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
C
*

1
2
Q
Q
*

Q
*
Q
,
Let T = Q/d, T * = Q * /d
(4.1)
C
C
*

1
2
T
T
*

T
*
T .
(4.2)


m
(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
2 T > T * to m = k + 1.
C
C
*

1
2
T
T
*

T
*
T

1
2
2 T
T
*
T
*

2 T
Setting
T
T
*
 x gives x

2 .
. (4.4)

T
*
T

2 T
T
*

C
C
*

1
2


2
1
2

2



1 .
06
, (4.5)
. (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
2
T T
1
T* 2T
2
2T 2T
1
T

Note that T
1
Note that 2 T
2 gives better cost than gives better cost than
T

T *
2
2 T

T * 2

2

1 / 2 
T
T i
*

2
1 / 2
T i
T
*
C i
C i
*

2 x i
 e ( x i
,
)


1
2

1
2
[ 2 x i x i

1
2

2
 x i ] ,

1
2
 x i

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 T i
*.
The relative cost increase can be expressed as:

C
C
*
 
 i
N 

1 i
N 

1
C i
C i
*
C i
*

C i
*
 i
N 

1
C i
*
( C i i
N 

1
C i
*
/ C i
*
)


C i
C i
*


 i
N 

1 w i e ( x 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 :
C i  e ( x i
)

1
2
( 2 x i

2
 x i ),

1 / 2
 x i

1 / 2 , (4.8)
C i
*
T i
/ T i
* 
2 x i values of x i
(

1
C i
*
/ 2

 i
N

1 x
C i i
*

1 / 2 ) The weights w i for the different
, 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

1

1
/
/

2 e (
2 x ) dF ( x )
C
*
.
(4.9)
Change q by multiplying by 2 y , 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 T i

2 x i T i
* 
2 m i q q

Let
2 x i T i
*
2 m i q ( y )


2
( x i q 2 y
,
 m i
)
T i
*
0
 y

1 , q ( 0 )

1
T i
' 
2 m i q ( y )

2 m i q 2 y 
2 m i
 y
2 x i T i
*
2 m i

2 x i
 y
T i
*

T i
( y )
 T i
' 
2 x i
 y
T *
T i
'

2 x i
 y

1
2
T * if if x i
 y

1 x i
 y

2
1
2
T i
C
C
*
T i
(
(
* y y
)
)





2
2 x i x i


1 /

1
2

/
 e
2 y
( x
 y y

1 y ) dF ( x )

1 /
1
2

/ 2
 e ( y x if if x i
 y

1
1
2 x i
 y

2
 y

1 ) dF ( x )
 y
1

1

/ 2
/ e (
2 u ) dF ( u
 y )
 y

1


1 /
/ 2 e (
2 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
C
0 min
 y

1
C
*
( y )
 
1
0



 y
1

1

/ 2
/ e (
2 u ) dF ( u
 y )
 y

1


1 /
/ e
2
2
( u ) dF ( u
 y

1 )



 dy

1

1

/
/
2
2 e ( u ) 

 u

1

/ 2 dF ( u
0
 y )
 u

1

1
/ dF ( u
2
 y

1 )


 du

1

1 /

/ 2
2 e ( u )


F (

1 / 2 )

F ( u )

F ( u )

F ( 1 / 2 )
 du

1

1

/
/
2
2 e ( u ) du

1
2 ln 2

1 .
02

The worst case will occur when the distribution
F ( x 2
 
1
A change of q will then not make any difference.

1 u
1/2
0 u = y -
1
2 y = u +
1
2
-1/2
 1 y

1
/ 2
/ 2 e ( u ) dF ( u
 y ) dy

0 u

1 / 2 e ( u ) dF ( u
 y ) du
1 y
Given y
Given u

0
1   1 y

1
/ 2
/ 2 e ( u ) dF ( u
 y ) dy


1

1
/
/
2  
2 0 u

1 / 2
 1

1
/
/
2
2 e ( u )

0 u

1 / 2 e ( u ) dF ( u
 dF ( u
 y ) du y ) du
  1

1
/
/
2
2 e ( u ) [

F ( u
 y )] u
0

1
2 du
  1

1
/
/
2
2 e ( u ) [ F ( u )

