Single Resource Revenue Management Problems withDependent De.docx

Single Resource Revenue Management Problems with
Dependent Demands c©
Guillermo Gallego
Spring 2013
Abstract
Providers of perishable capacity, such as airline seats and hotel
rooms, use market
segmentation tools to sell the same capacity to different
customers at difference prices.
They impose time-of-purchase and usage restrictions such as
Saturday Night stays on
low fares to increase demand while limiting bookings to prevent
demand cannibalization
from customers who book late and travel for business. Capacity
providers may also
bundle or unbundle ancillary services such as advance seat
selection, meals, mileage
accrual, and luggage handling to justify price differences for the
same basic service. We
will assume that the fare class structure (a menu of prices,
restrictions and ancillary
services) is given and that the capacity provider’s goal is to
allocate capacity among
fare classes to maximize expected revenues net of the cost of
the ancillary services.
We assume that customer demand is governed by a discrete
choice model and depends
on the products available for sale. Thus, if a customer does not

find his preferred
product, he may decide not to purchase or to purchase an
alternative product in the
offer set. The problem is to decide, at each state, which set of
products to make available
for sale. Since there are an exponential number of sets that can
be offered, we study
the structure of the optimization problem to define and
characterize efficient sets with
the purpose of reducing the dimensionality of the optimization
problem and to obtain
insights into the type of products, e.g., nested-by-fares, that
may be optimal to offer. We
provide a dynamic programming formulation for the revenue
management problem and
use fare and demand transformations to relate it to the
formulation of the independent
demand model. We present an upper bound on the value
function throughout the use
of approximate dynamic programming with affine functions. A
formulation that does
not allow fares to be opened once they are closed is also
presented. We then turn
our attention to static models, where the time element is
suppressed. We present an
optimal solution to the two fare problem and an efficient
heuristic for the multi-fare
case. Numerical examples show that the proposed heuristic
works almost as well as the
optimal solution for the restricted dynamic programming model
where fares cannot be
reopened once they are closed.
1

1 Introduction
Suppliers of perishable capacity often offer a menu of products
that vary in terms of price and
quality. If the products differ only on price we would expect
most, if not all, customers to
buy the lowest priced product that gives him a positive surplus
and to walk away otherwise.
When differences in quality are also present, customers do not
always buy the lowest priced
product. As an example, a customer may be willing to pay an
extra $50 for a room with an
ocean view or $29 for advance seat selection and priority
boarding.
When the difference between products in menu is small, there is
a real likelihood that a
customer will buy a different product if his preferred product is
not available. This is known
as demand recapture. Demand that is lost because a product is
not offered is called spilled
demand. Under the independent demand model, all demand for
an unavailable product is
spilled. In contrast, under a discrete choice model, part of this
demand may be recaptured
by other available products. In the context of revenue
management, the independent demand
model results in low protection levels for higher fare classes as
it ignores demand recapture.
This causes higher spill rates among the higher fare classes and
leads to a downward spiral in
revenues as estimates of demands for high fare classes become
lower over time as explained in
Cooper, Homem de Mello and Kleywegt [4].

Revenue mangers have been struggling for decades with the
problem of finding optimal
control mechanisms for fare class structures with dependent
demands. Many attempts have
been made by research groups in industry to cope with this
problem. At the same time aca-
demic research has moved vigorously to tackle the problem of
capacity allocation under choice
models. A key difference between is how the notion of time is
handled. Practitioners tend to
prefer to model time implicitly, e.g., through static models that
are extensions of Littlewood’s
rule and EMSR type heuristics. However, finding the right way
to extend Littlewood’s rule
proved to be more difficult than anticipated. The key
complication is that the marginal value
of capacity is more difficult to compute due to the fact that
protecting an additional unit of
capacity for later sale also alters the number of potential
customers (as some of their demand
may be recaptured). Nevertheless, it is possible to overcome
this difficulty and find static
heuristic policies that are easy to implement and perform almost
as well as dynamic policies.
Academic researchers tend to prefer models where time is
handled explicitly. In contrast to
the independent demand formulations, where static models are
relatively easier to understand
and implement, the opposite seems to be true for dependent
demand models. Indeed, dynamic
formulations of the dependent demand model are only
marginally more complicated to set
up. The problem is a bit more difficult because it requires
specifying a choice model and
identifying the collection of “efficient” sets of products that

may be offered at different states
of the system. Once the efficient sets are identified, either
exactly or approximately, the
problem can be reformulated to be essentially equivalent to that
of the independent demand
model.
In this chapter we will explore both formulations. We start by
introducing discrete choice
models as a tool to model demand dependencies. After
introducing choice models and provid-
ing a variety of examples, we discuss the sales and revenue
rates associated with each set and
give a formal definition of efficient sets. Essentially efficient
sets maximize the revenue rate
2
among all convex combinations of sets whose sales rate is
bounded by the sales rate of the
efficient set. The notion of efficient sets helps simplify the the
dynamic programming formu-
lations so only efficient sets need to be considered. Structural
results are presented that are
useful both in terms of computations and in understanding
dynamic policies. In particular,
we relate the formulation to that of the independent demand
model and to dynamic pricing
with finite price menus. We then present an upper bound on
revenues, and a variant of the
dynamic model where fares cannot be opened once they are
closed. We end the chapter with
a discussion of static models that handle time implicitly. As the
reader will see, these models

are complicated by the fact that changing the protection level
also changes the number of
potential customers for higher fare classes.
2 Discrete Choice Models
We will assume that the set of potential products to be offered
is N = {1, . . . ,n}. For any
subset S ⊂ N, denote by πj(S) the probability that a customer
will select j ∈ S with πj(S) = 0
if j ∈ S′ = {j ∈ N : j /∈ S}. Let π(S) =
∑
j∈ S πj(S) denote the probability of sale when subset
S is offered. The complement π0(S) = 1−π(S) denotes the
probability that a customer selects
an outside alternative. An outside alternative may mean either
that the customer does not
purchase or that he purchases from another vendor. Let S+ =
S∪ {0} denote the set of offered
products together with the outside alternative. Then πS+ (S) =
π(S) + π0(S) = 1. Notice
that this notation implies that the no-purchase alternative j = 0
is always available. For this
reason, it would be more appropriate to write πj(S+) for j ∈ S+.
However, we follow here the
convention of writing πj(S) instead of the more cumbersome
πj(S+) with the understanding
that the no-purchase alternative j = 0 is always available. This
choice of notation makes it
easier to refer to the set S ⊂ N of offered products, but the
reader needs to remember that
implicitly the no-purchase alternative is always available to the
customer. We will describe a
few commonly used choice models, define efficient sets, and

then move on to the problem of
dynamic capacity allocation.
One reason to look at discrete choice models is to better
understand how customers make
choices. Another important reason for our purposes is to find a
subset of products that
maximizes the expected profit or expected revenues among all S
⊂ N.
2.1 The Independent Demand Model (IDM)
Under this model πj(S) is independent of the offer set S
(containing j). Under the IDM all
the demand for fare j is lost (to the no-purchase alternative) if j
is removed from S. This
implies that there are non-negative constants, say v0,vj,j ∈ N,
such that
πj(S) =
vj
v0 + V (N)
j ∈ S
where V (N) =
∑
j∈ N vj ( the constants can be normalized so that v0 + V (N) =
1). In most
practical situations, it is reasonable to expect that some of the
demand for fare j may be
3

recaptured by other products in S. The IDM is pessimistic in the
sense that it ignores recap-
ture. This may lead to incorrect decisions in the context it is
used. In Revenue Management
applications it usually leads to offering too much capacity at
discounted fares. In assortment
planning, it may lead to offering excessive product variety.
2.2 The Basic Attraction Model and the Multinomial Logit
Model
The Basic Attraction Model (BAM) is a discrete choice model
where each fare j ∈ N has an
attraction value vj > 0 and v0 > 0 represents the attractiveness
of the no-purchase alternative.
The choice model is given by
πj(S) =
vj
v0 + V (S)
∀ j ∈ S, (1)
where V (S) =
∑
j∈ S vj. Consequently, products with higher attraction values are
more likely
to be selected.
The BAM was first proposed by Luce [12] who postulated two
choice axioms and demon-
strated that a discrete choice model satisfies the axioms if and
only if it is of the BAM form.
To adequately describe the Luce Axioms we need additional

notation. For any S ⊂ T we let
πS(T) =
∑
j∈ S πj(T) denote the probability that a customer selects a
product in S when the
set T is offered. Also, πS+ (T) = πS(T) + π0(T) = 1 −πT−S(T),
where set differences T −S
mean T ∩S′. The Luce Axioms can be written as:
• Axiom 1: If πi({i}) ∈ (0, 1) for all i ∈ T, then for any R ⊂ S+,
S ⊂ T
πR(T) = πR(S)πS+ (T).
• Axiom 2: If πi({i}) = 0 for some i ∈ T, then for any S ⊂ T
such that i ∈ S
πS(T) = πS−{i}(T −{i}).
Axiom 1 implies that the probability of selecting any set R ⊂
S+, when set T is offered, is
equal to the probability of selecting R when S is offered times
the probability of selecting S+
when T is offered assuming that S ⊂ T. Axiom 2 implies that if
alternative i has no probability
of being chosen, then it can be deleted from S without affecting
the choice probabilities (Luce
[13]).
The celebrated multinomial logit (MNL) model, see McFadden
[16], is a special case of
the BAM that arises from a random utility model. Under a
random utility model, each
product has random utility Ui, i ∈ N+ and the probability that j

∈ S is selected is given
by πj(S) = P(Uj ≥ Ui, i ∈ S+). In words, product j ∈ S is
selected if it gives as much
utility as any other product in S+. More specifically, McFadden
models the random utility of
product i, as Ui = µi + �i, where µi is the mean utility of
product i and depends on the price
and quality of product i. The term �i is modeled as an extreme
value distribution known as
a Gumbel random variable with parameter φ. The �is are
assumed to be independent and
4
identically distributed. We can think of �i as an idiosyncratic
variation on the mean utility
or as errors in measuring the utility. Under these assumptions,
πj(S) is of the BAM form
with vi = e
φµi, i ∈ N. The parameter φ is inversely related to the variance
of the �js. As
φ becomes large, the variance of the �js, becomes small, and
customers will gravitate to the
product in S with the largest mean utility µi (what is known as
the maximum utility model).
On the other hand, when φ becomes small the probability of
selecting any i ∈ S converges
to a uniform distribution where each product is equally likely to
be selected. This is because
when the variance is much larger than the mean, the customer
loses the ability to reliability
select products with higher mean utility.

2.3 The Generalized Attraction Model
There is considerable empirical evidence that the basic
attraction model (BAM) may be too
optimistic in estimating demand recapture probabilities when
the customer’s first choice is
not part of the offer set S. The BAM assumes that even if a
customer prefers j ∈ S′, he must
select among k ∈ S+. This ignores the possibility that the
customer may look for products
j ∈ S′ elsewhere or at a later time. As an example, suppose that
a customer prefers a certain
wine, and the store does not have it. The customer may then
either buy one of the wines
in the store, go home without purchasing, or drive to another
store and look for the specific
wine he wants. The BAM precludes the last possibility; it
implicitly assumes that the search
cost for an alternative source of product j ∈ S′ is infinity, or
equivalently that there is no
competition.
As an illustration, suppose that the consideration set is N = {1,
2} and that v0 = v1 =
v2 = 1, so πk({1, 2}) = 33.3̄% for k = 0, 1, 2. Under the BAM,
eliminating choice 2 results in
πk({1}) = 50% for k = 0, 1. Suppose, however, that product 2 is
available across town and
that the customer’s attraction for product 2 from the alternative
source is w2 = 0.5 ∈ [0,v2].
Then his choice set, when product 2 is not offered, is in reality
S = {1, 2′} with 2′ representing
product 2 in the alternative location with shadow attraction w2.
A customer who arrives to
the store has attractiveness v0 = v1 = 1 and w2 = 0.5, resulting
in

π0({1}) =
1.5
2.5
= 60%, π1({1}) =
1
2.5
= 40%.
This formulation may also help mitigate the optimism of the
BAM in inter-temporal models
where a choice j ∈ S′ may become available at a later time. In
this case w2 is the shadow
attraction of choice 2 discounted by time and the risk that it
may not be available in the
future.
To formally define the general attraction model (GAM),
introduced by Gallego, Ratliff
and Shebalov [9], we assume that in addition to the attraction
values vk,k ∈ N = {1, . . . ,n},
there are shadow attraction values wk ∈ [0,vk],k ∈ N such that
for any subset S ⊂ N
πj(S) =
vj
v0 + W(S′) + V (S)
j ∈ S, (2)
5

where W(R) =
∑
j∈ R wj for all R ⊂ N. For the GAM,
π0(S) =
v0 + W(S
′)
v0 + W(S′) + V (S)
is the probability of the no-purchase alternative. The case wk =
0,k ∈ N recovers the BAM,
while the case wk = vk,k ∈ N recovers the independent demand
model (IDM). As with the
BAM it is possible to normalize the parameters so that v0 = 1
when v0 > 0. The parsimonious
GAM (p-GAM) is given by wj = θvj ∀ j ∈ N for some θ ∈ [0,
1]. The p-GAM can serve to test
the competitiveness of the market, by testing the hypothesis H0
: θ = 0 or H0 : θ = 1 against
obvious alternatives to determine whether one is better off
deviating, respectively, from either
the BAM or the IDM.
There is an alternative, perhaps simpler, way of presenting the
GAM by using the following
transformation: ṽ0 = v0 + w(N) and ṽk = vk −wk,k ∈ N. For S
∈ N, let Ṽ (S) =
∑
j∈S ṽj.
With this notation the GAM becomes:

πj(S) =
vj
ṽ0 + Ṽ (S)
∀ j ∈ S and π0(S) = 1 −π(S). (3)
For S ⊂ T we will use the notation πS+ (T) =
∑
j∈ S+ πj(T).
Gallego, Ratliff and Shebalov [9] proposed the following
generalization to the Luce Axioms:
• Axiom 1’: If πi({i}) ∈ (0, 1) for all i ∈ T, then for any non-
negative R ⊂ S ⊂ T
πR(T)
πR(S)
= 1 −
∑
j∈ T−S
(1 −θj)πj(T)
for some set of values θj ∈ [0, 1],j ∈ N.
• Axiom 2: If πi({i}) = 0 for some i ∈ T, then for any S ⊂ T
such that i ∈ S
πS(T) = πS−{i}(T −{i}).
They call the set of Axioms 1’ and 2 the Generalized Luce
Axioms (GLA). The special
case θj = 0 for all j recovers the original Luce Axiom 1,

resulting in the BAM, while the
case θj = 1 for all j reduces to the IDM. Gallego et al [9]
establish the following connection
between the GLA and the GAM. The proof of Theorem 1 can be
found in the Appendix.
Theorem 1 A Discrete Choice Model satisfies the GLA if and
only if is of the GAM form.
We will now show that the GAM also arises as the limit of the
nested logit (NL) model.
The NL model was originally proposed in Domencich and
McFadden (1975) [5], and later
refined in McFadden(1978) [15], where it is shown that it
belongs to the class of Generalized
Extreme Value (GEV) family of models. Under the nested
choice model, customers first select
a nest and then an offering within the nest. The nests may
correspond to product categories
6
and the offerings within a nest may correspond to different
variants of the product category.
As an example, the product categories may be the different
modes of transportation (car vs.
bus) and the variants may be the different alternatives for each
mode of transportation (e.g.,
the blue and the red buses). As an alternative, a product
category may consist of a single
product, and the variants may be different offerings of the same
product by different vendors.
We are interested in the case where the random utility of the
different variants of a product

category are highly correlated. The NL model, allows for
correlations for variants in nest i
through the dissimilarity parameter γi. Indeed, if ρi is the
correlation between the random
utilities of nest i offerings, then γi =
√
1 −ρi is the dissimilarity parameter for the nest. The
case γi = 1 corresponds to uncorrelated products, and to a BAM.
The MNL is the special
case where the idiosyncratic part of the utilities are independent
and identically distributed
Gumbel random variables.
Consider now a NL model where the nests corresponds to
individual products offered in
the market. Customers first select a product, and then one of the
vendors offering the selected
product. Let Ok be the set of vendors offering product k, Sl be
the set of products offered
by vendor l, and vkl be the attraction value of product k offered
by vendor l. Under the NL
model, the probability that a customer selects product i ∈ Sj is
given by
πi(Sj) =
(∑
l∈ Oi v
1/γi
il
)γi
v0 +

∑
k
(∑
l∈ Ok v
1/γk
kl
)γk v
1/γi
ij∑
l∈ Oi v
1/γi
il
, (4)
where the first term is the probability that product i is selected
and the second term the
probability that vendor j is selected. Many authors, e.g.,
[Greene(1984)], use what is know
as the non-normalized nested models where vijs are not raised
to the power 1/γi. This is
sometimes done for convenience by simply redefining vkl ← v
1/γk
kl . There is no danger in using
this transformation as long as the γs are fixed. However, the
normalized model presented here
is consistent with random utility models, see [Train(2002)], and
we use the explicit formulation
because we will be taking limits as the γs go to zero.

It is easy to see that πi(Sj) is a BAM for each j, when γi = 1 for
all i. This case can be
viewed as a random utility model, where an independent
Gumbel random variable is associated
with each product and each vendor. Consider now the case
where γi ↓ 0 for all i. At γi = 0, the
random utilities of the different offerings of product i are
perfectly and positively correlated.
This makes sense when the products are identical and price and
location are the only things
differentiating vendors. When these differences are captured by
the deterministic part of the
utility, then a single Gumbel random variable is associated with
each product. In the limit,
customers select among available products and then buy from
the most attractive vendor. If
several vendors offer the same attraction value, then customers
select randomly among such
vendors. The next result shows that a GAM arises for each
vendor, as γi ↓ 0 for all i.
Theorem 2 The limit of (4) as γl ↓ 0 for all l is a GAM. More
precisely, there are attraction
values akj, ãkj ∈ [0,akj], and ã0j ≥ 0, such that
πi(Sj) =
aij
ã0j +
∑
k∈Sj ãkj
∀ i ∈ Sj ∀ j.
7

This shows that if customers first select the product and then
the vendor, then the NL
choice model becomes a GAM for each vendor as the
dissimilarity parameters are driven down
to zero. The proof of this result is in the Appendix. Theorem 2
justifies using a GAM when
some or all of the products offered by a vendor are also offered
by other vendors, and customers
have a good idea of the price and convenience of buying from
different vendors. The model
may also be a reasonable approximation when products in a nest
are close substitutes, e.g.,
when the correlations are high but not equal to one. Although
we have cast the justification
as a model that arises from external competition, the GAM also
arises as the limit of a NL
model where a firm has multiple versions of products in a nest.
At the limit, customers go
for the product with the highest attraction within each nest.
Removing the product with the
highest attraction from a nest shifts part of the demand to the
product with the second highest
attraction , and part to the no-purchase alternative.
Consequently a GAM of this form can
be used in Revenue Management to model buy up behavior
when there are no fences, and
customers either buy the lowest available fare for each product
or do not purchase.
2.4 Mixtures of BAMs
It can be shown that any discrete choice model that arrises from
a random utility model can

be approximated to any degree of accuracy by a mixture of
BAMs, see McFadden and Train
(2000) [17]. Unfortunately, as we will later see, the mixture of
logits model leads to difficulties
in optimization when used in RM or assortment planning.
2.5 Markov Driven Discrete Choice Models
Blanchet, Gallego and Goyal (2013), see [2], consider a Markov
Driven Discrete Choice Model
(MDDCM) that is characterized by the probability distribution
qi = πi(N), i ∈ N+ and by a
product substitution matrix Q = (qij), i,j ∈ N+. The vector q of
probabilities (known as first
choice probabilities) corresponding to the choice distribution
that arises when all the products
are available. The substitution probability qij is the probability
that a customer whose first
choice demand is i will substitute to product j when i is
unavailable, with qii = 0 for i 6= 0.
For i 6= 0, qi0 representing the probability that the customer
substitutes i for the no-purchase
alternative. We define q0i = 0 for all i 6= 0 and q00 = 1 as there
is no substitution from the
no-purchase alternative. Mathematically, we can define the
substitution matrix for i,j ∈ N,
i 6= j, via the equation qij = (πj(N − i) − qj)/qi, where N − i = {j
∈ N : j 6= i}. Notice that
qij is the rate at which customers substitute to j when the find i
unavailable. A discrete choice
model arises under the Markovian assumption that if j is also
unavailable, then the customer
(who originally preferred i) will now substitute according to qjk
for k 6= j. As we will explain
later, it is relatively easy to compute the sales rate π(S) and the
revenue rate r(S) associated

with any subset S ⊂ N. Moreover, it is also easy to find the
assortment that maximizes r(S).
One advantage of the MDDCM is that it is very easy to do
assortment planning and RM
optimization with it. Also, the MDDCM can be used to
accurately approximate the behavior
of a mixture of BAMs. This approximation, together with the
ease of optimization, allows for
efficient heuristics for RM and assortment planning for the
mixture of BAMs.
8
2.6 Revenues and Efficient Assortments
We assume without loss of generality that the products are
labeled so that their associated
fares are decreasing1 in j. More precisely, the fares are given by
p1 ≥ p2 ≥ . . . ≥ pn. For any
subset S ⊂ N, let r(S) =
∑
k∈ S pkπk(S) denote the expected revenue when set S is offered.
We will now present a number of examples, and for each
example, we will look into the
collection (π(S),r(S)),S ⊂ N, representing the sales probability
and the expected revenue
from offering set S ⊂ N. Certain of these sets will be special
and play an important role in
revenue management.
Example 1. (BAM) Assume that customers have linear
sensitivities to price and quality:

βp = −1 and βq = 1000. Then a product with price p and quality
q has mean utility
µ = βpp + βqq = −p + 1000q. If φ = .01 then the attractiveness
of the product is v =
exp(.01(−p + 1000q)). Table 1 shows the price and quality of
three different products as well
as µ and v. Table 2 shows the sale probability π(S) and the
expected revenue r(S) of all
the eight subsets, in increasing order of π(S) assuming that v0 =
0. Some subsets in Table 2
are in bold and correspond to the efficient sets: E0 = ∅ , E1 =
{1} and E2 = {1, 2}. We
will later give a precise definition of efficiency but roughly
speaking the efficient sets can be
graphically represented as those that lie in the least, increasing
concave majorant of the graph
(π(S),r(S)),S ⊂ N as can be seen in Figure 1, where a graph of
the (π(S),r(S)) in increasing
order of π(S) as well as the upper concave majorant that goes
through the efficient sets.
(p1,q1) (p2,q2) (p3,q3)
(1000, 1) (850, 0.9) (650, 0.5)
(µ1,v1) (µ2,v2) (µ3,v3)
(0, 1.0) (50, 1.65) (−150, 0.22)
Table 1: Parameters of MNL model with βp = −1, βq = 1000 and
v0 = 1
S π(S) r(S)
∅ 0% $0.00
{3} 18% $118.58
{1} 50% $500.00
{1,3} 55% $515.06
{2} 62% $529.09
{2,3} 65% $538.48

{1,2} 73% $658.15
{1,2,3} 74% $657.68
Table 2: Sets, Sale Rates and Revenue Rates (Efficient Sets are
in Bold)
Example 2. (GAM) We reconsider Example 1 with w1 = 0, w2 =
0.5 and w3 = 0.2. This
means that there are negative externalities associated with not
offering products 2 and 3. The
sales rates and revenue rates are given by Table 3 and also in
Figure 2. The efficient sets are
now E0 = ∅ and Ei = Si = {1, . . . , i} for i = 1, 2, 3. It is
interesting to contrast Tables 2
1We will use increasing and decreasing in the weak sense unless
noted otherwise.
9
$0.00
$100.00
$200.00
$300.00
$400.00

$500.00
$600.00
$700.00
0%
10%
20%
30%
40%
50%
60%
70%
80%
90%
100%
Re
ve
nu
e
Ra
te
Demand

Rate
Revenue
Efficient
Fron<er
Figure 1: Example 1: The inefficient sets are the diamonds
below the curve
and 3. For the BAM of Table 2 offering set S2 = {1, 2} results
in revenue r(S2) = $658.15.
However offering S2 = {1, 2} in Example 2 results in a
significant reduction of revenue since
more of the demand from Product 3 is lost under the GAM than
under the BAM. Notice that
offering set S3 = {1, 2, 3} hurts revenues, relative to offering
set S2, under the BAM but helps
under the GAM. This is because under the BAM the incremental
revenue from adding fare 3
to set S2 is smaller than the loss from demand cannibalization.
In contrast, under the GAM,
the incremental revenue from adding fare 3 to set S2 is larger
than the loss from demand
cannibalization.
S π(S) r(S)
∅ 0% $0.00
{3} 13% $84.17
{1} 37% $370.37
{1,3} 45% $420.48
{2} 58% $491.94
{2,3} 65% $538.48
{1,2} 69% $623.95
{1,2,3} 74% $657.68

Table 3: Sets, Sale Rates and Revenue Rates (Efficient Sets are
in Bold)
Figure 1 shows a graph of the (π(S),r(S)) in increasing order of
π(S) as well as the efficient
frontier (upper concave majorant that goes through the efficient
sets).
Example 3. (Mixtures of BAMs) Consider the following mixture
of BAMs with three products
and two customer classes taken from Rusmevichientong,
Shmoys and Topaloglu [18]. The fares
are p1 = 80,p2 = 40 and p3 = 30. An arriving customer may
follow one of two BAMs, each
with probability 0.5. BAM-1 has attractiveness (1, 5, 20, 1),
respectively, for the no-purchase
alternative and for the three products. BAM-2 has attractiveness
(5, 1, 50, 50), respectively,
for the no-purchase alternative and for the three products. Let
πi(S) and ri(S) denote,
respectively, the sales probability and the expected revenue
associated with BAM-i, for i =
1, 2. Given any offer set S, let π(S) = 0.5π1(S) + 0.5π2(S) be
the probability of sale for the
10
$-‐

$100.00
$200.00
$300.00
$400.00
$500.00
$600.00
$700.00

0%
10%
20%
30%
40%
50%
60%
70%
80%
90%
100%
Re
ve
nu
e
Ra
te
Demand
Rate
Revenue
Efficient
Fron=er
below the solid curve

mixture of the two BAMs and r(S) = 0.5r1(S) + 0.5r2(S) be the
expected revenue for the
mixture of the two BAMs. Table 4 shows (S,π(S),r(S)) for all S
⊂{1, 2, 3}. Notice that the
efficient sets are E0 = ∅ , E1 = {1} and E2 = {1, 3}. Figure 3
shows a graph of the (π(S),r(S))
in increasing order of π(S) as well as the efficient frontier.
S π(S) r(S)
∅ 0% $0.00
{1} 50% $40.00
{3} 70% $21.14
{1,3} 88% $44.82
{2} 93% $37.23
{1,2} 94% $41.65
{2,3} 95% $35.53
{1,2,3} 96% $39.66
Table 4: Sets, Sale Rates and Revenue Rates for Example 3
Here is another example of a mixture of BAMs where the
efficient sets are not nested.
Example 4. (Mixtures of BAMs) Consider the following mixture
with four products and
three customer classes. The fares are p1 = 11.50, p2 = 11.00, p3
= 10.80 and p4 = 10.75. The
attractiveness of the BAMs are, respectively, (1, 5, 2, 300, 1),
(1, 6, 4, 300, 1) and (1, 0, 1, 300, 7),
with the first component representing the attractiveness of the
no-purchase alternative. An
arriving customer has 1/6 probability of belonging to market
segment 1, 1/3 probability to

market segment 2 and 1/2 probability of belonging to market
segment 3. Table 5 lists the
efficient sets and the corresponding sales and revenue rates. The
efficient sets are E0 = ∅ ,
E1 = {1}, E2 = {1, 2}, E3 = {1, 4}, E4 = {1, 2, 4} and E5 = {1,
2, 3}. Notice that E3 does
not contain E2 and E5 does not contain E4. Moreover, this is an
instance where the number
of efficient sets m = 5 > n = 4.
11
$-‐
$5.00
$10.00
$15.00
$20.00

$25.00
$30.00
$35.00
$40.00
$45.00
$50.00
0%
10%
20%
30%
40%
50%
60%
70%
80%
90%
100%

Re
ve
nu
e
Ra
te
Demand
Rate
Revenue
Efficient
Fron=er
below the solid curve
S π(S) r(S)
∅ 0% $0.00
{1} 42.5% $4.88
{1,2} 69.9% $7.83
{1,4} 87.2% $9.65
{1,2,4} 89.8% $9.90
{1,2,3} 99.7% $10.77

