The Magnitude of Shortages (Out of Stock)

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Managing Flow Variability: Safety Inventory

The Newsvendor Problem

Ardavan Asef-Vaziri, Oct 2011

The Magnitude of Shortages (Out of Stock)

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What are the Reasons?

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Consumer Reaction

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Optimal Service Level: The Newsvendor Problem

Cost of ordering too much: holding cost, salvage Cost of ordering too

little: lost of sale, low service level

The decision maker balances the expected costs of ordering too much with the expected costs of ordering too little to determine the optimal order quantity.

How do we choose what level of service a firm should offer?

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News Vendor Model; Assimptions Demand is random Distribution of demand is known No initial inventory Set-up cost is zero Single period Zero lead time Linear costs

Purchasing (production) Salvage value Revenue Goodwill

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Demand Probability of Demand100 0.02110 0.05120 0.08130 0.09140 0.11150 0.16160 0.20170 0.15180 0.08190 0.05200 0.01

Cost =1800, Sales Price = 2500, Salvage Price = 1700Underage Cost = 2500-1800 = 700, Overage Cost = 1800-1700 = 100

What is probability of demand to be equal to 130? What is probability of demand to be less than or equal to 140?What is probability of demand to be greater than or equal to 140?What is probability of demand to be equal to 133?

0.090.02+0.05+0.08+0.09+0.11=

0.351-0.35+0.11= 0.76

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P(R ≥ Q ) = 1-P(R ≤ Q)+P(R = Q) R is quantity of demandQ is the quantity ordered

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Demand Probability of Demand100 0.002101 0.002102 0.002103 0.002104 0.002105 0.002106 0.002107 0.002108 0.002109 0.002

What is probability of demand to be equal to 116?What is probability of demand to be less than or equal to 116?What is probability of demand to be greater than or equal to 116?What is probability of demand to be equal to 113.3?

Demand Probability of Demand110 0.005111 0.005112 0.005113 0.005114 0.005115 0.005116 0.005117 0.005118 0.005119 0.005

0.0050.02+0.035 = 0.055

1-0.055+0.005 = 0.950

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What is probability of demand to be equal to 130?

Average Demand Probability of Demand100 0.02110 0.05120 0.08130 0.09140 0.11150 0.16160 0.20170 0.15180 0.08190 0.05200 0.01

0

What is probability of demand to be less than or equal to 145?

What is probability of demand to be greater than or equal to 145?

0.02+0.05+0.08+0.09+0.11 = 0.35

1-0.35 = 0.65P(R ≥ Q) = 1-P(R ≤ Q)

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Compute the Average Demand

Q i P( R =Q i )100 0.02110 0.05120 0.08130 0.09140 0.11150 0.16160 0.20170 0.15180 0.08190 0.05200 0.01

N

1i Demand Average )( ii QRPQ

Average Demand = +100×0.02 +110×0.05+120×0.08 +130×0.09+140×0.11 +150×0.16+160×0.20 +170×0.15 +180×0.08 +190×0.05+200×0.01Average Demand = 151.6

How many units should I have to sell 151.6 units (on average)? How many units do I sell (on average) if I have 100 units?

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Suppose I have ordered 140 Unities.On average, how many of them are sold? In other words, what is the

expected value of the number of sold units?

When I can sell all 140 units? I can sell all 140 units if R ≥ 140Prob(R ≥ 140) = 0.76The expected number of units sold –for this part- is(0.76)(140) = 106.4Also, there is 0.02 probability that I sell 100 units 2 unitsAlso, there is 0.05 probability that I sell 110 units5.5Also, there is 0.08 probability that I sell 120 units 9.6Also, there is 0.09 probability that I sell 130 units 11.7106.4 + 2 + 5.5 + 9.6 + 11.7 = 135.2

Deamand (Qi) 100 110 120 130 140 150 160 170 180 190 200Porbability 0.02 0.05 0.08 0.09 0.11 0.16 0.20 0.15 0.08 0.05 0.01Prob (R ≥ Qi) 1.00 0.98 0.93 0.85 0.76 0.65 0.49 0.29 0.14 0.06 0.01

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Suppose I have ordered 140 Unities.On average, how many of them are salvaged? In other words, what is

the expected value of the number of salvaged units?