F (

1
2
)] du
  1

1
/
/
2
2 e ( u ) F ( u ) du
Note F (

1
2
)

0

1/2
0
1 u u
 y

1
2 y
 u

1
2
-1/2


1 / y

1
2
/ 2 e ( u ) dF ( u
 y

1 ) dy
 1 u

1
2 e ( u ) dF ( u
 y

1 ) du y
Given y
Given u


1  
0

1 y

1
/ 2
/ 2 e ( u ) dF ( u

1

1 y
/ 2
/ 2   u
1

1 / 2
 y

1 ) dy e ( u ) dF ( u
 y

1 ) du
  1

1
/
/
2
2 e ( u )
 u
1

1 / 2 dF ( u
 y

1 ) du
 

1 y

1
/ 2
/ 2 e ( u ) [

F ( u )

F (
1
2
)] du
  1

1
/
/
2
2 e ( u ) [ 1

F ( u )] du
Note F (
1
2
)

1

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, h i
= holding cost per unit and time unit for item i
A i
= ordering or setup cost for item i , d i
= demand per time unit, p i s i
= production rate ( p i
> d i
),
= setup time in the production facility for item i, independent of the sequence of the items,
,
T i
= cycle time for item i (the batch quantity Q i
= T i d i
).

Define:
 i
= d i
/p i
,
 i
=
 i
T i
= production time per batch for item i excluding setup time,
 i
= s i
+
 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).
C i

A i
T i
 h i d i
( 1
  i
)
T i
2
. (4.11)
N
The problem is to minimize 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
T i
 h i d i
2 A i
( 1
  i
)
, (4.12)
C i

2 A i h i d i
( 1
  i
)
. (4.13)
Note that (4.11) and (4.12) are equivalent to (3.6) and
(3.7) if we replace T i by Q i
/d i
, and
 i by d i
/p i
.

Table 4.2 Independent solution of Bomberger’s problem.
Item
T i
C i

I
1 2 3 4 5 6 7 8 9 10
167.5
37.7
39.3
19.5
49.7
106.6
204.3
20.5
61.5
39.3
0.179
1.060
1.528
1.024
4.428
0.938
3.034
12.668
6.506
0.255
2.36
2.01
3.56
4.29
2.49
1.67
3.04
5.87
11.20
1.17
C
  i
10

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

Not feasible
=10.16 >9.30.

Does a feasible solution exist? If at least one setup time is positive an obvious necessary condition for a feasible solution to exist is i
N
 

1 i

1
.
Condition (4.14) is also sufficient for feasibility.
(4.14)
Given the assumption of a common cycle time, the problem now is to minimize
C
 i
N


1


A i
T
 h i d i
( 1
  i
)
T
2


, (4.15)

with respect to the constraint that the common cycle must be able to accommodate production lots of all items i
N
 

1 i
 i
N


1
( s i
  i
T )

T
.
(4.16)
T
 i
N 

1 s i
1
 i
N 

1
 i

T min a lower bound for the cycle time.
, (4.17)
Need large enough T to squeeze setups in the slack
1i
N 

1
 i

Disregard (4.17), from (4.15)
T
ˆ  i
N 

1
2 i
N 

1
A i h i d i
( 1
  i
)
.
(4.18)
Since (4.15) is convex in T the optimal solution,
T opt
 max (
ˆ
, T min
)
, (4.19)
For Bomberger’s problem, T
T opt

T
ˆ 
42 .
75 , 
42 .
75 min
= 31.86, and consequently,
. cost=41.17.
For problems where the individual cycle times are reasonably similar, the common cycle approach gives a very good approximation.

C i
* 
31 .
62

C
* 
41 .
17

I.
Dynamic programming model
Bomberger (1966)
Assuming
T i
= n i
W
W should be able to accommodate production of all items.

F i
w
i
i
N
W - w
1
2
i
F i

1
( w )
 min n i

C i
( n i
W )

F i
( w
  i
)

, (4.20) where C i
( n i
 i
= s i
+

W ) are the costs (4.11) for item i with T i i n i
W , and the integer n i
= n i
W , is subject to the constraint
 n

( w

)


1 i s i
/ i
W
.

