News Vendor Slides New New

–1– c©Samuel K. Eldersveld – OM301 – May 12, 2008

Inventory Decisions Under Uncertainty: Normal Demand

1. Review of Inventory issues

2. The newsvendor model

3. Demand forecasts

4. Performance Measures

5. Order quantities

Goals:

1. Describe inventory systems operationally.

2. Understand financial measures of inventory systems.

3. Formulate a model describing the behavior of an inventory system.

4. Derive an optimal policy with respect to the formulated model.

OM301 - L11 – p. 1/34


What can happen: Cellphones (NYTimes Article)

Source: “Cellphone envy lays Motorola low”, Brad Stone, New York Times, Business Day Section, page B1, 2/3/2007.

OM301 - L11 – p. 2/34


Some inventory items have short shelf lives

Perishable items (fresh food, flowers)

Trend-driven items (clothing and fashion items)

Items where technology is changes quickly (cell phones, DVD players, other electronics)

Seasonal items (Holiday cards, foods, ornaments, etc.)

If you keep an unsold item in inventory, it is stealing some of your capital.

If you don’t keep an unsold item in inventory you may have to sell it at a loss.

If you don’t order enough inventory, you lose the opportunity to make a sale.

OM301 - L11 – p. 3/34


A brief look at the fashion industry

Typical apparel companies average 5.0 inventory turns per year.

The Spanish Zara turns it’s inventory 10+ times per year.

Zara’s secrets? Good forecasts, small lot sizes, moving fast.

Zara has a 30 day production cycle.

U.S. companies take about 105-240 days for their production cycle.

Zara introduces new items ≈ every 2 weeks.

Customers can’t “wait for a discount” if the item will be gone the nexttime they visit.

ZaraAbercrombie & FitchArcadiaBenettonEl Corte InglésEtam DéveloppementFast RetailingFrench ConnectionThe GapH&MLimited BrandsMarks & SpencerNaf NafNew LookNEXTVivarte

Zara has low “inventory mismatch costs”

OM301 - L11 – p. 4/34


Single-period inventory model: O’Neill, Inc.

O’Neill, Inc. Sports apparel (wet suits) for water sports.

Designed in U.S.A. made in Mexico and Asia.

Two selling seasons (Fall, Spring)

Some products have a “long shelf life” (neoprene booties)

Some products are subject to “the whims of fashion”...

... they are difficult to sell at the end of the season. The “Hammer3/2” is one of these...

Hammer 3/2 Suit

OM301 - L11 – p. 5/34


Order cycle: O’Neill, Inc. “Hammer 3/2 Suit”

Three month lead time on ordering:

Orders must be submitted in Nov. (for spring sellingseason.)

“Point Estimate” of demand = 3200 units.O’Neill has a much better estimate of demand afterthe season starts (hot? cold? or just average?)...

... By then it is too late to order from Asianmanufacturers for delivery for this season.

Experience: forecast is off by 25% or more at leasthalf the time:

Half the time demand is between 75% and 125%of the original point estimate forecast.

Hammer 3/2 Suit

OM301 - L11 – p. 6/34


Economics: O’Neill, Inc. “Hammer 3/2 Suit”

Demand is forecast for 3,200 units sold in the spring,

O’Neill’s manufacturing cost = $110.

Sale price (direct, to retailer) = $180.

End-of-season discount price (direct, to retailer) = $90.

Hammer 3/2 Suit

How many units should O’Neill order?

1. Order at the forecast, point estimate of 3,200?

2. Order less than 3,200?

3. Order more than 3,200?

Order first, then observe the random variable of demand later

OM301 - L11 – p. 7/34


A problem similar to O’Neill’s

The news vendor must purchase newspapers at the start of the day before attempting tosell them at her designated street corner.

The news vendor pays $0.35 for each newspaper, sells each newspaper for $1.00 andeach unsold newspaper has a value of $0.01 at the end of the day.

Our news vendor orders a quantity y first, then observes (probabilistic) demand second.

There is a trade-off between ordering too many and too few:

1. If she orders too few, she could have sold more (an oppor-tunity loss: $0.65 each.)

2. If she orders too many, she will have left overs, (at a netloss of $0.34 each.)

The newsvendor must make a firm “bet” (how much inven-tory to order) before some random event occurs (demand).

