Transcript Newsvendor Model Chapter 9 utdallas.edu/~metin Learning Goals  Determine the optimal level of product availability – Demand forecasting – Profit maximization  Other measures such as a fill.

Newsvendor Model
Chapter 9
1
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Learning Goals
 Determine
the optimal level of product availability
– Demand forecasting
– Profit maximization

Other measures such as a fill rate
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2
O’Neill’s Hammer 3/2 wetsuit
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3
Hammer 3/2 timeline and economics
Economics:
Generate forecast
of demand and
submit an order
to TEC
•
•
•
Spring selling season
Nov Dec Jan
Feb Mar Apr May Jun
Receive order
from TEC at the
end of the
month

Each suit sells for p = $180
TEC charges c = $110/suit
Discounted suits sell for v = $90
Jul Aug
Left over
units are
discounted
The “too much/too little problem”:
– Order too much and inventory is left over at the end of the season
– Order too little and sales are lost.

Marketing’s forecast for sales is 3200 units.
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Newsvendor model implementation steps
 Gather
economic inputs:
– selling price,
– production/procurement cost,
– salvage value of inventory
 Generate
a demand model to represent demand
– Use empirical demand distribution
– Choose a standard distribution function
» the normal distribution,
» the Poisson distribution.
 Choose
an objective:
– maximize expected profit
– satisfy a fill rate constraint.
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 Choose
a quantity to order.
5
The Newsvendor Model:
Develop a Forecast
6
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Historical forecast performance at O’Neill
.
7000
6000
Actual demand
5000
4000
3000
2000
1000
0
0
1000
2000
3000
4000
5000
6000
7000
Forecast
Forecasts and actual demand for surf wet-suits from the previous season
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How do we know the actual
when the actual demand > forecast demand
 If
we underestimate the demand, we stock less than
necessary.
 The stock is less than the demand, the stockout occurs.
 Are the number of stockout units (= unmet demand)
observable, i.e., known to the store manager?
– Yes, if the store manager issues rain checks to customers.
– No, if the stockout demand disappears silently.
 No
implies demand filtering. That is, demand is known
exactly only when it is below the stock.
 Shall we order more than optimal to learn about demand
when the demand is filtered?
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Empirical distribution of forecast accuracy
Order by A/F ratio
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Error* A/F Ratio**
-50
1.56
37
0.69
-3
1.02
7
0.96
-42
1.25
5
0.97
-15
1.08
-47
1.17
-49
1.15
-207
1.54
-191
1.50
79
0.80
156
0.64
191
0.56
-183
1.42
85
0.81
10
0.98
354
0.25
-135
1.27
-220
1.36
286
0.56
-128
1.19
227
0.67
133
0.82
288
0.72
-492
1.46
499
0.59
-396
1.30
-342
1.23
-1314
1.60
1995
0.37
521
0.86
2817
0.57
100%
90%
80%
70%
Probability
Actual demand
Product description
Forecast
JR ZEN FL 3/2
90
140
EPIC 5/3 W/HD
120
83
JR ZEN 3/2
140
143
WMS ZEN-ZIP 4/3
170
163
HEATWAVE 3/2
170
212
JR EPIC 3/2
180
175
WMS ZEN 3/2
180
195
ZEN-ZIP 5/4/3 W/HOOD
270
317
WMS EPIC 5/3 W/HD
320
369
EVO 3/2
380
587
JR EPIC 4/3
380
571
WMS EPIC 2MM FULL
390
311
HEATWAVE 4/3
430
274
ZEN 4/3
430
239
EVO 4/3
440
623
ZEN FL 3/2
450
365
HEAT 4/3
460
450
ZEN-ZIP 2MM FULL
470
116
HEAT 3/2
500
635
WMS EPIC 3/2
610
830
WMS ELITE 3/2
650
364
ZEN-ZIP 3/2
660
788
ZEN 2MM S/S FULL
680
453
EPIC 2MM S/S FULL
740
607
EPIC 4/3
1020
732
WMS EPIC 4/3
1060
1552
JR HAMMER 3/2
1220
721
HAMMER 3/2
1300
1696
HAMMER S/S FULL
1490
1832
EPIC 3/2
2190
3504
ZEN 3/2
3190
1195
ZEN-ZIP 4/3
3810
3289
WMS HAMMER 3/2 FULL
6490
3673
* Error = Forecast - Actual demand
** A/F Ratio = Actual demand divided by Forecast
60%
50%
40%
30%
20%
10%
0%
0.00
0.25
0.50
0.75
1.00
1.25
1.50
1.75
A/F ratio
Empirical distribution function for the historical A/F ratios.
9
Normal distribution tutorial



