Newsvendor news boy

1
Managing Flow Variability: Safety Inventory
The Newsvendor Problem Ardavan Asef-Vaziri, Oct 2011
Marginal Profit:
Marginal Cost:
MP = p – c
MC = c - v
MP = 30 - 10 = 20
MC = 10-5 = 5
Analytical Solution for the Optimal Service Level
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)

2
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.

3
MC×Prob(R ≤ Q*) ≥ MP× [1 – Prob( R ≤ Q*)]
MP = p – c = Underage Cost = Cu
MC = c – v = Overage Cost = Co
ou
u
CC
c


MCMP
MP
QRP

 )( *
vp
cp
vccp
cp






MCMP
MP

MB
MB MC
Prob(R ≤ Q*) ≥

4
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
In Continuous Model where demand for example has Uniform
or Normal distribution
Demand Probability
1000 0.1
2000 0.15
3000 0.15
4000 0.2
5000 0.15
6000 0.15
7000 0.1
MCMP
MP
QRP

 )( *
ou
u
CC
c


vp
cp




5
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

6
Marginal Value: Uniform distribution
Suppose instead of a discreet
demand of
Demand Probability
Cumlat
Probab
1000 0.1 0.1
2000 0.15 0.25
3000 0.15 0.4
4000 0.2 0.6
5000 0.15 0.75
6000 0.15 0.9
7000 0.1 1
Pr{r ≤ Q*} = 0.80
We have a continuous demand uniformly distributed
between 1000 and 7000
1000 7000
How do you find Q?

7
Marginal Value: Uniform distribution
l=1000 u=7000
?
u-l=6000
1/60000.80
Q-l = Q-1000
(Q-1000)/6000=0.80
Q = 5800

8
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

9
Marginal Value: Normal Distribution
0
0.00005
0.0001
0.00015
0.0002
0.00025
0.0003
0.00035
0.0004
0.00045
0 2000 4000 6000 8000
4841
Probability of
excess inventory
Probability of
shortage
0.80
0.20
Given a service level, how do we calculate z?
From our normal table or
From Excel Normsinv(service level)

10
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
6.0
46
6
)( *





ou
u
CC
C
QRP

11
Additional Example - Solution
By convention, for the continuous demand distributions,
the results are rounded to the closest integer.
2533.06.0)(
**








 QQ
ZP
Look for P(x ≤ Z) = 0.6 in Standard Normal table or
for NORMSINV(0.6) in excel  0.2533
63533.62)2533.0(10602533.0*
 Q
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.

12
Qnew = 1.2 * 63 = 75.6 ~ 76 units
)6.1()
10
6076
()76()( *


 ZPZPRPQRP
Look for P(Z ≤ 1.6) = 0.6 in Standard Normal table
or for NORMSDIST(1.6) in excel  0.9452
10
12
212
12
9452.0)( * cc
vccp
cp
CC
C
QRP
ou
u 









55.2452.912  cc

13
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 9 10 11 12 13
P(R=Qi) 0.04 0.06 0.16 0.18 0.2 0.1 0.1 0.08 0.04 0.04
What is the optimum order quantity for Mac to minimize his
cost?

14
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
77.0)(
77.0
15.050.0
50.0
*)(
*






QRP
CC
C
QRP
ou
u
Probability
Cumulative
Probability
Qi P(R=Qi) F(Qi)
4 0.04 0.04
5 0.06 0.10
6 0.16 0.26
7 0.18 0.44
8 0.20 0.64
9 0.10 0.74
10 0.10 0.84
11 0.08 0.92
12 0.04 0.96
13 0.04 1.00
Q* = 10

15
More Example
Swell Productions (The Retailer) is sponsoring an outdoor conclave for
owners of collectible and classic Fords. The concession stand in the T-
Bird 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?
100600
100
)(






Q
LU
LQ
QRP
100 600Q
3
2
2080
4080
)( 






vP
cP
QRP
3
2
500
100

Q
43433.433
3
1300
Q

16
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.
5
3
100600
100400
)400( 






LU
LQ
RP
100 600Q
vvvP
cP
QRP








80
40
80
4080
)(
2003240  v
5
3
80
40

 v
33.13v

17
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?
80
300
44.0




QQ
z


3
2
2080
4080
)( 






vP
cP
QRP
P(R≤ Q) = 2/3 = 0.67
Probability is 0.67 find z
z = 0.43
2.335Q

18
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.
25.1
80
100
80
300400400







z
vvP
cP
QRP





80
40
)(
z = 1.25
P(z≤ Z) = 0.8944
3628.35 Q
8944.0
80
40

 v

19
More Example
Suppose the following table shows the probability of jerseys sales quantities.
Probability 0.05 0.10 0.30 0.20 0.20 0.15
Demand 100 200 300 400 500 600
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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