Chapter 7: Stochastic Inventory Model

Proportional Cost Models:
x: initial inventory,
y: inventory position (on hand + on order-backorder),
: random demand, () , (),
(y- )+: ending inventory position, N.B.L,
(y- ) : ending inventory position, B.L,
=1/(1+r) : discount factor,
ordering cost : c(y-x),
holding cost : h (y- )+
penalty cost : p( -y)+
salvage cost : - s(y- )+

Minimum cost f(x) satisfies:
f ( x)  min c( y  x)  (h  s )
y x

0

y
y x

( y   ) ( )d
  (  y) ( )d 
p
p c
 min c( y  x)  L( y )


y
( 2)
L(y) convex, L’() < -c (otherwise never order)
L′ eventually becomes positive
L' ( S )  c  0
Base Stock Policy
y * ( x)  max{ x, S }
q ( x) 
*

S  x,
0,
If x  S
otherwise
( 4)
c  (h  s ) ( y )  ( pp c )( 1   ( y ))  0
cu
pc
N .B.L  ( y ) 

( p  c )  ( c  h   s ) cu  c o
B.L
cu
p  c(1   )
( y) 

[ p  c(1   )]  (c  h  s ) cu  co
( 5a )
( 5b)
Example
c=$1, h=1¢ per month, =0.99, p=$2(NBL), p=$0.25(BL),
s=50 ¢, c+h- s=51.5 ¢,
NBL: p-c = 100 ¢, BL: p-c(1- )=24 ¢,

100
(i )  ( y ) 
 0.66 , y   1 (0.66 )
100  51 .5
24
(ii)  ( y ) 
 0.32 , y   1 (0.32 )
24  51 .5

Set up cost K
f ( x)  min K ( y  x)  c( y  x)  L( y)
y x



L(x) if we order nothing
K+c(S-x)+L(S) if we order upto S
If we order, L’(S)+c=0.
Use the cheaper of alternatives L(x) and K+ c(S-x)+L(S)
cost
L(x)+cx
K+c(S-x)+L(S)
K
K
c
L(x)
s
S
x
s
S
x
Two-bin or (s,S) policy

order S-x
if x ≤ s
order nothing
if x > s
Multiperiod models

f 2 ( x)  min c( y  x)  L( y )  
y x 

 f ( y   ) ( )d
0
1
Infinite Horizon (f1000 & f1001 cannot be different)

f ( x)  min c( y  x)  L( y )  
y x 


f ( y   ) ( )d 


0
( 9)
Taking derivative of {}
0  c  L' ( S )  

 f ' (S   ) ( )d
(10 )
0
If f convex, find S the base stock level, then for x ≤ S
f ( x )  c ( S  x )  L( S )  

 f (S   ) ( )d
(11)
0
We see from (11) that
f’(x)=-c for x ≤ S .
(12)
(10 ) reduces to
L' ( S )  c(1   )  0 B.L
(13)
Similarly for N.B.L
L' ( S )  c(1  ( S ))  0
(18)

Proportional costs:
L( y )  h

y
( y   ) ( )d  p
0


(  y ) ( )d
y
So that
L ' ( S )  ( h  p ) ( S )  p
(19 )
Substitute (19) into (13) and (18),
cu
pc
N .B.L :  ( S ) 

( p  c)  h  c(1   ) cu  co
(20 a )
cu
p  c(1   )
B.L :  ( S ) 

p  c(1   )  h  c(1   ) cu  co
(20 b)
cf. with (5) : - s replaced by - c
Remark :
Lead time, Setup cost  more complicate d, still (s, S) policy
Example 4:
Bread Demand ~ Uniform [1000,2000 ]. Makes profit of 20cents
per loaf if sold on time; otherwise, a store outlet sells at a loss of 5 cents
per loaf. Find the optimal daily number of loaves.
Solution :
cu  20 , co  5
cu
20
4
( S ) 


cu  co 5  20 5
S  1800
Example 5:
1
e
25
Demand distribution :  ( ) 


25 , K
 15, c  1,
3
z , z  0,
10
3
shortagecost : p(z)  z 2 , z  0
2
Solution :
Holding cost : h( z ) 
L( y ) 

y
h( y   ) ( )d 
3 1
10 25
p ( y   ) ( )d
y
0




y
( y   )e


25 d
0
 1882 .5e

y
25
 0.3 y  7.5

3 1
2 25

2 
( y   ) e 25 d

y
Intuition: The current period would be a separate one
period if we know what the next period would be willing
to pay for our leftover inventory. Assuming we are not
“overstocked”, every unit leftover will mean the next period will
order one less, thus saving c. So the next period should be
willing to pay c per unit in salvage for one leftover inventory.

