Managing Flow Variability: Safety Inventory
1
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

Managing Flow Variability: Safety Inventory
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.
2

Managing Flow Variability: Safety Inventory
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
3

Managing Flow Variability: Safety Inventory
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.
4
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.

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

Managing Flow Variability: Safety Inventory
F(z) = Prob( N(0,1) < z)
Risk
0.1
0.05
0.01
F(z)
0 z
Service level z value
0.9
0.95
1.28
1.65
0.99 2.33
0 z 0.01
0.5040
0.02
0.5080
0.03
0.5120
0.04
0.5160
0.05
0.5199
0.06
0.5239
0.07
0.5279
0.08
0.5319
0.09
0.5359
0.1
0.5438
0.5478
0.5517
0.5557
0.5596
0.5636
0.5675
0.5714
0.5753
0.2
0.5832
0.5871
0.5910
0.5948
0.5987
0.6026
0.6064
0.6103
0.6141
0.3
0.6217
0.6255
0.6293
0.6331
0.6368
0.6406
0.6443
0.6480
0.6517
0.4
0.6591
0.6628
0.6664
0.6700
0.6736
0.6772
0.6808
0.6844
0.6879
0.5
0.6950
0.6985
0.7019
0.7054
0.7088
0.7123
0.7157
0.7190
0.7224
0.6
0.7291
0.7324
0.7357
0.7389
0.7422
0.7454
0.7486
0.7517
0.7549
0.7
0.7611
0.7642
0.7673
0.7704
0.7734
0.7764
0.7794
0.7823
0.7852
0.8
0.7910
0.7939
0.7967
0.7995
0.8023
0.8051
0.8078
0.8106
0.8133
0.9
0.8186
0.8212
0.8238
0.8264
0.8289
0.8315
0.8340
0.8365
0.8389
1 0.8438
0.8461
0.8485
0.8508
0.8531
0.8554
0.8577
0.8599
0.8621
1.1
0.8665
0.8686
0.8708
0.8729
0.8749
0.8770
0.8790
0.8810
0.8830
1.2
0.8869
0.8888
0.8907
0.8925
0.8944
0.8962
0.8980
0.8997
0.9015
1.3
0.9049
0.9066
0.9082
0.9099
0.9115
0.9131
0.9147
0.9162
0.9177
1.4
0.9207
0.9222
0.9236
0.9251
0.9265
0.9279
0.9292
0.9306
0.9319
1.5
0.9345
0.9357
0.9370
0.9382
0.9394
0.9406
0.9418
0.9429
0.9441
1.6
0.9463
0.9474
0.9484
0.9495
0.9505
0.9515
0.9525
0.9535
0.9545
1.7
0.9564
0.9573
0.9582
0.9591
0.9599
0.9608
0.9616
0.9625
0.9633
1.8
0.9649
0.9656
0.9664
0.9671
0.9678
0.9686
0.9693
0.9699
0.9706
1.9
0.9719
0.9726
0.9732
0.9738
0.9744
0.9750
0.9756
0.9761
0.9767
2 0.9778
0.9783
0.9788
0.9793
0.9798
0.9803
0.9808
0.9812
0.9817
2.1
0.9826
0.9830
0.9834
0.9838
0.9842
0.9846
0.9850
0.9854
0.9857
2.2
0.9864
0.9868
0.9871
0.9875
0.9878
0.9881
0.9884
0.9887
0.9890
2.3
0.9896
0.9898
0.9901
0.9904
0.9906
0.9909
0.9911
0.9913
0.9916
2.4
0.9920
0.9922
0.9925
0.9927
0.9929
0.9931
0.9932
0.9934
0.9936
2.5
0.9940
0.9941
0.9943
0.9945
0.9946
0.9948
0.9949
0.9951
0.9952
2.6
0.9955
0.9956
0.9957
0.9959
0.9960
0.9961
0.9962
0.9963
0.9964
2.7
0.9966
0.9967
0.9968
0.9969
0.9970
0.9971
0.9972
0.9973
0.9974
2.8
0.9975
0.9976
0.9977
0.9977
0.9978
0.9979
0.9979
0.9980
0.9981
2.9
0.9982
0.9982
0.9983
0.9984
0.9984
0.9985
0.9985
0.9986
0.9986
3 0.9987
0.9987
0.9988
0.9988
0.9989
0.9989
0.9989
0.9990
0.9990
6