All normal distributions are characterized by two parameters,
mean = m and standard deviation = s
All normal distributions are related to the standard normal that has
mean = 0 and standard deviation = 1.
For example:
– Let Q be the order quantity, and (m, s) the parameters of the normal demand
forecast.
– Prob{demand is Q or lower} = Prob{the outcome of a standard normal is z
or lower}, where
z
Qm
s
or Q  m  z  s
– (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.)
– Look up Prob{the outcome of a standard normal is z or lower} in the
Standard Normal Distribution Function Table.
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Using historical A/F ratios to choose a
Normal distribution for the demand forecast
 Start
with an initial forecast generated from hunches,
guesses, etc.
– O’Neill’s initial forecast for the Hammer 3/2 = 3200 units.
 Evaluate
the A/F ratios of the historical data:
A/F ratio 
 Set
Actual demand
Forecast
the mean of the normal distribution to
Expectedactualdemand Expected A/F ratio  Forecast
 Set
the standard deviation of the normal distribution to
Standarddeviationof actualdem and
Standarddeviationof A/F ratios Forecast
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O’Neill’s Hammer 3/2 normal distribution forecast
Product description
JR ZEN FL 3/2
EPIC 5/3 W/HD
JR ZEN 3/2
WMS ZEN-ZIP 4/3
Forecast Actual demand
90
140
120
83
140
143
170
156
Error
-50
37
-3
14
A/F Ratio
1.5556
0.6917
1.0214
0.9176
…
…
…
… …
ZEN 3/2
ZEN-ZIP 4/3
WMS HAMMER 3/2 FULL
Average
Standard deviation
3190
3810
6490
1195
3289
3673
1995
521
2817
0.3746
0.8633
0.5659
0.9975
0.3690
Expectedactualdemand 0.9975 3200 3192
Standarddeviationof actualdemand 0.369  3200 1181
O’Neill should choose a normal distribution with mean 3192
and standard deviation 1181 to represent demand for the
Hammer 3/2 during the Spring season.
12
 Why not a mean of 3200?
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
Empirical vs normal demand distribution
1.00
0.90
Probability
.
0.80
0.70
0.60
0.50
0.40
0.30
0.20
0.10
0.00
0
1000
2000
3000
4000
5000
6000
Quantity
Empirical distribution function (diamonds) and normal distribution function with
13
mean 3192 and standard deviation 1181 (solid line)
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The Newsvendor Model:
The order quantity that maximizes
expected profit
14
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“Too much” and “too little” costs
 Co = overage cost
– The cost of ordering one more unit than what you would have ordered had you
known demand.
– In other words, suppose you had left over inventory (i.e., you over ordered). Co
is the increase in profit you would have enjoyed had you ordered one fewer
unit.
– For the Hammer 3/2 Co = Cost – Salvage value = c – v = 110 – 90 = 20
 Cu = underage cost
– The cost of ordering one fewer unit than what you would have ordered had you
known demand.
– In other words, suppose you had lost sales (i.e., you under ordered). Cu is the
increase in profit you would have enjoyed had you ordered one more unit.
– For the Hammer 3/2 Cu = Price – Cost = p – c = 180 – 110 = 70
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Balancing the risk and benefit of ordering a unit

Ordering one more unit increases the chance of overage
– Expected loss on the Qth unit = Co x F(Q), where F(Q) = Prob{Demand <= Q)

The benefit of ordering one more unit is the reduction in the chance
of underage:
– Expected benefit on the Qth unit = Cu x (1-F(Q))
.
70
Expected gain or loss
80
60
Expected marginal benefit
of understocking
50
40
30
Expected marginal loss
of overstocking
20
10
As more units are ordered,
 the expected benefit from
ordering one unit
decreases
 while the expected loss of
ordering one more unit
increases.
0
0
800
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1600
2400
3200
4000
4800
5600
6400
16
Expected profit maximizing order quantity

To minimize the expected total cost of underage and overage, order
Q units so that the expected marginal cost with the Qth unit equals
the expected marginal benefit with the Qth unit:
Co  F (Q)  Cu  1  F Q
Cu
Co  Cu