Table 5: Sets, Sale Rates and Revenue Rates for Example 4
3 Efficient Sets and Assortment Optimization
In this section we formally define efficient sets and show that
for certain choice models the
efficient sets have desirable properties such as being nested or
nested by fares. We also look
at assortment optimization, which is essentially the problem of
finding the assortment S ⊂ N
with highest revenue r(S). The reader not interested in the
technical details may skip this
section on first reading.
For any ρ ∈ [0, 1] consider the linear program
R(ρ) = max
∑
S⊂N
r(S)t(S) (5)
subject to
∑
S⊂N
π(S)t(S) ≤ ρ
∑
S⊂N
t(S) = 1
t(S) ≥ 0 ∀ S ⊂ N.
The linear program selects a convex combination of all possible
actions (subsets of N) to

maximize the expected revenue that can be obtained subject to
the bound ρ on the probability
of sale. Later we will see that ΛR(c/Λ) is an upper bound on the
expected revenue when we
12
expect Λ customers to arrive and the capacity is c. Thus the
ratio of capacity to potential
demand c/Λ will play the role of ρ.
The decision variables in the linear program (5) are the
proportion of time t(S) ≥ 0 that
each subset S ⊂ N is offered for sale. The following results
follow from the standard theory
of parametric linear programming.
Proposition 1 R(ρ) is increasing, concave, and piece-wise linear.
We can trace the efficient frontier (ρ,R(ρ)), 0 ≤ ρ ≤ 1, by
solving the linear program (5)
parametrically (perhaps using the dual formulation and column
generation), for all 0 ≤ ρ ≤ 1.
In some cases the column generation step may be NP-hard, but
for now we will assume that
it is possible to solve the problem for all ρ ∈ [0, 1]. Sets S ⊂ N
such that r(S) < R(π(S)) are
said to be inefficient as the lie below the efficient frontier
(ρ,R(ρ)) at ρ = π(S). Sets S such
that R(π(S)) = r(S) are said to be efficient as the pair (π(S),r(S))
lies in the efficient frontier
at ρ = π(S). Equivalently, a set S is efficient if t(S) = 1 is an
optimal solution to the linear
program for ρ = π(S), as then R(ρ) = r(S). For any choice model,

let C = {E0,E1, . . . ,Em}
be the collection of efficient sets. Let πj = π(Ej) and rj = r(Ej)
for all j ∈ M = {0, 1, . . . ,m}.
We will assume from now on that the efficient sets are sorted in
increasing order of πj,j ∈ M.
Let uj = (rj − rj−1)/(πj −πj−1) be the slope joining (πj−1,rj−1)
and (πj,rj). Because R(ρ)
is increasing concave and linear between the points (πj,rj), it
follows that u1 > u2 > ... >
um ≥ 0, and that
R(ρ) = rj−1 + uj(ρ−πj−1) ∀ ρ ∈ [πj−1,πj] ∀ j = 1, 2, . . . ,m.
Notice also that R(ρ) = R(πm) = rm for all ρ > πm, so the
definition of R(ρ) can be extended
to all ρ ≥ 0.The upper concave envelopes of Figures 1, 2 and 3
are all of these form.
When Ej−1 ⊂ Ej, then uj is the marginal contribution to revenue
of the fare or fares added
to Ej−1 to form Ej. It is possible to show, through dual
feasibility, that Ej+1 is a maximizer
of the ratio (r(S) − rj)/(π(S) −πj) over all sets S such that π(S) >
πj if the resulting ratio
is non-negative (otherwise Ej = Em is the last efficient set).
Let us reconsider the choice models of the previous section to
look more closely into the
efficient sets. For Example 1, the efficient sets are E0 = ∅ , E1 =
{1} and E2 = {1, 2}. For
Example 2, the efficient sets are E0 = ∅ , E1 = {1}, E2 = {1, 2}
and E3 = {1, 2, 3}, while for
Example 3, the efficient sets are E0 = ∅ , E1 = {1} and E2 = {1,
3}. Notice that in these
examples the efficient sets are nested, and in Examples 1 and 2
the efficient sets are also

nested-by-fare. In Example 3, the efficient sets are not nested-
by fare since the efficient set
E2 = {1, 3} skips product 2 with p2 > p3. Notice that the
number of efficient sets is at most
n in Examples 1,2 and 3. In contrast, the efficient sets of
Example 4 are E0 = ∅ , E1 = {1},
E2 = {1, 2}, E3 = {1, 4}, E4 = {1, 2, 4} and E5 = {1, 2, 3}, so
the efficient sets are not nested
and the number of efficient sets is greater than n.
Talluri and van Ryzin [19], who first identified efficient sets for
discrete choice models, used
a slightly different definition. Under their definition, a set S is
inefficient, if it is possible to
form a convex combination of other sets that result in a strictly
higher revenue with the same
sale probability or if it is possible to form a convex
combination of other sets that results in
13
a strictly lower sale probability but the same revenue. They
define efficient sets as those that
are not inefficient. The two definitions are essentially
equivalent, although it is possible for
our definition to include sets that would be deemed inefficient
by the definition in [19]. As an
example, suppose that rm = rm−1 with πm > πm−1. Then
(πm,rm) is in the efficient frontier,
but it is deemed inefficient by [19], because the same revenue
rm−1 = rm can be obtained
with a lower sale probability. This is an innocuous subtlety,
because in this case the efficient
set Em will not be selected by an optimization algorithm unless

we specify that we want to
provide better customer service without sacrificing revenue.
The following result is key in simplifying the optimization
problem resulting from the
dynamic programs in the next section. In essence it reduces the
optimization from 2n subsets
S ⊂ N to just m, the number of efficient sets.
Theorem 3 For any z ≥ 0,
max
S⊂N
[r(S) −zπ(S)] = max
S∈ C
[r(S) −zπ(S)] = max
j∈ M
[rj −zπj].
Proof: It is enough to show that any maximizer of [r(S)−zπ(S)]
must be efficient. Suppose
for a contradiction that r(T) − zπ(T) > r(S) − zπ(S) for all S 6=
T ⊂ N and that T is
not an efficient set. Then there exist a set of non-negative
weights t(S) : S ⊂ N such
that R(π(T)) =
∑
S⊂N t(S)r(S) > r(T) with
∑
S⊂N t(S) = 1 and

∑
S⊂N t(S)π(S) ≤ π(T).
Consequently,
r(T) −zπ(T) < R(π(T)) −zπ(T) ≤
∑
S⊂N
[r(S) −zπ(S)]t(S) < r(T) −zπ(T),
where the first inequality follows from the inefficiency of T, the
second from
∑
S⊂N t(S)π(S) ≤
π(T) and the third from the claimed optimality of T and the fact
that t(T) < 1 as T is not
efficient. Since the displayed equation is a contradiction, it
must be that T is efficient, and
consequently we can restrict the maximization to C resulting in
maxS∈ C[r(S) − zπ(S)] =
maxj∈ M [rj −zπj].
Theorem 3 tell us that to find the best set to offer when the
marginal cost is z, we should
search for an efficient set with the largest value of rj − zπj. Now
rj − zπj > rj−1 − zπj−1 if
and only if z < uj, where uj is the slope joining (πj−1,rj−1) and
(πj,rj). From Proposition 1
the uj’s are non-negative and decreasing in j = 1, . . . ,m. To
visualize this graphically, draw
the efficient points (πj,rj),j ∈ M and a line with slope z ≥ 0
from the origin. An optimal
solution is to select the efficient set with index

j(z) = max{j ∈ M : uj > z}.
From this it is clear that j(z) < m for all z ≥ 0 whenever rm =
rm−1, because um = 0 ≤ z.
This shows that it is indeed innocuous to use our more liberal
definition of efficient sets,
which is more consistent with the definition of efficiency used
in portfolio theory in Finance.
However, if z = 0, we may want to offer Em instead of Em−1
when the slope um = 0. This is
because more customers are served by offering Em resulting in
the same revenue as Em−1. To
do this, we just modify the definition of j(z) to j(z) = max{j ∈
M : uj ≥ z}.
14
Consider the linear program
RC(ρ) = max
∑
j∈ M
rjtj
subject to
∑
j∈ M
πjtj ≤ ρ
∑
j∈ M
tj = 1

tj ≥ 0 ∀ j ∈ M.
Clearly RC(ρ) ≤ R(ρ) for all ρ ∈ [0, 1] as the linear program
RC(ρ) is more restricted. The
next result shows that in fact RC(ρ) = R(ρ) so the optimization
problem can be reduces to
the collection C of efficient sets.
Theorem 4 RC(ρ) = R(ρ) for all ρ ∈ [0, 1].
Proof: Consider the dual of the linear program defining R(ρ) in
terms of variables z and β.
The problem is to minimize ρz +β subject to π(S)z +β ≥ r(S),
∀ S ⊂ N, z ≥ 0. The solution
in terms of β is to set β = maxS⊂N [r(S) − zπ(S)]. By Theorem
3, β = maxj∈ M [rj − zπj], so
the dual problem is reduced to minimizing ρz + β subject to πjz
+ β ≥ rj,j ∈ M. The result
follows by recognizing that this is the dual of the liner program
defining RC(ρ).
3.1 Nested-by-Fare Efficient Sets
A choice model is said to have the nested-by-fare property if p1
> p2 > .. . > pn and the
collection of efficient sets C is contained in the collection
{S0,S1, . . . ,Sn} where S0 = ∅ and
Sj = {1, . . . ,j} for j = 1, . . . ,n. The next theorem, due to
Talluri and van Ryzin [19],
provides necessary and sufficient conditions for the nested-by-
fare property to hold. This
result is important because it justifies in some important cases
the optimality of the nested-
by-fare structure and because it reduces the number of efficient
fares to at most n. The

theorem requires the following simple idea that compares the
revenue performance of two
n-dimensional vectors x and y with the same sums x[1,n] =
y[1,n] = ρ, where we use the
notation x[1, i] =
∑i
j=1 xj. It turns out that if the partial sums of x dominate the
partial
sums of y, then p′x ≥ p′y for all vectors p with decreasing
components. More precisely, if
x[1, i] ≥ y[1, i] for all i = 1, . . . ,n − 1, and p1 ≥ p2 ≥ . . . ≥ pn,
then
∑n
i=1 pixi ≥
∑n
i=1 piyi.
This idea is related to the powerful concept of majorization
popularized by Marshal and Olkin
[14]. Here we will use it to compare the revenues associated
with the sale probability vector
yi = πi(T), i = 1, . . . ,n induced by a set T ⊂ N to the sales
induced by a convex combination of
the nested sets S0,S1, . . . ,Sn. More precisely, consider a
convex combination α0,α1, . . . ,αn of
non-negative values adding to one such that
∑n
k=0 αkπ(Sk) = π(T) and let xi =
∑n
k=0 αkπi(Sk)

be the sale of product i induced by this combination. Then x[1,
i] ≥ y[1, i] i = 1, . . . ,n − 1
and x[1,n] = y[1,n] imply that the revenue from the convex
combination will dominate the
revenue under set T.
15
Theorem 5 (Taluri and van Ryzin [19])A choice model has the
nested-by-fare order property
if and only if π(S) is increasing in S, and for every subset T
there is a convex combination
of π(S0),π(S1), . . . ,π(Sn) with partial sums of sales that
dominate the partial sums of sales
under T with total sales equal to total sales under T .
When such a convex combination exists, it favors sales at
higher fares and thus results in
higher expected revenues. Talluri and van Ryzin [19] show that
the conditions of Theorem 5
are satisfied by the MNL and the independent demand model.
Here we show that the nested-
by-fare property holds more generally by showing that it holds
for the parsimonious GAM
model (P-GAM) with wi = θvi for all i ∈ N for some θ ∈ [0, 1].
This model reduces to the
BAM when θ = 0 and to the IDM when θ = 1. The proof of this
result is in the Appendix.
Proposition 2 The efficient sets are nested-by-fares for the P-
GAM.
While Theorem 5 assures us that C ⊂ {S0,S1, . . . ,Sn}, the

converse does not necessarily
hold as there may be some sets in the collection C that are not
efficient.
Proposition 3 Consider the P-GAM for a system with strictly
decreasing fares p1 > p2 >
... > pn. Then a necessary and sufficient condition for all nested
sets to be efficient is
pn ≥ (1 −θ)r(Sn−1).
Proof: Consider the plot (π(Sj),r(Sj)),j = 0, . . . ,n. For the P-
GAM model we can write
r(Sj) as a convex combination of r(Sj−1) and pj/(1−θ) of the
form r(Sj) = αjr(Sj−1) + (1−
αj)pj/(1 −θ), where αj = (ṽ0 + Ṽ (Sj−1))/(ṽ0 + Ṽ (Sj)).
Consequently r(Sj) ≥ r(Sj−1) if and
only if pj ≥ (1 − θ)r(Sj−1). Suppose now that pn ≥ (1 −
θ)r(Sn−1). We will first show that
the plot is increasing, or equivalently that pj ≥ (1 − θ)r(Sj−1)
for all j. Suppose that there
is a j such that pj < (1 − θ)r(Sj−1). Then pj/(1 − θ) < r(Sj) <
r(Sj−1) and consequently
pj+1 < pj < (1 −θ)r(Sj). We can then repeat the argument to
show that pk < (1 −θ)r(Sk−1)
for all k ≥ j, but for k = n, this contradicts pn ≥ (1 −θ)r(Sn−1).
Now, π(Sj) = αjπ(Sj−1) + (1−αj)/(1−θ). This allow us to write
the marginal revenue as
uj =
r(Sj) − r(Sj−1)
π(Sj) −π(Sj−1)
=
pj − (1 −θ)r(Sj−1)
1 − (1 −θ)π(Sj−1)

(6)
To establish concavity, it is enough to show that uj is
decreasing in j. By using the known
convex combinations of r(Sj) and of π(Sj) we see that
uj+1 = uj −
pj −pj+1
αj(1 −πj−1)
< uj
where the inequality follows from pj > pj+1.
16
For the BAM, Proposition 3 implies that all nested sets are
efficient as long as pn ≥ r(Sn−1).
At the other extreme, for the IDM all nested sets are efficient as
long as pn ≥ 0. Another
consequence is that for the P-GAM, the efficient sets are S0,S1,
. . . ,Sj∗ where
j∗ = max{j ∈ N : pj ≥ (1 −θ)r(Sj−1)}.
All fares pj,j > j
∗ are inefficient and need not be considered.
3.2 Assortment Optimization under the MDDCM
Suppose we want to an offer set S that maximizes r(S) − zπ(S)
over all S ⊂ N, where the
choice model is the MDDCM characterized by (q,Q). This

Markovian assumption greatly
simplifies the optimization problem and allows the use of
successive approximations to find
the optimal revenues and the optimal offer set under the
Markovian assumption. Indeed, let
g0i = pi − z for i = 1, . . . ,n, where for convenience we define
g00 = p0 = 0. Now consider
the successive approximation scheme gk+1 = max(Qgk,p),
where now p is the vector of fares
with p0 = 0. Blanchet, Gallego and Goyal [2] show that g
k is monotonically increasing and
converges to a vector g. The vector g represents the maximum
revenue that can be obtained
from each customer request under the Markov Chain discrete
choice model. Blanchet, Gallego
and Goyal [2] show that under the Markovian assumption an
optimal offer set is is given by
S = {i : gi = g0i}, and that q′g = r(S) −zπ(S). This method can
be used to identify efficient
sets by solving the problem for all z ≥ 0, resulting in Ej,j ∈ M.
The method can be used as
a heuristic to approximately compute the collection of efficient
sets for discrete choice models,
such as the mixture of BAMs, for which the optimization is NP-
hard.
4 Dynamic Capacity Allocation Models
In this section we consider models where time is considered
explicitly. Modeling time explicitly
allows for time-varying arrival rates and time-varying discrete
choice models. We assume that
customers arrive to the system according to a time heterogenous
Poisson process with intensity
λt, 0 ≤ t ≤ T where T is the length of the horizon, and t

represents the time-to-go. Then the
number of customers that arrive during the last t units of time,
say Nt, is a Poisson random
variable with mean Λt =
∫ t
0 λsds.
We assume that customers arriving at time t select product k
from the offered set, say
S with probability πkt(S) given by the discrete choice model
prevalent at time-to-go t. Let
V (t,x) denote the maximum expected revenue that can be
attained over the last t units of the
sale horizon with x units of capacity. Assume that at time t we
offer set S ⊂ N = {1, . . . ,n}
and keep this set of fares open for δt units of time. If λtδt << 1,
the probability that a
customer arrives and requests product k ∈ S is approximately
λtπkt(S)δt so
V (t,x) = max
S⊂N
∑
k∈ S
{λtδtπkt(S)[pk + V (t− δt,x− 1)] + (1 −λtδt
∑
k∈ S
πkt(S))V (t− δt,x)} + o(δt)
17

= V (t− δt,x) + λtδt max
S⊂N
∑
k∈ S
πkt(S)[pk − ∆V (t− δt,x)] + o(δt)
= V (t− δt,x) + λtδt max
S⊂N
[rt(S) − ∆V (t− δt,x)πt(S)] + o(δt) (7)
for x ≥ 1 with boundary conditions V (t, 0) = V (0,x) = 0, t ≥ 0,
x ∈ N where rt(S) =∑
k∈ S pkπkt(S), πt(S) =
∑
k∈ S πkt(S) and ∆V (t− δt,x) = V (t− δt,x) −V (t− δt,x− 1).
The value function V (t,x) is often computed approximately by
solving a discrete time
dynamic program based on (7) that involves rescaling time and
the arrival rates, using δt = 1,
and dropping the o(δt) term. Time can be rescaled by a positive
real number, say a, such
that T ← aT is an integer by setting λt ← 1aλt/a, πjt(S) ←
πj,t/a(S). The resulting dynamic
program is given by
V (t,x) = V (t− 1,x) + λt max
S⊂N
[rt(S) − ∆V (t− 1,x)πt(S)], (8)

with the same boundary conditions. Formulation (8) is due to
Talluri and van Ryzin [19].
It reduces to the Lee and Hersh [11] model when demands are
independent. Alternatively,
instead of rescaling (7) we can subtract V (t−δt,x) from both
sides of equation (7), divide by
δt and take limits as δt ↓ 0 to obtain the Hamilton-Jacobi-
Bellman equation:
∂V (t,x)
∂t
= λt max
S⊂N
[rt(S) − ∆V (t,x)πt(S)] (9)
with the same boundary conditions.
The generic optimization problem in formulations (8) and (9) is
of the form
max
S⊂N
[rt(S) −zπt(S)]
where z ≥ 0 is the marginal value of capacity. Since there are 2n
subsets S ⊂ N = {1, . . . ,n}
the optimization requires the evaluation of a large number of
subsets, and the problem has
to be solved for different values of z as the marginal value of
capacity changes with the state
(t,x) of the system. One may wonder if there is a way to
simplify the problem by reducing the
number of subsets that we need to look at. From Theorem 3, we

know that the optimization
problem can be reduces to maxj∈ Mt [rjt − zπjt], where Mt is the
index of the collection of
efficient sets at time t, rjt = rt(Ejt) and πjt = πt(Ejt) and Ejt is
the jth efficient for the
discrete choice model prevalent at time-to-go t. This greatly
simplifies the optimization step
in the dynamic programs (8) and (9). For convenience we will
refer to action j when offering
efficient set Ejt,j ∈ Mt. If the problem of finding the efficient
sets is NP-hard, then the
Markov Chain approximation suggested in [2] can be used for
positive values of z ≥ 0, to
heuristically identify a collection of sets that are nearly
efficient and then use this collection
in the calculation.
Using the notion of efficient sets, formulation (8) reduces to
V (t,x) = V (t− 1,x) + λt max
j∈ Mt
[rjt − ∆V (t− 1,x)πjt], (10)
18
while formulation (9) reduces to
∂V (t,x)
∂t
= λt max
j∈ Mt

[rjt − ∆V (t,x)πjt]. (11)
Just as in the case of the independent demand model, the
marginal value ∆V (t,x) is
increasing in t and decreasing in x. For formulation (10), let
a(t,x) = arg max
j∈ Mt
[rjt − ∆V (t− 1,x)πjt]. (12)
For formulation (11) the definition of a(t,x) is the same except
that we use ∆V (t,x) instead
of ∆V (t− 1,x) in the right hand side.
Notice that rjt − ∆V (t − 1,x) > rj−1,t − ∆V (t − 1,x), so offering
action j (set Ejt) is
better than offering action j−1 if and only if ujt =
(rjt−rj−1,t)/(πj,t−πj−1,t) > ∆V (t−1,x).
Moreover, since the ujts are non-negative and decreasing in j, it
follows that it is optimal to
offer action
a(t,x) = max{j : ujt > ∆V (t− 1,x)},
for formulation (8). The formula for a(t,x) for formulation (9)
requires ∆V (t,x) instead of
∆V (t− 1,x). The following result is valid for both formulations
(10) and (11).
Theorem 6 (Taluri and van Ryzin ([19])) It is optimal to offer
action j = a(t,x) (efficient
set Ea(t,x),t) at state (t,x). Moreover, a(t,x) is decreasing in t
and increasing in x over every
time interval where the choice model is time invariant.

As t increases ∆V (t− 1,x) increases and the optimal solution
shifts to efficient sets with
a smaller sale probability. In contrast, as x increases, ∆V
(t−1,x) decreases, and the optimal
solutions shifts to efficient sets with larger sale probability. In
general, we cannot say that
we close lower fares when t is large (or open lower fares when x
is large) because the efficient
sets need not be nested-by-fare. There is a large class of choice
models, however, for which
the efficient sets are nested-by-fare. We have shown in Section
3 that for the parsimonious
GAM (p-GAM) with wj = θvj,j ∈ N, θ ∈ [0, 1], the efficient
sets are of the form Ej = Sj =
{1, . . . ,j} for j = 0, . . . ,m ≤ n, so the efficient sets are nested-
by-fares. Since the IDM and
the BAM correspond to the cases θ = 0 and θ = 1, this implies
that the collection of efficient
sets are nested-by-fares for both the IDM and the BAM. If the
efficient sets are nested-by-
fare, then we can talk of opening and closing fares as the state
dynamics change with the
understanding that if a fare is open, then all higher fares will be
open at the same time.
4.1 Dynamic Pricing Formulation with a Finite Price Menu
While πjt and rjt are respectively, the sales rate and the revenue
rate under action j at time t,
the average fare per unit sold under action j is qjt = rjt/qjt,
where for convenience we define
q0t = 0 for E0 = ∅ . From the increasing concavity of R(ρ), it
follows that qjt is decreasing in
j ∈ Mt. This suggest that if t is large relative to x we should
offer action 1 as this maximizes

19
the revenue per unit of capacity. However, offering action 1
may result in very slow sales and
may lead to capacity spoilage. At the other extreme, offering
action m, maximizes the revenue
rate and this is optimal when capacity is abundant as this
maximizes the revenue over the
sales horizon. For other cases, a tradeoff is needed between the
average sales price qjt and
the demand rate λtπjt associated with action j that takes into
account the marginal value of
capacity at state (t,x). This leads to the following equivalent
pricing formulation:
Then
V (t,x) = V (t− 1,x) + max
j∈ Mt
λtπjt[qjt − ∆V (t− 1,x)]. (13)
The equivalent formulation for the continuous time model (11)
is
∂V (t,x)
∂t
= max
j∈ Mt
λtπjt[qjt − ∆V (t,x)] (14)
These formulation suggest that we are selecting among actions j
∈ Mt to maximize the

sales rate λtπjt times the average fare qjt net of the marginal
value of capacity. We can think
of λtπjt as the demand rate associated with average fare qjt,
which reduces dynamic capacity
allocation models to dynamic pricing models with finite price
menus.
5 Formulation as an Independent Demand Model
Recall from Chapter 1, that the discrete-time, independent
demand, formulation with n fares
pt1 > pt2 > .. . > ptn and demand rates λ1t, . . . ,λtn, is of the
form:
V (t,x) = V (t− 1,x) +
n∑
j=1
λjt[ptj − ∆V (t− 1,x)]+, (15)
with boundary conditions V (t, 0) = V (0,x) = 0. One may
wonder, whether it is possible to
transform formulation (10) for dependent demands into
formulation (15) for demand formula-
tion, by transforming data λt and (πjt,rjt), j = 0, 1, . . . ,m for
the dependent demand model
into data (p̂ jt, λ̂tj), j = 1, . . . ,m for the independent demand
model.
One reason to wish for this is that such a transformation would
allow computer codes
available to solve (15) to solve (10). Another may be to try to
use the transformed data as
input to static models and then use static solutions or heuristics
such as the EMSR-b.

While it is indeed possible to do the transformation from (10)
into (15), there are really
no computational benefits as the running time of both (10) and
(15) are O(nct). Moreover,
as we shall see later, using the transformed data as input for the
static model is fraught with
problems.
The data transformation is to set λ̂jt = λt[πjt −πj−1,t] and p̂ jt =
ujt = (rjt −rj−1,t)/(πjt −
πj−1,t) for j ∈ Mt. The reader can verify that
∑j
k=1 λ̂jt = λtπjt and that
∑j
k=1 λ̂jtp̂ jt = λtrjt.
This transformation is in effect creating artificial products with
independent demands λ̂jt and
20
prices p̂ jt for j ∈ Mt.
Theorem 7 Formulation (10) is equivalent to
V (t,x) = V (t− 1,x) +
∑
j∈ Mt
λ̂jt[p̂ jt − ∆V (t− 1,x)]+ (16)

with boundary conditions V (t, 0) = V (0,x) = 0.
Proof of Theorem 7: Let k(t,x) = max{j : p̂ jt > ∆V (t − 1,x)}. It
is therefore optimal
to accept all of the artificial products such that j ≤ k(t,x). Notice
that a(t,x) = k(t,x) on
account of p̂ jt = ujt. Aggregating artificial products 1, . . .
,a(t,x), results in action k(t,x) =
a(t,x), corresponding to efficient set Ea(t,x),t. More precisely,
∑
j∈ Mt
λ̂jt[p̂ jt − ∆V (t− 1,x)]+ =
∑
j≤k(t,x)
λ̂jt[p̂ jt − ∆V (t− 1,x)]
=
∑
j≤a(t,x)
λ̂jt[p̂ jt − ∆V (t− 1,x)]
= ra(t,x),t −πa(t,x),t)∆V (t− 1,x)
= max
j∈ Mt
[rjt −πjt∆V (t− 1,x)].
Consequently,
V (t,x) = V (t− 1,x) +
∑
j∈ Mt

λ̂jt[p̂ jt − ∆V (t− 1,x)]+ = V (t− 1,x) + λt max
j∈ Mt
[rjt −πjt∆V (t,x)].
The fare and demand transformations that map λt and (πjt,rjt),j
∈ Mt into (p̂ jt, λ̂jt),j ∈
M have appeared in many papers but the original ideas go back
to Kincaid and Darlin [10] as
documented in Walczack et al. [20]. The fact that the
transformation works for the dynamic
program has lead some practitioners to conclude that the
transformed data can be used as an
input to static models where demands are treated as
independent. In particular, Fiig et al. [6],
and Walczack et al. [20], proposed feeding the transformed data
and applying the EMSR-b
heuristic, treating demands as independent under the low-to-
high arrival pattern assumption.
While this is a tempting heuristic, it would be wrong to expect
that controls obtained this way
will be close to optimal if demand dependencies are ignored. In
particular, the low-to-high
demand arrival pattern, is tantamount to assuming that Poisson
demands with parameters
λ̂jt will arrive low-to-high, but this are artificial demands from
customers willing to buy under
action j but not under action j−1. When capacity is allocated to
this marginal customers, we
cannot prevent some degree of demand cannibalization from
customers willing to buy under
action j − 1 into some of the fares in action j. We will return to
this issue in Section 8.
21

6 Upper Bound on the Value Function
We will present an an upper bound on the value functions (7) an
(8) for the case where the
choice models are time invariant. The idea is based on
approximate dynamic programming
(ADP) using affine functions and it can be easily extended to
the case where the choice models
vary over time. It is well known that a dynamic program can be
solved as a mathematical pro-
gram by making the value function at each state a decision
variable. This leads to the formula-
tion V (T,c) = min F(T,c) subject to the constraints ∂F (t,x)/∂t ≥
λt[rj−∆F(t,x)πj] ∀ (t,x)
for all j ∈ M, where the decision variables are the class of non-
negative functions F(t,x) that
are differential in x with boundary conditions F(t, 0) = F(0,x) =
0 for all t ∈ [0,T] and all
x ∈ {0, 1, . . . ,c}.
While this formulation is daunting, it becomes easier once we
restrict the functions to
be of the affine form F̃(t,x) =
∫ t
0 βs(x)ds + xzt, zt ≥ 0. We will restrict ourselves to the
case βs(x) = βs for x > 0, βs(0) = 0, zt = z for t > 0 and z0 = 0.
With this restriction,
the partial derivative and marginal value of capacity have
simple forms and the boundary
conditions are satisfied. More precisely, ∂F̃(t,x)/∂t = βt for x >
0, ∆F̃(t,x) = z, for t > 0,

F̃(t, 0) = F̃(0, t) = 0. This reduces the program to Ṽ (T,c) = min
F̃(T,c) = min
∫T
0 βtdt + cz,
subject to βt ≥ λt[rj − zπj] ∀ j ∈ M. Since we have restricted the
set of functions F(t,x)
to be affine, it follows that V (T,c) ≤ Ṽ (T,c). Since this is a
minimization problem, the
optimal choice for βt is βt = λt maxj∈ M [rj − zπij] = λtg(z)
where g(z) = maxj∈ M [rj − zπj]
is a decreasing convex, non-negative, piecewise linear function
of z. Consequently the overall
problem reduces to
Ṽ (T,c) = min
z≥0
[
∫ T
0
λtg(z)dt + cz] = min
z≥0
[ΛTg(z) + cz], (17)
where ΛT =
∫T
0 λtdt is the aggregate arrival rate over the sales horizon [0,T].
Notice that
ΛTg(z) + cz is convex in z. If the discrete choice model is time
varying, then gt(z) =
maxj∈ Mt [rjt −zπjt], result in