Managing Flow Variability: Safety Inventory
7

Managing Flow Variability: Safety Inventory
z
x
8 z = (x-Average x)/(Standard Deviation of x) x = Average x +z (Standard Deviation of x)
LTD = Average lead time demand
σ
LTD
= Standard deviation of lead time demand
ROP = LTD + zσ
LTD
ROP = LTD + I safety

Managing Flow Variability: Safety Inventory
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, σ
LTD
= 5
Risk = 5%, SL = 95%  z = 1.65
I safety
= zσ
LTD
I safety
= 1.64 (5) = 8.2
ROP = LTD + I safety
ROP = 50 + 1.64(5) = 58.2
Risk
0.1
0.05
0.01
Service level z value
0.9
1.28
0.95
1.65
0.99 2.33
When Service level increases
I
Risk decreases z increases safety increases
9

Managing Flow Variability: Safety Inventory
10
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?
I safety
= 1.28(10) = 12.8
ROP = LTD + I safety
ROP = 75 + 12.8 = 87.8

Managing Flow Variability: Safety Inventory
11
Compute the service level at GE Lighting’s warehouse,
LTD = 20,000, s
LTD
= 5,000, and ROP = 24,000
ROP = LTD + I safety
 I safety
= 4,000 24000 = 2000 + I safety
I safety
= z s
LTD
4000 = z(5000) z = 4,000 / 5,000 = 0.8
SL= Prob (Z ≤ 0.8) from Normal Table

z = 0.8
Managing Flow Variability: Safety Inventory
12
Table returns probability z
0.00
Second digit after decimal
Given z 0.8
Up to the first digit after decimal
Probability
Probability = 0.7881
Service Level (SL) = 0.7881

Managing Flow Variability: Safety Inventory
13
SL = Prob (LTD ≤ ROP)
LTD is normally distributed
LTD = N(LTD, s
LTD
)
ROP = LTD + zσ
LTD
ROP = LTD + I s afety
I safety
= z s
LTD z = I safety
/ s
LTD
Then we go to table and find the probability

Managing Flow Variability: Safety Inventory
14

Managing Flow Variability: Safety Inventory
15
If Lead Time is fixed and Demand is variable
L: Lead Time
R: Demand per Period
R: Average Demand per Period
Average Demand During Lead Time LTD = L×R s
R
: Standard Deviation of Demand per Period
Standard Deviation of Demand During Lead Time = s
LTD s
LTD
=  L s
R
LTD = LR

Managing Flow Variability: Safety Inventory
16
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, s
R
= 3 tons per week
LTD = LR LTD = 2(50) = 100 s
LTD
=  L s
R s
LTD
=  2 3 = 4.24
ROP = LTD + Isafety = LTD + z s
LTD
ROP = 100 + 1.28 × 4.24
ROP = 100 + 5.43

Managing Flow Variability: Safety Inventory
17
If Demand is 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 s
L
: Standard Deviation of Lead Time
Standard Deviation of Demand During Lead Time = s
LTD s
LTD
= R s
L
LTD = LR

Managing Flow Variability: Safety Inventory
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.
18
Acceptable risk; 10%  z = 1.28
R: 50 tons, L = 2 weeks, s
L
= 0.5 week
LTD = LR LTD = 2(50) = 100 s
LTD
= R s
L s
LTD
= 50 ×0.5 = 25
ROP = LTD + Isafety = LTD + z s
LTD
ROP = 100 + 1.28 × 25 ROP = 100 + 32

