USA: +1(669)289-8683 
Sign in
I need elp with this questions ASAP!
Dynamic Pricing
and
Revenue
Management
IEOR 460
1
Spring 201
3
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 Gal ego’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
415
0
l� Deterministic Models at the level of IEOR 4003
¡� Corequisites: Stochastic Models IEOR 410
6
¡� 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 Saturda
y
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 8
5
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 b
i
denote the nested booking limit for class i.
b 1 ≥ b 2 ≥�≥ bn
l� Protection Levels:
y
�
i = b − b
, i
i
= ,
1 2, n −
1
1
1
+
l� Updates: If x units are sold in a transaction b
�
i ← max( bi − x
),
0
,
i = ,
1 2, n −1
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
2
7
8
8
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 wil 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
IEOR 4601 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
0.
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?
1
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.
2
Probability
and Dynamic Programming
Revie
w
for
Dynamic Pricing and Revenue
Managemen
t
• Poisson: http://en.wikipedia.org/wiki/Poisson_random_variable
–
Read Sections: 1,
2
, 3,
4
• Compound Poisson:
http://en.wikipedia.org/wiki/Compound_Poisson_distribution
• Poisson Process: http://en.wikipedia.org/wiki/Poisson_proces
s
• Compound Poisson Process:
http://en.wikipedia.org/wiki/Compound_Poisson_process
• Normal: http://en.wikipedia.org/wiki/Normal_random_variable
–
Read Sections 1,2.1,2.2,3,1,3.2,3.
3
• Brownian Motion: http://en.wikipedia.org/wiki/Wiener_process
–
Read sections 1, 2.
DP AS AN OPTIMIZATION METHODOLOGY
• Basic optimization problem
min g( u)
u∈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 problems 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
E
w
Demand at Period
k
k
Stock at
Period k
Stock at Period k +
1
Inventory
xk
S y s t e m
xk + 1 = xk + uk – wk
Stock Ordered at
Period k
Co s t o f P e rio d k
u k
c uk + r (xk + uk – wk)
• Discrete-time system
xk+1 = fk( xk, uk, wk) = xk + uk − wk
• Cost function that is additive over time
N − 1
E
gN ( xN ) +
gk( xk, uk, wk)
k=
0
N − 1
= E
cuk + r( xk + uk − wk) k=
0
• Optimization over policies: Rules/functions uk =
µk( xk) that map states to controls BASIC STRUCTURE OF STOCHASTIC DP
• Discrete-time system
xk+1 = fk( xk, uk, wk) , k = 0 , 1 , . . . , N − 1
− k: Discrete time
− xk: State; summarizes past information that is relevant for future optimization
− uk: Control; decision to be selected at time k from a given set
− wk: Random parameter (also called distur-bance or noise depending on the context)
− N: Horizon or number of times control is applied
• Cost function that is additive over time N − 1
E
gN ( xN ) +
gk( xk, uk, wk)
k=0
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 = {µ 0 , . . . , 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 x 0 is N − 1
Jπ( x 0) = E gN ( xN ) +
gk( xk, µk( xk) , wk) k=0
• Optimal cost function
J ∗( x 0) = min Jπ( x 0) π
• Optimal policy π∗ is one that satisfies Jπ∗( x 0) = J∗( x 0) PRINCIPLE OF OPTIMALITY
• Let π∗ = {µ∗ 0 , µ∗ 1 , . . . , µ∗N− 1 } be an optimal policy
• 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
N − 1
E
gN ( xN ) +
gk xk, µk( xk) , wk
k=
i
and the “tail policy” {µ∗, µ∗
i
i+1 , . . . , µ∗
N − 1 }
xi
Tail Subproblem
0
i
N
• Principle of optimality: The tail policy is optimal for the tail subproblem
• DP first solves ALL tail subroblems of final stage
• At the generic step, it solves ALL tail subproblems 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
E gk( xk, uk, wk)
uk∈Uk( xk) wk
+ Jk+1 fk( xk, uk, wk) , k = 0 , 1 , . . . , N − 1 .
• Then J 0( x 0), generated at the last step, is equal to the optimal cost J ∗( x 0). Also, the policy π∗ =
{µ∗ 0 , . . . , µ∗N− 1 }
where µ∗( x
k
k) minimizes in the right side above for
each xk and k, is optimal.
• Justification: Proof by induction that Jk( xk) is equal to J ∗( x
k
k), 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
with Cost Equal
to Terminal Cost
. . .
t
Artificial Terminal
. . .
N o d e
Initial State
s
. . .
Stage 0
Stage 1
Stage
2
. . . Stage N – 1
Stage 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 ∈ S
it
N
• 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) = aN , i ∈ S
it
N ,
Jk( i) = min
ak + Jk+1( j) , i ∈ Sk, k = 0 , . . . , N − 1 .
j∈S
ij
k+1
The optimal cost is J 0( 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 algorithm for the reverse problem):
˜
JN ( j) = a 0 , j ∈ S
sj
1 ,
˜
Jk( j) = min aN−k + ˜
Jk+1( i) , j ∈ SN−k+1
i∈S
ij
N−k
The optimal cost is ˜
J 0( t) = min i∈S
aN + ˜
J
N
it
1( i) .
• View ˜
Jk( 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 require 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.
J 0( i): Cost of the optimal path from i to t.
• DP algorithm:
Jk( i) = min
aij+ Jk+1( j) ,
k = 0 , 1 , . . . , N − 2 ,
j=1 ,…,N
with JN− 1( i) = ait, i = 1 , 2 , . . . , N.
EXAMPLE
State i
Destination
5
5
3
3
3
3
2
3
4
7
5
4
4
4
5
1
4
3
2
5
5
4.5
4.5
5.5
7
2
6
1
2
2
2
2
1
2
3
0 . 5
0
1
2
3
4
Stage k
(a)
(b)
JN− 1( i) = ait,
i = 1 , 2 , . . . , N,
Jk( i) = min
aij+ Jk+1( j) ,
k = 0 , 1 , . . . , N − 2 .
j=1 ,…,N
Document Outline
Probability Review
- Probability and Dynamic ProgrammingReview
Probability Review
<
p
>Si
n
gle Resou
r
ce Revenue Management with Independent Demand
s
c
Guillermo Gallego
Updated Spring 20
1
3
Abstrac
t
Providers of fixed perishable capacity, such as airline seats and hotel rooms use price discriminatio
n
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
approximatel
y
if the difference in fares is large so that demands are decoupled or if customers can find alternative
sources
of capacity for their preferred fare class. In some cases, however, part of the demand may be recaptured
by
other available fare classes. In such cases, the independent demand assumption is too strong and needs to
be
relaxed. We address this issue in a separate chapter where we discussed demand models based on
discrete
choice theory.
In this chapter, we present a variety of models that have been developed in industry and in academia.
There has been a preference in industry for models that suppress the time dimension and assume that the
arrival pattern of the fare classes is low-to-high. We call these class of models static to distinguish them
from the dynamic models, favored by academics, that model time explicitly. Both models have advantages
and disadvantages as we will soon see. Static models are relatively easy to understand. Also, good
heuristics
1
were developed before optimal solutions based on dynamic programming were discovered. Bringing in
the
time dimension helps deal with more general fare arrival patterns, but specifying the model requires
a
more
detailed estimation of demand. This Chapter starts with a review of the two fare class problem in §
2
where we
present a heuristic derivation of Littlewood’s rule via marginal analysis. Littlewood’s rule is formally
derived
in §2.3 where a formal DP for the two fare class problem is presented. The dynamic program for multiple
fare
classes is presented in §3. Commonly used heuristics are presented in §
4
and bounds on the optimal
expected
revenue are presented in §5. The dynamic model is presented in §6, for the Poisson case and for the
compound
Poisson case in §7, where each request is for a random demand size. In §
8
, we restrict fares so that they
cannot
be opened once they are closed.
2
Two Fare Classes: Marginal Analysis
The product can be sold either at the full-fare p1 or at a discounted-fare p2 < p1. The discounted-fare typically
has advance purchasing and usage restrictions. Let D1 and D2 denote respectively the random demand for
the two fare classes for a specific instance of the problem, e.g., for a specific flight for an airline or a
specific
night for a hotel.
We assume that all booked customers will actually travel. This avoids the need to overbook capacity and
allow us to focus on the problem of allocating capacity between the two fares. We will discuss how to
deal
with pre-travel cancellations and day-of-travel no shows in a separate chapter on overbooking models.
Fare class arrival order is an important part of the model. We assume what is commonly known as the
low-to-high fare class arrival order, which implies that demand for the discounted fare book earlier than
for
full-fare. This arrival pattern holds approximately in practice, and it is encouraged by advance purchase
restrictions imposed on lower fare classes. Notice that this is a worst case arrival pattern. Indeed, if full-
fare
class customers arrived first then we would accept them up to capacity and use residual capacity, if any,
to
satisfy demand from the discounted-fare class. We will relax the low-to-high fare order arrival
assumption
after we solve the multi-fare problem via dynamic programming.
