Profit Maximization in Air Cargo Overbooking ∗ by act50979

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             Profit Maximization in Air Cargo Overbooking


                                    Lama Moussawi and Metin Cakanyıldırım†
                                                            ¸

                                              School of Management
                                              P.O.Box 830688, SM 30
                                            University of Texas at Dallas
                                            Richardson, TX 75083-0688

                                                    August 10, 2005




                                                         Abstract
          This paper formulates a two-dimensional cargo overbooking problem with a profit maximization objec-
      tive. It provides a detailed formulation where the revenues are computed individually cargo by cargo. An
      aggregate formulation, requiring much less data to obtain, is shown to provide a lower bound for the detailed
      formulation. The aggregate formulation is solved under infinite and finite booking requests and is shown
      to yield the same optimal overbooking curve under both cases. Moreover, the optimal curve is a “box”,
      defined by only two numbers. Another easy-to-solve formulation is devised to construct an upper bound for
      the problem. We numerically show that the gap between lower and upper bound is small for a wide range
      of problems and hence conclude that the aggregate formulation is an effective approximation. Furthermore,
      the aggregate formulation can be solved efficiently and its simple solution “box” can be implemented by air
      cargo professionals.
      Keywords: Revenue Management, Transportation, Air Cargo, Overbooking.




∗
    Authors thank R. Kasilingam for introducing this problem to them.
†
    {lam018100, metin}@utdallas.edu
1    Introduction
    Cargo management is gaining importance as it is generating substantial revenues for airlines in the saturated
passenger markets. According to Boeing (World Air Cargo Forecast 2005) the cargo revenue, which represents
about 15% of the total traffic revenue, is expected to grow at an average annual rate of 6.2% for the next 20
years. According to the BTS (Bureau of Transportation Statistics 2004), cargo is one of the fastest growing
segments of the U.S. economy. The bureau also notes that air freight has experienced one of the fastest growth
in the cargo industry. Such expectations of growth are not new (Kasilingam 1996). As a growing market, air
cargo business in general deserves more attention. In particular, it is important to efficiently manage the cargo
bookings.
    Cargo booking process takes place as follows. Nowadays, many airlines accept reservations on the inter-
net. They allow certified shippers to make booking requests by logging onto a cargo booking website (e.g.
aacargo.com, unitedcargo.com), but the medium in which the reservation takes place is not very important for
the current paper. What is more important is the information exchanged between the shipper and the airline.
A shipper specifies to the best of his knowledge the information about the cargo to be shipped: the origin,
destination, weight and volume. Most often, shippers specify a time range for the shipment. When the time
range is very tight, it may include a single flight between the origin and the destination pair. This case is
similar to the case of a passenger purchasing a ticket on a particular flight. On the other hand, when the time
range is wide, it may include several flights departing at different times. Then the airlines typically attempt
to accomodate the cargo on the earliest flight departing within the given time range. Airlines assign a cargo
to a certain flight if the total cargo assigned to that flight, including the current cargo under consideration, is
within the overbooking limits. Then, we say that the booking request for the cargo is accepted. If the current
cargo brings the total cargos accepted for a flight above its overbooking limits, it cannot be accepted for that
flight. If the cargo cannot be accepted by any of the flights departing within the time range specified by the
shipper, alternative shipment times are often proposed by the airlines. This process repeats each day until the
departure.
    Not all cargo bookings made for a flight show up at the time of the departure. This can result in unutilized
capacity — called “spoilage” henceforth. The spoilage is a loss of potential revenue for airlines. In order to
avoid spoilage, airlines accept more bookings than the available capacity, i.e., they overbook. As a consequence,
however, the cargo that shows up at the departure may exceed the volume or the weight capacity. Then, some
of the showing up cargo is not loaded on the flight and it must be shipped with other flights or other cargo
carriers. The cargo that is shipped with alternative routes is called “offloaded”. Offloaded cargo, either on
another aircraft or another carrier, causes additional costs. Profit from a particular flight consists of the revenue
generated by the loaded cargo minus the costs incurred by offloading. Both the spoilage and the offload reduce
profits for airlines; the spoilage decreases the revenue while the offload increases the cost. Both the revenues
and the offload costs increase as the overbooking limits increase. In order to maximize their profits, airlines
seek to find the optimal overbooking limits beyond which no cargo is accepted. This paper aims to provide
a simple overbooking problem formulation, which would yield easily implementable overbooking limits in the
practice. We especially look for a parsimonious formulation because “so many airlines don’t have adequate

