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Single Resource Revenue Management with Independent Demands

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.

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

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.

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

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 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.

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 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 − r)

where F −1 denotes the inverse of F . In particular, if D

1 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 − 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

Fare Ratio

Dependence of protection level

r > 1

1 < µ1 and y1 decreases with σ1

r = 1

1 = µ1 independent of σ1

r < 1

1 > 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

W (y, c)

p2E min(c − y, D2) + p1E min(max(y, c − D2), D1)

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

E{p2 min(D2, c − y) + V1(max(c − D2, y))}.

(3)

y∈{0,1,...,c}

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.

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.

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) >

(5)

p1 + ρ

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

p2 − s

y1 = max y ∈ N : P (D1 ≥ y) >

(6)

p1 − s

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.

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

Wj(y, x)

y∈{0,1,...,x}

max

{pjE min(Dj, x − y) + EVj−1(max(x − Dj, y))} .

(8)

y∈{0,1,...,x}

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.

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

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.

E[Dj]

$100

$60

$40

101

$35

169

$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 (

E[D

j=1

j ])/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.

V1(c)

V2(c)

V3(c)

V4(c)

V5(c)

560%

1,500.0

3,426.8

100

280%

1,500.0

3,900.0

5,441.3

150

187%

1,500.0

3,900.0

5,900.0

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.

$120.00

$100.00

$80.00

DV_1(x)

DV_2(x)

$60.00

DV_3(x)

DV_4(x)

DV_5(x)

$40.00

$20.00

$-‐

100

120

140

160

Figure 1: ∆Vj(x), x = 1, . . . , 350, j = 1, 2, 3, 4, 5 for 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

∆V

j (x)

if x = 1, . . . , yj

j+1(x) =

E min(∆Vj(x − Dj+1), pj+1)

if x = yj + 1, . . ..

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

k−1

∆Vj+1(yj + k) = pj+1P (Dj+1 ≥ k) +

∆Vj(yj + k − i)P (Dj+1 = i)

k ∈ {1, . . . , c − yj}.

i=0

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 ≤ ¯

c and k(c) = ∞ for k > ¯

c. This may reflect the situation

where there is a fixed cost to leasing a resource with capacity ¯

c. The optimal choice is then to lease the

resource if Vn(¯

c) > K and not lease it otherwise.

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

optimal capacity

450

400

350

300

ity 250

acap 200

C 150

100

cost

Figure 2: Optimal Capacity as a Function of Cost for the Data of Example 3

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])

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.

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

yk,j = max{y ∈ N : P (Dk ≥ y) > rk,j}.

Then the EMSR-a heuristic will protect

j−1

k,j

k=1

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

j−1

= µ[1, j − 1] +

j−1

k Φ−1(1 − rk,j ),

k=1

where for any j, µ[1, j − 1] =

j−1 µ

k=1

k 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 D

k=1

k .

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

j−1

µk

pj−1 =

pk µ[1, j − 1]

k=1

and calls on Littlewood’s rule to obtain protection level

= max{y ∈ N : P (D[1, j − 1] ≥ y) > rb

}

j−1

j−1,j

where rb

= p

j−1,j

j / ¯

pj−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

= µ[1, j − 1] + σ[1, j − 1]Φ−1(1 − rb

j−1

j−1,j

Recall that for the Normal, variances are additive, so the standard deviation σ[1, j − 1] =

j−1 σ2.

k=1

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 = yh ≤ yh ≤ . . . ≤ yh

can be

n−1

computed exactly or via simulation. To compute the expected revenues exactly, let V h(x) be the expected profit

from x units of capacity from fares {j, . . . , 1} under the heuristic policy h. Then V h(x) = W h(min(x, yh

), x)

j−1

where for all y ≤ x we define W h(y, x) = p

(max(y, x − D

j E min(x − y, Dj ) + EV h

j−1

j )). The following result

is a direct consequence of Proposition 2 and its corollary.

Proposition 4

∆V h(x)

if x = 1, . . . , yh

∆V h (x) =

j+1

x−yh−1

= p

j+1P (Dj+1 ≥ x − yh) +

∆V h(x − i)P (D

i=0

j+1 = i)

if x > yh

Thus, if V h(x) has already been computed then V h (x) = V h(x) for x = 1, . . . , yh. For x > yh we can

j+1

use the recursion V h (x) = V h (x − 1) + ∆V h (x) starting with x = yh + 1 in conjunction with the second

j+1

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 (sh, . . . , sh) and revenues Rh =

under heuristic h. Averaging

i=1

ish

over all the values of Rh gives an estimate of V h(c). Simulated sales can be generated sequentially via sh =

min(Dj, (xj −yh

)+) starting with j = n and x

n−1

n = c and using the capacity update formula xj = xj+1 − sh

j+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 a(c) and V b(c) as well as V

5(c) for values of c ∈ {50, 100, 150, 200, 250, 300, 350}.