Under the low-to-high arrival pattern, discount-fare customers may exhaust capacity, say c, unless part
of it is protected for later-booking by full-fare customers. Consequently, booking limits (known as
discount
authorizations) are placed on the discount sales. Suppose we protect y ∈ {0, 1, . . . , c} units of
capacity
for
the full-fare demand, D1, before observing the actual demand for the discount-fare, D2. This results in a
booking limit c − y on the discounted-fare, so sales at the discounted-fare class are given by min(c − y,
D2
).
The remaining capacity is equal to c − min(c − y, D2)
=
max(y, c − D2) and it is all made available to the
full-fare class. Consequently, sales at the full fare equal min(max(y, c − D2), D1). The total expected
revenue
is
W (y, c) =
p2E min(c − y, D2) + p1E min(max(y, c − D2), D1)
and the goal is to find a protection level y that maximizes W (y, c). The extreme strategies y = 0 and y = c
correspond, respectively, to the case where no capacity is protected and all of the capacity is protected.
We
will later come back and discuss when these extreme strategies are optimal. In most cases, however, an
intermediate strategy is optimal.
The fare ratio r = p2/p1 plays an important role in determining optimal protection levels. If the ratio is
very small then we would be inclined to protect more capacity for the full-fare demand. If the ratio is
close
to one, we would be inclined to accept nearly all discount-fare requests since we can get almost the same
revenue, per unit of capacity, without risk. The distribution of full-fare demand is also important in
deciding
how many units to protect for that fare. If, P (D1 ≥ c) is very large, then it makes sense to protect the
entire
capacity for full-fare sales as it is likely that the provider can sell all of the capacity at the full-fare.
However,
if P (D1 ≥ c) is very low then it is unlikely that all the capacity can be sold at the full-fare, so fewer units
should be protected. It turns out that the demand distribution of the discount-fare D2 has no influence on
the
optimal protection level under our assumption that D2 and D1 are independent. A formula for the optimal
protection level, involving only P (D1 ≥ y) and r, was first proposed by Littlewood [14] in
19
72. His
arguments
were not formal; however, they were later
j
ustified by Bhatia and Prakesh [1] in 1973, and Richter [17]
in 1982.
2
One can obtain Littlewood’s formula intuitively by using marginal analysis. The advantage of marginal
analysis is that it allows us to quickly derive the solution for the two fare class problem. The marginal
analysis
argument goes as follows: Suppose we have y > 0 units of capacity, and that we receive a request for the
discounted-fare. Consider the marginal revenue associated with accepting and rejecting this request. If we
accept, we obtain p2. If we close down the discount-fare then we will be able to sell the yth unit at p1
only
if the full-fare demand D1 is at least as large as y, so it is intuitively optimal to reject the discount fare if
p1P (D1 ≥ y) > p2. This suggests that an optimal protection level y1 should be given by
:
y1 = max{y ∈ N : P (D1 ≥ y) > r},
(1)
where N = {0, 1, . . . , } is the set of non-negative integers. Equation (1) is known as Littlewood’s rule.
Example 1. Suppose D1 is Poisson with parameter
80
, the full fare is p1 = $100 and the discounted fare
is
p2 = $
60
, so r = 60/100 = 0.6. We are interested in the cumulative tail distribution P (D1 ≥ y) = 1 − P
(D1
≤
y − 1). Since most statistical software packages return the value of P (D1 ≤ y), we see that y1 satisfies
P (D1 ≤ y1 − 1) < 1 − r ≤ P (D1 ≤ y1). Since P (D1 ≤ 77) =< 0.4 ≤ P (D1 ≤ 78) we conclude that y1 = 78.
Consequently, if c = 200 then the booking limit for the discount fare is 122. However, if c < y1, then all units
should be protected for the full-fare resulting in a booking limit of zero.
Remarks:
• y
(c)
= min(y1, c) is also an optimal protection level. If y(c) = c, or equivalently if y1 ≥ c, then all the
capacity should be reserved for sale at the full-fare.
• The quantity b2 = max(c − y1, 0) is known as the optimal booking limit for the discount fare. It is the
maximum number of discount-fare customers that we will book.
• y1 is independent of the distribution of D2.
• If P (D1 ≥ y1 + 1) = r, then y1 + 1 is also optimal protection level, so both y1 and y1 + 1 result in the
same expected revenue. Protecting the y1 + 1 unit of capacity increases the variance of the revenue, but
it reduces the probability of rejecting requests from full-fare customers.
From Littlewood’s rule (1), we see that the extreme strategy y = 0 is optimal when P (D1 ≥ 1) ≤ r and
the extreme strategy y = c is optimal when P (D1 ≥ c) > r.
2.1
Continuous Demand Model
Although revenue management demands are actually discrete, continuous distributions can be easier to
wor
k
with and are often employed in practice. If we model D1 as a continuous random variable
with
cumulative
distribution function F1(y) = P (D1 ≤ y),
then
y1 = F −1(1 − r)
1
where F −1 denotes the inverse of F . In particular, if
D
1
1 is Normal with mean µ1 and standard deviation σ1
then
y1 = µ1 + σ1Φ−1(1 − r)
(2)
where Φ denotes the cumulative distribution function of the standard Normal random variable.
This formula allows for comparative statics as given in Table 1:
Example 2. Suppose that D1 is Normal with mean 80 and standard deviation 9, the full-fare is p1 =
$
10
0
and the discount-fare is p2 =
$60
. Then y1 = F −1(1 − 0.6) = 77.72 < 80 since r > 1/2. Notice that the
1
solution is quite close to that of Example 1. This is because a Poisson random variable with mean 80 can
be
√
well approximated by a normal with mean 80 and standard deviation
80
9.
3
Fare Ratio
Dependence of protection level
r > 1
y
2
1 < µ1 and y1 decreases with σ1
r = 1
y
2
1 = µ1 independent of σ1
r < 1
y
2
1 > µ1 and y1 increases with σ1
Table 1: Comparative Statics for Normal Full Fare Demand
2.2
Connection with the Newsvendor Problem
There is a close connection between the classical Newsvendor Problem and the two-fare Revenue
Management
Problem that we will briefly explore here. In the classical Newsvendor Problem a manager must decide
how
many units, say y, to stock for random sales D1 at p1 assuming a unit cost p2 < p1. The solution is to stock y1
units where y1 is the largest integer such that P (D1 ≥ y) > r = p2/p1. We can think of the two-fare
Revenue
Management Problem as a situation where capacity c is pre-decided, at a possible sub-optimal level,
there is
random demand D1 at ”salvage value” p1 > p2 that arrives after demand D2 at p2. The revenue
management
problem is to determine how many units to allow to be sold at p2. We know that the solution is to allow
(c − y1)+ units to book at p2, reserving max(y1, c − D2) units for sale at p1.
2.3
Two Fare Classes: Dynamic Programming
In this section we formulate and analyze the two fare class problem using dynamic programming and
present
a formal proof of the optimality of Littlewood’s rule. We will from now on refer to the full-fare class as
fare
class 1 and to the discounted fare class as fare class 2. Dynamic programming starts by solving the
problem
at the last stage, just before demand for fare class 1. Let
V
1(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) =
ma
x
E{p2 min(D2, c − y) + V1(max(c − D2, y))}.
(3)
y∈{0,1,…,c}
The key to Dynamic Programming is that it involves a recursive equation (3) linking the expected
revenues
V2(c), at stage 2, to the expected revenue function V1 at stage 1. To solve for V2(c) we first need to solve
for V1(y) for y ∈ {0, 1, . . . , c}. Before moving on to the multi-fare formulation we will provide a formal
proof of Littlewood’s rule (1), and discuss the quality of service implications of using Littlewood’s rule
under
competition.
4
2.4
Formal Proof of Littlewood’s Rule
For any function f (y) over the integers, let ∆f (y) = f (y) − f (y − 1). The following result will help us to
determine ∆V (y) and ∆W (y, c) = W (y, c) − W (y − 1, c).
Lemma 1 Let g(y) = EG(min(X, y)) where X is an integer valued random variable with E[X] < ∞ and G
is an arbitrary
function defined over the integers. Then
∆g(y) = ∆G(y)P (X ≥ y).
Let r(y) = ER(max(X, y)) where X is an integer valued random variable with E[X] < ∞ and R is an arbitrary
function defined over the integers. Then
∆r(y) = ∆R(y)P (X < y).
An application of the Lemma 1 yields the following proposition that provides the desired formulas for
∆V1(y) and ∆W (y, c).
Proposition 1
∆V1(y) = p1P (D1 ≥ y)
y ∈ {1, . . . , }
∆W (y, c) = [∆V1(y) − p2]P (D2 > c − y)
y ∈ {1, . . . , c}.
The proof of the Lemma 1 and Proposition 1 are relegated to the Appendix. With the help of Proposition 1
we can now formally establish the main result for the Two-Fare Problem.
Theorem 1 The function W (y, c) is unimodal in y and is maximized at y(c) = min(y1, c) where
y1 = max{y ∈ N : ∆V1(y) > p2}.
Moreover, V2(c) = W (y(c), c).
