Post on 17-Mar-2020
Revealing The Mystery of Safety Stock Calculations 1
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REVEALING THE MYSTERY OF SAFETY
STOCK CALCUTLATIONS –
SOME BASICS AND THE ROLE OF THE
“NORMAL” DISTRIBUTION
Revealing The Mystery of Safety Stock Calculations 2
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PART 1
Revealing the Mystery of Safety Stock Calculations - Some Basics
and the Role of the “Normal” Distribution By Ken Fordyce
1.0 Introduction
All of us experience the challenge of inventory every day – whether at home -how many cans of tuna to keep in the
cupboard- or as a critical component of effective management of your demand supply network. Although being a critical
component is constant, the amount attention among the trade press and vendors occurs in a cyclical fashion and 2015 is
certainly a high point in the cycle. A high level supply chain manager I knew referred to inventory policy as the Good, Bad,
and the Ugly – referring to the ability to meet client demand immediately; costs to hold inventory and potential
obsolescence; and un-coordinated hording. This is the first in a series of articles to reveal some of the mystery behind the
safety stock calculation – today we cover some basics and explain role of “normal” distribution.
Most of us use equations or formulas without a full understanding of why they work or more importantly when they work
well and when they do not? Not that we aren’t interested or do not consider it important, but the information is difficult to
find and typically written in a way only certain members of the “math tribe” can understand. There are members of this
tribe that are bilingual, the purpose of this blog is to reveal part of the mystery of the safety stock calculations with strange
symbols including when the “normal distribution” (figure 1) also called Z, bell, Gauss and other names students have
called this that cannot be said in polite company. This knowledge puts you in a better position to evaluate consultants and
software vendors; sadly, many (but not all) are using the equations without a full understanding.
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
de
ns
ity
z values
Fiigure 1: Graph of Standard Normal Distribution has mean of 0 and standard deivation of 1
Revealing The Mystery of Safety Stock Calculations 3
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2.0 Start Of An Adventure – Ice Cream Store – Rules Of Replacing Inventory & Daily Demand
The best way to take the mystery out of the “math” is to use a simple example. In fact the often the members of the “math
tribe” start with some examples on a white board and later create the equations with the strange symbols; erasing the
white board to eliminate the incriminating evidence.
Our business is small ice cream store that only sells its famous vanilla ice cream in five gallon units. When the store needs
to refill its supply of vanilla ice cream (in inventory speak - replenishment), the order can placed at night after the store
closes and the new supply is delivered well before the store opens the next day without fail. The store can, if it wants to,
reorder or replenish every night, every two nights … every 10 nights. In terms of quantity, the store can reorder between 1
and 9,999 units each time it places a replenishment decision. In inventory terminology:
The owner believes the probability of a certain demand on any day is as given in Table 1. For this article, we will accept is
as given. The complexity of generating this information is a topic for another time.
The “index” column is the identifier for each of the six possible demands. The “demand” column is the possible demand for
a specific day, which is 14 to 19. Probability is the probability of exactly that demand on a given day. For example, the
probability that daily demand is 16 is 0.08. Cumulative probability is the probability of demand for a specific day is the
value for that row or less. The probability that the demand is 15 or less is 0.45(=0.30 + 0.15) which equals the probability
the demand for the day is 15 (p=0.15). The “sum” row has the value 1 the sum of the probabilities for each possible
demand must be 1. Figure 2 is graph of the “PDF” – the probability distribution function – the probability of each demand.
index demand
probability
(PDF)
cumulative
probability
(CDF)
1 14 0.30 0.30
2 15 0.15 0.45
3 16 0.08 0.53
4 17 0.07 0.60
5 18 0.30 0.90
6 19 0.10 1.00
sum 1.00
Table 01: Probability of Each Demand
The replenishment lead time (time between an order is placed and the order arrives) is 0. Why? The order is placed
after the store closes and before it opens – 0 time in terms of meeting the needs of customers
The variability (uncertainty) of the lead time is also ZERO.
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Two natural questions are:
1. What is the average daily demand (called expected value in statistics)?
2. What is the average variation in demand (called standard deviation in statistics)?
Table 2 shows the calculation for average demand or expected demand. The value is 16.22. That is if the demand for each
day was independent of the demand from the previous day and the probability of demand did not change over time, then
if we collected demand for many days (say 100,000), added them up and divided by 100,000), the value would be 16.22 (or
very close). Here we have snuck in two other inventory terms:
Table 3 has the calculation for the average variation of demand from the “average demand” (16.22 in this case) – called the
standard deviation. This calculation is a bit tricky, so let’s take a look in detail. The first three columns are the same
columns as Table 1 and Table 2. The fourth column (demand – average) is the heart of the calculation of average variation.
