queation 1,4,5,6

I need full detail explaination.

rm_dep_demands_2013v1 removed due to copy rights issues.

IEOR 460

Homework 1: Due Wednesday, February 6

1. A coffee shop gets a daily allocation of 100 bagels. The bagels can be either sold
individually at $0.90 each or can be used later in the day for sandwiches. Each bagel
sold as a sandwich provides a revenue of $1.50 independent of the other ingredients.

a) Suppose that demand for bagel sandwiches is estimated to be Poisson with param-
eter 100. How many bagels would you reserve for sandwiches?

b) Compare the expected revenue of the solution of part a) to the expected revenue
of the heuristic that does not reserve capacity for sandwiches assuming that the
demand for individual bagels is Poisson with parameter 150?

c) Answer part a) if the demand for bagel sandwiches is normal with mean 100 and
standard deviation

2. Problem 2 in the textbook (page 173)

3. Problem 3 in the textbook (page 173)

[Hint: For problem 2 and 3, you might want to consider the NORMINV() function in
Excel.]

4. Suppose capacity is 120 seats and there are four fares. The demand distributions for
the different fares are given in the the following table.

Class Fare Demand Distribution
1 $200 Poisson(25)
2 $175 Poisson(30)
3 $165 Poisson(29)
4 $130 Poisson(30)

Determine the optimal protection levels. [Hints: The sum of independent Poisson ran-
dom variables is Poisson with the obvious choice of parameter to make the means match.
If D is Poisson with parameter λ then P(D = k + 1) = P(D = k)λ/(k + 1) for any non-
negative integer k. You might want to investigate the POISSON() function in Excel.]

5. Consider a parking lot in a community near Manhattan. The parking lot has 100 parking
spaces. The parking lot attracts both commuters and daily parkers. The parking lot
manager knows that he can fill the lot with commuters at a monthly fee of $180 each. The
parking lot manager has conducted a study and has found that the expected monthly
revenue from x parking spaces dedicated to daily parkers is approximated well by the
quadratic function R(x) = 300x − 1.5×2 over the range x ∈ {0, 1, . . . , 100}. Note:
Assume for the purpose of the analysis that parking slots rented to commuters cannot
be used for daily parkers even if some commuter do not always use their slots.

a) What would the expected monthly revenue of the parking lot be if all the capacity
is allocated to commuters?

b) What would the expected monthly revenue of the parking lot be if all the capacity
is allocated to daily parkers?

c) How many units should the parking manager allocate to daily parkers and how
many to commuters?

d) What is the expected revenue under the optimal allocation policy?

6. A fashion retailer has decided to remove a certain item of clothing from the racks in
one week to make room for a new item. There are currently 80 units of the item and
the current sale price is $150 per unit. Consider the following three strategies assuming
that any units remaining at the end of the week can be sold to a jobber at $30 per unit.

a) Keep the current price. Find the expected revenue under this strategy under the
assumption that demand at the current price is Poisson with parameter 50.

b) Lower the price to $90 per unit. Find the expected revenue under this strategy
under the assumption that demand at $90 is Poisson with parameter 120.

c) Keep the price at $150 but e-mail a 40% discount coupon for the item to a popula-
tion of price sensitive customers that would not buy the item at $150. The coupon
is valid only for the first day and does not affect the demand for the item at $150.
Compute the expected revenue under this strategy assuming that the you can con-
trol the number of coupons e-mailed so that demand from the coupon population
is Poisson with parameter x for values of x in the set {0, 5, 10, 15, 20, 25, 30, 35}. In
your calculations assume that demand from coupon holders arrives before demand
from customers willing to pay the full price. Assume also that you cannot deny
capacity to a coupon holder as long as capacity is available (so capacity cannot be
protected for customers willing to pay the full price). What value of x would you
select? You can assume, as in parts a) and b) that any leftover units are sold to
the jobber at $30 per unit.

DynamicPricing and
Revenue

Management

IEOR 4601 Spring 2013

Professor Guillermo Gallego
Class: Monday and Wednesday 11:40-12:55pm
Office Hours: Wednesdays 3:30-4:30pm
Office Location: 820 CEPSR
E-mail: gmg2@columbia.edu

Why Study Dynamic Pricing and Revenue
Management?

¡  Revenue Management had its origins in the
airline industry and is one of the most
successful applications of Operations
Research to decision making

¡  Pricing and capacity allocation decisions
directly impact the bottom line

¡  Pricing transparency and

competition

make
pricing and capacity allocation decisions
more difficult and more important

Applications of Dynamic Pricing and
Revenue Management

¡  Capacity allocation of limited, perishable,
resources to different fare

classes

l  Airlines, hotels, car rentals, cruises, travel

packages, tickets for events
¡  Design and pricing of products

l  Fare

restrictions

and pricing
l  Consumption and fulfillment options
l  Upgrades, downgrades and upsells

¡  Pricing under competition
l  Electronic-commerce

Objectives of this course
¡  Understand the critical tradeoffs and

decisions in Revenue Management
¡  Learn how to

l  Monitor and control product availability for single
and multiple resources

l  Overbook limited resources when customers
shows are random

l  Use upgrades, upsells and real options to improve

revenues

l  Price under competition
l  Improve on the current practice of Revenue

Management

“Physics should be explained as simply as possible, but
no simpler.”

Albert Einstein

Professor Gallego’s experience and
background on subject

¡  Author of seminal papers on dynamic
pricing and revenue management

¡  Winner of several prices from academia
and industry related to work on Revenue
Management

¡  Consultant for airlines and RM solution
providers

¡  Consultant for other users of Revenue
Management

Readings
¡  Class Notes

l  I will provide with notes of the different
topic we cover in class

¡  Textbook
l  R.L. Phillips, Pricing and Revenue

Optimization, Stanford University Press,
2005, ISBN 0-8047-4698-2.

¡  References
l  K.T. Talluri and G.J. Van Ryzin, The Theory

and Practice of Revenue Management,
Springer, 2005, ISBN 0-387-24376-3.

l  Assigned papers

Prerequisites and Grading

¡  Prerequisites
l  Probability and Statistics at the level of IEOR

4150
l  Deterministic Models at the level of IEOR 4003

¡  Corequisites: Stochastic Models IEOR 4106
¡  Grading

l  Assignments 20%
l  Midterm 35%
l  Final 45%

Introduction to Revenue Management

¡  Revenue Management refers to the
strategy and tactics used by perishable
capacity providers to allocate capacity to
different fare classes or market segments
to maximize expected revenues. (See
Chapter 6 in Phillips.)

¡  RM is often practice when
l  Sellers have fixed stock of perishable capacity
l  Customers book capacity prior to usage
l  Seller offers a sets of fare classes
l  Seller can change the availability of fare

classes

History of Revenue Management

¡  Prior to 1978 the Airline Industry was
heavily regulated

¡  In the early 80’s the industry was
deregulated to encourage new entrants

¡  Low-cost carriers such as People Express
started encroaching into key markets of
large carriers

¡  American Airline dilemma:
l  Match fares and lose money
l  Keep fares and lose customers

AA’s Response to People Express

¡  Ultimate Super Saver Discount Fare
l  Same fare as People Express
l  Passenger must buy at least two weeks

prior to departure
l  Stay at his destination over a Saturday

night
¡  AA restricted the number of

discount seats sold on each flight to
save seats for full-fare passengers

Rational and Impact of Strategy

¡  Product Design
l  Imposing restrictions that appealed to the

leisure segment without cannibalizing the
business segment

¡  Capacity Allocation
l  Carefully control capacity to maximize

revenues

¡  Strategy started in January 85
l  PE was struggling by March
l  PE was at the verge of bankruptcy by August

Post-mortem

¡  People Express was bought by
Texas Air for 10% of the market
value it had enjoyed a year before

¡ “We had great people, tremendous
value, terrific growth. We did a lot
of things right. But we didn’t get
our hands around the yield
management and automation
issues.” Donald Burr CEO of PE

RM: The System Context

¡  AA was based on a computerized
reservation system (CRS) called Sabre
developed in 1963. This system:
l  Replaced index cards to manage reservations
l  Sabre is also a GDS (global distribution

system) that allowed AA to distribute its
products and fares globally

¡  Other GDSs: Amadeus, Galileo, Worldspan.

RM Constraints Imposed by Systems

¡  AA’s used Sabre’s Computerized
Reservation System as the backbone to
Revenue Management
l  The reservation system served as a repository

of all the bookings that have been accepted for
future flights

l  The CRS also contains the controls that specify
how many bookings from different fare classes
the airline will accept on future flights

¡  Remember: RM systems were developed
in the context of existing CRSs.

Levels of Revenue Management

¡  Strategic:
l  Market segmentation (leisure vs business)
l  Product design (restrictions, fares, options)
l  Pricing (Static vs. Dynamic)

¡  Tactical:
l  Calculate and updated booking limits

¡  Booking Control:
l  Determine which booking to accept and which

to reject based on booking limits

Strategic Revenue Management

¡  Design low fares to increase sales without
cannibalizing full fare demand
l  Time of Purchase Restrictions

¡  Advance purchase requirements
l  Traveling Restrictions

¡  Saturday night stays
l  High cancellation and change penalties

¡  Other opportunities
l  Contingent options on capacity
l  Flexible and callable products

Tactical Revenue Management

¡  Resources
l  Units of capacity

¡  Seats on a flight
¡  Hotel capacity for a specific night

¡  Products
l  What consumers seek to purchase

¡  May involve one or more resources

¡  Fares
l  A combination of a price and a set of

restrictions

Tactical Definition of RM

¡  A supplier controls a set of resources with
fixed and perishable capacity, a portfolio
of products consisting of combinations of
one or more of the resources, and a set of
fare classes associated with each of the
products. The tactical revenue
management problem is to chose which
fare classes should be open and which
closed for sale at each moment to
maximize total revenue.

Components of Tactical RM

¡  Capacity Allocation
l  How many customers from different

fare classes should be allowed to book?
¡  Network Management

l  How should bookings be managed
across a network of resources?

¡  Overbooking
l  How many total bookings should be

accepted for a product when there are
cancellations and show uncertainty?

Booking Controls

¡  Limits on bookings for different fare
classes:
l  Example: An airline receives a B-class

request for 3 seats departing in two
weeks. The current B-class booking
limit is two seats. As a result, the
request is rejected

Booking Limits

¡  Nesting Controls:
l  Label the fare classes so that 1 is the highest

fare class and n is the lowest fare class. For any
i let bi denote the nested booking limit for class
i.

l  Protection Levels:

l  Updates: If x units are sold in a transaction

nbbb ≥≥≥ 21

1,2,1,11 −=−= + nibby ii 

1,2,1),0,max( −=−← nixbb ii 

Nested Booking Limits (Example)

Booking Limits Protection Levels Requests

1 2 3 4 5 1 2 3 4 5 Seats Class Action

1 100 73 12 4 0 27 88 96 100 100 2 5 Reject

2 100 73 12 4 0 27 88 96 100 100 5 2 Accept

3 95 68 7 0 0 27 88 95 95 95 1 2 Accept

4 94 67 6 0 0 27 88 94 94 94 1 4 Reject

5 94 67 6 0 0 27 88 94 94 94 3 3 Accept

6 91 64 3 0 0 27 88 91 91 91 4 3 Reject

7 91 64 3 0 0 27 88 91 91 91 2 3 Accept

8 89 62 1 0 0 27 88 89 89 89

Is RM successful?

¡  By most measures (revenues relative to
resources) it has been a success at most
major airline, hotel, rental car companies.
l  Why have major airlines have been losing

money?
¡  Costs are 25-30% higher per mile, so even though

larger carriers bring in about 25% more revenue
per mile the cost disadvantage is overwhelming

l  What can be done?
¡  Cost costs
¡  Improve RM systems

l  Big move from independent to dependent demand
models

RM and Price Discrimination

¡  Price discrimination exists when sales of identical goods or
services are transacted at different prices from the same
provider.

¡  A feature of monopolistic or oligopolistic markets where
market power can be exercised.

¡  Requires market segmentation and means to discourage
discount customers becoming resellers.
l  This is achieved by fences to keep segments separate.

¡  Price discrimination is more common in services where
resale is not possible.

¡  Price discrimination can also be seen when the
requirement of identical goods is relaxed.
l  Premium products have price differential not explained by production

costs.