(4.21) or w
 s i
 i n i
W
Note that the upper bound in (4.21) is equivalent to
F
N
( w ) = 0 for all w

0. F
0 i
 w .
( W ) gives the minimum costs when the basic period is equal to W .

 Minimize over W . Bomberger’s solution
C =36.65, W =40, n i
=1 for i

7, n
7
=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 n i and the basic period the following iterative procedure:
W according to
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,
( n i
= 2 mi , m i

0), to minimize the item costs (4.11).

3.
Given the multipliers n i
, minimize the total costs
C
 i
N


1


A i
/
W n i  h i d i
( 1
  i
) n i
W
2

 with respect to W.
W

2 i
N


1
A i
/ n i i
N


1 h i d i
( 1
  i
) n i
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 n i
 i
10
8 2 2 1 2 4 8 1 2 2
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
Basic period
7
8
5
6
3
4
1
2
4, 8, 2, 9
4, 8, 3, 5, 10, 1
4, 8, 2, 9
4, 8, 3, 5, 10, 6
4, 8, 2, 9
4, 8, 3, 5, 10, 7
4, 8, 2, 9
4, 8, 3, 5, 10, 6
Production time
22.93
22.30
22.93
21.18
22.93
22.55
22.93
21.18

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 = l
Q=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
T

1 /
1
 k


1
 t
Dt

/
Q /
Q

P
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 min
Q

0 h
2
( TD

Q )
 min
Q

0 h
2


1

Dt
Dt

/
DQ
Q

/
D
P
/ P

Q


.(4.23)
Q
* 
1

2 Dt
D / P
.
(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, h i
= holding cost per unit and time unit for item
A = setup cost for the group, i , a i d i
= setup cost for item i ,
= demand per time unit for item i ,
T i
 i
= cycle time for item i .
= h i d i
Assume all demands equal to one. Items are ordered so that a
1
/

1
 a
2
/

2

...
 a
N
/

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
T i

2 a i
 i
, (4.25) i.e., T
1 would be the smallest cycle time. Assume other cycle times of items 2, 3... N are integer multiples n i for item 1, of the cycle time
T i
 n i
T
1
, i = 2, 3,..., N.
(4.26)
Our objective is to minimize w.r.t T
1
, n
2
, n
3
, ... n
N the cost.
C

A
 a
1
 i
N


2 a i
T
1
/ n i

T
1
(

1
 i
N
 

2 i n i
)
2
Fix cost ith item holding cost=T i n i
 i
/2
,(4.27)

Given n
2
, n
3
... n
N
,
T
1
*
( n
2
, n
3
,..., n
N
)

2 ( A


1 a
1

 i
N


2 a i i
N
 

2 i n i
/ n i
)
, (4.28)
C
*
( n
2
, n
3
,..., n
N
)

2 ( A
 a
1
 i
N


2 a i
/ n i
)(

1
 i
N
 

2 i n i
)
. (4.29)
Note that T
1 n
2
, n
3
... n
N is not chosen according to (4.25). If we disregard to be integers, then from (4.29), n i
 a i
 i
( A

1
 a
1
) . (4.30)
From (4.30) and (4.29), the lower bound for the costs:
C

2 ( A
 a
1
)

1
 i
N


2
2 a i
 i .(4.31)

HEURISTIC
1. Determine start values of n
2
, n
3
... n
N
(4.30) to the closest positive integers.
by rounding
2. Determine the corresponding T
1 from (4.28).
3. Given T
1
, minimize (4.27) with respect to n
2
, n
3
... n
N
.
This means that we are choosing n integer satisfying i as the positive n i
( n i

1 )


2 a i i
T
1
2
 n i
( n i

1 )
Note : dC

0 gives n i
2 

2 a i i
T
1
2
; C is convex .
dn i
.
(4.32)
Return to Step 2, if any multiplier n i iteration.
has changed since the last

Example 4.1
N = 4 , A = 300, a
1
= a
2
= a
3
= a
4
= 50, h
1 d
= h
4
4
= 10, d
1
= 5000, d
2
= 100. As requested, a i
= 1000, d
3
/
 i
= 700, and is nondecreasing with i. When applying the heuristic we obtain
= h
2
= h
3
T n
2