0.02 probability that I sell 100 units. In that case 40 units are salvaged 0.02(40) = .80.05 probability to sell 110 30 salvaged 0.05(30)= 1.5 0.08 probability to sell 120 20 salvaged 0.08(20) = 1.60.09 probability to sell 130 10 salvaged 0.09(10) =0.9 0.8 + 1.5 + 1.6 + 0.9 = 4.8

Total number Sold 135.2 @ 700 = 94640Total number Salvaged 4.8 @ -100 = -480Expected Profit = 94640 – 480 = 94,160

Deamand (Qi) 100 110 120 130 140 150 160 170 180 190 200Porbability 0.02 0.05 0.08 0.09 0.11 0.16 0.20 0.15 0.08 0.05 0.01Prob (R ≥ Qi) 1.00 0.98 0.93 0.85 0.76 0.65 0.49 0.29 0.14 0.06 0.01

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Cumulative Probabilities

Qi P(R=Qi) P(R<Qi) P(R≥Qi)100 0.02 0 1110 0.05 0.02 0.98120 0.08 0.07 0.93130 0.09 0.15 0.85140 0.11 0.24 0.76150 0.16 0.35 0.65160 0.2 0.51 0.49170 0.15 0.71 0.29180 0.08 0.86 0.14190 0.05 0.94 0.06200 0.01 0.99 0.01

Probabilities

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Number of Units Sold, Salvages

Qi P(R=Qi) P(R<Qi) P(R≥Qi) Sold Salvage100 0.02 0 1 100 0110 0.05 0.02 0.98 109.8 0.2120 0.08 0.07 0.93 119.1 0.9130 0.09 0.15 0.85 127.6 2.4140 0.11 0.24 0.76 135.2 4.8150 0.16 0.35 0.65 141.7 8.3160 0.20 0.51 0.49 146.6 13.4170 0.15 0.71 0.29 149.5 20.5180 0.08 0.86 0.14 150.9 29.1190 0.05 0.94 0.06 151.5 38.5200 0.01 0.99 0.01 151.6 48.4

Probabilities Units Sold@700 Salvaged@-100

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Total Revenue for Different Ordering Policies

Qi P(R=Qi) P(R<Qi) P(R≥Qi) Sold Salvaged Sold Salvaged Total100 0.02 0 1 100 0 70000 0 70000110 0.05 0.02 0.98 109.8 0.2 76860 20 76840120 0.08 0.07 0.93 119.1 0.9 83370 90 83280130 0.09 0.15 0.85 127.6 2.4 89320 240 89080140 0.11 0.24 0.76 135.2 4.8 94640 480 94160150 0.16 0.35 0.65 141.7 8.3 99190 830 98360160 0.2 0.51 0.49 146.6 13.4 102620 1340 101280170 0.15 0.71 0.29 149.5 20.5 104650 2050 102600180 0.08 0.86 0.14 150.9 29.1 105630 2910 102720190 0.05 0.94 0.06 151.5 38.5 106050 3850 102200200 0.01 0.99 0.01 151.6 48.4 106120 4840 101280

Probabilities Units Revenue

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Denim Wholesaler; Marginal Analysis

The demand for denim is:

– 1000 with probability 0.10







Unit Revenue (p ) = 30Unit purchase cost (c )= 10Salvage value (v )= 5Goodwill cost (g )= 0

How much should we order?

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Marginal Analysis

Marginal analysis: What is the value of an additional unit ordered?

Suppose the wholesaler purchases 1000 units

What is the value of having the 1001st unit? Marginal Cost: The retailer must salvage the

additional unit and losses $5 (10 – 5). P(R ≤ 1000) = 0.1Expected Marginal Cost = 0.1(5) = 0.5

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Marginal Analysis

Marginal Profit: The retailer makes and extra profit of $20 (30 – 10)P(R > 1000) = 0.9 Expected Marginal Profit= 0.9(20) = 18MP ≥ MCExpected Value = 18-0.5 = 17.5By purchasing an additional unit, the expected profit increases by

$17.5

The retailer should purchase at least 1,001 units.

