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. 6
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Pinhal Novo 39.1 16.9 10.9 33.0 14.2


If all customers must be served (i.e. if the penalties pj are suf¬ciently high for
each j ∈ V2 ), variables zj , j ∈ V2 , can be assumed equal to 0. Hence, the location-
covering problem is a generalization of the well-known set covering problem and is
therefore NP-hard.
Several variants of the location-covering model can be used in practice. For exam-
ple, if ¬xed costs fi are equal for all potential sites i ∈ V1 , it can be convenient
to discriminate among all the solutions with the least number of open facilities the
one corresponding to the least total travelling time, or to the most equitable demand
distribution among the facilities. In the former case, let xij , i ∈ V1 , j ∈ V2 , be a
binary decision variable equal to 1 if customer j is served by facility i, 0 otherwise.
The problem can be modelled as follows:
Minimize
Myi + tij xij (3.60)
i∈V1 j ∈V2
i∈V1

subject to

j ∈ V2 ,
aij xij 1, (3.61)
i∈V1

|V2 |yi , i ∈ V1 ,
xij (3.62)
j ∈V2
yi ∈ {0, 1}, i ∈ V1 , (3.63)
xij ∈ {0, 1}, i ∈ V1 , j ∈ V2 , (3.64)

where M is a large positive constant. Constraints (3.61) guarantee that all the cus-
tomers j ∈ V2 are serviced, while constraints (3.62) ensure that if facility i ∈ V1 is
not set up (yi = 0), then no customer j ∈ V2 can be served by it.




TLFeBOOK
DESIGNING THE LOGISTICS NETWORK 115

Table 3.21 Distances (in kilometres) between municipalities of
the consortium in Portugal (Part II).

Macau Moita Montijo Palmela Pinhal Novo

Almada 39.0 29.5 38.8 34.3 39.1
Azenha 15.4 9.5 18.7 24.0 16.9
Carregosa 13.9 5.1 7.2 16.4 10.9
Corroios 32.9 23.4 32.7 28.2 33.0
Lavradio 16.1 7.4 16.6 26.2 14.2
Macau 0.0 9.1 12.0 6.9 1.7
Moita 9.1 0.0 10.6 11.7 7.2
Montijo 12.0 10.6 0.0 19.0 10.3
Palmela 6.9 11.7 19.0 0.0 8.8
Pinhal Novo 1.7 7.2 10.3 8.8 0.0



In Portugal, a consortium of 10 municipalities (Almada, Azenha, Carregosa, Cor-
roios, Lavradio, Macau, Moita, Montijo, Palmela, Pinhal Novo), located in the neigh-
bourhood of Lisbon, has decided to improve its ¬re-¬ghting service. The person
responsible for the project has established that each centre of the community must
be reached within 15 min from the closest ¬re station. Since the main aim is just to
provide a ¬rst help in case of ¬re, the decision maker has decided to assign a single
vehicle to each station. The annual cost of a station inclusive of the expenses of the
personnel is ‚¬198 000. It is assumed that the average travelling speed is 60 km/h
everywhere. In order to determine the optimal station location the location-covering
model (3.60)“(3.64) is used, where V1 = V2 = {Almada, Azenha, Carregosa, Cor-
roios, Lavradio, Macau, Moita, Montijo, Palmela, Pinhal Novo}. Time coef¬cients
tij , i ∈ V1 , j ∈ V2 , can be easily determined from the distances (in kilometres)
reported in Tables 3.20 and 3.21. Coef¬cients aij , i ∈ V1 , j ∈ V2 , were obtained
from Tables 3.20 and 3.21. The minimum number of ¬re stations turns out to be two.
The facilities are located in Almada and Moita. The ¬re station located in Almada
serves Corroios and Almada itself, the remaining ones are served by the ¬re station
located in Moita.

Another interesting variant of the location-covering model arises when one must
locate facilities to ensure double coverage of demand points. A classic case is ambu-
lance location when users are better protected if two ambulances are located within
their vicinity. If one of the two ambulances has to answer a call, there will remain one
ambulance to provide coverage.

3.6 Data Aggregation
Modelling and solving a location problem often requires a considerable amount of
data. For instance, the number of demand points of a large food producer often exceeds




TLFeBOOK
116 DESIGNING THE LOGISTICS NETWORK

several thousands. Similarly, the items marketed by a distribution ¬rm can be around
tens of thousands. Therefore, in order to keep computation times and hardware require-
ments acceptable, data have to be aggregated. This can be done mainly in two ways.

Demand aggregation. Customers can be aggregated according to their position,
service level or frequency of delivery. In the ¬rst case, demand points are clustered in
such a way that customers close to one another (e.g. customers having the same zip
code) are substituted by a customer zone. In the other cases, customers are aggregated
into classes.

Item aggregation. Items are aggregated into a suitable number of product groups.
This can be done according to their distribution pattern or their features (weight,
volume, shape, cost, etc.). In the former case, products manufactured by the same
plants and supplied to the same customers are treated as a single commodity. In
the latter case, similar items (e.g. variants of the same basic product model) are
aggregated.
Whatever the aggregation method, the reduced problem can be modelled with fewer
variables and constraints. In what follows, a demand aggregation method for the CPL
model (see Section 3.3.1) is analysed. The demand points of a subset S ‚ V2 are
aggregated in the following way:
xij = xik , i ∈ V1 , j ∈ S, k ∈ S.
Consequently, each point j ∈ S receives the same fraction of demand from each
facility i ∈ V1 . Let s be the customer zone resulting from the aggregation. Then the
CPL model becomes:
Minimize
cij xij + fi y i
i∈V1 j ∈(V2 \S)∪{s} i∈V1
subject to

xij = 1, j ∈ (V2 \ S) ∪ {s},
i∈V1

i ∈ V1 ,
dj xij qi yi ,
j ∈(V2 \S)∪{s}
1, i ∈ V1 , j ∈ (V2 \ S) ∪ {s},
xij
0
yi ∈ {0, 1}, i ∈ V1 ,
where
cis = i ∈ V1 ,
cij ,
j ∈S
and
ds = dj .
j ∈S




TLFeBOOK
DESIGNING THE LOGISTICS NETWORK 117

As a rule, the optimal solution value of the aggregated problem is worse than that
of the original problem. It is useful to evaluate an upper bound on the error made
— (a)
resulting from the aggregation procedure. Let zCPL and zCPL be the costs of the
optimal solutions of the original problem and of the aggregate problem, respectively.

