Managing Flow Variability: Safety Inventory
Analytical Solution for the Optimal Service Level
Marginal Profit:
MP = p – c
MP = 30 - 10 = 20
Marginal Cost:
MC = c - v
MC = 10-5 = 5
Suppose I have ordered Q units.
What is the expected cost of ordering one more units?
What is the expected benefit of ordering one more units?
If I have ordered one unit more than Q units, the probability of
not selling that extra unit is the probability demand to be less
than or equal to Q.
Since we have P( R ≤ Q).
The expected marginal cost =MC× P( R ≤ Q)
The Newsvendor Problem
Ardavan Asef-Vaziri, Oct 2011
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Managing Flow Variability: Safety Inventory
Analytical Solution for the Optimal Service Level
If I have ordered one unit more than Q units, the probability of
selling that extra unit is the probability of demand to be greater
than Q.
We know that P(R > Q) = 1- P(R ≤ Q).
The expected marginal benefit = MB× [1-Prob.( r ≤ Q)]
As long as expected marginal cost is less than expected
marginal profit we buy the next unit.
We stop as soon as: Expected marginal cost ≥ Expected
marginal profit.
The Newsvendor Problem
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Managing Flow Variability: Safety Inventory
Analytical Solution for the Optimal Service Level
MC×Prob(R ≤ Q*) ≥ MP× [1 – Prob( R ≤ Q*)]
Prob(R ≤ Q*) ≥
MB
MB  MC
MP = p – c = Underage Cost = Cu
MC = c – v = Overage Cost = Co
P( R  Q* ) 
MP
MP  MC
The Newsvendor Problem
cu

Cu  C o
MP
MP  MC
pc
pc


pccv pv
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Managing Flow Variability: Safety Inventory
Marginal Value: The General Formula
P(R ≤ Q*) ≥ Cu / (Co+Cu)
Cu / (Co+Cu) = (30-10)/[(10-5)+(30-10)] = 20/25 = 0.8
Order until P(R ≤ Q*) ≥ 0.8
P(R ≤ 5000) ≥ = 0.75 not > 0.8 still order
P(R ≤ 6000) ≥ = 0.9 > 0.8 Stop
Demand
1000
2000
3000
4000
5000
6000
7000
Probability
0.1
0.15
0.15
0.2
0.15
0.15
0.1
In Continuous Model where demand for example has Uniform
or Normal distribution
MP
cu
P( R  Q ) 

MP  MC
Cu  C o
*
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Ardavan Asef-Vaziri, Oct 2011
pc

pv
4
Managing Flow Variability: Safety Inventory
Type-1 Service Level
What is the meaning of the number 0.80?
80% of the time all the demand is satisfied.
–
Probability {demand is smaller than Q} =
–
Probability {No shortage} =
–
Probability {All the demand is satisfied from stock} = 0.80
The Newsvendor Problem
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Managing Flow Variability: Safety Inventory
Marginal Value: Uniform distribution
Suppose instead of a discreet
demand of
Demand
1000
2000
3000
4000
5000
6000
7000
Probability
0.1
0.15
0.15
0.2
0.15
0.15
0.1
Cumlat
Probab
0.1
0.25
0.4
0.6
0.75
0.9
1
We have a continuous demand uniformly distributed
between 1000 and 7000
1000
7000
Pr{r ≤ Q*} = 0.80
How do you find Q?
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Managing Flow Variability: Safety Inventory
Marginal Value: Uniform distribution
Q-l = Q-1000
?
0.80
l=1000
1/6000
u=7000
u-l=6000
(Q-1000)/6000=0.80
Q = 5800
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Managing Flow Variability: Safety Inventory
Marginal Value: Normal Distribution
Suppose the demand is normally distributed with a mean
of 4000 and a standard deviation of 1000.
What is the optimal order quantity?
Notice: F(Q) = 0.80 is correct for all distributions.
We only need to find the right value of Q assuming the
normal distribution.
P(z ≤ Z) = 0.8  Z= 0.842
Q = mean + z Standard Deviation  4000+841 = 4841
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Managing Flow Variability: Safety Inventory
Marginal Value: Normal Distribution
0.00045
0.0004
Probability of
excess inventory
0.00035
0.0003
Probability of
shortage
4841
0.00025
0.0002
0.00015
0.0001
0.80
0.00005
0
0
2000
4000
0.20
6000
8000
Given a service level, how do we calculate z?
From our normal table or
From Excel Normsinv(service level)
The Newsvendor Problem
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Managing Flow Variability: Safety Inventory
Additional Example
Your store is selling calendars, which cost you $6.00 and sell for $12.00
Data from previous years suggest that demand is well described by a
normal distribution with mean value 60 and standard deviation 10.
Calendars which remain unsold after January are returned to the
publisher for a $2.00 "salvage" credit. There is only one opportunity to
order the calendars. What is the right number of calendars to order?
MC= Overage Cost = Co = Unit Cost – Salvage = 6 – 2 = 4
MB= Underage Cost = Cu = Selling Price – Unit Cost = 12 – 6 = 6
Cu
6
P( R  Q ) 

 0. 6
Cu  C o 6  4
*
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Managing Flow Variability: Safety Inventory
Additional Example - Solution
Look for P(x ≤ Z) = 0.6 in Standard Normal table or
for NORMSINV(0.6) in excel  0.2533
Q*  
Q*  
P( Z 
)  0.6 
 0.2533


