OPSM 301 Operations Management
Class 17:
Inventory Management: the newsvendor
Koç University
Zeynep [email protected]
Single Period Inventory Control
Examples:– Style goods– Perishable goods (flowers, foods)– Goods that become obsolete (newspapers)– Services that are perishable (airline seats)
Demand Scenarios for a Jacket
Demand Scenarios
0%5%
10%15%20%25%30%
Sales
P
robabili
ty
Costs
Production cost per unit (C): $80 Selling price per unit (S): $125 Salvage value per unit (V): $20 Fixed production cost (F): $100,000 Q is production quantity, D demand
Profit =Revenue - Variable Cost - Fixed Cost + Salvage
Best Solution
Find order quantity that maximizes weighted average profit.
Question: Will this quantity be less than, equal to, or greater than average demand?
What to Make?
Question: Will this quantity be less than, equal to, or greater than average demand?
Average demand is 13,100 Look at marginal cost Vs. marginal profit
– if extra jacket sold, profit is 125-80 = 45– if not sold, cost is 80-20 = 60
So we will make less than average
Scenarios
Scenario One:– Suppose you make 12,000 jackets and
demand ends up being 13,000 jackets.– Profit = 125(12,000) - 80(12,000) - 100,000 = $440,000
Scenario Two:– Suppose you make 12,000 jackets and
demand ends up being 11,000 jackets.– Profit = 125(11,000) - 80(12,000) - 100,000 + 20(1000) = $
335,000
8
Scenarios and their probabilitiesDemand
Pro
du
ctio
n q
uan
tity
8000 10000 12000 14000 16000 1800011% 11% 28% 22% 18% 10%
5,000 $125,000.00 $125,000.00 $125,000.00 $125,000.00 $125,000.00 $125,000.00 $125,000
5,500 $147,500.00 $147,500.00 $147,500.00 $147,500.00 $147,500.00 $147,500.00 $147,500
6,000 $170,000.00 $170,000.00 $170,000.00 $170,000.00 $170,000.00 $170,000.00 $170,000
6,500 $192,500.00 $192,500.00 $192,500.00 $192,500.00 $192,500.00 $192,500.00 $192,500
7,000 $215,000.00 $215,000.00 $215,000.00 $215,000.00 $215,000.00 $215,000.00 $215,000
7,500 $237,500.00 $237,500.00 $237,500.00 $237,500.00 $237,500.00 $237,500.00 $237,500
8,000 $260,000.00 $260,000.00 $260,000.00 $260,000.00 $260,000.00 $260,000.00 $260,000
8,500 $230,000.00 $282,500.00 $282,500.00 $282,500.00 $282,500.00 $282,500.00 $276,725
9,000 $200,000.00 $305,000.00 $305,000.00 $305,000.00 $305,000.00 $305,000.00 $293,450
9,500 $170,000.00 $327,500.00 $327,500.00 $327,500.00 $327,500.00 $327,500.00 $310,175
10,000 $140,000.00 $350,000.00 $350,000.00 $350,000.00 $350,000.00 $350,000.00 $326,900
10,500 $110,000.00 $320,000.00 $372,500.00 $372,500.00 $372,500.00 $372,500.00 $337,850
11,000 $80,000.00 $290,000.00 $395,000.00 $395,000.00 $395,000.00 $395,000.00 $348,800
11,500 $50,000.00 $260,000.00 $417,500.00 $417,500.00 $417,500.00 $417,500.00 $359,750
12,000 $20,000.00 $230,000.00 $440,000.00 $440,000.00 $440,000.00 $440,000.00 $370,700
12,500 -$10,000.00 $200,000.00 $410,000.00 $462,500.00 $462,500.00 $462,500.00 $366,950
13,000 -$40,000.00 $170,000.00 $380,000.00 $485,000.00 $485,000.00 $485,000.00 $363,200
13,500 -$70,000.00 $140,000.00 $350,000.00 $507,500.00 $507,500.00 $507,500.00 $359,450
14,000 -$100,000.00 $110,000.00 $320,000.00 $530,000.00 $530,000.00 $530,000.00 $355,700
14,500 -$130,000.00 $80,000.00 $290,000.00 $500,000.00 $552,500.00 $552,500.00 $340,400
15,000 -$160,000.00 $50,000.00 $260,000.00 $470,000.00 $575,000.00 $575,000.00 $325,100
15,500 -$190,000.00 $20,000.00 $230,000.00 $440,000.00 $597,500.00 $597,500.00 $309,800
16,000 -$220,000.00 -$10,000.00 $200,000.00 $410,000.00 $620,000.00 $620,000.00 $294,500
16,500 -$250,000.00 -$40,000.00 $170,000.00 $380,000.00 $590,000.00 $642,500.00 $269,750
Average Profit
Expected Profit
Expected Profit
Expected Profit
$0
$100,000
$200,000
$300,000
$400,000
8000 12000 16000 20000
Order Quantity
Pro
fit
Expected Profit
Expected Profit
$0
$100,000
$200,000
$300,000
$400,000
8000 12000 16000 20000
Order Quantity
Pro
fit
Expected Profit
Expected Profit
$0
$100,000
$200,000
$300,000
$400,000
8000 12000 16000 20000
Order Quantity
Pro
fit
Important Observations
Tradeoff between ordering enough to meet demand and ordering too much
Several quantities have the same average profit Average profit does not tell the whole story
Question: 9000 and 16000 units lead to about the same average profit, so which do we prefer?
