4 COORDINATED REPLENISHMENTS


cyclic schedules
powers-of-two policies
4.1 Powers-of-two policies
Cycle times are restricted to be powers of two times a
certain basic period denoted as q . From (3.5) in
Chapter 3, we have:
C 1  Q Q* 
  * 
*
2Q
Q 
C
,
(4.1)
Let T = Q/d, T* = Q*/d
1  T T * 


T 
C* 2  T *
C
.
(4.2)
Lot sizing problem
T  2m q
(4.3)
Worst scenario:
Because the cost is convex, the worst possible error must
occur when two consecutive values of m, say m = k and m =
k + 1 give the same error. Let T < T* correspond to m = k, and
2T > T* to m = k + 1.
1  T T *  1  2T T * 
  *
  * 
*

2T
T  2T
2T 
C
C
T*
Setting
 x gives x  2 .
T
.
(4.4)
T * 2T
 *  2
T T
,

1 1
 
 2   1.06
*
2 2
C

(4.5)
C
.
(4.6)
Proposition 4.1 For a given basic period q, the maximum
relative cost increase of a powers-of-two policy is 6 percent.
C/C*
T
T2
T
T1
T*
2T2 2T 2T1
Note that T1 gives better cost than
T
T*
2
Note that 2T2 gives better cost than 2T  T * 2
Ti
1/ 2
2 
2
T*
Ti
1
1
xi
 2 ,   xi 
*
T
2
2
Ci
1 xi
1
1
 xi
 e( xi )  [2  2 ] ,   xi 
*
Ci
2
2
2
1 / 2
Suppose possible to change q
For a single item, the optimal solution is to
choose q equal to a power of two times T*.
For N items, we can, in general, not fit q
perfectly to all cycle times Ti*. The relative
cost increase can be expressed as:
N
C

*
C
 Ci
i 1
N
C
i 1


N

*
*
C
(
C
/
C
 i i i)
i 1
*
i
Ci*
*
C
 i
N
*
C
 i
i 1
 Ci

 C*
 i




N
 w e( x )
i 1
i
i
.
(4.7)
We know from (4.2) and (4.5) that for a given q, each C i / C *i
can be expressed as :
Ci
1 xi
xi
(4.8)

e
(
x
)

(
2

2
),
 1 / 2  x i  1 / 2,
i
*
2
Ci
Ti / Ti*  2 xi (1 / 2  x i  1 / 2) The weights wi for the different
values of xi , C*i / iN1 C*i can be seen as a probability distribution
F(x) on [-1/2, 1/2], i.e., F(-1/2) = 0 and F(1/2) = 1.
C
C
*

1/ 2
 e( x )dF( x )
1 / 2
.
(4.9)
Change q by multiplying by 2y , 0  y  1. This means that
a certain x is replaced by x + y for x + y ≤ ½, , and by x + y - 1
for x + y > ½.
Each Ti  2 Ti  2 q
xi
q
xi
2 Ti
mi
*
2
*
mi
( xi  mi )
Ti
*
2
y
Let q( y )  q 2 , 0  y  1, q(0)  1
Ti  2 q( y )  2 q 2  2
'
mi
mi
y
mi  y
xi
*
2 Ti
xi  y *
 2 Ti
mi
2
 Ti '  2 xi  y T *
Ti ( y )   T ' xi  y1
i
T*
 2  2
1
2
1
if xi  y 
2
if xi  y 
Ti ( y ) 
  xi  y1
*
2
Ti

2
C
C
*
xi  y
1
2
1
if xi  y 
2
if xi  y 
1 / 2 y
1/ 2
1 / 2
1 / 2 y
( y)   e( x  y)dF( x ) 
1/ 2
y 1 / 2
y 1 / 2
1 / 2
 e( x  y  1)dF( x )
  e(u )dF(u  y)   e(u )dF(u  y  1).
(4.10)
For a given distribution F(x) the minimum cost increase is
obtained by minimizing (4.10) with respect to 0 ≤ y ≤ 1.
Proposition 4.2 If we can change the basic period q, the
maximum relative cost increase of a powers-of-two
policy is 2 percent.
Proof
The average cost increase for 0 ≤ y ≤ 1 must be at least as large
as the minimum