Simply, for each demand, we subtract the average demand (16.22 for this example). We see the value for row 1 is -
2.22(=14-16.22). Perfectly logical, if we want to know the variation around the average – find each difference. Why not just
multiply each difference by the probability and add them? Not so fast, that value is ZERO. Observe we have negative and
positive values, we are interested in the size of the variation and therefore we need to make them all “positive’. We have
two options:
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
14 15 16 17 18 19
Figure 02: Probability for Each Possible Demand
average daily demand is 16.22average variation or standard deviation is 1.86
IID – independent and identically distributed
Stationary – the probability of demand does not change with time
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Long ago the math tribe decided to square the value. So we square each difference (the fifth column) and then multiply
each of these values with the probability (sixth column). This sum (3.47) is called the variance. We need one more step to
“undo” the squaring; we take the square root of the variation to get the standard deviation (1.86). Now you might ask –
why not just use the absolute value? There is logic in this. Later we will find the standard deviation has magical properties
when we want to know the distribution of total demand over “n” (2, 4, 10, …) days.
Table 02: Calculation of Average or Expected Demand
index demand probability
demand x
probability
1 14 0.30 4.20 = 14 x 0.30
2 15 0.15 2.25 = 15 x 0.15
3 16 0.08 1.28 = 16 x 0.08
4 17 0.07 1.19 = 17 x 0.07
5 18 0.30 5.40 = 18 x 0.30
6 19 0.10 1.90 = 19 x 0.10
sum 1.00 16.22
index demand probability demand - average
(demand -
average) 2
probability x
(demand -
average) 2
1 14 0.30 -2.22 = 14 - 16.22 4.93 1.48
2 15 0.15 -1.22 = 15 - 16.22 1.49 0.22
3 16 0.08 -0.22 = 16 - 16.22 0.05 0.00
4 17 0.07 0.78 = 17 - 16.22 0.61 0.04
5 18 0.30 1.78 = 18 - 16.22 3.17 0.95
6 19 0.10 2.78 = 19 - 16.22 7.73 0.77
sum 1.00 3.47
3.47
1.86 (= √3.47)
variance (σ2)
standard deviation (σ)
or average variation
Table 03: Calculation of Average Variation - Standard Deviation
Use absolute value to convert negative to positive
Square each value
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3.0 How Much Inventory To Have At The Start Of The Day If We Reorder Every Might?
In the previous sections we established the basic information about daily demand for vanilla ice cream and the rules
governing replenishment of inventory. Now we want to examine how much inventory to have on hand at the start of a
replenishment period – which is the number of days between reordering. We will start our examination with the simplest
case we reorder every night, which requires us to identify how vanilla ice cream to have on hand at the start of the day?
Options are the values 14 to 19. To decide which one, we need a definition of success & failure?
1. Success occurs when the amount of inventory on hand at the start of the day is sufficient to cover the demand for
the day
2. Failure occurs when the demand for the day exceeds the amount of inventory on hand at the start of the day
We now need the probability of success and failure for each inventory level option? The cumulative probability function
(CDF) provides this information. If we set the inventory level at the start of each day at 15, then success occurs when the
actual demand for the day is 15 or less (demand of 14 or 15). Failure occurs when demand for the day is more than 15
(demand of 16, 17, 18, or 19). The probability of success is the probability that demand less than or equal to 15, which is the
CDF for 15 which is 0.45(=0.30 + 0.15). The probability of failure is {1 – CDF for 15} = 1-0.45 = 0.55 = 0.08 + 0.07 + 0.30 +
0.10.
The term success and failure are intuitive and commonly used in probability and statistics. Alas, the inventory folks do not
use the terms success and failure, but the terms:
The fill rate and backorder rate for each option (14 to 19) is given in table04. This captures the risk for each option the
probability of having more ice cream in stock than needed to meet demand or not enough. . If start the day with 17 units
the chances of meeting all demand is 0.60 (fill rate) and not meeting all demand is 0.40 (backorder rate). What is the right
inventory level? That depends on other factors such as the cost to hold inventory, real cost of lost sales, scrap policy (can
left over ice cream be used tomorrow), etc. A topic for another time.
index
inventory
level at at
state of day
success or fill rate
= CDF
falure or back order
rate = (1-CDF)
1 14 0.30 0.70
2 15 0.45 0.55
3 16 0.53 0.47
4 17 0.60 0.40
5 18 0.90 0.10
6 19 1.00 0.00
Table 04: Fill rate and Backorder Rate for each Inventory level
option at start of day when replenishment occurs daily
Fill rate – probability of success, the probability all demand for the day is met from inventory
Back order rate – failure - probability demand for day cannot be met from inventory
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4.0 Going From Desired Inventory Level At Start Of Day To Safety Stock
In the previous section we reviewed the basics of deciding how much inventory to have on hand at the start of the day
(replenishment period) including the concept of fill rate and backorder rate. There was no mention of safety stock!
Safety stock is defined as the quantity of inventory (in this case five gallon containers of vanilla ice cream) a firm decides to
hold in inventory above the level of average demand. In our current example, it is the quantity in inventory at the start of
the day. Table 05 extends table 04 with the safety stock value with each option for inventory at the start of the day.
5.0 What-If We Replenish Every Other Day (Once Every Two Days)
Initially being able to replenish every day for the ice cream store seemed ideal, however having to be a the store a few
hours before opening to unload the inventory from the truck quickly became a tiring. The store owner decided to replenish
every other (every two days). This immediately generated a cascade of questions:
They key to answering these questions is creating the equivalent to Table 01 for two days. We need to know all possible
total demands on the ice cream show for a two day period and what is the probability of each of these values?