Property. The following relation holds,
— —
(a)
zCPL + µ,
zCPL
zCPL

where
dj r∈S cir
µ= ’ cij .
max (3.65)
r∈S dr
i∈V1
j ∈S

Proof. In Equation (3.65), r∈S cir represents the transportation cost when the whole
demand of customers in S is served by facility i ∈ V1 , while dj / r∈S dr represents
the fraction of the total demand of customers in S. The difference

’ cij
dj cir dr
r∈S r∈S

represents, therefore, the variation (positive, negative or zero) of the cost of assigning
the whole demand of customer j ∈ V2 to facility i ∈ V1 when the vertices of S are
aggregated. Therefore, Equation (3.65) expresses the worst-case error made in the
aggregation process as a sum of the maximum increases of the distribution costs of
the various demand points in S.
It can be shown that bound (3.65) is tight, i.e. there are instances such that the
aggregation error is nearly equal to µ.


If the demands of Brossard and Verdun are aggregated in the Goutte problem
(see Section 3.3, where the maximum distance between plants and sales districts is
assumed to be in¬nite), the worst-case error is, on the basis of Equation (3.65),
µ = µ1 + µ2 = 1578.92 Canadian dollars per year,

where
q1 (ci1 + ci6 )
µ1 = max ’ ci1 = 789.46 Canadian dollars per year,
(q1 + q6 )
i∈V1
q6 (ci1 + ci6 )
µ2 = max ’ ci6 = 789.46 Canadian dollars per year.
(q1 + q6 )
i∈V1




The above aggregation technique can be obviously extended to the case of K
disconnected subsets S1 , . . . , SK .




TLFeBOOK
118 DESIGNING THE LOGISTICS NETWORK

Property. An a priori upper bound on the error is given by
K
— —
(a)
+
zCPL zCPL zCPL µk ,
k=1

where
dj r∈Sk cir
µk = ’ cij .
max
r∈Sk dr
i∈V1
j ∈Sk



3.7 Questions and Problems
3.1 Illustrate why crude oil re¬neries are customarily located near home heating
and automotive fuel markets.
3.2 Describe how ¬xed costs fi , i ∈ V1 , in the CPL model should be computed
in order to take labour costs, property taxes and site-development costs into
account.
3.3 Your company has to close 20 of its 125 warehouses. Suppose the CPL hypothe-
ses hold. How would you de¬ne V1 ? What is the value of p?
3.4 Modify the CPL model to take into account that a subset of already existing
facilities V1 ⊆ V1 cannot be closed (but can be upgraded). Indicate the current
¬xed cost and capacity of facility i ∈ V1 as fi and qi , respectively. Moreover,
let fi and qi be the ¬xed cost and capacity if facility i ∈ V1 is upgraded,
respectively.
3.5 Borachera is a major Spanish wine wholesaler currently operating two CDCs in
Salamanca and Albacete, and a number of RDCs all over the Iberian peninsula.
In order to reduce the overall logistics cost, the company wishes to redesign its
distribution network by replacing its current RDCs with three (possibly new)
RDCs. Based on a preliminary qualitative analysis, an RDC should be located
in the Castilla-Leon region, either in Vallalolid, Burgos or Soria. A second RDC
should be located in the Extremadura region, either in Badajoz, Plasencia or
Caceres. Finally, the third RDC should be located in the Argon region, either
in Barbastro, Saragossa or Teruel. Transportation costs from RDCs to retailers
are charged to retailers. Formulate the Borachera problem as a modi¬ed CPL
model.
3.6 As illustrated in Section 3.3, when designing a distribution network it is cus-
tomary to remove excessively long transportation links from the model in order
to allow for a timely delivery to customers. How should the CPL Lagrangian
heuristic be modi¬ed in this case?
3.7 Modify the CPL Lagrangian heuristic for the case where demand is indivisible
(i.e. a customer demand must be satis¬ed by a single warehouse). Is the modi¬ed




TLFeBOOK
DESIGNING THE LOGISTICS NETWORK 119

heuristic still time-polynomial? How can it determine whether an instance of
the modi¬ed problem is feasible?
3.8 Modify the TEMC model in Section 3.4 for the case where a customer can
receive shipments from different DCs if these shipments are for different com-
modities. How should the proposed heuristics be changed in order to take this
modi¬cation into account?
3.9 Modify the location-covering model in order to determine, among the least-
cost solutions, the one associated with the most equitable demand distribution
among the facilities.
3.10 Data aggregation is useful in facility location even if the available algorithms
are able to solve the original problems. Why? (Hint: recall that demand forecasts
at the account and product levels are generally mediocre.)


3.8 Annotated Bibliography
An extensive literature review of strategic production-distribution models is presented
in the following article:
1. Vidal C and Goetschalckx M 1997 Strategic production-distribution models: a
critical review with emphasis on global supply chain models. European Journal
of Operational Research 98, 1“18.
An in-depth examination of location theory in a broader context is reported in the
following book:
2. Love RF, Morris JG and Wesolowsky GO 1988 Facilities Location. North-
Holland, Amsterdam.
A review of the most important location-routing problems is presented in:
3. Laporte G 1988 Location-routing problems. In Vehicle Routing: Methods and
Studies (ed. Golden BL and Assad AA). North-Holland, Amsterdam.