min * ( y)     e(u )dF(u  y)   e(u )dF(u  y  1)  dy
0 y 1 C
0  y 1 / 2
1 / 2

C
1  1/ 2
y 1 / 2
1
 u 1 / 2

  e(u )  dF(u  y)   dF(u  y  1)  du
1 / 2
u 1 / 2
 0

1/ 2
1/ 2
1/ 2
1 / 2
1 / 2
  e(u ) F(1 / 2)  F(u )  F(u )  F(1 / 2)  du   e(u )du 
1
2 ln 2
 1.02
The worst case will occur when the distribution
F(x) is uniform on 1 / 2  x  1 / 2 , see (4.9).
A change of q will then not make any difference.
u
1
1/2
y
0
1
2
1
y =u+
2
u = y-
-1/2
1/ 2

y 1 / 2

u 1/ 2
0
e(u )dF (u  y )dy
e(u)dF (u  y)du
1
Given y
Given u
1 1/ 2

0 y 1 / 2
1/ 2

e(u )dF (u  y )dy

u 1 / 2
e(u )dF (u  y )du
1 / 2 0
1/ 2

1 / 2
1/ 2

1 / 2
e(u ) 
u 1 / 2
0
dF (u  y )du
u
1
2
e(u )[ F (u  y )]0 du
1
  e(u )[ F (u )  F ( )] du
1 / 2
2
1/ 2
1/ 2

1 / 2
e(u )F (u )du
1
Note F ( )  0
2
u
1
1/2
1
2
1
yu
2
u  y
0
y
-1/2

y 1/ 2
1/ 2
1

1
u
2
e(u)dF (u  y  1)dy
e(u )dF (u  y  1)du
Given y
Given u

y 1 / 2

y 1 / 2 1
1
0 1 / 2

1 / 2
1/ 2

1 / 2

1/ 2

u 1 / 2
1 / 2
e(u )dF (u  y  1)du
1
e(u ) 
u 1 / 2
y 1 / 2
1 / 2
e(u )dF (u  y  1)dy
dF (u  y  1)du
1
e(u )[ F (u )  F ( )] du
2
e(u )[1  F (u )]du
1
Note F ( )  1
2
4.2 Production smoothing
Mixed integer program (MIP)
Manne (1958)
Billington et al. (1983)
Eppen and Martin (1987)
Shapiro (1993)
Objective: Low inventory cost and smooth capacity
utilization.
Simplification: ignore stochastic variations
ignore time-varying aspect
MIP not used in practice
When the demands for different items are relatively
stable, use cyclic schedules.
In a general case with many items and several
production facilities, it can be extremely difficult to
find suitable cyclic schedules.
It is also common in practice to smooth production
outside the inventory control system.
Each item ordered periodically. Ordering period
chosen to smoothen the load.
Example: 4 items, same demand
Item 1: 1,5,9 Item 2: 2,6,10
Item 3: 3,7,11 Item 4: 4,8,12
Periodic review: order up to S policy
Continuous review: large variation in production
load resulting in long and uncertain lead time.
4.2.1 The Economic Lot Scheduling Problem
(ELSP)
Cyclic schedules for a number of items with
constant demands.
Backorders not allowed.
Finite production rate
Single production facility
Notation:
N = number of items,
hi = holding cost per unit and time unit for item i,
Ai = ordering or setup cost for item i,
di = demand per time unit,
pi = production rate (pi > di),
si = setup time in the production facility for item i,
independent of the sequence of the items,
Ti = cycle time for item i (the batch quantity Qi = Tidi).
Define:
i = di/pi ,
i = i Ti = production time per batch for item i
excluding setup time,
i = si + i = total production time per batch for item i.
Table 4.1 N = 10
Bomberger (1966) Table 4.1 Bomberger’s
problem (time unit = one day).
Ai
Ti
Ci 
 h i d i (1   i )
Ti
2
. (4.11)
The problem is to minimize iN1 C i subject to the constraint
that all items should be produced in the common production
facility.
The optimal cycle time when disregarding
the capacity constraint is
Ti 
2A i
h i d i (1   i )
,
C i  2A i h i d i (1   i )
(4.12)
.
(4.13)
Note that (4.11) and (4.12) are equivalent to (3.6) and
(3.7) if we replace Ti by Qi/di , and i by di/pi .
Table 4.2 Independent solution of Bomberger’s
problem.
1
2
3
4
5
6
7
8
9
10
Ti
167.5
37.7
39.3
19.5
49.7
106.6
204.3
20.5
61.5
39.3
Ci
0.179
1.060
1.528
1.024
4.428
0.938
3.034
12.668
6.506
0.255
I
2.36
2.01
3.56
4.29
2.49
1.67
3.04
5.87
11.20
1.17
Item
C  10
i 1 C i  31 .62
, is a lower bound for the total costs.