6.0 Finding The Total Demand On The Ice Cream Shop For Two Days
With the convenient “IID” assumption that:
1. Probability of demand for any given is the same and follows Table 01
index
inventory
level at at
state of day
success or fill
rate = CDF
falure or back
order rate =
(1-CDF) safety stock
1 14 0.30 0.70 -2.22 = 14 - 16.22
2 15 0.45 0.55 -1.22 = 15 - 16.22
3 16 0.53 0.47 -0.22 = 16 - 16.22
4 17 0.60 0.40 0.78 = 17 - 16.22
5 18 0.90 0.10 1.78 = 18 - 16.22
6 19 1.00 0.00 2.78 = 19 - 16.22
average
demand16.22
Table 05: Fill rate, Backorder Rate, and Safety stock level for each
Inventory level option at start of day when replenishment occurs daily
How much inventory should be in the store at the start of the replenishment period (that is when the truck makes
it delivery)?
For a given inventory level at the start of the replenishment period what is the risk?
o What is the fill rate – probability at the end of the replenishment period (two days) the total demand can
be met from inventory?
o What is the back order rate – probability at the end of the day demand cannot be met from inventory?
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2. Demand for each day is independent of the demand on others days
This is an easy question to answer – which is why the math guys make this assumption and hide in the fine print.
Let’s start with identifying the possible demand options. On day 1 the demand might be 14 and day 2 it might any of the
values 14-19. We could write these as a pair, where the first member of the pair is demand for day 1 and the second
member of the pair is the demand for day 2. The pair (14/ 16) says the demand for day 1 is 14 and demand for day 2 is 16.
The same pattern applies if the demand for day 1 is 15. There are 36 possible pairs. There are 6 options for day 1 (values
14-19) and the same six options for day 2. 36 = 6 x 6 =62, where 2 is the number of days in the replenishment period. The
fancy math term for pairs is “tuble”.
Table06 lists all of the 36 pairs. The row refers to demand for day and column for day 2
Observe the pairs (14/16) and (15/15) each have the same total demand across two days. We only care about total demand
for two days, how it is split between the two days is not important to us (right now). Table07 provides the total for demand
for each cell.
The second part of our quest is finding the probability for each demand pair. Again the “IID” fine print makes this easier.
The probability for each pair is the probability of the demand on day 1 times the probability of demand on day2. For
example the probability of a demand of 16 is 0.08 (from table01) and the probability of a demand for 15 is 0.15. The
probability of the demand pair (16 / 15) is 0.0120 = 0.08 x 0.15. Table08 has all 36 probabilities.
014 015 016 017 018 019
014 14/14 15/14 16/14 17/14 18/14 19/14
015 14/15 15/15 16/15 17/15 18/15 19/15
016 14/16 15/16 16/16 17/16 18/16 19/16
017 14/17 15/17 16/17 17/17 18/17 19/17
018 14/18 15/18 16/18 17/18 18/18 19/18
019 14/19 15/19 16/19 17/19 18/19 19/19
De
man
d fo
r day 1
Demand for Day 2
Table06: All Possible Demand Pairs (day1/day2) for total demand for two days
Day1/Day2
014 015 016 017 018 019
014 28 29 30 31 32 33
015 29 30 31 32 33 34
016 30 31 32 33 34 35
017 31 32 33 34 35 36
018 32 33 34 35 36 37
019 33 34 35 36 37 38
Table07: Total Demand for Two Days for All Possible Demand Pairs
Tot =
Day1+Day2
Demand for Day 2
De
man
d fo
r day 1
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The last step to create the equivalent of table01 for the two day replenishment period is to aggregate the probabilities for
demand pairs with the same total. For example, the demand total 29 occurs twice (pairs 14/15 and 15/14). Therefore the
probability of total demand of 29 across two day is PRB (14/15) + PRB (15/14) = 0.0450 + 0.0450 = 0.0900. Table09 has all
unique total demand options for two days along with the associated probability for each possible total demand across the
replenishment period of two days. Figure03 plots the options and their probabilities.
Table09 and Figure03 have the probability distribution function for total demand on the ice cream store for the two day
replenishment period. Additionally it has the average “total demand” for two days and the standard deviation for “total
demand” for two days. Observe we have the term average and total next to each other – which can get confusing. From
table09 on any given two day period the possible total demand at the store has 11 options the values 28 to 38. If we
collected 100,000 observations on total demand for two days replenishment periods and calculate their average, the value
would be 32.44 (or close). 32.44 is the average total demand for a two day replenishment period. The average variation or
standard deviation for this total is 2.635.