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4

Solving Inventory Management
Problems

4.1 Introduction
Inventories are stockpiles of items (raw materials, components, semi-¬nished and
¬nished goods) waiting to be processed, transported or used at a point of the supply
chain. As pointed out in Chapter 1, there are a number of reasons to have invento-
ries, including improving service level, reducing overall logistics costs, coping with
randomness in customer demand and lead times, making seasonal items available all
year, speculating on price patterns, etc. At the same time, holding an inventory can
be very expensive (the annual cost can be 30% of the value of the materials kept in
stock, or even more). It is therefore crucial to manage inventories carefully.
This chapter deals with the solution of the most important inventory management
problems. Inventory management amounts to deciding for each stocking point in the
supply chain when to reorder and how much to order so that the expected annual cost
is minimized while meeting a given service level. A useful formula that can be used to
know whether a stocking point is well managed is the inventory turnover ratio (ITR),
de¬ned as the ratio between the annual sales priced at the value of the items in stock
and the average inventory investment. Of course, high ITRs generally correspond to
well-managed stocking points.


4.2 Relevant Costs
The costs relevant to the determination of an inventory policy can be classi¬ed into
four broad categories.

Procurement costs. Procurement costs are those associated with the acquisition of
goods. They typically include ¬xed costs (independent of the amount ordered) and
variable costs (dependent of the amount acquired). They may include

Introduction to Logistics Systems Planning and Control G. Ghiani, G. Laporte and R. Musmanno
© 2004 John Wiley & Sons, Ltd ISBN: 0-470-84916-9 (HB) 0-470-84917-7 (PB)




TLFeBOOK
122 SOLVING INVENTORY MANAGEMENT PROBLEMS

• a (¬xed) reorder cost (the cost of issuing and processing an order through the
purchasing and accounting departments if the goods are bought, or the cost for
setting up the production process if the goods are manufactured by the ¬rm);

• a purchasing cost or a (variable) manufacturing cost, depending on whether
the goods are bought from a supplier or manufactured by the ¬rm;

• a transportation cost, if not included in the price of the purchased goods; for the
sake of simplicity, we assume in the remainder of this chapter that ¬xed trans-
portation costs are included in the reorder cost, while variable transportation
costs are included in the purchasing cost;

• the cost of handling the goods at the receiving point.

Inventory holding costs. Inventory holding costs are incurred when materials are
stored for a period of time. They include the following.

• An opportunity (or capital) cost representing the return on investment the ¬rm
would have realized if money had been invested in a more pro¬table economic
activity (e.g. on the stock market) instead of inventory. This cost is generally
estimated on the basis of a standard banking interest rate.

• A warehousing cost. If the company runs its own warehouses, such costs include
space and equipment costs, personnel wages, insurance on inventories, mainte-
nance costs, energy costs and state taxes. Otherwise, warehousing costs amount
to the fee paid for storing the goods in third-party warehouses.

Shortage costs. Shortage costs are paid when customer orders are not met. Shortage
costs depend heavily on customer behaviour and are dif¬cult to evaluate accurately.
They can be classi¬ed as follows.

• Lost sales costs. A lost sale is likely to occur if the unavailable items can be
easily obtained from a competitor. Lost sales costs include the pro¬t that would
have made on the sale, and the negative effect that the shortage could have on
future sales.

• Back order costs. When goods are dif¬cult to replace, a shortage often results
in a delayed sale. Apart from the negative effect on future sales, a back order
could result in a penalty.

Obsolescence costs. Obsolescence costs arise when stocked items lose some of
their value over time. This happens, for example, when food inventories deteriorate,
clothing items in stock go out of fashion, or newspapers are unsold. The value of an
item at the end of its lifetime is usually referred to as its salvage value.




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SOLVING INVENTORY MANAGEMENT PROBLEMS 123

4.3 Classi¬cation of Inventory Management Models
Inventory management models can be classi¬ed according to a number of criteria.

Deterministic versus stochastic models. In a deterministic model, where demands,
prices and lead times are assumed to be known in advance, the focus is on balancing
different categories of costs (e.g. the ¬xed costs of ordering and the costs of holding
inventory). In a stochastic model, where some data are uncertain, it is impossible
to always satisfy all the demand. As a result, a constraint on customer service level
(stating, for example, that a customer is satis¬ed with a given probability) is usually
imposed.

Fast-moving items versus slow-moving items. Fast-moving items are those man-
ufactured and sold regularly, and include most products on the market. The main issue
when managing a fast-moving item inventory is to determine when stocking points
have to be replenished and how much to order. On the other hand, slow-moving items,
which are often spare parts of complex machineries, have a very low demand (e.g. a
few units every 10 or 20 years). As a rule, manufacturing a spare part several years
after the machinery is constructed is very expensive compared to producing it along
with the machinery. Hence the main issue is to determine the number of items that
have to be produced and stored at the beginning of the planning horizon.

Number of stocking points. Optimal inventory policies can often be derived ana-
lytically for single stocking point models, while multiple stocking point situations are
usually much harder to deal with.

Number of commodities. When holding a multicommodity inventory, joint costs
and constraints usually come into play. It is therefore not surprising that most multi-
commodity inventory management problems are NP-hard.

Instantaneous resupply versus noninstantaneous resupply. A stocking point can
be replenished almost instantaneously (e.g. as a consequence of a truck delivery) or
noninstantaneously (e.g. at a constant rate through a manufacturing process).