E[Dj]

$100

$60

$40

102

101

$35

171

166

169

$15

120

Table 4: Optimal and Heuristic Protection Levels for Example 3

V a(c)

V b(c)

5(c)

560%

3,426.8

100

280%

5,431.9

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

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.

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 U (c, D)

max

k xk

(11)

k=1

s.t.

≤

∀

k = 1, . . . , n

≤

k=1

≥

∀

k = 1, . . . , n.

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 U (c, D). For convenience, we

will denote this upper bound as V U (c) = EV U (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 U (c)

k E min(Dk , (c − D[1, k − 1])+)

(12)

k=1

pk (E min(D[1, k], c) − E min(D[1, k − 1], c))

k=1

(pk − pk+1)E min(D[1, k], c)

(13)

k=1

where for convenience we define pn+1 = 0. Moreover, since V U (c, D) is concave in D, it follows from Jensen’s

inequality that V U (c) = EV U (c, D) ≤ V U (c, µ) where V U (c, µ) is the solution to formulation (11) with

µ = E[D] instead of D. More precisely,

V U (c, µ)

max

k xk

(14)

k=1

s.t.

≤

µk

∀

k = 1, . . . , n

≤

k=1

≥

∀

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 U (c) ≤ V U (c, µ) =

k=1

k − 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) =

k=1

k E min(Dk , (c − D[k + 1, n])+).

Taking expectations we obtain

V L(c)

k E min(Dk , (c − D[k + 1, n])+)

(15)

k=1

pk (E min(D[k, n], c) − E min(D[k + 1, n], c))

k=1

(pk − pk−1)E min(D[k, n], c)

(16)

k=1

where p0 = 0. Notice that all of the terms in the sum are negative except for k = 1. Clearly V L(c) ≤ V

n(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 L(c) ≤ V

(17)

n(c) ≤ V U

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) =

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)) −

j=1

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 6 and 7 report V L(c), V

n(c), V U

n(c, µ) for the data of Examples 4 and 5, respectively.

Notice that V U (c) represents a significant improvement over the better known bound V

n(c, µ), particularly

for intermediate values of capacity. The spread V U (c) − V L(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.

V L(c)

(c)

V U (c, µ)

n(c)

V U

$42,728

$49,642

$53,039

$53,315

$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 6: Optimal Revenue and Bounds for Example 4.

V L(c)

(c)

V U n(c, µ)

n(c)

V U

$52,462

$67,505

$72,717

$73,312

$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 7: 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 U (c, D), where D is the uncensored demand. On the other hand, the

revenue based on not applying booking controls is just V L(c, D), so a measure of the revenue opportunity is

V U (c, D) − V L(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 U (c, D) − V L(c, D) is sometimes approximated by V U (c) − V L(c) to get an idea of the revenue opportunity.

Table 6 and 7 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 ˜

Vj(x) is

an approximation to Vj(x). Then a heuristic admission control rule can be obtained as follows:

yj = max{y ∈ N : ∆ ˜

Vj(y) > pj+1} j = 1, . . . , n − 1.

(18)

Suppose we approximate the value function Vj(x) by ˜

Vj(x) = θV L(x) + (1 − θ)V U (x) for some θ ∈ [0, 1]

and V L(x) and V U (x) are the bounds obtained in this section applied to n = j and c = x. Notice that

∆V L(x) = p

(x) =

j−1 (p

1P (D[1, j] ≥ x) +

k=2

k − pk−1)P (D[k, j] ≥ x), while ∆V U

k=1

k − pk+1)P (D[1, k] ≥

x) + pjP (D[1, j] ≥ x).

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, with the advance purchase restriction for fare j serving as a mechanism to end stage j. 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 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}.

Then the number of customers that arrive during the last t units of time and request product j, say Njt, is

Poisson with mean Λjt =

jsds. For simplicity we will write Λj , instead of ΛjT , to denote the expected

number of requests for fare j over the entire horizon [0, T ]. The low-to-high arrival pattern can be embedded

into the time varying model by dividing the selling horizon into n sub-intervals [tj−1, tj], j = 1, . . . , n with

tj = jT /n, and setting λjt = nΛj/T over t ∈ [tj−1, tj] and λjt = 0 otherwise.