Solution not feasible: Consider items
4, 8 and 9.
9.30
Item 9
Item 9
11.20
t
t+20.5
t+61.5
Item 4 & 8 have cycle times 19.5 and 20.5. Must
be able to produce one batch of #4 and #8 in
[t,t+20.5], or [t+11.20,t+20.5]. But the length of
available time is 9.30, while 4+8 =10.16 >9.30.
 Not feasible
Does a feasible solution exist? If at least one setup time
is positive an obvious necessary condition for a feasible
solution to exist is
N
 i  1
i 1
.
(4.14)
Condition (4.14) is also sufficient for feasibility.
Given the assumption of a common cycle time, the problem
now is to minimize
T
 Ai
C  
 h i d i (1   i ) 
2
i 1  T
N
,
(4.15)
with respect to the constraint that the common cycle
must be able to accommodate production lots of all
items
N
N
  i   (s i   i T)  T
i 1
i 1
.
(4.16)
N
T
s
i 1
N
i
1   i
 Tmin
i 1
,
(4.17)
a lower bound for the cycle time.
Need large enough T to squeeze setups in the slack
N
i
1- 
i 1
Disregard (4.17), from (4.15)
N
Tˆ 
2 Ai
i 1
N
 h d (1   )
i
i
i
. (4.18)
Since (4.15) is convex in T the optimal solution,
i 1
Topt  max (T̂, Tmin )
,
(4.19)
For Bomberger’s problem, Tmin = 31.86, and consequently,
Topt  Tˆ  42.75 , T̂  42 .75 . cost=41.17.
For problems where the individual cycle times are reasonably
similar, the common cycle approach gives a very good
approximation.  Ci*  31.62  C*  41.17
Two approaches for deriving better
solutions.
I.
Dynamic programming model
Bomberger (1966)
Assuming
Ti = niW
W should be able to accommodate production of
all items.
Fi(w) = minimum cost of producing items
i +1,i+2,..., N when the vailable capacity in
the basic period is w, i.e., W - w has been used for
items 1, 2,...,i.
Fi 1 ( w )  min C i (n i W)  Fi ( w   i )
, (4.20)
where Ci(niW) are the costs (4.11) for item i with Ti = niW,
i = si + iniW, and the integer ni is subject to the constraint
ni
1  n i  (w  s i ) / i W
.
(4.21)
or w  si  i niW
Note that the upper bound in (4.21) is equivalent to i  w.
FN(w) = 0 for all w  0. F0(W) gives the minimum costs when
the basic period is equal to W.

Minimize over W. Bomberger’s solution
C=36.65, W=40, ni=1 for i 7, n7 =3.

Serious Assumption: W should be able to
accommodate production of all items.
II. Heuristic
Doll and Whybark, 1973).
The procedure is to successively improve the
multipliers ni and the basic period W according to
the following iterative procedure:
1. Determine the independent solution and use the
shortest cycle time as the initial basic period W.
2. Given W, choose powers-of-two multipliers,
(ni = 2mi, mi  0), to minimize the item costs (4.11).
3. Given the multipliers ni , minimize the total costs
W
 Ai / ni
C  
 h i d i (1   i )n i

W
2 
i 1 
N
with respect to W.
N
W
2 Ai / n i
i 1
N
 h i d i (1   i )n i
i 1
4. Go back to Step 2 unless the procedure has converged. In
that case, check whether the obtained solution is feasible. If
the solution is infeasible, try to adjust the multipliers and then
go back to Step 3.
Compare with independent solution.
Apply the heuristic to Bomberger’s problem.
Table 4.3 Solution of Bomberger’s problem with
W = 23.42.
Item
1
2
3
4
5
6
7
8
9
10
ni
8
2
2
1
2
4
8
1
2
2
i
2.62
2.47
4.19
5.12
2.37
1.50
2.87
6.63
8.71
1.37
Table 4.4 Feasible production plan.
Items
Production
time
1
4, 8, 2, 9
22.93
2
4, 8, 3, 5, 10, 1
22.30
3
4, 8, 2, 9
22.93
4
4, 8, 3, 5, 10, 6
21.18
5
4, 8, 2, 9
22.93
6
4, 8, 3, 5, 10, 7
22.55
7
4, 8, 2, 9
22.93
8
4, 8, 3, 5, 10, 6
21.18
Basic period
In case of stochastic demand, one possible
approach is to first solve a deterministic
problem based on averages, and then try to
adapt the solution to the stochastic case by
adding suitable safety stocks.
4.2.2 Production smoothing and batch quantities
Adjust the batch quantities to obtain a
reasonably smooth load.
Karmarkar (1987, 1993). Axsäter (1980, 1986),
Bertrand (1985), and Zipkin (1986).
Consider a machine in a large multi-center shop.
D = average output of material (demand), units
per time unit,
P = average processing rate, units per time unit,
Q = batch quantity,
t = setup time,
T = average time in the system for a batch,
h = holding cost per unit and time unit after
processing.
Assume:
The batches arrive at the machine as a Poisson process
with rate l = D/Q. Thus Av demand=lQ=D.
The processing time is exponentially distributed.
average processing time for a batch is 1/k = t + Q/P.
Service rate = k. = l /k = Dt/Q + D/P.
The average time spent in the M/M/1 system is
1/ k
t  Q/P
T

