1 Managing Flow Variability: Safety Inventory Operations Management Session 23: Newsvendor Model.

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Managing Flow Variability: Safety Inventory

Operations Management

Session 23: Newsvendor Model

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Uncertain Demand

Uncertain Demand: What are the relevant trade-offs?

– Overstock

• Demand is lower than the available inventory

• Inventory holding cost

– Understock

• Shortage- Demand is higher than the available inventory

– Why do we have shortages?

– What is the effect of shortages?

Session 23 Operations Management 2

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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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What can be done to minimize shortages?

Better forecast

Produce to order and not to stock

– Is it always feasible?

Have large inventory levels

Order the right quantity

– What do we mean by the right quantity?


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Uncertain Demand

What is the objective?

– Minimize the expected cost (Maximize the expected profits).

What are the decision variables?

– The optimal purchasing quantity, or the optimal inventory

level.


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Optimal Service Level: The Newsvendor Problem

Cost of Holding Extra Inventory

Improved Service

Optimal Service Level under uncertainty

The Newsvendor ProblemThe 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

Assumptions– Demand is random– Distribution of demand is known– No initial inventory– Set-up cost is equal to 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 0.2170 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 140?What is probability of demand to be equal to 133?


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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 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


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What is probability of demand to be equal to 130?What is probability of demand to be less than or equal to 145?What is probability of demand to be greater than 145?

Average Demand Probability of Demand100 0.02110 0.05120 0.08130 0.09140 0.11150 0.16160 0.2170 0.15180 0.08190 0.05200 0.01


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Compute the Average Demand

X P(x=X)100 0.02110 0.05120 0.08130 0.09140 0.11150 0.16160 0.2170 0.15180 0.08190 0.05200 0.01

N

1i

)( Demand Average ii XxPX

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 x ≥ 140Prob(x ≥ 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 (X) 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(x ≥ X) 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 (X) 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(x ≥ X) 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

X P(x=X) P(x<X) P(x≥X)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

X P(x=X) P(x<X) P(x≥X) 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 0.2 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

X P(x=X) P(x<X) P(x≥X) 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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Example 2: Denim Wholesaler

The demand for denim is:

– 1000 with probability 0.1







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

Cost parameters:

How much should we order?Session 23 Operations Management 19

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Example 2: Marginal Analysis

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

Suppose the wholesaler purchases 1000 units

What is the value of the 1001st unit?


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Wholesaler purchases an additional unit

Case 1: Demand is smaller than 1001 (Probability 0.1)

– The retailer must salvage the additional unit and losses $5 (10 – 5)

Case 2: Demand is larger than 1001 (Probability 0.9)

– The retailer makes and extra profit of $20 (30 – 10)

Expected value = -(0.1*5) + (0.9*20) = 17.5


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What does it mean that the marginal value is positive?

– By purchasing an additional unit, the expected profit

increases by $17.5

The dealer should purchase at least 1,001 units.

Should he purchase 1,002 units?


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Case 1: Demand is smaller than 1002 (Probability 0.1)– The retailer must salvage the additional unit and losses $5 (10 – 5)

Case 2: Demand is larger than 1002 (Probability 0.9)– The retailer makes and extra profit of $20 (30 – 10)

Expected value = -(0.1*5) + (0.9*20) = 17.5


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Assuming that the initial purchasing quantity is between 1000 and

2000, then by purchasing an additional unit exactly the same

savings will be achieved.

Conclusion: Wholesaler should purchase at least 2000 units.


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Case 1: Demand is smaller than 2001 (Probability 0.25)– The retailer must salvage the additional unit and losses $5 (10 – 5)

Case 2: Demand is larger than 2001 (Probability 0.75)– The retailer makes and extra profit of $20 (30 – 10)

Expected value = -(0.25*5) + (0.75*20) = 13.75

What is the value of the 2001st unit?


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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


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We could continue calculating the marginal values

Demand ProbabilityCumlative Probability

Expected marginal cost

Expected marginal savings 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


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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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Net Marginal Benefit:

Net Marginal Cost:

MB = p – c

MC = c - v

MB = 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 if the demand is less than or equal to Q. Since we have P( D ≤ Q).