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. 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 will show

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). To see this notice that the probability that a customer arrives and requests one

unit of product j over the interval [t − δt, t]is

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

exp(−

λktδt) + o(δt) = 1 −

λktδt + o(δt).

k=j

Multiplying the two terms and collecting terms we obtain λjtδt + o(δt) as claimed.

Recall that some fares have embedded time-of-purchase restrictions. Let Nt ⊂ N = {1, . . . , n} to be the

set of allowable fares at time-to-go t. Usually Nt = N for large t, but low fares are dropped from Nt as the

time-of-purchase restrictions become binding.

We can now write

V (t, x)

λjtδt max(pj + V (t − δt, x − 1), V (t − δt, x)) + (1 −

λjtδt)V (t − δt, x) + o(δt)

j∈Nt

V (t − δt, x) + δt

λjt[pj − ∆V (t − δt, x)]+ + o(δt)

(19)

j∈Nt

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.

Subtracting V (t − δt, x) from both sides of equation (19), dividing by δt and taking the limit as δt ↓ 0, we

obtain the following equation, known as the Hamilton Jacobi Bellman (HJB) equation:

∂V (t, x) =

λjt[pj − ∆V (t, x)]+

(20)

∂t

j∈Nt

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).

While the value function can be computed by solving and pasting the differential equation (20), in practice

it is easier to understand and compute V (t, x) 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 T ← aT is an integer, and setting

λjt ← 1 λ

j,t/a, for t ∈ [0, aT ]. The scale factor a should be selected so that, after scaling,

j∈N

jt << 1,

e.g.,

j∈N

jt ≤ .01 for all t. The resulting dynamic program, after rescaling time, is given by

V (t, x) = V (t − 1, x) +

λjt[pj − ∆V (t − 1, x)]+.

(21)

j∈Nt

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) = {1, . . . , a(t, x)}, where

a(t, x) = max{j ∈ Nt : 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)}. For convenience we define

a(t, 0) = a(0, x) = 0 and A(t, 0) = A(0, x) = ∅. For each time-to-go t let the protection level for fares in

{1, . . . , j} be

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.

That intuition is that A(t, x) is decreasing in t because it is optimal to open fewer fares when we have more

time to sell capacity at higher fares. The intuition that A(t, x) is increasing in x is that it we may need open

more fares when we have more inventory. The intuition for yj(t) to be monotone in j is that we should protect

at least as many units for sales of fares in {1, . . . , j + 1} than for sales of fares in {1, . . . , j}, so yj+1(t) ≥ yj(t).

The intuition for yj(t) to be monotone in t is that with more time to sell, say t > t, we have the potential to

sell more from set {1, . . . , j} so at least as many units should be protected: yj(t ) ≥ yj(t).

6.1

A Pricing Formulation with Broader Interpretation

At any time t, let λt =

j=1

jt be the overall arrival rate at time t.

Define πjt =

k=1

kt/λt and

rjt =

k=1

k λkt/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 the consecutive set Sj = {1, . . . , j}. For convenience,

let π0t = r0t = 0 denote, respectively, the sales rate and the revenue rate associated with S0 = ∅. Now let

qjt = rjt/πjt be the average fare per unit sold when the offer set is Sj. If πjt = 0, we define qjt = 0. This

implies that πjt[qjt − ∆V (t − 1, x)] is zero whenever πjt = 0, e.g., when j = 0.

Let N + = N

t ∪ {0}. With this notation, we can write formulation (21) as

V (t, x)

V (t − 1, x) + λt max [rjt − πjt∆V (t − 1, x)]

j∈N +

V (t − 1, x) + λt max πjt[qjt − ∆V (t − 1, x)].

(22)

j∈N +

The reason to include 0 as a choice is that for j = 0, the term vanishes and this allow us to drop the positive

part that was present in formulation (21). The equivalent formulation for the continuous time model (20) is

∂V (t, x) = λt max πjt[qjt − ∆V (t, x)].

(23)

∂t

j∈N +

These formulation suggest that we are selecting among the actions S0, S1, . . . , Sn to maximize the sales rate

πjt times the average fare qjt net of the marginal value of capacity (∆V (t − 1, x) for model (22) and ∆V (t, x)

for model (23)). In essence, the problem has been reduced to a pricing problem with a finite price menu.