014 015 016 017 018 019
0.30 0.15 0.08 0.07 0.30 0.10
014 0.30
0.0900 =
0.30 x 0.30
0.0450 =
0.30 x 0.15
0.0240 =
0.30 x 0.08
0.0210 =
0.30 x 0.07
0.0900 =
0.30 x 0.30
0.0300 =
0.30 x 0.10
015 0.15
0.0450 =
0.15 x 0.30
0.0225 =
0.15 x 0.15
0.0120 =
0.15 x 0.08
0.0105 =
0.15 x 0.07
0.0450 =
0.15 x 0.30
0.0150 =
0.15 x 0.10
016 0.08
0.0240 =
0.08 x 0.30
0.0120 =
0.08 x 0.15
0.0064 =
0.08 x 0.08
0.0056 =
0.08 x 0.07
0.0240 =
0.08 x 0.30
0.0080 =
0.08 x 0.10
017 0.07
0.0210 =
0.07 x 0.30
0.0105 =
0.07 x 0.15
0.0056 =
0.07 x 0.08
0.0049 =
0.07 x 0.07
0.0210 =
0.07 x 0.30
0.0070 =
0.07 x 0.10
018 0.30
0.0900 =
0.30 x 0.30
0.0450 =
0.30 x 0.15
0.0240 =
0.30 x 0.08
0.0210 =
0.30 x 0.07
0.0900 =
0.30 x 0.30
0.0300 =
0.30 x 0.10
019 0.10
0.0300 =
0.10 x 0.30
0.0150 =
0.10 x 0.15
0.0080 =
0.10 x 0.08
0.0070 =
0.10 x 0.07
0.0300 =
0.10 x 0.30
0.0100 =
0.10 x 0.10
Table08: The probability of each demand pair (Day 1 / Day 2)
PRB =
Prb1xPRb2
Demand for Day 2 with probability
De
man
d fo
r day 1
with
pro
bab
ility
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index demand
probability
(PDF)
cumulative probability
(CDF)
aka fill rate or success
001 28 0.0900 0.0900
002 29 0.0900 0.1800
003 30 0.0705 0.2505
004 31 0.0660 0.3165
005 32 0.2074 0.5239
006 33 0.1612 0.6851
007 34 0.0829 0.7680
008 35 0.0580 0.8260
009 36 0.1040 0.9300
010 37 0.0600 0.9900
011 38 0.0100 1.0000
mean total demand for 2 days 32.440
variance total demand for 2 days 6.943
standard deviation total demand for 2 days 2.635
Table09: All Possible Options for Total Demand for Two Days with Probabilty
where demand for one day is specificed in Table01
0.0000
0.0500
0.1000
0.1500
0.2000
0.2500
28 29 30 31 32 33 34 35 36 37 38
Figure 03: Probability for each posssible total demand across 2 days
average daily demand is 32.44average variation or standard deviation is 2.635
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7.0 How Might The Store Owner Use The Information In Table09?
Remember fill rate is the probability of success that is the probability all demand during a replenishment period can be met
from inventory in stock. The fill rate is the cumulative probability.
If the owner wants a fill rate of 0.90, then we select the demand value with the smallest cumulative probability value greater
than or equal to 0.90. Table09 tells us this value is 36 with a cumulative probability value of 0.93. If the owner has 36 units
of vanilla ice cream in inventory at the start of the two day replenishment period, the chances of meeting all demand from
inventory is 0.93. If the owner goes with 35 units, the fill rate drops to 0.826.
What is the safety for a fill rate of 0.90? Safety stock is the amount in inventory at the start of the replenishment period
minus the average demand for this period. The average demand is 32.44, therefore safety stock is 3.56= (36-32.44).
Sometimes safety stock is stated in per day level. For this example safety stock per day for a fill rate of 0.90 is 1.78.
8.0 Gift 1 From Statistics God: Relationship Between Demand For 1 Day & Demand For 2 Days
The following fun facts from the statistics god will be helpful in the next few sections. They are provided without formal
proof, but a computation example is provided.
1. The expected total demand over “n” (in this case 2 days) is n x mean or average demand for a single day.
In section 6 we observe 32.44 = 2 x 16.22
2. The standard deviation of the distribution of total demand for n days is the square root of n times the standard
deviation of the distribution of demand for a single day
in section 6 we observe 2.635 = √2 x 1.8632 = 1.4142 x 1.8632
Remember the measure for average variation is the standard deviation. The critical observation is the average total
demand grows directly as n (number of days) grows. BUT, the average variation grows proportional to the square root of n.
A common measure to relate average variation to average value is the co-efficient of variation (CV) which is the standard
deviation divided by the average. Specific to the relationship between demand for 1 day and total demand for n days, the
following equations captures this relationship.
𝐶𝑉 = √𝑛 × 1.863
𝑛 × 16.22
As n gets larger this value continues to get smaller. For this example when n=1, CV is 0.1149. When n=10, CV is 0.0363.
CV is sometimes used to “measure” risk, we risk declines as n increases. We will also see as n increases the safety stock per
day declines for the same fill rate.
9.0 Finding The Total Demand On The Ice Cream Shop For Seven Days
Typically someone managing inventory is wants to know the following for a specified fill rate?
How demand do I need in inventory at the start of the replenishment period?
What is the safety stock required to meet this fill rate?
Revealing The Mystery of Safety Stock Calculations 12
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Our ice shop owner has decided to take a full week off and wants to know the distribution of total demand for seven days
to support a decision on how much inventory to have on hand at the start of the replenishment period. Extending the same
logic as we did with two days, Table10 has distribution of total demand for 7 days. Figure04 graphs the distribution.