OM301 - L11 – p. 8/34


The newsvendor single-period inventory model

Used to handle ordering perishable items with limited useful life.1. Items with short shelf-lives: Newspapers, magazines.2. Perishables: Fruit , flowers, baked goods, etc.

Items are not carried over from period toperiod.

Unsold items can be “salvaged”

There are three monetary inputs:

1. p = price (revenue) per unit sold.

2. c = cost per unit.

3. s = salvage value per unit.

Example product

Data Newspaper O’Neill Hammer 3/2

p Per-item sales revenue $1.00 $180c Manuf./acquisition cost 0.35 110s End-period salvage revenue 0.01 90

OM301 - L11 – p. 9/34


Steps for implementing the Newsvendor model

Step 1: Gather economic inputs:

p: Selling price,

c: production/procurement cost,

s: salvage value of inventory.

Step 2: Generate a demand model: (From historical data, expert opinion) Usea probability model for estimating demand. You may choose to use a stan-dard distribution function to represent demand, e.g. the normal distribu-tion, the Poisson distribution or base your demand on the use of historicaldata using an empirical demand distribution.

Step 3: Choose an objective: e.g. maximize expected profit

Step 4: Choose a quantity to order.

OM301 - L11 – p. 10/34


Building a forecast distribution of demandYou want demand (not just sales), you need a point estimate, a measure of uncertainty and the shape (type)of distribution of demand (uniform, normal, discrete).

What are the sources of forecast errors and what do they look like?

Forecasts and actual demand for surf wet-suits from the previous season

OM301 - L11 – p. 11/34


Normally distributed demand: Tutorial (or...refresher)

Suppose demand is normally distributed;

Suppose that mean demand = µ and standard deviation = σ.

Case 1: Suppose your boss chooses to order a quantity of items Q

What is the probability that demand is less than the amount Q?

P (demand ≤ Q) ≡ P (Z ≤ z1), where

z1 =Q − µ

σ

OM301 - L11 – p. 12/34


Normally distributed demand: Tutorial (or...refresher)

Example: Suppose µ = 300, σ = 50 and Q = 325,

What is the probability statement?

What is the probability demand is less than Q = 325 units.?

What is standardized z value?

z = 325−300

50= 0.5.

What is the probability? (Table: page 378).

Probability demand ≤ 325 units is Φ(0.5) = 0.6915.

OM301 - L11 – p. 13/34


Case 2: A 97.5% service level

Case 2: Suppose you want to choose a quantity to order Q so that you have a 97.5%

probability of satisfying all demand in the next period?

Example: Suppose µ = 300, σ = 50. You want

P (demand ≤ Q) = 0.975.

You know µ and σ and now a probability. You must calculate Q.

We need to find the z value for 0.975 (Table: page 378 z = 1.96)

Q = µ + z × σ = 300 + (1.96) × (50) = 398units.

Note:

z1 =Q − µ

σor Q = µ + z1 × σ

(The above are two ways to write the same equation, the first allows you to calculate z

from Q and the second lets you calculate Q from z.)

OM301 - L11 – p. 14/34


Another example using the normal distribution

F (Q) = Probability(demand ≤ Q)

P (d ≤ Q) = P (z ≤Q − µ

σ) ≡ Φ(z)

In this example suppose µ = 3192, σ = 1181and you desire to find the probability that demandis less than Q = 3664 units.

To determine P (d ≤ 3664) compute the zvalue:

z1 =3664 − 3192

1181= 0.400,

then

P (d ≤ 3664) = P (z ≤ 0.4) = Φ(0.4) = 0.6554 ,

using the tables, on pages 377–378 of Cachon.

OM301 - L11 – p. 15/34


Single-period inventory model: Costs of lost sales and overstockingThere are two resulting inventory costs from any ordering decision:

1. Stock-out cost (shortage cost) Potential lost profit due to stockout:

Cu = Revenue per unit - cost per unit = (p − c)

2. Mark-down cost (excess cost):

Co = original cost per unit - salvage value per unit = (c − s)

If the news vendor carries one too-many papers, she is left with a marginal profit

of her cost less the salvage value of the extra newspaper (Co). If the news vendor

carries one too-few papers she incurs the loss of profit corresponding to the price

of the newspaper less her cost (Cu).