Formulations (22) and (23) can be interpreted broadly as the problem of optimizing the expected revenue

from state (t, x) where there are a finite number of actions. These actions can be, as above, associated with

offering products in the sets S0, S1, . . . , Sn, but other interpretations are possible. For example, the different

actions can be associated with offering a product at different prices qjt, each price associated with a sales rate

πjt for all j ∈ N+. This turns the capacity allocation problem into a pricing problem with a finite price menu.

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 N+ will be the index corresponding to a collection of

sets that is efficient in a sense that will be made precise later.

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. 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) =

λjt

Pj(z)[zpj − ∆zV (t, x)]+,

(24)

∂t

j∈Nt

z=1

while the dynamic program (21) with compound Poisson demands is given by

∞

V (t, x) = V (t − 1, x) +

λjt

Pj(z)[zpj − ∆zV (t − 1, x)]+,

(25)

j∈Nt

z=1

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

as the terms z > x do not contribute to the sum given our convention that ∆

z=1

z V (t, x) = ∞ for

x > z. The optimal policies for the two programs are, respectively, to accept a size z request fare pj, j ∈ Nt,

if zpj ≥ ∆zV (t, x), and to accept a z request fare pj, j ∈ Nt, if zpj ≥ ∆zV (t − 1, x). The two policies should

largely coincide time is scaled correctly so that

j∈N

jt << 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

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 that may depend on the market served.

Example 4. 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 Nt = N for all t ∈ [0, 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 8

provides the values V (T, c) for c ∈ {50, 100, 150, 200, 250, 300, 350}. Table 8 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.

100

150

200

250

300

V (T, c)

$3,837

$6,463

$8,451

$10,241

$11,724

$12,559

∆V (t, x)

70.05

66.48

59.66

60.14

54.62

50.41

Table 8: Value function V (T, c) and marginal revenues ∆V (t, c) for Example 4: Compound Poisson

7.1

Static vs Dynamic Policies

Let Nj be the random number of request arrivals for fare j over the horizon [0, T ], Then Nj is Poisson with

parameter Λj =

jtdt. Suppose each arrival is of random size Zj . Then the aggregate demand, say Dj , for

fare j is equal to

Dj =

Zjk,

(26)

k=1

where Zjk is the size of the kth request. It is well known that E[Dj] = E[Nj]E[Zj] = ΛjE[Zj] and that

Var[Dj] = E[Nj]E[Z2] = Λ

], where E[Z2] is the second moment of Z

j E[Z 2

j . Notice that Jensen’s inequality

implies that the the variance to mean ratio E[Z2]/E[Z

j ] ≥ E[Zj ] ≥ 1.

In practice, demands D1, . . . , Dn are fed, under the low-to-high arrival assumption, into static policies to

compute Vj(c), j = 1, . . . , n and protection levels yj, j = 1, . . . , n − 1 using the dynamic program (8), or to

the EMSR-b heuristic to compute protection levels yb, . . . , yb

. Since the compound Poisson demands are

n−1

difficult to deal with numerically, practitioners often approximate the aggregate demands Dj by a Gamma

distribution with parameters αj and βj, such that αjβj = E[Nj]E[Zj] and αjβ2 = E[N

], yielding

j ]E[Z 2

αj = ΛjE[Zj]2/E[Z2], and β

]/E[Z

j = E[Z 2

j ].

We are interested in comparing the expected revenues obtained from static policies to those of dynamic

policies. More precisely, suppose that that demands are compound poisson and Dj is given by (26) for every

j = 1, . . . , n. Suppose that protection levels y1, y2, . . . , yn−1 are computed using the low-to-high static dynamic program (8) and let yb, yb, . . . , yb

be the protection levels computed using the EMSR-b heuristic. Protection

n−1

levels like these are often used in practice in situations where the arrival rates λjt, t ∈ [0, T ], j = 1, . . . , n, are not necessarily low-to-high. Two possible implementations are common. Under theft nesting a size z request

for fare class j as state (t, x) is accepted if x − z ≥ yj−1. This method is called theft nesting because the

remaining inventory x at time-to-go t is x = c − b[1, n] includes all bookings up to time-to-go t, including

bookings b[1, j − 1]. Standard nesting counts only bookings for lower fare classes and is implemented by

accepting a size z request for fare j at state (t, x) if x − z ≥ (yj−1 − b[1, j − 1])+, where b[1, j − 1] are the

observed bookings of fares [1, j − 1] up to state (t, x). When c > yj−1 > b[1, j − 1], this is equivalent to

accepting a request for z units for fare j if c − b[j, n] − z ≥ yj−1, or equivalently if b[j, n] + z ≤ c − yj−1, so

only bookings of low fares count. In practice, standard nesting works much better than theft nesting when

the arrival pattern is not low-to-high.