Rearrange terms in the above equation -> F (Q) 

The ratio Cu / (Co + Cu) is called the critical ratio.
Hence, to minimize the expected total cost of underage and
overage, choose Q such that we don’t have lost sales (i.e., demand
is Q or lower) with a probability that equals the critical ratio

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Expected cost minimizing order quantity with the
empirical distribution function


Inputs: Empirical distribution function table; p = 180; c = 110; v =
90; Cu = 180-110 = 70; Co = 110-90 =20
Evaluate the critical ratio:
Cu
70
Co  Cu



20  70
 0.7778
Look up 0.7778 in the empirical distribution function graph
Or, look up 0.7778 among the ratios:
– If the critical ratio falls between two values in the table, choose the one that
leads to the greater order quantity
…
24
25
26
Percentile
72.7%
75.8%
78.8%
…
…
1.25
1.27
1.30
…
…
212
635
1696
…
…
170
500
1300
…
…
…
HEATWAVE 3/2
HEAT 3/2
HAMMER 3/2
Forecast Actual demand A/F Ratio Rank
…
…
Product description
– Convert A/F ratio into the order quantity
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Q  Forecast * A / F  3200 *1.3  4160.
18
Hammer 3/2’s expected cost minimizing order
quantity using the normal distribution


Inputs: p = 180; c = 110; v = 90; Cu = 180-110 = 70; Co = 110-90 =20; critical
ratio = 0.7778; mean = m = 3192; standard deviation = s = 1181
Look up critical ratio in the Standard Normal Distribution Function Table:
z
0.5
0.6
0.7
0.8
0.9
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.6915
0.7257
0.7580
0.7881
0.8159
0.6950
0.7291
0.7611
0.7910
0.8186
0.6985
0.7324
0.7642
0.7939
0.8212
0.7019
0.7357
0.7673
0.7967
0.8238
0.7054
0.7389
0.7704
0.7995
0.8264
0.7088
0.7422
0.7734
0.8023
0.8289
0.7123
0.7454
0.7764
0.8051
0.8315
0.7157
0.7486
0.7794
0.8078
0.8340
0.7190
0.7517
0.7823
0.8106
0.8365
0.7224
0.7549
0.7852
0.8133
0.8389
– If the critical ratio falls between two values in the table, choose the greater z-statistic
– Choose z = 0.77

Convert the z-statistic into an order quantity:
Q  m  z s
 3192 0.77 1181 4101

Equivalently, Q = norminv(0.778,3192,1181) = 4096.003
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Another Example: Apparel Industry
How much to order? Parkas at L.L. Bean
Demand
D_i
4
5
6
7
8
9
10
11
12
13
14
15
16
17
Probabability
p_i
.01
.02
.04
.08
.09
.11
.16
.20
.11
.10
.04
.02
.01
.01
Cumulative Probability of demand Probability of demand
being this size or less, F()
greater than this size, 1-F()
.01
.99
.03
.97
.07
.93
.15
.85
.24
.76
.35
.65
.51
.49
.71
.29
.82
.18
.92
.08
.96
.04
.98
.02
.99
.01
1.00
.00
Expected demand is 1,026 parkas.
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Parkas at L.L. Bean
Cost per parka = c = $45
Sale price per parka = p = $100
Discount price per parka = $50
Holding and transportation cost = $10
Salvage value per parka = v = 50-10=$40
Profit from selling parka = p-c = 100-45 = $55
Cost of understocking = $55/unit
Cost of overstocking = c-v = 45-40 = $5/unit
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Optimal level of product availability
p = sale price; v = outlet or salvage price; c = purchase price
CSL = Probability that demand will be at or below order quantity
CSL later called in-stock probability
Raising the order size if the order size is already optimal
Expected Marginal Benefit of increasing Q= Expected Marginal Cost of Underage
=P(Demand is above stock)*(Profit from sales)=(1-CSL)(p - c)
Expected Marginal Cost of increasing Q = Expected marginal cost of overage
=P(Demand is below stock)*(Loss from discounting)=CSL(c - v)
Define Co= c-v; Cu=p-c
(1-CSL)Cu = CSL Co
CSL= Cu / (Cu + Co)
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Order Quantity for a Single Order
Co = Cost of overstocking = $5
Cu = Cost of understocking = $55
Q* = Optimal order size
Cu
55
CSL  P( Dem and Q ) 