Proof: Consider the expression in brackets for ∆W (y, c) and notice that the sign of ∆W (y, c) is
determined
by ∆V1(y) − p2 as P (D2 > c − y) ≥ 0.Thus W (y, c) ≥ W (y − 1, c) as long as ∆V1(y) − p2 > 0 and W (y,
c) ≤
W (y − 1, c) as long as ∆V1(y) − p2 ≤ 0. Since ∆V1(y) = p1P (D1 ≥ y) is decreasing1 in y, ∆V1(y) −
p2
changes
signs from + to − since ∆V1(0) − p2 = p1P (D1 ≥ 0) − p2 = p1 − p2 > 0 and limy→∞[∆V1(y) − p2] =
−p2.
This means that W (y, c) is unimodal in y. Then
y1 = max{y ∈ N : ∆V1(y) > p2}.
coincides with Littlewood’s rule (1).
When restricted to {0, 1, . . . , c}, W (y, c) is maximized at y(c) =
min(c, y1). Consequently, V2(c) = maxy∈{0,1,…,c} W (y, c) = W (y(c), c), completing the proof.
2.
5
Quality of Service, Spill Penalties, Callable Products and Salvage Values
Since max(y1, c − D2) units of capacity are available for fare class 1, at least one fare class 1 customer
will be
denied capacity when D1 > max(y1, c − D2). The probability of this happening is a measure of the quality
of
service to fare class 1, known as the full-fare spill rate. Brumelle et al. [4] have observed that
P (D1 > max(y1, c − D2)) ≤ P (D1 > y1) ≤ r < P (D1 ≥ y1).
(4)
They call P (D1 > y1) the maximal spill rate. Notice that if the inequality y1 ≥ c − D2 holds with high
probability, as it typically does in practice when D2 is large relative to c, then the spill rate approaches
the
1We use the term increasing and decreasing in the weak sense.
5
maximal flight spill rate which is, by design, close to the ratio r. High spill rates may lead to the loss of
full-fare customers to competition. To see this, imagine two airlines each offering a discount fare and a
full-
fare in the same market where the fare ratio r is high and demand from fare class 2 is high. Suppose
Airline
A practices tactically optimal Revenue Management by applying Littlewood’s rule with spill rates close
to
r. Airline B can protect more seats than recommended by Littlewood’s rule. By doing this Airline B will
sacrifice revenues in the short run but will attract some of the full-fare customers spilled by Airline A.
Over
time, Airline A may see a decrease in full-fare demand as a secular change and protect even fewer seats
for
full-fare passengers. In the meantime, Airline B will see an increase in full-fare demand at which time it
can
set tactically optimal protection levels and derive higher revenues in the long-run. In essence, Airline B
has
(correctly) traded discount-fare customers for full-fare customers with Airline A.
One way to cope with high spill rates and its adverse strategic consequences is to impose a penalty
cost
ρ for each unit of full-fare demand in excess of the protection level. This penalty is suppose to measure
the
ill-will incurred when capacity is denied to a full-fare customer. This results in a modified value function
V1(y) = p1E min(D1, y) − ρE[(D1 − y)+] = (p1 + ρ)E min(D1, y) − ρED1. From this it is easy to see that
∆V1(y) = (p + ρ)P (D1 ≥ y), resulting in
∆W (y, c) = [(p1 + ρ)P (D1 ≥ y) − p2]P (D2 > c − y)
and
p2
y1 = max y ∈ N : P (D1 ≥ y) >
.
(5)
p1 +
ρ
Notice that this is just Littlewood’s rule applied to fares p1 + ρ and p2, resulting in fare ratio p2/(p1 + ρ)
and,
consequently, lower maximal spill rates. Obviously this adjustment comes at the expense of having higher
protection levels and therefore lower sales at the discount-fare and lower overall revenues.
Consequently, an
airline that wants to protect its full-fare market by imposing a penalty on rejected full-fare demand does it
at
the expense of making less available capacity for the discount-fare and less expected revenue. One way to
avoid
sacrificing sales at the discount-fare and improve the spill rate at the same time is to modify the discount-
fare
by adding a restriction that allows the provider to recall or buy back capacity when needed. This leads to
revenue management with callable products; see Gallego, Kou and Phillips [12]. Callable products can
be
sold either by giving customers an upfront discount or by giving them a compensation if and when
capacity
is recalled. If managed correctly, callable products can lead to better capacity utilization, better service
to
full-fare customers and to demand induction from customers who are attracted to either the upfront
discount
or to the compensation if their capacity is recalled.
The value function V1(y) may also be modified to account for salvage values (also known as the
‘distressed
inventory problem’). Suppose there is a salvage value s < p2 on excess capacity after the arrival of the full-fare
demand (think of standby tickets or last-minute travel deals). We can handle this case by modifying V1(y)
to
account for the salvaged units. Then V1(y) = p1E min(D1, y) + sE(y − D1)+ = (p1 − s)E min(D1, y) + sy,
so
∆V1(y) = (p1 − s)P (D1 ≥ y) + s, resulting in
∆W (y, c) = [(p1 − s)P (D1 ≥ y) − (p2 − s)]P (D2 > c − y)
and
p2 − s
y1 = max y ∈ N : P (D1 ≥ y) >
.
(6)
p1 − s
Notice that this is just Littlewood’s rule applied to net fares p1 − s and p2 − s. This suggests that a
problem
with salvage values can be converted into a problem without salvage values by using net fares pi ← pi −
s,
i = 1, 2 and then adding cs to the resulting optimal expected revenue V2(c) in excess of salvage values.
3
Multiple Fare Classes: Exact Solution
In this section we present an exact solution to the muli-fare class problem using dynamic programming.
We
assume that the capacity provider has c units of perishable capacity to be allocated among n fares indexed
6
so pn < . . . < p1. Lower fares typically have severe time of purchase and traveling restrictions and may
have
restricted advanced selection that denies access to the more desirable capacity. Given the time-of-
purchase
restriction, it is natural to assume that demands for fare classes arrive in n stages, with fare class n
arriving
first, followed by n − 1, with fare class 1 arriving last. Let Dj denote the random demand for fare class
j ∈ N = {1, . . . , n}. We assume that, conditional on the given fares, the demands D1, . . . , Dn are
independent
random variables with finite means µj = E[Dj] j ∈ N . The independent assumption is approximately
valid
in situations where fares are well spread and there are alternative sources of capacity. Indeed, a customer
who finds his preferred fare closed is more likely to buy the same fare for an alternative flight (perhaps
with
a competing carrier) rather than buying up to the next fare class if the difference in fare is high. The case
of
dependent demands, where fare closures may result in demand recapture, will be treated in a different
chapter.
The use of Dynamic Programming for the multi-fare problem with discrete demands is due to Wollmer
[22].
Curry [6] derives optimality conditions when demands are assumed to follow a continuos distribution.
Brumelle
and McGill [5] allow for either discrete or continuous demand distributions and makes a connection with
the
theory of optimal stopping.
Let Vj(x) denote the maximum expected revenue that can be obtained from x ∈ {0, 1, . . . , c} units of
capacity from fare classes {j, . . . , 1}. The sequence of events for stage j are as follows:
1. Decide the protection level, say y ≤ x, for fares j − 1, j − 2, . . . , 1 thus allowing at most x − y units of
fare j demand.
2. The realization of the demand Dj occurs, and we observe min(Dj, x − y) sales at fare j.
3. The revenue pj min(Dj, x−y) is collected, and we proceed to the beginning of stage j −1 with a
remaining
capacity of max(x − Dj, y).
The revenue from this process is
Wj(y, x) = pjE min(Dj, x − y) + EVj−1 (max(x − Dj, y)) .
(7)
We can think of Wj(y, x) as the expected revenue from x units of capacity prior to seeing the demand for
fare
class j when up to x − y units are allowed to book at fare j and an optimal policy is followed thereafter.
This
leads to the dynamic programming recursion
Vj(x)
=
max
Wj(y, x)
y∈{0,1,…,x}
=
max
{pjE min(Dj, x − y) + EVj−1(max(x − Dj, y))} .
(8)
y∈{0,1,…,x}
The dynamic program simply states that the optimal value function is the sum of the expected revenues
from fare class j plus the expected revenues from fare classes j − 1, . . . , 1 evaluated at the protection
level that
maximizes this sum. Notice that once we are at the beginning of stage j − 1 we face a similar problem
over the
remaining j − 1 fare classes. Vj(x) is then the maximum expected revenue that can be obtained from x units
of
capacity for the j-fare problem. Consequently Vn(c) is the maximum expected revenue for the n-fare
problem
with capacity c. The recursion can be started with V0(x) = 0 if there are no salvage values or penalties
for
spill. Alternatively, the recursion can start with V1(x) = p1E min(D1, x) + sE[(x − D1)+] − ρE[(D1 −
x)+] for
x ≥ 0 if there is a salvage value s per unit of excess capacity and a penalty ρ per unit of fare class 1
demand
that is denied.
3.1
Structure of the Optimal Policy
In order to analyze the structure of the optimal policy, we begin by describing a few properties of the
value
function.
As a convention we set V0 ≡ 0. A function V (y) defined on y ∈ N is concave if ∆V (y) =
V (y) − V (y − 1) is decreasing in y ∈ N+.
7
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) > p
j+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 x
j
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.