1   1  Dt / Q  D / P
. (4.22)
The average cycle stock is approximated as Q/2.
Assume that the average holding cost per unit and time
unit for work-in-process is exactly half of the holding
cost h after the process. Av cycle stock = Q/2. Total holding
cost after the process=hQ/2.
Work-in-process TD=Av time in the system * Av demand

h
h  Dt  DQ / P
 Q 
min (TD  Q)  min 
Q0 2
Q0 2  1  Dt / Q  D / P

2Dt
Q 
1 D/ P
.(4.23)
*
.
(4.24)
For low values of D, Q* is essentially linear in D. For larger
values, Q* grows very rapidly.
4.3 Joint replenishments
4.3.1 A deterministic model
Setup costs:
Individual setup costs for each item, and a joint
setup cost for the whole group of items.
Reason: joint setup costs, quantity discounts,
coordinated transports.
constant continuous demand. No backorders.
batch quantities are constant. production time is
disregarded. No lead time or the lead time is same
for all items.
Notation:
N = number of items,
hi = holding cost per unit and time unit for item i,
A = setup cost for the group,
ai = setup cost for item i,
di = demand per time unit for item i,
Ti = cycle time for item i.
 i = h id i
Assume all demands equal to one. Items are
ordered so that a1/1  a2/ 2  ...  aN/N .
Note that increasing setup costs and decreasing
holding costs mean increasing lot sizes and cycle
times.
Approach 1. An iterative technique
If there were no joint setup cost
2a i
Ti 
i
, (4.25)
i.e., T1 would be the smallest cycle time. Assume other cycle
times of items 2, 3... N are integer multiples ni of the cycle time
for item 1,
Ti  n i T1
, i = 2, 3,..., N. (4.26)
Our objective is to minimize w.r.t T1, n2, n3, ... nN the cost.
N
C
A  a1   a i / n i
i 2
T1
Fix cost
N

T1 (1   i n i )
i 2
2
ith item holding cost=Tinii/2
,(4.27)
Given n2, n3... nN,
N
T1* (n 2 , n 3 ,..., n N ) 
2(A  a 1   a i / n i )
i2
, (4.28)
N
1    i n i
i2
N
N
i2
i2
C (n 2 , n 3 ,..., n N )  2(A  a 1   a i / n i )( 1    i n i )
*
. (4.29)
Note that T1 is not chosen according to (4.25). If we disregard
n2, n3... nN to be integers, then from (4.29),
ai
1
ni 
i (A  a 1 )
. (4.30)
From (4.30) and (4.29), the lower bound for the costs:
N
C  2(A  a 1 )1   2a i i
i 2
.(4.31)
HEURISTIC
1. Determine start values of n2, n3... nN by rounding
(4.30) to the closest positive integers.
2. Determine the corresponding T1 from (4.28).
3. Given T1, minimize (4.27) with respect to n2, n3... nN.
This means that we are choosing ni as the positive
integer satisfying
2ai
ni (ni  1) 
 ni (ni  1)
2
iT1
2ai
dC
2
Note :
 0 gives ni 
; C is convex.
2
dni
iT1
. (4.32)
Return to Step 2, if any multiplier ni has changed since the last
iteration.
Example 4.1
N = 4 , A = 300, a1 = a2 = a3 = a4 = 50, h1 = h2 = h3
= h4 = 10, d1 = 5000, d2 = 1000, d3 = 700, and
d4 = 100. As requested, ai/i is nondecreasing with
i. When applying the heuristic we obtain
50
5000  10
n2 
 1, n 3  1, n 4  3
1000  10 350
2  (350  50  50  50 / 3)
T1 
 0.1155
10  5000  10  1000  10  700  10  100  3
again n2 = 1, n3 = 1, n4 = 3, i.e., the algorithm has already
converged.
Approach 2. Roundy´s 98 percent approximation
the joint setups have cycle time T0  0,
Ti  2 ki T0 ,
i = 1, 2, ..., N, (4.33)
where ki is a nonnegative integer. Let a0 = A, and 0 = 0,
1 
 Ti
min    i
 ai