V
̄ (T,c) = min
z≥0
[∫ T
0
λtgt(z)dt + cz
]
,
where the objective function is also convex in z.
We can linearize (17) by introducing a new variable, say y, such
that y ≥ rj − zπj for all
j ∈ M and z ≥ 0, which results in the linear program:
Ṽ (T,c) = min
z≥0
[ΛTy + cz],
subject to ΛTy + ΛTπjzj ≥ ΛTrj j ∈ M
z ≥ 0,
where for convenience we have multiplied the constraints y +
πjz ≥ rj,j ∈ M by ΛT > 0.
22
The dual of this problem is given by
Ṽ (T,c) = ΛT max
∑

j∈ M
rjtj
subject to ΛT
∑
j∈ M
πjtj ≤ c
∑
j∈ M
tj = 1
tj ≥ 0 ∀ j ∈ M.
This linear program decides the proportion of time , tj ∈ [0, 1],
that each efficient set Ej is
offered to maximize the revenue subject to the capacity
constraint. The reader can verify that
Ṽ (T,c) is closely related to the functions RC(ρ) and R(ρ) used
to define efficient sets. The
following result establishes this relationship.
Proposition 4
Ṽ (T,c) = ΛTRC(c/ΛT ) = ΛTR(c/ΛT ).
Example 5 Suppose that p1 = 1000,p2 = 600 and a BAM with v0
= v1 = v2 = e
1. We will
assume that the aggregate arrival rate over the sales horizon
[0,T] = [0, 1] is ΛT = 40. We
know that for this problem the efficient sets are E0 = ∅ ,E1 =
{1} and E2 = {1, 2}. For c = 24,
the optimal solution is t1 = 6/15 and t2 = 9/15 with 8 units of

sales under action S1 and 16
units of sales under action S2 and revenue $20,800. Table 6
provides the upper bound Ṽ (T,c)
for different values of c. Notice that sales under action S1, first
increase and then decrease as
c increases. Table 6 also reports sales under fare 1 and fare 2,
and the reader can see that
sales under fare 1 first increase and then decrease.
c Ṽ (T,c) t1 t2 sales S1 sales S2 fare 1 sales fare 2 sales
12 12,000 0.6 0.0 12 0 12 0
16 16,000 0.8 0.0 16 0 16 0
20 20,000 1.0 0.0 20 0 20 0
22 20,400 0.7 0.3 14 8 18 4
24 20,800 0.4 0.6 8 16 16 8
26 21,200 0.1 0.9 2 24 14 12
28 21,333 0.0 1.0 0 26.6 13.3 13.3
32 21,333 0.0 1.0 0 26.6 13.3 13.3
Table 6: Upper Bound and Optimal Actions for Fluid Model
7 Uni-directional Dynamic Programming Models
Like the Lee and Hersh model, formulations (8) and (9),
implicitly assume that the capacity
provider can offer any subset of fares at any state (t,x). This
flexibility works well if customers
are not strategic. Otherwise, customers may anticipate the
possibility of lower fares and
23
postpone their purchases in the hope of being offered lower
fares at a later time. If customers

act strategically, the capacity provider may counter by imposing
restrictions that do not allow
lower fares to reopen once they are closed. Actions to limit
strategic customer behavior are
commonly employed by revenue management practitioners.
Let VS(t,x) be the optimal expected revenue when the state is
(t,x), we are allowed
to use any subset U ⊂ S of fares and fares cannot be offered
once they are closed. The
system starts at state (T,c) and S = N. If a strict subset U of S is
used then all fares in
U ′ = {j ∈ N : j /∈ U} are permanently closed and cannot be
offered at a later state regardless
of the evolution of sales. To obtain a discrete time counterpart
to (8), let
WU (t,x) = VU (t− 1,x) + λt[rt(U) −πt(U)∆VU (t− 1,x)].
Then the dynamic program is given by
VS(t,x) = max
U⊂S
WU (t,x) (18)
with boundary conditions VS(t, 0) = VS(0,x) = 0 for all t ≥ 0, x
∈ N and S ⊂ N. The goal
of this formulation is to find VN (T,c) and the corresponding
optimal control policy.
Notice that formally the state of the system has been expanded
to (S,t,x) where S is
the last offered set and (t,x) are, as usual, the time-to-go and the
remaining inventory. For-
mulation (8) requires optimizing over all subsets S ⊂ N, while

formulation (18) requires an
optimization over all subsets U ⊂ S for any given S ⊂ N.
Obviously the complexity of these
formulations is large if the number of fares is more than a
handful. Airlines typically have over
twenty different fares so the number of possible subsets gets
large very quickly. Fortunately,
in many cases we do not need to do the optimization over all
possible subsets. As we have
seen, the optimization can be reduced to the set of efficient
fares Ct at time t. For the p-GAM,
we know that Ct ⊂ {S0,S1, . . . ,Sn} where Sj = {1, . . . ,j} are
the nested-by-fares sets. For
the p-GAM, and any other model for which the nested-by-fare
property holds, the state of
the system reduces to (j,t,x) where Sj is the last offered set at
(t,x). For such models the
formulation (18) reduces to
Vj(t,x) = max
k≤j
Wk(t,x) (19)
where Vj(t,x) = VSj (t,x) and
Wk(t,x) = Vk(t− 1,x) + λt[rkt −πkt∆Vk(t− 1,x)],
rkt =
∑
l∈ Sk plπlt(Sk) and πkt =
∑
l∈ Sk πlt(Sk).
The structural results obtained for the Independent Demand

Model carry on to the p-
GAM model and for other models where the efficient sets have
the nested-by-fare property.
Let
aj(t,x) = max{k ≤ j : Wk(t,x) = Vj(t,x)}. (20)
Theorem 8 At state (j,t,x) it is optimal to offer set Saj (t,x) =
{1, . . . ,aj(t,x)}, where
aj(t,x), given by (20), is decreasing in t and increasing in x over
time intervals where the
choice model is time invariant.
24
The proof of this results follows the sample path arguments of
the corresponding proof in
the independent demand chapter. Clearly V1(t,x) ≤ V2(t,x) ≤
Vn(t,x) ≤ V (t,x) ≤ Ṽ (t,x) as
V (t,x) is not restricted, as Vj(t,x) is, to monotone offerings.
c Load Factor V3(T,c) V (T,c) Ṽ (T,c)
4 4.59 3,769 3,871 4,000
6 3.06 5,356 5,534 6,000
8 2.30 6,897 7,013 7,477
10 1.84 8,259 8,335 8,950
12 1.53 9,304 9,382 10,423
14 1.31 9,976 10,111 10,846
16 1.15 10,418 10,583 11,146
18 1.02 10,803 10,908 11,447
20 0.92 11,099 11,154 11,504
22 0.84 11,296 11,322 11,504
24 0.77 11,409 11,420 11,504

26 0.71 11,466 11,470 11,504
28 0.66 11,490 11,492 11,504
Table 7: Value Functions for Dynamic Allocation Policies
Example 6 Table 7 presents the value functions V3(T,c), V (T,c)
and the upper bound Ṽ (T,c)
that result from solving the dynamic programs (10) and (19) for
the MNL model with fares
p1 = $1, 000,p2 = $800,p3 = $500 with price sensitivity βp =
−0.0035, schedule quality
si = 200 for i = 1, 2, 3 with quality sensitivity βs = 0.005, and
an outside alternative with
p0 = $1, 100 and schedule quality s0 = 500, Gumbel parameter
φ = 1, arrival rate λ = 25
and T = 1. Recall that for the MNL model, the attractiveness vi
= e
φµi where µi is the
mean utility. In this case µi = βppi + βssi. The computations
were done with time rescaled
by a factor a = 25, 000. Not surprisingly V3(T,c) ≤ V (T,c) as
V3(T,c) constraints fares to
remain closed once they are closed for the first time. However,
the difference in revenues is
relatively small except for load factors around 1.15 where the
difference can be as large as
1.5% in revenues.
8 Static Capacity Allocation Models
We will assume that we are working with a choice model with
efficient sets that are nested:
E0 ⊂ E1 . . . ⊂ Em, even if they are not nested by fare. We
continue using the notation πj =∑
k∈ Ej πk(Ej) and rj =

∑
k∈ Ej pkπk(Ej), so the slopes uj = (rj−rj−1)/(πj−πj−1),j = 1, . . .
,m
are positive and decreasing. We will denote by qj = rj/πj the
average fare, conditioned on a
sale, under efficient set Ej (action j) for j > 1 and define q0 = 0.
Suppose that the total number of potential customers over the
selling horizon is a random
variable D. For example, D can be Poisson with parameter Λ.
Let Dj be the total demand if
only set Ej is offered. Then Dj is conditionally binomial with
parameters D and πj, so if D
is Poisson with parameter Λ, then Dj is Poisson with parameter
Λπj.
25
We will present an exact solution for the two fare class capacity
allocation problem and
a heuristic for the multi-fare case. The solution to the two fare
class problem is, in effect,
an extension of Littlewood’s rule for discrete choice models.
The heuristic for the multi-fare
problem applies the two fare result to each pair of consecutive
actions, say j and j + 1 just as
in the case of independent demands.
8.1 Two Fare Classes
While capacity is available, the capacity provider will offer
either action 2 associated with

efficient set E2 = {1, 2}, or action 1, associated with efficient
set E1 = {1}. If the provider
runs out of inventory he offers action 0, corresponding to E0 =
{0}, which is equivalent to
stop selling. We will assume action 2 will be offered first and
that y ∈ {0, . . . ,c} units of
capacity will be protected for sale under action 1.
Under action 2, sales are min(D2,c−y) and expected revenues
are q2E min(D2,c−y) as
q2 is the average fare per unit sold under action 2. Of the (D2 −
c + y)+ customers denied
bookings, a fraction β = π1/π2 will be willing to purchase under
action 1. Thus, the demand
under action 1 will be a conditionally binomial random variable,
say U(y), with a random
number (D2 − c + y)+ of trials and success probability β. The
expected revenue that results
from allowing up to c−y bookings under action 2 is given by
W2(y,c) = q1E min(U(y), max(y,c−D2)) + q2E min(D2,c−y),
where the first term corresponds to the revenue under action 1.
Notice that if D2 ≤ c − y
then U(y) = 0 since U(y) is a binomial random variable with
zero trials. Conditioning the
first term on the event D2 > c−y, allow us to write
W2(y,c) = q1E[min(U(y),y)|D2 ≥ c−y)]P(D2 > c−y) + q2E
min(D2,c−y).
The reader may be tempted to follow the marginal analysis idea
presented in Chapter 1 for the
independent demand case. In the independent demand case, the
marginal value of protecting
one more unit of capacity is realized only if the marginal unit is

sold. The counterpart here
would be P(U(y) ≥ y|D2 > c−y) and a naive application of
marginal analysis would protect
the yth unit whenever q1P(U(y) ≥ y|D2 > c−y) > q2.
However, with dependent demands, protecting one more unit of
capacity also increases
the potential demand under action 1 by one unit. This is because
an additional customer
is denied capacity under action 2 (when D2 > c − y) and this
customer may end up buying
a unit of capacity under action 1 even when not all the y units
are sold. Ignoring this
can lead to very different results in terms of protection levels.
The correct analysis is to
acknowledge that an extra unit of capacity is sold to the
marginal customer with probability
βP (U(y − 1) < y − 1|D2 > c−y). This suggest protecting the yth
unit whenever
q1[P(U(y) ≥ y|D2 > c−y) + βP (U(y − 1) < y − 1|D2 > c−y)] >
q2.
To simplify the left hand side, notice that conditioning on the
decision of the marginal customer
26
results in
P(U(y) ≥ y|D2 > c−y) = βP (U(y−1) ≥ y−1|D2 >
c−y)+(1−β)P(U(y−1) ≥ y|D2 > c−y).
Combining terms leads to protecting the yth unit whenever

q1[β + (1 −β)P(U(y − 1) ≥ y|D2 > c−y)] > q2.
Let
r = u2/q1 =
q2 −βq1
(1 −β)q1
, (21)
denote the critical fare ratio. In industry, the ratio r given by
(21) is know as fare adjusted
ratio, in contrast to the unadjusted ratio q2/q1 that results when
β = 0.
The arguments above suggests that the optimal protection level
can be obtained by se-
lecting the largest y ∈ {1, . . . ,c} such that P(U(y − 1) ≥ y|D2 >
c − y) > r provided that
P(U(0) ≥ 1|D2 ≥ c) > r and to set y = 0 otherwise.
To summarize, an optimal protection can be obtain by setting
y(c) = 0 if P(U(0) ≥ 1|D2 >
c) ≤ r and setting
y(c) = max{y ∈ {1, . . . ,c} : P(U(y − 1) ≥ y|D2 > c−y) > r}
otherwise. (22)
Formula (22) is due to Gallego, Lin and Ratliff [7]. This is
essentially a reinterpretation
of the main result in Brumelle et al. originally developed for
fares instead of actions.
One important observation is that for dependent demands the
optimal protection level

y(c) depends in c in a more complicated way than in the
independent demand model. For the
dependent demand model y(c) is first increasing and then
decreasing in c. The reason is that
for low capacity it is optimal to protect all the inventory for
sale under action 1. However, for
high capacity it is optimal to allocate all the capacity to action
2. The intuition is that action
2 has a higher revenue rate, so with high capacity we give up
trying to sell under action 1.
This is clearly seen in Table 6 of Example 5. Heuristic solutions
that propose protection levels
of the form min(yh,c), which are based on independent demand
logic, are bound to do poorly
when c is close to Λπ2. In particular, the heuristics developed
by Belobaba and Weatherford
[1], and more recently by Fiig et al. [6] and by Walczak et al.
[20] are of this form and are
therefore not recommended.
One can derive Littlewood’s rule for discrete choice models
(22) formally by analyzing
∆W2(y,c) = W2(y,c)−W2(y−1,c), the marginal value of
protecting the yth unit of capacity
for sale under action 1.
Proposition 5
∆W2(y,c) = [q1(β + (1 −β)P(U(y − 1) ≥ y|D2 > c−y) − q2]P(D2
> c−y). (23)
Moreover, the expression in brackets is decreasing in y ∈ {1, . .
. ,c}.
27

The proof of Proposition 5 is due to Gallego, Lin and Ratliff [7]
and can be found in the
Appendix. As a consequence, ∆W2(y,c) has at most one sign
change. If it does, then it must
be from positive to negative. W2(y,c) is then maximized by the
largest integer y ∈ {1, . . . ,c},
say y(c), such that ∆W2(y,c) is positive, and by y(c) = 0 if
∆W2(1,c) < 0. This confirms
Littlewood’s rule for discrete choice models (22).
8.2 Heuristic Protection Levels
In terms of computations, notice that
P(U(y − 1) ≥ y|D2 > c−y) = lim
L→∞
∑L
k=0 P(Bin(y + k,β) ≥ y)P(D2 = c + k + 1)
P(D2 > c−y)
,
where we have been careful to only include terms that can be
potentially positive in the
numerator. It is often enough to truncate the sum at a value, say
L, such that P(D2 >
L) ≤ �P(D2 > c − y) for some small �, as those terms do not
materially contribute to the
sum. While the computation of y(c) and V2(c) = W2(y(c),c) is
not numerically difficult, the
conditional probabilities involved may be difficult to
understand conceptually. Moreover, the
formulas do not provide intuition and do not generalize easily to

multiple fares. In this section
we develop a simple heuristic to find near-optimal protection
levels that provides some of the
intuition that is lacking in the computation of optimal
protection levels y(c). In addition, the
heuristic can easily be extended to multiple-fares.
The heuristic consists of approximating the conditional
binomial random variable U(y−1)
with parameters (D2−c+y−1)+ and β by its conditional
expectation, namely by (Bin(D2,β)−
β(c+ 1−y))+. Since Bin(D2,β) is just D1, the approximation
yields (D1−β(c+ 1−y))+. We
expect this approximation to be reasonable if ED1 ≥ β(c + 1−y).
In this case P(U(y−1) ≥
y|D2 > c−y) can be approximated by P(D1 ≥ (1 −β)y + β(c + 1)).
Let
yp = max{y ∈ N : P(D1 ≥ y) > r)}.
We think of yp = (1 − β)y + β(c + 1) as a pseudo-protection
level that will be modified
to obtain a heuristic protection level yh(c) when the
approximation is reasonable. Let y =
(yp −β(c + 1))/(1 −β). Then ED1 ≥ β(c + 1 −y)) is equivalent to
the condition
c < yp + d2
where d2 = E[D2] − E[D1] = Λ(π2 − π1) = Λπ2(1 − β), where
for convenience we write
β = β1 = π1/π2. Consequently, if c < y
p + d2 we set
yh(c) = max

{
y ∈ N : y ≤
yp −β(c + 1)
(1 −β)
}
∧ c.
When the condition c < yp +d2 fails, we set y
h(c) = 0. Thus, the heuristic will stop protecting
capacity for action 1 when c ≥ yp + d2. In a deterministic
setting yp = E[D1], so it stops
protecting when c ≥ E[D1] + d2 = E[D2], i.e., when there is
enough capacity to satisfy the
28
expected demand under action 2.
Notice that the heuristic involves three modifications to
Littlewood’s rule for independent
demands. First, instead of using the first choice demand D1,2
for fare 1, when both fares
are open, we use the stochastically larger demand D1 for fare 1,
when it is the only open
fare. Second, instead of using the ratio of the fares p2/p1 we use
the modified fare ratio
r = u2/q1 based on sell-up adjusted fare values. From this we
obtain a pseudo-protection level
yp that is then modified to obtain yh(c). Finally, we keep yh(c)
if c < yp +d2 and set y

h(c) = 0
otherwise. In summary, the heuristic involves a different
distribution, a fare adjustment, and a
modification to the pseudo-protection level. The following
example illustrates the performance
of the heuristic.
Example 7 Suppose that p1 = 1000,p2 = 600 and a BAM with v0
= v1 = v2 = e
1 and that
Λ = 40 as in Example 5. We report the optimal protection level
y(c), the heuristic protection
level yh(c), the upper bound V
̄ (c), the optimal expected
revenue V (c) of the uni-directional
formulation (18), the performance V2(c) of y(c) and the
performance V
h
2 (c) = W2(y
h(c),c)
of yh(c), and the percentage gap between (V2(c) − V h2
(c))/V2(c) in Table 8. Notice that the
performance of the statiheuristic, V h2 (c), is almost as good as
the performance V2(c) of the
optimal policy under the uni-directional formulation (18) where
fares classes cannot be opened
once they are closed.
c y(c) yh(c) V
̄ (c) V (c) V2(c) V
h
2 (c) Gap(%)
12 12 12 12,000 11,961 11,960 11,960 0.00

16 16 16 16,000 15,610 15,593 15,593 0.00
20 20 20 20,000 18,324 18,223 18,223 0.00
24 21 24 20,800 19,848 19,526 19,512 0.07
28 9 12 21,333 20,668 20,414 20,391 0.11
32 4 0 21,333 21,116 21,036 20,982 0.26
36 3 0 21,333 21,283 21,267 21,258 0.05
40 2 0 21,333 21,325 21,333 21,322 0.01
Table 8: Performance of Heuristic for Two Fare Example
8.3 Theft versus Standard Nesting and Arrival Patterns
The types of inventory controls used in the airline’s reservation
system along with the demand
order of arrival are additional factors that must be considered in
revenue management opti-
mization. If y(c) < c, we allow up to c−y(c) bookings under
action 2 with all sales counting
against the booking limit c−y(c). In essence, the booking limit
is imposed on action 2 (rather
than on fare 2). This is known as theft nesting. Implementing
theft nesting controls may be
tricky if a capacity provider needs to exert controls through the
use of standard nesting, i.e.,
when booking limits are only imposed on the lowest open fare.
This modification may be
required either because the system is built on the philosophy of
standard nesting or because
users are accustomed to thinking of imposing booking limits on
the lowest open fare. Here
we explore how one can adapt protection levels and booking
limits for the dependent demand
model to situations where controls must be exerted through
standard nesting.
29

A fraction of sales under action 2 correspond to sales under fare
p2. This fraction is given
by ω = π2(E2)/π2. So if booking controls need to be exerted
directly on the sales at fare p2,
we can set booking limit ω(c−y(c)) on sales at fare p2. This is
equivalent to using the larger
protection level
ŷ(c) = (1 −ω)c + ωy(c) (24)
for sales at fare 1. This modification makes implementation
easier for systems designed for
standard nesting controls and it performs very well under a
variety of demand arrival patterns.
It is possible to combine demand choice models with fare
arrival patterns by sorting cus-
tomers through their first choice demand and then assuming a
low-to-high demand arrival
pattern. For the two fare case, the first choice demands for fare
1 and fare 2 are Poisson
random variables with rates Λπ1(E2) and Λπ2(E2). Assume now
that customers whose first
choice demand is for fare 2 arrive first, perhaps because of
purchasing restrictions associated
with this fare. Customers whose first choice is fare 2 will
purchase this fare if available. They
will consider upgrading to fare 1 if fare 2 is not available. One
may wonder what kind of
control is effective to deal with this arrival pattern. It turns out
that setting protection level
ŷ(c) given by (24) for fare 1, with standard nesting, is optimal
for this arrival pattern and is

very robust to other (mixed) arrival patterns (Gallego, Li and
Ratliff [8]).
8.4 Multiple Fare Classes
For multiple fare classes, finding optimal protection levels can
be very complex. However, if
we limit our search to the best two consecutive efficient sets we
can easily adapt the results
from the two-fare class to deal with the multiple-fare class
problem. For any j ∈ {1, . . . ,n−1}
consider the problem of allocating capacity between actions j
(corresponding to efficient set
Ej) and action j + 1 (corresponding to efficient set Ej+1) where
action j + 1 is offered first.
In particular, suppose we want to protect y ≤ c units of capacity
for action j against action
j + 1. We will then sell min(Dj+1,c−y) units under action j + 1
at an average fare qj+1. We
will then move to action j with max(y,c−Dj+1) units of capacity
and residual demand Uj(y),
where Uj(y) is conditionally binomial with parameters (Dj+1 −
c + y)+ and βj = πj/πj+1.
Assuming we do not restrict sales under action j the expected
revenue under actions j + 1
and j will be given by
Wj+1(y,c) = qjE min(Uj(y), max(y,c−Dj+1)) + qj+1E
min(Dj+1,c−y). (25)
Notice that under action j we will either run out of capacity or
will run out of customers.
Indeed, if Uj(y) ≥ y then we run out of capacity, and if Uj(y) <
y then we run out of
customers. Let Wj+1(c) = maxy≤c Wj+1(y,c) and set W1(c) =
q1E min(D1,c). Clearly,

Vn(c) ≥ max
1≤j≤n
Wj(c), (26)
so a simple heuristic is to compute Wj(c) for each j ∈ {1, . . .
,n} and select j to maxi-
mize Wj(c). To find an optimal protection level for Ej against
Ej+1 we need to compute
∆Wj+1(y,c) = Wj+1(y,c)−Wj+1(y−1,c). For this we can repeat
the analysis of the two fare
case to show that an optimal protection level for action Ej
against action Ej+1 is given by
30
yj(c) = 0 if ∆Wj+1(1,c) < 0 and by
yj(c) = max{y ∈ {1, . . . ,c} : P(Uj(y − 1) ≥ y|Dj+1 > c−y) > rj},
(27)
where
rj =
uj+1
qj
=
qj+1 −βjqj
(1 −βj)qj
.

Alternatively, we can use the heuristic described in the two fare
section to approximate
Uj(y−1) by Dj−βj(c+ 1−y) and use this in turn to approximate
the conditional probability
in (27) by P(Dj ≥ y + β(c−y + 1)). This involves finding the
pseudo-protection level
y
p
j = max{y ∈ N : P(D1 ≥ y) > rj}.
If c < y
p
j + dj+1, then
yhj (c) = max
{
y ∈ N+ : y ≤
y
p
j −βj(c + 1)
1 −βj
}
∧ c, (28)
and set yh(c) = 0 if c ≥ ypj + dj+1.
We will let V hn (c) be the expected revenues resulting from
applying the protection levels.
Example 8 Consider now a three fare example with fares p1 =
1000,p2 = 800,p3 = 500,

schedule quality si = 200, i = 1, 2, 3, βp = −.0035, βs = .005, φ
= 1. Then v1 = .082,v2 =
0.165,v3 = 0.472. Assume that the outside alternative is a
product with price p0 = 1100 and
schedule quality s0 = 500 and that the expected number of
potential customers is Poisson with
parameter Λ = 25. Table 9 reports the protection levels yj(c)
and y
h
j (c) as well as V3(c) and
V h3 (c) for c ∈ {4, 6, . . . , 26, 28}. As shown in the Table, the
heuristic performs very well with
a maximum gap of 0.14% relative to V3(c) which was computed
through exhaustive search.
It is also instructive to see that V h3 (c) is not far from V3(c,T),
as reported in Table 7, for the
dynamic model. In fact, the average gap is less than 0.5% while
the largest gap is 1.0% for
c = 18.
Example 7 suggest that the heuristic for the static model works
almost as well as the
optimal dynamic program Vn(T,c) for the case where efficient
sets are nested-by-fare and
fares cannot be opened once they are closed for the first time.
Thus the multi-fare heuristic
described in this section works well to prevent strategic
customers from gaming the system
provided that the efficient fares are nested-by-fare as they are in
a number of important
applications. While the heuristic for the static model gives up a
bit in terms of performance
relative to the dynamic model, it has several advantages. First,
the static model does not

need the overall demand to be Poisson. Second, the static model
does not need as much detail
in terms of the arrival rates. These advantages are part of the
reason why people in industry
have a preference for static models, even thought dynamic
models are easier to understand,
easier to solve to optimality, and just as easy to implement.
31
c Load y1(c) y2(c) y
h
1 (c) y
h
2 (c) V3(c) V
h
3 (c) Gap(%)
4 4.59 4 4 4 4 3,769 3,769 0.00
6 3.06 3 6 3 6 5,310 5,310 0.00
8 2.30 1 8 1 8 6,845 6,845 0.00
10 1.84 0 10 0 10 8,217 8,217 0.00
12 1.53 0 12 0 12 9,288 9,288 0.00
14 1.31 0 14 0 14 9,971 9,971 0.00
16 1.15 0 13 0 14 10,357 10,354 0.02
18 1.02 0 9 0 10 10,700 10,694 0.05
20 0.92 0 5 0 6 11,019 11,019 0.00
22 0.84 0 4 0 2 11,254 11,238 0.14
24 0.77 0 3 0 0 11,391 11,388 0.03
26 0.71 0 2 0 0 11,458 11,450 0.08
28 0.66 0 2 0 0 11,488 11,485 0.03

Table 9: Performance of Heuristic for Three Fare Example
9 Acknowledgments
I acknowledge the feedback from my students and collaborators.
In particular, I would like
to recognize the contributions and feedback from Anrand Li,
Lin Li, Jing Dong and Richard
Ratliff.
32
10 Appendix
Proof of Theorem 1: If the choice model πj(S),j ∈ S satisfies
GAM then there exist constants
vi, i ∈ N+, ṽi ∈ [0,vi] i ∈ N and ṽ0 = v0 +
∑
j∈N (vj − ṽj) such that πR(S) = v(R)/(ṽ0 + Ṽ (S)
for all R ∈ S. For this choice model, the left hand side of
Axiom 1’ is (ṽ0 +Ṽ (S)/(ṽ0 +Ṽ (T)).
Let θj = 1−ṽj/vj. Then the right hand side of Axiom 1’ is given
by 1−
∑
j∈T−S(1−θj)vj/(ṽ0 +
Ṽ (T)) = (ṽ0 + Ṽ (S)/(ṽ0 + Ṽ (T)). This shows that the GAM
satisfies Axiom 1’. If πi({i}) = 0
then vi = 0 and consequently ṽi = 0. From this it follows that
πS−{i}(T −{i}) =

V (S)−vi
ṽ0+Ṽ (T)−ṽi
=
V (S)
ṽ0+Ṽ (T)
= πS(T), so Axiom 2 holds.
Conversely, suppose a choice model satisfies the GLA. Then by
selecting R = {i} ⊂ S ⊂
T = N we see from Axiom 1’ that
πi(S) =
πi(N)
1 −
∑
j/∈ S(1 −θj)πj(N)
.
Since π0(N) +
∑
j∈ N πj(N) = 1 the denominator can be written as π0(N) +
∑
j∈ N θjπj(N) +∑
j∈ S(1 −θj)πj(N), resulting in
πi(S) =
πi(N)
π0(N) +

∑
j∈ N θjπj(N) +
∑
j∈ S(1 −θj)πj(N)
.
Letting vj = πj(N) for all j ∈ N+, ṽ0 = v0 +
∑
j∈N θjvj and ṽj = (1 −θj)vj ∈ [0, 1] for j ∈ N
the choice model can be written as
πi(S) =
vi
ṽ0 + ṽ(S)
,
so it satisfies the GAM.
Proof of Theorem 2: Let N ′ be the universe of products. The set
of vendors is M =
{1, . . . ,m}. Let vij be the attraction value of product i offered
by vendor j, and let Sj ⊂ N ′
be the products offered by vendor j ∈ M. Let Oi = {j ∈ M : i ∈
Sj} be the set of vendors
offering product i, and N = {i ∈ N ′ : Oi 6= ∅ } be the collection
of products that are offered.
We consider a NL model where customers first select a product
and then a vendor. Let γi
be the dissimilarity parameter of nest i. Under the nested model,
a nest i with Oi 6= ∅ , is
selected with probability (∑

l∈ Oi v
1/γi
il
)γi
v0 +
∑
k
(∑
l∈ Ok v
1/γk
kl
)γk . (29)
The conditional probability that a customer who selects nest i
will buy the product from
vendor j ∈ Oi is given by
v
1/γi
ij∑
l∈ Oi v
1/γi
il
. (30)
Consequently, the probability that a customer selects product i
∈ Sj from vendor j is given
33

by (∑
l∈ Oi v
1/γi
il
)γi
v0 +
∑
k
(∑
l∈ Ok v
1/γk
kl
)γk v
1/γi
ij∑
l∈ Oi v
1/γi
il
.
We are interested in the limit of these probabilities as γk ↓ 0 for
all k. For this purpose
we remark that well know fact: limγk↓0(
∑

l∈ Ok v
1/γk
il )
γk = maxl∈ Oi vil. The limit, say pij, of
equation (30) is equal to 1 if vendor j is the most attractive, i.e.,
if vij > vil for all l ∈ Oi,
l 6= j; pij = 0, if vij < maxl∈ Oi vil; and pij = 1/m if vij =
maxl∈ Oi vil but vendor j is tied with
other m− 1 vendors.
Now,
πi(Sj) =
pij maxl∈ Oi vil
v0 +
∑
k maxl∈ Ok vkl
.
For all k ∈ N, define V −kj = maxl∈ Ok,l 6=j vkl and Vkj =
maxl∈ Ok∪ {j} vkl. Then, for i ∈ Sj,
πi(Sj) =
Vijpij
v0 +
∑
k∈ Sj Vkj +
∑
k∈S
̄ j V