4.4 Single Stocking Point: Single-Commodity
Inventory Models under Constant Demand Rate
A single stocking point has to service a single-commodity constant rate demand, and
places orders for the item from another facility. The planning horizon is assumed
to be in¬nite (for the ¬nite horizon case, see Problem 4.4). Since the demand rate
is constant, a minimal cost policy is to replenish on a periodic basis. Let d be the
constant demand rate, T the time lapse between two consecutive orders, q the order
size (i.e. the amount of product ordered at each replenishment). These quantities are




TLFeBOOK
124 SOLVING INVENTORY MANAGEMENT PROBLEMS
I(t)

T = q/d




m


r’d ’d
r




T1 T2 T3 T4

t


’s


Figure 4.1 Inventory level as a function of time.

such that
q = dT . (4.1)
Moreover, let c be the value of an item (assumed to be independent of the order size),
k the ¬xed reorder cost, h the holding cost per item and per time unit, u the shortage
cost per item, independent of the duration of the shortage, and v the shortage cost per
item and per time unit. The holding cost h can be expressed as a fraction p of c:

h = pc. (4.2)

The parameter p is a banking interest rate (measuring capital cost) increased to take
into account warehousing costs. Moreover, let I (t) be the inventory level at time t,
m the maximum inventory level, s the maximum shortage, tl the lead time, i.e. the
time lapse between when an order is placed and when the items are received (see
Chapter 1). The problem is to determine q (or, equivalently, T ) and s in such a way
that the overall average cost per time unit is minimum.


4.4.1 Noninstantaneous resupply
Let Tr 0 be the replenishment time, i.e. the time to make a replenishment, and r
the replenishment rate, i.e. the number of items per unit of time received during Tr .
The following relation holds:
q = rTr . (4.3)
The inventory level I (t) as a function of time t is shown in Figure 4.1. Dashed
lines represent the cumulative number of items arriving at the stocking point during
a replenishment (their slope is r). Since items are picked up at a rate d while a
replenishment takes place, the net number of items stocked per unit of time during
Tr is r ’ d. Finally, after a replenishment, the stock level decreases at a rate equal to




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SOLVING INVENTORY MANAGEMENT PROBLEMS 125

’d. Let T1 , T2 , T3 and T4 be the time the inventory level takes to go from ’s to 0,
from 0 to m, from m to 0 and from 0 to ’s, respectively.
The maximum inventory level m is given by
s + m = (r ’ d)Tr .
Therefore, from Equation (4.3),
m = (r ’ d)Tr ’ s = q(1 ’ d/r) ’ s.
The total average cost per time unit µ(q, s) is
1 ¯ ¯
µ(q, s) = (k + cq + hI T + us + v ST ). (4.4)
T
The quantity in parentheses in the right-hand side is the average cost per period,
given by the ¬xed and variable costs of a resupply (k and cq, respectively), and the
¯ ¯
holding cost hI T , plus the shortage costs (us and v ST ). The holding and shortage
¯ ¯
costs depend on the average inventory level I , and on the average shortage level S,
respectively:
m(T2 + T3 )
T
1 1
¯
I= I+ (t) dt = ,
T T 2
0
s(T1 + T4 )
T
1 1
¯
S= I’ (t) dt = ,
T T 2
0

where
’I (t)
I (t) if I (t) 0, if I (t) 0,
I+ (t) = and I’ (t) =
0 otherwise, 0 otherwise.

Moreover, since
s = (r ’ d)T1 ,
m = (r ’ d)T2 ,
m = dT3 ,
s = dT4 ,
the time lapses T1 , T2 , T3 and T4 are given by
s
T1 = ,
r ’d
m
T2 = ,
r ’d
m
T3 = ,
d
s
T4 = .
d




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126 SOLVING INVENTORY MANAGEMENT PROBLEMS
I(t)

T = q/d




m




t

Figure 4.2 Inventory level as a function of time when shortage is not allowed.

Consequently,
[q(1 ’ d/r) ’ s]2
m2
¯
I= = , (4.5)
2q(1 ’ d/r) 2q(1 ’ d/r)
s2
¯=
S . (4.6)
2q(1 ’ d/r)
Finally, using Equations (4.1), (4.5) and (4.6), Equation (4.4) can be rewritten as
h[q(1 ’ d/r) ’ s]2 vs 2
kd usd
µ(q, s) = + cd + + + . (4.7)
2q(1 ’ d/r) 2q(1 ’ d/r)
q q
If shortages are allowed, the minimum point (q — , s — ) of the convex function µ(q, s)
can be obtained by solving the equations:

= 0,
µ(q, s)
‚q q=q — ,s=s —

= 0.
µ(q, s)
‚s q=q — ,s=s —

As a result,
h+v (ud)2
2kd

q= ’ (4.8)
h(1 ’ d/r) h(h + v)
v
and
(hq — ’ ud)(1 ’ d/r)

s= . (4.9)
(h + v)
If shortages are not allowed (see Figure 4.2), Equation (4.7) can be simpli¬ed since
s = 0:
µ(q) = kd/q + cd + 2 hq(1 ’ d/r).
1
(4.10)




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SOLVING INVENTORY MANAGEMENT PROBLEMS 127
µ(q)




kd/q + hq(1 ’ d/r)/2


hq(1 ’ d/r)/2




kd/q

q
q*

Figure 4.3 Average costs as a function of q.


