Managing Flow Variability : Safety Inventory

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Managing Flow Variability: Safety Inventory


Forecasts Depend on: (a) Historical Data and (b) Market Intelligence.

Demand Forecasts and Forecast ErrorsSafety Inventory and Service Level

Optimal Service Level – The Newsvendor ProblemDemand and Lead Time VariabilityPooling Efficiency through Centralization and

AggregationShortening the Forecast HorizonLevers for Reducing Safety Inventory

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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.

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Within 200 time intervals, stockouts occur in 20. Probability of Stockout = # of stockout intervals/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

Service Level and Fill Rate

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μ and σ of Demand During Lead TimeDemand 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.

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The table will give you z

Given a 95% SL95% Probability Normal

table

Up to the first digitafterdecimal

Second digitafter decimal

Probability

z

1.6

0.05

Z = 1.65

z Value using TableGo to normal table, look inside the table. Find a

probability close to 0.95. Read its z from the corresponding row and column.

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The standard Normal Distribution F(z)

F(z)

z0

F(z) = Prob( N(0,1) < z)

Risk Service level z value0.1 0.9 1.280.05 0.95 1.650.01 0.99 2.33

z 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.090 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.53590.1 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.57530.2 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.61410.3 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.65170.4 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.68790.5 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.72240.6 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.75490.7 0.7611 0.7642 0.7673 0.7704 0.7734 0.7764 0.7794 0.7823 0.78520.8 0.7910 0.7939 0.7967 0.7995 0.8023 0.8051 0.8078 0.8106 0.81330.9 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315 0.8340 0.8365 0.83891 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.86211.1 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790 0.8810 0.88301.2 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.8980 0.8997 0.90151.3 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.91771.4 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.93191.5 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 0.9429 0.94411.6 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.95451.7 0.9564 0.9573 0.9582 0.9591 0.9599 0.9608 0.9616 0.9625 0.96331.8 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9699 0.97061.9 0.9719 0.9726 0.9732 0.9738 0.9744 0.9750 0.9756 0.9761 0.97672 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 0.98172.1 0.9826 0.9830 0.9834 0.9838 0.9842 0.9846 0.9850 0.9854 0.98572.2 0.9864 0.9868 0.9871 0.9875 0.9878 0.9881 0.9884 0.9887 0.98902.3 0.9896 0.9898 0.9901 0.9904 0.9906 0.9909 0.9911 0.9913 0.99162.4 0.9920 0.9922 0.9925 0.9927 0.9929 0.9931 0.9932 0.9934 0.99362.5 0.9940 0.9941 0.9943 0.9945 0.9946 0.9948 0.9949 0.9951 0.99522.6 0.9955 0.9956 0.9957 0.9959 0.9960 0.9961 0.9962 0.9963 0.99642.7 0.9966 0.9967 0.9968 0.9969 0.9970 0.9971 0.9972 0.9973 0.99742.8 0.9975 0.9976 0.9977 0.9977 0.9978 0.9979 0.9979 0.9980 0.99812.9 0.9982 0.9982 0.9983 0.9984 0.9984 0.9985 0.9985 0.9986 0.99863 0.9987 0.9987 0.9988 0.9988 0.9989 0.9989 0.9989 0.9990 0.9990

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Excel: Given Probability, Compute z

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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 demandROP = LTD + zσLTD

ROP = LTD + Isafety

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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 = 1.65Isafety = zσLTD Isafety = 1.64 (5) = 8.2ROP = LTD + Isafety

ROP = 50 + 1.64(5) = 58.2

When Service level increases Risk decreases z increases

Isafety increases

Risk Service level z value0.1 0.9 1.280.05 0.95 1.650.01 0.99 2.33

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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.

Example 2; total demand during lead time is variable

What is the Service Level?Service level = 1-risk of stockout = 1-0.1 = 0.9What is the corresponding z value? SL (90%) Probability of 90% z = 1.28Compute the safety stock?Isafety = 1.28(10) = 12.8ROP = LTD + Isafety

ROP = 75 + 12.8 = 87.8

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Service Level for a given ROP Example

Compute the service level at GE Lighting’s warehouse, LTD = 20,000, sLTD = 5,000, and ROP = 24,000ROP = 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

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Given z, Find the Probability

Given z

Table returns probability

Up to the first digitafterdecimal

Second digitafter decimal

Probability

z

0.8

0.00

z = 0.8 Probability = 0.7881Service Level (SL) = 0.7881

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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