Taxonomies of Price Discrimination

¡  First degree: requires selling at maximum willingness to
pay

¡  Second degree: quantity discounts (sellers not able to
differentiate consumer types)

¡  Third degree: Prices vary by attributes (e.g., senior
discounts)

¡  Fourth degree: Prices are the same but costs are different
(reverse discrimination)

¡  Alternative taxonomy:
l  Complete discrimination (like first degree)
l  Direct discrimination: seller conditions price on some attribute (like

third degree)
l  Indirect discrimination: the seller relies on some proxy such as

quantity discounts (like second degree)

RM and Price Discrimination

¡  Differentiating by time-of-purchase and imposing traveling
restrictions like Saturday night stays is a form of second
degree or indirect discrimination.

¡  Selling premium seats is another form of second degree or

indirect discrimination.
l  Eg., uncomfortable second class seats on trains to entice

wealthier people to purchase first class seats.
l  Advance seat selection, mileage accrual, use of lounge, and

priority boarding may be forms of second and/or fourth
degree discrimination.

Other Examples of Price Discrimination

¡  Retail price discrimination is in violation of the Robinson-
Patman Act (1936)

¡  Coupons
¡  Segmentation by age group and student status
¡  Discounts for members of certain occupations
¡  Employee discounts
¡  Retail incentives (rebates, seasonal discounts, quantity

discounts)
¡  Gender based examples
¡  College financial aid
¡  User-controlled price discrimination
¡  See http://en.wikipedia.org/wiki/Price_discrimination

Static and Dynamic Pricing

¡  Pricing is studied by people in Economics
and Marketing

¡  Economist look at equilibrium prices
¡  Marketing focuses on demand estimation
¡  We focus on more tactical aspects of

pricing
l  Customer arrival rates
l  Capacity constraints
l  And increasingly on choice models and

competition

Static Pricing

¡  d(p) demand at p
¡  z unit cost or dual of capacity constraint
¡  r(p,z) = (p-z)d(p) profit function
¡  Find p to maximize r(p,z)
¡  Is there a finite maximizer p(z)?
¡  Is p(z) monotone?
¡  Is r(z) = r(p(z),z) monotone? Convex?
¡  Multiple market segments with limited

price menus

Dynamic Pricing
¡  Finite sales horizon
¡  Customers arrive stochastically over time
¡  State (t,x)

l  t time-to-go
l  x remaining inventory

¡  What price p(t,x) should be charged at
state (t,x)?

¡  Are there simple and effective pricing
heuristics?

¡  What about strategic customers?
¡  What about competition?

Topics to be Covered

¡  Single Resource RM
l  Independent Demands

¡  Dynamic Programming, Bounds and Heuristics
l  Dependent Demands based on choice models

¡  Static and Dynamic Pricing
¡  Network RM

l  Independent Demands, Choice Models
¡  Overbooking
¡  Service Engineering

l  Design and pricing of service features

Useful Techniques you will Learn

¡  Dynamic Programming (DP)
l  Tool for sequential decision making
l  Optimal Control (continuous time DP)

¡  Approximate Dynamic Programming
l  Tool to approximate DPs

¡  Bounds and Heuristics Techniques
¡  Choice Modeling
¡  Game Theory

Probability
and Dynamic Programming

Review
for

Dynamic Pricing and Revenue
Management

Probability Review

• Poisson: http://en.wikipedia.org/wiki/Poisson_random_variable

– Read Sections: 1, 2, 3,

• Compound Poisson:
http://en.wikipedia.org/wiki/Compound_Poisson_distribution

• Poisson Process: http://en.wikipedia.org/wiki/Poisson_process
• Compound Poisson Process:

http://en.wikipedia.org/wiki/Compound_Poisson_process

• Normal: http://en.wikipedia.org/wiki/Normal_random_variable
– Read Sections 1,2.1,2.2,3,1,3.2,3.

• Brownian Motion: http://en.wikipedia.org/wiki/Wiener_process
– Read sections 1, 2.

DP AS AN OPTIMIZATION METHODOLOGY

• Basic optimization problem

min
u∈U

g(u)

where u is the optimization/decision variable, g(u)
is the cost function, and U is the constraint set

• Categories of problems:
− Discrete (U is finite) or continuous
− Linear (g is linear and U is polyhedral) or

nonlinear
− Stochastic or deterministic: In stochastic prob-

lems the cost involves a stochastic parameter
w, which is averaged, i.e., it has the form

g(u) = Ew
{
G(u, w)

}

where w is a random parameter.

• DP can deal with complex stochastic problems
where information about w becomes available in
stages, and the decisions are also made in stages
and make use of this information.

{ } ∑
{ } ∑ ( )

INVENTORY CONTROL EXAMPL

Inventory
S y s t e m

Stock Ordered at
Period k

Stock at Period k Stock at Period k + 1

Demand at Period k

xk + 1 = xk + uk – wk

u k
Co s t o f P e rio d k

c uk + r (xk + uk – wk)

• Discrete-time system

xk+1 = fk(xk, uk, wk) = xk + uk − wk
• Cost function that is additive over time

N−1
E gN (xN ) +

gk(xk, uk, wk)

k=0

N−1
= E cuk + r(xk + uk − wk)

k=0

• Optimization over policies: Rules/functions uk =
µk(xk) that map states to controls

BASIC STRUCTURE OF STOCHASTIC DP

• Discrete-time system

xk+1 = fk(xk, uk, wk), k = 0, 1, . . . , N −

− k: Discrete time
− xk: State; summarizes past information that

is relevant for future optimization
− uk: Control; decision to be selected at time

k from a given set
− wk: Random parameter (also called distur-

bance or noise depending on the context)
− N : Horizon or number of times control is

applied

• Cost function that is additive over time

{
gN (xN ) +

N−1∑
k=0

gk(xk, uk, wk)

}

Guillermo
Pencil

BASIC PROBLEM

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

Jπ(x0) = E

{
gN (xN ) +
N−1∑
k=0

gk(xk, µk(xk), wk)

}

• Optimal cost function

J∗(x0) = min
π

Jπ(x0)

• Optimal policy π∗ is one that satisfies

Jπ∗ (x0) = J∗(x0)

PRINCIPLE OF OPTIMALITY

• Let π∗ = {µ∗0, µ∗1, . . . , µ∗N−1} be an optimal pol-
icy

• Consider the “tail subproblem” whereby we are
at xi at time i and wish to minimize the “cost-to-
go” from time i to time N

E
{
gN (xN ) +

N−1∑
k=i

gk
(
xk, µk(xk), wk

)}

and the “tail policy” {µ∗i , µ∗i+1, . . . , µ∗N−1}

0 Ni

xi Tail Subproblem

• Principle of optimality: The tail policy is opti-
mal for the tail subproblem

• DP first solves ALL tail subroblems of final
stage

• At the generic step, it solves ALL tail subprob-
lems of a given time length, using the solution of
the tail subproblems of shorter time length

DP ALGORITHM

• Start with

JN (xN ) = gN (xN ),

and go backwards using

Jk(xk) = min
uk∈Uk(xk)

E
wk

{
gk(xk, uk, wk)

+ Jk+1
(
fk(xk, uk, wk)

)}
, k = 0, 1, . . . , N − 1.

• Then J0(x0), generated at the last step, is equal
to the optimal cost J∗(x0). Also, the policy

π∗ = {µ∗0, . . . , µ∗N−1}
where µ∗k(xk) minimizes in the right side above for
each xk and k, is optimal.

• Justification: Proof by induction that Jk(xk)
is equal to J∗k (xk), defined as the optimal cost of
the tail subproblem that starts at time k at state
xk.

• Note that ALL the tail subproblems are solved
in addition to the original problem, and the inten-
sive computational requirements.

DETERMINISTIC FINITE-STATE PROBLEM

Terminal Arcs

. . .

. . .
. . .

Initial State
s

t
Artificial Terminal
N o d e

with Cost Equal
to Terminal Cost

S t a g e 0 S t a g e 1 S t a g e 2 . . . S t a g e N – 1 S t a g e N

• States <==> Nodes
• Controls <==> Arcs
• Control sequences (open-loop) <==> paths
from initial state to terminal states

• ak : Cost of transition from state i ∈ Sk to state ij
j ∈ Sk+1 at time k (view it as “length” of the arc)
• aN : Terminal cost of state i ∈ SNit
• Cost of control sequence <==> Cost of the cor

responding path (view it as “length” of the path)

BACKWARD AND FORWARD DP ALGORITHMS

• DP algorithm:
JN (i) = aNit , i ∈ SN ,

Jk(i) = min
j∈Sk+1

[
akij +Jk+1(j)

]
, i ∈ Sk, k = 0, . . . , N−1.

The optimal cost is J0(s) and is equal to the
length of the shortest path from s to t.

• Observation: An optimal path s → t is also an
optimal path t → s in a “reverse” shortest path
problem where the direction of each arc is reversed
and its length is left unchanged.

• Forward DP algorithm (= backward DP algo-
rithm for the reverse problem):

J̃N (j) = a0sj , j ∈ S1,
J̃k(j) = min

i∈SN−k

[
aN−kij + J̃k+1(i)

]
, j ∈ SN−k+1

The optimal cost is J̃0(t) = mini∈SN
[
aNit + J̃1(i)

]
.

• View J̃k(j) as optimal cost-to-arrive to state j
from initial state s.

A NOTE ON FORWARD DP ALGORITHMS

• There is no forward DP algorithm for stochastic
problems.

• Mathematically, for stochastic problems, we
cannot restrict ourselves to open-loop sequences,
so the shortest path viewpoint fails.

• Conceptually, in the presence of uncertainty,
the concept of “optimal-cost-to-arrive” at a state
xk does not make sense. The reason is that it may
be impossible to guarantee (with prob. 1) that any
given state can be reached.

• By contrast, even in stochastic problems, the
concept of “optimal cost-to-go” from any state xk
makes clear sense.

GENERIC SHORTEST PATH PROBLEMS

• {1, 2, . . . , N, t}: nodes of a graph (t: the desti-
nation)

• aij : cost of moving from node i to node j
• Find a shortest (minimum cost) path from each
node i to node t

• Assumption: All cycles have nonnegative length.
Then an optimal path need not take more than N
moves

• We formulate the problem as one where we re-
quire exactly N moves but allow degenerate moves
from a node i to itself with cost aii = 0.

Jk(i) = optimal cost of getting from i to t in N−k moves.

J0(i): Cost of the optimal path from i to t.

• DP algorithm:
Jk(i) = min

j=1,…,N

[
aij +Jk+1(j)

]
, k = 0, 1, . . . , N−2,

with JN−1(i) = ait, i = 1, 2, . . . , N.

EXAMPLE

2
5 5

6 1

0 . 5
3

1
2
4

0 1 2 3 4

1
2
3
4
5

State i

S t a g e k

3 3 3 3

4 4 4 5

4 . 5 4 . 5 5 . 5 7

2 2 2 2

Destination
5

(a) (b)

JN−1(i) = ait, i = 1, 2, . . . , N,

Jk(i) = min
j=1,…,N

[
aij +Jk+1(j)

]
, k = 0, 1, . . . , N−2.

Probability and Dynamic ProgrammingReview

Probability Review

Single Resource Revenue Management with Independent Demands
c©

Guillermo Gallego

Updated Spring

Abstrac

Providers of fixed perishable capacity, such as airline seats and hotel rooms use price discrimination
to improve revenues; in practice, this discrimination is typically achieved by imposing booking and usage
restrictions or including ancillary services such as mileage accrual and luggage handling, to sell the same
capacity to different customers at different prices. We will assume that the set of fare classes (a menu
of prices, restrictions and ancillary services) is given, and that the capacity provider’s goal is to allocate
capacity among the different fare classes to maximize expected revenues. The problem of designing

and

pricing fare classes is treated in a separate chapter. We analyze the two fare class problem under the
assumption that the lower fare class books first. We use marginal analysis to informally derive Littlewood’s
rule and then show that Littlewood’s rule is in fact optimal. Spill rates, spill penalties and callable
products are discussed next. A dynamic programming formulation for the multiple fare class problem is
then introduced under the assumption that lower fare classes book first. Commonly used heuristics as well
as bounds on the value function are presented. Dynamic models that explicitly take time into account,
allow for more general fare arrival patterns and for randomness in the size of the requests. We compare
the performance of static and dynamic policies and find that dynamic policies have a real advantage when
the fare arrivals patterns are not low-to-high. We finalize the chapter with a model where fare classes are
not allowed to reopen after they are closed for the first time.