dL( y )
 75 .25 e
dy

y
25
 0. 3
dL( y )
c
 1  0.3  75 .25 e
dy y  S

S
25
0
 S  101 .5
cs  L( s )  K  cS  L( S )
s  1882 .5e

s
25
 0.3s  7.5  15  101 .5  1882 .5e
Succesive approximat ion :
s  80 .5
The optimal policy :
q

101.5  x
if x 80.5
0
otherwise

S
25
 0. 3S  7. 5
Multiperiod models: No Setup Cost

Begin with two periods
Demand D1, D2, i.i.d
Density: ()
L(y) = expected one period holding+ shortage penalty cost;
strictly convex with linear cost and () >0,
c
purchase cost /unit
c1(x1) optimal cost with 1 period to go;
c+L’(S1)=0
while S1 is the optimal base stock level.
c1 ( x1 ) 

L ( x1 )
if x1  S1
c ( S1  x1 )  L ( S1 ) if x1  S1
c1 ( x1 )  c1 ( y 2  D2 )


L ( y 2  D2 )
if y 2  D2  S1
c ( S1  y 2  D2 )  L ( S1 ) if y 2  D2  S1

 c ( y   ) ( )d

L( y   ) ( )d  
E (c1 ( x1 )) 
1 2
0
y 2  S1
0
2

y 2  S1
[c( S1  y 2   )  L2 ( S1 )] ( )d
c2 ( x2 )  min c2 ( y 2  x2 )  L( y 2 )  E[c1 ( x1 )]
y 2  x2
 S 2 basestock level with 2 periods to go

which is convex

Example: c=10, h=10, p=15 the demand density is
 ( )  

1
10
0
if 0   10
otherwise
Solution:
p  c 15  10 1
 ( S1 ) 


p  h 15  10 5
S1
Since  ( S1 )  , S1  2
10
10 15 (  z )
z10 ( z   )
L( z ) 
d 
d
z
0
10
10


 75  15 z  (5 / 4) z 2
E[c1 ( x1 )] 







y2  2
1
[75  15 ( y 2   )  (5 / 4)( y 2   ) ] d
0
10
10
2 1
[10 (2  y 2   )  75  15 * 2  (5 / 4)2 ] d
y2  2
10
y2  2
2 1
[75  15 ( y 2   )  (5 / 4)( y 2   ) ] d
0
10
10
1
[70  10 ( y 2   )] d
y2  2
10
2
 ( y 2 ) 3 / 24  ( y 2 ) 2 / 4  19 y 2 / 2  359 / 3
c2 ( x2 )  min 10 ( y 2  x2 )  75  15 y 2  (5 / 4) y 22
y 2  x2
 ( y 2 ) 3 / 24  ( y 2 ) 2 / 4  19 y 2 / 2  359 / 3

Take derivative with respect to y 2 , setting it equal to zero
d {}
1
 [29 / 2  2 S 2  ( S 2 ) 2 ]  0
dy2
8
S 2  5.42
Substituting S 2  5 and S 2  6 into c 2 (x 2 ) leads to a smaller
value with S 2  5.
The optimal policy :
q2 

q1 

5  x2
0
if x2  5
otherwise
2  x1 if x1  2
0
otherwise
Multi-Period Dynamic Inventory Model with
no Setup Cost
Cn(xn): n periods to go,
:
discount factor.
DP equations:
cn ( xn )  min c( y n  xn )  L( y n )  E[cn 1 ( y n  Dn )]
y n  xn
c0 ( x0 )  0
Properties :
1) S1  S 2  S 3 .......... ... S n 1  S n .........  S , where
L' ( S )  c(1 -  )  0; (Infinite horizon optimal)

2) c( x)  min c( y  x)  L( y )   c( y   ) ( )d 

y x
0
satisfied by c( x)  lim cn ( x)
n 
3) lim S n  S
n 
Multi-Period Dynamic Inventory Model with
Setup Cost

cn ( xn )  min K ( y n  xn )  c( y n  xn )  L( y n )   cn 1 ( y n   ) ( )d
y n  xn
K ( y n  x n ) 

K
0

0
if y n  xn
if y n  xn
If L(y) is convex, then find Sn .
The optimal (s n , Sn )policy :
qn 

S n  xn
0
if xn  S n
if xn  S n
Multi-Period Dynamic Inventory Model
with Lead Times
Lead time: 






f n (u n )  min  K ( y n  u n )  c( y n  u n )  
L( y n   ) ( )d   f n 1 ( y n   ) ( )d 
y n u n 
0
0

u n  inventory position (on hand  on order to arrive in next  - 1 periods)

Can transform to 0 lead time as follows :

 L( y )  



L( y   )  ( )d
0
infinite horizon
  L' ( S )  c(1   )  0
