Newsvendor (Newsboy) Model: A Brief Introduction
From sweatshirts (in the EOQ model introduction) to summer dresses. The big difference is that
while sweatshirts were continuous selling items, the demand for summer dresses is limited to
summer months. Once the summer season is over, the unsold dresses must be heavily discounted.
You are a local design firm that designs northwest-accented summer dresses, sources them from
China and sells them through retailers here.
The problem is that for a particular summer dress, total demand during the summer season is hard
to predict. All you can do is to make a guess, that is, develop a probability distribution of
demand. Let us generate our demand with the throw of a regular dice; it can be any number from
1 to 6, each with probability 1/6.
On the supply side, the lead time from your Chinese supplier is long. There is no possibility of
making multiple orders. You make one order before the summer season starts, sell as many as
you can during the season and then whatever is left is discounted. Let us say that per unit
purchase cost c is $80. For any units that you are able to sell per unit revenue r is $100. For the
units you are not able to sell during the season, let us say that you can discount them and are able
to sell them at a per unit salvage value s of $30.
The big decision is the order quantity S of dresses you should order from your Chinese supplier at
the beginning of the summer season.
The Trade-off
If you order a very large quantity, there is a bigger chance that you will not be able to sell all of
them. There will be excess units at the end of the season that you will have to discount. You will
lose money on them. On the other hand, if you order a small quantity, there is a bigger chance
that you will be short. That is, there will be some demand you will not be able to satisfy. You will
not be able to make as much money as you could have.
The order quantity decision resolves this trade-off between the expected cost of having excess
inventory and the expected cost of falling short. We will call the sum of these two costs as
Mismatch cost. The optimal order quantity will minimize mismatch cost.
Marginal Costs
To resolve this trade-off, we start with defining marginal costs of excess and shortage.
Marginal Cost of excess Ce is defined as the cost of having one unit excess. You bought this unit
for purchase cost of c=$80, were not able to sell it during the season and then had to discount it
down to the salvage value of s=$30. The cost to you is $80-$30=$50. That is, in this setting,
Ce=c-s.
Marginal Cost of shortage Cs is defined as the cost of having one unit short. Had you bought this
unit for purchase cost of c=$80, you would have been able to sell it during the season for a
revenue of r=$100. We say, that the cost to you for being one unit short is $100-$80=$20. That
is, in this setting, Cs=r-c.
Service Level
Service level is the chance that you will be able to meet all the demand in a single period
(summer season). Suppose you bought an order quantity S=3 units. Recall that demand is any
number between 1 and 6 with equal probability 1/6. In this case, you will be able to meet all the
demand only if demand is either 1 unit, 2 units or 3 units. That is, the probability that demand is
less than or equal to S=3 units. This probability is known as cumulative probability and is given
by the sum of the probabilities that demand is 1, demand is 2, and demand is 3 =
(1/6)+(1/6)+(1/6)=3/6=1/2. That is, if you buy S=3 units, you will provide a service level of 50%.
Here is a quick table to provide cumulative probabilities in our case:
Demand
1
2
3
4
5
6
Probability
1/6
1/6
1/6
1/6
1/6
1/6
Cumulative
1/6
1/6+1/6
2/6+1/6
3/6+1/6
4/6+1/6
5/6+1/6
Probability
=2/6
=3/6
=4/6
=5/6
=6/6=1
Optimal Service Level and Optimal Order Quantity
Single-period model tells us that, given the marginal costs of excess and shortage, Ce and Cs, the
optimal service level is given by (Cs/(Cs+Ce). In our case, the optimal service level is equal to
20/(20+50)= 0.2857.
Optimal order quantity S* is the minimum size of the order that will be able to provide the
optimal service level. Going by the above table, if you buy, for example, S=1, you will be able to
provide a service level of 1/6=0.1667 which is less than the optimal service level we wish to
provide. If we buy 2, service level is 2/6=0.3333 and we will be able to satisfy the optimal service
level requirement of 0.2857. Therefore S*=2.
Rule: compute optimal service level and find the minimum value of demand for which
cumulative probability, for the first time, equals or exceeds optimal service level. That is the
optimal order quantity.
Summary of Formulas for Continuous Demand: Normal Distribution
The demand distribution we considered above is a discrete distribution because demand can only
take a limited number of values. In some real settings, it is easier to work with the assumption
that the demand follows Normal distribution with a given mean and standard deviation. Normal
is a continuous distribution because demand can take any value. In this case, we can use the
following formulas:
Given per unit revenue r, per unit purchase cost c and per unit salvage value s:
Marginal cost of excess Ce=c-s; Marginal cost of shortage Cs=r-c.