Data Newspaper O’Neill Hammer 3/2p Per-item sales revenue $1.00 $180c Manuf./acquisition cost 0.35 110s End-period salvage revenue 0.01 90

Cu underage (opportunity) cost 0.65 70Co cost per unit of leftover inventory 0.34 20

OM301 - L11 – p. 16/34


Relationship of Q to expected gains and losses for the Hammer 3/2

Gain on a sold unit = $180 - $110 = $70

Loss on an unsold unit = $110 - $90 = $20

As Q increases, the marginal (per unit) expected gains decrease from $70 andmarginal expected losses increase to $20.

Marginal per unit expected gains and losses: Hammer 3/2

$ per unit

OM301 - L11 – p. 17/34


Balancing Expected Costs

Co Cu

Service Level

↑ Quantity →

Q

Balance point Q = optimal stocking level

Consider the “last” item in the ordering quantity Q.

If demand > Q, then we lose Cu in profits.

If demand < Q then we lose Co in profit.

Since we don’t know demand, we have to compute expected losses.

It makes sense to choose Q so that the marginal expected costs of too much inventory are equal to

the marginal expected cost of too little inventory.

The “expectation” of demand depends on the distribution.

OM301 - L11 – p. 18/34


Choosing Q: Special Case–Uniform Demand

Co Cu

Service Level

↑ Quantity →

Q

Balance point: Q = optimal stocking level

Service level is the probability that demand ≤ the stocking level.

Service level is the key to determining the optimal stocking level Q.

The critical value is = Cu

Cu+Co

Suppose our newsvendor has uniform demand between 200 and 400 papers.

For the newsvendor’s newspaper problem = Cu

Cu+Co

= 0.650.65+0.34

= 0.657.

For the uniform demand case she orders 200 + 0.657 × (400 − 200) = 331 papers

We order Q so that the probability not having enough (stockout) is the critical value

P (demand ≤ Q) = Cu

Cu+Co

OM301 - L11 – p. 19/34


Choosing Q: Case–Normally Distributed Demand

Suppose that our news vendor had normally distributed demand with mean demand µ = 200 and

standard deviation σ = 40 newspapers. What would be her optimal stocking level Q?

Demand Distribution

Demand

Den

sity

50 100 150 200 250 300 35050 100 150 200 250 300 350

Service Level =

P(STOCKOUT)=0.656

1. Cu = $1.00 − $0.35 = $0.65 per paper.

2. Co = $0.35 − $0.01 = $0.34 per papers.

3. Critical Ratio = Cu

Cu+Co

= 0.650.99

= 0.657

4. Using our table P (Z ≤ z) = SL gives z = 0.41.

OM301 - L11 – p. 20/34


Choosing Q: Case–Normally Distributed Demand

Suppose that our news vendor had normally distributed demand with mean demand µ = 200 and

standard deviation σ = 40 newspapers. What would be her optimal stocking level Q?

Demand Distribution

Demand

Den

sity

50 100 150 200 250 300 35050 100 150 200 250 300 350

Service Level =

P(STOCKOUT)=0.656

1. Cu = $1.00 − $0.35 = $0.65 per paper.

2. Co = $0.35 − $0.01 = $0.34 per papers.

3. Critical Ratio = Cu

Cu+Co

= 0.650.99

= 0.657

4. Using our table P (Z ≤ z) = SL gives z = 0.41.

5. The optimal stocking level is thus

Q = µ + z × σ

= 200 + (0.41)(40)

= 216 papers.

OM301 - L11 – p. 20/34


Case–Normal distribution method: finding the profit-maximizing QStep 1: Gather your inputs (here for the Hammer 3/2 example:)

1. Standard normal table (pgs. 377–378).

2. p = 180; c = 110; s = 90; Cu = 180 − 110 = 70; Co = 110 − 90 = 20

3. µ = 3192 Point estimate–the forecast of demand.

4. σ = 1181 Estimate of standard deviation of the forecast.

Step 2: Evaluate the critical ratio:Cu

Cu+Co= 70

70+20= 0.7778

Step 3: Lookup 0.7778 Look up critical ratio in the Standard Normal Distribution Function Table (page377–378) If the critical ratio falls between two values in the table, choose the greater z-statistic For thisproduct, we choose z = 0.77 (From the table we find that z = 0.77 → P (z ≤ 0.77) = 0.7794.)