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data to implement a [cargo] RM [Revenue Management] system” according to Talluri and van Ryzin (2004).
Therein lies the significance of the current paper.
   In this paper, we adopt a pricing structure currently used in business-to-business (B2B) transactions. This
is because “long-term customer relations take priority [in cargo RM]” where “a few important customers . . . ship
large volumes” as stated in Talluri and van Ryzin (2004). Indeed, the cargo airlines mostly work for businesses
such as Dell, Texas Instruments, and General Electrics. Airlines charge the shippers depending on both the
volume and the weight of the cargo. This pricing structure, although natural for cargos, differentiates cargo
overbooking from passenger overbooking. Another critical distinctive feature of cargo overbooking deals with
when the revenues are collected from the shippers. Unlike the passenger case, the cargo revenue is often
collected when the cargo shows up instead of when it is booked. A shipper usually does not pay a cargo carrier
if the shipper books space for the cargo but changes his opinion and does not ship the cargo. We call this as
“B2B pricing”. The B2B pricing occurs as a result of the contracts between the cargo carriers and the shippers.
Since the shippers tend to be big businesses, they have some power to dictate a pricing structure favorable to
them. Because of the competition among the cargo airlines, many cargo airlines accept such structures. This
paper studies the cargo overbooking problem with B2B pricing. Our solution can potentially be applied to the
maritime shippers where cargo is shipped via ships/tankers.

1.1   Literature review

   Overbooking area has the longest research history among all the other areas of revenue management (McGill
and van Ryzin, 1999, and Talluri and van Ryzin, 2004). Unlike cargo overbooking, extensive research has been
done on passenger overbooking. Rothstein (1985), McGill and van Ryzin (1999) and Talluri and van Ryzin
(2004) survey the research in revenue management, which mostly deals with passenger overbooking. We first
briefly review the passenger overbooking and then focus on cargo overbooking.
   Passenger Overbooking: Because of no-shows, passenger airlines may be forced to fly with some empty
seats, which results in spoilage. However, it is also possible that all or most of the overbooked demand shows
up. This leads to a situation, where there are more passengers than the seats at the time of departure. In
that case, some passengers are offloaded from their original flights. This is not an uncommon situation, so the
airline management justifies it to the passengers via in-flight magazines, e.g., Arpey (2004).
   We briefly survey some of the static and dynamic models studied in the literature. Static models find
an overbooking limit in each period independent of the limits in the previous or next periods. One of the
early static optimization models is developed by Beckmann (1958). He develops a model to minimize the sum
of expected spoilage and offload costs. Shlifer and Vardi (1975) consider three cases of airline overbooking:
single-leg flight carrying a single type of passenger, single-leg flight carrying two types of passengers, and a
two-leg flight. Bodily and Pfeifer (1992) obtain a formula for what the optimal probability of having offloaded
passengers should be, and specialize this formula for Binomial distribution as suggested by Thompson (1961).
   Dynamic models are proposed by Alstrup (1986), Belobaba (1989), Chatwin (1998) and Subramanian et al.
(1999). They all assume a single-leg flight. Alstrup (1986) and Subramanian et al. (1999) model the booking
process as a nonhomogeneous Markov decision process. Belobaba (1989) extends the model he developed for