8
Example 3. Suppose there are five different fare classes. We assume the demand for each of the fares is
Poisson. The fares and the expected demands are given in the first two columns of Table 2. The third
column
includes the optimal protection levels for fares 1, 2, 3 and 4.
j
pj
E[Dj]
yj
1
$
100
15
14
2
$
60
40
54
3
$40
50
101
4
$35
55
16
9
5
$15
1
20
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
E[D
j=1
j ])/c = 280/c.
These results should be intuitive. Greater revenue potential is seen as
capacity increases (since potentially more demand can be accepted). Further, the effect of restrictions on
discounted fares is apparent in the pattern of revenue across classes; e.g. revenue V2(50) through V5(50)
is
$3,426.8 because fare classes 3,4, and 5 are rationed since y2 = 54 > c = 50 units are protected for fare 1
and
2. However, V1(
350
) through V5(350) vary from $1,500 to $9,625 because there is sufficient capacity to
accept
sales in all fare classes.
c
D
F
V1
(c)
V2(c)
V3(c)
V4(c)
V
5(c)
50
560%
1,500.0
3,426.8
3,426.8
3,426.8
3,426.8
100
2
80%
1,500.0
3,900.0
5,441.3
5,441.3
5,441.3
150
187%
1,500.0
3,900.0
5,900.0
7,188.7
7,188.7
200
140%
1,500.0
3,900.0
5,900.0
7,824.6
8,159.1
250
112%
1,500.0
3,900.0
5,900.0
7,825.0
8,909.1
300
93%
1,500.0
3,900.0
5,900.0
7,825.0
9,563.9
350
80%
1,500.0
3,900.0
5,900.0
7,825.0
9,625.0
Table 3: Expected Revenues Vj(c) and Demand Factors
Figure 1 shows the marginal value as a function of the remaining resources for the data of Example 3.
$120.00
$100.00
$80.00
DV_1(x)
DV_2(x)
$60.00
DV_3(x)
DV_4(x)
DV_5(x)
$40.00
$20.00
$-‐
0
20
40
60
80
100
120
140
160
Figure 1: ∆Vj(x), x = 1, . . . , 350, j = 1, 2, 3, 4, 5 for Example 3
9
3.2
Speeding up the Computation of the Value Function
While the value functions Vj(x), j ∈ {1, . . . , n}, x ∈ {1, . . . c}, can be computed recursively there are
some
tricks to speed up the computations. Here we focus on how to efficiently update ∆Vj+1(x) from ∆Vj(x).
The
key idea is to express ∆Vj+1(x) for x > yj in terms of previously computed values of ∆Vj(x). The proof of
Proposition 2 is in the Appendix.
Proposition 2
∆
V
∆V
j (x)
if x = 1, . . . , yj
j+1
(x) =
E min(∆Vj(x − Dj+1), pj+1)
if x = yj + 1, . . ..
Since ∆Vj+1(x) = ∆Vj(x) for x ≤ yj, we only need to worry about ∆Vj+1(x) for x > yj. The following
corollary to Proposition 2 makes the formula for ∆Vj+1(x), x > yj more explicit.
Corollary 1
k−1
∆Vj+1(yj + k) = pj+1P (Dj+1 ≥ k)
+
∆Vj(yj + k − i)P (D
j+1 = i)
k ∈ {1, . . . , c − yj}.
i=0
3.3
Linear and Convex Procurement Costs
Suppose that capacity c can be procured at a linear cost kc before observing demands for the n fares: pn <
pn−1 < . . . < p1. How much capacity should be procured? The objective is to find c to maximize Πn(c, k) =
Vn(c) − kc. Let c(k) be the smallest optimizer of Πn(c, k) as a function of the marginal cost k. The
following
Proposition characterizes c(k), shows that c(k) is decreasing in k and relates c(pj+1) to protection level
yj for
j < n.
Proposition 3 The optimal procurement quantity at linear cost kc is given by
c(k) = max{c ∈ N : ∆Vn(c) > k}.
Moreover, c(k) is decreasing in k, and yj = c(pj+1) for all j ∈ {1, . . . , n − 1}.
Clearly c(0) = ∞ since ∆Vn(c) ≥ ∆V1(c) = p1P (D1 ≥ c) > 0 for all c ∈ N . At the other extreme,
c(p1) = y0 = 0 since ∆Vn(1) = p1P (D1 ≥ 1) < p1 = ∆Vn(0), so no capacity would be purchased if k ≥ p1.
If the cost of capacity k(c) is increasing convex then Π(c, k(c)) is concave in c and
c(k) = max{c ∈ N : ∆Vn(c) − ∆k(c) > 0}.
Consider the convex cost function k(c) = K if c ≤
¯
c and k(c) = ∞ for k > ¯
c. This may reflect the situation
where there is a fixed cost to leasing a resource with capacity ¯
c. The optimal choice is then to lease the
resource if Vn(¯
c) > K and not lease it otherwise.
3.4
Relaxing the Monotonicity of Fares
We continue to assume that demands arrive in the order Dn, Dn−1, . . . , D1 but will relax the assumption
that
the fares are monotone p1 > p2 > . . . > pn. All of the results work as stated, except the monotonicity of the
protection levels, if we redefine ∆Vj(0) = max{p1, . . . , pj}. It is also possible to skip some of the
optimization
10
optimal capacity
450
400
350
300
ity 250
acap 200
C
150
100
50
0
0
10
20
30
40
50
60
70
80
90
100
cost
Figure 2: Optimal Capacity as a Function of Cost for the Data of Example 3
steps as it is clear that yj−1 = 0 whenever pj > max(p1, . . . , pj−1) since it is optimal to allow all
bookings at
fare pj. The reader is referred to Robinson [18] for more details. As an example, suppose that p3 < p2 >
p1.
Then V1(x) = p1E min(D1, x) and at stage 2 the decision is y1 = 0, so
V2(x) = p2E min(D2, x) + p1E min(D1, (x − D2)+).
Notice that
∆V2(x) = p2P (D2 ≥ x) + p1P (D2 < x ≤ D[1, 2])
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}.
11
Then the EMSR-a heuristic will protect
j−1
ya
=
y
j−1
k,j
k=1
units of capacity for fares j − 1, . . . , 1 against fare j.
In particular, if Dk is Normal with mean
µk
and standard deviation σk, then
j−1
ya
= µ[1, j − 1] +
σ
j−1
k Φ−1(1 − rk,j ),
k=1
where for any j, µ[1, j − 1] =
j−1 µ
k=1
k and sums over empty sets are zero.
Notice that the EMSR-a heuristic involves j − 1 calls to Littlewood’s rule to find the protection level for
fares j − 1, . . . , 1. In contrast, the EMSR-b heuristic is based on a single call to Littlewood’s rule for
each
protection level. However, using the EMSR-b heuristic requires the distribution of D[1, j − 1] =
j−1 D
k=1
k .
This typically requires computing a convolution but in some cases, such as the Normal or the Poisson, the
distribution of D[1, j − 1] can be easily obtained (because sums of independent Normal or Poisson
random
variables are, respectively, Normal or Poisson). The distribution of D[1, j − 1] is used together with the
weighted average fare
j−1
µk
¯
pj−1 =
pk µ[1, j − 1]
k=1
and calls on Littlewood’s rule to obtain protection level
y
b
= max{y ∈ N : P (D[1, j − 1] ≥ y) > rb
}
j−1
j−1,j
where rb
=
p
j−1,j
j / ¯
pj−1. Notice that the weighted average fare assumes that a proportion µk/µ[1, j − 1] of the
protected capacity will be sold at fare pk, k = 1, . . . , j − 1. In the special case when demands are Normal
we
obtain
yb
= µ[1, j − 1] + σ[1, j − 1]Φ−1(1 − rb
).
j−1
j−1,j
Recall that for the Normal, variances are additive, so the standard deviation σ[1, j − 1] =
j−1 σ2.
k=1
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 = yh ≤ yh ≤ . . . ≤ yh
can be
0
1
n−1
computed exactly or via simulation. To compute the expected revenues exactly, let V h(x) be the expected
profit
j
from x units of capacity from fares {j, . . . , 1} under the heuristic policy h. Then V h(x) = W h(min(x, yh
), x)
j
j
j−1
where for all y ≤ x we define W h(y, x) = p
(max(y, x − D
j
j E min(x − y, Dj ) + EV h
j−1
j )). The following result
is a direct consequence of Proposition 2 and its corollary.
Proposition 4
∆V h(x)
if x = 1, . . . , yh
j
j
∆V h (x) =
j+1
x−yh−1
= p
j
j+1P (Dj+1 ≥ x − yh) +
∆V h(x − i)P (D
j
i=0
j
j+1 = i)
if x > yh
j
Thus, if V h(x) has already been computed then V h (x) = V h(x) for x = 1, . . . , yh. For x > yh we can
j
j+1
j
j
j
use the recursion V h (x) = V h (x − 1) + ∆V h (x) starting with x = yh + 1 in conjunction with the second
j+1
j+1
j+1
j
part of Proposition 4.
12
To estimate the expected revenue and other measures of performance, such as the variance, we can also
use Monte Carlo simulation. Suppose we generate many random copies of simulated demands (D1, . . . ,
Dn).