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Marginal Analysis

Should he purchase 1,002 units?Marginal Cost: $5 salvage P(R ≤ 1001) = 0.1 Expected Marginal Cost = 0.5Marginal Profit: $20 profit P(R >1002) = 0.9 18 Expected Marginal Profit = 18Expected Value = 18-0.5 = 17.5

Conclusion:

Wholesaler should purchase at least 2000 units.

Assuming that the initial purchasing quantity is between 1000 and 2000, then by purchasing an additional unit exactly the same savings will be achieved.

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Marginal Analysis

Marginal analysis: What is the value of an additional unit ordered?

Suppose the retailer purchases 2000 units

What is the value of having the 2001st unit? Marginal Cost: The retailer must salvage the

additional unit and losses $5 (10 – 5). P(R ≤ 2000) = 0.25Expected Marginal Cost = 0.25(5) = 1.25

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Marginal Analysis

Marginal Profit: The retailer makes and extra profit of $20 (30 – 10)P(R > 2000) = 0.75 Expected Marginal Profit= 0.75(20) = 15MP ≥ MCExpected Value = 15-1.25 = 13.75By purchasing an additional unit, the expected profit increases by

$13.75

The retailer should purchase at least 2,001 units.

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Marginal Analysis

Should he purchase 2,002 units?Marginal Cost: $5 salvage P(R ≤ 2001) = 0.25 Expected Marginal Cost = 1.25Marginal Profit: $20 profit P(R >2002) = 0.75 Expected Marginal Profit = 15Expected Value = 18-0.5 = 13.75

Conclusion:

Wholesaler should purchase at least 3000 units.

Assuming that the initial purchasing quantity is between 2000 and 3000, then by purchasing an additional unit exactly the same savings will be achieved.

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Marginal Analysis

Why does the marginal value of an additional unit decrease, as the purchasing quantity increases?

– Expected cost of an additional unit increases

– Expected savings of an additional unit decreases

Demand ProbabilityCumulative Probability

Expected Marginal Cost

Expected Marginal Profit

Expected Marginal Value

1000 0.10 0.1 0.50 18 17.502000 0.15 0.25 1.25 15 13.753000 0.15 0.40 2.00 12 10.004000 0.20 0.60 3.00 8 5.005000 0.15 0.75 3.75 5 1.256000 0.15 0.90 4.50 2 -2.507000 0.10 1.00 5.00 0 -5.00

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Marginal Analysis

What is the optimal purchasing quantity?

– Answer: Choose the quantity that makes marginal value: zero

Quantity

Marginal value

1000 2000 3000 4000 5000 6000 7000 8000

17.5

13.75

10

5

1.3

-2.5

-5

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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)

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Analytical Solution for the Optimal Service LevelIf 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.

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MC×Prob(R ≤ Q*) ≥ MP× [1 – Prob( R ≤ Q*)]

Analytical Solution for the Optimal Service Level

MP = p – c = Underage Cost = Cu

MC = c – v = Overage Cost = Co

ou

u

CCc

MCMPMPQRP

)( *

vpcp

vccpcp

MCMPMP

MBMB MCProb(R ≤ Q*) ≥

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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 ProbabilityCumlative Probability Marginal Value

1000 0.1 0.1 0.5 18.0 17.502000 0.15 0.25 1.3 15.0 13.753000 0.15 0.4 2.0 12.0 10.004000 0.2 0.6 3.0 8.0 5.005000 0.15 0.75 3.8 5.0 1.256000 0.15 0.9 4.5 2.0 -2.507000 0.1 1 5.0 0.0 -5.00

MCMPMPQRP

)( *

ou

u

CCc

vpcp

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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

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Marginal Value: Uniform distribution

Suppose instead of a discreet demand of

Demand ProbabilityCumlative Probability Marginal Value

1000 0.1 0.1 0.5 18.0 17.502000 0.15 0.25 1.3 15.0 13.753000 0.15 0.4 2.0 12.0 10.004000 0.2 0.6 3.0 8.0 5.005000 0.15 0.75 3.8 5.0 1.256000 0.15 0.9 4.5 2.0 -2.507000 0.1 1 5.0 0.0 -5.00

Pr{r ≤ Q*} = 0.80

We have a continuous demand uniformly distributed between 1000 and 7000

1000 7000

How do you find Q?