                                                       2
seat inventory control to find the booking limits for fare classes. Chatwin (1998) assumes that passengers get
a partial refund in case of cancelations. He gives conditions for the optimality of a booking-limits policy for a
multi-period problem.
   Cargo Overbooking: Kasilingam (1996) discusses the main differences of passenger and cargo RM. In
cargo RM, the cargo is at least two dimensional (weight and volume) and can be shipped along any route
as long as it reaches its destination within the specified delivery time. Kasilingam also notes that the cargo
capacity available for sale is not known in advance on aircrafts carrying passengers. Talluri and van Ryzin
(2004) goes further by saying that “the decisions for both passenger and cargo are interrelated and ideally
should be coordinated by a single RM system”. The practice of cargo RM is far from such coordination so
we aim our models for aircrafts reserved only for cargo transportation. Kasilingam (1997) develops cargo
overbooking models that minimize the sum of the offload and spoilage costs for a given distribution of the
capacity and the show-up rate (the ratio of the showing up cargo to the booked cargo), the offload cost per
unit of capacity and the spoilage cost per unit of capacity. However, his models are single-dimensional.
   Luo et al. (2005) present the first two dimensional model for cargo overbooking but it has a cost minimization
objective as opposed to the profit maximization objective in the current paper. They introduce the concept
of an overbooking curve, which is also used in the current paper. They also restrict the optimal curve to be a
box determined by only two numbers. Therefore, unlike a curve which requires infinitely many pairs to define,
the box solution is easier to implement in the cargo RM practice, which currently suffers from the lack of
optimization applications.
   Cooper et al. (2005) consider a cargo booking problem with a maximization objective, where the volume of
each accepted cargo at the the time of the booking is random. In their multiperiod model, the number of booked
cargos of a certain type is known in each period. But because of random volumes, there may be spoilages and
offloads. They use the same revenue function as in the current paper but a linear offload cost function as in
Luo et al. (2005). They develop a heuristic based on approximating the value function and perform numerical
analysis to show that their heuristic outperforms the first-come first-booked policy. However, theirs is not an
overbooking problem.

1.2   Contribution and implementation of the paper

   In this paper, we formulate and solve a two-dimensional cargo overbooking problem with a nonlinear profit
and offload cost. To the best of our knowledge, this problem has not been formulated before, although cargo
RM is an important area. We provide a detailed and an aggregate formulation. The detailed formulation
considers cargos individually, but it is difficult to construct (due to extensive data needs) and solve. Therefore,
we approximate the detailed objective with the aggregate objective, which turns out to be a lower bound. The
aggregate formulation is, by design, parsimonious in its data needs. Moreover, the optimal overbooking curve
of the aggregate problem is easy to find and it turns out to be a box. A box is easier, than an arbitrary curve,
to communicate and implement. Therein lies the strength of our aggregate model. Furthermore, we develop
an upper bound, which is used numerically to show that the aggregate model is a good approximation of the
detailed model.


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    Besides being parsimonious and yielding a box as an optimal solution, our aggregate model yields the same
box regardless of the distribution of the total booking requests. This insensitivity is particularly important after
noting that airlines often do not keep the data necessary to estimate the distribution of the boking requests.
That is, databases do not record the booking requests that are not accepted. Thus, all the booking request data
are filtered by the aircraft capacity (or the previously used overbooking curves). It is possible to learn about
booking requests with censored data, where overbooking limits can be kept high for estimation purposes; see
filtered demand models of Ding et al. (2002) and Bensoussan et al. (2005), but such a sophisticated procedure
in cargo RM may be an overkill. The beauty of the aggregate model is its simplicity due to the avoidance of
estimation in the presence of filtered data.
    When the implementation of an air cargo model is the issue, we have to understand that the industry does
not record some of the necessary data and does not yet seem to be ready for sophisticated models. Considering
these two limitations, we provide an aggregate model, which runs with limited data, whose optimal solution
is easy to obtain (e.g. by using standard Microsoft Excel functions) and communicate. We believe that these
aspects of our model make it implementable in practice.
    The remainder of the paper is organized as follows. Section 2 discusses the revenue and offload cost functions
that we consider in our formulation. In Section 3, we provide detailed, aggregate and upper bound formulations.
This is ensued by introducing and analyzing the aggregate revenue problem in Section 4. The upper bound
problem is presented and analyzed in Section 5. We then pursue a performance analysis of the aggregate and
upper bound problems in Section 6. A brief conclusion is provided in Section 7. The proofs that do not appear
immediately after the associated lemmas are relegated to the appendix to maintain the fluency of the paper.