1

50
1000

10
5000

10
350

1 , n
3

1 , n
4

3
10

5000
2


( 350
10


50
1000


50
10


50
700
/ 3 )

10

100

3

0 .
1155 again n
2
= 1, n converged.
3
= 1, n
4
= 3, i.e., the algorithm has already

Approach 2. Roundy´s 98 percent approximation the joint setups have cycle time T
0

0,
T i where k i

2 k i T
0 , i = 1, 2, ..., N, (4.33) is a nonnegative integer. Let a
0
= A, and

0
= 0,
T
0
, min
T
1
,..., T
N i
N


0


 i
T
2 i  a i
T
1 i


, (4.34) subject to the constraints (4.33). replacing (4.33) by
i

0 , i = 1, 2, ..., N.
(4.35)

The resulting solution will give a lower bound for the costs. I lagrangian relaxation l
1
, l max
2
,..., l
N
T
0 min
, T
1
,..., T
N i
N


0



 i
T i
2
 a i
1
T i



 i
N
 l i
(


1
T
0
T i
) where l i define are nonnegative.

0

1

=

0

1

2
- 2 l i
N
 l

1
1
,
.
i
, (4.36)
. (4.37)
.

N
=

N
- 2 l
N
.

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.
T
0 min
, T
1
,..., T
N i
N


0


 i
T
2 i  a i
1
T i


(4.38) without any constraints on the cycle times.
rounding the cycle times of the relaxed problem,
T i

2 m i 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,
Since

0
T i

T i

1
, i = 1, 2, ..., N.
(4.40)
= 0 we will always aggregate items 0 and 1. After aggregation we have an item with cost parameters A + a
1

1
. Next we check whether a
2
/

2
< (A + a and
1
)/

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, a
1
50000, h
2
= a
2
= 10000, h
3
= a
3
= a
4
= 50,
= 7000, and h
4 h
1
=
= 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, h i
= holding cost per unit and time unit for item i , b
1, i
= shortage cost per unit and time unit for item i ,
A = setup cost for the group, a i
= setup cost for item i,
L i
= 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.
C i
( 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,
C ( T )

A / T
 i
N


1
C i
( T ) (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.
( S i
, 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
7
8
5
6
2
3
1
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
3
4
1
2
Installation stock inventory position
5
2
3
5
Echelon stock inventory position
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
.......
2 1
3
Raw material
Final product
Increasing batch sizes
Figure 5.7 Serial inventory system with N installations.

Q n
= batch quantity at installation n.
Q n
 j n
Q n

1
, (5.1) where j n
Notation: is a positive integer.
IP i n
= installation inventory position at installation n,
IP n e
= IP n i 
IP i n

1

...

IP
1 i
= echelon stock inventory position at installation n,
R i n
= installation stock reorder point at installation n,
R e n
=echelon stock reorder point at installation n.

initial inventory positions IP n i 0 and IP n e 0 satisfying
R i n

IP n i 0 
R i n

Q n
, R e n

IP e 0 n

R e n

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:
IP i n
0 
R i n is an integer multiple of Q n-1
.
All demands at installation n are for multiples of Q n-1 all replenishments are also multiples of Q n-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
IP n e  k n


1
( R i k

Q k
) .
(5.3)
Unit demand. Orders would be triggered exactly at the same time. When (n-1) orders Q n-1
, its inventory reaches the level
R n-1
+Q n-1
. At this point, level n inventory goes from R
R n
. And then n orders.
n
+Q n-1 to
R e n

R i n
 k n

1


1
( R i k

Q k
) .
(5.4)
If is not a multiple of Q less than Q n-1
IP n i 0  n-1
, change by an amount i
.So that is a multiple of Q n-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, R i
1