T0 ,T1 ,..., TN i  0 
2
Ti 
, (4.34)
subject to the constraints (4.33). replacing (4.33) by
N
Ti  T0
,
i = 1, 2, ..., N.
(4.35)
The resulting solution will give a lower bound for the
costs. I lagrangian relaxation
1 N
 Ti
   i  a i    l i (T0  Ti )
2
Ti  i 1
i 0 
N
max
min
l1 ,l 2 ,..., l N T0 ,T1 ,..., TN
,
where li are nonnegative.
define
N
0   0  2  l i
(4.36)
i 1
1 = 1 - 2l1,
.
.
.
N = - 2l .
N
N
(4.37)
the optimal solution must have all i  0. The
optimal solution of the relaxed problem can be
obtained by solving N + 1 independent classical
lot sizing problems.
1 
 Ti
 ai

  i
2
Ti 
i 0 
N
min
T0 ,T1 ,..., TN
(4.38)
without any constraints on the cycle times.
rounding the cycle times of the relaxed problem,
Ti  2 mi q
(4.39)
for some number q > 0.
From Proposition 4.1, if q is given, the
maximum cost increase is at most 6 percent. If
we adjust q to get a better approximation the
cost increase is at most 2 percent according to
Proposition 4.2. We have now obtained
Roundy’s solution of the problem.
I. A simpler technique instead of Lagrangian
Relaxation
From (4.34) and (4.35) the optimal cycle times in the
relaxed problem are nondecreasing with i,
Ti  Ti1
,
i = 1, 2, ..., N.
(4.40)
Since 0 = 0 we will always aggregate items 0 and 1. After
aggregation we have an item with cost parameters A + a1 and
1. Next we check whether a2/2 < (A + a1)/ 1. If this is the
case the aggregate item should include also item 2, etc. When
no more aggregations are possible we can optimize the
resulting aggregate items individually.
Example 4.2
N = 4, A = 300, a1 = a2 = a3 = a4 = 50, h1 =
50000, h2 = 10000, h3 = 7000, and h4 = 1000.
Both of the approaches considered assume
constant demand, but can also be used in case
of stochastic demand.
4.3.2 A stochastic model
independent, stationary, integer demand,
complete backordering.
N = number of items,
hi = holding cost per unit and time unit for item i,
b1, i = shortage cost per unit and time unit for
item i,
A = setup cost for the group,
ai = setup cost for item i,
Li = constant lead-time for item i.
Viswanathan (1997)
First step: disregard the joint setup cost and consider
the items individually for a suitable grid of review
periods T. For each review period, determine the
optimal individual (s, S) policies for all items and the
corresponding average costs.
Ci(T) = average costs per time unit for item i when
using the optimal individual (s, S) policy with a review
interval of T time units.
Second step: determine the review period T
by minimizing,
N
C(T)  A / T   C i (T)
i 1
(4.41)
Note that the actual costs are lower than the
costs according to (4.41), since the major
setup cost A is not incurred at reviews where
none of the items are ordered.
can-order policies.
(Si, Q) policy.
5 MULTI-ECHELON SYSTEMS
5.1 Inventory systems in distribution and production
A
B
Distribution: Warehouse
Store
Production: Subassembly
Final product
Figure 5.1 An inventory system with two coupled inventories.
Reduces the length and uncertainty of lead times.
5.1.1 Distribution inventory systems
Distribution system or arborescent system
Serial system
Central warehouse
Retailers
Figure 5.2 Distribution inventory system.
5.1.2 Production inventory systems
convergent flow
Figure 5.3 An assembly system.
It is, in general, considerably easier to deal with serial
systems than with other types of multi-echelon systems.
Raw material are low volume items. Higher setup cost at
earlier stages, large bathes.
4
2
7
1
5
8
6
3
Figure 5.4 A general multi-echelon inventory system.
Figure 5.5 Bill of material corresponding to the inventory
system in Figure 5.4.
5.2 Different ordering systems
5.2.1 Installation stock reorder point policies
(R, Q) policies; special case R=S-1, Q=1(S
policy with discrete units.)
Installation stock (R, Q) policy (on hand +
outstanding orders - backorders)
(s, S) policy is not the optimal policy for a multiechelon system
KANBAN policy
Smoothing aspect and local decentralized
controls.
Example 5.1
Table 5.1 Installation and echelon stock inventory
positions in Figure 5.6.
Item
Installation stock inventory
position
Echelon stock
inventory position
1
2
3
4
5
2
3
5
5
7
3
28
echelon stock reorder point policy will generally use
larger reorder points than an installation stock policy
to achieve similar control.
5 + 2*5 + 2*2 + 3*3=28
5 + 2*7 +3*3=28
5.2.3 Comparison of installation and echelon stock
policies
.......
N
3
2
1
Final product
Raw material
Increasing batch sizes
Figure 5.7 Serial inventory system with N installations.
Qn = batch quantity at installation n.
Q n  j n Q n 1
, (5.1)
where jn is a positive integer.
Notation:
IPni = installation inventory position at installation n,
IPne= IPni  IPni 1  ...  IP1i
= echelon stock inventory position at installation n,
R in= installation stock reorder point at installation n,
R en =echelon stock reorder point at installation n.
initial inventory positions IPni 0and
R in  IPni0  R in  Q n ,
IPne 0satisfying
R en  IPne0  R en  Q n
(5.2)
An installation stock policy is always nested. Installation n
may order only if (n-1) has just ordered. Echelon IP at n is
only changed by final demand at 1 and replenishment order at
n.
Assuming:
IPni 0  R in is
an integer multiple of Qn-1.
All demands at installation n are for multiples of Qn-1
all replenishments are also multiples of Qn-1
Proposition 5.1 An installation stock reorder point
policy can always be replaced by an equivalent echelon stock
reorder point policy.
Proof When n orders, it means 1,2,…,n-1 has ordered. Then
IPne
n
  (R ik  Q k )
.
(5.3)
k 1
Unit demand. Orders would be triggered exactly at the same
time. When (n-1) orders Qn-1, its inventory reaches the level
Rn-1+Qn-1. At this point, level n inventory goes from Rn+Qn-1 to
Rn. And then n orders.
R en