2    Two Dimensional Revenue and Offload Cost
    We present the pricing structure commonly found in the air cargo management practice. First, we define
the inverse density d for a cargo,
                                                      Volume in m3
                                               d :=                 .
                                                      Weight in ton
Note that the inverse density with units in m3 /ton is the reciprocal of the physical density. Airline companies
also specify a standard inverse density ds , which is currently 5 m3 /ton for all aircrafts at one of the leading
cargo carriers. Standard density represents the ideal ratio of the volume to weight for cargos such that the
aircraft capacity is used efficiently. Cargos whose inverse densities are drastically different than the standard
inverse density consume too much of one dimension of the capacity without utilizing the other dimension.
Standard density is employed to avoid such uneven capacity usage.
    Choice of the standard inverse density is a challenging issue on its own. A separate revenue optimization
model can be designed for each aircraft type to tackle this issue. Since each aircraft has different volume and
weight capacities, the standard inverse density should vary from one aircraft to another. However, aircraft
type dependent standard densities lead to different prices for the same cargo when shipped on different types
of aircraft. From a customer relationship management view, a carrier can use only one standard density for the
entire fleet. Industrial practice supports this conclusion; currently only one standard density (commonly based


                                                          4
on Boeing 747 freight airplanes) is used. For our discussion, the standard density is assumed to be given.
    Currently airlines combine weight and volume of a cargo to compute a “chargeable weight”, which is used
to compute the revenue from the cargo. The chargeable weight is max{Weight, Volume/ds }. Then, the revenue
obtained from a cargo depends on its volume and weight (or inverse density), and is given as

                               Revenue(Volume, Weight) = a max{Volume/ds , Weight},
                  Revenue(Volume, Inverse Density=d) = a max{Volume/ds , Volume/d},

where a is the price per chargeable weight. In this paper, we are focusing on a single flight, so that the travel
distance is constant and is already factored in a. A carrier charges about a =$5000 for transatlantic flights
when the weight is in tons and the volume is in metercubes. For example, consider sending a desktop which
can be probably fit into a box of 0.3 × 0.5 × 0.6 = 0.09 m3 and may weigh 6 kg. For this box, the chargeable
weight is max{0.09/5, 0.006} = 0.018 so the price of this shipment would be about $90. In all likelihood, B2B
transactions involve substantial discounts, e.g., Dell should pay less than $90 for shipping a desktop. We do
not study price discounts in this paper.
    When a showing up cargo cannot be accommodated due to lack of either volume or weight capacity, that
cargo is offloaded to be carried by alternative flights. These alternative flights, generally longer and inefficient,
lead to extra costs for airlines. It is conceivable that offload costs also vary depending on the chargeable weight
as revenues do but with a different multiplier b,

                        Offload cost(Volume, Weight) = b max{Volume/ds , Weight},

where b can be larger or smaller than a. In this equation, weight and density both refer to the offloaded
weight and density. For the rest of the paper, we arrange the volume units so that ds = 1. This is achived by
considering volume in units of 5 m3 .
    In this paper, we maximize the profit for single-leg flights. We consider the profit per flight, which is the
revenue minus the offload cost. We remark that no operational costs (e.g., fuel costs, crew salaries), which
are sunk costs at the time of overbooking, appear in our profit expressions. We justify this by noting that
operational costs typically do not depend on the volume and weight loaded on to an aircraft.


3    Formulations of Cargo Overbooking
    In this section, we discuss the detailed revenue function, aggregate revenue function, bounds on the detailed
revenue, and the offload cost. Finally, we lay out the aggregate and upper bound problems that will be analyzed
in the forthcoming sections.
    The term reading day is coined in the airline industry for the number of days remaining until the departure
of a particular flight. All of our formulations deal with finding the overbooking limits on a particular reading
day for a particular flight, so the reading day and the flight are fixed during our analysis. The random variable
B denotes our best estimate of the total booking requests with the information available on a particular reading
day. B includes all the requests made while the flight is open (until the departure). As seen from the same


                                                        5
reading day, θ is the density of the showing up cargo. Both B and θ can differ from one reading day to another.
To emphasize that B can potentially depend on θ, we write B(θ), i.e., the magnitude of booking requests can
depend on their density. Show up rate ξ is the ratio of the showing up cargo to the bookings just before the
departure. Since we are focusing only on a single reading day in our analysis, we do not need to attach reading
day indices to any of our variables. Table 1 lists important notations. Our notations deal with the showing up
cargo when its density is θ. Also, the showing up cargo naturally depends on the overbooking curve. That is
why Sv , Sw , N and N are all parameterized by (r, θ). However, we often suppress (r, θ) for brevity.