Optimal service level = Cs / (Cs+Ce)
Given a normally distributed demand with given mean and standard deviation
compute z = spreadsheet function Normsinv (required service level)
Order quantity that can provide required service level = mean + z*standard deviation
Alternatively, given an order quantity S, the service level it can provide =
Spreadsheet function = Normdist (S, mean, standard dev., TRUE)
For Normal distribution, we can also compute the following:
Expected shortage = Std. Dev.*{ Normdist(z,0,1,false) -z +z Normdist(z,0,1,true)}
Expected excess = S – mean + Expected shortage
Expected mismatch cost = Cs*Expected shortage + Ce*Expected excess
Expected profit = (r-c)* mean - Expected mismatch cost
If quick access to a spreadsheet is not available, the following table provides z values (in first
column) for corresponding service level values (in second column).
Newsvendor Model: Practice Problems
1. Suzie’s café sells fresh muffins every day. The price for the blueberry muffin is $3
each, and it costs Suzie $.50 to prepare each day. All muffins left at the end of each day
are donated, which gives Suzie a tax benefit of $.1 each. From her experience, Suzie
knows that
If she prepares 2 muffins, then she can sell all of them.
If she prepares 3 muffins, then she can sell all of them with probability 0.9.
If she prepares 4 muffins, then she can sell all of them with probability 0.65.
If she prepares 5 muffins, then she can sell all of them with probability 0.5.
If she prepares 6 muffins, then she can sell all of them with probability 0.32.
If she prepares 7 muffins, then she can sell all of them with probability 0.08.
If she prepares 8 or more muffins, she will not be able to sell all of them.
Optimally, how many muffins should Suzie prepare everyday?
2. Joe’s Slop House is a little hole in the wall restaurant that serves slow-cooked barbecue
beef sandwiches. The beef for the sandwiches cooks overnight, so each evening Joe
decides how much beef to cook for the next day. The beef costs $2 a pound and Joe sells
each sandwich (filled with 1 pound of beef) for $4. At the end of the day, any leftover
beef is given to the neighborhood church for their soup kitchen. The tax deduction for
this is worth $0.50 a pound. Daily demand for the beef can be approximated by a normal
distribution with a mean of 80 pounds and a standard deviation of 20 pounds. How many
pounds of beef should Joe cook for the next day?
Newsvendor Model: Practice Problems Solutions
1. Suzie’s café sells fresh muffins every day. The price for the blueberry muffin is $3
each, and it costs Suzie $.50 to prepare each day. All muffins left at the end of each day
are donated, which gives Suzie a tax benefit of $.1 each. From her experience, Suzie
knows that
If she prepares 2 muffins, then she can sell all of them.
If she prepares 3 muffins, then she can sell all of them with probability 0.9.
If she prepares 4 muffins, then she can sell all of them with probability 0.65.
If she prepares 5 muffins, then she can sell all of them with probability 0.5.
If she prepares 6 muffins, then she can sell all of them with probability 0.32.
If she prepares 7 muffins, then she can sell all of them with probability 0.08.
If she prepares 8 or more muffins, she will not be able to sell all of them.
Optimally, how many muffins should Suzie prepare everyday?
Costs 0.50, sells for 3.00 and if not sold, earns a tax benefit of 0.10.
Therefore marginal costs of shortage and excess are:
Cs=3-0.5=2.5, Ce=0.5-0.1=0.4,
Optimal SL = Cs/(Cs+Ce) = 2.5/2.9 = 0.86
Note that what is given is that if she prepares x, what is the probability that the demand is
greater than or equal to x which is the same as the probability
that the demand is greater than x-1. For example, if she can sell all 3 muffins with
probability 0.9 that means that the demand is greater than 2 with probability 0.9. This
gives us the first column. Second column is just 1 minus the number in first column.
x
Probability of demand
Probability that demand is less than
greater than x
or equal to x (cumulative probability)
1
1
0
2
0.9
0.1
3
0.65
0.35
4
0.5
0.5
5
0.32
0.68
6
0.08
0.92
7
0
1
6 is the smallest number for which cumulative probability 0.92 goes beyond the required
service level 0.86. So she should optimally prepare 6 muffins everyday.
2. Joe’s Slop House is a little hole in the wall restaurant that serves slow-cooked barbecue
beef sandwiches. The beef for the sandwiches cooks overnight, so each evening Joe
decides how much beef to cook for the next day. The beef costs $2 a pound and Joe sells
each sandwich (filled with 1 pound of beef) for $4. At the end of the day, any leftover
beef is given to the neighborhood church for their soup kitchen. The tax deduction for
this is worth $0.50 a pound. Daily demand for the beef can be approximated by a normal
distribution with a mean of 80 pounds and a standard deviation of 20 pounds. How many
pounds of beef should Joe cook for the next day?
Let the unit of analysis be “pound”. For each pound that is short, the restaurant loses $2
in potential revenue, so the underage cost Cs=2. For each pound that is above the
demand, the restaurant spends $2 but can recuperate $0.5, so the cost of excess is Ce=1.5.
Therefore, optimal SL = Cs/(Cs+Ce) = 2/3.5 = 0.5714.
z = spreadsheet function Normsinv(0.5714)=0.18.
80+0.18*20 = 83.6 pounds every day.
Therefore Joe should cook