OM301 F11 Inventory News Vendor

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OPSM 301 Operations Management Class 17: Inventory Management: the newsvendor Koç University Zeynep Aksin [email protected] 

Transcript of OM301 F11 Inventory News Vendor

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OPSM 301 Operations Management

Class 17:

Inventory Management: thenewsvendor 

Koç University

Zeynep Aksin

[email protected] 

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Single Period Inventory Control

Examples:

 ± Style goods

 ± Perishable goods (flowers, foods) ± Goods that become obsolete (newspapers)

 ± Services that are perishable (airline seats)

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Demand Scenarios for a Jacket

Demand Scenarios

0%

5%

10%

15%

20%

25%

30%

   8   0   0   0

  1   0   0   0

   0

  1   2   0   0

   0

  1  4   0   0

   0

  1   6   0   0

   0

  1   8   0   0

   0

Sales

    P  r  o   b  a   b   i   l   i   t  y

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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

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Best Solution

Find order quantity that maximizesweighted average profit.

Question: Will this quantity be less than,equal to, or greater than averagedemand?

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What to Make?

Question: Will this quantity be lessthan, 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

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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 anddemand ends up being 11,000 jackets. ± Profit = 125(11,000) - 80(12,000) - 100,000 + 20(1000) = $

335,000

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8

Scenarios and their probabilitiesDemand

   P  r  o   d  u  c   t   i  o  n

  q  u  a  n   t   i   t  y

8000 10000 12000 14000 16000 18000

11% 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 $260,0008,500  $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

ProfitExpected

Profit

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Expected Profit

Expected Profit

$0

$100,000

$200,000

$300,000

$400,000

8000 12000 16000 20000

Order Quantity

      P     r     o      f      i      t

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Expected Profit

Expected Profit

$0

$100,000

$200,000

$300,000

$400,000

8000 12000 16000 20000

Order Quantity

      P     r     o      f      i      t

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Expected Profit

Expected Profit

$0

$100,000

$200,000

$300,000

$400,000

8000 12000 16000 20000

Order Quantity

      P     r     o      f      i      t

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Important Observations

Tradeoff between ordering enough to meet demand andordering too much

Several quantities have the same average profit

 Average profit does not tell the whole story

Question: 9000 and 16000 unitslead to about the same average

profit, so which do we prefer?

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Probability of Outcomes

0%

20%

40%

60%

80%

100%

  -   3  0  0

  0  0  0

  -  1  0  0

  0  0  0

  1  0  0  0

  0  0

   3  0  0  0

  0  0

   5  0  0  0

  0  0

Cost

     P    r    o     b    a     b     i     l     i     t

Q=9000

Q=16000

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Key Insights from this Model

The optimal order quantity is not necessarily equal to averageforecast 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 andthen decreases

 As production quantity increases, risk increases. In other words, the probability of large gains and of large lossesincreases

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Example

M ean 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 

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Single Period Inventory Control

Economics of the Situation Known:

1. Demand > Stock --> Underage (under stocking) Cost

C u = Cost of foregone profit, loss of goodwill

2. Demand < Stock --> Overage (over stocking) Cost

C o = Cost of excess inventory

C o = 10 and  C u = 20 How much would you

order? M ore than 3.85 or less than 3.85?

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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  

C o = 10 and  C u = 20 

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Generalization of the Incremental Analysis

ChancePoint 

Stock n-1

DecisionPoint 

Stock n

Base Case

nth unit needed

nth unit not needed

Pr{Demand u n}

Pr{Demand e n-1}

CashFlowCu

-Co

0

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Generalization of the Incremental Analysis

ChancePoint 

Stock n-1

DecisionPoint 

Stock n

Base Case

Expected Cash FlowCu Pr{Demand u n} -Co Pr{Demand e n-1}

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Generalization of the Incremental Analysis

Order the nth unit if C u Pr{Demand u n} - C o Pr{Demand e n-1} >= 0

or 

C u (1-Pr{Demand e n-1}) - C o Pr{Demand e n-1} >= 0or 

C u -C u Pr{Demand e n-1} -C o Pr{Demand e n-1} >= 0

or 

Pr{Demand e n-1} =< C u /( C o +C u)

Then order n units, where n is the greatest number thatsatisfies the above inequality.