The average total demand for seven days is 113.540 and the standard deviation is 4.930. The CV value is 0.043. This table
provides the core risk information for the owner to make a decision about how much inventory to have on hand at the start
of the replenishment period.
Assume our owner still desires a replenishment rate of 0.90. Table10 informs the store needs 120 units of vanilla in stock at
the start of the 7 day replenishment period and will achieve a fill rate of 0.9192 92% of the time all demand for these
seven days will be met from stock.
What about safety stock. The average total demand for seven days is 113.54. The safety stock is 6.46(=120-113.54). The
safety stock per day is 0.923 = 6.46/7. This is lower than the safety stock per day for the two day replenishment period.
10.0 Gift 2 From Statistics God: As N Gets Large, All Paths Lead To Normal
If we compare the basic shape of Figure04 with the shape of Figure01, we see the irregular shape (called multi-modal,
multiple high peaks) found in figures 02 and 04, is beginning to morph into a “normal” shape.
This is an example of the Central Limit Theorem (CLT) at work – as n gets sufficiently large, the total demand over n days,
given the IID conditions, will morph to a normal distribution where the mean of this normal distribution is n times the
average daily demand and the standard deviation is the square root of n times the standard deviation of demand for a
single day. Figure 05 shows the normal distribution when the mean is 113.540 and the standard deviation is 4.930.
11.0 The magic question how large does n need to be to use the normal as a reasonable approximation of the
distribution of total demand over n days?
It depends on how “ugly” the daily distribution is, the level of precision required, and what fill rate is desired. Now you
might have heard the magic number “30”. If n is greater than 30, then then assuming the normal distribution is correct –
“TILT” – not correct. The value of 30 has to do whether one uses the student T distribution or normal distribution for
confidence intervals. That is not relevant to this discussion. Of course it sounds nice and the consultant or software
vendor might attempt to use this when you ask them. If so it might be time for a new source of support
In Part 2 we will incorporate variation of replenishment time into our analysis.
Revealing The Mystery of Safety Stock Calculations 13
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index demand probability (PDF)
cumulative probability (CDF)
aka fill rate or success
001 98 0.0002 0.0002
002 99 0.0008 0.0010
003 100 0.0016 0.0025
004 101 0.0025 0.0051
005 102 0.0049 0.0100
006 103 0.0090 0.0190
007 104 0.0137 0.0327
008 105 0.0184 0.0510
009 106 0.0260 0.0771
010 107 0.0363 0.1134
011 108 0.0456 0.1590
012 109 0.0524 0.2114
013 110 0.0615 0.2729
014 111 0.0715 0.3445
015 112 0.0769 0.4213
016 113 0.0767 0.4980
017 114 0.0769 0.5749
018 115 0.0767 0.6516
019 116 0.0713 0.7229
020 117 0.0616 0.7845
021 118 0.0530 0.8375
022 119 0.0454 0.8829
023 120 0.0363 0.9192
024 121 0.0267 0.9459
025 122 0.0192 0.9652
026 123 0.0139 0.9790
027 124 0.0093 0.9883
028 125 0.0055 0.9938
029 126 0.0031 0.9969
030 127 0.0017 0.9986
031 128 0.0009 0.9995
032 129 0.0004 0.9999
033 130 0.0001 1.0000
034 131 0.0000 1.0000
035 132 0.0000 1.0000
036 133 0.0000 1.0000
mean total demand for 7 days 113.540
standard deviation total demand for 7 days 4.930
Table10: All Possible Options for Total Demand for Two Days with Probabilty
where demand for one day is specificed in Table01
Revealing The Mystery of Safety Stock Calculations 14
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0.0000
0.0100
0.0200
0.0300
0.0400
0.0500
0.0600
0.0700
0.0800
0.0900
98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133
Figure 04: Probability for each posssible total demand across 7 days
average daily demand is 113.540average variation or standard deviation is 4.930
0.00
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
den
sit
y
z values
Fiigure 05: Graph of Normal Distribution has mean of 113.54 and standard deivation of 4.93
Revealing The Mystery of Safety Stock Calculations 15
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11.0 How Do We Get To The Safety Stock Equation- 𝒔𝒂𝒇𝒆𝒕𝒚 𝒔𝒕𝒐𝒄𝒌 = 𝒁𝒇𝒓 × √𝒏 × 𝝈
Our ice cream store owner wants to investigate a replenishment period of 30 days and is willing to use the normal
distribution to approximate the distribution of total demand over the 30 day periods. From statistics gift 1, we calculate:
Figure05 has the normal approximation of the distribution of total demand for the replenishment period of 30 days with a
mean of 486.6 and standard deviation of 10.205.
The owner now asks how much inventory is needed to be in stock at the start of the replenishment period to have a fill rate
of 0.90. Hmmm – before we looked at the table to find the demand value with the smallest cumulative probability value
greater than 0.90. We have no such table. The normal is a continuous probability distribution, not a discrete distribution.
In a continuous distribution the list of options for demand is all values between a start and end value. To adjust for this, we
can only talk about the probability of demand occurring between a start value and end value and this probability is the area
under the curve. The total area under the curve is 1, so any subsection of area has a value between 0 and 1. The smaller the
area, the lower the probability.
