R = N(4000,1000) Overage Cost = Marginal Cost = MC = 10-6 = 4 Underage Cost = Marginal Benefit = p-c + v = 30-10 +1 = 21 Optimal Service Level = P(R≤ROP) = MB/(MB+MC) SL* = 21/25 = 0.84 Z(0.84) =

  24. Optimal Service Level: The Newsvendor Problem Probability of excess inventory Probability of shortage 0.99 0.84 0.16 ROP = LTD + Isafety = LTD + zσLTD ROP = 4000+0.99(1000) ROP = 4999 Risk = 16%

  25. Centralization and ROP There are N warehouses. Each with lead time demand of LTD and with standard deviation of lead time demand of σLTD. If demand in each warehouse is independent of demand in other warehouses. If they order all together and have a centralized safety stock then The average demand during lead time for all the warehouses is N(LTD). The standard deviation of the lead time demand for all warehouses is (σLTD)

  26. Centralization and ROP SL = 95% Isafety each = 1.65(10) Isafetyeach = 16.5 Isafetyall = 33 Warehouse A Warehouse B Demand N(80,10) Demand N(80,10) Warehouse A Warehouse B SL = 95% Isafetyall = 1.65(14.14) Isafetyall = 23.33 Demand N(160,)=N(160,14.14)

  27. Independent Lead time demands at two locations GE lighting with 7 warehouses. LTD for each warehouse has mean of 20,000 units and StdDev of 5,000units and. Compute total Isafety at SL= 95% service level for centralized and decentralized systems. Isafety= 1.65×5000= 8250 is not 7(5000) ( 13250

  28. independent Lead time demands at N locations In Waiting Line; Centralization , or Polling, leads to (i) flow time reduction and (ii) throughput improvement. • In Inventory; Centralization leads to reduction in (i) cycle inventory, (ii) safety inventory, and (iii) flow time. • If centralization reduces inventory, why doesn’t everybody do it? • Longer response time • Higher shipping cost • Less understanding of customer needs • Less understanding of cultural, linguistics, and regulatory barriers These disadvantages may reduce the demand.

  29. Principle of Aggregation and polling Inventory • Inventory benefits  due to principle of aggregation. • Statistics: Standard deviation of sum of random variables is less than the sum of the individual standard deviations. • Physical consolidation is not essential, as long as available inventory is shared among various locations  Polling Inventory • Virtual Centralization • Specialization • Component Commonality • Delayed Differentiation • Product Substitution

  30. Virtual Centralization Virtual Centralization: inventory polling is facilitated using information regarding availability of goods and subsequent transshipment between locations. • Location A • Exceeds Available stock Location B Less than Available stock 1. Information about product demand and availability must be available at both locations 2. Shipping the product from one location to a customer at another location must be fast and cost effective polling is achieved by keeping the inventories at decentralized locations.

  31. Component Commonality • We discussed aggregating demand across various geographic locations, either physical or virtual • Aggregating demand across various products has the same benefits. • Computer manufacturers: offer a wide range of models, but few components are used across all product lines. • Replace Make-to-stock with make Make-to-Order • Commonality + MTO: • Commonality: Safety inventory of the common components much less than safety inventory of unique components stored separately. • MTO: Inventory cost is computed in terms of WIP cost not in terms of finished good cost (which is higher).

  32. Postponement (Delayed Differentiation) • Forecasting Characteristic: Forecasts further into the future tends to be less accurate than those of more imminent events. • Since shorter-range forecasts are more accurate, operational decisions will be more effective if supply is postponed closer to the point of actual demand. • Two Alternative processes (each activity takes one week) • Alternative A: (1) Coloring the fabric, (2) assembling T-shirts • Alternative B: (1) Assembling T-shirts, (2) coloring the fabric • No changes in flow time. Alternative B postponed the color difference until one week closer to the time of sale. Takes advantage of the forecasting characteristic: short-Range forecast more accurate.

  33. Postponement (Delayed Differentiation) • Two advantages: Taking advantage of two demand forecasting characteristics • Commonality Advantage: At week 0; Instead of forecast for each individual item, we forecast for aggregates item – uncolored T-shirt. Forecast for aggregate demand is more accurate than forecast for individual item. It is easier to more accurately forecast total demand for different colored T-shirts for next week than the week after the next. • Postponement Advantage: Instead of forecasting for each individual items two weeks ahead, we do it at week 1. Shorter rang forecasts are more accurate. It is easier to more accurately forecast demand for different colored T-shirts for next week than the week after the next.

  34. Lessons Learned Levers for Reducing Safety Inventory • Reduce demand variability through improved forecasting • Reduce replenishment lead time • Reduce variability in replenishment lead time • poll safety inventory for multiple locations or products • Exploit product substitution • Use common components • Postpone product-differentiation processing until closer to the point of actual demand

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