Probability of Outcomes
0%
20%
40%
60%
80%
100%
Cost
Pro
ba
bilit
y
Q=9000
Q=16000
Key Insights from this Model
The optimal order quantity is not necessarily equal to average forecast demand
The optimal quantity depends on the relationship between marginal profit and marginal cost
Fixed cost has no impact on production quantity, only on whether to produce or not
As order quantity increases, average profit first increases and then decreases
As production quantity increases, risk increases. In other words, the probability of large gains and of large losses increases
Example
Mean demand=3.85 How much would you order?
Demand Probability
1 0.10
2 0.15
3 0.20
4 0.20
5 0.15
6 0.10
7 0.10
Total 1.00
Single Period Inventory Control
Economics of the Situation Known:
1. Demand > Stock --> Underage (under stocking) Cost
Cu = Cost of foregone profit, loss of goodwill
2. Demand < Stock --> Overage (over stocking) Cost
Co = Cost of excess inventory
Co = 10 and Cu = 20 How much would you
order? More than 3.85 or less than 3.85?
Incremental AnalysisProbability Probability Incremental
Incremental that incremental that incremental Expected
Demand Decision unit is not needed unit is needed Contribution
1 First 0.00 1.00 -10(0.00)+20(1.00)
=20
2 Second 0.10 0.90 -10(0.10)+20(0.90)
=17
3 Third 0.25 0.75 12.5
4 Fourth 0.45 0.55 6.5
5 Fifth 0.65 0.35 0.5
6 Sixth 0.80 0.20 -4
7 Seventh 0.90 0.10 -7
Co = 10 and Cu = 20
Generalization of the Incremental Analysis
ChancePoint
Stock n-1
DecisionPoint
Stock n
Base Case
nth unit needed
nth unit not needed
Pr{Demand n}
Pr{Demand n-1}
Cash FlowCu
-Co
0
Generalization of the Incremental Analysis
ChancePoint
Stock n-1
DecisionPoint
Stock n
Base Case
Expected Cash FlowCu Pr{Demand n} -Co Pr{Demand n-1}
Generalization of the Incremental Analysis
Order the nth unit ifCu Pr{Demand n} - Co Pr{Demand n-1} >= 0
or
Cu (1-Pr{Demand n-1}) - Co Pr{Demand n-1} >= 0
or
Cu - Cu Pr{Demand n-1} -Co Pr{Demand n-1} >= 0
or
Pr{Demand n-1} =< Cu /(Co +Cu)
Then order n units, where n is the greatest number that satisfies the above inequality.
Incremental Analysis
IncrementalDemand Decision Pr{Demand n-1} Order the unit?1 First 0.00 YES
2 Second 0.10 YES3 Third 0.25 YES4 Fourth 0.45 YES5 Fifth 0.65 YES6 Sixth 0.80 NO -7 Seventh 0.90 NO
Cu /(Co +Cu)=20/(10+20)=0.66
Order quantity n should satisfy:P(Demand n-1) Cu /(Co +Cu)< P(Demand n)
Order Quantity for Single Period, Normal Demand
Find the z*: z value such that F(z)= Cu /(Co +Cu)
Optimal order quantity is:
Do we order more or less than the mean if:– Cu > Co ?– Cu < Co ?
*zQ
Example 1: Single Period Model
Our college basketball team is playing in a tournament game this weekend. Based on our past experience we sell on average 2,400 shirts with a standard deviation of 350. We make 10TL on every shirt we sell at the game, but lose 5TL on every shirt not sold. How many shirts should we make for the game?
Transform
X = N(mean,s.d.) to
z = N(0,1)
z = (X - mean) / s.d.
F(z) = Prob( N(0,1) < z)
Transform back, knowing z*:
X* = mean + z*s.d.