R
1 e
For installation n > 1, immediately after ordering
IP n i 
IP n e 
IP n e

1

R n e 
Q n

R n e

1

Q n

1
(5.5)
Therefore,
R
1 i 
R
1 e
, R n i 
IP n i 
Q n

R n e 
R e n

1

Q n

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, Q
1
= 50, Q
2 final demand at installation 1 =50. lead-time at
=100. installation 1 is one, at installation 2 is 0.5. No shortages allowed, holding costs at installation 1 are higher than at installation 2.
50
LT=0.5, R
2 e =25+50=75,
IP
2
0 (0.5)=0.
0 1 2 3 4 Time
Stock on hand at installation 2
0
LT=1, R
1 e =50, IP
1 i (0.5)=75.
50
1 2 3 4
Stock on hand at installation 1
5 Time
Figure 5.8 Inventory development in the optimal solution.
Echelon Policy is more general. At t=0.5
, IP
2 i =0,IP
1 i =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 2 3 4 5 6 7 8
Lead-time = 1 Gross requirements
Order quantity = 25 Scheduled receipts
Safety stock
= 5
Projected inventory
Net requirements
10 25 10
25
22 12 37 12 2 -18 -23 -23 -33
3
20
20
5
5
10
10

Table 5.4 MRP record for item 1 with planned orders in periods 3 and 5.
Item 1
Lead-time = 1
Order quantity = 25
Safety stock
= 5
Period 1 2 3 4 5 6 7 8
Gross requirements 10 25 10 20 5 10
Scheduled receipts 25
Projected inventory 22 12 37 12 27 7 27 27 17
Planned orders 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.

Table 5.5 MRP record for item 1 with a safety time.
Item 1
Lead-time = 1
Order quantity = 25
Safety stock
= 5
Safety time
= 1
Period
Gross requirements
Scheduled receipts
1 2 3 4 5 6 7 8
10 25 10 20 5 10
25
22 12 37 37 27 32 27 27 17 Projected inventory
Planned orders 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, A
1
= 20, A
2
= 80, h
1
= 5, h
2
= 4.

I.Treat the two installations independently
C
1
 h
1
Q
1
2

A
1 d
Q
1
Q
1

2 A
1 d h
1

8
,
.
(5.7)
C
1

2 A
1 dh
1

40 .
Q
2
 k Q
1 , (5.8) where k is a positive integer.

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.

C
2
 h
2
( k

1 ) Q
1
2

A
2 d k Q
1 k
* 
1
Q
1
2 A
2 d h
2
, (5.9)
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.
C
2
= 56, C = C
1
+ C
2
= 40 + 56 = 96.

II. Treat the two installations together
C

C
1

C
2

( h
1

( k

1 ) h
2
)
Q
1
2

( A
1

A
2
) k d
Q
1
. (5.10)
Alternatively, use the echelon holding costs e
1 and e
2
= h
2
,
= h
1
- h
2
,
C
1 e  e
1
Q
1
2

A
1 d
Q
1
, (5.11)
C e
2
 e
2 kQ
1
2

A
2 d kQ
1
.
C

C
1 e 
C e
2

( e
1
 k e
2
)
Q
1
2

( A
1

A
2
) k d
Q
1
(5.10) and (5.13) are equivalent.
(5.12)
, (5.13)

Q
1

2 ( A
1 e
1

A
2
) d k
 k e
2
.
C ( k )

2 ( A
1

A
2
) d ( e
1 k
 k e
2
) .
(5.14)
(5.15)
C
2
( k )

2 d ( A
1 e
1

A
2 e
2

A
1 e
2 k

A
2 e
1
) k
(5.16) k
* 
A
2 e
1
A
1 e
2 .
(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´ optimal to choose k = k´ if k * /k´
 k * < k´ + 1, it is

(k ´ + 1)/k * , otherwise k = k´+ 1.

Example 5.5
d = 8, A e
1
= h
1
1
- h
= 20, A
2
= 1, e
2
2
= 80, h
1
= h
2
= 5, h
2
= 4.
= 4. From (5.17), k * = 1.
Applying (5.14) and (5.15), Q
*
1

17 .
89
,
Q
*
2
 kQ
*
1

Q
*
1

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.
A

1
= A
1
+ A
2
/k , (5.18) h

1
= e
1
+ ke
2
. (5.19)

Stochastic Inventory Model
 Proportional Cost Models: x : initial inventory, y : inventory position (on hand + on order-backorder),

: random demand,

(

) ,

(

),
( y-
( y-

) + : ending inventory position, N.B.L,

) : 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 y
 x
 c ( y
  p min y
 x
  s
 c (
 y

(
 y


 x )
 y )

