Download Lec 5. Inventory Control with Stochastic Demand (Feb 9, 2011)

Inventory Control with
Stochastic Demand
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Lecture Topics
 Week 1
 Week 2
 Week 3
 Week 4
 Week 5
 Week 6
 Week 7
Introduction to Production Planning and
Inventory Control
Inventory Control – Deterministic Demand
Inventory Control – Stochastic Demand
Inventory Control – Stochastic Demand
Inventory Control – Stochastic Demand
Inventory Control – Time Varying Demand
Inventory Control – Multiple Echelons
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Lecture Topics (Continued…)
 Week 8
 Week 9
 Week 12
 Week 13
 Week 14
 Week 10
 Week 11
 Week 15
Production Planning and Scheduling
Production Planning and Scheduling
Managing Manufacturing Operations
Managing Manufacturing Operations
Managing Manufacturing Operations
Demand Forecasting
Demand Forecasting
Project Presentations
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 Demand per unit time is a random variable X with mean
E(X) and standard deviation s
 Possibility of overstocking (excess inventory) or
understocking (shortages)
 There are overage costs for overstocking and shortage
costs for understocking
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Types of Stochastic Models
 Single period models
 Fashion goods, perishable goods, goods with short
lifecycles, seasonal goods
 One time decision (how much to order)
 Multiple period models
 Goods with recurring demand but whose demand
varies from period to period
 Inventory systems with periodic review
 Periodic decisions (how much to order in each
period)
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Types of Stochastic Models (continued…)
 Continuous time models
 Goods with recurring demand but with variable
inter-arrival times between customer orders
 Inventory system with continuous review
 Continuous decisions (continuously deciding on
how much to order)
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Example
 If l is the order replenishment lead time, D is demand per unit time,
and r is the reorder point (in a continuous review system), then
Probability of stockout = P(demand during lead time  r)
 If demand during lead time is normally distributed with mean E(D)l,
then choosing r = E(D)l leads to
Probability of stockout = 0.5
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The Newsvendor Model
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Assumptions of the Basic Model
 A single period
 Random demand with known distribution
 Cost per unit of leftover inventory (overage cost)
 Cost per unit of unsatisfied demand (shortage cost)
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
Objective: Minimize the sum of expected shortage
and overage costs

Tradeoff: If we order too little, we incur a shortage
cost; if we order too much we incur a an overage
cost
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Notation
X  demand (in units), a random variable.
G (x )  P(X  x ), cumulative distribution function of demand
(assumed to be continuous)
d
g (x )  G (x )  probability density function of demand.
dx
co  cost per unit left over after demand is realized.
cs  cost per unit of shortage.
Q  Order (or production quantity); a decision variable
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The Cost Function
Y (Q )  expected overage cost + shortage cost
 co E  units over   cs E  units short 
 Q  X if Q  X
N o  Number of units over  
otherwise
0
 max(Q  X , 0)  [Q  X ]+
 X  Q if Q  X
N S  Number of units short  
otherwise
0
 max( X  Q, 0)  [ X  Q ]
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The Cost Function (Continued…)
Y (Q )  co E [ N o ]  cs E [ N S ]


0
0
 co  max Q  x,0 g ( x )dx  cs  max x  Q ,0 g ( x )dx
Q

0
Q
 co  (Q  x ) g ( x )dx  cs  ( x  Q ) g ( x )dx
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Leibnitz’s Rule

a1 (Q )
a1 (Q ) Q [ f ( x, Q )]dx 
da2 (Q )
da1 (Q )
f ( a2 (Q ), Q )
 f ( a1 (Q ), Q )
dQ
dQ
d
dQ
a2 ( Q )
f ( x, Q )dx 
a2 ( Q )
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The Optimal Order Quantity
Q

Y (Q )
 co  g ( x )dx  cs   g ( x )dx  coG (Q )  cs (1  G (Q ))  0
0
Q
Q
cs
G(Q ) 
co  cs
The optimal solution satisfies
cs
G(Q )  Pr( X  Q ) 
co  cs
*
*
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The Exponential Distribution
The Exponential distribution with parameters l
G( x)  1  el x
l e  l x ,
g ( x)  
0,
1
E( X ) 
x0
x0
l
Var ( X ) 
1
l2
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The Exponential Distribution
(Continued…)
G (Q )  1  e  lQ
cs
G (Q*) 

cs  c0
 1  e  lQ*    Q * 
 log(1   )
l
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Example
Scenario:
 Demand for T-shirts has the exponential distribution
with mean 1000 (i.e., G(x) = P(X  x) = 1- e-x/1000)
 Cost of shirts is $10.
 Selling price is $15.
 Unsold shirts can be sold off at $8.
Model Parameters:
 cs = 15 – 10 = $5
 co = 10 – 8 = $2
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Example (Continued…)
Solution:
G (Q )  1  e
*