R in
n 1
  (R ik  Q k )
k 1
.
(5.4)
If IPni 0  R in is not a multiple of Qn-1, change R in by an amount
i0
i
less than Qn-1 .So that IPn  R n is a multiple of Qn-1.
Proposition 5.2 An echelon stock reorder point policy
which is nested can always be replaced by an equivalent
installation stock reorder point policy.
Proof
For installation 1, R1i  R1e
For installation n > 1, immediately after ordering
IPni  IPne  IPne1  Rne  Qn  Rne1  Qn 1
(5.5)
Therefore,
R1i  R1e , Rni  IPni  Qn  Rne  Rne1  Qn 1 n  1
(5.6)

Example 5.2 Consider a serial system with N = 3
installations and batch quantities Q1 = 5, Q2 = 10,
and Q3 = 20. Assume that the initial inventory
positions are , , and . Assume furthermore that the
demand for item 1 is one unit per time unit.
Consider first an installation stock policy with
reorder points , , and Note that our assumptions
that Qn as well as are multiples of Qn-1 are
satisfied. It is easy to check that installation 1 will
order at times 5, 10, 15,..... The demand at
installation 2 is consequently 5 units at each of
these times. This means that installation 2 will
order at times 10, 20, 30,.... Demands for 10 units
at installation 3 at these times will trigger orders at
times 20, 40, 60,.....

Using (5.4) we obtain the equivalent echelon
stock policy as , , and Recall that when
considering the echelon stock the inventory
positions at all installations are reduced by the
final demand, i.e., by one unit each time unit,
and not by the internal system orders. The initial
inventory positions are , , and . It is easy to verify
that the orders will be triggered at the same
times as with the installation stock policy.