The Standard Normal Distributionz 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09
0.0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.53590.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.57530.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.61410.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.65170.4 0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.68790.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.72240.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.75490.7 0.7580 0.7611 0.7642 0.7673 0.7704 0.7734 0.7764 0.7794 0.7823 0.78520.8 0.7881 0.7910 0.7939 0.7967 0.7995 0.8023 0.8051 0.8078 0.8106 0.81330.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315 0.8340 0.8365 0.83891.0 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.86211.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790 0.8810 0.88301.2 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.8980 0.8997 0.90151.3 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.91771.4 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.93191.5 0.9332 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 0.9429 0.94411.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.95451.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 0.9608 0.9616 0.9625 0.96331.8 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9699 0.97061.9 0.9713 0.9719 0.9726 0.9732 0.9738 0.9744 0.9750 0.9756 0.9761 0.97672.0 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 0.98172.1 0.9821 0.9826 0.9830 0.9834 0.9838 0.9842 0.9846 0.9850 0.9854 0.98572.2 0.9861 0.9864 0.9868 0.9871 0.9875 0.9878 0.9881 0.9884 0.9887 0.98902.3 0.9893 0.9896 0.9898 0.9901 0.9904 0.9906 0.9909 0.9911 0.9913 0.99162.4 0.9918 0.9920 0.9922 0.9925 0.9927 0.9929 0.9931 0.9932 0.9934 0.99362.5 0.9938 0.9940 0.9941 0.9943 0.9945 0.9946 0.9948 0.9949 0.9951 0.99522.6 0.9953 0.9955 0.9956 0.9957 0.9959 0.9960 0.9961 0.9962 0.9963 0.99642.7 0.9965 0.9966 0.9967 0.9968 0.9969 0.9970 0.9971 0.9972 0.9973 0.99742.8 0.9974 0.9975 0.9976 0.9977 0.9977 0.9978 0.9979 0.9979 0.9980 0.99812.9 0.9981 0.9982 0.9982 0.9983 0.9984 0.9984 0.9985 0.9985 0.9986 0.99863.0 0.9987 0.9987 0.9987 0.9988 0.9988 0.9989 0.9989 0.9989 0.9990 0.99903.1 0.9990 0.9991 0.9991 0.9991 0.9992 0.9992 0.9992 0.9992 0.9993 0.99933.2 0.9993 0.9993 0.9994 0.9994 0.9994 0.9994 0.9994 0.9995 0.9995 0.99953.3 0.9995 0.9995 0.9995 0.9996 0.9996 0.9996 0.9996 0.9996 0.9996 0.9997
F(z)
z0
Example 1: Single Period Model
Our college basketball team is playing in a tournament game this weekend. Based on our past experience we sell on average 2,400 shirts with a standard deviation of 350. We make 10 TL on every shirt we sell at the game, but lose 5 TL on every shirt not sold. How many shirts should we make for the game?
Cu = $10 and Co = $5; P ≤ $10 / ($10 + $5) = .667
Z.667 = .4 (from standard normal table or using NORMSINV() in Excel)
therefore we need 2,400 + .4(350) = 2,540 shirts
Example 2: Finding Cu and Co
A textile company in UK orders coats from China. They buy a coat from 250€ and sell for 325€. If they cannot sell a coat in winter, they sell it at a discount price of 225€. When the demand is more than what they have in stock, they have an option of having emergency delivery of coats from Ireland, at a price of 290.
The demand for winter has a normal distribution with mean 32,500 and std dev 6750.
How much should they order from China??
Example 2: Finding Cu and Co
A textile company in UK orders coats from China. They buy a coat from 250€ and sell for 325€. If they cannot sell a coat in winter, they sell it at a discount price of 225€. When the demand is more than what they have in stock, they have an option of having emergency delivery of coats from Ireland, at a price of 290.
The demand for winter has a normal distribution with mean 32,500 and std dev 6750.
How much should they order from China??
Cu=75-35=40Co=25F(z)=40/(40+25)=40/65=0.61z=0.28 q=32500+0.28*6750=34390
Example 3: Single Period Inventory Management Problem
Manufacturing cost=60TL,
Selling price=80TL, Discounted price (at the end of the season)=50TL
Market research gave the following probability distribution for demand.
Find the optimal q, expected number of units sold for this orders size, and expected profit, for this order size.
Demand Probability500 0.10600 0.2700 0.2800 0.2900 0.101000 0.101100 0.10
P(D<=n-1)00.10.30.50.70.80.9
Cu=20 Co=10P(D<=n-1)<=20/30=0.66
<=0.66 q=800
For q=800:E(units sold)=710E(profit)=13,300