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Marginal Value: Uniform distribution

l=1000 u=7000

?

u-l=6000

1/60000.80

Q-l = Q-1000

(Q-1000)/6000=0.80Q = 5800

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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

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Marginal Value: Normal Distribution

00.00005

0.00010.00015

0.00020.00025

0.00030.00035

0.00040.00045

0 2000 4000 6000 8000

4841

Probability of

excess inventoryProbability of

shortage

0.80

0.20

Given a service level, how do we calculate z?From our normal table orFrom Excel Normsinv(service level)

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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

6.046

6)( *

ou

u

CCCQRP

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Additional Example - Solution

By convention, for the continuous demand distributions, the results are rounded to the closest integer.

2533.06.0)(**

QQZP

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.

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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

1012

212129452.0)( * cc

vccpcp

CCCQRP

ou

u

55.2452.912 cc

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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 13P(R=Qi) 0.04 0.06 0.16 0.18 0.2 0.1 0.1 0.08 0.04 0.04

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Additional Example

a) How many units are sold if we have ordered 7 units

There is 0.18 + 0.20 + 0.10 + 0.10 + 0.08 + 0.04 + 0.04 = 0.74

There is 0.74 probability that demand is greater than or equal to 7.

There is 0.16 probability that demand is equal to 6.



The expected number of units sold is

0.74(7) + 0.16 (6) + 0.06 (5) + 0.04 (4) = 6.6

Qi 4 5 6 7 8 9 10 11 12 13P(R=Qi) 0.04 0.06 0.16 0.18 0.2 0.1 0.1 0.08 0.04 0.04

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Additional Example

b) How many units are salvaged?

7-6.6 = 0.4. Alternatively, we can compute it directly

There is 0.74 probability that we salvage 7 – 7 = 0 units

There is 0.16 probability that we salvage 7- 6 = 1 units

There is 0.06 probability that we salvage 7- 5 = 2 units

There is 0.04 probability that we salvage 7-4 = 3 units

The expected number of units salvaged is

0.74(0) + 0.16 (1) + 0.06 (2) + 0.04 (3) = 0.4 and 7-0.4 = 6.6 sold

Qi 4 5 6 7 8 9 10 11 12 13P(R=Qi) 0.04 0.06 0.16 0.18 0.2 0.1 0.1 0.08 0.04 0.04

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Additional Example

c) Compute the total profit if we order 7 units.

Out of 7 units, 6.6 sold, 0.4 salvaged.

P = 75, c= 25, v=10.

Profit per unit sold = 75-25 = 50

Cost per unit salvaged = 25-10 = 15

Total Profit = 6.6(50) + 0.4(15) = 333 - 9 = 324

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Additional Example

d) Compute the expected Marginal profit of ordering the 8th unit.

MP = 75-25 = 50

P(R ≥ 8) = 0.2 + 0.1 + 0.1 + 0.08 + 0.04 + 0.04 = 0.56

Expected Marginal profit = 0.56(50) = 28

d) Compute the expected Marginal cost of ordering the 8th unit.

MC = 25 – 10 = 15

P(R ≤ 7) = 1-0.56 = 0.44

Expected Marginal cost = 0.44(15) = 6.6

Qi 4 5 6 7 8 9 10 11 12 13P(R=Qi) 0.04 0.06 0.16 0.18 0.2 0.1 0.1 0.08 0.04 0.04

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Overage Cost = Co = Unit Cost – Salvage = 0.25 – 0.1 = 0.15Underage Cost = Cu = Selling Price – Unit Cost = 0.75 – 0.25 = 0.50

77.0)(

77.015.050.0

50.0*)(

*

QRP

CCCQRP

ou

u ProbabilityCumulative Probability

Qi P(R=Qi) F(Qi)4 0.04 0.045 0.06 0.106 0.16 0.267 0.18 0.448 0.20 0.649 0.10 0.74

10 0.10 0.8411 0.08 0.9212 0.04 0.9613 0.04 1.00

Q* = 10

e) What is the optimum order quantity for Mac to minimize his cost?

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The Magnitude of Shortages (Out of Stock)

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