# question 1,4,5,6

i need help

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

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

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

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

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 and Dynamic ProgrammingReview
• 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

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

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

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

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

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

¡�  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

¡�  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

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

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

¡�  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

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