To answer the owner’s question we need find the demand value so the area under the curve from left to this point is 0.90
and from this point to the right the area is 0.10. Figure07 identifies this value as 499.68. This is how much stock must be
on hand at the start of the replenishment period. The safety stock is 13.08(=499.68 – 486.6).
0.00
0.01
0.01
0.02
0.02
0.03
0.03
0.04
0.04
0.05
den
sit
y
z values
Fiigure 06: Graph of Normal Distribution has mean of 486.6 and standard deivation of 10.205
Average total demand for 30 days is 486.6 = 30 x 16.22
Standard deviation for 30 days is 10.205 = √30 x 1.863 = 5.477 x 1.863
Revealing The Mystery of Safety Stock Calculations 16
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How did we get the value of 499.68?
For a given mean and standard deviation, the inventory level for a given refill rate (success probability) is: mean + {Z(for
refill rate)x standard_ deviation}. Where Z is the value for the standard normal (figure 01) which has 90% of the area to the
left of this value and 10% to the right. This value can be found in table or statistics software. In Excel the syntax is
norm.s.inv (0.90) and the Z value is 1.2816. We have:
Putting the math symbols back in we get 𝑠𝑎𝑓𝑒𝑡𝑦 𝑠𝑡𝑜𝑐𝑘 = 𝑍𝑓𝑟 × √𝑛 × 𝜎, where:
0.00
0.01
0.01
0.02
0.02
0.03
0.03
0.04
0.04
0.05
den
sit
y
z values
Fiigure 07: Graph Normal Dist mean 486.6,standard deivation10.205, repenishment rate 0.90
0.90
0.10
499.68
499.68 = 488.6 + 1.2815 x 10.205
o If we put back in the average daily demand and standard deviation we get
499.68 = (30 x 16.22) + 1.2815 x √30 x 1.863
o Since safety stock is the desired inventory level minus average demand, the safety stock is
13.08 = 1.2815 x √30 x 1.863
Z is the value of Z for the specified fill rate
n is the number of days in the replenishment period
σ is the standard deviation of the daily demand
Revealing The Mystery of Safety Stock Calculations 17
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12.0 Final Comment
The material presented in the previous sections provides unravels the mystery of one part of inventory management -
managing the risk of not meeting demand even when there is no uncertainty in replenishment. It sets out the core
questions, defines terms, develops the core calculations, and identifies when the normal distribution will be work and how
it will work.
Part 2 will demonstrate how variability in replenishment lead time is incorporated into the mix of calculations. Part 3 will
look at a simple network of inventory points.
Inventory is a huge and critical area that cannot be treated separate from other areas of S&OP such as demand
management and supply planning – key areas include demand variability, demand classes, and smoothing or load leveling
production. In fact one director of manufacturing I know saw inventory management, especially the flexibility associated
with line side stocking and build to forecast, as key tools for smoothing production and absorbing variations in yields.
These are topics for later sections.
PART 2
Revealing the Mystery of Safety Stock Calculations – Incorporating
Variation in Replenishment Lead Time
1.0 Introduction
In part 1, using the ice cream store example as our primary vehicle, we provided insight into the basics of understanding
total inventory required to cover demand for a period n (1, 2, …30 …) days accounting for variability in the daily demand.
From this, we identified what portion of this demand would be classified as safety stock and how to use this information to
make an informed decision. Last, we demonstrated when n is sufficiently large, then the normal distribution is a good
approximation of total demand over n days independent of the distribution of daily demand applying the 8th wonder of the
world – Central Limit Theorem (CLT).
Throughout all of part 1, the assumption made was there was ZERO variation in replenishment lead time. The condition
was, a replenishment order was placed at the end of the day after the store was closed and the owner was certain the new
5 gallon cans of vanilla ice cream would be in the freezer before the first customer arrived the next day. The purpose of part
2 of the mystery tale is “how to handle uncertainty or variability in the lead time!
2.0 Quick Review Of Ice Cream Business – Rules Of Replacing Inventory & Daily Demand
Our business is small ice cream store that only sells its famous vanilla ice cream in five gallon units. When the store needs
to refill its supply of vanilla ice cream, the order can placed at night after the store closes and the new supply is delivered
well before the store opens the next day without fail. The store can, if it wants to, reorder or replenish every night, every
two nights … every 10 nights. In terms of quantity, the store can reorder between 1 and 9,999 units each time it places a
replenishment decision. The owner believes the probability of a certain demand on any day is as given in Table 1. For this
article, we will accept is as given. The complexity of generating this information is a topic for another time.