Q*    0.2533  60  10(0.2533)  62.533  63
By convention, for the continuous demand distributions,
the results are rounded to the closest integer.
Suppose the supplier would like to decrease the unit cost in
order to have you increase your order quantity by 20%. What is
the minimum decrease (in $) that the supplier has to offer.
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Managing Flow Variability: Safety Inventory
Additional Example - Solution
Qnew = 1.2 * 63 = 75.6 ~ 76 units
76  60
P( R  Q )  P( R  76)  P( Z 
)  P( Z  1.6)
10
*
Look for P(Z ≤ 1.6) = 0.6 in Standard Normal table
or for NORMSDIST(1.6) in excel  0.9452
Cu
pc
12  c 12  c
P( R  Q )  0.9452 



Cu  Co p  c  c  v 12  2
10
*
12  c  9.452  c  2.55
The Newsvendor Problem
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Managing Flow Variability: Safety Inventory
Additional Example
On consecutive Sundays, Mac, the owner of your local newsstand,
purchases a number of copies of “The Computer Journal”. He pays 25
cents for each copy and sells each for 75 cents. Copies he has not sold
during the week can be returned to his supplier for 10 cents each. The
supplier is able to salvage the paper for printing future issues. Mac has
kept careful records of the demand each week for the journal. The
observed demand during the past weeks has the following distribution:
Qi
4
5
6
7
8
P(R=Qi) 0.04 0.06 0.16 0.18 0.2
9
0.1
10
0.1
11 12 13
0.08 0.04 0.04
What is the optimum order quantity for Mac to minimize his
cost?
The Newsvendor Problem
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Managing Flow Variability: Safety Inventory
Additional Example - Solution
Overage Cost = Co = Unit Cost – Salvage = 0.25 – 0.1 = 0.15
Underage Cost = Cu = Selling Price – Unit Cost = 0.75 – 0.25 = 0.50
Cu
0.50
P( R  Q*) 

 0.77
Cu  Co 0.50  0.15
P( R  Q* )  0.77
Qi
4
5
6
7
8
9
10
11
12
13
Probability
P(R=Qi)
0.04
0.06
0.16
0.18
0.20
0.10
0.10
0.08
0.04
0.04
Cumulative
Probability
F(Qi)
0.04
0.10
0.26
0.44
0.64
0.74
0.84
0.92
0.96
1.00
Q* = 10
The Newsvendor Problem
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Managing Flow Variability: Safety Inventory
More Example
Swell Productions (The Retailer) is sponsoring an outdoor conclave for
owners of collectible and classic Fords. The concession stand in the TBird area will sell clothing such as official Thunderbird racing jerseys.
Suppose the probability of jerseys sales quantities is uniformly (and
continuously) distributed between 100 and 600. Suppose P= $80, c= $40,
and v=$20. How many Jerseys Swell Production orders? distributed
with mean of 300 and standard deviation of 80. Suppose P= $80, c=
$40, and v=$20. How many Jerseys Swell Production orders?
P( R  Q) 
P  c 80  40 2


P  v 80  20 3
100
P( R  Q) 
The Newsvendor Problem
QL
Q  100

U  L 600  100
Q  100 2

500
3
Q
Q
Ardavan Asef-Vaziri, Oct 2011
600
1300
 433.33  434
3
15
Managing Flow Variability: Safety Inventory
More Example
Suppose the probability of jerseys sales quantities is uniformly (and
continuously) distributed between 100 and 600. Suppose P= $80 and c=
$40, but the salvage value is negotiable. Compute the salvage value
such that Swell Production orders 400 units.
P( R  Q) 
P  c 80  40
40


P  v 80  v 80  v
100
P( R  400) 
Q
600
Q  L 400  100 3


U  L 600  100 5
40
3

80  v 5
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240  3v  200
v  13.33
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Managing Flow Variability: Safety Inventory
More Example
Suppose the probability of jerseys sales quantities is normally
distributed with mean of 300 and standard deviation of 80. Suppose P=
$80, c= $40, and v=$20. How many Jerseys Swell Production orders?
P(R≤ Q) = 2/3 = 0.67
P  c 80  40 2
P( R  Q) 


P  v 80  20 3
Probability is 0.67 find z
z = 0.43
z  0.44 
Q


Q  300
80
Q  335.2
The Newsvendor Problem
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Managing Flow Variability: Safety Inventory
More Example
Suppose the probability of jerseys sales quantities is normally
distributed with mean of 300 and standard deviation of 80. Suppose P=
$80 and c= $40, but the salvage value is negotiable. Compute the
salvage value such that Swell Production orders 400 units.
P( R  Q) 
z
400  

Pc
40

P  v 80  v
z = 1.25
400  300 100


 1.25
80
80
P(z≤ Z) = 0.8944
40
 0.8944
80  v
Q  35.28  36
The Newsvendor Problem
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Managing Flow Variability: Safety Inventory
More Example
Suppose the following table shows the probability of jerseys sales quantities.
Probability 0.05
Demand
100
0.10
200
0.30
300
0.20
400
0.20
500
0.15
600
Suppose P= $80 and c= $40, but the salvage value is negotiable. Compute
the minimal salvage value such that Swell Production orders 400 units.
As long as P(R≤ Q) ≥ (P-c)/(P-v) we order more than Q.
If we want to order 400, then
At 300 we must have P(R≤ 300) < (P-c)/(P-v), P(R≤ 300) < 40/(80-v), and
At 400 we must have P(R≤ 400) ≥ 40/(80-v).
At 400 we must have 0.05+0.10+0.30+0.20 = 0.65 ≥ 40/(80-v)
52-0.65v ≥ 40  18.5 ≥ v
At 400 we must have 0.05+0.10+0.30 =
0.45 ≥ 40/(80-v)
The smaller the v, the smaller the right hand side. If v= 0, the RHS is 0.5.
18.5 ≥ V ≥ 0
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