                           Notation               Description
                           kv , k w               volume and weight cargo capacity of the aircraft
                           ξ                      show up rate with p.d.f. g(.) and c.d.f. G(.)
 Input parameters          θ                      physical cargo density with p.d.f. h(.) and c.d.f. H(.)
                           B(θ)                   magnitude of booking requests with p.d.f. fθ (.) and c.d.f. Fθ (.)
 Decision variable         r(.)                   overbooking limit curve
                           Sv (r, θ), Sw (r, θ)   volume and weight of the showing up cargo
 Consequential variables   N (r, θ)               number of cargos that show up at the departure
                           N (r, θ)               number of cargos that are loaded at the departure

                                           Table 1: Important notations.

   For a particular flight, the profit is simply the revenue generated from the shipped cargo minus the offload
cost. The revenue depends on the weight and the volume of the cargo that is loaded at the departure. This is
different than the revenue that passenger airlines generate, where passengers are charged for the ticket prices
at the time of reservation. Moreover, most of the cargo transportation is based on a B2B relationship where
airlines deal with a few big shippers, which ship large volumes; see Talluri and van Ryzin (2004) where this
special relationship is recognized. The relationship between a shipper and a cargo airline is typically governed
by long term contracts. In this B2B relationship, contracts stipulate that shippers are not responsible for
reserved but unused capacity so they do not pay for such capacity. Clearly, such contracts favor shippers.
However, these contracts exist, because shippers are as powerful negotiators as airlines. Thus, a cargo flight
generates revenue only from the cargos loaded on an aircraft. In practice, events often occur in the following
sequence: First, the portion of the showing up cargo that fits into the aircraft is loaded. The cargo that cannot
be loaded must be sent on another aircraft. The airline incurs the cost for sending all of the offloaded cargo
on another aircraft, which may be owned by the same or another airline.


   Detailed Revenue: We now obtain the detailed revenue formulation, which is based on the volume and
weight of each loaded cargo. For the ith cargo, we respectively denote by wi and vi the weight and volume,
and denote by θi the density. Loaded cargos are indexed by {1, .., N } and offloaded cargos are indexed by
{N + 1, .., N }. The detailed revenue can therefore be written as
                           N                                                               N                         N
    Detailed Revenue = a                      I
                                 max{vi , wi }1   i
                                                        vj ≤kv ,       i
                                                                             wj ≤kw   =a         max{vi , wi } = a         vi max{1, θi }
                                                  j=1                  j=1
                           i=1                                                             i=1                       i=1

      I
where 1 is the indicator function. The following assumptions are needed to establish simple bounds for this

                                                                   6
revenue.


Same density assumption: We assume that the loaded cargo has the same density θ as the showing up cargo.
This assumption is statistically established with real-life data in Luo et al. (2005). The assumption can be
written as
                                  Total weight showing up   Total loaded weight   Total offloaded weight
                Density =                                 =                     =
                                  Total volume showing up   Total loaded volume   Total offloaded volume
or
                                                                   N              N               N
                                                  Sw               i=1 wi         i=1 wi          i=N +1 wi
                                            θ=       =             N
                                                                            =     N
                                                                                            =     N
                                                                                                            ,                                   (1)
                                                  Sv               i=1 vi         i=1 vi          i=N +1 vi

where the equalities can be interpreted as the equality of random variables in distribution. Expression (1)
simply states that the density of the showing up cargo is equal to the density of the loaded cargo, which in
turn is equal to the density of the offloaded cargo.