Notice that the expected revenue, say V s(T, c), resulting from applying the static protection levels y1, . . . , yn−1

with theft nesting is not, in general, equal to Vn(c), the optimal expected revenue when the arrivals are low-

to-high. Similarly, the expected revenue, say V b(T, c), resulting from applying the EMSR-b protection

levels

yb, . . . , yb

with theft nesting is not, in general, equal to V b(c), the expected revenue when the arrivals are

n−1

low-to-high.

The next proposition shows that V s(T, x) ≤ V (T, x), where 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

x−(yj−1−b[1,j−1])+

V s(t, x)

λjt[

Pj(z)(zpj + V s(t − 1, x − z))

j∈Nt

z=1

Qj(x − (yj−1 − b[1, j − 1])+)V s(t − 1, x)] + (1 −

λjt)V s(t − 1, x)

j∈Nt

x−(yj−1−b[1,j−1])+

≤

λjt[

Pj(z)(zpj + V (t − 1, x − z))

j∈Nt

z=1

Qj(x − (yj−1 − b[1, j − 1])+)V (t − 1, x)] + (1 −

λjt)V (t − 1, x)

j∈Nt

x−(yj−1−b[1,j−1])+

V (t − 1, x) +

λjt

Pj(z)(zpj − ∆zV (t − 1, x))

j∈Nt

z=1

∞

≤

V (t − 1, x) +

λjt

Pj(z)(zpj − ∆zV (t − 1, x))+

j∈Nt

z=1

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).

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).

We have come a long way in this chapter and have surveyed most of the models for the independent

demand case. Practitioners and proponents of static models, have numerically compared the performance of

static vs dynamic policies. Diwan [7], for example, compares the performance of the EMSR-b heuristic against

the performance of the dynamic formulation for Poisson demands (21) even for cases where the aggregate

demands Dj, j = 1, . . . , n are not Poisson. Not surprisingly, this heuristic use of (21) can underperform relative

to the EMSR-b heuristic. However, as seen in Proposition 7, the expected revenue under the optimal dynamic

program (25) is always at least as large as the expected revenue generated by any heuristic, including the

EMSR-b. In addition, the dynamic program does not require assumptions about the arrival being low-to- high

as the EMSR-b does. Even so, the EMSR-b heuristic performs very well when the low-to-high assumptions

hold. However, when the low-to-high assumptions are relaxed, then the performance of the EMSR-b heuristic

suffers relative to that of the dynamic program as illustrated by the following example.

Example 5. Consider again the data of Example 4 with uniform arrival rates. Table 9 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.

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 9: Sub-Optimality of EMSR-b with Standard Nesting

7.2

Bounds on V (T, c)

We will now briefly show that the upper bound V U (c) for V

n(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 U (c) =

EVn(c, D) =

k=1

k − 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.

Monotonic Fare Offerings

The dynamic programs (20) and (21) and their 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) = {1, . . . , 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 third parties that specialize

on inter-temporal fare arbitrage. To avoid this 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) when we can offer any consecutive subset of open fares Sk = {1, . . . , k}, k ≤ j and are

not allowed to reopen fares once they are closed. Let Wk(t, x) be the expected revenue from accepting fares

Sk at state (t, x) and then following an optimal policy. More precisely,

Wk(t, x)

λit[pi + Vk(t − 1, x − 1)] + (1 −

λit)Vk(t − 1, x)

i=1

Vk(t − 1, x) + rkt − πkt∆Vk(t − 1, x)

Vk(t − 1, x) + πkt[pkt − ∆Vk(t − 1, x)],

where ∆Vk(t, x) = Vk(t, x) − Vk(t, x − 1), where πkt =

i=1

it and rkt =

i=1

iλit and pkt = rkt/πkt when

πkt > 0 and pkt = 0 otherwise.

Then Vj(t, x) satisfies the dynamic program

Vj(t, x) = max Wk(t, x) = max{Wj(t, x), Vj−1(t, x)}

(27)

k≤j

with the boundary conditions Vj(t, 0) = Vj(0, x) = 0 for all t ≥ 0 and all x ∈ N for all j = 1, . . . , n. Notice

that the optimization is over consecutive subsets Sk = {1, . . . , k}, k ≤ j. It follows immediately that Vj(t, x)

is monotone increasing in j. An equivalent version of (27) 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 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

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 Sj. Let

Aj(t, x) = {1, . . . , 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

Zj = {(t, x) : Vj(t, x) = Vj−1(t, x)}. We can think of Zj as the stopping set for fare j as it is optimal to close

down fare j upon entering set Zj. For each t let yj(t) = max{x ∈ N : (t, x) ∈ Zj+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.