For each copy we compute sales (sh, . . . , sh) and revenues Rh =
n
p
under heuristic h. Averaging
1
n
i=1
ish
i
over all the values of Rh gives an estimate of V h(c). Simulated sales can be generated sequentially via sh
=
n
j
min(Dj, (xj −yh
)+) starting with j = n and x
.
n−1
n = c and using the capacity update formula xj = xj+1 − sh
j+1
Example 3 (continued) We have applied the EMSR-a and EMSR-b heuristics to the data of Example 3.
Table
4 repeats the data and reports the heuristic protection levels ya, yb as well as the optimal protection
levels.
Table 5 reports V a(c) and V b(c) as well as V
5
5
5(c) for values of c ∈ {50, 100, 150, 200, 250, 300, 350}.
j
pj
E[Dj]
ya
yb
y
j
j
j
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 a(c)
V b(c)
V
5
5
5(c)
50
560%
3,426.8
3,426.8
3,426.8
100
280%
5,4
31
.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
n
V U (c, D)
=
max
p
n
k
xk
(11)
k=1
s.t.
xk
≤
Dk
∀
k = 1, . . . , n
n
xk
≤
c
k=1
xk
≥
0
∀
k = 1, . . . , n.
13
Clearly, for each realization of D, advance knowledge results in revenues that are at least as high as the
optimal
dynamic policy that does not have perfect foresight. As a result, Vn(c) ≤ EV U (c, D). For convenience,
we
n
will denote this upper bound as V U (c) = EV U (c, D).
n
n
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:
n
V U (c)
=
p
n
k E min(Dk , (c − D[1, k − 1])+)
(12)
k=1
n
=
pk (E min(D[1, k], c) − E min(D[1, k − 1], c))
k=1
n
=
(pk − pk+1)E min(D[1, k], c)
(13)
k=1
where for convenience we define pn+1 = 0. Moreover, since V U (c, D) is concave in D, it follows from
Jensen’s
n
inequality that V U (c) = EV U (c, D) ≤ V U (c, µ) where V U (c, µ) is the solution to formulation (11)
with
n
n
n
n
µ = E[D] instead of D. More precisely,
n
V U (c, µ)
=
max
p
n
k xk
(14)
k=1
s.t.
xk
≤
µk
∀
k = 1, . . . , n
n
xk
≤
c
k=1
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 U (c) ≤ V U (c, µ) =
n
(p
n
n
k=1
k − pk+1) min(µ[1, k], c).
A lower bound can be obtained by assuming a low to high arrival pattern with zero protection levels.
This gives rise to sales min(Dk, (c − D[k + 1, n])+) at fare k = 1, . . . , n and revenue lower bound V L(c,
D) =
n
p
k=1
k E min(Dk , (c − D[k + 1, n])+).
Taking expectations we obtain
n
V L(c)
=
p
n
k E min(Dk , (c − D[k + 1, n])+)
(15)
k=1
n
=
pk (E min(D[k, n], c) − E min(D[k + 1, n], c))
k=1
n
=
(pk − pk−1)E min(D[k, n], c)
(16)
k=1
where p0 = 0. Notice that all of the terms in the sum are negative except for k = 1. Clearly
V L(c) ≤ V
n
n(c)
since the expected revenue is computed under sub-optimal protection levels. The above arguments justify
the
main result of this section.
14
Proposition 5
V L(c) ≤ V
(c) ≤ V U (c, µ)
(17)
n
n(c) ≤
V U
n
n
Of course, the bounds require the computation of E min(D[1, k], c), k = 1, . . . , n. However, this is often
an easy computation. Indeed, if D[1, k] is any non-negative integer random variable then E min(D[1, k], c)
=
c
P (D[1, k] ≥ j). If D[1, k] is Normal we can take advantage of the fact that E min(Z, z) = z(1 − Φ(z)) −
j=1
φ(z) when Z is a standard Normal random variable and φ is the standard Normal density function. If
follows
that if D[1, k] is Normal with mean µ and variance σ2, then
E min(D[1, k], c) = µ + σE min(Z, z) = µ + σ [z(1 − Φ(z)) − φ(z)]
where z = (c − µ)/σ.
Tables 6 and 7 report V L(c), V
(c) and V
n
n(c), V U
n
n(c, µ) for the data of Examples 4 and 5, respectively.
Notice that V U (c) represents a significant improvement over the better known bound V
n
n(c, µ), particularly
for intermediate values of capacity. The spread V U (c) − V L(c) between the lower and upper bound is a
n
n
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 L(c)
V
(c)
V U (c, µ)
n
n(c)
V U
n
n
80
$42,7
28
$49,642
$53,039
$53,315
90
$48,493
$54,855
$58,293
$58,475
100
$54,415
$60,015
$63,366
$63,815
110
$60,393
$65,076
$68,126
$69,043
120
$66,180
$69,801
$72,380
$74,243
130
$71,398
$73,9
26
$75,923
$79,443
140
$75,662
$77,252
$78,6
18
$82,563
150
$78,751
$79,6
17
$80,456
$82,563
160
$80,704
$81,100
$81,564
$82,563
Table 6: Optimal Revenue and Bounds for Example 4.
c
V L(c)
V
(c)
V U n(c, µ)
n
n(c)
V U
n
80
$52,462
$67,505
$72,717
$73,312
90
$61,215
$74,003
$79,458
$80,302
100
$70,136
$79,615
$85,621
$87,292
110
$78,803
$84,817
$91,1
22
$92,850
120
$86,728
$89,963
$95,819
$98,050
130
$93,446
$94,869
$99,588
$103,250
140
$98,630
$99,164
$102,379
$106,370
150
$102,209
$102,418
$104,251
$106,370
160
$104,385
$104,390
$105,368
$106,370
Table 7: Optimal Revenue and Bounds for Example 5.
5.1
Revenue Opportunity Model
The bounds presented here can help with the so called Revenue Opportunity Model (ROM). The revenue
opportunity is the spread between the optimal revenue, obtained by hindsight using the estimated
uncensored
15
demand, and the revenue that results from not applying booking controls. Demand uncensoring refers to a
sta-
tistical technique that attempts to estimate actual demand from the observed sales which may be
constrained
by booking limits. The ex-post optimal revenue is a hindsight optimization and is equivalent to our perfect
foresight model, resulting in revenue V U (c, D), where D is the uncensored demand. On the other hand,
the
n
revenue based on not applying booking controls is just V L(c, D), so a measure of the revenue opportunity
is
n
V U (c, D) − V L(c, D). The achieved revenue opportunity is the difference between the actual revenue
from
n
n
applying optimal or heuristic controls and the lower bound. The ratio of the achieved revenue opportunity
to
the revenue opportunity is often called the percentage achieved revenue opportunity. The revenue
opportunity
V U (c, D) − V L(c, D) is sometimes approximated by V U (c) − V L(c) to get an idea of the revenue
opportunity.
n
n
n
n
Table 6 and 7 shows there is significant revenue opportunity, particularly for c ≤ 140. Thus, one use for
the
ROM is to identify situations where RM has the most potential so that more effort can be put where is
most
needed. The ROM has also been used to show the benefits of using leg-based control versus network-
based
controls. The reader is refer to Chandler and Ja ([8]) and to Temath et al. ([21]) for further information on
the uses of the ROM.
5.2
Bounds Based Heuristic
It is common to use an approximation to the value function as a heuristic. To do this, suppose that
˜
Vj(x) is
an approximation to Vj(x). Then a heuristic admission control rule can be obtained as follows:
˜
yj = max{y ∈ N : ∆ ˜
Vj(y) > pj+1} j = 1, . . . , n − 1.
(18)
Suppose we approximate the value function Vj(x) by ˜
Vj(x) = θV L(x) + (1 − θ)V U (x) for some θ ∈ [0, 1]
j
j
and V L(x) and V U (x) are the bounds obtained in this section applied to n = j and c = x. Notice that
j
j
∆V L(x) = p
(p
(x) =
j−1 (p
j
1P (D[1, j] ≥ x) +
j
k=2
k − pk−1)P (D[k, j] ≥ x), while ∆V U
j
k=1
k − 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
λ
jt, 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
t
Poisson with mean Λjt =
λ
0
jsds. For simplicity we will write Λj , instead of ΛjT , to denote the expected
number of requests for fare j over the entire horizon [0, T ]. The low-to-high arrival pattern can be
embedded
into the time varying model by dividing the selling horizon into n sub-intervals [tj−1, tj], j = 1, . . . , n with
tj = jT /n, and setting λjt = nΛj/T over t ∈ [tj−1, tj] and λjt = 0 otherwise.
Let V (t, x) denote the maximum expected revenue that can be attained over the last t units of the sale
horizon with x units of capacity. We will develop both discrete and continuous time dynamic programs to
compute V (t, x). To construct a dynamic program we will need the notion of functions that go to zero
faster
16
than their argument. More precisely, we say that a function g(x) is o(x) if limx↓0 g(x)/x = 0. We will show
that the probability that over the interval [t − δt, t] there is exactly one request and the request is for
product
j is of the form λjtδt + o(δt). To see this notice that the probability that a customer arrives and requests
one
unit of product j over the interval [t − δt, t]is
λjtδ exp(−λjtδt) + o(δt) = λjtδt[1 − λjtδt] + o(δt) = λjtδt + o(δt),
while the probability that there are no requests for the other products over the same interval is
exp(−
λktδt) + o(δt) = 1 −
λktδt + o(δt).
k=j
k=j
Multiplying the two terms and collecting terms we obtain λjtδt + o(δt) as claimed.