Assume then that we change the echelon
stock reorder point at installation 3 to This
will not change the orders at installations 1
and 2, but the orders at installation 3 will be
triggered 2 time units earlier, i.e., at times 18,
38, 58,.... Recall that the echelon stock is
reduced by one unit at a time. The resulting
echelon stock policy is not nested and it is
impossible to get the same control by an
installation stock policy.
Propositions 5.1 and 5.2 show that in any serial
inventory system an installation stock policy is
simply a special case of an echelon stock policy.
This is also true for assembly systems.
Installation policy: Only local information needed.
The higher generality of an echelon stock policy
can be an advantage in certain situations.
Example 5.3
a serial system (Figure 5.7), N = 2, Q1 = 50, Q2 =100.
final demand at installation 1 =50. lead-time at
installation 1 is one, at installation 2 is 0.5. No shortages
allowed, holding costs at installation 1 are higher than at
installation 2.
LT=0.5, R2e =25+50=75,
IP20(0.5)=0.
50
0
1
2
3
4
Time
Stock on hand at installation 2
0
LT=1, R1e=50, IP1i(0.5)=75.
50
1
2
3
4
5
Time
Stock on hand at installation 1
Figure 5.8 Inventory development in the optimal solution.
Echelon Policy is more general. At t=0.5- , IP2i=0,IP1i=25.
The dominance of echelon stock policies for
serial and assembly systems does not carry
over to distribution systems.
5.2.4 Material Requirements Planning
periodic review, rolling horizon
Master Production Schedule (MPS).
External demands of other items
A bill of material for each item specifying all of its
immediate components and their numbers per unit
of the parent.
Inventory status for all items
Constant lead-times for all items.
Rules for safety stocks and batch quantities.
MPS. A production or program of final products.
Must cover total system lead time.
1
2(1)
3(1)
Figure 5.9 Considered product structure.
Table 5.2 Net requirements of item 1.
Item 1
Period
1
Lead-time = 1
Gross
requirements
10
Order quantity = 25
Scheduled receipts
Safety stock
=5
Projected
inventory
Net requirements
2
3
4
5
6
25 10
20
5
2
-18
-23
3
20
5
7
8
10
25
22 12 37 12
-23
-33
10
Table 5.4 MRP record for item 1 with planned orders in
periods 3 and 5.
Item 1
Period
1
Lead-time = 1
Gross requirements
10
Order quantity = 25
Scheduled receipts
Safety stock
=5
Projected inventory
Planned orders
2
3
4
5
6
7
25 10 20 5
8
10
25
22 12 37 12 27 7 27 27 17
25
25
reorder point = the lead-time demand plus the safety stock
minus one
Figure 5.10
Material requirements
planning for items 1,
2, and 3 in Figure 5.9.
safety time
Table 5.5 MRP record for item 1 with a safety time.
Item 1
Period
1
Lead-time = 1
Gross requirements
10
Order quantity = 25
Scheduled receipts
Safety stock
=5
Projected
inventory
Safety time
=1
Planned orders
2
3
4
5
6
25 10 20
5
7
8
10
25
22 12 37 37 27 32 27 27 17
25
25
Figure 5.11 Product structures.
Figure 5.12 Gross requirements from different sources.
MRP is often referred to as a push system since
orders are in a sense triggered in anticipation of
future needs. Reorder point systems and KANBAN
systems are similarly said to be pull systems
because orders are triggered when downstream
installations need them.
The MRP logic is simple. Yet the computational
effort can be very large if there are thousands of
items and complex multi-level product
structures.
“nervousness”
DRP, Distribution Requirements Planning.
Manufacturing Resource Planning- MRP II
Rough Cut Capacity Planning (RCCP)
Capacity Requirements Planning (CRP)
Enterprise Resource Planning (ERP)
Figure 5.13 Manufacturing Resource Planning.
5.2.5 Ordering system dynamics
bullwhip effect, Forrester (1961)
decentralized installation stock policies in multiechelon systems can yield very large demand
variations early in the material flow, even though the
final demand is very stable.
Note: both the information delays and the problems
of large demand variations at upstream facilities are
largely due to long lead-times and large batch
quantities.
5.3 Order quantities
Assumptions:
demand is known and deterministic. In case of
stochastic demand, replace the stochastic
demand by its mean and use a deterministic
model when determining batch quantities.
all lead-times are zero.
5.3.1 A simple serial system with constant demand
2
1
Figure 5.14 A simple serial two-echelon inventory system.
Example 5.4
item 1 is produced from one unit of the component 2.
d = 8, A1 = 20, A2 = 80, h1 = 5, h2 = 4.
I.Treat the two installations independently
C1  h 1
Q1 
Q1
d
 A1
2
Q1
2A1d
8
h1
.
(5.7)
,
C1  2A1dh1  40
Q 2  k Q1
where k is a positive integer.
.
, (5.8)
Figure 5.15 illustrates the behavior of the inventory
levels for k = 3.
Echelon stock, item 2
Installation stock, item 2
Time
Installation stock = echelon stock, item 1
Time
Figure 5.15 Inventory levels for k = 3.
(k  1)Q1
d
C2  h 2
 A2
2
k Q1
1
k 
Q1
*
,
(5.9)
2A 2 d
h2
If k* < 1 it is optimal to choose k = 1. If k* > 1. Let k´ be the
largest integer less or equal to k*, i.e., k´  k* < k´ + 1, it is
optimal to choose k = k´ if k*/k´  (k´ + 1)/k*, otherwise
k = k´+ 1.
k*  2.24, k = 2. C2 = 56, C = C1 + C2 = 40 + 56 = 96.
C  C1  C2  (h1  (k  1)h 2 )
Q1
A
d
 (A1  2 )
2
k Q1
. (5.10)
Alternatively, use the echelon holding costs e1 = h1 - h2,
and e2 = h2,
C1e  e1
C e2
Q1
d
 A1
2
Q1
kQ 1
d
 e2
 A2
2
kQ 1
C  C1e
 C e2
,
(5.11)
.
(5.12)
Q1
A2 d
 (e1  k e 2 )
 ( A1 
)
2
k Q1
(5.10) and (5.13) are equivalent.
, (5.13)
Q1 
A2
)d
k
e1  k e 2
2(A1 
.
A2
C(k )  2(A1 
) d ( e1  k e 2 )
k
(5.14)
.
A 2 e1
C (k)  2d(A1e1  A 2 e 2  A1e 2 k 
)
k
2
k* 
A 2 e1
A1e 2
(5.15)
(5.16)
.
(5.17)
If k* < 1 it is optimal to choose k = 1. If k* > 1. Let k´ be the
largest integer less or equal to k*, i.e., k´  k* < k´ + 1, it is
optimal to choose k = k´ if k*/k´  (k´ + 1)/k*, otherwise k =
k´+ 1.
Example 5.5
d = 8, A1 = 20, A2 = 80, h1 = 5, h2 = 4.
e1 = h1 - h2 = 1, e2 = h2 = 4. From (5.17), k* = 1.
Applying (5.14) and (5.15), Q1*  17 .89 , Q *2  kQ 1*  Q1*  17 .89
and C*  89.44, about 7 percent lower than the costs obtained
in Example 5.4.
(5.14) and (5.15) are essentially equivalent to the
corresponding expressions (3.3) and (3.4) for the classical
economic order quantity model.
A1 = A1 + A2/k ,
(5.18)
h 1 = e1 + ke2.
(5.19)
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 s
 min c( y  x)  L( y )