−
kj
,
where S
̄ j = {i ∈ N : i /∈ Sj}. The selection probability can be
written as
πi(Sj) =
Vijpij
v0 +
∑
k∈ N V
−
kj +
∑
k∈ Sj (Vkj −V
−
kj )
,
or equivalently, as
πi(Sj) =
Vijpij
v0 +
∑
k∈ N V
−

kj +
∑
k∈ Sj (Vkj −V
−
kj )pkj
,
since Vkj − V −kj = (Vkj − V
−
kj )pkj for all k. This is because pkj < 1 implies Vkj − V
−
kj = 0.
This shows that πi(Sj) corresponds to a GAM with ã0j = v0 +
∑
k∈ N V
−
kj , akj = pkjVkj, and
ãkj = (Vkj −V −kj )pkj.
Proof of Proposition 2: Notice that π(S) is increasing in S for
the GAM since V (S) ≥
Ṽ (S) ≥ 0 are both increasing in S. Since the P-GAM is a special
case π(S) is increasing in S.
Now, consider a set T that is not nested-by-fare and let ρ =
π(T). Then there exist an index
k such that πk < ρ ≤ πk+1. Let tk = (πk+1 −ρ)/(πk+1 −πk), tk+1
= 1 − tk and tj = 0 for all
other j. Then

∑n
j=1 πjtj = tkπk + tk+1πk+1 = ρ. We will show that the vector x,
defined by
xj = tkπj(Sk) + tk+1πj(Sk+1),j = 1, . . . ,n majorizes the vector
πj(T), j = 1, . . .n. To do this
we will need to establish the following equation:
tkπi(Sk) + tk+1πi(Sk+1) =
vi
ṽ0 + Ṽ (T)
∀ i ≤ k. (31)
The result then follows, because for j ≤ k, equation (31) implies
that
x[1,j] =
v(Sj)
ṽ0 + Ṽ (T)
≥ πSj (T),
34
as not all elements in Sj need to be in T. On the other hand, for
j > k, we have
x[1,j] = π(T) ≥ πSj (T),
since again, not all elements in Sj need to be in T.
To verify equation (31) we will show that tk and tk+1 can be
written as

tk =
ṽ0 + ṽ(Sk)
ṽ0 + ṽ(T)
v(Sk + 1) −v(T)
v(Sk + 1) −v(Sk)
,
and
tk+1 =
ṽ0 + ṽ(Sk+1)
ṽ0 + ṽ(T)
v(T) −v(Sk)
v(Sk + 1) −v(Sk)
,
as then (31) follows from simple algebra. We know from
arguments in the proof of Proposi-
tion 2, that
πk+1 −πk = πk+1(Sk+1)
ṽ0
ṽ0 + ṽ(Sk)
,
so showing the result for tk is equivalent to showing that
πk+1 −ρ =

v(Sk + 1) −v(T)
ṽ0 + ṽ(Sk + 1)
ṽ0
ṽ0 + ṽ(T)
,
which follows from easily from the fact that ṽj = (1−θ)vj for all
j. The proof for tk+1 follows
along similar lines.
Proof of Proposition 5: We need to compute the difference
between E[min(U(y), max(y,c−
D2)) and E[min(U(y − 1), (y − 1,c− D2)]. Since both quantities
are zero when c − D2 ≥ y
and if c − D2 < y the difference reduces to E[min(U(y),y)] −
E[min(U(y − 1),y − 1)]. By
expressing the expectation as the sum of tail probabilities and
conditioning on the demand
from the potential customer we see that
E[min(U(y),y)] =
y∑
z=1
P(U(y) ≥ z)
= P(U(y) ≥ y) + β
y−1∑
z=1
P(U(y − 1) ≥ z − 1) + (1 −β)
y−1∑
z=1

P(U(y − 1) ≥ z).
We now use the fact that E[min(U(y − 1),y − 1)] =
∑y−1
z=1 P(U(y − 1) ≥ z), to express the
difference as
E[min(U(y),y)] −E[min(U(y − 1),y − 1)] = P(U(y) ≥ y) + βP
(U(y − 1) < y − 1).
Since P(U(y) ≥ y) = βP (U(y − 1) ≥ y − 1) + (1 −β)P(U(y − 1) ≥
y), we conclude that
E[min(U(y),y)]−E[min(U(y−1),y−1)] = [β+(1−β)P(U(y−1) ≥
y|D2 > c−y)]P(D2 > c−y).
35
On the other hand, E min(c−y,D2) −E min(c + 1 −y,D2)] =
−P(D2 > c−y). Therefore,
∆W2(c,y) = [q1(β + (1 −β)P(U(y − 1) ≥ y|D2 > c−y) − q2)]P(D2
> c−y)
as claimed.
36
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of Majorization and Its Ap-
plications. Academic Press, New York, NY.
[15] McFadden, D. (1978) . Modelling the choice of residential
location, in A. Karlquist,
ed., Spatial Interaction Theory and Planning Models, North
Holland Publishing Co.,
Amsterdam, chapter 25, pp. 7596.
[16] McFadden, D. (1980). “Econometric models for
probabilistic choice among products.”
The Journal of Business. 53, 13-29.
[17] McFadden, D., K. Train. (2000). “Mixed mnl models for
discrete response.” Journal of
applied Econometrics. 15, 447470.
[18] Rusmevichientong, P., Shmoys, D. and H. Topaloglu.
(2010). “Assortment Optimization
with Mixtures of Logits.” Working paper, Cornell University.
[19] Talluri, K and G. van Ryzin. (2004) “Revenue Management
Under a General Discrete
Choice Model of Consumer Behavior.” Management Science,
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[Train(2002)] Train, K. 2002. Discrete Choice Methods with
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[20] Walczack, D., S. Mardan, and Kallesen, R. 2010.
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the marginal expected revenue data transformation: A note on
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Management, 9, 94-109.
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Page 1Page 2Page 3Page 4
Overview
Introduction
Neoclassical Choice Model
Luce Model
Random Utility Models
Applications to Revenue Management
Discrete Choice Models
Guillermo Gallego
Department of Industrial Engineering and Operations Research
Columbia University
Spring 2013
Guillermo Gallego Discrete Choice Models
Overview
Introduction
Luce Model

Outline
I Introduction
I Neoclassical Theory
I Luce Model
I Random Utility Models
I Applications to Revenue Management
Overview
Introduction
Luce Model
Outline
I Introduction
I Luce Model

Overview
Introduction
Luce Model
Outline
I Introduction
I Luce Model
Overview
Introduction

Luce Model
Outline
I Introduction
I Luce Model
Overview
Introduction
Luce Model
Outline
I Introduction

I Luce Model
Overview
Introduction
Luce Model
Outline
I Introduction
I Luce Model

Overview
Introduction
Luce Model
Introduction
Choice Modeling is a general purpose tool for making
probabilistic
predictions about human decision making behavior. It is
regarded
as the most suitable method for estimating consumers
willingness
to pay in multiple dimensions. The Nobel Prize for economics
was
awarded to Daniel McFadden in 2000 for his work in the area.
We will consider three theories about about how a rational
decision
maker decides among a discrete number of alternatives:
I Neoclassical choice model
I Luce choice model
I Random utility choice models
Overview
Introduction

Luce Model
Introduction
probabilistic
regarded
willingness
was
decision
I Luce choice model
Overview
Introduction

Luce Model
Introduction
probabilistic
regarded
willingness
was
decision
I Luce choice model
Overview
Introduction
Luce Model

This theory assumes there is a preference operator, say � to
compare alternatives in a choice set S. The operator is assumed
to
satisfy the following three properties:
I i � i ∀ i ∈ N
I i � j and j � k implies i � k
I i � j or j � i ∀ i, j ∈ N
Overview
Introduction
Luce Model
to
I i � i ∀ i ∈ N

I i � j or j � i ∀ i, j ∈ N
Overview
Introduction
Luce Model
to
I i � i ∀ i ∈ N
I i � j or j � i ∀ i, j ∈ N
Overview
Introduction
Luce Model

Theorem: There exists a function U : N →< such that i � i if
and only if U(i) ≥ U(j). Moreover, U is unique up to positive
affine transformations.
Decision makers act to maximize expected utility.
If U is concave then EU(W ) ≤ U(EW ). Justifies insurance.
Difficult to specify utility functions
Overview
Introduction
Luce Model
The Luce Model (1959)
Let πi (S) be the probability of choosing i ∈ S ⊂ N and let
πT (S) =
∑
i∈ T

πi (S) for all T ⊂ S.
Luce proposed the following axioms:
I If there is an j ∈ S such that
πi ({i, j}) = 0 then πT (S) = πT−i (S − i).
I If πi ({i, j}) ∈ (0, 1) for all i, j, then πi (S) = πi (T )πT (S) for
all i ∈ T .
Overview
Introduction
Luce Model
Let πi (S) be the probability of choosing i ∈ S ⊂ N and let
πT (S) =
∑
i∈ T
πi (S) for all T ⊂ S.
Luce proposed the following axioms:
I If there is an j ∈ S such that

πi ({i, j}) = 0 then πT (S) = πT−i (S − i).
I If πi ({i, j}) ∈ (0, 1) for all i, j, then πi (S) = πi (T )πT (S) for
all i ∈ T .
Overview
Introduction
Luce Model
Theorem: There is a function v : N →<+ such that for all
i ∈ S ⊂ N
πi (S) =
vi
V (S)
where V (S) =
∑
j∈ S
vj

Overview
Introduction
Luce Model
Random utility models assume that the the decision maker can
perfectly discriminate among alternatives but the decision
maker
may have incomplete information. The utility for alternative i is
modeled as Ui = µi + �i where �i is a mean zero random
variable.
Then
πi (S) = P(Ui ≥ Uj ∀ j ∈ S)
To put the model to good use we need to make assumptions
about
the deterministic term µi and the distribution of the random
term
�i for each i ∈ S.
Overview
Introduction

Luce Model
Modeling the Deterministic Term
If each alternative can be characterized by a vector, say x, of
attributes then we can model the deterministic part of the utility
as a linear function of a vector of attributes, e.g.,
µi = β
′xi
Common attributes used in Airline Revenue Management
include
the fare, fare restrictions, travel time, number of stops, carrier,
type of plane, etcetera. Some of the attributes may be in log
space, e.g., log of fares.
Overview
Introduction
Luce Model
Modeling the Random Term

I Uniform random term: Linear model
I Normal random term: Probit model
I Gumbel random term: Multinomial Logit model (our focus).
A Gumbel random variable X has cdf
F (x|φ) = e−e
−(γ+φx)
−∞ < x < ∞
with mean zero and variance σ2 = π
2
6φ2
, where γ = 0.57721 is
the Euler constant.
Overview
Introduction
Luce Model

F (x|φ) = e−e
−(γ+φx)
−∞ < x < ∞
2
6φ2
, where γ = 0.57721 is
the Euler constant.
Overview
Introduction
Luce Model

F (x|φ) = e−e
−(γ+φx)
−∞ < x < ∞
2
6φ2
, where γ = 0.57721 is
the Euler constant.
Overview
Introduction
Luce Model
Multinomial Logit Model
Theorem: If the error terms are independent Gumbel-φ then
πi (S) = Pr(Ui ≥ Uj ∀ j ∈ S) =

vi
V (S)
where vi = e
φµi , ∀ i ∈ S.
Consistent with Luce’s model; with neoclassical when φ is large
Winbugs: http://www.mrc-
bsu.cam.ac.uk/bugs/winbugs/contents.shtml
NLOGIT: http://www.limdep.com/
Overview
Introduction
Luce Model
Independence of Irrelevant Alternatives (IIA)
The MNL suffers from the IIA property:
πi (S)
πj (S)
=
πi (T )

πj (T )
∀ i, j ∈ S ⊂ T
Example: S = {c,b1} and T = C = {c,b1,b2}, with
v(c) = v(b1) = v(b2) = 1. Then
πc (S) = 1/2 πc (T ) = 1/3.
Overview
Introduction
Luce Model
Nested Multinomial Logit Model
Under this model, the choice set is partitioned into non-
overlapping
subsets S = S1 ∪ . . .∪ Sk and for i ∈ Sj , the choice is modeled
as
πi (S) = πi (Sj )πSj (S)
where each component on the right is modeled by a MNL
model.
Example: S1 = {c},S2 = {b1,b2}, πS1 (S) = πS2 (S) = 1/2, then
πc (S) = 1/2 and πb1 (S) = πb2 (S) = 1/4.

Overview
Introduction
Luce Model
Mixture of MNLs
I All random utility models can be approximated to any degree
of accuracy by a mixture of MNLs.
I The mixture is specified by (θk,vk ) where θk is the
probability
that an arriving customer belongs to market segment
k = 1, . . . ,K , and vk is the vector of weights vk (a) a ∈ C is
the vector of attraction values that governs the MNL for
market segment k.
I The problem with the mixture model is that it is difficult to
estimate the parameters and it is difficult to do optimization,
e.g., to find the assortment that maximizes revenues.
Overview
Introduction

Luce Model
I There set of products is N = {1, . . . ,n}.
I The no purchase alternative i = 0 is always available.
I For each S ⊂ N
πi (S) =
vi
v0 + V (S)
π0(S) =
v0
v0 + V (S)
.
I There is a fare pj associated with each product j ∈ N.
I r(S) =
∑
i∈ S piπi (S) is the expected revenue associated with
S ⊂ N.
I Often interested in finding S ⊂ N to maximize r(S) − zπ(S),
where π(S) =
∑

j∈ S π(S).
dynamic pricing session 6.notebook
1
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2
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Page 1Page 2
Single Resource Revenue Management with Independent
Demands
c©
Guillermo Gallego
Updated Spring 2013
Abstract
Providers of fixed perishable capacity, such as airline seats and

hotel rooms use price discrimination
to improve revenues; in practice, this discrimination is typically
achieved by imposing booking and usage
restrictions or including ancillary services such as mileage
accrual and luggage handling, to sell the same
capacity to different customers at different prices. We will
assume that the set of fare classes (a menu
of prices, restrictions and ancillary services) is given, and that
the capacity provider’s goal is to allocate
capacity among the different fare classes to maximize expected
revenues. The problem of designing and
pricing fare classes is treated in a separate chapter. We analyze
the two fare class problem under the
assumption that the lower fare class books first. We use
marginal analysis to informally derive Littlewood’s
rule and then show that Littlewood’s rule is in fact optimal.
Spill rates, spill penalties and callable
products are discussed next. A dynamic programming
formulation for the multiple fare class problem is
then introduced under the assumption that lower fare classes
book first. Commonly used heuristics as well
as bounds on the value function are presented. Dynamic models
that explicitly take time into account,
allow for more general fare arrival patterns and for randomness
in the size of the requests. We compare
the performance of static and dynamic policies and find that
dynamic policies have a real advantage when
the fare arrivals patterns are not low-to-high. We finalize the
chapter with a model where fare classes are
not allowed to reopen after they are closed for the first time.
1 Introduction
This chapter considers the simplest and best known revenue
management problem, the single resource, inde-
pendent demand problem. We assume that the capacity provider

is trying to maximize the expected revenues
from a sunk investment in c units of capacity. We assume that
capacity is sold through a reservation system
and that capacity cannot be modified or replenished during the
booking horizon. We also assume that unsold
capacity has no salvage value. Later we will see that the zero
salvage value assumption is made without loss
of generality as any problem with positive salvage value can be
transformed into a problem with zero salvage
value. We assume that the set of fare classes (a menu of prices
and restrictions) is given, and that the demands
for the different fare classes are statistically independent. In
particular, we assume that if a customer finds his
preferred fare class closed, he will leave the system without
purchasing. This assumption holds approximately
if the difference in fares is large so that demands are decoupled
or if customers can find alternative sources
of capacity for their preferred fare class. In some cases,
however, part of the demand may be recaptured by
other available fare classes. In such cases, the independent
demand assumption is too strong and needs to be
relaxed. We address this issue in a separate chapter where we
discussed demand models based on discrete
choice theory.
In this chapter we present a variety of models that have been
developed in industry and in academia.
There has been a preference in industry for models that suppress
the time dimension and assume that the
arrival pattern of the fare classes is low-to-high. We call these
class of models static to distinguish them
from the dynamic models, favored by academics, that model
time explicitly. Both models have advantages
and disadvantages as we will soon see. Static models are
relatively easy to understand. Also, good heuristics

1
were developed before optimal solutions based on dynamic
programming were discovered. Bringing in the
time dimension helps deal with more general fare arrival
patterns, but specifying the model requires a more
detailed estimation of demand. This chapter starts with a review
of the two fare class problem in §2 where we
present a heuristic derivation of Littlewood’s rule via marginal
analysis. Littlewood’s rule is formally derived
in §2.3 where a formal DP for the two fare class problem is
presented. The dynamic program for multiple
fare classes is presented in §3. Commonly used heuristics are
presented in §4 and bounds on the optimal
expected revenue are presented in §5. The dynamic model for
the Poisson case is presented in §6, and for the
compound Poisson case in §7, where each request is for a
random demand size. In this section we also provide
a numerical example comparing static to dynamic policies. We
remark that static policies are near optimal
under fixed arrival patterns, such as low-to-high fare arrivals,
but can significantly underperform dynamic
policies for more general demand arrival patterns. In §9, we
restrict fares so that they cannot be opened once
they are closed. This last model helps to discourage strategic
customers from waiting for a better fare.
2 Two Fare Classes: Marginal Analysis
The product can be sold either at the full-fare p1 or at a
discounted-fare p2 < p1. The discounted-fare typically
has advance purchasing and usage restrictions such as Saturday
night stays. Let D1 and D2 denote respectively
the random demand for the two fare classes for a specific

instance of the problem, e.g., for a specific flight for
an airline or a specific night for a hotel.
We assume that all booked customers will actually travel. This
avoids the need to overbook capacity and
allow us to focus on the problem of allocating capacity between
the two fares. We will discuss how to deal
with pre-travel cancellations and day-of-travel no shows in a
separate chapter on overbooking models.
Fare class arrival order is an important part of the model. We
assume what is commonly known as the
low-to-high fare class arrival order, which implies that demand
for the discounted-fare class book before the
demand for the full-fare class. This arrival pattern holds
approximately in practice, and it is encouraged by
advance purchase restrictions imposed on lower fare classes.
Notice that this is a worst case arrival pattern.
Indeed, if full-fare class customers arrived first then we would
accept them up to capacity and use the residual
capacity, if any, to satisfy demand from the discounted-fare
class. We will relax the low-to-high fare order
arrival assumption in Section 3.4, after we solve the multi-fare
problem via dynamic programming.
Under the low-to-high arrival pattern, discount-fare customers
may exhaust capacity, say c, unless part
of it is protected for later-booking by full-fare customers.
Consequently, booking limits (known as discount
authorizations) are placed on the discount sales. Suppose we
protect y ∈ {0, 1, . . . ,c} units of capacity for
the full-fare demand, D1, before observing the actual demand
for the discount-fare, D2. This results in a
booking limit c−y on the discounted-fare, so sales at the
discounted-fare class are given by min(c−y,D2).
The remaining capacity is equal to c − min(c − y,D2) = max(y,c

− D2), and it is all made available to the
full-fare class. Consequently, sales at the full fare equal
min(max(y,c−D2),D1). The total expected revenue
is
W(y,c) = p2E min(c−y,D2) + p1E min(max(y,c−D2),D1)
and the goal is to find a protection level y that maximizes
W(y,c). The extreme strategies y = 0 and y = c
correspond, respectively, to the case where no capacity is
protected and all of the capacity is protected. We
will later come back and discuss when these extreme strategies
are optimal. In most cases, however, an
intermediate strategy is optimal.
The fare ratio r = p2/p1 plays an important role in determining
optimal protection levels. If the ratio is
very small then we would be inclined to protect more capacity
for the full-fare demand. If the ratio is close
to one, we would be inclined to accept nearly all discount-fare
requests since we can get almost the same
revenue, per unit of capacity, without risk. The distribution of
full-fare demand is also important in deciding
how many units to protect for that fare. If, P(D1 ≥ c) is very
large, then it makes sense to protect the entire
capacity for full-fare sales as it is likely that the provider can
sell all of the capacity at the full-fare. However,
if P(D1 ≥ c) is very low then it is unlikely that all the capacity
can be sold at the full-fare, so fewer units
should be protected. It turns out that the demand distribution of
the discounted-fare class has no influence
2

on the optimal protection level under our assumption that D2
and D1 are independent. A formula for the
optimal protection level, involving only P(D1 ≥ y) and r, was
first proposed by Littlewood [14] in 1972. His
arguments were not formal; however, they were later justified
by Bhatia and Prakesh [1] in 1973, and Richter
[17] in 1982.
One can obtain Littlewood’s formula intuitively by using
marginal analysis: Suppose we have y > 0 units
of capacity, and a request for a unit of capacity at the
discounted-fare arrives. If we accept, we obtain p2.
If we close down the discount-fare then we will be able to sell
the yth unit at p1 only if the full-fare demand
D1 is at least as large as y, so it is intuitively optimal to reject
the discount fare if p1P(D1 ≥ y) > p2. This
suggests that an optimal protection level y1 should be given by:
y1 = max{y ∈ N : P(D1 ≥ y) > r}, (1)
where N = {0, 1, . . . ,} is the set of non-negative integers.
Equation (1) is known as Littlewood’s rule.
Example 1. Suppose D1 is Poisson with parameter 80, the full
fare is p1 = $100 and the discounted fare is
p2 = $60, so r = 60/100 = 0.6. We are interested in the
cumulative tail distribution P(D1 ≥ y) = 1−P(D1 ≤
y − 1). Since most statistical software packages return the value
of P(D1 ≤ y), we see that y1 satisfies
P(D1 ≤ y1 − 1) < 1 − r ≤ P(D1 ≤ y1). Since P(D1 ≤ 77) =< 0.4 ≤
P(D1 ≤ 78) we conclude that y1 = 78.
Consequently, if c = 200 then the booking limit for the discount
fare is 122. However, if c < y1, then all units
should be protected for the full-fare resulting in a booking limit
of zero.

Remarks:
• y(c) = min(y1,c) is also an optimal protection level. If y(c) =
c, or equivalently if y1 ≥ c, then all the
capacity should be reserved for sale at the full-fare.
• The quantity b2 = max(c−y1, 0) is known as the optimal
booking limit for the discount fare. It is the
maximum number of discount-fare customers that we will book.
• y1 is independent of the distribution of D2.
• If P(D1 ≥ y1 + 1) = r, then y1 + 1 is also optimal protection
level, so both y1 and y1 + 1 result in the
same expected revenue. Protecting the y1 + 1 unit of capacity
increases the variance of the revenue, but
it reduces the probability of rejecting requests from full-fare
customers.
From Littlewood’s rule (1), we see that the extreme strategy y =
0 is optimal when P(D1 ≥ 1) ≤ r and
the extreme strategy y = c is optimal when P(D1 ≥ c) > r.
2.1 Continuous Demand Model
Although revenue management demands are actually discrete,
continuous distributions can be easier to work
with and are often employed in practice. If we model D1 as a
continuous random variable with cumulative
distribution function F1(y) = P(D1 ≤ y), then
y1 = F
−1
1 (1 −r)
where F−11 denotes the inverse of F. In particular, if D1 is

Normal with mean µ1 and standard deviation σ1
then
y1 = µ1 + σ1Φ
−1(1 −r) (2)
where Φ denotes the cumulative distribution function of the
standard Normal random variable.
This formula allows for comparative statics as given in Table 1:
Example 2. Suppose that D1 is Normal with mean 80 and
standard deviation 9, the full-fare is p1 = $100
and the discount-fare is p2 = $60. Then y1 = F
−1
1 (1 − 0.6) = 77.72 < 80 since r > 1/2. Notice that the
3
Fare Ratio Dependence of protection level
r > 1
2
y1 < µ1 and y1 decreases with σ1
r = 1
2
y1 = µ1 independent of σ1
r < 1
2

y1 > µ1 and y1 increases with σ1
Table 1: Comparative Statics for Normal Full Fare Demand
solution is quite close to that of Example 1. This is because a
Poisson random variable with mean 80 can be
well approximated by a normal with mean 80 and standard
deviation
√
80 ' 9.
2.2 Connection with the Newsvendor Problem
There is a close connection between the classical Newsvendor
Problem and the two-fare Revenue Management
Problem that we will briefly explore here. In the classical
Newsvendor Problem a manager must decide how
many units, say y, to stock for random sales D1 at p1 assuming
a unit cost p2 < p1. The solution is to stock y1
units where y1 is the largest integer such that P(D1 ≥ y) > r =
p2/p1. We can think of the two-fare Revenue
Management Problem as a situation where capacity c is pre-
decided, at a possible sub-optimal level, there is
random demand D1 at ”salvage value” p1 > p2 that arrives after
demand D2 at p2. The revenue management
problem is to determine how many units to allow to be sold at
p2. We know that the solution is to allow
(c−y1)+ units to book at p2, reserving max(y1,c−D2) units for
sale at p1.
2.3 Two Fare Classes: Dynamic Programming
In this section we formulate and analyze the two fare class
problem using dynamic programming and present a

formal proof of the optimality of Littlewood’s rule. We will
from now on refer to the full-fare class as fare class
1 and to the discounted fare class as fare class 2. This will be
helpful in formulating the dynamic program
and to extend the analysis to multiple fares. The idea is to start
solving the problem at the last stage, just
before demand for fare class 1 arrives. Let V1(y) be the optimal
expected revenue that can be obtained from
fare class 1 when capacity is y. Since it is optimal to allow fare
class 1 customers to book all of the available
capacity, sales are equal to min(D1,y) and the optimal expected
revenue is
V1(y) = p1E min(D1,y).
Our next task is to find V2(c), the optimal expected revenue
that can be obtained from c units of ca-
pacity. Suppose that y ∈ {0, 1, . . . ,c} units are protected for
fare class 1 demand. This results in revenues
p2 min(D2,c − y) from sales to fare class 2 and remaining
inventory max(c − D2,y) available for fare class
1. Notice that we can obtain expected revenue EV1(max(c −
D2,y)) from this inventory from fare class 1
customers. Then
= E{p2 min(D2,c−y) + V1(max(c−D2,y))}
is the expected revenue associated with protecting y ∈ {0, 1, . . .
,c} units for the full-fare. V2(c) can be obtained
by maximizing W(y,c) over y. More precisely,
V2(c) = max
y∈ {0,1,...,c}
E{p2 min(D2,c−y) + V1(max(c−D2,y))}. (3)

The key to Dynamic Programming is that it involves a recursive
equation (3) linking the expected revenues
4
V2(c), at stage 2, to the expected revenue function V1 at stage
1. To solve for V2(c) we first need to solve
for V1(y) for y ∈ {0, 1, . . . ,c}. Before moving on to the multi-
fare formulation we will provide a formal
proof of Littlewood’s rule (1), and discuss the quality of service
implications of using Littlewood’s rule under
competition.
2.4 Formal Proof of Littlewood’s Rule
For any function f(y) over the integers, let ∆f(y) = f(y) − f(y −
1). The following result will help us to
determine ∆V (y) and ∆W(y,c) = W(y,c) −W(y − 1,c).
Lemma 1 Let g(y) = EG(min(X,y)) where X is an integer valued
random variable with E[X] < ∞ and G
is an arbitrary function defined over the integers. Then
∆g(y) = ∆G(y)P(X ≥ y).
Let r(y) = ER(max(X,y)) where X is an integer valued random
variable with E[X] < ∞ and R is an arbitrary
function defined over the integers. Then
∆r(y) = ∆R(y)P(X < y).
An application of the Lemma 1 yields the following proposition
that provides the desired formulas for

∆V1(y) and ∆W(y,c).
Proposition 1
∆V1(y) = p1P(D1 ≥ y) y ∈ {1, . . . ,}
∆W(y,c) = [∆V1(y) −p2]P(D2 > c−y) y ∈ {1, . . . ,c}.
The proof of the Lemma 1 and Proposition 1 are relegated to the
Appendix. With the help of Proposition 1
we can now formally establish the main result for the Two-Fare
Problem.
Theorem 1 The function W(y,c) is unimodal in y and is
maximized at y(c) = min(y1,c) where
y1 = max{y ∈ N : ∆V1(y) > p2}.
Moreover, V2(c) = W(y(c),c).
Proof: Consider the expression in brackets for ∆W(y,c) and
notice that the sign of ∆W(y,c) is determined
by ∆V1(y) −p2 as P(D2 > c−y) ≥ 0.Thus W(y,c) ≥ W(y − 1,c) as
long as ∆V1(y) −p2 > 0 and W(y,c) ≤
W(y−1,c) as long as ∆V1(y)−p2 ≤ 0. Since ∆V1(y) = p1P(D1 ≥
y) is decreasing1 in y, ∆V1(y)−p2 changes
signs from + to − since ∆V1(0) − p2 = p1P(D1 ≥ 0) − p2 = p1 −
p2 > 0 and limy→∞[∆V1(y) − p2] = −p2.
This means that W(y,c) is unimodal in y. Then
y1 = max{y ∈ N : ∆V1(y) > p2}.
coincides with Littlewood’s rule (1). When restricted to {0, 1, .
. . ,c}, W (y,c) is maximized at y(c) =
min(c,y1). Consequently, V2(c) = maxy∈ {0,1,...,c}W(y,c) =
W(y(c),c), completing the proof.