(

) d
 x )

( h
  s )

0 y
( y
L ( y )


 
)

(

) d

( 2 )


L ( y ) convex, L ’(0) < c (otherwise never order)
L ′ eventually becomes positive
L ' ( S )
 c

0 ( 4 )
Base Stock Policy y
*
( x ) q
*
( x )


 max{
S
 x
0 ,
, x , S }
If x

S otherwise

c

( h
  s )

( y )

( p p
  s
)(

1
 
( y ))
N .
B .
L
B .
L

0

( y )

( y )

 p
 c
( p
 c )

( c
 h
  s )
 c u c u
 c o
[ p
 p
 c ( 1
 c ( 1
 
)]

( c

)
 h
  s )
 c u c u
 c o
( 5 a )
( 5 b )
 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 ¢,
( i )
( ii )

( y )

100
100

51 .
5

0 .
66 , y
  
1
( 0 .
66 )

( y )

24
24

51 .
5

0 .
32 , y
  
1
( 0 .
32 )

 Set up cost K f ( x )
 min y
 x

K

( y
 x )
 c ( y
 x )

L ( y )

 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 s S
L (x) 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 y
 x
 c ( y
 x )

L ( y )
  
0
 f
1
( y
 
)

(

) d

Infinite Horizon ( f
1000
& f
1001 f ( x )
 min y
 x c ( y
 x )

L ( y )
  
0
 f ( y cannot be different)
 
)

(

) d

( 9 )

Taking derivative of {}
0
 c

L ' ( S )
  
0
 f ' ( S
 
)

(

) d

If f convex, find S the base stock level, then f ( x )
 c ( S
 x )

L ( S )
  
0
 f ( S
 
)

(

) d

( 11 )
It is possible to show that f’ ( x )=c for x ≤ S
(12)
( 10 ) which reduce
L ' ( S )
 c ( 1
 
) to

0 B.L
similarly for N.B.L
L ' ( S )
 c ( 1
  
( S ))

0
(13)
( 18 )

 Proportional costs:
L ( y )
 h

0 y
( y
 
)

(

) d
  p
 y

(
  y )

(

) d

So that
L ' ( S )

( h
 p )

( S )
 p ( 19 )
Substitute (19) into (13) and (18),
N .
B .
L :

( S )

B .
L :

( S )

( p
 c )
 p h
 c
 c ( 1
 
)
 p
 p
 c ( 1
  c ( 1
 
)
)
 h
 c ( 1
 
) c u c
 u c o
 c u c u
 c o
( 20 a )
( 20 b )
Remark :
Lead time, Setup cost
 more complicate d, still (s, S) policy

Example 4: c u

20 , c o

5

(
S
S )


1800 c u c u
 c o

20
5

20

4
5
Example 5:

(

)

1
25 e


25 , K
L ( y )


15 ,

0 y h ( y
 
)

(

) d
 c

1 , h ( z )

3
10 z , z
  y
 p ( y
 
)

(

) d


0, p(z)

3
10
1
25

0 y
( y
 
) e


25 d
 
3
2
1
25
 y

( y
 
)
2 e


25 d


3
2 z 2 , z

0

1882 .
5 e
 y
25

0 .
3 y

7 .
5

dL ( y )
 
75 .
25 e
 y
25

0 .
3 dy c
 dL ( y )

S dy y

S

101 .
5

1

0 .
3

75 .
25 e

Cs

L ( s )

K
 cS

L ( S )
S
25

0 s

1882 .
5 e

S
25

0 .
3 s

7 .
5

15

101 .
5

1882 .
5 e

S
25

0 .
3 S

7 .
5
Succesive approximat ion : s

80 .
5
The optimal q


0
101 .
5
 x policy if x

80 .
5 otherwise
:

Multiperiod models: No Setup Cost
 Begin with two periods
Demand D
1
, D
2
, i.i.d
Density:

(

)
L ( y ) = expected one period holding+ shortage penalty cost; strictly convex with linear cost and

(

) >0, c purchase cost /unit c
1
(x
1
) optimal cost with 1 period to go; c+L’(y
1
0 )=0 while y
1
0 is the optimal base stock level.

c
1
( x
1
)
 L ( x
1
) c ( y
1
0  x
1
)