Hence, a single equation has to be solved:

d
= 0,
µ(q)
dq q=q —

Finally, the optimal order size q — is (see Figure 4.3)

2kd
q— = . (4.11)
h(1 ’ d/r)


Golden Food distributes tinned foodstuff in Great Britain. In a warehouse located in
Birmingham, the demand rate d for tomato pur´ e is 400 pallets a month. The value of
e
a pallet is c = £2500 and the annual interest rate p is 14.5% (including warehousing
costs). Issuing an order costs £30. The replenishment rate r is 40 pallets per day.
Shortages are not allowed. The holding cost is given by
h = 0.145 — 2500 = £362.5 per year per pallet
= £30.2 per month per pallet.
Therefore, from Equation (4.11),

2 — 30 — 400
q— = = 39.9 ≈ 40 pallets,
30.2[1 ’ 400/(40 — 20)]

where it is assumed that the number of workdays in a month equals 20 (hence the




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128 SOLVING INVENTORY MANAGEMENT PROBLEMS
I(t)



T = q/d


m




t



’s


Figure 4.4 Inventory level as a function of time in the instantaneous resupply case.



demand rate d is 20 pallets per workday). Finally, from Equations (4.1) and (4.3),
T — = 40/400 = 1/10 month = 2 workdays,
Tr— = 40/40 = 1 workday.




4.4.2 Instantaneous resupply

If resupply is instantaneous, the optimal inventory policy can be obtained by Equations
(4.8), (4.9) and (4.11), taking into account that r ’ ∞. If shortages are allowed (see
Figure 4.4), then

h + v 2kd (ud)2

q= ’ ,
h(h + v)
v h
hq — ’ ud

s= .
(h + v)

If shortages are not allowed (see Figure 4.5), Equation (4.10) becomes

µ(q) = kd/q + cd + 2 hq
1
(4.12)

and the optimal order size is given by

2kd
q— = . (4.13)
h




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SOLVING INVENTORY MANAGEMENT PROBLEMS 129
I(t)




T = q/d



m=q




t

Figure 4.5 Inventory level as a function of time in the EOQ model.

This is the classical economic order quantity (EOQ) model introduced by F. W. Harris
in 1913. The total cost per time unit of an EOQ policy is

µ(q — ) = 2kdh + cd.
Optimal policies with no backlogging (in particular, the EOQ policy) satisfy the
zero inventory ordering (ZIO) property, which states that an order is received exactly
when the inventory level falls to zero.

Al-Bufeira Motors manufactures spare parts for aircraft engines in Saudi Arabia. Its
component Y02PN, produced in a plant located in Jiddah, has a demand of 220 units
per year and a unit production cost of $1200. Manufacturing this product requires
a time-consuming set-up that costs $800. The current annual interest rate p is 18%,
including warehousing costs. Shortages are not allowed. The holding cost is
h = 0.18 — 1200 = 216 dollars/year per unit.
Therefore, from Equation (4.13),
q — = 40.37 units,
and, from Equation (4.1),
T — = 40.37/220 = 0.18 years = 66.8 days.
The total cost is given by Equation (4.12),
hq — 800 — 220 216 — 40.37
kd
+ = + = 8719.63 dollars per year,
q— 2 40.37 2
plus
cd = 220 — 1200 = 264 000 dollars per year.




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130 SOLVING INVENTORY MANAGEMENT PROBLEMS
I(t)




m




l




t


’s

tl


Figure 4.6 Reorder point.


In practice, the values of q — and of T — must be rounded as explained in Section 4.12.



4.4.3 Reorder point
An order has to be issued tl time instants before the inventory level equals ’s. In case
resupplies are instantaneous, a replenishment is needed every time the inventory level
equals the following reorder point l,
l = (tl ’ tl /T T )d ’ s,
where tl /T is the number of periods included in the lead time (see Figure 4.6).

In the Al-Bufeira Motors problem, a set-up has to be planned seven days in advance.
Assuming T = 67 days, the reorder point l is
220
l = (7 ’ 7/67 — 67) ’ 0 ≈ 4 units.
365




4.5 Single Stocking Point: Single-Commodity
Inventory Models under Deterministic
Time-Varying Demand Rate
When orders are placed in advance, demands can be assumed to be deterministic,
yet time-varying. Let 1, . . . , TH be a ¬nite and discrete time horizon. In addition, let




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SOLVING INVENTORY MANAGEMENT PROBLEMS 131

dt , t = 1, . . . , TH , be the demand at time period t, k the ¬xed reorder cost, and h the
holding cost. The problem is to decide how much to order in each time period in such
a way that the sum of reorder costs plus holding costs is minimized. No backlogging
is allowed. In 1958, H. M. Wagner and T. M. Whithin formulated this problem as
follows. The decision variables are the amount qt , t = 1, . . . , TH , ordered at the
beginning of time period t, the inventory level It , t = 1, . . . , TH , at the end of time
period t; in addition let yt , t = 1, . . . , TH , be a decision variable equal to 1 if an
order is placed in time period t, 0 otherwise.
Minimize
(kyt + hIt ) (4.14)
t=1,...,TH
subject to
It = It’1 + qt ’ dt , t = 1, . . . , TH , (4.15)
t = 1, . . . , TH ,
qt yt dr , (4.16)
r=t,...,n
I0 = 0, (4.17)
t = 1, . . . , TH ,
It 0,
t = 1, . . . , TH ,
qt 0,
yt ∈ {0, 1}, t = 1, . . . , TH ,
where the objective function (4.14) is the total cost. Equations (4.15) are the inventory-
balance constraints, inequalities (4.16) state that for each time period t = 1, . . . , TH ,
qt is zero if yt is zero, and equation (4.17) speci¬es the initial inventory.
2
An optimal solution of the Wagner“Within model can be obtained in O(TH ) time
through a dynamic programming procedure. This algorithm is based on the following
theoretical results.

Proposition. Any optimal policy satis¬es the ZIO property, i.e.
qt It’1 = 0, t = 1, . . . , TH .
Proof. If It’1 = δ > 0 and qt > 0, the total cost can be reduced by setting It’1 := 0,
qt := qt + δ, and modifying the remaining variables accordingly.