2.5 Quality of Service, Spill Penalties, Callable Products and
Salvage Values
Since max(y1,c−D2) units of capacity are available for fare
class 1, at least one fare class 1 customer will be
denied capacity when D1 > max(y1,c−D2). The probability of
this happening is a measure of the quality of
1We use the term increasing and decreasing in the weak sense.
5
service to fare class 1, known as the full-fare spill rate.
Brumelle et al. [4] have observed that
P(D1 > max(y1,c−D2)) ≤ P(D1 > y1) ≤ r < P(D1 ≥ y1). (4)
They call P(D1 > y1) the maximal spill rate. Notice that if the
inequality y1 ≥ c − D2 holds with high
probability, as it typically does in practice when D2 is large
relative to c, then the spill rate approaches the
maximal flight spill rate which is, by design, close to the ratio
r. High spill rates may lead to the loss of
full-fare customers to competition. To see this, imagine two
airlines each offering a discount fare and a full-
fare in the same market where the fare ratio r is high and
demand from fare class 2 is high. Suppose Airline
A practices tactically optimal Revenue Management by applying
Littlewood’s rule with spill rates close to
r. Airline B can protect more seats than recommended by
Littlewood’s rule. By doing this Airline B will
sacrifice revenues in the short run but will attract some of the
full-fare customers spilled by Airline A. Over
time, Airline A may see a decrease in full-fare demand as a

secular change and protect even fewer seats for
full-fare passengers. In the meantime, Airline B will see an
increase in full-fare demand at which time it can
set tactically optimal protection levels and derive higher
revenues in the long-run. In essence, Airline B has
(correctly) traded discount-fare customers for full-fare
customers with Airline A.
One way to cope with high spill rates and its adverse strategic
consequences is to impose a penalty cost
ρ for each unit of full-fare demand in excess of the protection
level. This penalty is suppose to measure the
ill-will incurred when capacity is denied to a full-fare customer.
This results in a modified value function
V1(y) = p1E min(D1,y) − ρE[(D1 − y)+] = (p1 + ρ)E min(D1,y)
− ρED1. From this it is easy to see that
∆V1(y) = (p + ρ)P(D1 ≥ y), resulting in
∆W(y,c) = [(p1 + ρ)P(D1 ≥ y) −p2]P(D2 > c−y)
and
y1 = max
{
y ∈ N : P(D1 ≥ y) >
p2
p1 + ρ
}
. (5)
Notice that this is just Littlewood’s rule applied to fares p1 + ρ
and p2, resulting in fare ratio p2/(p1 + ρ) and,
consequently, lower maximal spill rates. Obviously this

adjustment comes at the expense of having higher
protection levels and therefore lower sales at the discount-fare
and lower overall revenues. Consequently, an
airline that wants to protect its full-fare market by imposing a
penalty on rejected full-fare demand does it at
the expense of making less available capacity for the discount-
fare and less expected revenue. One way to avoid
sacrificing sales at the discount-fare and improve the spill rate
at the same time is to modify the discount-fare
by adding a restriction that allows the provider to recall or buy
back capacity when needed. This leads to
revenue management with callable products; see Gallego, Kou
and Phillips [12]. Callable products can be
sold either by giving customers an upfront discount or by giving
them a compensation if and when capacity
is recalled. If managed correctly, callable products can lead to
better capacity utilization, better service to
full-fare customers and to demand induction from customers
who are attracted to either the upfront discount
or to the compensation if their capacity is recalled.
The value function V1(y) may also be modified to account for
salvage values (also known as the ‘distressed
inventory problem’). Suppose there is a salvage value s < p2 on
excess capacity after the arrival of the full-fare
demand (think of standby tickets or last-minute travel deals).
We can handle this case by modifying V1(y) to
account for the salvaged units. Then V1(y) = p1E min(D1,y) +
sE(y −D1)+ = (p1 −s)E min(D1,y) + sy, so
∆V1(y) = (p1 −s)P(D1 ≥ y) + s, resulting in
∆W(y,c) = [(p1 −s)P(D1 ≥ y) − (p2 −s)]P(D2 > c−y)
and
y1 = max

{
y ∈ N : P(D1 ≥ y) >
p2 −s
p1 −s
}
. (6)
Notice that this is just Littlewood’s rule applied to net fares p1
−s and p2 −s. This suggests that a problem
with salvage values can be converted into a problem without
salvage values by using net fares pi ← pi − s,
i = 1, 2 and then adding cs to the resulting optimal expected
revenue V2(c) in excess of salvage values.
6
3 Multiple Fare Classes: Exact
Solution
In this section we present an exact solution to the muli-fare
class problem using dynamic programming. We
assume that the capacity provider has c units of perishable
capacity to be allocated among n fares indexed

so pn < ... < p1. Lower fares typically have severe time of
purchase and traveling restrictions and may have
restricted advanced selection that denies access to the more
desirable capacity. Given the time-of-purchase
restriction, it is natural to assume that demands for fare classes
arrive in n stages, with fare class n arriving
first, followed by n − 1, with fare class 1 arriving last. Let Dj
denote the random demand for fare class
j ∈ N = {1, . . . ,n}. We assume that, conditional on the given
fares, the demands D1, . . . ,Dn are independent
random variables with finite means µj = E[Dj] j ∈ N. The
independent assumption is approximately valid
in situations where fares are well spread and there are
alternative sources of capacity. Indeed, a customer
who finds his preferred fare closed is more likely to buy the
same fare for an alternative flight (perhaps with
a competing carrier) rather than buying up to the next fare class
if the difference in fare is high. The case of
dependent demands, where fare closures may result in demand
recapture, will be treated in a different chapter.
The use of Dynamic Programming for the multi-fare problem
with discrete demands is due to Wollmer [22].
Curry [6] derives optimality conditions when demands are
assumed to follow a continuos distribution. Brumelle
and McGill [5] allow for either discrete or continuous demand

distributions and makes a connection with the
theory of optimal stopping.
Let Vj(x) denote the maximum expected revenue that can be
obtained from x ∈ {0, 1, . . . ,c} units of
capacity from fare classes {j, . . . , 1}. The sequence of events
for stage j are as follows:
1. Decide the protection level, say y ≤ x, for fares j − 1,j − 2, . .
. , 1 thus allowing at most x−y units of
fare j demand.
2. The realization of the demand Dj occurs, and we observe
min(Dj,x−y) sales at fare j.
3. The revenue pj min(Dj,x−y) is collected, and we proceed to
the beginning of stage j−1 with a remaining
capacity of max(x−Dj,y).
The revenue from this process is
Wj(y,x) = pjE min(Dj,x−y) + EVj−1 (max(x−Dj,y)) . (7)
We can think of Wj(y,x) as the expected revenue from x units of
capacity prior to seeing the demand for fare

class j when up to x−y units are allowed to book at fare j and an
optimal policy is followed thereafter. This
leads to the dynamic programming recursion
Vj(x) = max
y∈ {0,1,...,x}
Wj(y,x)
= max
y∈ {0,1,...,x}
{pjE min(Dj,x−y) + EVj−1(max(x−Dj,y))} . (8)
The dynamic program simply states that the optimal value
function is the sum of the expected revenues
from fare class j plus the expected revenues from fare classes
j−1, . . . , 1 evaluated at the protection level that
maximizes this sum. Notice that once we are at the beginning of
stage j−1 we face a similar problem over the
remaining j−1 fare classes. Vj(x) is then the maximum expected
revenue that can be obtained from x units of
capacity for the j-fare problem. Consequently Vn(c) is the
maximum expected revenue for the n-fare problem
with capacity c. The recursion can be started with V0(x) = 0 if

there are no salvage values or penalties for
spill. Alternatively, the recursion can start with V1(x) = p1E
min(D1,x) + sE[(x−D1)+]−ρE[(D1 −x)+] for
x ≥ 0 if there is a salvage value s per unit of excess capacity
and a penalty ρ per unit of fare class 1 demand
that is denied.
7
3.1 Structure of the Optimal Policy
In order to analyze the structure of the optimal policy, we begin
by describing a few properties of the value
function. As a convention we set V0 ≡ 0. A function V (y)
defined on y ∈ N is concave if ∆V (y) =
V (y) −V (y − 1) is decreasing in y ∈ N+.
Lemma 2 For any j ≥ 1,
a) ∆Vj(y) = Vj(y) −Vj(y − 1) is decreasing in y ∈ N+, so the
marginal value of capacity is diminishing.
b) ∆Vj(y) is increasing in j ∈ {1, . . . ,n} so the marginal value

of capacity increases when we have more
stages to go.
The proof of Lemma 2 is in the Appendix. Using the Lemma we
can characterize an optimal policy as
stated in the following theorem. For the purpose of simplifying
notation we will extend the definition of ∆Vj(y)
to y = 0 by setting ∆Vj(0) = ∆V1(0) = p1 just as we did for j =
1.
Theorem 2 The function Wj(y,x) is unimodal in y and it is
maximized at min(yj−1,c), where the nested
protection levels 0 = y0 ≤ y1 ≤ y2 ≤ ···≤ yn−1 are given by
yj = max{y ∈ N : ∆Vj(y) > pj+1} j = 1, . . . ,n− 1. (9)
The optimal value functions are given by
Vj(x) = Wj(min(x,yj−1),x) j = 1, . . . ,n, x ∈ N . (10)
Moreover, Vj(x) is concave in x ∈ N for each j = 1, . . . ,n.
Proof: An algebraic argument similar to that used to justify
Littlewood’s rule for n = 2, reveals that for
y ∈ {1, . . . ,x}

∆Wj(y,x) = Wj(y,x) −Wj(y − 1,x) = [∆Vj−1(y) −pj] P(Dj > x−y).
Let yj−1 = max{y ∈ N : ∆Vj−1(y) > pj}. By part a) of Lemma 2,
∆Vj−1(y) is decreasing in y so ∆Vj−1(y)−
pj > 0 for all y ≤ yj−1 and ∆Vj−1(y)−pj ≤ 0 for all y > yj−1.
Consequently, if x ≤ yj−1 then ∆Wj(y,x) ≥ 0
for all y ∈ {1, . . . ,x} implying that Vj(x) = Wj(x,x).
Alternatively, if x > yj−1 then ∆Wj(y,x) ≥ 0
for y ∈ {1, . . . ,yj} and ∆Wj(y,x) ≤ 0 for y ∈ {yj−1 + 1, . . . ,x}
implying Vj(x) = Wj(yj−1,x). Since
∆Vj(x) = ∆Vj−1(x) on x ≤ yj−1, it follows that ∆Vj(yj−1) =
∆Vj−1(yj−1) > pj > pj+1, so yj ≥ yj−1. The
concavity of Vj(x) is is equivalent to ∆Vj(x) decreasing in x,
and this follows directly from part a) of Lemma 2.
Remarks:
1. Notice that the unconstrained protection level yj−1 is
independent of the demands Dk, k ≥ j as observed
before in the two fare setting (Littlewood’s Rule).
2. We can think of yj,j = 1, . . . ,n− 1 as the unconstrained
protection levels. If we start stage j with xj
units of capacity, the constrained protection level for fares {j−1,

. . . , 1} is min(xj,yj−1). Thus capacity
is made available to fare j only if xj > yj−1.
3. The policy is implemented as follows. At stage n we start
with xn = c units of inventory, and we protect
yn−1(xn) = min(xn,yn−1) units of capacity for fares {n−1, . . . ,
1} by allowing up to (xn−yn−1)+ units
to be sold at fare pn. Since min(Dn, (xn−yn−1)+) units are sold
during stage n, we start stage n−1 with
xn−1 = xn − min(Dn, (xn − yn−1)+). We protect yn−2(xn−1) =
min(xn−1,yn−2) units of capacity for
fares {n−2, . . . , 1} and thus allow up to (xn−1−yn−2)+ units of
capacity to be sold at pn−1. The process
8
continues until we reach stage one with x1 units of capacity and
allow (x1 − y0)+ = (x1 − 0)+ = x1
to be sold at p1. Assuming discrete distributions, the
computational requirement to solve the dynamic
program for the n stages has been estimated by Talluri and van
Ryzin [20] to be of order O(nc2).

4. The concavity of Vn(c) is helpful if capacity can be procured
at a linear or convex cost because in this
case the problem of finding an optimal capacity level is a
concave problem in c.
Example 3. Suppose there are five different fare classes. We
assume the demand for each of the fares is
Poisson. The fares and the expected demands are given in the
first two columns of Table 2. The third column
includes the optimal protection levels for fares 1, 2, 3 and 4.
j pj E[Dj] yj
1 $100 15 14
2 $60 40 54
3 $40 50 101
4 $35 55 169
5 $15 120
Table 2: Five Fare Example with Poisson Demands: Data and
Optimal Protection Levels
Table 3 provides the expected revenues for different capacity
levels as well as the corresponding demand
factors (

∑5
j=1 E[Dj])/c = 280/c. These results should be intuitive. Greater
revenue potential is seen as
capacity increases (since potentially more demand can be
accepted). Further, the effect of restrictions on
discounted fares is apparent in the pattern of revenue across
classes; e.g. revenue V2(50) through V5(50) is
$3,426.8 because fare classes 3,4, and 5 are rationed since y2 =
54 > c = 50 units are protected for fare 1 and
2. However, V1(350) through V5(350) vary from $1,500 to
$9,625 because there is sufficient capacity to accept
sales in all fare classes.
c DF V1(c) V2(c) V3(c) V4(c) V5(c)
50 560% 1,500.0 3,426.8 3,426.8 3,426.8 3,426.8
100 280% 1,500.0 3,900.0 5,441.3 5,441.3 5,441.3
150 187% 1,500.0 3,900.0 5,900.0 7,188.7 7,188.7
200 140% 1,500.0 3,900.0 5,900.0 7,824.6 8,159.1
250 112% 1,500.0 3,900.0 5,900.0 7,825.0 8,909.1
300 93% 1,500.0 3,900.0 5,900.0 7,825.0 9,563.9
350 80% 1,500.0 3,900.0 5,900.0 7,825.0 9,625.0
Table 3: Expected Revenues Vj(c) and Demand Factors

Figure 1 shows the marginal value as a function of the
remaining resources for the data of Example 3.
3.2 Speeding up the Computation of the Value Function
While the value functions Vj(x), j ∈ {1, . . . ,n}, x ∈ {1, . . .c},
can be computed recursively there are some
tricks to speed up the computations. Here we focus on how to
efficiently update ∆Vj+1(x) from ∆Vj(x). The
key idea is to express ∆Vj+1(x) for x > yj in terms of previously
computed values of ∆Vj(x). The proof of
Proposition 2 is in the Appendix.
Proposition 2
∆Vj+1(x) =
{
∆Vj(x) if x = 1, . . . ,yj
E min(∆Vj(x−Dj+1),pj+1) if x = yj + 1, . . ..
9

$-‐
$20.00
$40.00
$60.00
$80.00

$100.00
$120.00
0
20
40
60
80
100
120
140
160
DV_1(x)

DV_2(x)
DV_3(x)
DV_4(x)
DV_5(x)
Figure 1: ∆Vj(x),x = 1, . . . , 350, j = 1, 2, 3, 4, 5 for Example 3
Since ∆Vj+1(x) = ∆Vj(x) for x ≤ yj, we only need to worry
about ∆Vj+1(x) for x > yj. The following
corollary to Proposition 2 makes the formula for ∆Vj+1(x),x >
yj more explicit.
Corollary 1
∆Vj+1(yj + k) = pj+1P(Dj+1 ≥ k) +
k−1∑

i=0
∆Vj(yj + k − i)P(Dj+1 = i) k ∈ {1, . . . ,c−yj}.
3.3 Linear and Convex Procurement Costs
Suppose that capacity c can be procured at a linear cost kc
before observing demands for the n fares: pn <
pn−1 < ... < p1. How much capacity should be procured? The
objective is to find c to maximize Πn(c,k) =
Vn(c) −kc. Let c(k) be the smallest optimizer of Πn(c,k) as a
function of the marginal cost k. The following
Proposition characterizes c(k), shows that c(k) is decreasing in
k and relates c(pj+1) to protection level yj for
j < n.
Proposition 3 The optimal procurement quantity at linear cost
kc is given by
c(k) = max{c ∈ N : ∆Vn(c) > k}.
Moreover, c(k) is decreasing in k, and yj = c(pj+1) for all j
∈ {1, . . . ,n− 1}.
Clearly c(0) = ∞ since ∆Vn(c) ≥ ∆V1(c) = p1P(D1 ≥ c) > 0 for

all c ∈ N . At the other extreme,
c(p1) = y0 = 0 since ∆Vn(1) = p1P(D1 ≥ 1) < p1 = ∆Vn(0), so
no capacity would be purchased if k ≥ p1.
If the cost of capacity k(c) is increasing convex then Π(c,k(c))
is concave in c and
c(k) = max{c ∈ N : ∆Vn(c) − ∆k(c) > 0}.
Consider the convex cost function k(c) = K if c ≤ c̄ and k(c) = ∞
for k > c̄ . This may reflect the situation
where there is a fixed cost to leasing a resource with capacity c̄ .
The optimal choice is then to lease the
resource if Vn(c̄ ) > K and not lease it otherwise.
10
0
50

100
150
200
250
300
350
400
450
0

10
20
30
40
50
60
70
80
90
100
C
ap
ac
it
y
cost
optimal capacity
Figure 2: Optimal Capacity as a Function of Cost for the Data
of Example 3

3.4 Relaxing the Monotonicity of Fares
We continue to assume that demands arrive in the order
Dn,Dn−1, . . . ,D1 but will relax the assumption that
the fares are monotone p1 > p2 > ... > pn. All of the results
work as stated, except the monotonicity of the
protection levels, if we redefine ∆Vj(0) = max{p1, . . . ,pj}. It is
also possible to skip some of the optimization
steps as it is clear that yj−1 = 0 whenever pj > max(p1, . . .
,pj−1) since it is optimal to allow all bookings at
fare pj. The reader is referred to Robinson [18] for more details.
As an example, suppose that p3 < p2 > p1.
Then V1(x) = p1E min(D1,x) and at stage 2 the decision is y1 =
0, so
V2(x) = p2E min(D2,x) + p1E min(D1, (x−D2)+).
Notice that
∆V2(x) = p2P(D2 ≥ x) + p1P(D2 < x ≤ D[1, 2])
so
y2 = max{y ∈ N : ∆V2(y) > p3}.
Notice that the capacity protected for fare 2 is higher than it

would be if there was no demand at fare 1.
4 Multiple Fare Classes: Commonly Used Heuristics
Several heuristics, essentially extensions of Littlewood’s rule,
were developed in the 1980’s. The most important
heuristics are known as EMSR-a and EMSR-b, where EMSR
stands for expected marginal seat revenue. Credit
for these heuristics is sometimes given to the American Airlines
team working on revenue management problems
shortly after deregulation. The first published account of these
heuristics appear in Simpson [19] and Belobaba
[2], [3]. For a while, some of these heuristics were even thought
to be optimal by their proponents until optimal
policies based on dynamic programming were discovered in the
1990’s. By then heuristics were already part of
implemented systems, and industry practitioners were reluctant
to replace them with the solutions provided
by dynamic programming algorithms. There are several reasons
for this. First, people feel more comfortable
with something they understand. Also, the performance gap
between the heuristics and the optimal dynamic
program tends to be small. Finally, there is a feeling among
some users that the heuristics may be more robust
to estimates of the mean and variance of demand.

Version a of the heuristic, EMSR-a, is based on the idea of
adding protection levels produced by applying
Littlewood’s rule to successive pairs of classes. At state j, we
need to decide how much capacity to protect
for fares j − 1, . . . , 1. We can use Littlewood’s rule to decide
how much capacity to protect for fare k demand
against fare j for k = j − 1, . . . , 1 and then add the protection
levels. More precisely, let rk,j = pj/pk and set
11
yk,j = max{y ∈ N : P(Dk ≥ y) > rk,j}.
Then the EMSR-a heuristic will protect
yaj−1 =
j−1∑
k=1
yk,j

units of capacity for fares j − 1, . . . , 1 against fare j.
In particular, if Dk is Normal with mean µk and standard
deviation σk, then
yaj−1 = µ[1,j − 1] +
j−1∑
k=1
σkΦ
−1(1 −rk,j),
where for any j, µ[1,j − 1] =
∑j−1
k=1 µk and sums over empty sets are zero.
Notice that the EMSR-a heuristic involves j − 1 calls to
Littlewood’s rule to find the protection level for
fares j − 1, . . . , 1. In contrast, the EMSR-b heuristic is based
on a single call to Littlewood’s rule for each
protection level. However, using the EMSR-b heuristic requires
the distribution of D[1,j − 1] =
∑j−1
k=1 Dk.

This typically requires computing a convolution but in some
cases, such as the Normal or the Poisson, the
distribution of D[1,j − 1] can be easily obtained (because sums
of independent Normal or Poisson random
variables are, respectively, Normal or Poisson). The distribution
of D[1,j − 1] is used together with the
weighted average fare
p̄ j−1 =
j−1∑
k=1
pk
µk
µ[1,j − 1]
and calls on Littlewood’s rule to obtain protection level
ybj−1 = max{y ∈ N : P(D[1,j − 1] ≥ y) > r
b
j−1,j}

where rbj−1,j = pj/p̄ j−1. Notice that the weighted average fare
assumes that a proportion µk/µ[1,j−1] of the
protected capacity will be sold at fare pk,k = 1, . . . ,j − 1. In
the special case when demands are Normal we
obtain
ybj−1 = µ[1,j − 1] + σ[1,j − 1]Φ
−1(1 −rbj−1,j).
Recall that for the Normal, variances are additive, so the
standard deviation σ[1,j − 1] =
√∑j−1
k=1 σ
2
k.
4.1 Evaluating the Performance of Heuristics
While heuristic protection levels are easy to compute,
evaluating them is as hard as solving for the optimal
policy. The expected return of a heuristic policy based on
protection levels 0 = yh0 ≤ yh1 ≤ . . . ≤ yhn−1 can be
computed exactly or via simulation. Let V hj (x) be the
expected revenue of using the heuristic with x units

of inventory before observing the demand for fare class j. We
will show how to compute V hj+1(x) assuming
that we have already computed V hj (x). We will do this by
determining a relationship between ∆V
h
j+1(x) and
∆V hj (x), using the following result.
Proposition 4
∆V hj+1(x) =
{
∆V hj (x) if x = 1, . . . ,y
h
j
= pj+1P(Dj+1 ≥ x−yhj ) +
∑x−yhj −1
i=0 ∆V

h
j (x− i)P(Dj+1 = i) if x > y
h
j
12
Thus, if V hj (x) has already been computed then V
h
j+1(x) = V
h
j (x) for x = 1, . . . ,y
h
j . For x > y
h
j we can
use the recursion V hj+1(x) = V
h

j+1(x− 1) + ∆V
h
j+1(x) starting with x = y
h
j + 1 in conjunction with the second
part of Proposition 4.
To estimate the expected revenue and other measures of
performance, such as the variance, we can also
use Monte Carlo simulation. Suppose we generate many random
copies of simulated demands (D1, . . . ,Dn).
For each copy we compute sales (sh1, . . . ,s
h
n) and revenues R
h =
∑n
i=1 pis
h
i under heuristic h. Averaging

over all the values of Rh gives an estimate of V hn (c).
Simulated sales can be generated sequentially via s
h
j =
min(Dj, (xj−yhn−1)+) starting with j = n and xn = c and using
the capacity update formula xj = xj+1−shj+1.
Example 3 (continued) We have applied the EMSR-a and
EMSR-b heuristics to the data of Example 3. Table
4 repeats the data and reports the heuristic protection levels ya,
yb as well as the optimal protection levels.
Table 5 reports V a5 (c) and V
b
5 (c) as well as V5(c) for values of c ∈ {50, 100, 150, 200, 250,
300, 350}.
j pj E[Dj] y
a
j y
b
j yj

1 $100 15 14 14 14
2 $60 40 53 54 54
3 $40 50 97 102 101
4 $35 55 171 166 169
5 $15 120
Table 4: Optimal and Heuristic Protection Levels for Example 3
c DF V a5 (c) V
b
5 (c) V5(c)
50 560% 3,426.8 3,426.8 3,426.8
100 280% 5,431.9 5,441.3 5,441.3
150 187% 7,184.4 7,188.6 7,188.7
200 140% 8,157.3 8,154.4 8,159.1
250 112% 8,907.3 8,901.4 8,909.1
300 93% 9,536.5 9,536.0 9,563.9
350 80% 9,625.0 9,625.0 9,625.0
Table 5: Performance of Heuristics for Example 3
As seen in Table 5, both the EMSR-a and the EMSR-b heuristic
perform very well against Poisson demands

under a low-to-high arrival pattern. The heuristics continue to
perform well if demands are compound Poisson
and aggregate demands are approximated by the use of a
Gamma distribution. However, EMSR based
heuristics can significantly underperform relative to models that
allow more general fare arrival rates. We will
have an opportunity to revisit this issue in Section 7.
Two more examples are presented below.
Example 4. From Wollmer [22]. There are 4 classes, and the
demand is assumed to be Normally distributed.
Table 6 shows the demand data and the protection levels under
optimal, EMSR-a, and EMSR-b policies.
There are considerable discrepancies between the protection
levels, and as expected, these discrepancies are
more severe in the earlier stages.
The expected revenues of the optimal policy and the two
heuristics, estimated through a simulation study
with 500,000 replications, are shown in Table 7. Capacity is
varied from 80 to 160 to create the demand factor
(ratio of total mean demand to capacity) in the range of 1.7 to
0.85. Note that the optimal protection levels
remain the same even when we change the capacity! The

percentage sub-optimality is also reported. For this
example, EMSR-a is slightly better than EMSR-b, but both
perform quite well.
Example 5. From Wollmer [22]. The demand parameters and the
lowest and highest fare are as in Example 4,
but the other fares are more evenly spaced. Such spacing is
more typical of what is found in practice. The
corresponding protection levels are shown in Table 8. Table 9
shows the expected revenues for the heuristics in
Example 5. In this case, both heuristics perform very well,
especially EMSR-b. Ratliff reports that EMSR-b
13
Demand Parameters Protection Levels (yj)
j pj µj σj OPT EMSR-a EMSR-b
1 1,050 17.3 5.8 16.7 16.7 16.7
2 567 45.1 15.0 42.5 38.7 50.9
3 534 39.6 13.2 72.3 55.7 83.2
4 520 34.0 11.3
Table 6: Normal demand parameters and protection levels for

Example 4 under the optimal, EMSR-a, and
EMSR-b policies.
Dmd Std EMSR-a EMSR-b
c Factor Vn(c) Err. % Sub % Sub
80 1.70 49,642 5.57 0.33% 0.43%
90 1.51 54,855 5.59 0.24% 0.52%
100 1.36 60,015 5.77 0.13% 0.44%
110 1.24 65,076 6.36 0.06% 0.34%
120 1.13 69,801 7.66 0.02% 0.21%
130 1.05 73,926 9.73 0.01% 0.10%
140 0.97 77,252 12.17 0.00% 0.04%
150 0.91 79,617 14.64 0.00% 0.01%
160 0.85 81,100 16.69 0.00% 0.00%
Table 7: Revenue performance for Example 4
usually provides improved performance on real world problems,
especially ones involving nested inventory
controls [16].
Demand Statistics Protection Levels (yj)
j pj µj σj OPT EMSR-a EMSR-b

1 1,050 17.3 5.8 9.7 9.8 9.8
2 950 45.1 15.0 54.0 50.4 53.2
3 699 39.6 13.2 98.2 91.6 96.8
4 520 34.0 11.3
Table 8: Model parameters and protection levels for Example 5
under optimal, EMSR-a, and EMSR-b policies.
5 Bounds, Revenue Opportunity Model, and New Heuristics
In this section we develop bounds on Vn(c) which may be
useful in evaluating the potential of applying revenue
management solutions. To obtain an upper bound, consider the
perfect foresight problem where the demand
vector D = (D1, . . . ,Dn) is known in advance. This demand
knowledge allows us to optimally allocate capacity
by solving the following knapsack type problem
V Un (c,D) = max
n∑
k=1
pkxk (11)

s.t. xk ≤ Dk ∀ k = 1, . . . ,n
n∑
k=1
xk ≤ c
xk ≥ 0 ∀ k = 1, . . . ,n.
14
Dmd Std EMSR-a EMSR-b
c Factor Vn(c) Error % Sub % Sub
80 1.70 67,505 9.29 0.10% 0.00%
90 1.51 74,003 10.89 0.06% 0.00%
100 1.36 79,615 13.27 0.40% 0.02%
110 1.24 84,817 13.30 0.35% 0.02%
120 1.13 89,963 13.58 0.27% 0.01%
130 1.05 94,860 14.73 0.15% 0.01%
140 0.97 99,164 17.06 0.06% 0.01%
150 0.91 102,418 20.08 0.01% 0.00%

160 0.85 104,390 22.85 0.00% 0.00%
Table 9: Revenue performance for Example 5
Clearly, for each realization of D, advance knowledge results in
revenues that are at least as high as the optimal
dynamic policy that does not have perfect foresight. As a result,
Vn(c) ≤ EV Un (c,D). For convenience, we
will denote this upper bound as V Un (c) = EV
U
n (c,D).
The solution to (11) can be written explicitly as xk = min(Dk, (c
− D[1,k − 1])+),k = 1, . . . ,n where
for convenience we define D[1, 0] = 0. The intuition here is that
we give priority to higher fares so fare
k ∈ {1, . . . ,n} gets the residual capacity (c − D[1,k − 1])+. The
expected revenue can be written more
succinctly after a few algebraic calculations:
V Un (c) =
n∑
k=1

pkE min(Dk, (c−D[1,k − 1])+) (12)
=
n∑
k=1
pk (E min(D[1,k],c) −E min(D[1,k − 1],c))
=
n∑
k=1
(pk −pk+1)E min(D[1,k],c) (13)
where for convenience we define pn+1 = 0. Moreover, since V
U
n (c,D) is concave in D, it follows from Jensen’s
inequality that V Un (c) = EV
U
n (c,D) ≤ V Un (c,µ) where V Un (c,µ) is the solution to
formulation (11) with

µ = E[D] instead of D. More precisely,
V Un (c,µ) = max
n∑
k=1
pkxk (14)
s.t. xk ≤ µk ∀ k = 1, . . . ,n
n∑
k=1
xk ≤ c
xk ≥ 0 ∀ k = 1, . . . ,n.
The linear program (14) is known as the fluid model or the
deterministic capacity allocation problem. It is
essentially a knapsack problem whose solution can be given in
closed form xk = min(µk, (c−µ[1,k−1])+) for
all k = 1, . . . ,n. Consequently, V Un (c) ≤ V Un (c,µ) =

∑n
k=1(pk −pk+1) min(µ[1,k],c).
A lower bound can be obtained by assuming a low to high
arrival pattern with zero protection levels.
This gives rise to sales min(Dk, (c−D[k + 1,n])+) at fare k = 1, .
. . ,n and revenue lower bound V L(c,D) =∑n
k=1 pkE min(Dk, (c−D[k + 1,n])
+).
Taking expectations we obtain
15
V Ln (c) =
n∑
k=1
pkE min(Dk, (c−D[k + 1,n])+) (15)
=

n∑
k=1
pk (E min(D[k,n],c) −E min(D[k + 1,n],c))
=
n∑
k=1
(pk −pk−1)E min(D[k,n],c) (16)
where p0 = 0. Notice that all of the terms in the sum are
negative except for k = 1. Clearly V
L
n (c) ≤ Vn(c)
since the expected revenue is computed under sub-optimal
protection levels. The above arguments justify the
main result of this section.
Proposition 5
V Ln (c) ≤ Vn(c) ≤ V

U
n (c) ≤ V
U
n (c,µ) (17)
Of course, the bounds require the computation of E
min(D[1,k],c), k = 1, . . . ,n. However, this is often
an easy computation. Indeed, if D[1,k] is any non-negative
integer random variable then E min(D[1,k],c) =∑c
j=1 P(D[1,k] ≥ j). If D[1,k] is Normal we can take advantage of
the fact that E min(Z,z) = z(1−Φ(z))−
φ(z) when Z is a standard Normal random variable and φ is the
standard Normal density function. If follows
that if D[1,k] is Normal with mean µ and variance σ2, then
E min(D[1,k],c) = µ + σE min(Z,z) = µ + σ [z(1 − Φ(z)) −φ(z)]
where z = (c−µ)/σ.
Tables 10 and 11 report V Ln (c),Vn(c),V
U
n (c) and Vn(c,µ) for the data of Examples 4 and 5,
respectively.