Managing Flow Variability: Safety Inventory
19
R: demand rate per period
R: Average demand rate
σ
R
: Standard deviation of demand
L: lead time
L: Average lead time
σ
L
: Standard deviation of the lead time
LTD: demand during the lead time (a random variable)
LTD: Average demand during the lead time
σ
LTD
: Standard deviation of the demand during lead time
𝐿𝑇𝐷 = 𝐿𝑅 𝜎
𝐿𝑇𝐷
= 𝐿𝜎 2
𝑅
+𝑅 2 𝜎 2
𝐿

Managing Flow Variability: Safety Inventory
20
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
σ
σ
L
R
: Standard deviation of demand = 1581
L: Average lead time = 10 days
: Standard deviation of the lead time = 2 days
𝐿𝑇𝐷 = 𝐿𝑅 = 10(2000) = 20000 𝜎
𝐿𝑇𝐷
= 𝐿𝜎 2
𝑅
+𝑅 2 𝜎
𝐿
2 = 10(1581 𝜎
𝐿𝑇𝐷
=6402.78
𝑧
95%
=1.65
2 ) + (2000 2 )(2 2 )
𝐼𝑠𝑎𝑓𝑒𝑡𝑦 = 𝑧𝜎
𝐿𝑇𝐷
=1.65(6402.78) = 10565

Managing Flow Variability: Safety Inventory
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)
21

Managing Flow Variability: Safety Inventory
22
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
0.875
1.15
Probability of shortage
0.125
R =N(150,15)
ROP =
LTD + Isafety
= LTD + zσ
LTD
= 150+1.15(15)
Isafety = 17.25 = 18
ROP = 168
Risk = 12.5%

Managing Flow Variability: Safety Inventory
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) =
23

Managing Flow Variability: Safety Inventory
24
Probability of excess inventory
0.99
Probability of shortage
0.84
0.16
ROP = LTD + Isafety = LTD + zσ
LTD
ROP = 4000+0.99(1000)
ROP = 4999
Risk = 16%

Managing Flow Variability: Safety Inventory
There are N warehouses. Each with lead time demand of LTD and with standard deviation of lead time demand of σ
LTD.
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 𝑁 (σ
LTD)
25

Warehouse A
Demand
N(80,10)
Managing Flow Variability: Safety Inventory
26
Warehouse B
SL = 95%
Isafety each = 1.65(10)
Isafety each = 16.5
Isafety all = 33
Demand
N(80,10)
Warehouse A Warehouse B SL = 95%
Isafety all = 1.65(14.14)
Isafety all = 23.33
Demand
N(160, 𝟐𝟎𝟎 )=N(160,14.14)

Managing Flow Variability: Safety Inventory
Independent Lead time demands at two locations
I
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. safety
= 1.65
× 5000= 8250 𝐼
𝐷𝑒𝑐𝑒𝑛𝑡𝑟𝑎𝑙𝑖𝑧𝑒𝑑 𝑠𝑎𝑓𝑒𝑡𝑦
= 7 8250 = 57,750 𝜎
𝐶𝑒𝑛𝑡𝑟𝑎𝑙𝑖𝑧𝑒𝑑
𝐿𝑇𝐷 is not 7(5000) 𝜎 𝐶𝑒𝑛𝑡𝑟𝑎𝑙𝑖𝑧𝑒𝑑
𝐿𝑇𝐷
= 𝑁 ( 𝜎
𝐿𝑇𝐷
) = 7 5000 = 2.65 5000 𝜎 𝐶𝑒𝑛𝑡𝑟𝑎𝑙𝑖𝑧𝑒𝑑
𝐿𝑇𝐷
= 13250
𝐼
𝐶𝑒𝑛𝑡𝑟𝑎𝑙𝑖𝑧𝑒𝑑 𝑠𝑎𝑓𝑒𝑡𝑦
= 21862.5
27

Managing Flow Variability: Safety Inventory independent Lead time demands at N locations
28
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.

Managing Flow Variability: Safety 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.
29
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

Managing Flow Variability: Safety Inventory
30
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.

Managing Flow Variability: Safety Inventory
31
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).

Managing Flow Variability: Safety Inventory
32
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.

Managing Flow Variability: Safety Inventory
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 Tshirts for next week than the week after the next.
33
 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.

Managing Flow Variability: Safety Inventory
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
34