Q
1000
cs
5


 0.714
co  c s 2  5
Q *  1,253
Sensitivity:
If co = $10 (i.e., shirts must be discarded) then
G (Q* )  1  e
Q*  405

Q
1000

cs
5

 0.333
co  cs 10  5
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The Normal Distribution
The Normal distribution with parameters m and s, N(m, s)
1
( x  m ) 2
g ( x) 
exp[
],
2
2s
s 2
E( X )  m
 x  
Var ( X )  s 2
• If X has the normal distribution N(m, s), then (X-m)/s has
the standard normal distribution N(0, 1).
• The cumulative distributive function of the Standard
normal distribution is denoted by F.
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The Normal Distribution (Continued…)
G(Q*)=

Pr(X Q*)= 

Pr[(X - m)/s  (Q* - m)/s] = 

Let Y = (X - m)/s, then Y has the the standard Normal
distribution
Pr[(Y  (Q* - m)/s] = F[(Q* - m)/s] = 
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The Normal Distribution (Continued…)
F((Q* - m)/s) = 

Define z such that F(z)

Q* = m + zs
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The Optimal Cost for Normally
Distributed Demand
If Q  Q * , then it can be shown that
Y (Q * )  (cs  co )s ( z ),
1
 z2
where  ( z ) 
exp[
]
2
2
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The Optimal Cost for Normally
Distributed Demand
Both the optimal order quantity and the optimal cost
increase linearly in the standard deviation of demand.
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Example
 Demand has the Normal distribution with mean m = 10,000
and standard deviation s = 1,000
 cs = 1
 co = 0.5    0.67
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Example
 Demand has the Normal distribution with mean m = 10,000
and standard deviation s = 1,000
 cs = 1
 co = 0.5    0.67
Q* = m + zs
From a standard normal table, we find that z0.67 = 0.44
Q* = m + sz  10,000  0.44(1,000)  10,440
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Service Levels
 Probability of no stockout
cs
Pr( X  Q ) 

co  cs
 Fill rate
E[min(Q, X )] E[ X ]  E[max( X  Q,0)]
E[ N s ]

 1
E[ X ]
E[ X ]
E[ X ]
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Service Levels
 Probability of no stockout
cs
Pr( X  Q ) 

co  cs
 Fill rate
E[min(Q, X )] E[ X ]  E[max( X  Q,0)]
E[ N s ]

 1
E[ X ]
E[ X ]
E[ X ]
Fill rate can be significantly higher than
the probability of no stockout
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Discrete Demand
X is a discrete random variable
Y (Q )  co E[ N o ]  cs E[ N S ]
 co  x 0 max Q  x,0 Pr( X  x )  cs  x 0 max x  Q,0Pr( X  x )


 co  x 0 (Q  x ) Pr( X  x )  cs  x Q ( x  Q ) Pr( X  x )
Q

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Discrete Demand (Continued)
The optimal value of Q is the smallest integer that satisfies
Y (Q  1)  Y (Q)  0
This is equivalent to choosing the smallest integer Q that
satisfies
cs
 x1 P( X  x)  c  c
s
o
Q
or equivalently
cs
Pr( X  Q ) 
cs  co
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The Geometric Distribution
The geometric distribution with parameter  , 0    1
P ( X  x )   x (1   ).
E[ X ] 

1 
Pr( X  x )   x
Pr( X  x )  1   x 1
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The Geometric Distribution
The optimal order quantity Q* is the smallest integer that
satisfies
cs
Pr( X  Q ) 
cs  co
*
 1 
Q* 1
cs

 Q* 
cs  co
co


ln[
]

cs  co 
*
Q  

 ln[  ] 


co
]
cs  co
1
ln[  ]
ln[
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Extension to Multiple Periods
The news-vendor model can be used to a solve a
multi-period problem, when:
 We face periodic demands that are independent and
identically distributed (iid) with distribution G(x)
 All orders are either backordered (i.e., met eventually)
or lost
 There is no setup cost associated with producing an
order
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Extension to Multiple Periods
(continued…)
In this case
 co is the cost to hold one unit of inventory in stock for
one period
 cs is either the cost of backordering one unit for one
period or the cost of a lost sale
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Handling Starting Inventory/backorders
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Handling Starting Inventory/backorders
S0 : Starting inventory position
S: order up to level,
S  S0 : order quantity

Y ( S )  co E[( S  X )  ]  cs E[( X  S )  ]

cs
The optimal order-up-to level satisfies Pr( X  S ) 
.
cs  c0
*

The optimal policy: order nothing if S0  S * , otherwise order S * - S0 .
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