Notice that V Un (c) represents a significant improvement over
the better known bound Vn(c,µ), particularly
for intermediate values of capacity. The spread V Un (c) − V Ln
(c) between the lower and upper bound is a
gauge of the potential improvements in revenues from using an
optimal or heuristic admission control policy.
When capacity is scarce relative to the potential demand, then
the relative gap is large, and the potential
for applying revenue management solutions is also relatively
large. This is because significant improvements
in revenues can be obtained from rationing capacity to lower
fares. As capacity increases, the relative gap
decreases indicating that less can be gained by rationing
capacity. At very high levels of capacity it is optimal
to accept all requests, and at this point there is nothing to be
gained from the use of an optimal admission
control policy.
c V Ln (c) Vn(c) V
U
n (c) V
U
n (c,µ)

80 $42,728 $49,642 $53,039 $53,315
90 $48,493 $54,855 $58,293 $58,475
100 $54,415 $60,015 $63,366 $63,815
110 $60,393 $65,076 $68,126 $69,043
120 $66,180 $69,801 $72,380 $74,243
130 $71,398 $73,926 $75,923 $79,443
140 $75,662 $77,252 $78,618 $82,563
150 $78,751 $79,617 $80,456 $82,563
160 $80,704 $81,100 $81,564 $82,563
Table 10: Optimal Revenue and Bounds for Example 4.
16
c V Ln (c) Vn(c) V
U
n (c) V
Un(c,µ)
80 $52,462 $67,505 $72,717 $73,312
90 $61,215 $74,003 $79,458 $80,302

100 $70,136 $79,615 $85,621 $87,292
110 $78,803 $84,817 $91,122 $92,850
120 $86,728 $89,963 $95,819 $98,050
130 $93,446 $94,869 $99,588 $103,250
140 $98,630 $99,164 $102,379 $106,370
150 $102,209 $102,418 $104,251 $106,370
160 $104,385 $104,390 $105,368 $106,370
Table 11: Optimal Revenue and Bounds for Example 5.
5.1 Revenue Opportunity Model
The bounds presented here can help with the so called Revenue
Opportunity Model (ROM). The revenue
opportunity is the spread between the optimal revenue, obtained
by hindsight using the estimated uncensored
demand, and the revenue that results from not applying booking
controls. Demand uncensoring refers to a sta-
tistical technique that attempts to estimate actual demand from
the observed sales which may be constrained
by booking limits. The ex-post optimal revenue is a hindsight
optimization and is equivalent to our perfect
foresight model, resulting in revenue V Un (c,D), where D is
the uncensored demand. On the other hand, the

revenue based on not applying booking controls is just V Ln
(c,D), so a measure of the revenue opportunity is
V Un (c,D) − V Ln (c,D). The achieved revenue opportunity is
the difference between the actual revenue from
applying optimal or heuristic controls and the lower bound. The
ratio of the achieved revenue opportunity to
the revenue opportunity is often called the percentage achieved
revenue opportunity. The revenue opportunity
V Un (c,D)−V Ln (c,D) is sometimes approximated by V Un
(c)−V Ln (c) to get an idea of the revenue opportunity.
Table 10 and 11 shows there is significant revenue opportunity,
particularly for c ≤ 140. Thus, one use for the
ROM is to identify situations where RM has the most potential
so that more effort can be put where is most
needed. The ROM has also been used to show the benefits of
using leg-based control versus network-based
controls. The reader is refer to Chandler and Ja ([8]) and to
Temath et al. ([21]) for further information on
the uses of the ROM.
5.2 Bounds Based Heuristic
It is common to use an approximation to the value function as a
heuristic. To do this, suppose that Ṽj(x) is
an approximation to Vj(x). Then a heuristic admission control

rule can be obtained as follows:
ỹj = max{y ∈N : ∆Ṽj(y) > pj+1} j = 1, . . . ,n− 1. (18)
Suppose we approximate the value function Vj(x) by Ṽj(x) = θV
L
j (x) + (1 − θ)V
U
j (x) for some θ ∈ [0, 1]
and V Lj (x) and V
U
j (x) are the bounds obtained in this section applied to n = j and
c = x. Notice that
∆V Lj (x) = p1P(D[1,j] ≥ x) +
∑j
k=2(pk−pk−1)P(D[k,j] ≥ x), while ∆V
U
j (x) =
∑j−1
k=1(pk−pk+1)P(D[1,k] ≥

x) + pjP(D[1,j] ≥ x).
6 Multiple Fare Classes with Arbitrary Fare Arrival Patterns
So far we have suppressed the time dimension; the order of the
arrivals has provided us with stages that are a
proxy for time. In this section we consider models where time is
considered explicitly. There are advantages
of including time as part of the model as this allows for a more
precise formulation of the customer arrival
process. For example, we can relax the low-to-high arrival
assumption and allow for overlapping or concurrent
17
arrival rates. On the other hand, the flexibility advantage comes
at the cost of estimating arrival rates for
each of the fare classes over the sales horizon. If arrival rates
are not estimated accurately, then adding
the time dimension may hurt rather than help performance. In
addition, the formulations presented in this
section assumes that demand for each fare class follows a

Poisson process, whereas our earlier models based
on sequential fare arrivals do not have this restriction. We will
extend the formulation in this section to the
case of compound Poisson in §7.
We assume that customers arrive to the system according to a
time heterogeneous Poisson process with
intensity λjt, 0 ≤ t ≤ T where T is the length of the horizon, t
represents the time-to-go and j ∈ {1, . . . ,n}.
This means that the number of customers that arrive during the
last t units of time and request product j,
say Njt, is Poisson with parameter Λjt =
∫ t
0
λjsds. We will use the shorthand notation Λj = ΛjT to denote
the total expected demand for fare j ∈ {1, . . . ,n}.
Let V (t,x) denote the maximum expected revenue that can be
attained over the last t units of time from x
units of capacity. We will develop both discrete and continuous
time dynamic programs to compute V (t,x). To
construct a dynamic program we will need the notion of
functions that go to zero faster than their argument.

More precisely, we say that a function g(x) is o(x) if limx↓0
g(x)/x = 0. We now argue that the probability
that over the interval [t − δt,t] there is exactly one request and
the request is for product j is of the form
λjtδt + o(δt). Let Nj(t− δt,t] denote the number of requests for
product j over the interval (t− δt,t]. Then
Nj(t− δt,t] is Poisson with parameter
∫ t
t−δt λjsds = λjtδt + o(δt). Then
P(Nj(t− δt,t] = 1) = λjtδ exp(−λjtδt) + o(δt) = λjtδt[1 −λjtδt] +
o(δt) = λjtδt + o(δt),
while the probability that there are no requests for the other
products over the same interval is
P(Nk(t−δt,t] = 0,∀ k 6= j) = exp(−
∑
k 6=j
λktδt) + o(δt) = 1 −
∑
k 6=j

λktδt + o(δt).
Multiplying the two terms and collecting terms we obtain λjtδt
+ o(δt) as claimed.
We can now write
V (t,x) =
∑
j∈ Mt
λjtδt max(pj + V (t− δt,x− 1),V (t− δt,x)) + (1 −
∑
j∈ Mt
λjtδt)V (t− δt,x) + o(δt)
= V (t− δt,x) + δt
∑
j∈ Mt
λjt[pj − ∆V (t−δt,x)]+ + o(δt) (19)
with boundary conditions V (t, 0) = 0 and V (0,x) = 0 for all x ≥
0, where ∆V (t,x) = V (t,x)−V (t,x−1) for

x ≥ 1 and t ≥ 0. Here Mt ⊂{0, 1, . . . ,n} is the set of allowable
fares at time-to-go t. Usually Mt = {1, . . . ,n}
for large t, but low fares are dropped from Mt as the time-of-
purchase restrictions become binding.
The first term of (19) corresponds to the arrival of one request
for product j and the decision is made
between accepting the request obtaining pj + V (t − δt,x − 1)
and rejecting it obtaining V (t − δt,x). The
second term corresponds to not obtaining a request resulting in
revenue V (t−δt,x). Subtracting V (t−δt,x)
from both sides of equation (19), dividing by δt and taking the
limit as δt ↓ 0, we obtain the Hamilton Jacobi
Bellman (HJB) equation:
∂V (t,x)
∂t
=
∑
j∈ Mt
λjt[pj − ∆V (t,x)]+ (20)
with the same boundary conditions. The equation tells us that

the rate at which V (t,x) grows with t is the
weighted sum of the positive part of the fares net of the
marginal value of capacity ∆V (t,x) at state (t,x).
We can think of the right hand side as the profit rate in excess
of the marginal cost ∆V (t,x).
The value function (20) can be accurately computed by solving
and pasting the HJB equation. In practice,
V (t,x) is computed approximately by using a discrete time
dynamic programming formulation. A discrete
time dynamic programming formulation emerges from (19) by
rescaling time, setting δt = 1, and dropping
the o(δt) term. This can be done by selecting a > 1, so that aT is
an integer, and setting λjt ← 1aλj,t/a,
18
for t ∈ [0,aT]. The scale factor a should be sufficiently large so
that after scaling,
∑
j∈ Mt λjt << 1, e.g.,∑
j∈ Mt λjt ≤ .01 for all t ∈ [0,T], with T ← aT. The resulting

discrete time dynamic program is given by
V (t,x) = V (t− 1,x) +
∑
j∈ Mt
λjt[pj − ∆V (t− 1,x)]+ ∀ t ∈ {1, . . . ,T} and x ∈ {1, . . . ,c}, (21)
with the same boundary conditions. Computing V (t,x) via (21)
is quite easy and fairly accurate if time is
scaled appropriately. For each t, the complexity is order O(n)
for each x ∈ {1 . . . ,c} so the complexity per
period is O(nc), and the overall computational complexity is
O(ncT).
A formulation equivalent to (21) was first proposed by Lee and
Hersh [13], who also show that ∆V (t,x) is
increasing in t and decreasing in x. The intuition is that the
marginal value of capacity goes up if we have more
time to sell and goes down when we have more units available
for sale. From the dynamic program (21), it is
optimal to accept a request for product j when pj ≥ ∆V (t−1,x)
or equivalently, when pj + V (t−1,x−1) ≥
V (t−1,x), i.e., when the expecte revenue from accepting the
request exceeds the expected revenue of denying

the request. Notice that if it is optimal to accept a request for
fare j, then it is also optimal to accept a request
for any higher fare. Indeed, if pk ≥ pj and pj ≥ ∆V (t − 1,x),
then pk ≥ ∆V (t − 1,x). Assuming that the
fares are ordered: p1 ≥ p2 ≥ . . . ≥ pn, then it is optimal to
accept all fares in the active set
A(t,x) = {j ∈ Mt : j ≤ a(t,x)}, where
a(t,x) = max{j : pj ≥ ∆V (t− 1,x)} t ≥ 1,x ≥ 1,
and to reject all fares in the complement R(t,x) = {j ∈ {1, . . .
,n} : j /∈ A(t,x)}.
We can also express the policy in terms of dynamic protection
levels yj(t),j = 1, . . . ,n− 1, given by
yj(t) = max{x : a(t,x) = j},
so if x ≤ yj(t) then fares j + 1 and higher should be closed.
Proposition 6 The active set A(t,x) is decreasing in t and
increasing in x. Moreover, yj(t) is increasing in
j and increasing in t.

Proof: Both results follow directly from the fact that ∆V (t,x) is
increasing in t and decreasing in x.
Notice that it may be optimal to reopen a fare after it is closed.
Indeed, suppose that a(t,x) = j and there
is an absence of sales over the interval (s,t], then at state (s,x),
a(s,x) ≥ a(t,x), and more fares may open
if a(s,x) > a(t,x). In §9 will discuss a modification of this
formulation that keeps fares closed once they are
closed for the first time, as such formulations are more robust in
the presence of strategic customers that may
prefer to wait for a lower fare.
6.1 A Pricing Formulation with Broader Interpretation
In this section we recast the dynamic programs (20) and (21) as
dynamic pricing problems with finite price
menus. The idea is that the active sets A(t,x) are of the form Sjt
= {k ∈ Mt : k ≤ j}. Associated
with each of these active sets is a probability of sale, say πjt
and an average fare, say qjt. To see this, let
λt =
∑
j∈ Mt λjt denote the overall arrival rate for request at time t for

fares in Mt. Define πjt =
∑
k∈ Sjt λkt/λt
and rjt =
∑
k∈ Sjt pkλkt/λt. We can think of πjt and rjt as the probability of
sale and the average revenue
rate, per arriving customer, when we offer all the fares in Sjt for
all j ∈ Nt. . Now let qjt = rjt/πjt be the
average fare per unit sold when the offer set is Sjt,j ∈ Mt. For
convenience we let q0t = π0t = r0t = 0 for
19
j = 0 and M+t = Mt ∪ {0}. With this notation, we can write
formulation (21) as
V (t,x) = V (t− 1,x) + λt max
j∈ M+

t
[rjt −πjt∆V (t− 1,x)]
= V (t− 1,x) + max
j∈ M+
t
λtπjt[qjt − ∆V (t− 1,x)]. (22)
The equivalent formulation for the continuous time model (20)
is
∂V (t,x)
∂t
= max
j∈ M+
t
λtπjt[qjt − ∆V (t,x)]. (23)
Formulations (22) and (23) can be interpreted broadly as the

problem of optimizing the expected revenue
from state (t,x) over a finite price menu qjt,j ∈ M+t with
associated demand rate λtπjt,j ∈ Mt. We will come
back to this pricing formulation when we discuss the dynamic
capacity allocation problem with dependent
demands. We will see there that essentially the same pricing
formulation works for dependent demands. For
the case of dependent demands, the set M+t will be the index
corresponding to a collection of efficient sets or
assortments.
7 Compound Poisson Demands
The formulations of the dynamic programs (20) and (21),
implicitly assume that each request is for a single
unit. Suppose instead, that each arrival is for a random number
of units. More specifically, suppose that
request for fare j are of random size Zj, and that the probability
mass function Pj(z) = P(Zj = z),z ≥ 1 is
known for each j. As before, we assume independent demands
for the different fare classes j ∈ N. We seek
to generalize the dynamic programs (20) and (21) so that at each
state (t,x) we can decide whether or not to
accept a fare pj request of size Zj = z. The expected revenue
from accepting the request is zpj +V (t−1,x−z)

and the expected revenue from rejecting the request is V (t−
1,x). Let ∆zV (t,x) = V (t,x) −V (t,x− z) for
all z ≤ x and ∆zV (t,x) = ∞ if z > x. The purpose is to assure
ourselves that a request that exceeds the
current capacity will always be rejected. We can think of ∆zV
(t,x) as a the sum of the the z marginal values
∆V (t,x) + ∆V (t,x− 1) + . . . + ∆V (t,x−z + 1).
The dynamic program (20) with compound Poisson demands is
given by
∂V (t,x)
∂t
=
∑
j∈ Mt
λjt
x∑
z=1
Pj(z)[zpj − ∆zV (t,x)]+, (24)

while the discrete time dynamic program (21) with compound
Poisson demands is given by
V (t,x) = V (t− 1,x) +
∑
j∈ Mt
λjt
x∑
z=1
Pj(z)[zpj − ∆zV (t− 1,x)]+, (25)
with boundary conditions V (t, 0) = V (0,x) = 0. Notice that the
sums in (24) and (25) can be changed to
∑∞
z=1
instead of
∑x
z=1 as the terms z > x do not contribute to the sum given our
convention that ∆zV (t,x) = ∞ for
z > x. The optimal policies for the two programs are,

respectively, to accept a size z request fare pj, j ∈ Mt,
if zpj ≥ ∆zV (t,x), and to accept a z request fare pj, j ∈ Mt, if
zpj ≥ ∆zV (t− 1,x). The two policies should
largely coincide if time is scaled correctly so that
∑
j∈ Mt λjt << 1 for all t ∈ [0,T].
For compound Poisson demands, we can no longer claim that
the marginal value of capacity ∆V (t,x) is
decreasing in x, although it is still true that ∆V (t,x) is
increasing in t. To see why ∆V (t,x) is not monotone in
x, consider a problem where the majority of the requests are for
two units and request are seldom for one unit.
Then the marginal value of capacity for even values of x may be
larger than the marginal value of capacity for
20
odd values of x. Consequently, some of the structure may be
lost. For example, it may be optimal to accept a
request of a single unit of capacity when x is odd, but not if x is
even, violating the monotonicity of ∆V (t,x).

However, even if some of the structure is lost, the computations
involved to solve (25) are straightforward as
long as the distribution of Zj is known. Airlines, for example,
have a very good idea of the distribution of Zj
for different fare classes.
Example 6. Consider again the data of Examples 3 with fares p1
= $100,p2 = $60,p3 = $40,p4 = $35 and
p5 = $15 with independent compound Poisson demands, with
uniform arrival rates λ1 = 15,λ2 = 40,λ3 =
50,λ4 = 55,λ5 = 120 over the horizon [0, 1]. We will assume
that Mt = {1, . . . ,n} for all t ∈ [0,T], with
T = 1. The aggregate arrival rates are given by Λj = λjT = λj for
all j. We will assume that the distribution
of the demand sizes is given by P(Z = 1) = 0.65,P(Z = 2) =
0.25,P(Z = 3) = 0.05 and P(Z = 4) = .05 for
all fare classes. Notice that E[Z] = 1.5 and E[Z2] = 2.90, so the
variance to mean ratio is 1.933. We used the
dynamic program (25) with a rescaled time horizon T ← aT = 2,
800, and rescaled arrival rates λj ← λj/a for
all j. Table 12 provides the values V (T,c) for c ∈ {50, 100, 150,
200, 250, 300, 350}. Table 12 also provides the
values ∆V (t,x) for t = 207 in the rescaled horizon for x ∈ {1, . .
. , 6} to illustrate the behavior of the policy.
The reader can verify that at state (t,x) = (208, 3) it is optimal

to accept a request for one unit at fare p2,
and to reject the request if it is for two units. Conversely, if the
state is (t,x) = (208, 4) then it is optimal to
reject a request for one unit at fare p2, and to accept the request
if it is for two units. The reason for this is
that the value of ∆V (t,x) is not monotone decreasing at x = 4.
c 50 100 150 200 250 300
V (T,c) $3,837 $6,463 $8,451 $10,241 $11,724 $12,559
x 1 2 3 4 5 6
∆V (t,x) 70.05 66.48 59.66 60.14 54.62 50.41
Table 12: Value function V (T,c) and marginal revenues ∆V
(t,c) for t = 208, Example 6: Compound Poisson
8 Static vs Dynamic Policies
In this section we compare the performance of static policies
obtained using the dynamic program (8) to the
performance of dynamic policies for compound Poisson
demands based on (25). To our knowledge, this is the
first time these two policies are compared. Diwan [7], for
example, compares the performance of the EMSR-b
heuristic for static policies against the performance of the

dynamic formulation for Poisson demands (21). Not
surprisingly, he finds that this heuristic use of (21) can
underperform static policies when demands are not
Poisson. A more compelling exercise is to compare static
policies against dynamic policies for the compound
Poisson case as we do here. It is possible to formally proof that
dynamic policies do at least as well, so
computational studies should concentrate on measuring the gap
between static and dynamic policies. It turns
out that the gap is fairly small when the arrival rates are low-to-
high, or the arrival rates follow a prescribed
order that is consistent with the computations of the protection
levels. This is particularly true if the static
policy is implemented by frequently resolving the problem
during the sales horizon.
Comparing static and dynamic policies when the fare arrival
rates do not follow a specific is more difficult
because revenues depend heavily on how the static protection
levels are implemented. Two possible implemen-
tations are possible. Under theft nesting a request of size z for
fare j is accepted if x−z ≥ yj−1, where x os the
current inventory and yj−1 is the protection level for fares {1, .
. . ,j − 1}. This method is called theft nesting
because the remaining inventory x at time-to-go t deducts all

previous bookings regardless of fare class. In
contrast, standard nesting is implemented by accepting a size z
request for fare j if x−z ≥ (yj−1−b[1,j−1])+,
where b[1,j−1] are the observed bookings of fares [1,j−1] at
state (t,x). In practice, standard nesting works
much better than theft nesting when the fare arrival pattern is
not low-to-high. This makes sense because
under standard nesting we do not insist on protecting yj−1 units
for fares {1, . . . ,j} even though we have
already booked b[1,j − 1] units of these fares. Thus, we use
standard nesting in comparing static policies
versus dynamic policies. The performance of static policies
suffer relative to that of the dynamic program (25)
as illustrated in Example 7 based on the data of Example 6.
21
Example 7 Consider again the data of Example 6 with uniform
arrival rates. Table 13 compares the perfor-
mance V (T,c) of the compound poisson formulation (25) to the
performance of the EMSR-b under standard
nesting. Part of the gap between V b(T,c) and V (T,c) can be
reduced by frequently recomputing the booking

limits applying the EMSR-b heuristic during the sales horizon.
However, this is not enough to overcome the
disadvantage of the EMSR-b heuristic when arrivals are not
low-to-high.
c 50 100 150 200 250 300
V b(T,c) $3,653 $6,177 $8,187 $9,942 $11,511 $12,266
V (T,c) $3,837 $6,463 $8,451 $10,241 $11,724 $12,559
Gap 4.8% 4.4% 3.1% 2.9% 1.8% 2.3%
Table 13: Sub-Optimality of EMSR-b with Standard Nesting
The next proposition shows that V s(T,x) ≤ V (T,x), where V
s(T,x) is the expected revenue resulting from
applying a static booking policy with standard nesting, and V
(T,x) is the optimal expected revenue for the
compound Poisson Dynamic Program. The same simple proof
can be used to show that V b(T,x) ≤ V (T,x).
In fact, a proof is hardly needed as we are comparing heuristics
to optimal dynamic policies.
Proposition 7
V s(T,x) ≤ V (T,x) ∀ x ∈ {0, 1, . . . ,c}.

Proof: Clearly for V s(0,x) = V (0,x) = 0 so the result holds for t
= 0, for all x ∈ {0, 1, . . . ,c}. Suppose
the result holds for time-to-go t− 1, so V s(t− 1,x) ≤ V (t− 1,x)
for all x ∈ {0, 1, . . . ,c}. We will show that
it also holds for time-to-go t. If a request of size z arrives for
fare class j, at state (t,x), the policy based on
protection levels y1, . . . ,yn−1 will accept the request if x−z ≥
(yj−1−b[1,j−1])+ and will rejected otherwise.
In the following equations, we will use Qj(z) to denote P(Zj >
z). We have
V s(t,x) =
∑
j∈ Nt
λjt[
x−(yj−1−b[1,j−1])+∑
z=1
Pj(z)(zpj + V
s(t− 1,x−z))
+ Qj(x− (yj−1 − b[1,j − 1])+)V s(t− 1,x)] + (1 −
∑

j∈ Nt
λjt)V
s(t− 1,x)
≤
∑
j∈ Nt
λjt[
x−(yj−1−b[1,j−1])+∑
z=1
Pj(z)(zpj + V (t− 1,x−z))
+ Qj(x− (yj−1 − b[1,j − 1])+)V (t− 1,x)] + (1 −
∑
j∈ Nt
λjt)V (t− 1,x)
= V (t− 1,x) +
∑
j∈ Nt

λjt
x−(yj−1−b[1,j−1])+∑
z=1
Pj(z)(zpj − ∆zV (t− 1,x))
≤ V (t− 1,x) +
∑
j∈ Nt
λjt
∞∑
z=1
Pj(z)(zpj − ∆zV (t− 1,x))+
= V (t,x),
where the first equation follows from the application of the
protection level policy, the first inequality follows
from the inductive hypothesis V s(t − 1,x) ≤ V (t − 1,x). The
second equality collects terms, the second

inequality follows because we are taking positive parts, and the
last equality from the definition of V (t,x).
While we have shown that V s(T,c) ≤ V (T,c), one may wonder
whether there are conditions where equality
holds. The following results answers this question.
Corollary 2 If the Dj’s are independent Poisson random
variables and the arrivals are low-to-high then
Vn(c) = V
s(T,c) = V (T,c).
22
Proof: Notice that if the Djs are Poisson and the arrivals are
low-to-high, then we can stage the arrivals
so that λjt = nE[Dj]/T over t ∈ (tj−1, tj] where tj = jT/n for j =
1, . . . ,n. We will show by induction in
j that Vj(x) = V (tj,x). Clearly y0 = 0 and V1(x) = p1E
min(D1,x) = V (t1,x) assuming a sufficiently large
rescale factor. Suppose, by induction, that Vj−1(x) = V (tj−1,x).
Consider now an arrival at state (t,x) with

t ∈ (tj−1, tj]. This means that an arrival, if any, will be for one
unit of fare j. The static policy will accept this
request if x−1 ≥ yj−1, or equivalently if x > yj−1. However, if x
> yj−1, then ∆(t−1,x) ≥ ∆V (tj−1,x) ≥ pj,
because ∆V (t,x) is increasing in t and because yj−1 = max{y :
∆Vj−1(y) > pj} = max{y : ∆V (tj−1,x) > pj},
by the inductive hypothesis. Conversely, if the dynamic
program accepts a request, then pj ≥ ∆V (t,x) and
therefore x > yj−1 on account of ∆V (t,x) ≥ ∆V (tj−1,x).
8.1 Bounds on V (T, c)
We will now briefly show that the upper bound V Un (c) for
Vn(c), developed in Section 5 for the static multi-fare
model is still valid for V (T,c). The random revenue associated
with the perfect foresight model is Vn(c,D)
and can be obtained by solving the linear program (11). Notice
that for all sample paths, this revenue is
at least as large as the revenue for the dynamic policy. Taking
expectations we obtain V (T,c) ≤ V Un (c) =
EVn(c,D) =
∑n
k=1(pk − pk+1)E min(D[1,k],c), where for convenience pn+1 =
0. Moreover, since dynamic

policies do at least as well as static policies, the lower bounds
obtained in Section 5 also apply to dynamic
policies.
9 Monotonic Fare Offerings
The dynamic programs (20) and (21) and their compound
Poisson counterparts (22) and (23), all implicitly
assume that fares can be opened and closed at any time. To see
how a closed fare may reopen, suppose that
a(t,x) = j so set A(t,x) = {k ∈ Mt : k ≤ j} is offered at state
(t,x), but an absence of sales may trigger
fare/action j + 1 to open as a(s,x) increases as the time-to-go s
decreases. This can lead to the emergence
of customers or third parties that specialize on inter-temporal
fare arbitrage. To avoid this, the capacity
provider may commit to a policy of never opening fares once
they are closed. To handle monotonic fares
requires modifying the dynamic programming into something
akin to the dynamic program (8) where time
was handled implicitly. Let Vj(t,x) be the maximum expected
revenue from state (t,x) that can be obtained
by offering any set Sit = {k ∈ Mt,k ≤ i}, i ≤ j, so that closed
fares are not allowed to reopen. Let Wk(t,x)

be the expected revenue from accepting fares Skt at state (t,x)
and then following an optimal policy. More
precisely,
Wk(t,x) =
∑
i∈ Skt
λit[pi + Vk(t− 1,x− 1)] + (1 −
∑
i∈ Skt
λit)Vk(t− 1,x)
= Vk(t− 1,x) + λt[rkt −πkt∆Vk(t− 1,x)]
= Vk(t− 1,x) + λtπkt[qkt − ∆Vk(t− 1,x)],
where ∆Vk(t,x) = Vk(t,x) −Vk(t,x− 1), and the quantities λt,πkt
and qkt are as defined in §6.1.
Then Vj(t,x) satisfies the dynamic program
Vj(t,x) = max
k≤j

Wk(t,x) = max{Wj(t,x),Vj−1(t,x)} (26)
with the boundary conditions Vj(t, 0) = Vj(0,x) = 0 for all t ≥ 0
and all x ∈ N for all j = 1, . . . ,n. It
follows immediately that Vj(t,x) is monotone increasing in j. An
equivalent version of (26) for the case n = 2
can be found in Weng and Zheng [23]. The complexity to
compute Vj(t,x),x = 1, . . . ,c for each j is O(c)
so the complexity to compute Vj(t,x),j = 1, . . . ,n,x = 1, . . . ,c
is O(nc). Since there are T time periods
the overall complexity is O(ncT). While computing Vj(t,x)
numerically is fairly simple, it is satisfying to
know more about the structure of optimal policies as this gives
both managerial insights and can simplify
computations. The proof of the structural results are intricate
and subtle, but they parallel the results for the
23
dynamic program (8) and (21). The following Lemma is the
counterpart to Lemma 2 and uses sample path
arguments based on ideas in [23] to extend their results from n
= 2 to general n. The proof can be found in

the Appendix.
Lemma 3 For any j ≥ 1,
a) ∆Vj(t,x) is decreasing in x ∈ N+, so the marginal value of
capacity is diminishing.
b) ∆Vj(t,x) is increasing in j ∈ {1, . . . ,n} so the marginal value
of capacity increases when we have more
stages to go.
c) ∆Vj(t,x) is increasing in t, so the marginal value of capacity
increases as the time-to-go increases.
Let
aj(t,x) = max{k ≤ j : Wk(t,x) = Vj(t,x)}.
In words, aj(t,x) is the index of the lowest open fare that is
optimal to post at state (t,x) if we are allowed
to use any fares in Sjt. Let
Aj(t,x) = {k ∈ Mt : k ≤ aj(t,x)}.
Then Aj(t,x) is the optimal set of fares to open at state (j,t,x).
Clearly Vi(t,x) = Vj(t,x) for all i ∈

{aj(t,x), . . . ,j}. The following Lemma asserts that aj(t,x) is
monotone decreasing in t (it is optimal to have
fewer open fares with more time-to-go and the same inventory),
monotone increasing in x (it is optimal to
have more open fares with more inventory and the same time-to-
go) and monotonically increasing in j.
Lemma 4 aj(t,x) is decreasing in t and increasing in x and j.
Moreover, aj(t,x) = k < j implies ai(t,x) = k
for all i ≥ k.
It is possible to think of the policy in terms of protection levels
and in terms of stopping sets. Indeed, let
Tj = {(t,x) : Vj(t,x) = Vj−1(t,x)}. We can think of Tj as the
stopping set for fare j as it is optimal to close
down fare j upon entering set Tj. For each t, let yj(t) = max{x
∈ N : (t,x) ∈ Tj+1}. We can think of yj(t) as
the protection level for fares in Sj against higher fares. The
following result is the counterpart to Theorem 2.
Theorem 3 • Aj(t,x) is decreasing in t and increasing in x and j.
• T1 ⊂ T2 ⊂ . . . ⊂ Tn.
• yj(t) is increasing in t and in j.

