Transcript Newsvendor problem and demand uncertainty

Inventory Models
Uncertain Demand: The Newsvendor Model
Background: expected value
A fruit seller example
Undamaged mango
Damaged mango
Profit
$4
$1
Probability
80%
20%
What is the expected profit for a stock of 100 mangoes ?
0.8 x 100 ($4) + 0.2 x 100 x ($1) = 320 + 20 = $340
random variable: ai
probability: pi
Expected value = a1 p1 + a2 p2 + … + ak pk = Si = 1,,k aipi
Probabilistic models: Flower seller example
Wedding bouquets:
Selling price: $50 (if sold on same day), $ 0 (if not sold on that day)
Cost = $35
number of bouquets
probability
3
4
5
6
7
8
9
0.05
0.12
0.20
0.24
0.17
0.14
0.08
How many bouquets should he make each morning
to maximize the expected profit?
Probabilistic models: Flower seller example..
number of bouquets
probability
3
4
5
6
7
8
9
0.05
0.12
0.20
0.24
0.17
0.14
0.08
CASE 1: Make 3 bouquets
probability( demand ≥ 3) = 1
Exp. Profit = 3x50 – 3x35 = $45
CASE 2: Make 4 bouquets
if demand = 3, then revenue = 3x $50 = $150
if demand = 4 or more, then revenue = 4x $50 = $200
prob = 0.05
prob = 0.95
Exp. Profit = 150x0.05 + 200x0.95 – 4x35 = $57.5
Probabilistic models: Flower seller example
Compute expected profit for each case 
number of bouquets
probability
Expected profit
3
4
5
6
7
8
9
0.05
0.12
0.20
0.24
0.17
0.14
0.08
45
57.5
64
60.5
45
21
-10
Making 5 bouquets will maximize expected profit.
Probabilistic models: definitions
number of bouquets
probability
3
4
5
6
7
8
9
0.05
0.12
0.20
0.24
0.17
0.14
0.08
Discrete random variable
Probability (sum of all likelihoods = 1)
Continuous random variable:
Example, height of people in a city
-4
-3140
-2
150
-1
160
0
170
1
180
2
190
3
200
4
Probability density function (area under curve = integral over entire range = 1)
Probabilistic models: normal distribution function
Standard normal distribution curve: mean = 0, std dev. = 1
P( a≤ x ≤ b) = ab f(x) dx
-4
-3
-2
-1
0
a
1
2
3
4
b
Property:
normally distributed random variable x,
mean = m, standard deviation = s,
Corresponding standard random variable: z = (x – m)/ s
z is normally distributed, with a m = 0 and s = 1.
The Newsvendor Model
Assumptions:
- Plan for single period inventory level
- Demand is unknown
- p(y) = probability( demand = y), known
- Zero setup (ordering) cost
Example: Mrs. Kandell’s Christmas Tree Shop
Order for Christmas trees must be placed in Sept
Cost per tree: $25
Price per tree:
$55 before Dec 25
$15 after Dec 25
If she orders too few, the unit shortage cost is cu = 55 – 25 = $30
If she orders too many, the unit overage cost is co = 25 – 15 = $10
Past
Data
Sales
22
24
26
28
30
32
34
36
Probability
.05
.10
.15
.20
.20
.15
.10
.05
How many trees should she order?
Stockout and Markdown Risks
1. Mrs. Kandell has only one chance to order
until the sales begin: no information to revise the forecast;
after the sales start: too late to order more.
2. She has to decide an order quantity Q now
D total demand before Christmas
F(x) the demand distribution,
D > Q  stockout, at a cost of: cu (D – Q)+ = cu max{D –Q, 0}
D < Q  overstock, at a cost of co (Q–D)+ = co max{Q – D, 0}
Key elements of the model
1. Uncertain demand
2. One chance to order (long) before demand
3. ( order > demand OR order < demand)  COST
Model development
Stockout cost = cu max{D –Q, 0}
Overstock cost = co max{Q – D, 0}
Total cost = G(Q) = cu (D – Q)+ + co (Q – D)+
Expected cost, E( G(Q) ) = E(cu (D – Q)+ + co (Q – D)+)
= cu E(D – Q)+ + co E(Q – D)+


[c ( x  Q)
x 0
u


 co (Q  x) ]P( x) 

[c ( x  Q)
x Q
u

Q
]P( x)  [co (Q  x)  ]P( x)
x 0
Model Development: generalization
Suppose Demand  a continuous variable
++ good approximation when number of possibilities is high
-- difficult to generate probabilities, but…
++ probability distribution can be guessed
E (G(Q)) 

Q

[
c
(
x

Q
)
]
P
(
x
)

[
c
(
Q

x
)
]P( x)
 u
 o

x Q
x 0

Q
g (Q)  E ( G (Q)) 
c
x 0
0
(Q  x) P( x) dx 
c
x Q
u
( x  Q) P( x) dx
Model solution

Q
g (Q)  E ( G (Q)) 
c
0
(Q  x) P( x) dx 
x 0
Minimize g(Q) 
c
u
( x  Q) P( x) dx
x Q
d g (Q)
0
dQ
Q


d 
c0 (Q  x) P( x) dx   cu ( x  Q) P( x) dx  0


dQ  x 0
x Q

• g(Q) is a convex function: it has a unique minimum
• when g(Q) is at minimum value, F(Q) = cu/(cu + co)
The Critical Ratio
Solution to the Newsvendor problem:
cu
dg(Q)
 0  F (Q*) 
dQ
c0  cu
β = cu /(co + cu ) is called the critical ratio
b  relative importance of stockout cost vs. markdown cost
Mrs. Kandell’s Problem, solved:
cu = 55 – 25 = $30
Past
Data
D 22
Probability 0.05
F (D ) 0.05
co = 25 – 15 = $10
24
0.1
0.15
26
0.15
0.3
28
0.2
0.5
β = cu /(co + cu ) = 30/(30 + 10) = 0.75
NOTE:
30
0.2
0.7
32
0.15
0.85
34
0.1
0.95
36
0.05
1
 optimum ≈ 31
E(D) = 22x 0.05 + 24 x 0.1 + … + 36 x 0.05 = 29
Newsvendor model: effect of critical ratio
D 22
Probability 0.05
F (D ) 0.05
24
0.1
0.15
26
0.15
0.3
28
0.2
0.5
30
0.2
0.7
32
0.15
0.85
34
0.1
0.95
36
0.05
1
β = cu /(co + cu ) = 30/(30 + 10) = 0.75  optimum: 31
b
b
 overstock cost less significant  order more
 overstock cost dominates  order less
Summary
When demand is uncertain, we minimize expected costs
newsvendor model: single period, with over- and under-stock costs
Critical ratio determines the optimum order point
Critical ratio affects the direction and magnitude of order quantity
Concluding remarks on inventory control
Inventory costs lead to success/failure of a company
Example: Dell Inc.
“Dell's direct model enables us to keep low component inventories
that enable us to give customers immediate savings when
component prices are reduced, ...
Because of our inventory management, Dell is able to offer some
of the newest technologies at low prices while our competitors struggle
to sell off older products.”
Drive to reduce inventory costs was main motivation for
Supply Chain Management
next: Quality Control