1 Introduction

This chapter considers the simplest and best known revenue management problem, the single resource, inde-
pendent demand problem. We assume that the capacity provider is trying to maximize the expected revenues
from a sunk investment in c units of capacity. We assume that capacity is sold through a reservation system
and that capacity cannot be modified or replenished during the booking horizon. We also assume that unsold
capacity has no salvage value. Later we will see that the zero salvage value assumption is made without loss
of generality as any problem with positive salvage value can be transformed into a problem with zero salvage
value. We assume that the set of fare classes (a menu of prices and restrictions) is given, and that the demands
for the different fare classes are statistically independent. In particular, we assume that if a customer finds his
preferred fare class closed, he will leave the system without purchasing. This assumption holds approximately
if the difference in fares is large so that demands are decoupled or if customers can find alternative sources
of capacity for their preferred fare class. In some cases, however, part of the demand may be recaptured by
other available fare classes. In such cases, the independent demand assumption is too strong and needs to be
relaxed. We address this issue in a separate chapter where we discussed demand models based on discrete
choice theory.

In this chapter, we present a variety of models that have been developed in industry and in academia.
There has been a preference in industry for models that suppress the time dimension and assume that the
arrival pattern of the fare classes is low-to-high. We call these class of models static to distinguish them
from the dynamic models, favored by academics, that model time explicitly. Both models have advantages
and disadvantages as we will soon see. Static models are relatively easy to understand. Also, good heuristics

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 §

and bounds on the optimal expected
revenue are presented in §

. The dynamic model is presented in §

, for the Poisson case and for the compound
Poisson case in §

, where each request is for a random demand size. In §

, we restrict fares so that they cannot
be opened once they are closed.

2 Two Fare Classes: Marginal Analysis

The product can be sold either at the full-fare p1 or at a discounted-fare p2 < p1. The discounted-fare typically has advance purchasing and usage restrictions. Let D1 and D2 denote respectively the random demand for the two fare classes for a specific instance of the problem, e.g., for a specific flight for an airline or a specific night for a hotel.

We assume that all booked customers will actually travel. This avoids the need to overbook capacity and
allow us to focus on the problem of allocating capacity between the two fares. We will discuss how to deal
with pre-travel cancellations and day-of-travel no shows in a separate chapter on overbooking models.

Fare class arrival order is an important part of the model. We assume what is commonly known as the
low-to-high fare class arrival order, which implies that demand for the discounted fare book earlier than for
full-fare. This arrival pattern holds approximately in practice, and it is encouraged by advance purchase
restrictions imposed on lower fare classes. Notice that this is a worst case arrival pattern. Indeed, if full-fare
class customers arrived first then we would accept them up to capacity and use residual capacity, if any, to
satisfy demand from the discounted-fare class. We will relax the low-to-high fare order arrival assumption
after we solve the multi-fare problem via dynamic programming.

Under the low-to-high arrival pattern, discount-fare customers may exhaust capacity, say c, unless part
of it is protected for later-booking by full-fare customers. Consequently, booking limits (known as discount
authorizations) are placed on the discount sales. Suppose we protect y ∈ {0, 1, . . . ,c} units of capacity for
the full-fare demand, D1, before observing the actual demand for the discount-fare, D2. This results in a
booking limit c−y on the discounted-fare, so sales at the discounted-fare class are given by min(c−y,D2).
The remaining capacity is equal to c − min(c − y,D2) = max(y,c − D2) and it is all made available to the
full-fare class. Consequently, sales at the full fare equal min(max(y,c−D2),D1). The total expected revenue
is

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 [

] in 1

72. His arguments
were not formal; however, they were later justified by Bhatia and Prakesh [1] in

73, and Richter [

] 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 y

th unit at p1 only
if the full-fare demand D1 is at least as large as y, so it is intuitively optimal to reject the discount fare if
p1P(D1 ≥ y) > p2. This suggests that an optimal protection level y1 should be given by:

y1 = max{y ∈N : P(D1 ≥ y) > r}, (1)

where N = {0, 1, . . . ,} is the set of non-negative integers. Equation (1) is known as Littlewood’s rule.

Example 1. Suppose D1 is Poisson with parameter 80, the full fare is p1 = $

0 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 =

0 then the booking limit for the discount fare is

2. However, if c < y1, then all units should be protected for the full-fare resulting in a booking limit of zero.

Remarks:

• y(c) = min(y1,c) is also an optimal protection level. If y(c) = c, or equivalently if y1 ≥ c, then all the
capacity should be reserved for sale at the full-fare.

• The quantity b2 = max(c−y1, 0) is known as the optimal booking limit for the discount fare. It is the
maximum number of discount-fare customers that we will book.

• y1 is independent of the distribution of D2.

• If P(D1 ≥ y1 + 1) = r, then y1 + 1 is also optimal protection level, so both y1 and y1 + 1 result in the
same expected revenue. Protecting the y1 + 1 unit of capacity increases the variance of the revenue, but
it reduces the probability of rejecting requests from full-fare customers.

From Littlewood’s rule (1), we see that the extreme strategy y = 0 is optimal when P(D1 ≥ 1) ≤ r and
the extreme strategy y = c is optimal when P(D1 ≥ c) > r.

2.1 Continuous Demand Model

Although revenue management demands are actually discrete, continuous distributions can be easier to work
with and are often employed in practice. If we model D1 as a continuous random variable with cumulative
distribution function F1(y) = P(D1 ≤ y), then

y1 = F
−1
1 (1 −r)

where F−

denotes the inverse of F. In particular, if D1 is Normal with mean µ1 and standard deviation σ1
then

y1 = µ1 + σ1Φ
−1(1 −r) (2)

where Φ denotes the cumulative distribution function of the standard Normal random variable.

This formula allows for comparative statics as given in Table 1:

Example 2. Suppose that D1 is Normal with mean 80 and standard deviation 9, the full-fare is p1 = $100
and the discount-fare is p2 = $60. Then y1 = F

−1
1 (1 − 0.6) = 77.72 < 80 since r > 1/2. Notice that the

solution is quite close to that of Example 1. This is because a Poisson random variable with mean 80 can be
well approximated by a normal with mean 80 and standard deviation

√
80 ‘ 9.

Fare Ratio Dependence of protection level

r > 1
2

y1 < µ1 and y1 decreases with σ1

r = 1
2

y1 = µ1 independent of σ1

r < 1 2

y1 > µ1 and y1 increases with σ1

Table 1: Comparative Statics for Normal Full Fare Demand

2.2 Connection with the Newsvendor Problem

There is a close connection between the classical Newsvendor Problem and the two-fare Revenue Management
Problem that we will briefly explore here. In the classical Newsvendor Problem a manager must decide how
many units, say y, to stock for random sales D1 at p1 assuming a unit cost p2 < p1. The solution is to stock y1 units where y1 is the largest integer such that P(D1 ≥ y) > r = p2/p1. We can think of the two-fare Revenue
Management Problem as a situation where capacity c is pre-decided, at a possible sub-optimal level, there is
random demand D1 at ”salvage value” p1 > p2 that arrives after demand D2 at p2. The revenue management
problem is to determine how many units to allow to be sold at p2. We know that the solution is to allow
(c−y1)+ units to book at p2, reserving max(y1,c−D2) units for sale at p1.

2.3 Two Fare Classes: Dynamic Programming

In this section we formulate and analyze the two fare class problem using dynamic programming and present
a formal proof of the optimality of Littlewood’s rule. We will from now on refer to the full-fare class as fare
class 1 and to the discounted fare class as fare class 2. Dynamic programming starts by solving the problem
at the last stage, just before demand for fare class 1. Let V1(y) be the optimal expected revenue that can be
obtained from fare class 1 when capacity is y. Since it is optimal to allow fare class 1 customers to book all
of the available capacity, sales are equal to min(D1,y) and the optimal expected revenue is

V1(y) = p1E min(D1,y).

Our next task is to find V2(c), the optimal expected revenue that can be obtained from c units of ca-
pacity. Suppose that y ∈ {0, 1, . . . ,c} units are protected for fare class 1 demand. This results in revenues
p2 min(D2,c − y) from sales to fare class 2 and remaining inventory max(c − D2,y) available for fare class
1. Notice that we can obtain expected revenue EV1(max(c − D2,y)) from this inventory from fare class 1
customers. Then

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
y∈{0,1,…,c}

E{p2 min(D2,c−y) + V1(max(c−D2,y))}. (3)

The key to Dynamic Programming is that it involves a recursive equation (3) linking the expected revenues
V2(c), at stage 2, to the expected revenue function V1 at stage 1. To solve for V2(c) we first need to solve
for V1(y) for y ∈ {0, 1, . . . ,c}. Before moving on to the multi-fare formulation we will provide a formal
proof of Littlewood’s rule (1), and discuss the quality of service implications of using Littlewood’s rule under
competition.

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)

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) >

p2
p1 + ρ

}
. (5)

Notice that this is just Littlewood’s rule applied to fares p1 + ρ and p2, resulting in fare ratio p2/(p1 + ρ) and,
consequently, lower maximal spill rates. Obviously this adjustment comes at the expense of having higher
protection levels and therefore lower sales at the discount-fare and lower overall revenues. Consequently, an
airline that wants to protect its full-fare market by imposing a penalty on rejected full-fare demand does it at
the expense of making less available capacity for the discount-fare and less expected revenue. One way to avoid
sacrificing sales at the discount-fare and improve the spill rate at the same time is to modify the discount-fare
by adding a restriction that allows the provider to recall or buy back capacity when needed. This leads to
revenue management with callable products; see Gallego, Kou and Phillips [12]. Callable products can be
sold either by giving customers an upfront discount or by giving them a compensation if and when capacity
is recalled. If managed correctly, callable products can lead to better capacity utilization, better service to
full-fare customers and to demand induction from customers who are attracted to either the upfront discount
or to the compensation if their capacity is recalled.

The value function V1(y) may also be modified to account for salvage values (also known as the ‘distressed
inventory problem’). Suppose there is a salvage value s < p2 on excess capacity after the arrival of the full-fare demand (think of standby tickets or last-minute travel deals). We can handle this case by modifying V1(y) to account for the salvaged units. Then V1(y) = p1E min(D1,y) + sE(y −D1)+ = (p1 −s)E min(D1,y) + sy, so ∆V1(y) = (p1 −s)P(D1 ≥ y) + s, resulting in

∆W(y,c) = [(p1 −s)P(D1 ≥ y) − (p2 −s)]P(D2 > c−y)

and
y1 = max
{
y ∈N : P(D1 ≥ y) >

p2 −s
p1 −s

}
. (6)

Notice that this is just Littlewood’s rule applied to net fares p1 −s and p2 −s. This suggests that a problem
with salvage values can be converted into a problem without salvage values by using net fares pi ← pi − s,
i = 1, 2 and then adding cs to the resulting optimal expected revenue V2(c) in excess of salvage values.

3 Multiple Fare Classes: Exact Solution

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 [

].
Curry [6] derives optimality conditions when demands are assumed to follow a continuos distribution. Brumelle
and McGill [5] allow for either discrete or continuous demand distributions and makes a connection with the
theory of optimal stopping.

Let Vj(x) denote the maximum expected revenue that can be obtained from x ∈ {0, 1, . . . ,c} units of
capacity from fare classes {j, . . . , 1}. The sequence of events for stage j are as follows:

1. Decide the protection level, say y ≤ x, for fares j − 1,j − 2, . . . , 1 thus allowing at most x−y units of
fare j demand.

2. The realization of the demand Dj occurs, and we observe min(Dj,x−y) sales at fare j.

3. The revenue pj min(Dj,x−y) is collected, and we proceed to the beginning of stage j−1 with a remaining
capacity of max(x−Dj,y).

The revenue from this process is

Wj(y,x) = pjE min(Dj,x−y) + EVj−1 (max(x−Dj,y)) . (7)

We can think of Wj(y,x) as the expected revenue from x units of capacity prior to seeing the demand for fare
class j when up to x−y units are allowed to book at fare j and an optimal policy is followed thereafter. This
leads to the dynamic programming recursion

Vj(x)

= max
y∈{0,1,…,x}

Wj(y,x)

= max
y∈{0,1,…,x}

{pjE min(Dj,x−y) + EVj−1(max(x−Dj,y))} . (8)

The dynamic program simply states that the optimal value function is the sum of the expected revenues
from fare class j plus the expected revenues from fare classes j−1, . . . , 1 evaluated at the protection level that
maximizes this sum. Notice that once we are at the beginning of stage j−1 we face a similar problem over the
remaining j−1 fare classes. Vj(x) is then the maximum expected revenue that can be obtained from x units of
capacity for the j-fare problem. Consequently Vn(c) is the maximum expected revenue for the n-fare problem
with capacity c. The recursion can be started with V0(x) = 0 if there are no salvage values or penalties for
spill. Alternatively, the recursion can start with V1(x) = p1E min(D1,x) + sE[(x−D1)+]−ρE[(D1 −x)+] for
x ≥ 0 if there is a salvage value s per unit of excess capacity and a penalty ρ per unit of fare class 1 demand
that is denied.

