OPTIMAL POLICIES FOR A MULTIECHELON INVENTORY PROBLEM
ANDREW J. CLARK
AND
HERBERT SCARF
October 1959
Presented By İsmail Koca
What is the Paper About?
• Determining optimal purchasing quantities
separately for all levels of installations in a
multi-installation model where the model
consists of several installations, say 1,2,...,N
with installation 1 receiving stock from 2,
with 2 receiving stock from 3, etc.
Model Type 1
N
N-1
Model Type 2
2
1
DIFFERENCE FROM OTHER WORKS
Previous works:
• determine the optimal purchasing quantities
at a single installation
This work:
• considers multi-installation models’
optimal purchasing quantities where lead
time is also affected by the availability of
stock at the supplier installation.
APPROACH FOR SOLUTION
1. Define a cost function for each
configuration of stock at the various
installations, and in transit from one
installation to another
2. From defined cost function do a recursive
computation for the optimal provisioning
policy.
...
 in practice this computation of a sequence
of functions of at least N variables is
difficult and impractical
 this work simplifies this computation
without compromising the optimality of
the solution by several very plausible
assumptions
ASSUMPTIONS
• Demand originates in the system at the
lowest installation
• Purchasing, shipment costs are linear. No
setup cost except the last installation
...
• Holding and shortage costs are assumed to
be function of echelon stock which is the
stock at that level plus all all other stock in
the system which is actually at a lower level
or in transit to lower level.
• Each echelon backlogs excess demand
COST FUNCTION
c(z): cost of purchasing an amount of z
: lead time
(t):density function of demand (may differ
from period to period) (t: demand)
h: holding cost per unit
p:shortage cost per unit
x:stock on hand at the beginning of the period
Cost during the period (exclusive
purchase cost):


hx + p (t  x) (t )dt ,
L(x) =

p
x
(
t

x
)

(
t
)
dt
,

o
x>0
x0
Discounted cost function for n
periods
• x1: stock on hand
• wj :units to be delivered j periods in the
future
• : discount factor
Cn ( x1 , w1 , w2 ,...., w 1 )  Minc( z )  L( x1 )
z 0

   Cn 1 ( x1  w1  t , w2 ,...., w 1 , z ) (t )dt
0

• In the above function it is assumed that all
excess demand is backlogged until the
necessary stock becomes available
• Minimizing value of z is the optimal
purchase quantity for the given stock
configuration
An open form of the above formula is:

C n ( x1 , w1 , w2 ,...., w 1 )  L( x1 )    L( x1  w1  t ) (t )dt  ...
0

 1
 
 ... L(x1  w1  ...  w 1  t1  ...  t  1 ) (t1 )
0
0
...dt1 ...  f n ( x1  ...  w 1 )
and fn is:


0
0
f n (u )  Min c( y  u )     ... L( y  t1  ...  t  ) (t1 )
y u

... (t  )dt1 ...dt    f n 1 ( y  t ) (t )dt
0

order policy...
• The discounted cost is assumed to be
convex
– True if holding and shortage costs are linear
• Then there exists a sequence of critical
numbers (Sn , sn )
• Order if x1 +.....+w-1 < sn and with an
amount of Sn-(x1 +.....+w-1)
Example:
• Results are shown by means of an example
2 installations
Lead time: 2 periods
x1:stock on hand at installation 1
w1:stock to be delivered one period in the future
x2:echelon 2 stock
L(x1): one period costs at installation 1
~
L(x2): one period costs at echelon 2
Optimal policy for the lowest
installation
• For n>2 the discounted cost for the example
is:

C n ( x1 , w1 )  L( x1 )    L( x1  w1  t ) (t )dt  f n ( x1  w1 )
0
and fn(u) satisfy:
f n (u )  Min c( y  u )   2   L( y  t1  t 2 ) (t1 ) (t 2 ) dt1 dt 2
y u

   f n 1 ( y  t ) (t ) dt
0

~x
if x1 + w1 < n
• minimizing value is ~
xn
~
• if x1 + w1 < x n ordering occurs and the
minimum cost will be:
c1 ( ~
x n , u )   2   L( ~
x n  t1  t 2 ) (t1 ) (t 2 ) dt1 dt 2

   f n 1 ( ~
x n  t ) (t ) dt
0
~x
if x2< n
• we are only be to ship x2 – (x1 + w1) and
therefore the minimum cost will be:
c1 ( x 2 , u )   2   L( x 2  t1  t 2 ) (t1 ) (t 2 )dt1 dt 2

   f n 1 ( x 2  t ) (t )dt
0
...
• This cost is larger than the previous one. The
insufficiency of stock level at 2 caused
additional cost, which is one period loss to
be charged to this echelon. The difference is:
c1 ( x 2 , ~x 2 )   2   L( x 2  t1  t 2 )  L( ~x 2  t1  t 2 ) (t1 ) (t 2 )dt1 dt2

    f n 1 ( x 2  t )  f n 1 ( ~x 2  t ) (t )dt
0
~
x

• if 2 x 2
and zero if x 2  ~x 2
...
• By adding this one period loss to the second
echelon, the optimal policy is then
computed using the formula given.
• Also as the previous function is a convex
function of x2 the optimal policy for second
echelon is will be of type (S,s)
Optimality of the formula for
model type 1:
N
N-1
2
1
• There is a sequence of functions gn(x2), with
~
g1(x2) = L ( x 2 ) , such that
C n ( x1 , w1 , x 2 )  C n ( x1 , w1 )  g ( x 2 )
...
• where n(x2) is:


0
0
2

L( x 2  t  y )  L( ~
x n  t  y ) (t ) ( y )dtdy

    f n 1 ( x 2  t )  f n 1 ( ~
x 2  t ) (t )dt
0
~
for x 2  x n and zero for x 2  ~x n
• is a function of x2 alone
and…
g n ( x2 )  Min
z 0


~
c( z )  L ( x2 )  ( x2 )    g n1 ( x2  z  t ) (t )dt 
0
• The solution of this equation provides us
with the optimal policy for the entire system
Optimality of the formula for
model type 2:
A3
B2
A2
C1
B1
A1
•
The assumptions for the previous model is
retained for this model
There are 2 points:
1. It is not permitted to exchange stock
between any arbitrary pair of installations
as it does not run contrary to what is in
practice
2. If all requests cannot be satisfied because
of insufficient stock at a higher echelon,
how is the available stock to be rationed
among the requesting installations?
• After some iterations similar to the first
model we see that Cn (x1,x2,x3) cannot be
broken down in the form of
C n ( x1 , x 2 , x3 )  C ( x1 )  C ( x 2 )  g n ( x3 )
1
n
2
n
1
2
~
~
x3  x n  x n
• If this was the case at the end of the solution
of the minimization problem than there is so
far so good solution.
• The problem arises when the equation
above is as:
1
2
~
~
x3  x n  x n
…
• This is the case pointed out in Point 2
above. In order to allocate the insufficient
stock between the sub installations we
should solve the minimization problem for
the cost function.
• The solution of the minimization problem is
good when we lower installations are not
out of balance. And this is expected to occur
rather frequently
• So the approximation to is an
excellent for this model..
EXTENSIONS
• If the demand at echelons are dependent to
each other we use joint density function of
them in formulas instead of (t).
• This does not change the general structure
of the formulas, so the minimization for the
new function works similar to the model 1’s
solution. Then gives good solution.