Step 4: Convert z score into the order quantity:

Q = µ + z × σ = 3192 + 0.77 × 1181 = 4101.

OM301 - L11 – p. 21/34


The Expected Loss Function-Discrete Case

Since demand is a random variable, for most choices of Q there will be a positive

probability that actual demand exceeds your choice.

This is the expected shortfall of units for your order quantity Q.

Note, even if Q > µ, this quantity may be nontrivially large.

OM301 - L11 – p. 22/34


Lost Sales Explained

Suppose demand is discrete, Q∗ = 120 and demand (D) can be any one of {0, 10, 20, . . . , 200}

Associated with each choice of D is

a probability (e.g. P (D = 140)).

When D ≤ 120 there is no loss.

When D > 120 loss is (D − Q)

Expected Lost Sales =

10 × P (D = 130)+20 × P (D = 140)+

.

.

....

.

.

.80 × P (D = 200) =

200X

x=130

(x − Q) × P (D = x).

OM301 - L11 – p. 23/34


Example: Loss Function for Discrete DemandThe left columns are the distribution of demand.

The Loss is calculated for an order of size Q = 120

L(Q) = Expected Lost sales (the sum of the left-hand column) = 7.777 units.

OM301 - L11 – p. 24/34


The Expected Loss Function - Normal Case

Assume demand is normally distributed.

For any choice of Q there will be a positive probability that demand > Q.

The expected number shortfall of units is a positive number for normal demand.

The loss function is defined for every choice of Q

L(Q) = Loss Function for choice Q = L(Q, µ, σ) =

∫∞

Q

(x − Q)φ(x)dx,

where φ(x) is the normal density with mean µ and standard deviation σ.

Luckily, we do not have to perform the integration above, since

L(Q) = L(z) × σ,

where we convert Q to find z and then look up L(z) in a table (pages 379–380)...

OM301 - L11 – p. 25/34


Loss Function with the Normal DistibutionSuppose demand is a normally distributed random variable with µ = 5000 σ = 1000.

Your boss chooses order quantity Q = 5900. What is Expected Lost Sales?

!4 !2 0 2 4

0.0

0.1

0.2

0.3

0.4

Demand Distribution mean=0, std.dev.=1

Q=Quantity

De

nsity

UnshadedArea=0.8159

Loss=L(.9)=0.1004

z = 0.9

z = Q−µσ

= 5900−5000

1000= 0.9

From page 380 of the text, we find that L(z) = L(0.9) = 0.1004,

Expected Lost Sales = L(z) × σ = (0.1004) × (1000) = 100.4.

OM301 - L11 – p. 26/34


Newsvendor model performance measures

For any order quantity we would like to evaluate the following performance measures:

1. Expected lost sales: The average numberof units demand exceeds the order quan-tity

2. Expected sales: The average number ofunits sold.

3. Expected leftover inventory: The averagenumber of units left over at the end of theseason.

4. Expected profit

5. Expected fill rate: The fraction of demandthat is satisfied immediately

6. In-stock probability: Probability all de-mand is satisfied

7. Stockout probability: Probability some de-mand is lost

OM301 - L11 – p. 27/34


1.Expected Lost Sales of Hammer 3/2s with Q = 3500Definition: The average number of units demand exceeds the order quantity:

e.g., if demand is 3800 and Q = 3500, then lost sales is 300 units.e.g., if demand is 3200 and Q = 3500, then lost sales is 0 units.

Expected lost sales is the average over all possible demand outcomes. If demand is normally distributed:

Step 1: Normalize the order quantity to find its z-statistic.

z =Q − µ

σ=

3500 − 3192

1181= 0.26

Step 2: Look up in the Standard Normal Loss Function Table (page 379–380 of the text) the expected lostsales for a standard normal distribution with that z-statistic: L(0.26) = 0.2824

you can find L(z) in Excel, by plugging in your value for z into (e.g. substitute 0.26 for z):

=Normdist(z,0,1,0)-z*(1-Normsdist(z))

Step 3: Evaluate lost sales for the actual normal distribution:

Expected Lost sales = σ × L(z) = 1181 × 0.2824 = 334.