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.

j pj E[Dj] yj
1 $100

14
2 $60 40 54
3 $40 50 101
4 $35 55

9
5 $15 120

Table 2: Five Fare Example with Poisson Demands: Data and Optimal Protection Levels

Table 3 provides the expected revenues for different capacity levels as well as the corresponding demand
factors (

∑5
j=1 E[Dj])/c =

0/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
$3,4

.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 $9,6

because there is sufficient capacity to accept
sales in all fare classes.

c DF V1(c) V2(c) V3(c) V4(c) V5(c)
50 560% 1,500.0 3,426.8 3,426.8 3,426.8 3,426.8

100 280% 1,500.0 3,900.0 5,441.3 5,441.3 5,441.3
150

7% 1,500.0 3,900.0 5,900.0 7,188.7 7,188.7
200 140% 1,500.0 3,900.0 5,900.0 7,8

.6 8,159.1
250 112% 1,500.0 3,900.0 5,900.0 7,825.0 8,909.1

0 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.

$-‐

$20.00

$40.00

$60.00

$80.00

$100.00

$120.00

0
20
40
60
80
100
120
140
16

DV_1(x)

DV_2(x)

DV_3(x)

DV_4(x)

DV_5(x)

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

3.2 Speeding up the Computation of the Value Function

While the value functions Vj(x), j ∈ {1, . . . ,n}, x ∈ {1, . . .c}, can be computed recursively there are some
tricks to speed up the computations. Here we focus on how to efficiently update ∆Vj+1(x) from ∆Vj(x). The
key idea is to express ∆Vj+1(x) for x > yj in terms of previously computed values of ∆Vj(x). The proof of
Proposition 2 is in the Appendix.

Proposition 2

∆Vj+1(x) =

{
∆Vj(x) if x = 1, . . . ,yj
E min(∆Vj(x−Dj+1),pj+1) if x = yj + 1, . . ..

Since ∆Vj+1(x) = ∆Vj(x) for x ≤ yj, we only need to worry about ∆Vj+1(x) for x > yj. The following
corollary to Proposition 2 makes the formula for ∆Vj+1(x),x > yj more explicit.

Corollary 1

∆Vj+1(yj + k) = pj+1P(Dj+1 ≥ k)

k−1∑
i=0

∆Vj(yj + k − i)P(Dj+1 = i) k ∈{1, . . . ,c−yj}.

3.3 Linear and Convex Procurement Costs

Suppose that capacity c can be procured at a linear cost kc before observing demands for the n fares: pn < pn−1 < ... < p1. How much capacity should be procured? The objective is to find c to maximize Πn(c,k) = Vn(c) −kc. Let c(k) be the smallest optimizer of Πn(c,k) as a function of the marginal cost k. The following Proposition characterizes c(k), shows that c(k) is decreasing in k and relates c(pj+1) to protection level yj for j < n.

Proposition 3 The optimal procurement quantity at linear cost kc is given by

c(k) = max{c ∈N : ∆Vn(c) > k}.

Moreover, c(k) is decreasing in k, and yj = c(pj+1) for all j ∈{1, . . . ,n− 1}.

Clearly c(0) = ∞ since ∆Vn(c) ≥ ∆V1(c) = p1P(D1 ≥ c) > 0 for all c ∈ N . At the other extreme,
c(p1) = y0 = 0 since ∆Vn(1) = p1P(D1 ≥ 1) < p1 = ∆Vn(0), so no capacity would be purchased if k ≥ p1.

If the cost of capacity k(c) is increasing convex then Π(c,k(c)) is concave in c and

c(k) = max{c ∈N : ∆Vn(c) − ∆k(c) > 0}.

Consider the convex cost function k(c) = K if c ≤ c̄ and k(c) = ∞ for k > c̄. This may reflect the situation
where there is a fixed cost to leasing a resource with capacity c̄. The optimal choice is then to lease the
resource if Vn(c̄) > K and not lease it otherwise.

3.4 Relaxing the Monotonicity of Fares

100

150

200

250

300

350

400

450

0
10
20
30
40
50
60
70
80
90
100

C

ap
ac

it
y

cost

optimal capacity

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

steps as it is clear that yj−1 = 0 whenever pj > max(p1, . . . ,pj−1) since it is optimal to allow all bookings at
fare pj. The reader is referred to Robinson [18] for more details. As an example, suppose that p3 < p2 > p1.
Then V1(x) = p1E min(D1,x) and at stage 2 the decision is y1 = 0, so

V2(x) = p2E min(D2,x) + p1E min(D1, (x−D2)+).

Notice that
∆V2(x) = p2P(D2 ≥ x) + p1P(D2 < x ≤ D[1, 2])

so
y2 = max{y ∈N : ∆V2(y) > p3}.

Notice that the capacity protected for fare 2 is higher than it would be if there was no demand at fare 1.

4 Multiple Fare Classes: Commonly Used Heuristics

Several heuristics, essentially extensions of Littlewood’s rule, were developed in the 1980’s. The most important
heuristics are known as EMSR-a and EMSR-b, where EMSR stands for expected marginal seat revenue. Credit
for these heuristics is sometimes given to the American Airlines team working on revenue management problems
shortly after deregulation. The first published account of these heuristics appear in Simpson [19] and Belobaba
[2], [3]. For a while, some of these heuristics were even thought to be optimal by their proponents until optimal
policies based on dynamic programming were discovered in the 1990’s. By then heuristics were already part of
implemented systems, and industry practitioners were reluctant to replace them with the solutions provided
by dynamic programming algorithms. There are several reasons for this. First, people feel more comfortable
with something they understand. Also, the performance gap between the heuristics and the optimal dynamic
program tends to be small. Finally, there is a feeling among some users that the heuristics may be more robust
to estimates of the mean and variance of demand.

Version a of the heuristic, EMSR-a, is based on the idea of adding protection levels produced by applying
Littlewood’s rule to successive pairs of classes. At state j, we need to decide how much capacity to protect
for fares j − 1, . . . , 1. We can use Littlewood’s rule to decide how much capacity to protect for fare k demand
against fare j for k = j − 1, . . . , 1 and then add the protection levels. More precisely, let rk,j = pj/pk and set

yk,j = max{y ∈N : P(Dk ≥ y) > rk,j}.

Then the EMSR-a heuristic will protect

yaj−1 =

j−1∑
k=1

yk,j

units of capacity for fares j − 1, . . . , 1 against fare j.

In particular, if Dk is Normal with mean µk and standard deviation σk, then

yaj−1 = µ[1,j − 1] +
j−1∑
k=1

σkΦ
−1(1 −rk,j),

where for any j, µ[1,j − 1] =
∑j−1

k=1 µk and sums over empty sets are zero.

Notice that the EMSR-a heuristic involves j − 1 calls to Littlewood’s rule to find the protection level for
fares j − 1, . . . , 1. In contrast, the EMSR-b heuristic is based on a single call to Littlewood’s rule for each
protection level. However, using the EMSR-b heuristic requires the distribution of D[1,j − 1] =

∑j−1
k=1 Dk.

This typically requires computing a convolution but in some cases, such as the Normal or the Poisson, the
distribution of D[1,j − 1] can be easily obtained (because sums of independent Normal or Poisson random
variables are, respectively, Normal or Poisson). The distribution of D[1,j − 1] is used together with the
weighted average fare

p̄j−1 =

j−1∑
k=1

pk
µk

µ[1,j − 1]

and calls on Littlewood’s rule to obtain protection level

ybj−1 = max{y ∈N : P(D[1,j − 1] ≥ y) > r
b
j−1,j}

where rbj−1,j = pj/p̄j−1. Notice that the weighted average fare assumes that a proportion µk/µ[1,j−1] of the
protected capacity will be sold at fare pk,k = 1, . . . ,j − 1. In the special case when demands are Normal we
obtain

ybj−1 = µ[1,j − 1] + σ[1,j − 1]Φ
−1(1 −rbj−1,j).

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

k=1 σ
2
k.

4.1 Evaluating the Performance of Heuristics

While heuristic protection levels are easy to compute, evaluating them is as hard as solving for the optimal
policy. The expected return of a heuristic policy based on protection levels 0 = yh0 ≤ yh1 ≤ . . . ≤ yhn−1 can be
computed exactly or via simulation. To compute the expected revenues exactly, let V hj (x) be the expected profit

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

h
j−1),x)

where for all y ≤ x we define Whj (y,x) = pjE min(x−y,Dj) + EV
h
j−1(max(y,x−Dj)). The following result

is a direct consequence of Proposition 2 and its corollary.

Proposition 4

∆V hj+1(x) =

{
∆V hj (x) if x = 1, . . . ,y

h
j

= pj+1P(Dj+1 ≥ x−yhj ) +
∑x−yhj −1

i=0 ∆V
h
j (x− i)P(Dj+1 = i) if x > y

h
j

Thus, if V hj (x) has already been computed then V
h
j+1(x) = V

h
j (x) for x = 1, . . . ,y

h
j . For x > y

h
j we can

use the recursion V hj+1(x) = V
h
j+1(x− 1) + ∆V

h
j+1(x) starting with x = y

h
j + 1 in conjunction with the second

part of Proposition 4.

To estimate the expected revenue and other measures of performance, such as the variance, we can also
use Monte Carlo simulation. Suppose we generate many random copies of simulated demands (D1, . . . ,Dn).
For each copy we compute sales (sh1, . . . ,s

h
n) and revenues R

h =
∑n

i=1 pis
h
i under heuristic h. Averaging

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

min(Dj, (xj−yhn−1)+) starting with j = n and xn = c and using the capacity update formula xj = xj+1−shj+1.

Example 3 (continued) We have applied the EMSR-a and EMSR-b heuristics to the data of Example 3. Table
4 repeats the data and reports the heuristic protection levels ya, yb as well as the optimal protection levels.
Table 5 reports V a5 (c) and V

b
5 (c) as well as V5(c) for values of c ∈{50, 100, 150, 200, 250, 300, 350}.

j pj E[Dj] y
a
j y

b
j yj

1 $100 15 14 14 14
2 $60 40 53 54 54
3 $40 50 97 102 101
4 $35 55 171 166 169
5 $15 120

Table 4: Optimal and Heuristic Protection Levels for Example 3

c DF V a5 (c) V
b
5 (c) V5(c)

50 560% 3,426.8 3,426.8 3,426.8
100 280% 5,4

.9 5,441.3 5,441.3
150 187% 7,184.4 7,188.6 7,188.7
200 140% 8,157.3 8,154.4 8,159.1
250 112% 8,907.3 8,901.4 8,909.1
300 93% 9,536.5 9,536.0 9,563.9
350 80% 9,625.0 9,625.0 9,625.0

Table 5: Performance of Heuristics for Example 3

As seen in Table 5, both the EMSR-a and the EMSR-b heuristic perform very well against Poisson demands
under a low-to-high arrival pattern. The heuristics continue to perform well if demands are compound Poisson
and aggregate demands are approximated by the use of a Gamma distribution. However, EMSR based
heuristics can significantly underperform relative to models that allow more general fare arrival rates. We will
have an opportunity to revisit this issue in Section 7.

5 Bounds, Revenue Opportunity Model, and New Heuristics

In this section we develop bounds on Vn(c) which may be useful in evaluating the potential of applying revenue
management solutions. To obtain an upper bound, consider the perfect foresight problem where the demand
vector D = (D1, . . . ,Dn) is known in advance. This demand knowledge allows us to optimally allocate capacity
by solving the following knapsack type problem

V Un (c,D) = max

n∑
k=1

pkxk (11)

s.t. xk ≤ Dk ∀ k = 1, . . . ,n

n∑
k=1

xk ≤ c

xk ≥ 0 ∀ k = 1, . . . ,n.

Clearly, for each realization of D, advance knowledge results in revenues that are at least as high as the optimal
dynamic policy that does not have perfect foresight. As a result, Vn(c) ≤ EV Un (c,D). For convenience, we
will denote this upper bound as V Un (c) = EV

U
n (c,D).

The solution to (11) can be written explicitly as xk = min(Dk, (c − D[1,k − 1])+),k = 1, . . . ,n where
for convenience we define D[1, 0] = 0. The intuition here is that we give priority to higher fares so fare
k ∈ {1, . . . ,n} gets the residual capacity (c − D[1,k − 1])+. The expected revenue can be written more
succinctly after a few algebraic calculations:

V Un (c) =

n∑
k=1

pkE min(Dk, (c−D[1,k − 1])+) (12)

=
n∑
k=1

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

=
n∑
k=1

(pk −pk+1)E min(D[1,k],c) (13)

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

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

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

V Un (c,µ) = max

n∑
k=1

pkxk (14)

s.t. xk ≤ µk ∀ k = 1, . . . ,n

n∑
k=1
xk ≤ c
xk ≥ 0 ∀ k = 1, . . . ,n.

