#10
Solutions
Newsvendor Practice Model
BUAD311 Operations Management
1) Consider a newsvendor problem and we are now thinking of increasing the order
quantity from Q to Q+1. The potential benefit of ordering this one unit is that you could
meet the demand of one customer that we would otherwise lose. What is the potential
loss of ordering this one unit?
If the demand realization is such that this one unit is unsold, we have to salvage it under
cost, or incur holding cost.
2) In a typical newsvendor problem what do we know about the demand?
We dont know the exact demand but we do know the distribution of the demand.
3) Solve the newsvendor problem. What is the optimal order quantity?
Probability
Value
0.2
1
0.1
2
0.1
3
0.2
4
0.3
5
0.1
6
purchase cost c = 15
selling price p = 25
salvage value v = 10
Pr[demand < 1]=0 Q: Is this (strictly) less than (p-c)/(p-v)=0.66..? A: Yes.
Pr[demand < 2]=0.2 Q: Is this (strictly) less than 0.66..? A: Yes.
Pr[demand < 3]=0.3 Q: Is this (strictly) less than 0.66..? A: Yes.
Pr[demand < 4]=0.4 Q: Is this (strictly) less than 0.66..? A: Yes.
Pr[demand < 5]=0.6 Q: Is this (strictly) less than 0.66..? A: Yes.
Pr[demand < 6]=0.9 Q: Is this (strictly) less than 0.66..? A: No.
Therefore the optimal order quantity = 5.
4) Demand distribution is continuous uniform between 10 and 50. Costs are as follows:
c=15
p=20
v=0
a. What is the optimal target inventory level?
20
b. If we follow the optimal policy, what is the probability that we dont meet the
entire customer demand?
75% (Isnt this number a little too high?)
5) Demand distribution is normal with mean 100 and standard deviation 5. Costs are as
follows:
c=15
p=20
v=0
a. What is the optimal target inventory level?
(p-c)/(p-v) = 0.25. Z-value for 0.25, from the standard normal table, is
0.675 ( I took the average of 0.67 and 0.68.). The optimal target level
is therefore 100-0.675*5 = 96.625.
b. If we follow the optimal policy, what is the probability that we dont meet the
entire customer demand?
75%
6) Demand distribution is normal with mean 200 and standard deviation 5. Costs are as
follows:
c=10
p=20
v=5
a. What is the optimal target inventory level?
(p-c)/(p-v) = 0.666 Z-value is 0.43. Target level is 200+0.43*5 = 202.15
b. Suppose we have 10 units of the product on hand. What is the order quantity?
We know that the optimal policy is order-up-to type. If we have 10 units on
hand for whatever reason, we order 202.15-10 = 192.15 units.
NOTE: In case demand distribution is continuous, dont round your answer.
7) This probability mass function is from the example we used in class. You are asked to
check again whether it makes sense to move from 4 to 5. Recall that c=15, p=25, v=10.
Probability
Demand
0.2
1
0.2
2
0.2
3
0.2
4
0.2
5
What is the marginal benefit of ordering the 5th unit when you are at 4 (This number
should be non-positive.)
We are thinking of moving from 4 to 5. This additional unit costs you 15 dollars.
Probability of selling this one unit is 0.2, so expected revenue from this one unit is
0.2*25 = 5 dollars. Probability of this one unit unsold is 0.8, so expected salvage
is 0.8*10 = 8. Marginal benefit of buying this additional unit, therefore, is
15+5+8= 2.
8) Consider the newsvendor problem where p=20 and v=10. Demand is continuous, and
uniformly distributed between 200 and 300. If the optimal target level is 260, compute
the purchase cost.
Optimal order quantity of 260 implies that the optimal fractile is 0.6.
(20-c)/(20-10) = 0.6 solves c=14.
9) Demand is N (50, 5 2 ) and optimal fractile (p-c)/(p-v) turned out to be 0.6.
a. What is the optimal target level?
From the standard normal table we get z = 0.255. Target level is
50+0.255*5 = 51.275
b. This fractile 0.6, or 60%, is called service level. Explain in English what it means.
Probability that we meet the entire demand is 60%.
10) In any newsvendor type of problem with continuous demand distribution, when c
increases (and other things equal), you order:
a. more
b. less
c. It depends.
b.less: As c increases, (p-c)/(p-v) decreases. Given the distribution, the smaller the
fractile, the smaller the target inventory level.
11) Demand is N (50, 5 2 ) and the optimal target level turned out 65. What is the service
level (i.e., optimal fractile)?
(65-50)/5 = 3
From the table we get 0.9987 (or 99.87%)
12) Wholemark is a mail order business that sells one popular New Year greeting card
once a year. The cost of the paper on which the card is printed is $0.05 per card, and
the cost of printing is $0.15 per card. The company receives $2.15 per card sold.
Since the cards have the current year printed on them, unsold cards have no salvage
value. Their customers are from the four areas Los Angeles, Santa Monica,
Hollywood, and Pasadena. Based on past data, the number of customers from each of
the four regions is normally distributed with mean 2,000 and standard deviation 500.
(Assume these four are independent.) What is the optimal production quantity for the
card?
(2.15-0.2)/(2.15-0)=90.69%.
From the standard normal table, Z-value is 1.325.
Combined has demand mean 2000*4=8000, and standard deviation
500*sqrt(4)=1000.
Using the above, the optimal production quantity is 8000+1.325*1000=9325.
13) Solve the following newsvendor problem.
