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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 Expected stockout = Expected Demand – Expected Sales

  4. μ and 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. Forecasts should be accompanied by a measure of forecast error Forecast and a Measure of Forecast Error

  6. Demand During Lead Time Inventory Demand during LT Time Lead Time

  7. ROP when Demand During Lead Time is Fixed LT

  8. Demand During Lead Time is Variable LT

  9. Demand During Lead Time is Variable Inventory Time

  10. Quantity A large demand during lead time Average demand during lead time ROP Safety stock Time LT Safety Stock Safety stock reduces risk of stockout during lead time

  11. Quantity ROP Time LT Safety Stock

  12. Re-Order Point: ROP Demand during lead time has Normal distribution. If we order when the inventory on hand is equal to the average demand during the lead time; then there is 50% chance that the demand during lead time is less than our inventory. However, there is also 50% chance that the demand during lead time is greater than our inventory, and we will be out of stock for a while. We usually do not like 50% probability of stock out We can accept some risk of being out of stock, but we usually like a risk of less than 50%.

  13. Risk of a stockout Average demand Safety stock z-scale Safety Stock and ROP Service level Probability of no stockout ROP Quantity 0 z Each Normal variable x is associated with a standard Normal Variable z x is Normal (Average x , Standard Deviation x)  z is Normal (0,1)

  14. Risk of a stockout Average demand Safety stock z Values Service level Probability of no stockout ROP Quantity 0 z z-scale • There is a table for z which tells us • Given anyprobability of not exceeding z. What is the value of z • Given anyvalue forz. What is the probability of not exceeding z

  15. 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

  16. 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

  17. Excel: Given Probability, Compute z

  18. 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

  19. 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= 1.65 (5) = 8.3 ROP= LTD + Isafety ROP = 50 + 1.65(5) = 58.3 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

  20. 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 = zσLTD = 1.28(10) = 12.8 ROP = LTD + Isafety ROP = 75 + 12.8 = 87.8

  21. 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 + Isafety 24000 = 20000 + Isafety Isafety = 4,000 Isafety= z sLTD 4000 = z(5000) z = 4,000 / 5,000 = 0.8 SL= Prob (Z ≤ 0.8) from Normal Table

  22. 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

  23. Excel: Given z, Compute Probability

  24. Service Level for a given ROP • SL= Prob (LTD ≤ ROP) • LTD is normally distributed • LTD = N(LTD, sLTD) • ROP = LTD + Isafety • ROP = LTD + zσLTD • The recording does not cover the last 3 lines of this slide. • Isafety = z σLTD • z = Isafety/sLTD • Then we go to table and find the probability

  25. 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

  26. 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

  27. 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

  28. 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

  29. 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 =

  30. 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

  31. Optimal Service Level: The News Vendor 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 = SL*= P(LTD ≤ ROP) = Cu/(Cu+Co) Or in NVP Terminology SL* = P(R≤ Q*) = Cu/(Cu+Co)

  32. Optimal Service Level: The Newsvendor Problem Underage cost = Marginal Benefit =Cu = 2500-1800 = 700 Overage Cost = Marginal Cost = Co = 1800-1700 = 100 SL* = Cu/(Cu+Co) SL* = 700/800 = 0.875 Probability of excess inventory Probability of shortage LTD =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

  33. The Newsvendor Problem Terminology In the NVP terminology, the following notations is used R: demand during the single period of study R =N(R, σR) Q* : What we order (only once) P(R ≤ Q*) = Cu/(Cu+Co) = 0.875 Compute z value related to Cu/(Cu+Co) = 0.875 probability. z (0.875) = 1.15 z = (Q* - R)/σR 1.15 = (Q*-150)/15 Q* = 168

  34. 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 Goodwill cost is the present value of all the future lost sales due to creating bad reputation. LTD = N(4000,1000) Overage Cost = Marginal Cost = Co = c-v = 10-6 = 4 Underage Cost = Marginal Benefit = p-c+g= 30-10 +1 = 21 SL*= Cu/(Cu+Co) SL*= 21/25 = 0.84  Z(0.84) =

  35. 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%

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