The expected marginal cost =MC× P( D ≤ Q)

If I have ordered one unit more than Q units, the probability of selling that extra unit is if the demand is greater than Q. We know that P(D>Q) = 1- P( D≤ Q).

The expected marginal benefit = MB× [1-Prob.( D ≤ Q)]


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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

MC×Prob(D ≤ Q*) ≥ MB× [1 – Prob(D ≤ Q*)]

MB

MB MCProb(D ≤ Q*) ≥


MB = p – c = Underage Cost = Cu

MC = c – v = Overage Cost = Co

ou

u

CC

c

MCMB

MBQDP

)( *

vp

cp

vccp

cp


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Marginal Value: The General Formula

P(D ≤ Q*) ≥ Cu / (Co+Cu)

Cu / (Co+Cu) = (30-10)/[(10-5)+(30-10)] = 20/25 = 0.8

Order until P(D ≤ Q*) ≥ 0.8

P(D ≤ 5000) ≥ = 0.75 not > 0.8 still order

P(D ≤ 6000) ≥ = 0.9 > 0.8 Stop

Order 6000 units


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

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In Continuous Model where demand for example has Uniform or Normal distribution

MCMB

MBQDP

)( *

ou

u

CC

c

vp

cp


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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{D ≤ 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)*1/6000=0.80Q = 5800


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Type-1 Service Level

What is the meaning of the number 0.80?

F(Q) = (30 – 10) / (30 – 5) = 0.8

– Pr {demand is smaller than Q} =

– Pr {No shortage} =

– Pr {All the demand is satisfied from stock} = 0.80

It is optimal to ensure that 80% of the time all the demand is

satisfied.


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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.


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Marginal Value: Normal Distribution

0

0.00005

0.0001

0.00015

0.0002

0.00025

0.0003

0.00035

0.0004

0.00045

0 1000 2000 3000 4000 5000 6000 7000 8000

Series1

4841

Probability of excess inventory

Probability of shortage

0.80

0.20


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Recall that:

F(Q) = Cu / (Co + Cu) = Type-1 service level


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Is it correct to set the service level to 0.8?

Shouldn’t we aim to provide 100% serviceability?


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What is the optimal purchasing quantity?

0

0.00005

0.0001

0.00015

0.0002

0.00025

0.0003

0.00035

0.0004

0.00045

0 1000 2000 3000 4000 5000 6000 7000 8000

Series15282

Probability of excess inventory

Probability of shortage

0.90


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How do you determine the service level?

For normal distribution, it is always optimal to have:

Mean + z*Standard deviation

µ + z

The service level determines the value of Z

z is the level of safety stock

+z is the base stock (order-up-to level)


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Given a service level, how do we calculate z?

From our normal table or

From 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 You cannot predict demand for the calendars with certainty.

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?


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Additional Example - Solution

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

CC

CQDP

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


2533.06.0)(**

QQZP

Look for P(Z ≤ z) = 0.6 in Standard Normal table or for

NORMSINV(0.6) in excel 0.2533

63533.62)2533.0(10602533.0* Q

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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.

Qnew = 1.2 * 63 = 75.6 ~ 76 units


)6.1()10

6076()76()( *

ZPzPDPQDP

Look for P(Z ≤ 1.6) = 0.6 in Standard Normal table or for

NORMSDIST(1.6) in excel 0.9452

10

12

212

129452.0)( * cc

vccp

cp

CC

CQDP

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:

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

Quantity Q 4 5 6 7 8 9 10 11 12 13Probability p(D=Q) 0.04 0.06 0.16 0.18 0.20 0.10 0.10 0.08 0.04 0.04


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Overage Cost = Co = Unit Cost – Salvage = 0.25 – 0.1 = 0.15

Underage Cost = Cu = Selling Price – Unit Cost = 0.75 – 0.25 = 0.50

77.0)(

77.015.050.0

50.0*)(

*

QDP

CC

CQDP

ou

uProbability

Cumulative Probability

Q p(D=Q) F(Q)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 1.00The critical ratio, 0.77, is between Q = 9 and Q = 10.

Remember from the marginal analysis explanation that the results are rounded up. Because at 9 still it is at our benefit to order one more.

So Q* = 10

1 Managing Flow Variability: Safety Inventory Operations Management Session 23: Newsvendor Model.

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