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Incremental Analysis

I ncremental 

Demand Decision Pr{Demand e n-1} Order the unit?1 First 0.00 YES  

2 Second 0.10 YES  

3 Third 0.25 YES  

4 Fourth 0.45 YES  

5 Fifth 0.65 YES  

6 Sixth 0.80 NO -

7 Seventh 0.90 NO

C u /( C o +C u)=20/(10+20)=0.66

Order quantity n should satisfy:

P(Demand e n-1) e C u   /( C o +C u )< P(Demand e n) 

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Order Quantity for Single Period, Normal Demand

Find the z*: z value such that F(z)= C u   /( C o +C u )

Optimal order quantity is:

Do we order more or less than the mean if: ± C

u

> Co

?

 ± Cu < Co ?

WQ*

 zQ !

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Example 1: Single Period Model

Our college basketball team is playing in atournament game this weekend. Based on our pastexperience we sell on average 2,400 shirts with astandard deviation of 350. We make 10TL on everyshirt we sell at the game, but lose 5TL on every shirtnot sold. How many shirts should we make for thegame?

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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 0.5359

0.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.5753

0.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.6141

0.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.6517

0.4 0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.6879

0.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.7224

0.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.7549

0.7 0.7580 0.7611 0.7642 0.7673 0.7704 0.7734 0.7764 0.7794 0.7823 0.7852

0.8 0.7881 0.7910 0.7939 0.7967 0.7995 0.8023 0.8051 0.8078 0.8106 0.8133

0.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315 0.8340 0.8365 0.8389

1.0 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.8621

1.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790 0.8810 0.8830

1.2 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.8980 0.8997 0.9015

1.3 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.9177

1.4 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.9319

1.5 0.9332 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 0.9429 0.9441

1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.9545

1.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 0.9608 0.9616 0.9625 0.9633

1.8 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9699 0.9706

1.9 0.9713 0.9719 0.9726 0.9732 0.9738 0.9744 0.9750 0.9756 0.9761 0.9767

2.0 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 0.9817

2.1 0.9821 0.9826 0.9830 0.9834 0.9838 0.9842 0.9846 0.9850 0.9854 0.9857

2.2 0.9861 0.9864 0.9868 0.9871 0.9875 0.9878 0.9881 0.9884 0.9887 0.9890

2.3 0.9893 0.9896 0.9898 0.9901 0.9904 0.9906 0.9909 0.9911 0.9913 0.9916

2.4 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 0.9952

2.6 0.9953 0.9955 0.9956 0.9957 0.9959 0.9960 0.9961 0.9962 0.9963 0.9964

2.7 0.9965 0.9966 0.9967 0.9968 0.9969 0.9970 0.9971 0.9972 0.9973 0.9974

2.8 0.9974 0.9975 0.9976 0.9977 0.9977 0.9978 0.9979 0.9979 0.9980 0.9981

2.9 0.9981 0.9982 0.9982 0.9983 0.9984 0.9984 0.9985 0.9985 0.9986 0.9986

3.0 0.9987 0.9987 0.9987 0.9988 0.9988 0.9989 0.9989 0.9989 0.9990 0.9990

3.1 0.9990 0.9991 0.9991 0.9991 0.9992 0.9992 0.9992 0.9992 0.9993 0.9993

3.2 0.9993 0.9993 0.9994 0.9994 0.9994 0.9994 0.9994 0.9995 0.9995 0.9995

3.3 0.9995 0.9995 0.9995 0.9996 0.9996 0.9996 0.9996 0.9996 0.9996 0.9997

F(z)

 z 0

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Example 1: Single Period Model

Our college basketball team is playing in atournament game this weekend. Based on our pastexperience we sell on average 2,400 shirts with astandard deviation of 350. We make 10 TL on everyshirt we sell at the game, but lose 5 TL on every shirtnot sold. How many shirts should we make for thegame?

C u = $10 and C o = $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

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Example 2: Finding Cu and Co

 A textile company in UK orders coats from China. They buy a coat from250¼ and sell for 325¼. If they cannot sell a coat in winter, they sell itat a discount price of 225¼. When the demand is more than whatthey 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 32,500and std dev 6750.

How much should they order from China??

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Example 2: Finding Cu and Co

 A textile company in UK orders coats from China. They buy a coat from250¼ and sell for 325¼. If they cannot sell a coat in winter, they sell it ata discount price of 225¼. When the demand is more than what theyhave 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 andstd 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

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Example 3: Single Period InventoryManagement 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 Probability 

500 0.10  

600 0.2  

700 0.2  

800 0.2  900 0.10  

1000 0.10  

1100 0.10  

P(D<=n-1)00.10.3

0.50.70.80.9

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