 0.917
Cu  Co 55  5
*
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Optimal Order Quantity
1.2
0.917
1
0.8
Cumulative
Probability
0.6
0.4
0.2
0
4 5 6 7 8 9 10 11 12 13 14 15 16 87
Optimal Order Quantity = 13(‘00)
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Parkas at L.L. Bean
 Expected
demand = 10 (‘00) parkas
 Expected profit from ordering 10 (‘00) parkas = $499

Approximate Expected profit from ordering 1(‘00) extra
parkas if 10(’00) are already ordered
= 100.55.P(D>=1100) - 100.5.P(D<1100)
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Parkas at L.L. Bean
Additional Expected
Expected
Expected Marginal
100s
Marginal Benefit Marginal Cost Contribution
11th
5500.49 = 2695 500.51 = 255 2695-255 = 2440
12th
5500.29 = 1595 500.71 = 355 1595-355 = 1240
13th
5500.18 = 990
500.82 = 410 990-410 = 580
14th
5500.08 = 440
500.92 = 460 440-460 = -20
15th
5500.04 = 220
500.96 = 480 220-480 = -260
16th
5500.02 = 110
500.98 = 490 110-490 = -380
17th
5500.01 = 55
500.99 = 495 55-495 = -440
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Revisit L.L. Bean as a Newsvendor Problem

Total cost by ordering Q units:
– C(Q) = overstocking cost
+
understocking cost
Q

0
Q
C(Q)  Co  (Q  x) f ( x)dx  Cu  ( x  Q) f ( x)dx
dC(Q)
 Co F (Q)  Cu (1  F (Q))  0
dQ
Marginal cost of overage at Q* - Marginal cost of underage at Q* = 0
Cu
F (Q )  P( D  Q ) 
Co  Cu
*
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*
27
Safety Stock
Inventory held in addition to the expected
demand is called the safety stock
The expected demand is 1026 parkas but
we order 1300 parkas.
So the safety stock is 1300-1026=274 parka.
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The Newsvendor Model:
Performance measures
29
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Newsvendor model performance measures

For any order quantity we would like to evaluate the following
performance measures:
– Expected lost sales
» The average number of units demand exceeds the order quantity
– Expected sales
» The average number of units sold.
– Expected left over inventory
» The average number of units left over at the end of the season.
– Expected profit
– Expected fill rate
» The fraction of demand that is satisfied immediately
– In-stock probability
» Probability all demand is satisfied
– Stockout probability
» Probability some demand is lost
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Expected (lost sales=shortage)

ESC is the expected shortage in a season

ESC is not a percentage, it is the number of units, also see next page
Demand- Q if
Shortage 
0
if

Demand Q
Demand Q
ESC  E(max{Demand in a season  Q,0})

ESC 
 (x - Q)f(x)dx
x Q
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Inventory and Demand during a season
0
Q
Inventory
0
Season
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Upside
down
Q
Inventory=Q-D
D, Demand
During
A Season
Demand
During a
Season
0
32
0
Q
Upside
down
0
Shortage
=D-Q
Q
Shortage
Season
Demand
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0
D: Demand During s Season
Shortage and Demand during a season
33
Expected shortage during a season

First let us study shortage during the lead time
Expectedshortage E (max{D  Q,0})


 ( D  Q) f
D
( D)dD
wheref D is pdf of D.
D Q

Ex:
 d1  9 with prob p1  1/4 


Q  10, D  d 2  10 with prob p2  2/4, ExpectedShort age?
 d  11 with prob p  1/4 
3
 3

3
Expectedshortage  max{0,(d i  Q)} pi 
i 1
11
 (d  Q)}P( D  d )
d 10
1
2
1 1
 max{0,(9 - 10)}  max{0,(10- 10)}  max{0,(11- 10)} 
4
4
4 4
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Expected shortage during a season
 Ex:
Q  10, D  Uniform(6,12), ExpectedShortage?
D 12
 1  102

1
1  D2
1  122
Expectedshortage  ( D  10) dD  
 10D
 
 10(12)   
 10(10) 
6
6 2
6 2
 6 2

D 10
D 10
 2/6
12
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Expected lost sales of Hammer 3/2s with Q = 3500
Normal demand with mean 3192, standard deviation=1181
– Step 1: normalize the order quantity to find its z-statistic.
z
Qm
s