Revealing The Mystery of Safety Stock Calculations 18
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The “index” column is the identifier for each of the six possible demands. The “demand” column is the possible
demand for a specific day, which is 14 to 19. Probability is the probability of exactly that demand on a given
day. For example, the probability that daily demand is 16 is 0.08. Cumulative probability is the probability of
demand for a specific day is the value for that row or less. The probability that the demand is 15 or less is
0.45(=0.30 + 0.15) which equals the probability the demand for the day is 15 (p=0.15). The “sum” row has the
value 1 the sum of the probabilities for each possible demand must be 1. Figure 2 is graph of the “PDF” – the
probability distribution function – the probability of each demand. The average daily demand is 16.22 and
average variation (standard deviation) is 1.86
3.0 Replenishing Inventory Every Three Days with 0.90 Fill Rate
index demand
probability
(PDF)
cumulative
probability
(CDF)
1 14 0.30 0.30
2 15 0.15 0.45
3 16 0.08 0.53
4 17 0.07 0.60
5 18 0.30 0.90
6 19 0.10 1.00
sum 1.00
Table 01: Probability of Each Demand
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
14 15 16 17 18 19
Figure 02: Probability for Each Possible Demand
average daily demand is 16.22average variation or standard deviation is 1.86
Revealing The Mystery of Safety Stock Calculations 19
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After working the store’s analytics team, the owner had decided to replenish inventory every third day with a fill rate of
0.90. Table11 and figure08 have the distribution of total demand for three days using computational mechanism described
in part 1. This information, along with estimates for inventory holding cost, reorder costs (transport, people, etc.) shelve
life, cost for unmet demand, and others were analyzed with other models to support this decision.
index demand
probability
(PDF)
cumulative probability
(CDF)
aka fill rate or success
001 42 0.0270 0.0270
002 43 0.0405 0.0675
003 44 0.0419 0.1094
004 45 0.0439 0.1532
005 46 0.1111 0.2643
006 47 0.1257 0.3900
007 48 0.1004 0.4904
008 49 0.0841 0.5745
009 50 0.1266 0.7011
010 51 0.1131 0.8143
011 52 0.0654 0.8796
012 53 0.0393 0.9189
013 54 0.0420 0.9609
014 55 0.0291 0.9900
015 56 0.0090 0.9990
016 57 0.0010 1.0000
mean total demand for 3 days 48.660
variance total demand for 3 days 10.415
standard deviation total demand for 3 days 3.227
Table11: All Possible Options for Total Demand for Three Days with Probabilty
where demand for one day is specificed in Table01
Revealing The Mystery of Safety Stock Calculations 20
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3.0 Replenishing Inventory Every Three Days Accounting for Uncertainty in Delivery
With the basic decisions in place for inventory and the establishment of a regular process for replenishing inventory in an
orderly fashion, the owner now asks the next logical question in the journey
What if the delivery of ice cream is delayed one day?
He goes to his ace analytics team, asks them to the analysis, and provide him a solution he can understand - no voodoo
equations and hand waving – all by Friday.
The lead of his analytics team, having read the great works of Gene Woolsey, comes up with the following solution.
1. In the vast majority of the time, when the order is placed at night, it arrives the next morning. Since an order is placed
one every three days, the key calculation is finding the probability distribution of total demand for 3 days.
2. If the truck is delayed one day, that is the equivalent of having to know the probability distribution of total demand for
four days. This can be calculated using the same method already established. Table12 and Figure09 have this
information in Appendix 1.
3. The question becomes how to combine this information in a way that can be understood and works (this is called
mixture problem in the literature – as well going under some other names).
0.0000
0.0200
0.0400
0.0600
0.0800
0.1000
0.1200
0.1400
42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57
Figure 08: Probability for each posssible total demand across 3 days
average total demand is 48.66average variation or standard deviation is 3.227
Revealing The Mystery of Safety Stock Calculations 21
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4. The lead starts with the assumption the truck will arrive on time 80% (0.80) of the time. It will arrive one day late 20%
(0.20) of the time.
5. Step 1 – the lead multiplies the probability distribution of total demand for 3 days by 0.80 and the probability
distribution of total demand for four days by 0.20. This information is displayed in Tables 13 and 14. He quickly notices
the column total for the probability of each event multiplied by 0.80 in Table 13 is 0.80, similarly 0.20 in Table14.
6. After a bit of thought, the lead realizes whether the truck arrives on time or one day late is independent of the demand
on the store.
7. The lead reframes the question as follows: What is distribution of demand on the store just after the replenishment
delivery has arrived which brings the total inventory to target level until the next arrival of inventory, where the next
arrival is either in three days or four day where the probability of three days is 0.80 and four days is 0.20.
8. This distribution, that is the probability of different levels demand or this period with a variable end point (3 days or 4
days), is simply the weighted average of the probability distribution of total demand for three days with its equivalent
for 4 days.
9. Step 2 is simply noting all possible options is the unique set of demand values between Table13 and Tabl14 and
combining (adding) probabilities where a value is in both tables.
10. For example the values 42 to 55 can only be found in Table13. The values 58 to 76 can only be found in Table 14. The
value 56 and 57 are in both tables.
11. Table 15 and Figure 10 has the results of this combination or mixture work.
This story has a happy ending, the owner is happy and now trusts the analytics lead to finish the work – there some other
substantial complexities – coordinating safety stock decisions across the network, intermittent demand, impact of cycle
time and time between reordering, and others – future topics.