• If x ≤ yj(t) then Vi(t,x) = Vj(t,x) for all i > j.
Proof: The properties of Aj(t,x) follow from the properties of
aj(t,x) established in Lemma 4. Tj = {(t,x) :
aj(t,x) < j}. From Lemma 4, aj(t,x) < j implies that ai(t,x) < i
for all i > j, so Tj ⊂ Ti for all i > j. This
implies that yj(t) is increasing in j for any t ≥ 0. If t′ > t, then
aj+1(t′,yj(t)) ≤ aj+1(t,yj(t)) < j + 1, so
yj(t
′) ≥ yj(t). Since yj(t) ≤ yi(t) for all i > j, then x ≤ yj(t) implies
Vi+1(t,x) = Vi(t,x) for all i ≥ j and
therefore Vi(t,x) = Vj(t,x) for all i > j.
The policy is implemented as follows: The starting state is
(n,T,c) as we can use any of the fares {1, . . . ,n},
we have T units of time to go and c is the initial inventory. At
any state (j,t,x) we post fares Aj(t,x) =
{1, . . . ,aj(t,x)}. If a unit is sold during period t the state is
updated to (aj(t,x), t−1,x−1) since all fares in
the set Aj(t,x) are allowed, the time-to-go is t− 1 and the
inventory is x− 1. If no sales occur during period
t the state is updated to (aj(t,x), t− 1,x). The process continues
until either t = 0 or x = 0.

24
Example 8. Consider Example 1 again with 5 fares p1 = $100,p2
= $60,p3 = $40,p4 = $35 and p5 = $15
with independent Poisson demands with means Λ1 = 15, Λ2 =
40, Λ3 = 50, Λ4 = 55 and Λ5 = 120 and
T = 1. The scaling factor was selected so that
∑5
i=1 Λi/a < .01 resulting in T ← aT = 2, 800. We also
assume that the arrival rates are uniform over the horizon [0,T],
i.e., λj = Λj/T. In Table 14 we present
the expected revenues Vj(T,c),j = 1, . . . , 5 and V (T,c) for c ∈
{50, 100, 150, 200, 250}. The first row is V5(c)
from Example 1. Notice that V5(c) ≤ V5(T,c). This is because
we here we are assuming uniform, rather than
low-to-high arrivals. V (T,c) is even higher because we have the
flexibility of opening and closing fares at will.
While the increase in expected revenues [V (T,c) −V5(T,c)] due
to the flexibility of opening and closing fares
may be significant for some small values of c (it is 1.7% for c =

50), attempting to go for this extra revenue
may invite strategic customers or third parties to arbitrage the
system.
c 50 100 150 200 250 300 350
V5(c) 3,426.8 5,441.3 7,188.7 8,159.1 8,909.1 9,563.9 9,625.0
V (T,c) 3,553.6 5,654.9 7,410.1 8,390.6 9,139.3 9,609.6 9,625.0
V5(T,c) 3,494.5 5,572.9 7,364.6 8,262.8 9,072.3 9,607.2
9,625.0
V4(T,c) 3,494.5 5,572.9 7,364.6 7,824.9 7,825.0 7,825.0
7,825.0
V3(T,c) 3,494.5 5,572.9 5,900.0 5,900.0 5,900.0 5,900.0
5,900.0
V2(T,c) 3,494.5 3,900.0 3,900.0 3,900.0 3,900.0 3,900.0
3,900.0
V1(T,c) 1,500.0 1,500.0 1,500.0 1,500.0 1,500.0 1,500.0
1,500.0
Table 14: Expected Revenues V (T,c) with uniform arrival rates
To obtain a continuous time formulation, we can use the same
logic that lead to (20) to obtain
∂Vj(t,x)

∂t
=
{
rjt −πjt∆Vj(t,x) if (t,x) /∈ Zj−1
∂Vj−1(t,x)
∂t
if (t,x) ∈ Zj−1
(27)
with the same boundary conditions.
9.1 Mark-up and Mark-down Policies
We now go back to the broader pricing interpretation coupled
with the monotonic fare formulation (26). In
many applications the price menu pjt,j = 1, . . . ,n is time
invariant, but the associated sales rates πjt,j =
1, . . . ,n are time varying. In addition, we will assume that
there is a price p0t such that π0t = 0 for all t.
This technicality helps with the formulation as a means of
turning off demand when the system runs out of

inventory. The case p1t ≥ p2t ≥ . . . ≥ pnt and π1t ≤ π2t ≤ . . . ≤
πnt is known as the mark-up problem, while
the case p1t ≤ p2t ≤ . . . ≤ pnt and π1t ≥ π2t ≥ . . . ≥ πnt is
known as the mark-down problem. The former
model is relevant in Revenue Management while the second is
relevant in Retailing.
For the RM formulation, the problem can be viewed as
determining when to mark-up (switch from action
j to j − 1). The optimal mark-up times are random as they
depend on the evolution of sales under the
optimal policy. Suppose that the current state is (j,t,x), so the
last action was j, the time-to-go is t and the
inventory is x. We want to determine whether we should
continue using action j or switch to action j − 1.
We know that if x > yj−1(t), then we should keep action j and if
x ≤ yj−1(t) then we should close action
j. Let Tj = {(t,x) : x ≤ yj−1(t)}, then it is optimal to stop action
j upon first entering set Tj. Notice that
a mark-up occurs when the current inventory falls below a
curve, so low inventories trigger mark-ups, and
mark-ups are triggered by sales. The retailing formulation also
has a threshold structure, but this time a
mark-down is triggered by inventories that are high relative to a
curve, so the optimal timing of a mark-down

is triggered by the absence of sales. Both the mark-up and the
mark-down problems can be studied from the
point of view of stopping times. We refer the reader to Feng and
Gallego [9], [10], and Feng and Xiao [11] and
reference therein for more on the markup and markdown
problems.
25
10 Acknowledgments
I acknowledge the feedback from my students and collaborators.
In particular, I would like to recognize the
contributions and feedback from Anran Li, Lin Li, and Richard
Ratliff.
26
11 Appendix
Proof of Lemma 1: Taking expectations yields g(y) = G(y)P(X ≥

y) +
∑
j≤y−1 G(j)P(X = j) and
g(y−1) = G(y−1)P(X ≥ y) +
∑
j≤y−1 G(j)P(X = j). Taking the difference yields ∆g(y) =
∆G(y)P(X ≥ y).
Taking expectations yields r(y) = R(y)P(X < y) +
∑
j≥y R(j)P(X = j) and r(y − 1) = R(y − 1)P(X <
y) +
∑
j≥y R(j)P(X = j). Taking the difference we see that ∆r(y) =
∆R(y)P(X < y).
Proof of Proposition 1: Let G(y) = p1y, then V1(y) = g(y) =
EG(min(D1,y)), so ∆V1(y) = ∆g(y) =
p1P(D1 ≥ y). This establishes the first part of the Proposition.
To establish the second part of the Proposition
we use the first part of Lemma 1 to show that p2E
min(D2,c−y)−p2E min(D2,c−y + 1) = −p2P(D2 > c−y)
and the second part of the Lemma 1 to show that EV1(max(x −

D2,y)) − EV1(max(x − D2,y − 1)) =
∆V1(y)P(D2 > c−y). The second part of the Proposition then
follows from putting the two parts together. To
see the first part, let r(y) = p2c−p2E max(c−D2,y), then ∆r(y) =
p2E min(D2,c−y)−p2E min(D2,c−y+1) =
∆R(y)P(c − D2 < y) where R(y) = −p2y, so ∆r(y) = −p2P(c − D2
< y) = −p2P(D2 > c − y). Now let
R(y) = V1(y), then ∆r(y) = ∆V1(y)P(c−D2 < y) = ∆V1(y)P(D2 >
c−y) completing the proof.
Proof of Lemma 2: We will prove the result by induction on j.
The result holds for j = 1 since ∆V1(y) =
p1P(D1 ≥ y) is decreasing in y and clearly ∆V1(y) = p1P(D1 ≥
y) ≥ ∆V0(y) = 0. Assume that the result is
true for Vj−1. It follows from the dynamic programming
equation (8) that
Vj(x) = max
y≤x
{Wj (y,x))} ,
where for any y ≤ x, Wj(y,x) = E [pj min{Dj,x−y}] + E [Vj−1
(max{x−Dj,y})].
An application of 1 shows that for y ∈ {1, . . . ,x}

∆Wj(y,x) = Wj(y,x) −Wj(y − 1,x) = [∆Vj−1(y) −pj] P(Dj > x−y).
Since ∆Vj−1(y) is decreasing in y (this is the inductive
hypothesis), we see that Wj(y,x) ≥ Wj(y − 1,x)
if ∆Vj−1(y) > pj and Wj(y,x) ≤ Wj(y − 1,x) if ∆Vj−1(y) ≤ pj.
Consider the expression
yj−1 = max{y ∈ N : ∆Vj−1(y) > pj} (28)
where the definition of ∆Vj(y) is extend to y = 0 for all j by
setting ∆Vj(0) = p1. If yj−1 ≤ x then
Vj(x) = max
y≤x
Wj(y,x) = Wj(yj−1,x)
On the other hand, if x < yj−1 then
Vj(x) = max
y≤x
Wj(y,x) = Wj(x,x).

In summary,
Vj(x) = Wj(min(x,yj−1),x)
=
{
Vj−1(x), if x ≤ yj−1
E [pj min{Dj,x−yj−1}] + E [Vj−1 (max{x−Dj,yj−1})] if x >
yj−1
27
Computing ∆Vj(x) = Vj(x) −Vj(x− 1) for x ∈ N results in:
∆Vj(x) =
{
∆Vj−1(x), if x ≤ yj−1
E min(pj, ∆Vj−1(x−Dj)) if x > yj−1
(29)

We will now use this result to show that ∆Vj(x) is itself
decreasing in x. Since ∆Vj(x) = ∆Vj−1(x) for
x ≤ yj−1 and ∆Vj−1(x) is decreasing in x we only need to worry
about the case x > yj−1.
However, in this case
∆Vj(x) = E min(pj, ∆Vj−1(x−Dj))
is decreasing in x since ∆Vj−1(x) is itself decreasing in x.
We now show that ∆Vj(x) ≥ ∆Vj−1(x). For x > yj−1 we have
min(pj, ∆Vj−1(x−Dj)) ≥ min(pj, ∆Vj−1(x)) =
∆Vj−1(x) where the inequality follows since ∆Vj−1(x) is
decreasing in x and the equality since x > yj−1. Tak-
ing expectations we see that ∆Vj(x) ≥ ∆Vj−1(x) on x > yj−1
while ∆Vj(x) = ∆Vj−1(x) on x ≤ yj−1.
Proof of Proposition 2: We first show that ∆Vj+1(x) = ∆Vj(x)
for x = 1, . . . ,yj. This follows since
Vj+1(x) = Wj+1(x,x) = Vj(x) in this range as no capacity is
made available for fare pj+1 when x ≤ yj. For
x > yj
Vj+1(x) = pj+1E min(x−yj,Dj+1) + EVj(max(x−Dj+1,yj)

= pj+1
x−yj∑
k=1
Pr(Dj+1 ≥ k) + Vj(yj)Pr(Dj+1 > x−yj)
+
x−yj∑
k=0
Vj(x−k)Pr(Dj+1 = k).
Consequently, for x > yj
∆Vj+1(x) = pj+1Pr(Dj+1 ≥ x−yj) +
x−yj−1∑
k=0
∆Vj(x−k)Pr(Dj+1 = k) (30)
follows from Vj(x − 1) = Vj(yk) for x = yj + 1. Since pj+1 <
∆Vj(y) for y ≤ yj, we can write ∆Vj+1(x) =∑∞

k=0 min(pj+1, ∆Vj(x−k))P(Dj+1 = k) = E min(pj+1,
∆Vj(x−Dj+1)).
Proof of Proposition 3: Since Πn(c,k) is the difference of a
concave and a linear function, Πn(c,k) it is
itself concave. The marginal value of adding the cth unit of
capacity is ∆Vn(c) − k so the cth unit increase
profits as long as ∆Vn(c) > k. Therefore, the smallest optimal
capacity is given by c(k). (Notice that c(k) + 1
may be also optimal if ∆Vn(c(k) + 1) = k.) c(k) is decreasing in
k since ∆Vn(c) is decreasing in c. Suppose
that k = pj+1. To establish c(pj+1) = yj it is enough to show that
that ∆Vn(yj) > pj+1 ≥ ∆Vn(yj + 1). By
definition yj = max{y ∈ N : ∆Vj(y) > pj+1} so ∆Vj(yj) > pj+1 ≥
∆Vj(yj + 1). Since it is optimal to protect
up to yj units of capacity for sale at fares j,j − 1, . . . , 1, it
follows that Vn(c) = Vj(c) for all c ≤ yj, and
consequently ∆Vn(yj) = ∆Vj(yj) > pj+1. Now ∆Vn(yj + 1) can
be written as a convex combination of pj+1
and ∆Vj(yj + 1) ≤ pj+1 which implies that ∆Vn(yj + 1) ≤ pj+1,
completing the proof.
Proof of Lemma 3: We will first show part that ∆Vj(t,x) is
decreasing in x which is equivalent to showing
that 2Vj(t,x) ≥ Vj(t,x + 1) + Vj(t,x− 1)] for all x ≥ 1. Let A be

an optimal admission control rule starting
from state (t,x + 1) and let B be an optimal admission control
rule starting from (t,x− 1). These admission
control rules are mappings from the state space to subsets Sk =
{1, . . . ,k},k = 0, 1, . . . ,j where S0 = ∅ is
the optimal control whenever a system runs out of inventory.
Consider four systems: two starting from state
(t,x), using control rules A′ and B′, respectively, and one each
starting from (t,x + 1) and (t,x − 1), using
control rule A and B, respectively. Our goal is to specify
heuristic control rules A′ and B′ that together make
the expected revenues of the two systems starting with (t,x) at
least as large as the expected revenues from
the systems starting at (t,x + 1) and (t,x− 1). This will imply
that 2Vj(t,x) ≥ Vj(t,x + 1) + Vj(t,x− 1).
28
We will use the control rules A′ = A ∩ B and B′ = A ∪ B until
the first time, if ever, the remaining
inventory of the system (t,x) controlled by A′ is equal to the
remaining inventory of the system (t,x + 1)
controlled by A. This will happen the first time, if ever, there is

a sale under A and not under A′, i.e., a sale
under A but not under B. Let t′ be the first time this happens, if
it happens before the end of the horizon,
and set t′ = 0 otherwise. If t′ > 0 then we apply policy A′ = A
and B′ = B over s ∈ [0, t′). We claim that the
expected revenue from the two systems starting with (t,x) is the
same as the expected revenue from the other
two systems. This is because the sales and revenues up to, but
before t′, are the same in the two systems.
At t′ sales occur only for the system (t,x) controlled by B′ and
the system (t,x + 1) controlled by A and the
revenues from the two sales are identical. After the sales at t′,
the inventory of the system (t,x) controlled by
A′ becomes identical to the inventory of the system (t,x+ 1)
controlled by A while the inventory of the system
(t,x) controlled by B′ becomes identical to the inventory of the
system (t,x− 1) controlled by B. Since the
policy switches to A′ = A and B′ = B then sales and revenues
are the same over [0, t′). If t′ = 0 then the
sales of the two systems are the same during the entire horizon.
It remains to verify that inventories don’t become negative.
Prior to time t′ the systems remain balance
in the sense that system (t,x) governed by A′ always has one
unit of inventory less than system (t,x + 1)

governed by A and system (t,x) governed by B′ has one more
unit of inventory than system (t,x−1) governed
by B. Thus the only two systems that could potential run out of
inventory before t′ are A′ and B.
Since sales under A′ = A∩B are more restricted than sales under
B, the inventory of system (t,x) governed
by A′ will always be at least one unit since at most x − 1 units
of sale are allowed under B. Therefore the
only way the system can run out of inventory is if system (t,x−
1) runs out of inventory under B before t′.
However, in this case sales would stop under systems A′ and B,
while sales will continue under B′ = A and A
so revenues will continue to be the same until the first sale
under A at which point we reached t′. This shows
that even if the system (t,x − 1) runs out of inventory under B
the two systems continue to have the same
revenues over the entire horizon. Consequently 2∆Vj(t,x) ≥
Vj(t,x + 1) + Vj(t,x− 1) for all x ≥ 1.
To show that ∆Vj(t,x) is increasing in j it is enough to show that
Vj(t,x) + Vj−1(t,x− 1) ≥ Vj(t,x− 1) + Vj−1(t,x).
To do this we again use a sample path argument. Let A be an

optimal admission control rule for the system
(j,t,x−1) and B be an admission control rule for the system (j−1,
t,x) Let A′ and B′ be heuristic admission
rules applied, respectively, to the systems (j,t,x) and (j−1,
t,x−1). Our goal is to exhibit heuristics A′ and B′
such that when applied to the systems (j,t,x) and (j−1, t,x−1)
they generate as much revenue as the applying
A to (j,t,x−1) and B to (j−1, t,x). This will imply that Vj(t,x) +
Vj−1(t,x−1) ≥ Vj(t,x−1) + Vj−1(t,x).
Let A′ = A ∪ B and B′ = A ∩ B and let t′ be the first time there
is a sale under A ∪ B without a
corresponding sale in A, so there is a sale under B but not under
A. If t′ = 0 then the revenues of the sets
of two systems are equal. If t′ > 0 switch at that point to the
policy A′ = A and B′ = B. Then sales and
revenues under both sets of two systems are equal up to t′. At t′
there are sales for the system (j,t,x) and
(j−1, t,x−1) that generate the same revenues. Moreover, the
inventories of the two sets of two systems have
the same inventories immediately after the sale at t′. Since the
policy then switches to A′ = A and B′ = B
then sales and revenues are the same for the two set of systems
over s ∈ [0, t′). The only system in danger to
run out of inventory is system (j,t,x) under A′ = A∪ B, but that

system has the same number of sales as the
system (j,t,x− 1) under A up to t′. Therefore the system (j,t,x)
has at least one unit of inventory up to t′.
To show that ∆Vj(t,x) is increasing in t it is enough to show that
Vj(t,x) + Vj(t− 1,x− 1) ≥ Vj(t,x− 1) + Vj(t− 1,x).
To do this we again use a sample path argument. Let A be an
optimal admission control rule for the system
(t,x − 1) and B be an optimal admission control rule for the
system (t − 1,x) Let A′ and B′ be heuristic
admission rules applied, respectively, to the systems (t,x) and
(t− 1,x− 1). Our goal is to exhibit heuristics
A′ and B′ such that when applied to the systems (t,x) and
(t−1,x−1) they generate as much revenue as the
applying A to (t,x−1) and B to (t−1,x). This will imply that
Vj(t,x)+Vj(t−1,x−1) ≥ Vj(t,x−1)+Vj(t−1,x).
Let A′ = A∪ B and B′ = A∩B and let t′ be the first time there is
a sale under A′ without a corresponding
29

sale in A, so there is a sale under B but not under A. If t′ = 0
then the revenues of the sets of two systems are
equal. If t′ > 0 switch at that point to the policy A′ = A and B′ =
B. Then sales and revenues under both
sets of two systems are equal up to t′. At t′ there are sales for
the system (t,x) and (t−1,x) that generate the
same revenues. Moreover, the inventories of the two sets of two
systems have the same inventories immediately
after the sale at t′. Since the policy then switches to A′ = A and
B′ = B then sales and revenues are the
same for the two set of systems over s ∈ [0, t′). The only
system in danger to run out of inventory is system
(t− 1,x− 1) under B′ = A∪ B, but that system has the same
number of sales as the system (t− 1,x) under
B up to t′. Therefore the system (t− 1,x− 1) has at least one unit
of inventory up to t′.
Proof of Lemma 4: We will first show that aj(t,x) can also be
characterized as aj(t,x) = max{k ≤ j :
pk ≥ ∆Vk(t − 1,x)}. The result will then follow from Lemma 3.
First notice that if aj(t,x) = k < j
then Vi(t,x) = Vk(t,x) for all i ∈ {k,. . . ,j}. Moreover, aj(t,x) =
k < j implies that Wk(t,x) > Wk+1(t,x).
Consequently 0 > Wk+1(t,x)−Wk(t,x) =
(pk+1−∆Vk+1(t−1,x))λk+1, so pk+1 < ∆Vk+1(t−1,x).

Conversely,
if pk ≥ ∆Vk(t−1,x) then Wk(t,x)−Wk−1(t,x) ≥ (pk
−∆Vk(t−1,x))λk ≥ 0 so Wk(t,x) ≥ Wk−1(t,x). With
the new characterization we now turn to the monotonicity of
aj(t,x) = max{k ≤ j : pk ≥ ∆Vk(t − 1,x)}.
The monotonicity with respect to j is obvious because it
expands the set over which we are maximizing.
To see the monotonicity with respect to t, notice that ∆Vk(t,x) ≥
∆Vk(t − 1,x) so k is excluded from
the set whenever ∆Vk(t − 1,x) ≤ pk < ∆Vk(t,x). To see the
monotonicity with respect to x, notice that
∆Vk(t − 1,x + 1) ≤ ∆Vk(t,x) ≤ pk implies that k contributes
positively at state (t − 1,x + 1) whenever it
contributes at (t− 1,x).
30
12 Terminology
Item Description
c capacity

p price
s salvage value on excess capacity after the arrival of full-fare
demand (e.g. last-
minute travel specials).
x remaining capacity, i.e. x ∈ {0, 1, ...,c}
E expected value operator
j fare class identifier where price decreases with increasing
class index, i.e. p1 ≥
p2 ≥ .. ≥ pj
pj demand weighted average fare class price
D fare class demand
D[1,j] =
∑j
k=1 Dk aggregate demand for fare classes 1, . . . ,j
y protection level for a fare class
ya protection level obtained using EMSRa heuristic

yb protection level obtained using EMSRb heuristic
W(y,c) revenue given protection level y and capacity c
r ratio of discount to full-fare price
F cumulative density function (CDF) of continuous demand
F−1 inverse density function of continuous demand
b booking limit for sales of discount fares
Vj(x) optimal expected revenue for classes {1, 2, ...,j} from
capacity x
ρ penalty cost for each unit of full-fare demand that is rejected
sj observed sales for fare j in Monte Carlo simulation
Sk Set of highest k fares {1, . . . ,k}
a(t,x) highest opened fare class at state (t,x)
Zj random demand size per request of fare j.
Tj the stopping set for fare j

λjt arrival rate of fare j demand at time-to-go t
Λjt expected demand for fare j over [0, t].
Table 15: Summary of Terminology and Notation Used
References
[1] Bathia, A. V. and S. C. Prakesh (1973) “Optimal Allocation
of Seats by Fare,” Presentation by TWA
Airlines to AGIFORS Reservation Study Group.
[2] Belobaba, P. P. (1987) “Airline Yield Management An
Overview of Seat Inventory Control,” Transporta-
tion Science, 21, 63–73.
[3] Belobaba, P. P. (1989) “Application of a Probabilistic
Decision Model to Airline Seat Inventory Control,”
Operations Research, 37, 183–197.
[4] Brumelle, S. L., J. I. McGill, T. H. Oum, M. W. Tretheway
and K. Sawaki (1990) “Allocation of Airline
Seats Between Stochastically Dependent Demands,”
Transporation Science, 24, 183-192.

31
[5] Brumelle, S. L. and J. I. McGill (1993) “Airline Seat
Allocation with Multiple Nested Fare Classes,”
Operations Research, 41, 127-137.
[6] Curry, R. E. (1990) “Optimal Airline Seat Allocation with
Fare Nested by Origins and Destinations,”
Transportation Science, 24, 193–204.
[7] Diwan, S. (2010) “Performance of Dynamic Programming
Methods in Airline Revenue Management.”
M.S. Dissertation MIT. Cambridge, MA.
[8] Chandler, S. and Ja, S. (2007) Revenue Opportunity
Modeling at American Airlines. AGIFORS Reser-
vations and Yield Management Study Group Annual Meeting
Proceedings; Jeju, Korea.
[9] Feng, Y. and G. Gallego (1995) “Optimal Starting Times of
End-of-Season Sales and Optimal Stopping
Times for Promotional Fares,” 41, 1371-1391.

[10] Feng, Y. and G. Gallego (2000) “Perishable asset revenue
management with Markovian time-dependent
demand intensities,” Management Science, 46(7): 941-956.
[11] Feng, Y. and B. Xiao (2000) “Optimal policies of yield
management with multiple predetermined prices,”
Operations Research, 48(2): 332-343.
[12] Gallego, G., S. Kou and Phillips, R. (2008) “Revenue
Management of Callable Products,” Management
Science, 54(3): 550564.
[13] Lee. C. T. and M. Hersh. (1993) “A Model for Dynamic
Airline Seat Inventory Control with Multiple
Seat Bookings,” Transportation Science, 27, 252–265.
[14] Littlewood, K. (1972) “Forecasting and Control of
Passenger Bookings.” 12th AGIFORS Proceedings:
Nathanya, Israel, reprinted in Journal of Revenue and Pricing
Management, 4(2): 111-123.
[15] Pfeifer, P. E. (1989) “The Airline Discount Fare Allocation
Problem,” Decision Science, 20, 149–157.
[16] Ratliff, R. (2005) “Revenue Management Demand

Distributions”, Presentation at AGIFORS Reservations
and Yield Management Study Group Meeting, Cape Town,
South Africa, May 2005.
[17] Richter, H. (1982) “The Differential Revenue Method to
Determine Optimal Seat Allotments by Fare
Type,” In Proceedings 22nd AGIFORS Symposium 339-362.
[18] Robinson, L.W. (1995) “Optimal and Approximate Control
Policies for Airline Booking With Sequential
Nonmonotonic Fare Classes,” Operations Research, 43, 252-
263.
[19] Simpson, R. (1985) “Theoretical Concepts for
Capacity/Yield Management,” In Proceedings 25th AGI-
FORS Symposium 281-293.
[20] Talluri, K and G. van Ryzin (2004) “The Theory and
Practice of Revenue Management,” Springer Sci-
ence+Business Media, New York, USA, pg. 42.
[21] Tematha, C., S. Polt, and and L. Suhi. (2010) “On the
robustness of the network-based revenue oppor-
tunity model,” Journal of Revenue and Pricing Management ,
Vol. 9, 4, 341355.