• Z1 ⊂ Z2 ⊂ . . . ⊂ Zn.

• 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. Zj = {(t, x)

aj(t, x) < j}. From Lemma 4, aj(t, x) < j implies that ai(t, x) < i for all i > j, so Zj ⊂ Zi for all i > j. This 23

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.

Example 6. 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

i=1

i/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 10 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. As such, it is not generally

recommended in practice.

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

V3(T, c)

3,494.5

5,572.9

5,900.0

V2(T, c)

3,494.5

3,900.0

V1(T, c)

1,500.0

Table 10: 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)

rjt − πjt∆Vj(t, x)

if (t, x) /

∈ Zj−1

(28)

∂t

∂Vj−1(t,x)

if (t, x) ∈ Z

∂t

j−1

with the same boundary conditions.

8.1

Mark-up and Mark-down Policies

We now go back to the broader pricing interpretation coupled with the monotonic fare formulation (27). 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 Zj = {(t, x) : x ≤ yj−1(t)}, then it is optimal to stop action j upon first entering set Zj. 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.

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.

Appendix

Proof of Lemma 1. Notice that g(y) = G(y)P (X ≥ y) +

G(j)P (X = j), while g(y − 1) = G(y −

j≤y−1

1)P (X ≥ y) +

G(j)P (X = j). Taking the difference yields ∆g(y) = G(y)P (X ≥ y). Notice that

j≤y−1

r(y) = R(y)P (X < y) +

R(j)P (X = j) while r(y − 1) = R(y − 1)P (X < y) +

R(j)P (X = j).

j≥y

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 above result by induction on j. The result is true 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 {Wj (y, x))} ,

y≤x

where for any y ≤ x,

Wj(y, x) = E [pj min {Dj, x − y}] + E [Vj−1 (max {x − Dj, y})]

A little work reveals 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}

(29)

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 Wj(y, x) = Wj(yj−1, x)

y≤x

On the other hand, if x < yj−1 then

Vj(x) = max Wj(y, x) = Wj(x, x).

y≤x

In summary,

Vj(x)

Wj(min(x, yj−1), x)

j−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

Computing ∆Vj(x) = Vj(x) − Vj(x − 1) for x ∈ N results in:

∆V

j−1(x),

if x ≤ yj−1

j (x)

(30)

E min(pj, ∆Vj−1(x − Dj))

if x > yj−1

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)

x−yj

pj+1

P r(Dj+1 ≥ k) + Vj(yj)P r(Dj+1 > x − yj)

k=1

x−yj

Vj(x − k)P r(Dj+1 = k).

k=0

Consequently, for x > yj

x−yj −1

∆Vj+1(x) = pj+1P r(Dj+1 ≥ x − yj) +

∆Vj(x − k)P r(Dj+1 = k)

(31)

k=0

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) =

∞

min(p

k=0

j+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).

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

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).

Terminology

Item

Description

capacity

price

salvage value on excess capacity after the arrival of full-fare demand (e.g. last-

minute travel specials).

remaining capacity, i.e. x ∈ {0, 1, ..., c}

expected value operator

fare class identifier where price decreases with increasing class index, i.e. p1 ≥

p2 ≥ .. ≥ pj

demand weighted average fare class price

fare class demand

protection level for a fare class

protection level obtained using EM SRa heuristic

protection level obtained using EM SRb heuristic

W (y, c)

revenue given protection level y and capacity c

ratio of discount to full-fare price

cumulative density function (CDF) of continuous demand

F −1

inverse density function of continuous demand

booking limit for sales of discount fares

Vj(x)

optimal expected revenue for classes {1, 2, ..., j} from capacity x

penalty cost for each unit of full-fare demand that is rejected

observed sales for fare j in Monte Carlo simulation

Set of highest k fares {1, . . . , k}

a(t, x)

highest opened fare class at state (t, x)

the stopping set for fare j (where it is optimal to close down fare j upon entering

set Zj)

λjt

arrival rate of fare j demand at time-to-go t

Λjt

expected demand for fare j over [0, t].

Table 11: Summary of Terminology and Notation Used

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