Corollary. In an optimal policy, the amount ordered at each time period is the total
demand of a set of consecutive subsequent periods.
The algorithm is as follows. Let G = (V , A) be a directed acyclic graph, where
V = {1, . . . , TH , TH + 1} is a vertex set and A = {(t, t ) : t = 1, . . . , TH , t =
t + 1, . . . , TH + 1} is an arc set. With each arc (t, t ) is associated the cost of ordering
in time period t to satisfy the demands in time periods t, t + 1, . . . , t ’ 1:

gtt = k + h (r ’ t)dr .
r=t,...,t ’1




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132 SOLVING INVENTORY MANAGEMENT PROBLEMS

Then the shortest path between vertices 1 and TH + 1 corresponds to a least-cost
inventory policy.

Sao Vincente Chemical is a Portuguese company producing lubricants. In the fol-
lowing year, its product Serrado Oil is expected to have a demand of 720, 1410, 830
and 960 pallets in winter, spring, summer and autumn, respectively. Manufacturing
this product requires a time-consuming set-up that costs ‚¬8900. The current annual
interest rate p is 7.5%, including warehousing costs. The variable production cost
amounts to $350 per pallet while the initial inventory is zero. The holding cost is
h = 0.075 — 350/4 = 6.56 euros/season per unit.
Let t = 1, 2, 3, 4 represent the winter, spring, summer and autumn periods, respec-
tively. By solving the Wagner“Within model, it follows that the optimal policy is
to produce at the beginning of winter and summer of the next year. In particular,
y1 = y3 = 1, y2 = y4 = 0, q1 = 2130, q2 = 0, q3 = 1790, q4 = 0, I1 = 1410,
I2 = 0, I3 = 960, I4 = 0. Total holding and set-up costs amount to ‚¬33 353.



4.6 Models with Discounts
In the previous sections it has been assumed that the value of an item is always constant
(equal to c). In practice, quantity discounts offered by suppliers, or economies of scale
in the manufacturing processes, make the value of an item dependent on the order
size q. In the following, quantity-discounts-on-all-units and incremental quantity
discounts are examined under the EOQ hypothesis. Let f (q) be the value of q items.
If the value of an item is independent of q, then f (q) = cq. Otherwise, f (q) is a
concave nondecreasing function.


4.6.1 Quantity-discounts-on-all-units
The function f (q) is assumed to be piecewise linear (see Figure 4.7),
f (q) = ci q, i = 1, 2, . . . ,
qi’1 q < qi ,
where q0 = 0, q1 , . . . are known discount breaks (qi < qi+1 , i = 1, 2, . . . ), f (q0 ) =
0, and ci > ci+1 , i = 1, 2 . . . . Hence, if the order size q is included between discount
breaks qi’1 and qi , the value of every item is ci , i = 1, 2, . . . . It is worth noting
that, depending on ci coef¬cients, f (q) can be greater than f (q ) for q < q (see
Figure 4.7). In practice, the effective cost function is
f (q) = min{ci q, ci+1 qi }, i = 1, 2, . . . .
qi’1 q < qi ,
The total average cost function µ(q) can be written as
µ(q) = µi (q), i = 1, 2, . . . ,
qi’1 q < qi ,




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SOLVING INVENTORY MANAGEMENT PROBLEMS 133

f (q)




q
q0 q1 q2 q3


Figure 4.7 The value of q items: quantity-discounts-on-all-units.


where, as hi = pci , i = 1, 2, . . . ,

µi (q) = kd/q + ci d + 2 hi q, i = 1, 2, . . . .
1
(4.18)

Then, the optimal order size q — can be obtained through the following procedure.

Step 1. By imposing

dµi (qi )
= 0, i = 1, 2, . . . ,
dqi qi =qi

determine the order size qi , i = 1, 2, . . . , that minimizes µi (q),

2kd
qi = i = 1, 2, . . . .
, (4.19)
hi

Let ±
qi’1 , if qi < qi’1 ,


qi = qi , i = 1, 2, . . . . (4.20)
if qi’1 qi qi ,


qi , if qi > qi ,

Step 2. Compute the optimal solution q — = qi—— , where

i — = arg min {µ(qi— )}.
i=1,2,...




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134 SOLVING INVENTORY MANAGEMENT PROBLEMS


Maliban runs more than 200 stationery outlets in Spain. The ¬rm buys its products
from a restricted number of suppliers and stores them in a warehouse located near
Sevilla. Maliban expects to sell 3000 boxes of the Prince Arthur pen during the next
year. The current annual interest rate p is 30%. Placing an order costs ‚¬50. The
supplier offers a box at ‚¬3, if the amount bought is less than 500 boxes. The price
is reduced by 1% if 500“2000 boxes are ordered. Finally, if more than 2000 boxes
are ordered, an additional 0.5% discount is applied. Then, by using Equations (4.19)
and (4.20),

2 — 50 — 3000 —
q1 = = 577.35 boxes, q1 = 500 boxes,
0.30 — 3
2 — 50 — 3000 —
q2 = = 580.26 boxes, q2 = 580.26 boxes,
0.30 — 2.97
2 — 50 — 3000 —
q3 = = 581.73 boxes, q3 = 2000 boxes.
0.30 — 2.955
By comparing the corresponding annual average costs given by Equations (4.18),

the optimal order size is q — = 580 boxes (≈ q2 ), corresponding to an annual cost of
‚¬9427.