[22] Wollmer, R. D. (1992) “An Airline Seat Management
Model for a Single Leg Route When Lower Fare
Classes Book First,” Operations Research, 40, 26-37.
[23] Zhao, W. and Y-S Zheng (2001) “A Dynamic Model for
Airline Seat Allocation with Passenger Diversion
and No-Shows,” Transportation Science,35: 80-98.
32
4601 session 5.notebook
1
February 12, 2013
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1
February 12, 2013
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RM session 3.notebook
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Probability
and Dynamic Programming
Review
for
Dynamic Pricing and Revenue
Management

Probability Review
• Poisson:
http://en.wikipedia.org/wiki/Poisson_random_variable
– Read Sections: 1, 2, 3,4
• Compound Poisson:
http://en.wikipedia.org/wiki/Compound_Poisson_distribution
• Poisson Process: http://en.wikipedia.org/wiki/Poisson_process
• Compound Poisson Process:
http://en.wikipedia.org/wiki/Compound_Poisson_process
• Normal: http://en.wikipedia.org/wiki/Normal_random_variable
– Read Sections 1,2.1,2.2,3,1,3.2,3.3
• Brownian Motion:
http://en.wikipedia.org/wiki/Wiener_process
– Read sections 1, 2.

DP AS AN OPTIMIZATION METHODOLOGY
• Basic optimization problem
min
u∈ U
g(u)
where u is the optimization/decision variable, g(u)
is the cost function, and U is the constraint set
• Categories of problems:
− Discrete (U is finite) or continuous
− Linear (g is linear and U is polyhedral) or
nonlinear
− Stochastic or deterministic: In stochastic prob-
lems the cost involves a stochastic parameter
w, which is averaged, i.e., it has the form
g(u) = Ew
{
G(u, w)

}
where w is a random parameter.
• DP can deal with complex stochastic problems
where information about w becomes available in
stages, and the decisions are also made in stages
and make use of this information.
{ } ∑
{ } ∑ ( )
INVENTORY CONTROL EXAMPLE
Inventory
S y s t e m
Stock Ordered at
Period k
Stock at Period k Stock at Period k + 1

Demand at Period k
xk
wk
xk + 1 = xk + uk - wk
u k
Co s t o f P e rio d k
c uk + r (xk + uk - wk)
• Discrete-time system
xk+1 = fk(xk, uk, wk) = xk + uk − wk
• Cost function that is additive over time
N−1
E gN (xN ) + gk(xk, uk, wk)
k=0

N−1
= E cuk + r(xk + uk − wk)
k=0
• Optimization over policies: Rules/functions uk =
µk(xk) that map states to controls
BASIC STRUCTURE OF STOCHASTIC DP
• Discrete-time system
xk+1 = fk(xk, uk, wk), k = 0, 1, . . . , N − 1
− k: Discrete time
− xk: State; summarizes past information that
is relevant for future optimization
− uk: Control; decision to be selected at time
k from a given set
− wk: Random parameter (also called distur-

bance or noise depending on the context)
− N : Horizon or number of times control is
applied
• Cost function that is additive over time
E
{
gN (xN ) +
N−1∑
k=0
gk(xk, uk, wk)
}
Guillermo
Pencil
BASIC PROBLEM

• System xk+1 = fk(xk, uk, wk), k = 0, . . . , N −1
• Control constraints uk ∈ U (xk)
• Probability distribution Pk(· | xk, uk) of wk
• Policies π = {µ0, . . . , µN−1}, where µk maps
states xk into controls uk = µk(xk) and is such
that µk(xk) ∈ Uk(xk) for all xk
• Expected cost of π starting at x0 is
Jπ(x0) = E
{
gN (xN ) +
N−1∑
k=0
gk(xk, µk(xk), wk)
}
• Optimal cost function
J∗ (x0) = min
π

Jπ(x0)
• Optimal policy π∗ is one that satisfies
Jπ∗ (x0) = J∗ (x0)
PRINCIPLE OF OPTIMALITY
• Let π∗ = {µ∗ 0, µ∗ 1, . . . , µ∗ N−1} be an optimal pol-
icy
• Consider the “tail subproblem” whereby we are
at xi at time i and wish to minimize the “cost-to-
go” from time i to time N
E
{
gN (xN ) +
N−1∑
k=i

gk
(
xk, µk(xk), wk
)}
and the “tail policy” {µ∗ i , µ∗ i+1, . . . , µ∗ N−1}
0 Ni
xi Tail Subproblem
• Principle of optimality: The tail policy is opti-
mal for the tail subproblem
• DP first solves ALL tail subroblems of final
stage
• At the generic step, it solves ALL tail subprob-
lems of a given time length, using the solution of
the tail subproblems of shorter time length

DP ALGORITHM
• Start with
JN (xN ) = gN (xN ),
and go backwards using
Jk(xk) = min
uk∈ Uk(xk)
E
wk
{
gk(xk, uk, wk)
+ Jk+1
(
fk(xk, uk, wk)
)}
, k = 0, 1, . . . , N − 1.
• Then J0(x0), generated at the last step, is equal

to the optimal cost J∗ (x0). Also, the policy
π∗ = {µ∗ 0, . . . , µ∗ N−1}
where µ∗ k(xk) minimizes in the right side above for
each xk and k, is optimal.
• Justification: Proof by induction that Jk(xk)
is equal to J∗ k (xk), defined as the optimal cost of
the tail subproblem that starts at time k at state
xk.
• Note that ALL the tail subproblems are solved
in addition to the original problem, and the inten-
sive computational requirements.
DETERMINISTIC FINITE-STATE PROBLEM
Terminal Arcs
. . .
. . .

. . .
Initial State
s
t
Artificial Terminal
N o d e
with Cost Equal
to Terminal Cost
S t a g e 0 S t a g e 1 S t a g e 2 . . . S t a g e N - 1 S t a g
e N
• States <==> Nodes
• Controls <==> Arcs
• Control sequences (open-loop) <==> paths
from initial state to terminal states
• ak : Cost of transition from state i ∈ Sk to state ij
j ∈ Sk+1 at time k (view it as “length” of the arc)
• aN : Terminal cost of state i ∈ SNit
• Cost of control sequence <==> Cost of the cor-

responding path (view it as “length” of the path)
BACKWARD AND FORWARD DP ALGORITHMS
• DP algorithm:
JN (i) = aNit , i ∈ SN ,
Jk(i) = min
j∈ Sk+1
[
akij +Jk+1(j)
]
, i ∈ Sk, k = 0, . . . , N−1.
The optimal cost is J0(s) and is equal to the
length of the shortest path from s to t.
• Observation: An optimal path s → t is also an
optimal path t → s in a “reverse” shortest path

problem where the direction of each arc is reversed
and its length is left unchanged.
• Forward DP algorithm (= backward DP algo-
rithm for the reverse problem):
J̃N (j) = a0sj , j ∈ S1,
J̃k(j) = min
i∈ SN−k
[
aN−kij + J̃k+1(i)
]
, j ∈ SN−k+1
The optimal cost is J̃0(t) = mini∈ SN
[
aNit + J̃1(i)
]
.
• View J̃k(j) as optimal cost-to-arrive to state j

from initial state s.
A NOTE ON FORWARD DP ALGORITHMS
• There is no forward DP algorithm for stochastic
problems.
• Mathematically, for stochastic problems, we
cannot restrict ourselves to open-loop sequences,
so the shortest path viewpoint fails.
• Conceptually, in the presence of uncertainty,
the concept of “optimal-cost-to-arrive” at a state
xk does not make sense. The reason is that it may
be impossible to guarantee (with prob. 1) that any
given state can be reached.
• By contrast, even in stochastic problems, the
concept of “optimal cost-to-go” from any state xk
makes clear sense.

GENERIC SHORTEST PATH PROBLEMS
• {1, 2, . . . , N, t}: nodes of a graph (t: the desti-
nation)
• aij : cost of moving from node i to node j
• Find a shortest (minimum cost) path from each
node i to node t
• Assumption: All cycles have nonnegative length.
Then an optimal path need not take more than N
moves
• We formulate the problem as one where we re-
quire exactly N moves but allow degenerate moves
from a node i to itself with cost aii = 0.
Jk(i) = optimal cost of getting from i to t in N−k moves.
J0(i): Cost of the optimal path from i to t.
• DP algorithm:
Jk(i) = min
j=1,...,N

[
aij +Jk+1(j)
]
, k = 0, 1, . . . , N−2,
with JN−1(i) = ait, i = 1, 2, . . . , N.
EXAMPLE
2
7 5
2
5 5
6 1
3
0 . 5
3

1
2
4
0 1 2 3 4
1
2
3
4
5
State i
S t a g e k
3 3 3 3

4 4 4 5
4 . 5 4 . 5 5 . 5 7
2 2 2 2
Destination
5
(a) (b)
JN−1(i) = ait, i = 1, 2, . . . , N,
Jk(i) = min
j=1,...,N
[
aij +Jk+1(j)
]
, k = 0, 1, . . . , N−2.
Probability and Dynamic ProgrammingReviewProbability
Review

RM session 2.notebook
1
February 12, 2013
Page 1
Single Resource Revenue Management with Independent
Demands
c©
Guillermo Gallego
Updated Spring 2013
Abstract
Providers of fixed perishable capacity, such as airline seats and
hotel rooms use price discrimination
to improve revenues; in practice, this discrimination is typically
achieved by imposing booking and usage

restrictions or including ancillary services such as mileage
accrual and luggage handling, to sell the same
capacity to different customers at different prices. We will
assume that the set of fare classes (a menu
of prices, restrictions and ancillary services) is given, and that
the capacity provider’s goal is to allocate
capacity among the different fare classes to maximize expected
revenues. The problem of designing and
pricing fare classes is treated in a separate chapter. We analyze
the two fare class problem under the
assumption that the lower fare class books first. We use
marginal analysis to informally derive Littlewood’s
rule and then show that Littlewood’s rule is in fact optimal.
Spill rates, spill penalties and callable
products are discussed next. A dynamic programming
formulation for the multiple fare class problem is
then introduced under the assumption that lower fare classes
book first. Commonly used heuristics as well
as bounds on the value function are presented. Dynamic models
that explicitly take time into account,
allow for more general fare arrival patterns and for randomness
in the size of the requests. We compare
the performance of static and dynamic policies and find that
dynamic policies have a real advantage when
the fare arrivals patterns are not low-to-high. We finalize the

chapter with a model where fare classes are
not allowed to reopen after they are closed for the first time.
1 Introduction
This chapter considers the simplest and best known revenue
management problem, the single resource, inde-
pendent demand problem. We assume that the capacity provider
is trying to maximize the expected revenues
from a sunk investment in c units of capacity. We assume that
capacity is sold through a reservation system
and that capacity cannot be modified or replenished during the
booking horizon. We also assume that unsold
capacity has no salvage value. Later we will see that the zero
salvage value assumption is made without loss
of generality as any problem with positive salvage value can be
transformed into a problem with zero salvage
value. We assume that the set of fare classes (a menu of prices
and restrictions) is given, and that the demands
for the different fare classes are statistically independent. In
particular, we assume that if a customer finds his
preferred fare class closed, he will leave the system without
purchasing. This assumption holds approximately
if the difference in fares is large so that demands are decoupled
or if customers can find alternative sources

of capacity for their preferred fare class. In some cases,
however, part of the demand may be recaptured by
other available fare classes. In such cases, the independent
demand assumption is too strong and needs to be
relaxed. We address this issue in a separate chapter where we
discussed demand models based on discrete
choice theory.
In this chapter, we present a variety of models that have been
developed in industry and in academia.
There has been a preference in industry for models that suppress
the time dimension and assume that the
arrival pattern of the fare classes is low-to-high. We call these
class of models static to distinguish them
from the dynamic models, favored by academics, that model
time explicitly. Both models have advantages
and disadvantages as we will soon see. Static models are
relatively easy to understand. Also, good heuristics
1
were developed before optimal solutions based on dynamic
programming were discovered. Bringing in the

time dimension helps deal with more general fare arrival
patterns, but specifying the model requires a more
detailed estimation of demand. This Chapter starts with a
review of the two fare class problem in §2 where we
present a heuristic derivation of Littlewood’s rule via marginal
analysis. Littlewood’s rule is formally derived
in §2.3 where a formal DP for the two fare class problem is
presented. The dynamic program for multiple fare
classes is presented in §3. Commonly used heuristics are
presented in §4 and bounds on the optimal expected
revenue are presented in §5. The dynamic model is presented in
§6, for the Poisson case and for the compound
Poisson case in §7, where each request is for a random demand
size. In §8, we restrict fares so that they cannot
be opened once they are closed.
2 Two Fare Classes: Marginal Analysis
The product can be sold either at the full-fare p1 or at a
discounted-fare p2 < p1. The discounted-fare typically
has advance purchasing and usage restrictions. Let D1 and D2
denote respectively the random demand for
the two fare classes for a specific instance of the problem, e.g.,
for a specific flight for an airline or a specific
night for a hotel.

We assume that all booked customers will actually travel. This
avoids the need to overbook capacity and
allow us to focus on the problem of allocating capacity between
the two fares. We will discuss how to deal
with pre-travel cancellations and day-of-travel no shows in a
separate chapter on overbooking models.
Fare class arrival order is an important part of the model. We
assume what is commonly known as the
low-to-high fare class arrival order, which implies that demand
for the discounted fare book earlier than for
full-fare. This arrival pattern holds approximately in practice,
and it is encouraged by advance purchase
restrictions imposed on lower fare classes. Notice that this is a
worst case arrival pattern. Indeed, if full-fare
class customers arrived first then we would accept them up to
capacity and use residual capacity, if any, to
satisfy demand from the discounted-fare class. We will relax the
low-to-high fare order arrival assumption
after we solve the multi-fare problem via dynamic
programming.
Under the low-to-high arrival pattern, discount-fare customers
may exhaust capacity, say c, unless part

of it is protected for later-booking by full-fare customers.
Consequently, booking limits (known as discount
authorizations) are placed on the discount sales. Suppose we
protect y ∈ {0, 1, . . . ,c} units of capacity for
the full-fare demand, D1, before observing the actual demand
for the discount-fare, D2. This results in a
booking limit c−y on the discounted-fare, so sales at the
discounted-fare class are given by min(c−y,D2).
The remaining capacity is equal to c − min(c − y,D2) = max(y,c
− D2) and it is all made available to the
full-fare class. Consequently, sales at the full fare equal
min(max(y,c−D2),D1). The total expected revenue
is
and the goal is to find a protection level y that maximizes
W(y,c). The extreme strategies y = 0 and y = c
correspond, respectively, to the case where no capacity is
protected and all of the capacity is protected. We
will later come back and discuss when these extreme strategies
are optimal. In most cases, however, an
intermediate strategy is optimal.
The fare ratio r = p2/p1 plays an important role in determining

optimal protection levels. If the ratio is
very small then we would be inclined to protect more capacity
for the full-fare demand. If the ratio is close
to one, we would be inclined to accept nearly all discount-fare
requests since we can get almost the same
revenue, per unit of capacity, without risk. The distribution of
full-fare demand is also important in deciding
how many units to protect for that fare. If, P(D1 ≥ c) is very
large, then it makes sense to protect the entire
capacity for full-fare sales as it is likely that the provider can
sell all of the capacity at the full-fare. However,
if P(D1 ≥ c) is very low then it is unlikely that all the capacity
can be sold at the full-fare, so fewer units
should be protected. It turns out that the demand distribution of
the discount-fare D2 has no influence on the
optimal protection level under our assumption that D2 and D1
are independent. A formula for the optimal
protection level, involving only P(D1 ≥ y) and r, was first
proposed by Littlewood [14] in 1972. His arguments
were not formal; however, they were later justified by Bhatia
and Prakesh [1] in 1973, and Richter [17] in 1982.
2

One can obtain Littlewood’s formula intuitively by using
marginal analysis. The advantage of marginal
analysis is that it allows us to quickly derive the solution for the
two fare class problem. The marginal analysis
argument goes as follows: Suppose we have y > 0 units of
capacity, and that we receive a request for the
discounted-fare. Consider the marginal revenue associated with
accepting and rejecting this request. If we
accept, we obtain p2. If we close down the discount-fare then
we will be able to sell the y
th unit at p1 only
if the full-fare demand D1 is at least as large as y, so it is
intuitively optimal to reject the discount fare if
p1P(D1 ≥ y) > p2. This suggests that an optimal protection level
y1 should be given by:
y1 = max{y ∈ N : P(D1 ≥ y) > r}, (1)
where N = {0, 1, . . . ,} is the set of non-negative integers.
Equation (1) is known as Littlewood’s rule.
Example 1. Suppose D1 is Poisson with parameter 80, the full
fare is p1 = $100 and the discounted fare is

p2 = $60, so r = 60/100 = 0.6. We are interested in the
cumulative tail distribution P(D1 ≥ y) = 1−P(D1 ≤
y − 1). Since most statistical software packages return the value
of P(D1 ≤ y), we see that y1 satisfies
P(D1 ≤ y1 − 1) < 1 − r ≤ P(D1 ≤ y1). Since P(D1 ≤ 77) =< 0.4 ≤
P(D1 ≤ 78) we conclude that y1 = 78.
Consequently, if c = 200 then the booking limit for the discount
fare is 122. However, if c < y1, then all units
should be protected for the full-fare resulting in a booking limit
of zero.
Remarks:
• y(c) = min(y1,c) is also an optimal protection level. If y(c) =
c, or equivalently if y1 ≥ c, then all the
capacity should be reserved for sale at the full-fare.
• The quantity b2 = max(c−y1, 0) is known as the optimal
booking limit for the discount fare. It is the
maximum number of discount-fare customers that we will book.
• y1 is independent of the distribution of D2.
• If P(D1 ≥ y1 + 1) = r, then y1 + 1 is also optimal protection
level, so both y1 and y1 + 1 result in the

same expected revenue. Protecting the y1 + 1 unit of capacity
increases the variance of the revenue, but
it reduces the probability of rejecting requests from full-fare
customers.
From Littlewood’s rule (1), we see that the extreme strategy y =
0 is optimal when P(D1 ≥ 1) ≤ r and
the extreme strategy y = c is optimal when P(D1 ≥ c) > r.
2.1 Continuous Demand Model
Although revenue management demands are actually discrete,
continuous distributions can be easier to work
with and are often employed in practice. If we model D1 as a
continuous random variable with cumulative
distribution function F1(y) = P(D1 ≤ y), then
y1 = F
−1
1 (1 −r)
where F−11 denotes the inverse of F. In particular, if D1 is
Normal with mean µ1 and standard deviation σ1
then

y1 = µ1 + σ1Φ
−1(1 −r) (2)
where Φ denotes the cumulative distribution function of the
standard Normal random variable.
This formula allows for comparative statics as given in Table 1:
Example 2. Suppose that D1 is Normal with mean 80 and
standard deviation 9, the full-fare is p1 = $100
and the discount-fare is p2 = $60. Then y1 = F
−1
1 (1 − 0.6) = 77.72 < 80 since r > 1/2. Notice that the
solution is quite close to that of Example 1. This is because a
Poisson random variable with mean 80 can be
well approximated by a normal with mean 80 and standard
deviation
√
80 ' 9.
3

Fare Ratio Dependence of protection level
r > 1
2
y1 < µ1 and y1 decreases with σ1
r = 1
2
y1 = µ1 independent of σ1
r < 1
2
y1 > µ1 and y1 increases with σ1
Table 1: Comparative Statics for Normal Full Fare Demand
2.2 Connection with the Newsvendor Problem
There is a close connection between the classical Newsvendor
Problem and the two-fare Revenue Management

Problem that we will briefly explore here. In the classical
Newsvendor Problem a manager must decide how
many units, say y, to stock for random sales D1 at p1 assuming
a unit cost p2 < p1. The solution is to stock y1
units where y1 is the largest integer such that P(D1 ≥ y) > r =
p2/p1. We can think of the two-fare Revenue
Management Problem as a situation where capacity c is pre-
decided, at a possible sub-optimal level, there is
random demand D1 at ”salvage value” p1 > p2 that arrives after
demand D2 at p2. The revenue management
problem is to determine how many units to allow to be sold at
p2. We know that the solution is to allow
(c−y1)+ units to book at p2, reserving max(y1,c−D2) units for
sale at p1.
2.3 Two Fare Classes: Dynamic Programming
In this section we formulate and analyze the two fare class
problem using dynamic programming and present
a formal proof of the optimality of Littlewood’s rule. We will
from now on refer to the full-fare class as fare
class 1 and to the discounted fare class as fare class 2. Dynamic
programming starts by solving the problem
at the last stage, just before demand for fare class 1. Let V1(y)
be the optimal expected revenue that can be

obtained from fare class 1 when capacity is y. Since it is
optimal to allow fare class 1 customers to book all
of the available capacity, sales are equal to min(D1,y) and the
optimal expected revenue is
V1(y) = p1E min(D1,y).
Our next task is to find V2(c), the optimal expected revenue
that can be obtained from c units of ca-
pacity. Suppose that y ∈ {0, 1, . . . ,c} units are protected for
fare class 1 demand. This results in revenues
p2 min(D2,c − y) from sales to fare class 2 and remaining
inventory max(c − D2,y) available for fare class
1. Notice that we can obtain expected revenue EV1(max(c −
D2,y)) from this inventory from fare class 1
customers. Then
= E{p2 min(D2,c−y) + V1(max(c−D2,y))}
is the expected revenue associated with protecting y ∈ {0, 1, . . .
,c} units for the full-fare. V2(c) can be obtained
by maximizing W(y,c) over y. More precisely,
V2(c) = max

y∈ {0,1,...,c}
E{p2 min(D2,c−y) + V1(max(c−D2,y))}. (3)
The key to Dynamic Programming is that it involves a recursive
equation (3) linking the expected revenues
V2(c), at stage 2, to the expected revenue function V1 at stage
1. To solve for V2(c) we first need to solve
for V1(y) for y ∈ {0, 1, . . . ,c}. Before moving on to the multi-
fare formulation we will provide a formal
proof of Littlewood’s rule (1), and discuss the quality of service
implications of using Littlewood’s rule under
competition.
4
2.4 Formal Proof of Littlewood’s Rule
For any function f(y) over the integers, let ∆f(y) = f(y) − f(y −
1). The following result will help us to
determine ∆V (y) and ∆W(y,c) = W(y,c) −W(y − 1,c).
Lemma 1 Let g(y) = EG(min(X,y)) where X is an integer valued

random variable with E[X] < ∞ and G
is an arbitrary function defined over the integers. Then
∆g(y) = ∆G(y)P(X ≥ y).
Let r(y) = ER(max(X,y)) where X is an integer valued random
variable with E[X] < ∞ and R is an arbitrary
function defined over the integers. Then
∆r(y) = ∆R(y)P(X < y).
An application of the Lemma 1 yields the following proposition
that provides the desired formulas for
∆V1(y) and ∆W(y,c).
Proposition 1
∆V1(y) = p1P(D1 ≥ y) y ∈ {1, . . . ,}
∆W(y,c) = [∆V1(y) −p2]P(D2 > c−y) y ∈ {1, . . . ,c}.
The proof of the Lemma 1 and Proposition 1 are relegated to the
Appendix. With the help of Proposition 1
we can now formally establish the main result for the Two-Fare
Problem.

Theorem 1 The function W(y,c) is unimodal in y and is
maximized at y(c) = min(y1,c) where
y1 = max{y ∈ N : ∆V1(y) > p2}.
Moreover, V2(c) = W(y(c),c).
Proof: Consider the expression in brackets for ∆W(y,c) and
notice that the sign of ∆W(y,c) is determined
by ∆V1(y) −p2 as P(D2 > c−y) ≥ 0.Thus W(y,c) ≥ W(y − 1,c) as
long as ∆V1(y) −p2 > 0 and W(y,c) ≤
W(y−1,c) as long as ∆V1(y)−p2 ≤ 0. Since ∆V1(y) = p1P(D1 ≥
y) is decreasing1 in y, ∆V1(y)−p2 changes
signs from + to − since ∆V1(0) − p2 = p1P(D1 ≥ 0) − p2 = p1 −
p2 > 0 and limy→∞[∆V1(y) − p2] = −p2.
This means that W(y,c) is unimodal in y. Then
y1 = max{y ∈ N : ∆V1(y) > p2}.
coincides with Littlewood’s rule (1). When restricted to {0, 1, .
. . ,c}, W (y,c) is maximized at y(c) =
min(c,y1). Consequently, V2(c) = maxy∈ {0,1,...,c}W(y,c) =
W(y(c),c), completing the proof.
2.5 Quality of Service, Spill Penalties, Callable Products and

Salvage Values
Since max(y1,c−D2) units of capacity are available for fare
class 1, at least one fare class 1 customer will be
denied capacity when D1 > max(y1,c−D2). The probability of
this happening is a measure of the quality of
service to fare class 1, known as the full-fare spill rate.
Brumelle et al. [4] have observed that
P(D1 > max(y1,c−D2)) ≤ P(D1 > y1) ≤ r < P(D1 ≥ y1). (4)
They call P(D1 > y1) the maximal spill rate. Notice that if the
inequality y1 ≥ c − D2 holds with high
probability, as it typically does in practice when D2 is large
relative to c, then the spill rate approaches the
1We use the term increasing and decreasing in the weak sense.
5
maximal flight spill rate which is, by design, close to the ratio
r. High spill rates may lead to the loss of
full-fare customers to competition. To see this, imagine two

airlines each offering a discount fare and a full-
fare in the same market where the fare ratio r is high and
demand from fare class 2 is high. Suppose Airline
A practices tactically optimal Revenue Management by applying
Littlewood’s rule with spill rates close to
r. Airline B can protect more seats than recommended by
Littlewood’s rule. By doing this Airline B will
sacrifice revenues in the short run but will attract some of the
full-fare customers spilled by Airline A. Over
time, Airline A may see a decrease in full-fare demand as a
secular change and protect even fewer seats for
full-fare passengers. In the meantime, Airline B will see an
increase in full-fare demand at which time it can
set tactically optimal protection levels and derive higher
revenues in the long-run. In essence, Airline B has
(correctly) traded discount-fare customers for full-fare
customers with Airline A.
One way to cope with high spill rates and its adverse strategic
consequences is to impose a penalty cost
ρ for each unit of full-fare demand in excess of the protection
level. This penalty is suppose to measure the
ill-will incurred when capacity is denied to a full-fare customer.
This results in a modified value function
V1(y) = p1E min(D1,y) − ρE[(D1 − y)+] = (p1 + ρ)E min(D1,y)

− ρED1. From this it is easy to see that
∆V1(y) = (p + ρ)P(D1 ≥ y), resulting in
∆W(y,c) = [(p1 + ρ)P(D1 ≥ y) −p2]P(D2 > c−y)
and
y1 = max
{
y ∈ N : P(D1 ≥ y) >
p2
p1 + ρ
}
. (5)
Notice that this is just Littlewood’s rule applied to fares p1 + ρ
and p2, resulting in fare ratio p2/(p1 + ρ) and,
consequently, lower maximal spill rates. Obviously this
adjustment comes at the expense of having higher
protection levels and therefore lower sales at the discount-fare
and lower overall revenues. Consequently, an
airline that wants to protect its full-fare market by imposing a

penalty on rejected full-fare demand does it at
the expense of making less available capacity for the discount-
fare and less expected revenue. One way to avoid
sacrificing sales at the discount-fare and improve the spill rate
at the same time is to modify the discount-fare
by adding a restriction that allows the provider to recall or buy
back capacity when needed. This leads to
revenue management with callable products; see Gallego, Kou
and Phillips [12]. Callable products can be
sold either by giving customers an upfront discount or by giving
them a compensation if and when capacity
is recalled. If managed correctly, callable products can lead to
better capacity utilization, better service to
full-fare customers and to demand induction from customers
who are attracted to either the upfront discount
or to the compensation if their capacity is recalled.
The value function V1(y) may also be modified to account for
salvage values (also known as the ‘distressed
inventory problem’). Suppose there is a salvage value s < p2 on
excess capacity after the arrival of the full-fare
demand (think of standby tickets or last-minute travel deals).
We can handle this case by modifying V1(y) to
account for the salvaged units. Then V1(y) = p1E min(D1,y) +
sE(y −D1)+ = (p1 −s)E min(D1,y) + sy, so

∆V1(y) = (p1 −s)P(D1 ≥ y) + s, resulting in
∆W(y,c) = [(p1 −s)P(D1 ≥ y) − (p2 −s)]P(D2 > c−y)
and
y1 = max
{
y ∈ N : P(D1 ≥ y) >
p2 −s
p1 −s
}
. (6)
Notice that this is just Littlewood’s rule applied to net fares p1
−s and p2 −s. This suggests that a problem
with salvage values can be converted into a problem without
salvage values by using net fares pi ← pi − s,
i = 1, 2 and then adding cs to the resulting optimal expected
revenue V2(c) in excess of salvage values.
3 Multiple Fare Classes: Exact

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