The linear program (14) is known as the fluid model or the deterministic capacity allocation problem. It is
essentially a knapsack problem whose solution can be given in closed form xk = min(µk, (c−µ[1,k−1])+) for
all k = 1, . . . ,n. Consequently, V Un (c) ≤ V Un (c,µ) =

∑n
k=1(pk −pk+1) min(µ[1,k],c).

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

k=1 pkE min(Dk, (c−D[k + 1,n])
+).

Taking expectations we obtain

V Ln (c) =

n∑
k=1

pkE min(Dk, (c−D[k + 1,n])+) (15)

=
n∑
k=1

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

=
n∑
k=1

(pk −pk−1)E min(D[k,n],c) (16)

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

since the expected revenue is computed under sub-optimal protection levels. The above arguments justify the
main result of this section.

Proposition 5
V Ln (c) ≤ Vn(c) ≤ V

U
n (c) ≤ V

U
n (c,µ) (17)

j=1 P(D[1,k] ≥ j). If D[1,k] is Normal we can take advantage of the fact that E min(Z,z) = z(1−Φ(z))−
φ(z) when Z is a standard Normal random variable and φ is the standard Normal density function. If follows
that if D[1,k] is Normal with mean µ and variance σ2, then

E min(D[1,k],c) = µ + σE min(Z,z) = µ + σ [z(1 − Φ(z)) −φ(z)]

where z = (c−µ)/σ.

Tables 6 and 7 report V Ln (c),Vn(c),V
U
n (c) and Vn(c,µ) for the data of Examples 4 and 5, respectively.

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

c V Ln (c) Vn(c) V
U
n (c) V

U
n (c,µ)

80 $42,728 $49,642 $53,039 $53,315
90 $48,493 $54,855 $58,

3 $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 $75,9

$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.

c V Ln (c) Vn(c) V
U
n (c) V

Un(c,µ)
80 $52,462 $67,505 $72,717 $73,312
90 $61,

5 $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

demand, and the revenue that results from not applying booking controls. Demand uncensoring refers to a sta-
tistical technique that attempts to estimate actual demand from the observed sales which may be constrained
by booking limits. The ex-post optimal revenue is a hindsight optimization and is equivalent to our perfect
foresight model, resulting in revenue V Un (c,D), where D is the uncensored demand. On the other hand, the
revenue based on not applying booking controls is just V Ln (c,D), so a measure of the revenue opportunity is
V Un (c,D) − V Ln (c,D). The achieved revenue opportunity is the difference between the actual revenue from
applying optimal or heuristic controls and the lower bound. The ratio of the achieved revenue opportunity to
the revenue opportunity is often called the percentage achieved revenue opportunity. The revenue opportunity
V Un (c,D)−V Ln (c,D) is sometimes approximated by V Un (c)−V Ln (c) to get an idea of the revenue opportunity.
Table 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 Ṽj(x) is
an approximation to Vj(x). Then a heuristic admission control rule can be obtained as follows:

ỹj = max{y ∈N : ∆Ṽj(y) > pj+1} j = 1, . . . ,n− 1. (18)

Suppose we approximate the value function Vj(x) by Ṽj(x) = θV
L
j (x) + (1 − θ)V

U
j (x) for some θ ∈ [0, 1]

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

∆V Lj (x) = p1P(D[1,j] ≥ x) +
∑j

k=2(pk −pk−1)P(D[k,j] ≥ x), while ∆V
U
j (x) =

∑j−1
k=1(pk −pk+1)P(D[1,k] ≥

x) + pjP(D[1,j] ≥ x).

6 Multiple Fare Classes with Arbitrary Fare Arrival Patterns

So far we have suppressed the time dimension; the order of the arrivals has provided us with stages that are
a proxy for time, 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 =
∫ t
0
λjsds. 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(−
∑
k 6=j

λktδt) + o(δt) = 1 −
∑
k 6=j

λktδt + o(δt).

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

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) =
∑
j∈Nt

λjtδt max(pj + V (t−δt,x− 1),V (t−δt,x)) + (1 −
∑
j∈Nt

λjtδt)V (t− δt,x) + o(δt)

= V (t−δt,x) + δt
∑
j∈Nt

λjt[pj − ∆V (t− δt,x)]+ + o(δt) (19)

with boundary conditions V (t, 0) = 0 and V (0,x) = 0 for all x ≥ 0, where ∆V (t,x) = V (t,x)−V (t,x−1) for
x ≥ 1 and t ≥ 0.

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)

∂t
=

∑
j∈Nt

λjt[pj − ∆V (t,x)]+ (20)

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 ← 1aλj,t/a, for t ∈ [0,aT]. The scale factor a should be selected so that, after scaling,

∑
j∈Nt λjt << 1,

e.g.,
∑

j∈Nt λjt ≤ .01 for all t. The resulting dynamic program, after rescaling time, is given by

V (t,x)

= V (t− 1,x) +
∑
j∈Nt

λjt[pj − ∆V (t− 1,x)]+. (21)

with the same boundary conditions. Computing V (t,x) via (21) is quite easy and fairly accurate if time is
scaled appropriately. For each t, the complexity is order O(n) for each x ∈ {1 . . . ,c} so the complexity per
period is O(nc), and the overall computational complexity is O(ncT).

A formulation equivalent to (21) was first proposed by Lee and Hersh [13], who also show that ∆V (t,x) is
increasing in t and decreasing in x. The intuition is that the marginal value of capacity goes up if we have more
time to sell and goes down when we have more units available for sale. From the dynamic program (21), it is
optimal to accept a request for product j when pj ≥ ∆V (t−1,x) or equivalently, when pj + V (t−1,x−1) ≥
V (t−1,x), i.e., when the expecte revenue from accepting the request exceeds the expected revenue of denying
the request. Notice that if it is optimal to accept a request for fare j, then it is also optimal to accept a request

for any higher fare. Indeed, if pk ≥ pj and pj ≥ ∆V (t−1,x), then pk ≥ ∆V (t−1,x). Assuming that the fares
are ordered: p1 ≥ p2 ≥ . . . ≥ pn, then it is optimal to accept all fares in the active set A(t,x) = {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 =
∑n

j=1 λjt be the overall arrival rate at time t. Define πjt =
∑j

k=1 λkt/λt and

rjt =
∑j

k=1 pkλkt/λt. We can think of πjt and rjt as the probability of sale and the average revenue rate,
per arriving customer, when we offer all the fares in 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+t = Nt ∪{0}. With this notation, we can write formulation (21) as

V (t,x)

= V (t− 1,x) + λt max
j∈N+

[rjt −πjt∆V (t− 1,x)]

= V (t− 1,x) + λt max
j∈N+
t

πjt[qjt − ∆V (t− 1,x)]. (22)

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)

∂t
= λt max

j∈N+
t

πjt[qjt − ∆V (t,x)]. (23)

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.

7 Compound Poisson Demands

The formulations of the dynamic programs (20) and (21), implicitly assume that each request is for a single
unit. Suppose instead, that each arrival is for a random number of units. More specifically, suppose that
request for fare j are of random size Zj, and that the probability mass function Pj(z) = P(Zj = z),z ≥ 1 is
known for each j. As before, we assume independent demands for the different fare classes j ∈ N. We seek
to generalize the dynamic programs (20) and (21) so that at each state (t,x) we can decide whether or not to
accept a fare pj request of size Zj = z. The expected revenue from accepting the request is zpj +V (t−1,x−z)
and the expected revenue from rejecting the request is V (t− 1,x). Let ∆zV (t,x) = V (t,x) −V (t,x− z) for
all z ≤ x and ∆zV (t,x) = ∞ if z > x. We can think of ∆zV (t,x) as a the sum of the the z marginal values
∆V (t,x) + ∆V (t,x− 1) + . . . + ∆V (t,x−z + 1).

The dynamic program (20) with compound Poisson demands is given by

∂V (t,x)
∂t
=
∑
j∈Nt

λjt

∞∑

z=1

Pj(z)[zpj − ∆zV (t,x)]+, (24)

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

V (t,x) = V (t− 1,x) +
∑
j∈Nt
λjt

∞∑
z=1

Pj(z)[zpj − ∆zV (t− 1,x)]+, (25)

with boundary conditions V (t, 0) = V (0,x) = 0. Notice that the sums in (24) and (25) can be changed to
∑x

z=1

instead of
∑∞

z=1 as the terms z > x do not contribute to the sum given our convention that ∆zV (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∈Nt λjt << 1 for all t ∈ [0,T].

For compound Poisson demands, we can no longer claim that the marginal value of capacity ∆V (t,x) is
decreasing in x, although it is still true that ∆V (t,x) is increasing in t. To see why ∆V (t,x) is not monotone in
x, consider a problem where the majority of the requests are for two units and request are seldom for one unit.
Then the marginal value of capacity for even values of x may be larger than the marginal value of capacity for
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.

c 50 100 150 200 250 300
V (T,c) $3,837 $6,463 $8,451 $10,241 $11,724 $12,559

x 1 2 3 4 5 6
∆V (t,x) 70.05 66.48 59.66 60.14 54.62 50.41

Table 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 =
∫ T
0
λjtdt. Suppose each arrival is of random size Zj. Then the aggregate demand, say Dj, for

fare j is equal to

Dj =

Nj∑
k=1

Zjk, (26)

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[Z

2
j ] = ΛjE[Z

2
j ], where E[Z

2
j ] is the second moment of Zj. Notice that Jensen’s inequality

implies that the the variance to mean ratio E[Z2j ]/E[Zj] ≥ 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 yb1, . . . ,y

b
n−1. Since the compound Poisson demands are

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 αjβ

2
j = E[Nj]E[Z

2
j ], yielding

αj = ΛjE[Zj]
2/E[Z2j ], and βj = E[Z

2
j ]/E[Zj].

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 yb1,y

b
2, . . . ,y

b
n−1 be the protection levels computed using the EMSR-b heuristic. Protection

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
yb1, . . . ,y

b
n−1 with theft nesting is not, in general, equal to V

b
n (c), the expected revenue when the arrivals are

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

V s(t,x) =
∑
j∈Nt

λjt[

x−(yj−1−b[1,j−1])+∑
z=1

Pj(z)(zpj + V
s(t− 1,x−z))

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

λjt)V
s(t− 1,x)

≤
∑
j∈Nt

λjt[
x−(yj−1−b[1,j−1])+∑
z=1

Pj(z)(zpj + V (t− 1,x−z))

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

λjt)V (t− 1,x)

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

Pj(z)(zpj − ∆zV (t− 1,x))

≤ V (t− 1,x) +
∑
j∈Nt

λjt
∞∑
z=1

Pj(z)(zpj − ∆zV (t− 1,x))+

= V (t,x),

where the first equation follows from the application of the protection level policy, the first inequality follows
from the inductive hypothesis V s(t − 1,x) ≤ V (t − 1,x). The second equality collects terms, the second
inequality follows because we are taking positive parts, and the last equality from the definition of V (t,x).

While we have shown that V s(T,c) ≤ V (T,c), one may wonder whether there are conditions where equality
holds. The following results answers this question.

Corollary 2 If the Dj’s are independent Poisson random variables and the arrivals are low-to-high then
Vn(c) = V

s(T,c) = V (T,c).

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.

c 50 100 150 200 250 300
V b(T,c) $3,653 $6,177 $8,187 $9,942 $11,511 $12,266
V (T,c) $3,837 $6,463 $8,451 $10,241 $11,724 $12,559

Gap 4.8% 4.4% 3.1% 2.9% 1.8% 2.3%

Table 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 Un (c) for Vn(c), developed in Section 5 for the static multi-fare
model is still valid for V (T,c). The random revenue associated with the perfect foresight model is Vn(c,D)
and can be obtained by solving the linear program (11). Notice that for all sample paths, this revenue is
at least as large as the revenue for the dynamic policy. Taking expectations we obtain V (T,c) ≤ V Un (c) =
EVn(c,D) =

∑n
k=1(pk − pk+1)E min(D[1,k],c), where for convenience pn+1 = 0. Moreover, since dynamic

policies do at least as well as static policies, the lower bounds obtained in Section 5 also apply to dynamic
policies.

8 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) =

k∑

i=1

λit[pi + Vk(t− 1,x− 1)] + (1 −
k∑

i=1

λit)Vk(t− 1,x)

= 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 =
∑k

i=1 λit and rkt =
∑j

i=1 piλit and pkt = rkt/πkt when
πkt > 0 and pkt = 0 otherwise.