4.6.2 Incremental quantity discounts
The function f (q) is assumed to depend on q as follows (see Figure 4.8),
f (q) = f (qi’1 ) + ci (q ’ qi’1 ), i = 1, 2, . . . ,
qi’1 q < qi , (4.21)
where q0 = 0, q1 , . . . are known discount breaks (qi < qi+1 , i = 1, 2, . . . ), f (q0 ) =
0, and ci > ci+1 , i = 1, 2 . . . . Consequently, if the order size q is included between
discount breaks qi’1 and qi , the value of (q ’ qi’1 ) items is ci , the value of (qi’1 ’
qi’2 ) items is ci’1 , etc. The average total cost function µ(q) is
µ(q) = µi (q), i = 1, 2, . . . ,
qi’1 q < qi ,
where, on the basis of Equation (4.12),
kd f (q)d f (q) q
µi (q) = + +p i = 1, 2, . . . .
,
q q q2
Using Equation (4.21), µi (q), i = 1, 2, . . . , can be rewritten as
µi (q) = kd/q + [f (qi’1 ) + ci (q ’ qi’1 )]d/q
+ 2 p[f (qi’1 ) + ci (q ’ qi’1 )], i = 1, 2, . . .
1
(4.22)
The optimal order size q — can be computed through a procedure very similar to
that used in the previous case.




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SOLVING INVENTORY MANAGEMENT PROBLEMS 135
f (q)




q
q0 q1 q2 q3


Figure 4.8 The value of q items: incremental quantity discounts.

Step 1. Determine the value qi , i = 1, 2, . . . , that minimizes µi (q) by imposing
that
dµi (qi )
= 0, i = 1, 2, . . . .
dqi qi =q
i

Hence
2d[k + f (qi’1 ) ’ ci qi’1 ]
qi = i = 1, 2, . . . .
, (4.23)
pci

If qi ∈ [qi’1 , qi ], then let µi (qi ) = ∞, i = 1, 2, . . . .
/
Step 2. Compute the optimal solution q — = qi — , where

i — = arg min {µ(qi )}.
i=1,2,...



If Maliban (see the previous problem) applies an incremental quantity discount
policy, then, by using Equation (4.23),

2 — 3000 — 50
q1 = = 577.4 boxes,
0.30 — 3
2 — 3000[50 + (3 — 500) ’ (2.97 — 500)]
q2 = = 661.6 boxes,
0.30 — 2.97
2 — 3000{50 + [(3 — 500) + (2.97 — 1500)] ’ (2.955 — 2000)}
q3 =
0.30 — 2.955
= 801.9 boxes.




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136 SOLVING INVENTORY MANAGEMENT PROBLEMS


Consequently, as q1 > 500 and q3 < 2000, the optimal order size is q — =
662 boxes (≈ q2 ), corresponding to an annual average cost, given by Equation (4.22),
equal to ‚¬9501.67.



4.7 Single Stocking Point: Multicommodity
Inventory Models
When several commodities are kept in stock, their inventory policies are intertwined
because of common constraints and joint costs, as we now discuss in two separate
cases. In the ¬rst case, a limit is placed on the total investment in inventories, or on the
warehouse space. In the second case, commodities share joint ordering costs. For the
sake of simplicity, both analysis will be performed under the EOQ model hypotheses.


4.7.1 Models with capacity constraints
Let n be the number of commodities in stock and qj , j = 1, . . . , n, the amount of
commodity j ordered at each replenishment. The inventory management problem can
be formulated as follows.
Minimize
µ(q1 , . . . , qn ) (4.24)
subject to

g(q1 , . . . , qn ) b, (4.25)
q1 , . . . , qn 0, (4.26)

where the objective function (4.24) is the total average cost per time unit. Under the
EOQ hypothesis, the objective function µ(q1 , . . . , qn ) can be written as
n
µ(q1 , . . . , qn ) = µj (qj ),
j =1

where, on the basis of Equation (4.12),

µj (qj ) = kj dj /qj + cj dj + 2 hj qj , j = 1, . . . , n,
1


and quantities kj , dj , cj , hj , j = 1, . . . , n, are the ¬xed reorder cost, the demand
rate, the value, and the holding cost of item j , respectively. As is customary, hj =
pj cj , j = 1, . . . , n, where pj is the interest rate of commodity j .
Equation (4.25) is a side constraint (referred to as a capacity constraint) represent-
ing both a budget constraint or a warehouse constraint. It can usually be considered




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SOLVING INVENTORY MANAGEMENT PROBLEMS 137

as linear,
n
aj qj b, (4.27)
j =1

where aj , j = 1, . . . , n, and b are constants. As a result, problem (4.24)“(4.26)
has to be solved through iterative methods for NLP problems, such as the conjugate
gradient method. Alternatively, the following simple heuristic can be used if the
capacity constraint is linear and the interest rates are identical for all the commodities
(pj = p, j = 1, . . . , n).
Step 1. Using Equation (4.13), compute the EOQ order sizes qj , j = 1, . . . , n:

2kj dj
qj = j = 1, . . . , n.
, (4.28)
pcj

If the capacity constraint (4.27) is satis¬ed, STOP, the optimal order size for each
product j, j = 1, . . . , n, has been determined.
Step 2. Increase the interest rate p of a δ quantity to be determined. Then, the order
sizes become
2kj dj
qj (δ) = , j = 1, . . . , n. (4.29)
(p + δ)cj
Determine the value δ — satisfying the relation,
n
aj qj (δ — ) = b.
j =1

Hence,
n 2
2kj dj
1

δ= ’ p.
aj (4.30)
b cj
j =1

Insert δ — in Equations (4.29) in order to determine the order sizes qj , j = 1, . . . , n.
¯

New Frontier distributes knapsacks and suitcases in most US states. Its most suc-
cessful models are the Preppie knapsack and the Yuppie suitcase. The Preppie knap-
sack has a yearly demand of 150 000 units, a value of $30 and a yearly holding cost
equal to 20% of its value. The Yuppie suitcase has a yearly demand of 100 000 units,
a value of $45 and a yearly holding cost equal to 20% of its value. In both cases,
placing an order costs $250. The company™s management requires the average capital
invested in inventories does not exceed $75 000. This condition can be expressed by
the following constraint,
30q1 /2 + 45q2 /2 75 000,