OM301 - L11 – p. 28/34


Performance measures 2–5 follow from expected lost sales2. Expected Sales (the average number of units sold)

= µ− Expected lost sales = 3192 − 334 = 2858

3. Expected leftover inventory (The average number of units left over at the end of the season)

= Q− Expected sales = 3500 − 2858 = 642.

4. Expected Profit:

= (Price - Cost) × (Expected Sales) −(Cost - Salvage Value) × (Expected Leftover Inventory)

= ($70 × 2858) −($20 × 642)

= $187, 221

5. Expected fill rate (The fraction of demand that is satisfied immediately):

=Expected Sales

Expected Demand=

Expected Sales

µ

= 1 −Expected lost sales

µ

=2858

3192

= 89.6%

Note, the above equations hold for any demand distribution.

OM301 - L11 – p. 29/34


Performance measures (6–7): Service related performance

6. In-stock probability (Probability all demand is satisfied)= F (Q) = Φ(z):

Evaluate the z-statistic for the order quantity :

z =Q − µ

σ=

3500 − 3192

1181= 0.26

Look up Φ(z) in the Std. Normal Distribution Function Table (pages 377-378)

Φ(0.26) = 60.26%

Note:The in-stock probability is not the same as the fill rate

7. Stockout probability (Probability some demand is

lost) = 1 − F (Q)

= 1−In-stock probability

= 1 − 0.6026 = 39.74%

OM301 - L11 – p. 30/34


Choosing Q subject to a minimum in-stock probabilitySuppose we wish to find the order quantity for the Hammer 3/2 that minimizes left over inventory while

generating at least a 99% in-stock probability.

Step 1: Using pages 377–378 of the textbook, find the z-statistic that yields the target in-stock probability.

In the Standard Normal Distribution Function Table we find Φ(2.32) = 0.9898 and

Φ(2.33) = 0.9901. Choose z = 2.33 to satisfy our in-stock probability constraint.

.

.

....

.

.

....

.

.

....

.

.

....

.

.

....

.

.

.

Step 2: Convert the z-statistic into an order quantity for the actual demand distribution.

Q = µ + z × σ = 3192 + 2.33 × 1181 = 5944

OM301 - L11 – p. 31/34


Choosing Q subject to a minimum fill rate constraintSuppose we wish to find the order quantity for the Hammer 3/2 that minimizes left over inventory whilegenerating at least a 99% fill rate.

Step 1: Find the lost sales with a standard normal distribution that yields the target fill rate.

L(z) =µ

σ× (1 − Fill rate) =

„

3192

1181

«

× (1 − 0.99) = 0.0270.

Step 2: Using pages 379–380 of the textbook, find the z-statistic that yields the lost sales found in step 1.From the Standard Normal Loss Function Table,

L(1.53) = 0.0274 and L(1.54) = 0.0267

Choose the higher z-statistic, z = 1.54

.

.

....

.

.

....

.

.

....

.

.

....

.

.

....

.

.

.

Step 3: Convert the z-statistic into an order quantity for the actual demand distribution.

Q = µ + z × σ = 3192 + 1.54 × 1181 = 5011.

OM301 - L11 – p. 32/34


Newsvendor Summary

Too Little Inventory Too Much Inventory

The model can be applied to settings in which....

1. There is a single order/production/replenishment opportunity.

2. Demand is uncertain.

3. There is a “too much-too little” challenge:If demand exceeds the order quantity, sales are lost.If demand is less than the order quantity, there is left over inventory.

OM301 - L11 – p. 33/34


Newsvendor Summary II

Firm must have a demand model that includes an expected demand and

uncertainty in that demand.

With the normal distribution, uncertainty in demand is cap-tured with the standard deviation parameter.

For maximizing expected profit, find Q so that “the probability that

demand is less than the order quantity” equals the critical ratio defined

by price, salvage value and cost:

P (demand ≤ Q) =Underage

Underage + Overage.

The expected profit maximizing order quantity balances the too much-too little costs.

Check your HW hand work: NewsvendorDoubleCheck.xls

Spreadsheet is downloadable from blackboard (“Course Materials”).

OM301 - L11 – p. 34/34

Documents

News Vendor Slides New New