Then Vj(t,x) satisfies the dynamic program

Vj(t,x) = max
k≤j

Wk(t,x) = max{Wj(t,x),Vj−1(t,x)} (

)

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 more
stages to go.

c) ∆Vj(t,x) is increasing in t, so the marginal value of capacity increases as the time-to-go increases.

Let
aj(t,x) = max{k ≤ j : Wk(t,x) = Vj(t,x)}.

In words, aj(t,x) is the index of the lowest open fare that is optimal to post at state (t,x) if we are allowed
to use any fares in 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

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

∑5
i=1 Λi/a < .01 resulting in T ← aT = 2, 800. We also

assume that the arrival rates are uniform over the horizon [0,T], i.e., λj = Λj/T. In Table 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.

c 50 100 150 200 250 300 350
V5(c) 3,426.8 5,441.3 7,188.7 8,159.1 8,909.1 9,563.9 9,625.0

V (T,c) 3,553.6 5,654.9 7,410.1 8,390.6 9,139.3 9,609.6 9,625.0
V5(T,c) 3,494.5 5,572.9 7,364.6 8,262.8 9,072.3 9,607.2 9,625.0
V4(T,c) 3,494.5 5,572.9 7,364.6 7,824.9 7,825.0 7,825.0 7,825.0
V3(T,c) 3,494.5 5,572.9 5,900.0 5,900.0 5,900.0 5,900.0 5,900.0
V2(T,c) 3,494.5 3,900.0 3,900.0 3,900.0 3,900.0 3,900.0 3,900.0
V1(T,c) 1,500.0 1,500.0 1,500.0 1,500.0 1,500.0 1,500.0 1,500.0

Table 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)

∂t
=

{
rjt −πjt∆Vj(t,x) if (t,x) /∈ Zj−1
∂Vj−1(t,x)

∂t
if (t,x) ∈ Zj−1

(28)

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.

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.

9 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.

10 Appendix

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

j≤y−1 G(j)P(X = j), while g(y − 1) = G(y −
1)P(X ≥ y) +

∑
j≤y−1 G(j)P(X = j). Taking the difference yields ∆g(y) = G(y)P(X ≥ y). Notice that

r(y) = R(y)P(X < y) + ∑

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

j≥y R(j)P(X = j).
Taking the difference we see that ∆r(y) = ∆R(y)P(X < y).

Proof of Proposition 1. Let G(y) = p1y, then V1(y) = g(y) = EG(min(D1,y)), so ∆V1(y) = ∆g(y) =
p1P(D1 ≥ y). This establishes the first part of the Proposition. To establish the second part of the Proposition
we use the first part of Lemma 1 to show that p2E min(D2,c−y)−p2E min(D2,c−y + 1) = −p2P(D2 > c−y)
and the second part of the Lemma 1 to show that EV1(max(x − D2,y)) − EV1(max(x − D2,y − 1)) =
∆V1(y)P(D2 > c−y). The second part of the Proposition then follows from putting the two parts together. To
see the first part, let r(y) = p2c−p2E max(c−D2,y), then ∆r(y) = p2E min(D2,c−y)−p2E min(D2,c−y+1) =
∆R(y)P(c − D2 < y) where R(y) = −p2y, so ∆r(y) = −p2P(c − D2 < y) = −p2P(D2 > c − y). Now let
R(y) = V1(y), then ∆r(y) = ∆V1(y)P(c−D2 < y) = ∆V1(y)P(D2 > c−y) completing the proof.

Proof of Lemma 2: We will prove the 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
y≤x

{Wj (y,x))} ,

where for any y ≤ x,

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

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
y≤x

Wj(y,x) = Wj(yj−1,x)

On the other hand, if x < yj−1 then

Vj(x) = max
y≤x

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

In summary,

Vj(x) = Wj(min(x,yj−1),x)

{
Vj−1(x), if x ≤ yj−1
E [pj min{Dj,x−yj−1}] + E [Vj−1 (max{x−Dj,yj−1})] if x > yj−1

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

∆Vj(x) =

{
∆Vj−1(x), if x ≤ yj−1
E min(pj, ∆Vj−1(x−Dj)) if x > yj−1

(30)

We will now use this result to show that ∆Vj(x) is itself decreasing in x. Since ∆Vj(x) = ∆Vj−1(x) for
x ≤ yj−1 and ∆Vj−1(x) is decreasing in x we only need to worry about the case x > yj−1.

However, in this case
∆Vj(x) = E min(pj, ∆Vj−1(x−Dj))

is decreasing in x since ∆Vj−1(x) is itself decreasing in x.

We now show that ∆Vj(x) ≥ ∆Vj−1(x). For x > yj−1 we have min(pj, ∆Vj−1(x−Dj)) ≥ min(pj, ∆Vj−1(x)) =
∆Vj−1(x) where the inequality follows since ∆Vj−1(x) is decreasing in x and the equality since x > yj−1. Tak-
ing expectations we see that ∆Vj(x) ≥ ∆Vj−1(x) on x > yj−1 while ∆Vj(x) = ∆Vj−1(x) on x ≤ yj−1.

Proof of Proposition 2. We first show that ∆Vj+1(x) = ∆Vj(x) for x = 1, . . . ,yj. This follows since
Vj+1(x) = Wj+1(x,x) = Vj(x) in this range as no capacity is made available for fare pj+1 when x ≤ yj. For
x > yj

Vj+1(x) = pj+1E min(x−yj,Dj+1) + EVj(max(x−Dj+1,yj)

= pj+1

x−yj∑
k=1

Pr(Dj+1 ≥ k) + Vj(yj)Pr(Dj+1 > x−yj)

x−yj∑

k=0

Vj(x−k)Pr(Dj+1 = k).

Consequently, for x > yj

∆Vj+1(x) = pj+1Pr(Dj+1 ≥ x−yj) +
x−yj−1∑

k=0

∆Vj(x−k)Pr(Dj+1 = k) (31)

follows from Vj(x − 1) = Vj(yk) for x = yj + 1. Since pj+1 < ∆Vj(y) for y ≤ yj, we can write ∆Vj+1(x) =∑∞ k=0 min(pj+1, ∆Vj(x−k))P(Dj+1 = k) = E min(pj+1, ∆Vj(x−Dj+1)).

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

Proof of Lemma 3: We will first show part that ∆Vj(t,x) is decreasing in x which is equivalent to showing
that 2Vj(t,x) ≥ Vj(t,x + 1) + Vj(t,x− 1)] for all x ≥ 1. Let A be an optimal admission control rule starting
from state (t,x + 1) and let B be an optimal admission control rule starting from (t,x− 1). These admission
control rules are mappings from the state space to subsets Sk = {1, . . . ,k},k = 0, 1, . . . ,j where S0 = ∅ is
the optimal control whenever a system runs out of inventory. Consider four systems: two starting from state
(t,x), using control rules A′ and B′, respectively, and one each starting from (t,x + 1) and (t,x − 1), using
control rule A and B, respectively. Our goal is to specify heuristic control rules A′ and B′ that together make
the expected revenues of the two systems starting with (t,x) at least as large as the expected revenues from
the systems starting at (t,x + 1) and (t,x− 1). This will imply that 2Vj(t,x) ≥ Vj(t,x + 1) + Vj(t,x− 1).

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).

11 Terminology

Item Description

c capacity

p price

s salvage value on excess capacity after the arrival of full-fare demand (e.g. last-
minute travel specials).

x remaining capacity, i.e. x ∈{0, 1, …,c}
E expected value operator

j fare class identifier where price decreases with increasing class index, i.e. p1 ≥
p2 ≥ .. ≥ pj

pj demand weighted average fare class price

D fare class demand

y protection level for a fare class

ya protection level obtained using EMSRa heuristic

yb protection level obtained using EMSRb heuristic

W(y,c) revenue given protection level y and capacity c

r ratio of discount to full-fare price

F cumulative density function (CDF) of continuous demand

F−1 inverse density function of continuous demand

b booking limit for sales of discount fares

Vj(x) optimal expected revenue for classes {1, 2, …,j} from capacity x
ρ penalty cost for each unit of full-fare demand that is rejected

sj observed sales for fare j in Monte Carlo simulation

Sk Set of highest k fares {1, . . . ,k}
a(t,x) highest opened fare class at state (t,x)

Zj 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

References

[1] Bathia, A. V. and S. C. Prakesh (1973) “Optimal Allocation of Seats by Fare,” Presentation by TWA
Airlines to AGIFORS Reservation Study Group.

[2] Belobaba, P. P. (1987) “Airline Yield Management An Overview of Seat Inventory Control,” Transporta-
tion Science, 21, 63–73.

[3] Belobaba, P. P. (1989) “Application of a Probabilistic Decision Model to Airline Seat Inventory Control,”
Operations Research, 37, 183–197.

[4] Brumelle, S. L., J. I. McGill, T. H. Oum, M. W. Tretheway and K. Sawaki (1990) “Allocation of Airline
Seats Between Stochastically Dependent Demands,” Transporation Science, 24, 183-192.

[5] Brumelle, S. L. and J. I. McGill (1993) “Airline Seat Allocation with Multiple Nested Fare Classes,”
Operations Research, 41, 127-137.

[6] Curry, R. E. (1990) “Optimal Airline Seat Allocation with Fare Nested by Origins and Destinations,”
Transportation Science, 24, 193–204.

[7] Diwan, S. (2010) “Performance of Dynamic Programming Methods in Airline Revenue Management.”
M.S. Dissertation MIT. Cambridge, MA.

[8] Chandler, S. and Ja, S. (2007) Revenue Opportunity Modeling at American Airlines. AGIFORS Reser-
vations and Yield Management Study Group Annual Meeting Proceedings; Jeju, Korea.

[9] Feng, Y. and G. Gallego (1995) “Optimal Starting Times of End-of-Season Sales and Optimal Stopping
Times for Promotional Fares,” 41, 1371-1391.

[10] Feng, Y. and G. Gallego (2000) “Perishable asset revenue management with Markovian time-dependent
demand intensities,” Management Science, 46(7): 941-956.

[11] Feng, Y. and B. Xiao (2000) “Optimal policies of yield management with multiple predetermined prices,”
Operations Research, 48(2): 332-343.

[12] Gallego, G., S. Kou and Phillips, R. (2008) “Revenue Management of Callable Products,” Management
Science, 54(3): 550564.

[13] Lee. C. T. and M. Hersh. (1993) “A Model for Dynamic Airline Seat Inventory Control with Multiple
Seat Bookings,” Transportation Science, 27, 252–265.

[14] Littlewood, K. (1972) “Forecasting and Control of Passenger Bookings.” 12th AGIFORS Proceedings:
Nathanya, Israel, reprinted in Journal of Revenue and Pricing Management, 4(2): 111-123.

[15] Pfeifer, P. E. (1989) “The Airline Discount Fare Allocation Problem,” Decision Science, 20, 149–157.

[16] Ratliff, R. (2005) “Revenue Management Demand Distributions”, Presentation at AGIFORS Reservations
and Yield Management Study Group Meeting, Cape Town, South Africa, May 2005.

[17] Richter, H. (1982) “The Differential Revenue Method to Determine Optimal Seat Allotments by Fare
Type,” In Proceedings 22nd AGIFORS Symposium 339-362.

[18] Robinson, L.W. (1995) “Optimal and Approximate Control Policies for Airline Booking With Sequential
Nonmonotonic Fare Classes,” Operations Research, 43, 252-263.

[19] Simpson, R. (1985) “Theoretical Concepts for Capacity/Yield Management,” In Proceedings 25th AGI-
FORS Symposium 281-293.

[20] Talluri, K and G. van Ryzin (2004) “The Theory and Practice of Revenue Management,” Springer Sci-
ence+Business Media, New York, USA, pg. 42.

[21] Tematha, C., S. Polt, and and L. Suhi. (2010) “On the robustness of the network-based revenue oppor-
tunity model,” Journal of Revenue and Pricing Management , Vol. 9, 4, 341355.

[22] Wollmer, R. D. (1992) “An Airline Seat Management Model for a Single Leg Route When Lower Fare
Classes Book First,” Operations Research, 40, 26-37.

[23] Zhao, W. and Y-S Zheng (2001) “A Dynamic Model for Airline Seat Allocation with Passenger Diversion
and No-Shows,” Transportation Science,35: 80-98.

IEOR

: Dynamic Pricing and Revenue Management
Midterm

010

Professor Guillermo Gallego

INSTRUCTIONS:

1. Write all your answers on the question sheets. Answers written on blue books will not be
evaluated. Show and explain your work. Points will be deducted for solutions that are not
clearly explained.