y
( 2)
L(y) convex, L’(0) < -c (otherwise never order)
L′ eventually becomes positive
L' ( S )  c  0
( 4)
Base Stock Policy
y * ( x)  max{ x, S }
q ( x) 
*

S  x,
0,
If x  S
otherwise
c  (h  s ) ( y )  ( pp s )( 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 y-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
f ( x )  c ( S  x )  L( S )  

 f (S   ) ( )d
(11)
0
It is possible to show that
f’(x)=-c for x ≤ S
(12)
which reduce to
L' ( S )  c(1   )  0 B.L
similarly for N.B.L
(13)
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),
N .B.L : ( S ) 
cu
pc

( p  c)  h  c(1   ) cu  co
cu
p  c(1   )
B.L : ( S ) 

p  c(1   )  h  c(1   ) cu  co
(20 a)
(20b)
Remark :
Lead time, Setup cost  more complicate d, still (s, S) policy
Example 4:
cu  20 , co  5
cu
20
4
( S ) 


cu  co 5  20 5
S  1800
Example 5:
1
 ( )  e
25
L( y ) 

y


25 , K
3
3 2
 15, c  1, h( z ) 
z, z  0, p(z)  z , z  0
10
2
h( y   ) ( )d 
0

3 1
10 25

 p( y   ) ( )d
y

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
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’(y10)=0
while y10 is the optimal base stock level.
L( x )
if
c1 ( x1 )   10
0
c
(
y

x
)

L
(
y
)
1
1
1

c1 ( x1 )  c1 ( y 2  D2 )


x1  y10
if x1  y10
L ( y 2  D2 )
if y 2  D2  y10
c ( y10  y 2  D2 )  L ( y10 ) if y 2  D2  y10
E (c1 ( x1 )) 


 c (y
0

1
y 2  y10
0
2
  ) ( )d
L( y 2   ) ( )d 


0
0
[
c
(
y

y


)

L
(
y
1
2
2 1 )] ( ) d
0
y 2  y1
c2 ( x2 )  min c2 ( y 2  x2 )  L( y 2 )  E[c1 ( x1 )]
y 2  x2
 y 02 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:
 ( y10 )
p  c 15  10 1



p  h 15  10 5
0
y
Since  ( y10 )  1 , y10  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 0 2
0
 [29 / 2  2 y 2  ( y 2 ) ]  0
dy2
8
y 20  5.42
Substituting y 02  5 and y 02  6 into c 2 (x 2 ) leads to a smaller
value with y 02  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;

2) c( x)  min c( y  x)  L( y )   c( y   ) D ( )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: l

f n (u n )  min  K ( y n  u n )  c( y n  u n )   l
yn u n 
u n  inventory position
can transform to 0 lead time as follows :
l
  ( y)  
l


0
L( y   ) Dl ( )d
infinite horizon
 l ' ( S )  c(1   )  0


0
L( y n   ) Dl ( )d



f n 1 ( y n   ) D ( )d 


0