Recall that some fares have embedded time-of-purchase restrictions. Let Nt ⊂ N = {1, . . . , n} to be the
set of allowable fares at time-to-go t. Usually Nt = N for large t, but low fares are dropped from Nt as the
time-of-purchase restrictions become binding.
We can now write
V (t, x)
=
λjtδt max(pj + V (t − δt, x − 1), V (t − δt, x)) + (1 −
λjtδt)V (t − δt, x) + o(δt)
j∈Nt
j∈Nt
=
V (t − δt, x) + δt
λjt[pj − ∆V (t − δt, x)]+ + o(δt)
(19)
j∈Nt
with boundary conditions V (t, 0) = 0 and V (0, x) = 0 for all x ≥ 0, where ∆V (t, x) = V (t, x) − V (t, x −
1) for
x ≥ 1 and t ≥ 0.
Subtracting V (t − δt, x) from both sides of equation (19), dividing by δt and taking the limit as δt ↓ 0, we
obtain the following equation, known as the Hamilton Jacobi Bellman (HJB) equation:
∂V (t, x) =
λjt[pj − ∆V (t, x)]+
(20)
∂t
j∈Nt
with the same boundary conditions. The equation tells us that the rate at which V (t, x) grows with t is the
weighted sum of the positive part of the fares net of the marginal value of capacity ∆V (t, x) at state (t, x).
While the value function can be computed by solving and pasting the differential equation (20), in practice
it is easier to understand and compute V (t, x) using a discrete time dynamic programming formulation. A
discrete time dynamic programming formulation emerges from (19) by rescaling time, setting δt = 1, and
dropping the o(δt) term. This can be done by selecting a > 1, so that T ← aT is an integer, and setting
λjt ← 1 λ
λ
a
j,t/a, for t ∈ [0, aT ]. The scale factor a should be selected so that, after scaling,
j∈N
jt << 1,
t
e.g.,
λ
j∈N
jt ≤ .01 for all t. The resulting dynamic program, after rescaling time, is given by
t
V (t, x) =
V (t − 1, x) +
λjt[pj − ∆V (t − 1, x)]+.
(21)
j∈Nt
with the same boundary conditions. Computing V (t, x) via (21) is quite easy and fairly accurate if time is
scaled appropriately. For each t, the complexity is order O(n) for each x ∈ {1 . . . , c} so the complexity
per
period is O(nc), and the overall computational complexity is O(ncT ).
A formulation equivalent to (21) was first proposed by Lee and Hersh [13], who also show that
∆V (t, x)
is
increasing in t and decreasing in x. The intuition is that the marginal value of capacity goes up if we have
more
time to sell and goes down when we have more units available for sale. From the dynamic program (21),
it is
optimal to accept a request for product j when pj ≥ ∆V (t − 1, x) or equivalently, when pj + V (t − 1, x −
1) ≥
V (t − 1, x), i.e., when the expecte revenue from accepting the request exceeds the expected revenue of
denying
the request. Notice that if it is optimal to accept a request for fare j, then it is also optimal to accept a
request
17
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
λ
λ
j=1
jt be the overall arrival rate at time t.
Define πjt =
j
k=1
kt/λt and
rjt =
j
p
k=1
k λkt/λt.
We can think of πjt and rjt as the probability of sale and the average revenue rate,
per arriving customer, when we offer all the fares in the consecutive set Sj = {1, . . . , j}. For
convenience,
let π0t = r0t = 0 denote, respectively, the sales rate and the revenue rate associated with S0 = ∅. Now let
qjt = rjt/πjt be the average fare per unit sold when the offer set is Sj. If πjt = 0, we define qjt = 0. This
implies that πjt[qjt − ∆V (t − 1, x)] is zero whenever πjt = 0, e.g., when j = 0.
Let N + = N
t
t ∪ {0}. With this notation, we can write formulation (21) as
V (t, x)
=
V (t − 1, x) + λt max [rjt − πjt∆V (t − 1, x)]
j∈N +
t
=
V (t − 1, x) + λt max πjt[qjt − ∆V (t − 1, x)].
(22)
j∈N +
t
The reason to include 0 as a choice is that for j = 0, the term vanishes and this allow us to drop the
positive
part that was present in formulation (21). The equivalent formulation for the continuous time model (20)
is
∂V (t, x) = λt max πjt[qjt − ∆V (t, x)].
(23)
∂t
j∈N +
t
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
18
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) =
λjt
Pj(z)[zpj − ∆zV (t, x)]+,
(24)
∂t
j∈Nt
z=1
while the dynamic program (21) with compound Poisson demands is given by
∞
V (t, x) = V (t − 1, x) +
λjt
Pj(z)[zpj − ∆zV (t − 1, x)]+,
(25)
j∈Nt
z=1
with boundary conditions V (t, 0) = V (0, x) = 0. Notice that the sums in (24) and (25) can be changed to
x
z=1
∞
instead of
as the terms z > x do not contribute to the sum given our convention that ∆
z=1
z V (t, x) = ∞ for
x > z. The optimal policies for the two programs are, respectively, to accept a size z request fare pj, j ∈
Nt,
if zpj ≥ ∆zV (t, x), and to accept a z request fare pj, j ∈ Nt, if zpj ≥ ∆zV (t − 1, x). The two policies
should
largely coincide time is scaled correctly so that
λ
j∈N
jt << 1 for all t ∈ [0, T ].
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)
19
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
$11,7
24
$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
T
parameter Λj =
λ
0
jtdt. Suppose each arrival is of random size Zj . Then the aggregate demand, say Dj , for
fare j is equal to
Nj
Dj =
Zjk,
(26)
k=1
where Zjk is the size of the kth request. It is well known that E[Dj] = E[Nj]E[Zj] = ΛjE[Zj] and that
Var[Dj] = E[Nj]E[Z2] =
Λ
], where E[Z2] is the second moment of Z
j
j E[Z 2
j
j
j . Notice that Jensen’s inequality
implies that the the variance to mean ratio E[Z2
]/E[Z
j
j ] ≥ E[Zj ] ≥ 1.
In practice, demands D1, . . . , Dn are fed, under the low-to-high arrival assumption, into static policies to
compute Vj(c), j = 1, . . . , n and protection levels yj, j = 1, . . . , n − 1 using the dynamic program (8), or
to
the EMSR-b heuristic to compute protection levels
yb, . . . , yb
. Since the compound Poisson demands are
1
n−1
difficult to deal with numerically, practitioners often approximate the aggregate demands Dj by a Gamma
distribution with parameters αj and βj, such that αjβj = E[Nj]E[Zj] and αjβ2 = E[N
], yielding
j
j ]E[Z 2
j
αj = ΛjE[Zj]2/E[Z2], and β
]/E[Z
j
j = E[Z 2
j
j ].
We are interested in comparing the expected revenues obtained from static policies to those of dynamic
policies. More precisely, suppose that that demands are compound poisson and Dj is given by (26) for
every
j = 1, . . . , n. Suppose that protection levels y1, y2, . . . , yn−1 are computed using the low-to-high static
dynamic program (8) and let yb, yb, . . . , yb
be the protection levels computed using the EMSR-b heuristic. Protection
1
2
n−1
levels like these are often used in practice in situations where the arrival rates λjt, t ∈ [0, T ], j = 1, . . . ,
n, are not necessarily low-to-high. Two possible implementations are common. Under theft nesting a size
z request
for fare class j as state (t, x) is accepted if x − z ≥ yj−1. This method is called theft nesting because the
remaining inventory x at time-to-go t is x = c − b[1, n] includes all bookings up to time-to-go t, including
bookings b[1, j − 1]. Standard nesting counts only bookings for lower fare classes and is implemented by
accepting a size z request for fare j at state (t, x) if x − z ≥ (yj−1 − b[1, j − 1])+, where b[1, j − 1] are the
observed bookings of fares [1, j − 1] up to state (t, x). When c > yj−1 > b[1, j − 1], this is equivalent to
accepting a request for z units for fare j if c − b[j, n] − z ≥ yj−1, or equivalently if b[j, n] + z ≤ c − yj−1,
so
only bookings of low fares count. In practice, standard nesting works much better than theft nesting when
the arrival pattern is not
low-to-high.
Notice that the expected revenue, say V s(T, c), resulting from applying the static protection levels y1, . . .
, yn−1
with theft nesting is not, in general, equal to Vn(c), the optimal expected revenue when the arrivals are
low-
to-high. Similarly, the expected revenue, say V b(T, c), resulting from applying the EMSR-b protection
levels
yb, . . . , yb
with theft nesting is not, in general, equal to V b(c), the expected revenue when the arrivals are
1
n−1
n
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).
20
In fact, a proof is hardly needed as we are comparing heuristics to optimal dynamic
policies.
Proposition 7
V s(T, x) ≤ V (T, x) ∀x ∈ {0, 1, . . . , c}.