2. Write your name, sign the honor pledge and indicate your degree program.

HONOR PLEDGE: I pledge that I have neither given nor received unauthorized aid during this
examination.

Name: Signature: Degree Program :

1. Static Model Independent Demands For the multi-fare problem with low-to-high arrivals
we have shown that the optimal protection levels for fares {j, . . . , 1} is given by

yj = max{y ∈N : ∆Vj(y) > pj+1} j = 1, . . . ,n− 1.

a) What is the definition of ∆Vj(y)?

b) What is the interpretation of ∆Vj(y)?

c) Is ∆Vj(y) monotone increasing in y?

d) Is ∆Vj(y) monotone increasing in j?

e) Write an explicit expression for ∆V1(y)?

f) Is this formula true or false?: ∆V2(x) = max{∆V1(x),E min(p1P(D[1, 2] ≥ x),p2)}

g) For general j what is the approximation for ∆Vj(y) implied by the EMSR-b heuristic?

2. Dynamic Allocation Model: Independent Demands Consider the dynamic capacity
allocation model

V (t,x) = V (t− 1,x) +
n∑

j=1

λjt[pj − ∆V (t− 1,x)]+.

a) What is λjt?

b) What is the interpretation of ∆V (t− 1,x)?

c) Which fares will you accept at state (t,x)?

d) What happens to a(t,x) = max{j : pj ≥ ∆V (t− 1,x)} as t increases?

e) What happens to a(t,x) as x increases?

f) List three disadvantages of this model?

3. Dynamic Model: Independent Demands Monotonic Fares Consider the dynamic ca-
pacity allocation model

Vj(t,x) = max
k≤j

Wk(t,x)

where Wk(t,x) =
∑k

i=1 λit[pi + Vk(t− 1,x− 1)] + (1 −
∑k

i=1 λit)Vk(t− 1,x).

a) What is the interpretation of Vj(t,x)?

b) What is the interpretation of Wk(t,x)?

c) Show that Wk(t,x) = Vk(t− 1,x) +
∑k

i=1 λit[pi − ∆Vk(t− 1,x)].

d) What happens to fares {k + 1, . . . ,j} if Vj(t,x) = Wk(t,x) for some k < j?

e) Suppose Vj(t,x) = Wk(t,x) for k < j. What can you say about Vj−1(t,x)?

f) How would you modify V1(t,x) to impose a penalty for denying capacity to high fare
customers?

4. Choice Models and Dynamic Model with Dependent Demands Consider a choice
model where for each set S ⊂ Sn = {1, . . . ,n}, π(S) =

∑
i∈S πi(S) and r(S) =

∑
i∈S piπi(S).

a) What is the interpretation of π(S)?

b) What is the interpretation of r(S)?

c) What is the interpretation of q(S) = r(S)/π(S)?

d) Give formulas for πj,rj and qj and uj = (rj −rj−1)/(πj −πj−1) for S = Sj = {1, . . . ,j}
for the MNL model.

e) Consider the dynamic program V (t,x) = V (t−1,x) + λt max0≤k≤n[rk −πk∆V (t−1,x)]
for the MNL model. For what states (t,x) are you better off offering Sj rather than Sj−1?

5. Static Model with Dependent Demands Consider a choice model as in Problem 4 and
suppose that there are D potential customers, where D is a random variable taking positive
integer values. Suppose you want to allocate capacity between sets Sj+1 and Sj by protecting
y units of capacity for sales under set Sj.

a) What is the total demand, say Dj+1, for fares {1, . . . ,j + 1} when set Sj+1 is offered?

b) What are the total sales under action Sj+1 if c units of capacity are available and y units
are protected for sales under set Sj?

c) What is the expected revenue under action Sj+1?

d) Write an expression for sales under Sj if Dj+1 < c−y, if Dj+1 > c−y.

e) Write an expression for the expected revenue under action Sj when Dj+1 > c−y

f) Write an expression for the expected revenue under both actions Sj and Sj+1.

IEOR

: Dynamic Pricing and Revenue Management
Midterm

010

Professor Guillermo Gallego

INSTRUCTIONS:

1. Write all your answers on the question sheets. Answers written on blue books will not be
evaluated. Show and explain your work. Points will be deducted for solutions that are not
clearly explained

2. Write your name, sign the honor pledge and indicate your degree program.

HONOR PLEDGE: I pledge that I have neither given nor received unauthorized aid during this
examination.

Name: Signature: Degree Program :

1. Static Model Independent Demands For the multi-fare problem with low-to-high arrivals
we have shown that the optimal protection levels for fares {j, . . . , 1} is given by

yj = max{y ∈N : ∆Vj(y) > pj+1} j = 1, . . . ,n− 1.

a) What is the definition of ∆Vj(y)? Definition: ∆Vj(y) = Vj(y) −Vj(y − 1) for y ≥ 1.
b) What is the interpretation of ∆Vj(y)? Interpretation: Expected marginal seat value.

c) Is ∆Vj(y) monotone increasing in y?

No: It is decreasing in y.

d) Is ∆Vj(y) monotone increasing in j?

Yes.

e) Write an explicit expression for ∆V1(y)?

∆V1(y) = p1P(D1 ≥ y).
f) Is this formula true or false?: ∆V2(x) = max{∆V1(x),E min(p1P(D[1, 2] ≥ x),p2)}

Yes

g) For general j what is the approximation for ∆Vj(y) implied by the EMSR-b heuristic?

∆Vj(y) ‘ qjP(D[1,j] ≥ y) where qj =
∑j
k=1 pkE[Dk]/ED[1,j].

2. Dynamic Allocation Model: Independent Demands Consider the dynamic capacity
allocation model

V (t,x) = V (t− 1,x) +
n∑
j=1

λjt[pj − ∆V (t− 1,x)]+.

a) What is λjt? It is the arrival rate of fare j at time t.

b) What is the interpretation of ∆V (t− 1,x)?
It is the marginal value of the xth unit of capacity at time t− 1.

c) Which fares will you accept at state (t,x)?

Accept those fares in the set {j : pj ≥ ∆V (t− 1,x)}.
d) What happens to a(t,x) = max{j : pj ≥ ∆V (t− 1,x)} as t increases?

As the time-to-go increases a(t,x) decreases.

e) What happens to a(t,x) as x increases?

As capacity increases a(t,x) increases.

f) List three disadvantages of this model?

Allows fares to open and reopen. Assumes Poisson arrival rates. Assumes independent
demands.

. Dynamic Model: Independent Demands Monotonic Fares Consider the dynamic ca-
pacity allocation model

Vj(t,x) = max
k≤j

Wk(t,x)

where Wk(t,x) =
∑k
i=1 λit[pi + Vk(t− 1,x− 1)] + (1 −

∑k
i=1 λit)Vk(t− 1,x).

a) What is the interpretation of Vj(t,x)?

It is the maximum expected revenue from state (t,x) where only fares {j, . . . , 1} are
allowed and when closing a fare forces it to remain closed until the end of the horizon.

b) What is the interpretation of Wk(t,x)?

It is the maximum expected revenue of using action Sk at state (t,x) and then following
an optimal policy thereafter.

c) Show that Wk(t,x) = Vk(t− 1,x) +
∑k
i=1 λit[pi − ∆Vk(t− 1,x)].

This is simple algebra reorganizing the terms in the definition of Wk(t,x).

d) What happens to fares {k + 1, . . . ,j} if Vj(t,x) = Wk(t,x) for some k < j? All such fares are closed and will remain closed.

e) Suppose Vj(t,x) = Wk(t,x) for k < j. What can you say about Vj−1(t,x)?

Then Vj−1(t,x) = Vk(t,x).

f) How would you modify V1(t,x) to impose a penalty for denying capacity to high fare
customers?

I would write V1(t,x) = p1E min(D1(t),x) −ρE(D1(t) −x)+.

4. Choice Models and Dynamic Model with Dependent Demands Consider a choice
model where for each set S ⊂ Sn = {1, . . . ,n}, π(S) =

∑
i∈S πi(S) and r(S) =

∑
i∈S piπi(S).

a) What is the interpretation of π(S)?

It is the probability that the customer selects an item in S rather than the outside
alternative.

b) What is the interpretation of r(S)?

It is the expected revenue when S is offered.

c) What is the interpretation of q(S) = r(S)/π(S)?

It is the average fare assuming something is sold.

d) Give formulas for πj,rj and qj and uj = (rj −rj−1)/(πj −πj−1) for S = Sj = {1, . . . ,j}
for the MNL model.

πj = π(Sj),rj = r(Sj). For the MNL model uj =
pj−rj−1
1−πj−1

e) Consider the dynamic program V (t,x) = V (t−1,x) + λt max0≤k≤n[rk −πk∆V (t−1,x)]
for the MNL model. For what states (t,x) are you better off offering Sj rather than Sj−1?

For states (t,x) such that ∆V (t,x) ≤ uj.

a) What is the total demand, say Dj+1, for fares {1, . . . ,j + 1} when set Sj+1 is offered? It
is binomial with parameters D and πj+1.

b) What are the total sales under action Sj+1 if c units of capacity are available and y units
are protected for sales under set Sj? min(c−y,Dj+1).

c) What is the expected revenue under action Sj+1?

It is qj+1E min(c−y,Dj+1).
d) Write an expression for sales under Sj if Dj+1 < c−y, if Dj+1 > c−y.

If Dj+1 < c−y then we run out of customers and sales under Sj will be zero. Otherwise sales under Dj will be min(y,Uj(y)) where Uj(y) is binomial with parameters (Dj+1 − c + y)+, θj = πj/πj+1.

e) Write an expression for the expected revenue under action Sj when Dj+1 > c − y It is
qjE min(y,Uj(y)).

f) Write an expression for the expected revenue under both actions Sj and Sj+1.

Just add: qj+1E min(c−y,Dj+1) + qjE min(y,Uj(y)).

IEOR

: Dynamic Pricing and Revenue Management
Midterm

011

Professor Guillermo Gallego

INSTRUCTIONS:

1. Write all your answers on the question sheets. Answers written on blue books will not be
evaluated. Show and explain your work. Points will be deducted for solutions that are not
clearly explained

2. Write your name, sign the honor pledge and indicate your degree program.

HONOR PLEDGE: I pledge that I have neither given nor received unauthorized aid during this
examination.

Name: Signature: Degree Program :

1. Static Model Independent Demands Consider the two fare problem with capacity c, fares
p2 < p1 and independent demand D2 and D1 with fare 2 demand arriving first.

a) Suppose you have perfect foresight and can anticipate the value of D1. How many units
would you protect for fare 1?

b) Give an expression for the optimal revenue revenue for the two fare problem assuming
perfect foresight.

c) Give an expression for the optimal revenue for the n-fare perfect foresight model.

d) How would you modify the formula for the case n = 2 if there is known demand D0 at
salvage value s < p2?

2. Dynamic Allocation Model: Independent Demands Consider the dynamic capacity
allocation model

V (t,x) = V (t− 1,x) +
n∑

j=1

λjt[pj − ∆V (t− 1,x)]+.

Let πtj =
∑j

k=1 λkt and rtj =
∑j

k=1 pkλkt.

a) Argue that fare j should be closed at state (t,x) whenever

∆V (t− 1,x) > pj.

Used this to show that the optimal set of fares that should be open at state (t,x) is
Sa(t,x) = {1, . . . ,a(t,x)} where a(t,x) = max{j : pj ≥ ∆V (t− 1,x)}.

b) What happens to the set of open fares Sa(t,x) if x increases? If t increases?

c) Let Vn(c) be the value function for the independent demand model that suppresses time
and assumes that the fares arrive from low-to-high. Consider now a dynamic model with
arrival rates λtj = E[Dj]/T, 0 ≤ t ≤ T for all j = 1, . . . ,n. Argue intuitively why the
value function V (T,c) for this model should be at least as large as Vn(c).

d) Suppose you are an airline that leases planes on a day to day basis. For a certain route
you are offered the choice of leasing a plane with capacity c = 100 or a plan where half of
the time you operate an aircraft with capacity 90 and half of the time one with capacity
110. If the two plans are equally costly to lease and operate, which plan would you prefer?

Since V (T,c) is concave in c it is better to have a plane with capacity c = 100.

. Efficient Sets Finding efficient fares for a general choice model requires solving parametric
linear programs with exponentially many variables, one for each of the 2n subsets of {1, . . . ,n}.
A heuristic is given below. The heuristic assumes that for any subset S ⊂{1, . . . ,n} we know
π(S) =

∑
k∈S πk(S) and r(S) =

∑
k∈S pkπk(S), with π(∅) = r(∅) = 0.