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Excel: Given z, Compute Probability

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If Lead Time is fixed and Demand is variable

μ and σ of Demand Per Period and Fixed Lead Time

L: Lead TimeR: Demand per PeriodR: Average Demand per PeriodAverage Demand During Lead Time LTD = L×R R : Standard Deviation of Demand per PeriodStandard Deviation of Demand During Lead Time = LTD

LTD = L R LTD = LR

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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

μ and σ of demand per period and fixed Lead Time

L= 2 weeks, R= 50 tons per week, R = 3 tons per weekLTD = LR LTD = 2(50) = 100

LTD = L R LTD =2 3 = 4.24 ROP = LTD + Isafety = LTD + z LTD

ROP = 100 + 1.28 × 4.24 ROP = 100 + 5.43

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If Demand is fixed and Lead Time is variable

μ and σ of Lead Time and Fixed Demand per Period

R: Demand per PeriodL: Lead TimeL: Average Lead TimeAverage 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

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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 = LR LTD = 2(50) = 100

LTD = R L

ROP = 100 + 1.28 × 25 ROP = 100 + 32

ROP = LTD + Isafety = LTD + z LTD

LTD = 50 ×0.5 = 25

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Both Demand and Lead Time are Variable

R: demand rate per period R: Average demand rateσR: Standard deviation of demand

L: lead time L: Average lead timeσL: 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

=

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Both Demand and Lead Time are Variable

Lead time has mean of 10 days and a stddev of 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 daysσL: Standard deviation of the lead time = 2 days = 10(2000) = 20000

==6402.78 =1.65

=1.65(6402.78) = 10565

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Optimal Service Level: The Newsvendor ProblemAn 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 = 1700Underage Cost = Marginal Benefit = p-c = 2500-1800 = 700Overage Cost = Marginal Cost = c-v = 1800-1700 = 100Optimal Service Level = P(R≤ROP) = MB/(MB+MC)

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Optimal Service Level: The Newsvendor ProblemUnderage Cost =Cu = 2500-1800 = 700Overage Cost = Co = 1800-1700 = 100Optimal Service Level = P(R≤ROP) = MB/(MB+MC)SL = 700/800 = 0.875

1.15

Probability of excess inventory

0.875

Probability of shortage

0.125

R =N(150,15)ROP = LTD + Isafety= LTD + zσLTD= 150+1.15(15)Isafety = 17.25 = 18ROP = 168Risk = 12.5%

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Optimal Service Level: The Newsvendor ProblemDemand 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 = 1R = N(4000,1000)Overage Cost = Marginal Cost = MC = 10-6 = 4Underage Cost = Marginal Benefit = p-c + v = 30-10 +1 = 21Optimal Service Level = P(R≤ROP) = MB/(MB+MC)SL* = 21/25 = 0.84 Z(0.84) =

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Optimal Service Level: The Newsvendor Problem

0.99Probability of

excess inventory

0.84

Probability of shortage

0.16

ROP = LTD + Isafety = LTD + zσLTD

ROP = 4000+0.99(1000)ROP = 4999Risk = 16%

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Centralization and ROPThere 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)

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Centralization and ROP

Demand N(80,10)

Warehouse A

Demand N(80,10)

Warehouse B

SL = 95%Isafety each =

1.65(10)Isafety each = 16.5Isafety all = 33

Demand N(160,)=N(160,14.14)

Warehouse A

Warehouse B

SL = 95%Isafety all =

1.65(14.14)Isafety all = 23.33

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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 5,000 units and. Compute total Isafety at SL= 95% service level for centralized and decentralized systems. Isafety = 1.65×5000= 8250

𝐼 𝑠𝑎𝑓𝑒𝑡𝑦𝐷𝑒𝑐𝑒𝑛𝑡𝑟𝑎𝑙𝑖𝑧𝑒𝑑=7 (8250 )=57,750 is not 7(5000)(

13250

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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

barriersThese disadvantages may reduce the demand.

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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

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Virtual Centralization

Location AExceeds Available

stock1. 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

Location BLess than Available

stock

Virtual Centralization: inventory polling is facilitated using information regarding availability of goods and subsequent transshipment between locations.

polling is achieved by keeping the inventories at decentralized locations.

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Component CommonalityWe 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).

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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

fabricNo 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.

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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.

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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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Managing Flow Variability : Safety Inventory