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138 SOLVING INVENTORY MANAGEMENT PROBLEMS


where it is assumed, as a precaution, that the average inventory level is the sum
of the average inventory levels of the two items. The EOQ order sizes, given by
Equation (4.28),

2 — 250 — 150 000
q1 = = 3535.53 units,
0.2 — 30
2 — 250 — 100 000
q2 = = 2357.02 units,
0.2 — 45
do not satisfy the budget constraint. Applying the conjugated gradient method starting
from the initial values (q1 , q2 ) = (1, 1), the following solution is obtained after 300
iterations,
q1 = 2500 units,
¯
q2 = 1666.66 units,
¯
whose total cost is $9 045 000. Applying the heuristic procedure, by using Equa-
tion (4.30), the same solution is obtained. In effect,
2
2 — 250 — 150 000 45 2 — 250 — 100 000
1 30

δ= + ’ 0.2
75 000 2 30 2 45
= 0.2,
hence,

2 — 250 — 150 000
q1 =
¯ = 2500 units,
(0.2 + 0.2) 30

2 — 250 — 100 000
q2 =
¯ = 1666.66 units.
(0.2 + 0.2) 45




4.7.2 Models with joint costs
For the sake of simplicity, we assume in this section that only two commodities
are kept in the inventory. Let k1 and k2 be the ¬xed costs for reordering the two
commodities at different moments in time, and let k1’2 be the ¬xed cost for ordering
both commodities at the same time (k1’2 < k1 + k2 ). In addition, let T1 and T2 be the
time lapses between consecutive replenishments of commodities 1 and 2, respectively
(see Figure 4.9). Then,
q1 = d1 T1 , (4.31)
q2 = d2 T2 . (4.32)




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SOLVING INVENTORY MANAGEMENT PROBLEMS 139
I(t)
T2

q2




T1

q1




t

Figure 4.9 Inventory level as a function of time in case of synchronized orders.


The periodicity of a joint replenishment policy is

T = max{T1 , T2 }.

In each period T , the orders issued for the two items are

N1 = T /T1 ,
N2 = T /T2 .

N1 and N2 are positive integer numbers, one of them being equal to 1 (in the situation
depicted in Figure 4.9, N1 = 3 and N2 = 1). During each period T , two items are
ordered simultaneously exactly once. Moreover, Nj ’ 1 single orders are placed for
each item j, j = 1, 2. Hence, the total average cost per time unit is

µ(T , N1 , N2 )
k1’2 + (N1 ’ 1)k1 + (N2 ’ 1)k2 h 1 d1 T h2 d2 T
= + c1 d1 + c2 d2 + + . (4.33)
T 2N1 2N2
By solving the equation,

= 0,
µ(T , N1 , N2 )
‚T T =T —

the value T — (N1 , N2 ) that minimizes µ(T , N1 , N2 ) is obtained,

2N1 N2 [k1’2 + (N1 ’ 1)k1 + (N2 ’ 1)k2 ]
T — (N1 , N2 ) = , (4.34)
h1 d1 N2 + h2 d2 N1

as a function of N1 and N2 .




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140 SOLVING INVENTORY MANAGEMENT PROBLEMS


Shamrock Microelectronics Ltd is an Irish company which assembles printed cir-
cuit boards (PCBs) for a number of major companies in the appliance sector. The Y 23
PCB has an annual demand of 3000 units, a value of ‚¬30 and a holding cost equal to
20% of its value. The Y 24 PCB has an annual request of 5000 units, a value of ‚¬40
and a holding cost equal to 25% of its value. The cost of issuing a joint order is ‚¬300
while ordering a single item costs ‚¬250. If no joint orders are placed, the order sizes
are, according to Equation (4.13),

2 — 250 — 3000

q1 = = 500 units,
0.2 — 30
2 — 250 — 5000

q2 = = 500 units.
0.25 — 40
From Equations (4.31) and (4.32):
T1— = 500/3000 = 1/6,
T2— = 500/5000 = 1/10.
This means that Shamrock would issue 1/T1— = 6 orders per year of Y 23 PCB and
1/T2— = 10 orders per year of the Y 24 PCB. Since
250 — 3000 0.2 — 30 — 500

µ1 (q1 ) = + 30 — 3000 +
500 2
= 93 000 euros per year,
250 — 5000 0.25 — 40 — 500

µ2 (q2 ) = + 40 — 5000 +
500 2
= 205 000 euros per year,
the average annual cost is ‚¬298 000 per year. If a joint order is placed and N1 = 1,
N2 = 2, the periodicity of joint orders is, according to Equation (4.34),

2 — 1 — 2 — (300 + 250)
T— = = 0.16.
0.2 — 30 — 3000 — 2 + 0.25 — 40 — 5000 — 1
Shamrock would issue 1/T — = 6.25 joint orders per year. The annual average cost,
computed through Equation (4.33), is equal to
300 + 250 0.2 — 30 — 3000 — 0.16
µ(T — , 1, 2) = + 30 — 3000 + 40 — 5000 +
0.16 2
0.25 — 40 — 5000 — 0.16
+
2—2
= 296 877.5 euros per year.




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SOLVING INVENTORY MANAGEMENT PROBLEMS 141

4.8 Stochastic Models
Inventory problems with uncertain demand or lead times have quite a complex math-
ematical structure. In this section, a restricted number of stochastic models are illus-
trated. First, the classical Newsboy Problem, where a one-shot reorder decision has
to be made, is examined. Then, (s, S) policies are introduced for a variant of the
Newsboy Problem. Finally, the most common inventory policies used by practition-
ers (namely, the reorder level method, the reorder cycle method, the (s, S) method
and the two-bin technique) are reviewed and compared. The ¬rst three policies make

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