Algorithm

– Step 1. Set S0 = ∅ and find the set of size k = 1 that maximizes

r(S) −r(∅)
π(S) −π(∅)

Label this set S1 and set k = 2.

– Step 2. While k ≤ n do: find the set of size k that maximizes

r(S) −r(Sk−1)
π(S) −π(Sk−1)

label the set Sk and set k = k + 1.

S π(S) r(S)
∅ 0% $0
{1} 10.0% $10.00
{2} 15.0% $12.00
{3} 25.0% $15.00
{12} 20.0% $17.50
{13} 30.0% $21.00
{23} 35.00% $23.50
{123} 42.50% $31.00

Table 1: Data for Problem 3

a) Apply the heuristic to the data set in Table 1 and find the sets S1,S2,S3.

b) Suppose someone claims that set {1, 3} is efficient. Prove them wrong by finding a convex
combination of sets S2 and S3 (obtained by the algorithm) that consumes capacity with
probability 30% but produces revenue higher than $21.00.

c) What convex combination of S1,S2,S3 would you use to maximize revenues if you can
consume capacity with probability 15%? I would put weights .5, .5 and 0. I would then
consume capacity with probability .15 and have 13.75 in revenues.

4. Static Model Dependent Demands Suppose there are two fares p1 > p2 for a flight with
capacity c = 100 and that demands are derived from a basic attraction model (BAM) where
fare 1 has attractiveness v1 = 1, fare 2 has attractiveness v2 = 1 and the outside alternative
has attractiveness v0 = 1.

a) Suppose that the number of potential customers is Poisson with parameter λ = 200.
What is the distribution of demand under action S1 = {1}? Under action S2 = {1, 2}?

b) Suppose you offer action S2. What is the probability that an arriving customer will buy
fare 1? fare 2?

c) What is the average revenue per arriving customer under action S2? What is the average
revenue per sale under action S2?

d) Suppose we protect y units for action S1. Write an expression for the expected revenue
under action S2 before it is closed?

e) What is the distribution of demand under action S1 once action 2 is closed? Give this in
terms of D2, c and y.

f) What is the expected revenue under action S1?

5. Dynamic Choice Models with Dependent Demands Consider a choice model where for
each set S ⊂ Sn = {1, . . . ,n}, π(S) =

∑
i∈S πi(S) and r(S) =

∑
i∈S piπi(S).

a) Write down the formula for the value function V (t,x) assuming you do not know which
sets are efficient.

b) At state (t,x), what problem do you need to solve to figure out which set to offer?

c) Suppose the efficient fares are S0,S1, . . . ,Sn and that you are given a graph of (πj,rj),j =
1, . . . ,n where πj = π(Sj) and rj = r(Sj). How would you visually obtain the optimal
set to offer at state (t,x)?

d) Continue with the logic of part d) to argue why you may need to open fares as t decreases
for fixed x and close fares as x decreases for fixed t.

6. Deterministic Static Pricing with Two Market Segments

Suppose there are two market segments (leisure and business) and that demand from the
business segment is D1 = 60 at p1 = $1, 000. Suppose further that the demand from the
leisure segment is deterministic and price sensitive and given by D2(p) = 600 −p for values of
p ∈{0, 1, . . . , 600}. Assume capacity is c = 100.

a) Suppose that demand from the leisure segment arrives before demand from the business
segment. How many units will you protect for the business sector? How would you set
the discounted fare? How much money would you make under this scheme?

b) Reconsider the revenue calculations of part a) under the assumption that demand from
both segments arrive at the same time with both segments competing for the capacity
allocated at the discounted fare. Assume that each customer willing to buy the discounted
fare is equally likely to get it. Business customers unable to obtain capacity at the
discounted fare will then buy capacity at p1. Hint: Remember to apply the booking limit
on sales at the discounted fare. Is there a better booking limit?

c) Suppose now that we set the discounted fare at p = 450 under the assumptions of part
b). How many leisure customers will be willing to buy at this fare? How many business
customers would be willing to buy at this fare?

to the business segment with probability 60/210 = 2/7.

d) Suppose that each customer willing to buy at p = 450 is equally likely to get it and
that capacity is non transferable. How much capacity will you make available for the
discounted fare to maximize revenues assuming that business customer unable to secure
capacity at $450 will buy at $1,000? How much money would you make?

e) What happens to the solution of part d) if capacity is transferable and there is a secondary
market where leisure customers who secured capacity at $450 can sell it to business
customers for $800?

IEOR

: Dynamic Pricing and Revenue Management
Midterm

011

Professor Guillermo Gallego

INSTRUCTIONS:

1. Write all your answers on the question sheets. Answers written on blue books will not be
evaluated. Show and explain your work. Points will be deducted for solutions that are not
clearly explained

2. Write your name, sign the honor pledge and indicate your degree program.

HONOR PLEDGE: I pledge that I have neither given nor received unauthorized aid during this
examination.

Name: Signature: Degree Program :

1. Static Model Independent Demands Consider the two fare problem with capacity c, fares
p2 < p1 and independent demand D2 and D1 with fare 2 demand arriving first.

a) Suppose you have perfect foresight and can anticipate the value of D1. How many units
would you protect for fare 1?

Answer: y = min(c,D1)

b) Give an expression for the optimal revenue revenue for the two fare problem assuming
perfect foresight.

Answer: p1 min(D1,c) + p2 min(D2, (c−D1)+)
c) Give an expression for the optimal revenue for the n-fare perfect foresight model.

Answer:
∑n

k=1 pk min(Dk, (c−D[1,k − 1])
+)

d) How would you modify the formula for the case n = 2 if there is known demand D0 at
salvage value s < p2?

Answer: p1 min(D1,c) + p2 min(D2, (c−D1)+) + s min(D0, (c−D1 −D2)+)

2. Dynamic Allocation Model: Independent Demands Consider the dynamic capacity
allocation model

V (t,x) = V (t− 1,x) +
n∑

j=1

λjt[pj − ∆V (t− 1,x)]+.

a) Argue that fare j should be closed at state (t,x) whenever

∆V (t− 1,x) > pj.

Used this to show that the optimal set of fares that should be open at state (t,x) is
Sa(t,x) = {1, . . . ,a(t,x)} where a(t,x) = max{j : pj ≥ ∆V (t− 1,x)}.
Answer: If a fare with ∆V (t−1,x) > pj is accepted it will contributes negatively towards
the expected revenue. The only fares that contribute positievly are those in Sa(t,x).

b) What happens to the set of open fares Sa(t,x) if x increases? If t increases?

Answer: As x increases a(t,x) increases so more fare should be open. If t increases then
a(t,x) decreases so fewer fares should be open.

Answer: The arrival rates are now uniform so we have an opportunity to sell higher fares
from the outset. This should intuitvely improve revenues.

Since V (T,c) is concave in c it is better to have a plane with capacity c = 100.

∑
k∈S πk(S) and r(S) =

∑
k∈S pkπk(S), with π(∅) = r(∅) = 0.

Algorithm

– Step 1. Set S0 = ∅ and find the set of size k = 1 that maximizes

r(S) −r(∅)
π(S) −π(∅)

Label this set S1 and set k = 2.

– Step 2. While k ≤ n do: find the set of size k that maximizes

r(S) −r(Sk−1)
π(S) −π(Sk−1)

label the set Sk and set k = k + 1.

S π(S) r(S)
∅ 0% $0
{1} 10.0% $10.00
{2} 1

.0% $12.00
{3} 25.0% $15.00
{12} 20.0% $17.50
{13} 30.0% $21.00
{23} 35.00% $23.50
{123} 42.50% $31.00

Table 1: Data for Problem 3

a) Apply the heuristic to the data set in Table 1 and find the sets S1,S2,S3.

Answer: The procedure selects first S1 = {1}, then S2 = {1, 2} and finally S3 = {1, 2, 3}.
b) Suppose someone claims that set {1, 3} is efficient. Prove them wrong by finding a convex

combination of sets S2 and S3 (obtained by the algorithm) that consumes capacity with
probability 30% but produces revenue higher than $21.00.

Answer: Given weight 5/9 to S2 and weigth 4/9 to S3 results in revenue 23.5 and con-
sumption 30%, so this combination is better than offering {1, 3}.

a) Suppose that the number of potential customers is Poisson with parameter λ = 200.
What is the distribution of demand under action S1 = {1}? Under action S2 = {1, 2}?
Answer: Demand under action 1 is Poisson with rate λv1/(v0 +v1) = 100. Demand under
action 2 is Poisson with rate λ(v1 + v2)/(v0 + v1 + v2) = 400/3.

b) Suppose you offer action S2. What is the probability that an arriving customer will buy
fare 1? fare 2?

Answer: The probability that an arriving customer will buy fare 1 is v1/v[0, 2] = 1/3, the
probability that he will buy fare 2 is v2/v[0, 2] = 1/3.

c) What is the average revenue per arriving customer under action S2? What is the average
revenue per sale under action S2?

Answer: The average revenue per arriving customer is r2 = (p1 + p2)/3. The average
revenue per sale is q2 = (p1 + p2)/2.

d) Suppose we protect y units for action S1. Write an expression for the expected revenue
under action S2 before it is closed?

Answer: Expected revenues are q2E min(c−y,S2).
e) What is the distribution of demand under action S1 once action 2 is closed? Give this in

terms of D2, c and y.

Answer: Demand is conditionally binomial with parameters (D2+y−c)+ and θ = π1/π2 =
3/4.

f) What is the expected revenue under action S1? Answer:

p1E min(Bin(D2 + y − c)+,θ), max(y,c−D2)).

5. Dynamic Choice Models with Dependent Demands Consider a choice model where for
each set S ⊂ Sn = {1, . . . ,n}, π(S) =

∑
i∈S πi(S) and r(S) =

∑
i∈S piπi(S).

a) Write down the formula for the value function V (t,x) assuming you do not know which
sets are efficient.

Answer:
V (t,x) = V (t− 1,x) + max

S⊂N
[r(S) −π(S)∆V (t− 1,x)].

b) At state (t,x), what problem do you need to solve to figure out which set to offer?

Answer: We need to find the set S ⊂ N that maximizes r(S) − ∆V (t− 1,x)π(S).
c) Suppose the efficient fares are S0,S1, . . . ,Sn and that you are given a graph of (πj,rj),j =

1, . . . ,n where πj = π(Sj) and rj = r(Sj). How would you visually obtain the optimal
set to offer at state (t,x)?

Answer: I would draw the line α∆V (t−1,x) and then look at the different points πj and
look at which point the difference between rj and ∆V (t− 1,x)πj is the largest.

d) Continue with the logic of part d) to argue why you may need to open fares as t decreases
for fixed x and close fares as x decreases for fixed t.

Answer: If t decreases without a sale then the marginal value goes down and that would
shift the optimal j to the right resulting in more open fares. If x decreases through sales
then the marginal value goes up and that shifts the optimal to the left resulting in fewer
open fares.

6. Deterministic Static Pricing with Two Market Segments

Answer: I would protect y = 60 units of capacity for the business fare and would price
the leisure segment at p = 560. At that price I would sell the 40 units of capacity at
the leisure segment making $22,400 from that segment plus $60,000 from the business
segment for a total of $82,400.

discounted fare will then buy capacity at p1. Hint: Remember to apply the booking limit
on sales at the discounted fare. Is there a better booking limit?

Answer: If both demands arrive at the same time we would see 60 customers from the
business segment and 40 from the leisure segment willing to purchase capacity at the
discounted fare. Since only 40 units are available at this fare, there will be 16 expected
sales to leisure and 24 to business customers. That will leave us with 36 customers buying
at 1000 so the total revenue is $58,400. If the booking limit is zero then 60 tickets are
sold at $1,000 improving the revenues to $60,000.

Answer: At p = 450, there will be 150 leisure customers and 60 business customers willing
to buy at this fare. Thus a low fare ticket will go to the leisure segment with probability
15/21 = 5/7 and to the business segment with probability 60/210 = 2/7.

Answer: I would make 56 units of capacity available at $450. Then 40 units will go to
the leisure segment and 16 to the business segment. This would leave me with 44 units
of capacity and 44 business customers to which I can sell at fare p1 = $1000. The total
results in revenue $69,200

e) What happens to the solution of part d) if capacity is transferable and there is a secondary
market where leisure customers who secured capacity at $450 can sell it to business
customers for $800?

Answer: In this case all 56 units sold at $450 go to the business segment, leaving us with
only 4 customers willing to buy capacity at p1 = $1, 000. The revenues then drop to
$29,200.