Proof: Clearly for V s(0, x) = V (0, x) = 0 so the result holds for t = 0, for all x ∈ {0, 1, . . . , c}. Suppose
the result holds for time-to-go t − 1, so V s(t − 1, x) ≤ V (t − 1, x) for all x ∈ {0, 1, . . . , c}. We will show
that it also holds for time-to-go t. If a request of size z arrives for fare class j, at state (t, x), the policy
based on
protection levels y1, . . . , yn−1 will accept the request if x − z ≥ (yj−1 − b[1, j − 1])+ and will rejected
otherwise.
In the following equations, we will use Qj(z) to denote P (Zj > z). We have
x−(yj−1−b[1,j−1])+
V s(t, x)
=
λjt[
Pj(z)(zpj + V s(t − 1, x − z))
j∈Nt
z=1
+
Qj(x − (yj−1 − b[1, j − 1])+)V s(t − 1, x)] + (1 −
λjt)V s(t − 1, x)
j∈Nt
x−(yj−1−b[1,j−1])+
≤
λjt[
Pj(z)(zpj + V (t − 1, x − z))
j∈Nt
z=1
+
Qj(x − (yj−1 − b[1, j − 1])+)V (t − 1, x)] + (1 −
λjt)V (t − 1, x)
j∈Nt
x−(yj−1−b[1,j−1])+
=
V (t − 1, x) +
λjt
Pj(z)(zpj − ∆zV (t − 1, x))
j∈Nt
z=1
∞
≤
V (t − 1, x) +
λjt
Pj(z)(zpj − ∆zV (t − 1, x))+
j∈Nt
z=1
=
V (t, x),
where the first equation follows from the application of the protection level policy, the first inequality
follows
from the inductive hypothesis V s(t − 1, x) ≤ V (t − 1, x). The second equality collects terms, the second
inequality follows because we are taking positive parts, and the last equality from the definition of V (t,
x).
While we have shown that V s(T, c) ≤ V (T, c), one may wonder whether there are conditions where
equality
holds. The following results answers this question.
Corollary 2 If the Dj’s are independent Poisson random variables and the arrivals are low-to-high then
Vn(c) = V s(T, c) = V (T, c).
Proof: Notice that if the Djs are Poisson and the arrivals are low-to-high, then we can stage the arrivals
so that λjt = nE[Dj]/T over t ∈ (tj−1, tj] where tj = jT /n for j = 1, . . . , n. We will show by induction in
j that Vj(x) = V (tj, x). Clearly y0 = 0 and V1(x) = p1E min(D1, x) = V (t1, x) assuming a sufficiently
large
rescale factor. Suppose, by induction, that Vj−1(x) = V (tj−1, x). Consider now an arrival at state (t, x)
with
t ∈ (tj−1, tj]. This means that an arrival, if any, will be for one unit of fare j. The static policy will accept
this
request if x − 1 ≥ yj−1, or equivalently if x > yj−1. However, if x > yj−1, then ∆(t − 1, x) ≥ ∆V (tj−1, x) ≥
pj, because ∆V (t, x) is increasing in t and because yj−1 = max{y : ∆Vj−1(y) > pj} = max{y : ∆V (tj−1, x)
> pj},
by the inductive hypothesis. Conversely, if the dynamic program accepts a request, then pj ≥ ∆V (t, x) and
therefore x > yj−1 on account of ∆V (t, x) ≥ ∆V (tj−1, x).
We have come a long way in this chapter and have surveyed most of the models for the independent
demand case. Practitioners and proponents of static models, have numerically compared the performance
of
static vs dynamic policies. Diwan [7], for example, compares the performance of the EMSR-b heuristic
against
the performance of the dynamic formulation for Poisson demands (21) even for cases where the aggregate
21
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 U (c) for V
n
n(c), developed in Section 5 for the static multi-fare
model is still valid for V (T, c). The random revenue associated with the perfect foresight model is Vn(c,
D)
and can be obtained by solving the linear program (11). Notice that for all sample paths, this revenue is
at least as large as the revenue for the dynamic policy. Taking expectations we obtain V (T, c) ≤ V U (c) =
n
EVn(c, D) =
n
(p
k=1
k − pk+1)E min(D[1, k], c), where for convenience pn+1 = 0.
Moreover, since dynamic
policies do at least as well as static policies, the lower bounds obtained in Section 5 also apply to
dynamic
policies.
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,
k
k
Wk(t, x)
=
λit[pi + Vk(t − 1, x − 1)] + (1 −
λit)Vk(t − 1, x)
i=1
i=1
=
Vk(t − 1, x) + rkt − πkt∆Vk(t − 1, x)
=
Vk(t − 1, x) + πkt[pkt − ∆Vk(t − 1, x)],
where ∆Vk(t, x) = Vk(t, x) − Vk(t, x − 1), where πkt =
k
λ
p
i=1
it and rkt =
j
i=1
iλit and pkt = rkt/πkt when
πkt > 0 and pkt = 0 otherwise.
22
Then Vj(t, x) satisfies the dynamic program
Vj(t, x) = max Wk(t, x) = max{Wj(t, x), Vj−1(t, x)}
(27)
k≤j
with the boundary conditions Vj(t, 0) = Vj(0, x) = 0 for all t ≥ 0 and all x ∈ N for all j = 1, . . . , n. Notice
that the optimization is over consecutive subsets Sk = {1, . . . , k}, k ≤ j. It follows immediately that Vj(t,
x)
is monotone increasing in j. An equivalent version of (27) for the case n = 2 can be found in Weng and
Zheng [23]. The complexity to compute Vj(t, x), x = 1, . . . , c for each j is O(c) so the complexity to
compute
Vj(t, x), j = 1, . . . , n, x = 1, . . . , c is O(nc). Since there are T time periods the overall complexity is
O(ncT ).
While computing Vj(t, x) numerically is fairly simple, it is satisfying to know more about the structure of
optimal policies as this gives both managerial insights and can simplify computations. The proof of the
structural results are intricate and subtle, but they parallel the results for the dynamic program (8) and
(21).
The following Lemma is the counterpart to Lemma 2 and uses sample path arguments based on ideas in
[23]
to extend their results from n = 2 to general n. The proof can be found in the Appendix.
Lemma 3 For any j ≥ 1,
a) ∆Vj(t, x) is decreasing in x ∈ N+, so the marginal value of capacity is diminishing.
b) ∆Vj(t, x) is increasing in j ∈ {1, . . . , n} so the marginal value of capacity increases when we have
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
23
implies that yj(t) is increasing in j for any t ≥ 0. If t > t, then aj+1(t , yj(t)) ≤ aj+1(t, yj(t)) < j + 1, so
yj(t ) ≥ yj(t). Since yj(t) ≤ yi(t) for all i > j, then x ≤ yj(t) implies Vi+1(t, x) = Vi(t, x) for all i ≥ j and
therefore Vi(t, x) = Vj(t, x) for all i > j.
The policy is implemented as follows: The starting state is (n, T, c) as we can use any of the fares {1, . . .
, n},
we have T units of time to go and c is the initial inventory. At any state (j, t, x) we post fares Aj(t, x) =
{1, . . . , aj(t, x)}. If a unit is sold during period t the state is updated to (aj(t, x), t − 1, x − 1) since all
fares in the set Aj(t, x) are allowed, the time-to-go is t − 1 and the inventory is x − 1. If no sales occur
during period
t the state is updated to (aj(t, x), t − 1, x). The process continues until either t = 0 or x = 0.
Example 6. Consider Example 1 again with 5 fares p1 = $100, p2 = $60, p3 = $40, p4 = $35 and p5 =
$15
with independent Poisson demands with means Λ1 = 15, Λ2 = 40, Λ3 = 50, Λ4 = 55 and Λ5 = 120 and
T = 1. The scaling factor was selected so that
5
Λ
i=1
i/a < .01 resulting in T ← aT = 2, 800.
We also
assume that the arrival rates are uniform over the horizon [0, T ], i.e., λj = Λj/T . In Table 10 we present
the expected revenues Vj(T, c), j = 1, . . . , 5 and V (T, c) for c ∈ {50, 100, 150, 200, 250}. The first row
is
V5(c)
from Example 1. Notice that V5(c) ≤ V5(T, c). This is because we here we are assuming uniform, rather
than
low-to-high arrivals. V (T, c) is even higher because we have the flexibility of opening and closing fares
at
will. While the increase in expected revenues [V (T, c) − V5(T, c)] due to the flexibility of opening and
closing
fares may be significant for some small values of c (it is 1.7% for c = 50), attempting to go for this extra
revenue may invite strategic customers or third parties to arbitrage the system. As such, it is not generally
recommended in practice.
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)
rjt − πjt∆Vj(t, x)
if (t, x) /
∈ Zj−1
=
(28)
∂t
∂Vj−1(t,x)
if (t, x) ∈ Z
∂t
j−1
with the same boundary conditions.
8.1
Mark-up and Mark-down Policies
We now go back to the broader pricing interpretation coupled with the monotonic fare formulation (27).
In
many applications the price menu pjt, j = 1, . . . , n is time invariant, but the associated sales rates πjt, j =
1, . . . , n are time varying. In addition, we will assume that there is a price p0t such that π0t = 0 for all t.