Demand
Probability
5
10 15 20 25 30
5% 20% 25% 25% 15% 10%
Purchase cost: c= $5.00
Selling price: p=$17.00
Salvage value: v=$1.00
a) Based on the demand distribution and costs above, how many units of the
product should we order?
(p-c)/(p-v)=17-5)/(17-1)=0.75
The optimal quantity is 20. (25 is also fine.)
b) If we implement your policy in part a, what is the expected profit?
Suppose we order 20.
If demand is 5 (probability 0.05), we will sell 5 units and have 15 leftover.
If demand is 10 (probability 0.2), we will sell 10 units and have 10 leftover.
If demand is 15 (probability 0.25), we will sell 15 units and have 5 leftover.
If demand is 20 or more (probability 0.5), we will sell all the 20 units.
The expected profit is
0.05*(5*17+15*1) + 0.2*(10*17+10*1) + 0.25*(15*17+5*1) + 0.5*(20*17)
20*5 = 176
c) According to part a, what is the probability we will meet the entire customer
demand?
75% (or 90%, depending on your answer in part a.)
14) Question 20, page 581 in the textbook.
CO = 8 4 = $4
CU = 20 8 = $12
The service level is: 12/ (12+4) = 0.75
Demand
Probability Cumulative Probability
300
0.05
0.05
400
0.10
0.15
500
0.40
0.55
600
0.30
0.85
700
0.10
0.95
800
0.05
1.00
She should produce 600 T-shirts.
15) XYZ Bakery bakes fresh pies every morning. The daily demand for its apple pies is a
random variable with (discrete) distribution, based on past experience, given by:
Demand
Probability
5
10 15 20 25 30
10% 20% 25% 25% 15% 5%
Each apple pie costs the Bakery $6.75 to make and is sold for $17.99. Unsold
apple pies at the end of the day are purchased by a nearby soup kitchen for 99
cents each. Assume no goodwill cost.
a) If the company decided to bake 15 apple pies each day, what would be her
expected profit?
If demand is 5 (probability 0.1), we will sell 5 apple pies and have 10
leftover.
If demand is 10 (probability 0.2), we will sell 10 apple pies and have 5
leftover.
If demand is at least 15 (probability 0.7), we will sell all the 15 apple pies.
The expected profit is
(5*0.1+10*0.2+15*0.7)*17.99+(10*0.1+5*0.2)*0.9915*6.75=134.6
b) Based on the demand distribution above, how many apple pies should the
company bake each day to maximize her expected profit?
Cu = 17.99 6.75 = 11.24, Co = 6.75 0.99 = 5.76
Cu
17.99 6.75
=
= 0.6612
We need to have service level F (Q) =
C u + C o 17.99 0.99
The optimal apple pies quantity is 20. (For anything less than 20, the
service level is no more than 0.1+0.2+0.25=0.55<0.66.)
c) The company decides to use normal distribution to approximate demand
and decide the optimal quantity. Based on the demand distribution given
above, what is the average daily demand and standard deviation for apple
pies?
The average demand is
5*0.1+10*0.2+15*0.25+20*0.25+25*0.15+30*0.05=16.5.
Variance =
(5-16.5)^2*0.1+(10-16.5)^2*0.2+(15-16.5)^2*0.25+(20-16.5)^2*0.25
+(25-16.5)^2*0.15+(30-16.5)^2*0.05=45.25.
STD2 = Variance
STD = 6.73
d) Using the mean and standard deviation, the company can approximate
demand using a normal distribution. Based on this normal demand
distribution, how many apple pies should the company bake each day to
maximize the expected profit?
We still need to have service level
Cu
17.99 6.75
F (Q) =
=
= 0.6612
C u + C o 17.99 0.99
If we use normal distribution, then G(z)=0.6612 => z=0.40
Q = + z = 16.5 + 6.73 * 0.40 = 19.19
The company should bake 19 apple pies (you have full credit if you round
up and choose to bake 20 apple pies).
16) Sallys Silk Screening produces specialty T-shirts that are primarily sold at
special events. She is trying to decide how many to produce for an upcoming
event. During the event, Sally can sell T-shirts for $20 apiece. However, when the
event ends, any unsold T-shirts are sold for $4 apiece. It costs Sally $8 to make a
specialty T-shirt. Sallys estimate of demand is the following:
DEMAND
300
400
500
600
700
800
PROBABILITY
.05
.10
.40
.30
.10
.05
a. What is the service rate (or optimal fractile)?
CO = 8 4 = $4
CU = 20 8 = $12
The service rate or service level is: 12/ (12+4) = 0.75
b. How many T-shirts should she produce for the upcoming event?
Demand
Probability Cumulative Probability
300
0.05
0.05
400
0.10
0.15
500
0.40
0.55
600
0.30
0.85
700
0.10
0.95
800
0.05
1.00
She should produce 600 T-shirts.
17) You are a newsvendor selling San Pedro Times every morning. Before you get to
work, you go to the printer and buy the days paper for $0.25 a copy. You sell a
copy of San Pedro Times for $1.00. Daily demand is distributed normally with
mean = 250 and standard deviation = 50. At the end of each morning, any leftover
copies are worthless and they go to a recycle bin.
a) How many copies of San Pedro Times should you buy each morning? (The z
table can be found at the end of the quiz.)
(1.00-0.25)/(1.00-0) = 0.75. Z-value for 75% is 0.67.
250 + 0.67 * 50 = 283.5
b) Based on part (a), what is the probability that you will run out of stock?
25%