L ( y
1
0
) if x
1 if
 x y
1
0
1
 y
1
0 c
1
( x
1
)


E ( c
1
( x
1
)) c
1 
L c
(
(
( y y
1
0 y
2
2



D y
2
2

)
D
2
D
2
)
)

L ( y
1
0
) if
 
0
 c
1
( y
2 y
2 if
 y
D
2
2


D y
2
0
1
 y
1
0
 
)

(

) d
 c
2
( x
2
)

 
0 y
2 min y
2
 x
2
 c

2 y
1
0
(
L ( y
2 y
2

  x
2
)

)

(

L (
) d
 y
2
)

   y
2
 y
1
0
[ c (
E [ c
1
( x
1
)] y
1
0

 y
2
 
)

L
2
( y
1
0
)]

(

) d
 which is convex
 y
0
2 basestock level with 2 periods to go

 Example: c =10, h =10, p =15 the demand density is




1
10
0 if 0
  
10 otherwise
 Solution:

( y
1
0
)
 p
 c p
 h

15

10
15

10

1
5
Since
L ( z )


(
 z
10 y
1
0
)
15 (


10 y
1
0
10
 z )
, d
 y
1
0


2

0 z 10 ( z
10
 
) d


75

15 z

( 5 / 4 ) z
2

E [ c
1
( x
1
)]
 
0 y
2

2
[ 75

15 ( y
2
 
)

( 5 / 4 )( y
2
 
)
2
1
]
10 d


  10 y
2

2
[ 10 ( 2
 y
2

0 y
2

2
[ 75

15 ( y
2
 
)

75

15 * 2

( 5 / 4 ) 2
2
1
]
10 d

 
)

( 5 / 4 )( y
2
 
)
2
1
]
10 d

  10 y
2

2
[ 70

10 ( y
2
 
)]
1
10 d
 c
2
( x
2
)

 y
3
( y min

2 x
2
)
2

10
/
(
24 y
2

( y
2
)
2

/ 4

19 y
2 x
2
)

75

15 y
2
/ 2

359

( 5 / 4 ) y
2
2
/ 3

( 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 {}

[

29 / 2

2 y
2
0 dy
2

1
8
( y
2
0
)
2
]

0 y
2
0 
5 .
42
Substituti ng value with y
0
2 y
0
2

5 and y
0
2

5.

6 into c
2
(x
2
) leads to a smaller
The optimal policy : q
2
 5

0
 x
2 if x
2
 otherwise
5 q
1
 2

0
 x
1 if x
1

2 otherwise

Multi-Period Dynamic Inventory Model with no Setup Cost
C n
( x n
): n periods to go,

: discount factor.
DP equations: c n
( x n
)
 y min
 x n n
 c ( y n c
0
( x
0
)

0
 x n
)

L ( y n
)
 
E [
 c n

1
( y n

D n
) ]

Properties :
1) S
1
L' ( S

)
S

2
 c(1
S
3
-

..........
)

0;
...
S n

1

S n
.........

S , where
2) c ( x )
 min y
 x
 c ( y
 x )

L ( y )
  
0
 c ( y
 
)

D
(

) d

 satisfied by
3 ) c ( x )
 n lim
  c n
( x )

S n lim
 
S n

Multi-Period Dynamic Inventory Model with
Setup Cost c n
( x n
)

K

( y n
 y min
 n x n

K
 x n
)


K
0
( y n if if
 x n
)
 y n y n

 x n x n c ( y n
 x n
)

L ( y n
)
 
0
  c n

1
( y n
 
)

(

) d

If L(y) is convex, then find S n
.
The optimal (s n
, S n
)policy : q n
 S

0 n
 x n if if x n x n


S n
S n

Lead time: l f n
( u n
) u n
 y min
 u n n



K

( y n
 inventory position
 u n
)
 c ( y n
 u n
)
  l 
0

L ( y n
 
)
 l
D
(

) d
   
0
 f n

1
( y n
 
)

D
(

) d
 can transf orm to 0 lead time as follows :
 l 
( y )
  l 
0

L ( y
 
)
 l
D
(

) d
 infinite horizon
 l 
' ( S )
 c ( 1
 
)

0