3500  3192
 0.26
1181
– Step 2: Look up in the Standard Normal Loss Function Table the expected lost
sales for a standard normal distribution with that z-statistic: L(0.26)=0.2824 see
Appendix B table on p.380 of the textbook
» or, in Excel L(z)=normdist(z,0,1,0)-z*(1-normdist(z,0,1,1)) see Appendix D on p.389
– Step 3: Evaluate lost sales for the actual normal distribution:
Expected lost sales  s  L( z)  1181 0.2824  334
Keep 334 units in mind, we shall repeatedly use it
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Measures that follow expected lost sales

Demand=Sales+Lost Sales
D=min(D,Q)+max{D-Q,0} or min(D,Q)=D- max{D-Q,0}
Expected sales = m - Expected lost sales
= 3192 – 334 = 2858

Inventory=Sales+Left Over Inventory
Q=min(D,Q)+max{Q-D,0} or max{Q-D,0}=Q-min(D,Q)
Expected Left Over Inventory = Q - Expected Sales
= 3500 – 2858 = 642
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Measures that follow expected lost sales
Expected total underage and overage cost with (Q=3500)
=70*334 + 20*642
Expected profit   Price-Cost   Expected sales 
  Cost-Salvage value   Expected left over inventory 
  $70  2858    $20  642   $187, 221
What is the relevant objective? Minimize the cost or maximize the profit?
Hint: What is profit + cost? It is 70*3192=Cu*μ, which is a constant.
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Type I service measure: Instock probability = CSL
Instock probability: percentage of seasons without a stock out
For exampleconsider10 seasons:
11 0 111 0 1 0 1
InstockProbability 
10
Write0 if a season has stockout,1 otherwise
InstockProbability  0.7
InstockProbability  0.7  Probability thata single season has sufficientinventory
[Sufficient inventory] [Demandduring a season  Q]
InstockPro
bability  P(Demand Q)
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Instock Probability with Normal Demand
N(μ,σ) denotes a normal demand with mean μ and standard deviation σ
PN ( m , s )  Q 
 Normdist(Q, m , s ,1)
 P N ( m , s )  m  Q  m  T akingout themean
 N (m ,s )  m Q  m 
 P

 Dividing by theStDev
s
s 

Qm 

 P N (0,1) 
 Obtainingstandardnormaldistribution
s 

Qm

 Normdist
,0,1,1
 s

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Example: Finding Instock probability for given Q
μ = 2,500 /week; s= 500; Q = 3,000;
Instock probability if demand is Normal?
Instock probability = Normdist((3,000-2,500)/500,0,1,1)
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Example: Finding Q for given Instock probability
μ = 2,500/week; s= 500; To achieve Instock
Probability=0.95, what should be Q?
Q = Norminv(0.95, 2500, 500)
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Type II Service measure: Fill rate
Recall:
Expected sales = m - Expected lost sales = 3192 – 334 = 2858
Expected sales
Expected sales

Expected demand
m
Expected lost sales 2858
 1

m
3192
 89.6%
Expected fill rate 
Is this fill rate too low?
Well, lost sales of 334 is with Q=3500, which is less than optimal.
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Service measures of performance
100%
90%
80%
Expected fill
rate
70%
60%
50%
In-stock probability
CSL
40%
30%
20%
10%
0%
0
1000
2000
3000
4000
5000
6000
7000
Order quantity
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Service measures: CSL and fill rate are different
inventory
CSL is 0%, fill rate is almost 100%
0
time
inventory
0
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CSL is 0%, fill rate is almost 0%
time
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Summary

Determine the optimal level of product availability
– Demand forecasting
– Profit maximization / Cost minimization

Other measures
–
–
–
–
–
–
–
Expected shortages = lost sales
Expected left over inventory
Expected sales
Expected cost
Expected profit
Type I service measure: Instock probability = CSL
Type II service measure: Fill rate
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Homework Question:
A newsvendor can price a call option
 Q1:
Suppose that you own a simple call option for a
stock with a strike price of Q. Suppose that the price of
the underlying stock is D at the expiration time of the
option. If D <Q, the call option has no value.
Otherwise its value is D-Q. What is the expected value
of a call option, whose strike price is $50, written for a
stock whose price at the expiration is normally
distributed with mean $51 and standard deviation $10?
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