Revealing The Mystery of Safety Stock Calculations 22
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index demand probability (PDF)
0.80 x
probability
001 42 0.0270 0.0216
002 43 0.0405 0.0324
003 44 0.0419 0.03348
004 45 0.0439 0.0351
005 46 0.1111 0.088848
006 47 0.1257 0.100548
007 48 0.1004 0.0803296
008 49 0.0841 0.0672792
009 50 0.1266 0.1013088
010 51 0.1131 0.0905144
011 52 0.0654 0.052296
012 53 0.0393 0.031416
013 54 0.0420 0.0336
014 55 0.0291 0.02328
015 56 0.0090 0.0072
016 57 0.0010 0.0008
SUM 1.0000 0.8000
Table13: All Possible Options for Total Demand for Three
Days with Probabilty where demand for one day is
specificed in Table01, then multiplied by 0.80
Revealing The Mystery of Safety Stock Calculations 23
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index demand
probability
(PDF)
0.20 x
probability
001 56 0.0081 0.00162
002 57 0.0162 0.00324
003 58 0.0208 0.004158
004 59 0.0246 0.004914
005 60 0.0542 0.010836
006 61 0.0757 0.015131
007 62 0.0775 0.015507
008 63 0.0755 0.015094
009 64 0.1051 0.021028
010 65 0.1155 0.023101
011 66 0.0953 0.019058
012 67 0.0748 0.014955
013 68 0.0780 0.015608
014 69 0.0694 0.013871
015 70 0.0441 0.00882
016 71 0.0252 0.005047
017 72 0.0194 0.003887
018 73 0.0136 0.002728
019 74 0.0057 0.001136
020 75 0.0012 0.00024
021 76 0.0001 0.00002
SUM 1.0000 0.2000
Table14: All Possible Options for Total Demand for Four
Days with Probabilty where demand for one day is
specificed in Table01, then multiplied by 0.20
Revealing The Mystery of Safety Stock Calculations 24
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index demand
probability
(PDF)
cumulative probability (CDF)
aka fill rate or success
001 42 0.0216 0.0216
002 43 0.0324 0.0540
003 44 0.0335 0.0875
004 45 0.0351 0.1226
005 46 0.0888 0.2114
006 47 0.1005 0.3120
007 48 0.0803 0.3923
008 49 0.0673 0.4596
009 50 0.1013 0.5609
010 51 0.0905 0.6514
011 52 0.0523 0.7037
012 53 0.0314 0.7351
013 54 0.0336 0.7687
014 55 0.0233 0.7920
015 56 0.0088 0.8008
016 57 0.0040 0.8049
017 58 0.0042 0.8090
018 59 0.0049 0.8139
019 60 0.0108 0.8248
020 61 0.0151 0.8399
021 62 0.0155 0.8554
022 63 0.0151 0.8705
023 64 0.0210 0.8915
024 65 0.0231 0.9146
025 66 0.0191 0.9337
026 67 0.0150 0.9486
027 68 0.0156 0.9643
028 69 0.0139 0.9781
029 70 0.0088 0.9869
030 71 0.0050 0.9920
031 72 0.0039 0.9959
032 73 0.0027 0.9986
033 74 0.0011 0.9997
034 75 0.0002 1.0000
035 76 0.0000 1.0000
mean total demand for replenisment period 51.904
standard deviation total demand replenishment 7.294
Table15: All Possible Options for Total Demand Between Replenishment, when the probabity it
is 3 days is 0.80 and 4 days is 0.20
Revealing The Mystery of Safety Stock Calculations 25
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0.0000
0.0200
0.0400
0.0600
0.0800
0.1000
0.1200
42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76
Figure 10: Probability for each posssible total demand across 3 or 4 days
average total demand is 51.90average variation or standard deviation is 7.294probability of 3 days is 0.80, four days 0.20
Revealing The Mystery of Safety Stock Calculations 26
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index demand
probability
(PDF)
cumulative probability
(CDF)
aka fill rate or success
001 56 0.0081 0.0081
002 57 0.0162 0.0243
003 58 0.0208 0.0451
004 59 0.0246 0.0697
005 60 0.0542 0.1238
006 61 0.0757 0.1995
007 62 0.0775 0.2770
008 63 0.0755 0.3525
009 64 0.1051 0.4576
010 65 0.1155 0.5732
011 66 0.0953 0.6684
012 67 0.0748 0.7432
013 68 0.0780 0.8213
014 69 0.0694 0.8906
015 70 0.0441 0.9347
016 71 0.0252 0.9599
017 72 0.0194 0.9794
018 73 0.0136 0.9930
019 74 0.0057 0.9987
020 75 0.0012 0.9999
021 76 0.0001 1.0000
mean total demand for 3 days 64.880
variance total demand for 3 days 13.886
standard deviation total demand for 3 days 3.726
Table12: All Possible Options for Total Demand for Four Days with Probabilty
where demand for one day is specificed in Table01
Revealing The Mystery of Safety Stock Calculations 27
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0.0000
0.0200
0.0400
0.0600
0.0800
0.1000
0.1200
0.1400
56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76
Figure 09: Probability for each posssible total demand across 4 days
average total demand is 64.88average variation or standard deviation is 3.726
Revealing The Mystery of Safety Stock Calculations 28
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