This technicality helps with the formulation as a means of turning off demand when the system runs out of
inventory. The case p1t ≥ p2t ≥ . . . ≥ pnt and π1t ≤ π2t ≤ . . . ≤ πnt is known as the mark-up problem,
while
the case p1t ≤ p2t ≤ . . . ≤ pnt and π1t ≥ π2t ≥ . . . ≥ πnt is known as the mark-down problem. The former
model is relevant in Revenue Management while the second is relevant in Retailing.
For the RM formulation, the problem can be viewed as determining when to mark-up (switch from action
j to j − 1). The optimal mark-up times are random as they depend on the evolution of sales under the
optimal policy. Suppose that the current state is (j, t, x), so the last action was j, the time-to-go is t and the
inventory is x. We want to determine whether we should continue using action j or switch to action j − 1.
We know that if x > yj−1(t), then we should keep action j and if x ≤ yj−1(t) then we should close action
24
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.
25
10
Appendix
Proof of Lemma 1. Notice that g(y) = G(y)P (X ≥ y) +
G(j)P (X = j), while g(y − 1) = G(y −
j≤y−1
1)P (X ≥ y) +
G(j)P (X = j). Taking the difference yields ∆g(y) = G(y)P (X ≥ y). Notice that
j≤y−1
r(y) = R(y)P (X < y) +
R(j)P (X = j) while r(y − 1) = R(y − 1)P (X < y) +
R(j)P (X = j).
j≥y
j≥y
Taking the difference we see that ∆r(y) = ∆R(y)P (X < y).
Proof of Proposition 1. Let G(y) = p1y, then V1(y) = g(y) = EG(min(D1, y)), so ∆V1(y) = ∆g(y) =
p1P (D1 ≥ y). This establishes the first part of the Proposition. To establish the second part of the
Proposition
we use the first part of Lemma 1 to show that p2E min(D2, c − y) − p2E min(D2, c − y + 1) = −p2P (D2 >
c − y)
and the second part of the Lemma 1 to show that EV1(max(x − D2, y)) − EV1(max(x − D2, y − 1)) =
∆V1(y)P (D2 > c−y). The second part of the Proposition then follows from putting the two parts together.
To
see the first part, let r(y) = p2c−p2E max(c−D2, y), then ∆r(y) = p2E min(D2, c−y)−p2E min(D2, c−y+1)
=
∆R(y)P (c − D2 < y) where R(y) = −p2y, so ∆r(y) = −p2P (c − D2 < y) = −p2P (D2 > c − y). Now let
R(y) = V1(y), then ∆r(y) = ∆V1(y)P (c − D2 < y) = ∆V1(y)P (D2 > c − y) completing the proof.
Proof of Lemma 2: We will prove the above result by induction on j. The result is true for j = 1 since
∆V1(y) = p1P (D1 ≥ y) is decreasing in y and clearly ∆V1(y) = p1P (D1 ≥ y) ≥ ∆V0(y) = 0. Assume that
the
result is true for Vj−1. It follows from the dynamic programming equation (8) that
Vj(x) = max {Wj (y, x))} ,
y≤x
where for any y ≤ x,
Wj(y, x) = E [pj min {Dj, x − y}] + E [Vj−1 (max {x − Dj, y})]
A little work reveals that for y ∈ {1, . . . , x}
∆Wj(y, x) = Wj(y, x) − Wj(y − 1, x) = [∆Vj−1(y) − pj] P (Dj > x − y).
Since ∆Vj−1(y) is decreasing in y (this is the inductive hypothesis), we see that Wj(y, x) ≥ Wj(y − 1, x)
if ∆Vj−1(y) > pj and Wj(y, x) ≤ Wj(y − 1, x) if ∆Vj−1(y) ≤ pj.
Consider the expression
yj−1 = max{y ∈ N : ∆Vj−1(y) > pj}
(29)
where the definition of ∆Vj(y) is extend to y = 0 for all j by setting ∆Vj(0) = p1. If yj−1 ≤ x then
Vj(x) = max Wj(y, x) = Wj(yj−1, x)
y≤x
On the other hand, if x < yj−1 then
Vj(x) = max Wj(y, x) = Wj(x, x).
y≤x
In summary,
Vj(x)
=
Wj(min(x, yj−1), x)
V
=
j−1(x),
if x ≤ yj−1
E [pj min {Dj, x − yj−1}] + E [Vj−1 (max {x − Dj, yj−1})]
if x > yj−1
26
Computing ∆Vj(x) = Vj(x) − Vj(x − 1) for x ∈ N results in:
∆V
∆V
j−1(x),
if x ≤ yj−1
j (x)
=
(30)
E min(pj, ∆Vj−1(x − Dj))
if x > yj−1
We will now use this result to show that ∆Vj(x) is itself decreasing in x. Since ∆Vj(x) = ∆Vj−1(x) for
x ≤ yj−1 and ∆Vj−1(x) is decreasing in x we only need to worry about the case x > yj−1.
However, in this case
∆Vj(x) = E min(pj, ∆Vj−1(x − Dj))
is decreasing in x since ∆Vj−1(x) is itself decreasing in x.
We now show that ∆Vj(x) ≥ ∆Vj−1(x). For x > yj−1 we have min(pj, ∆Vj−1(x−Dj)) ≥ min(pj, ∆Vj−1(x))
=
∆Vj−1(x) where the inequality follows since ∆Vj−1(x) is decreasing in x and the equality since x > yj−1.
Tak-
ing expectations we see that ∆Vj(x) ≥ ∆Vj−1(x) on x > yj−1 while ∆Vj(x) = ∆Vj−1(x) on x ≤ yj−1.
Proof of Proposition 2. We first show that ∆Vj+1(x) = ∆Vj(x) for x = 1, . . . , yj. This follows since
Vj+1(x) = Wj+1(x, x) = Vj(x) in this range as no capacity is made available for fare pj+1 when x ≤ yj.
For
x > yj
Vj+1(x)
=
pj+1E min(x − yj, Dj+1) + EVj(max(x − Dj+1, yj)
x−yj
=
pj+1
P r(Dj+1 ≥ k) + Vj(yj)P r(Dj+1 > x − yj)
k=1
x−yj
+
Vj(x − k)P r(Dj+1 = k).
k=0
Consequently, for x > yj
x−yj −1
∆Vj+1(x) = pj+1P r(Dj+1 ≥ x − yj) +
∆Vj(x − k)P r(Dj+1 = k)
(31)
k=0
follows from Vj(x − 1) = Vj(yk) for x = yj + 1. Since pj+1 < ∆Vj(y) for y ≤ yj, we can write ∆Vj+1(x) =
∞
min(p
k=0
j+1, ∆Vj (x − k))P (Dj+1 = k) = E min(pj+1, ∆Vj (x − Dj+1)).
Proof of Proposition 3. Since Πn(c, k) is the difference of a concave and a linear function, Πn(c, k) it is
itself concave. The marginal value of adding the cth unit of capacity is ∆Vn(c) − k so the cth unit increase
profits as long as ∆Vn(c) > k. Therefore, the smallest optimal capacity is given by c(k). (Notice that c(k)
+ 1
may be also optimal if ∆Vn(c(k) + 1) = k.) c(k) is decreasing in k since ∆Vn(c) is decreasing in c.
Suppose
that k = pj+1. To establish c(pj+1) = yj it is enough to show that that ∆Vn(yj) > pj+1 ≥ ∆Vn(yj + 1). By
definition yj = max{y ∈ N : ∆Vj(y) > pj+1} so ∆Vj(yj) > pj+1 ≥ ∆Vj(yj + 1). Since it is optimal to
protect
up to yj units of capacity for sale at fares j, j − 1, . . . , 1, it follows that Vn(c) = Vj(c) for all c ≤ yj, and
consequently ∆Vn(yj) = ∆Vj(yj) > pj+1. Now ∆Vn(yj + 1) can be written as a convex combination of
pj+1
and ∆Vj(yj + 1) ≤ pj+1 which implies that ∆Vn(yj + 1) ≤ pj+1, completing the proof.
Proof of Lemma 3: We will first show part that ∆Vj(t, x) is decreasing in x which is equivalent to
showing
that 2Vj(t, x) ≥ Vj(t, x + 1) + Vj(t, x − 1)] for all x ≥ 1. Let A be an optimal admission control rule starting
from state (t, x + 1) and let B be an optimal admission control rule starting from (t, x − 1). These
admission
control rules are mappings from the state space to subsets Sk = {1, . . . , k}, k = 0, 1, . . . , j where S0 = ∅
is
the optimal control whenever a system runs out of inventory. Consider four systems: two starting from
state
(t, x), using control rules A and B , respectively, and one each starting from (t, x + 1) and (t, x − 1), using
control rule A and B, respectively. Our goal is to specify heuristic control rules A and B that together
make
the expected revenues of the two systems starting with (t, x) at least as large as the expected revenues
from
the systems starting at (t, x + 1) and (t, x − 1). This will imply that 2Vj(t, x) ≥ Vj(t, x + 1) + Vj(t, x − 1).
27
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
28
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).
29
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
p
demand weighted average fare class price
j
D
fare class demand
y
protection level for a fare class
ya
protection level obtained using EM SRa heuristic
yb
protection level obtained using EM SRb 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.
30
[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.
31