Divisible cargo assumption: When there are many cargos on the aircraft, the effect of discrete cargo size is
                                                                                      N           N                       N
negligible. To illustrate this, consider the case when                                i=1 wi /    i=1 vi   ≤ kw /kv and   i=1 vi   ≥ kv . Since the
density is small, the volume capacity is violated before the weight capacity. Thus, the total volume loaded is
     N
            I
     i=1 vi 1   i
                j=1   vj ≤kv .   We split each cargo into m pieces and let m → ∞ to obtain

                                   N                               mN
                                                                         vi
                                            I
                                         vi 1   i
                                                      vj ≤kv   =            1
                                                                            I   mi vj       −→ kv as m −→ ∞.                                    (2)
                                                j=1                      m      j=1 m ≤kv
                                   i=1                             i=1

That is, the loaded cargo converges to the volume limit as the cargo size becomes smaller. Moreover, the
total weight loaded is kv Sw /Sv by (1)-(2). If some cargo is offloaded, the loaded cargo has volume and weight
(kv , kv Sw /Sv ) or (kw Sv /Sw , kw ), depending on whether Sw /Sv ≤ kw /kv or Sw /Sv ≥ kw /kv . This assumption
approximates the volume and weight loaded on the aircraft. It slightly overestimates the revenue.
      The divisible cargo assumption helps us to avoid the difficulty of dealing with a stochastic knapsack prob-
lem. This problem has been studied in the area of passenger revenue management to address the issue of seat
inventory control with mutliple bookings; see van Slyke and Young (2000) and Brumelle and Walczak (2003).
However, these models do not consider mutli-dimensional requirements present in the cargo problems.


      The showing up volume and weight can now be written in terms of the cargo density θ, the magnitude
of the total booking requests B(θ), and the overbooking limit curve r(θ). When the booking requests have
a density θ and a magnitude larger than r(θ), only r(θ) is accepted. Hence, the accepted bookings have the
magnitude equal to min{r(θ), B(θ)}. Consequently, the showing up volume and weight respectively are
                                                1                                         θ
                  Sv (r, θ) = min{r(θ), B(θ)} √      ξ and Sw (r, θ) = min{r(θ), B(θ)} √       ξ.                                               (3)
                                               1+θ 2                                    1 + θ2
We borrow these showing up volume and weight expressions from Luo et al. (2005). We recall that showing up
                                                                         N                 N
volume and weight can also be expressed as                               i=1 vi   and      i=1 wi .   However, we shall often use expressions in

                                                                                  7
(3), which are reminiscent of polar coordinate representations.


   Aggregate Revenue: We consider three different cases, each having a different revenue implication. The
first case occurs when the showing up volume Sv and weight Sw are respectively smaller than capacities kv and
kw . The loaded volume and weight in this case are (Sv , Sw ). In the second case, the showing up volume Sv
exceeds the volume capacity kv , only kv and the corresponding weight kv Sw /Sv are loaded. The third case is
similar to the second case. All the cases are depicted in Figure 1. By the divisible cargo assumption, either the
loaded volume is kv , or the loaded weight is kw , or the loaded volume and weight are the showing up volume
and weight. By the same density assumption and combining the three cases, we obtain
                                                     N            N
       (Loaded volume, Loaded weight) =                    vi ,            wi
                                                     i=1          i=1
                                                                                                                      
                                                  (Sv , Sw )                         if Sv ≤ kv and Sw ≤ kw ,         
                                               =   (kv , kv Sw /Sv )                  if Sw /Sv ≤ kw /kv and Sv ≥ kv ,   .    (4)
                                                                                                                      
                                                   (kw Sv /Sw , kw )                  if Sw /Sv ≥ kw /kv and Sw ≥ kw




                                  Case III:
                                  Weight capacity
                                  caused offload
                                  Sw/Sv ≥ kw/kv,
                    Weight        Sw ≥ k w

                                                                  Case II:
                                                                  Volume capacity
                                                                  caused offload
                             kw                                   S /S ≤ kw/kv ,
                                  Case I:                (Sv, Sw) w v
                                  No offload                      Sv ≥ k v
                                  Sv ≤ k v                   offload
                                  Sw ≤ k w           (kv Sw/Sv, kv)
                                           loaded

                                                kv
                                                                           Volume

             Figure 1: The three cases considered in expressing the revenue and the offload cost.

From the loaded volume and weight in (4), we can immediately write the aggregate revenue.

                                                            N              N
    Aggregate Revenue(Sv , Sw ; kv , kw ) = a max                   vi ,         wi
                                                           i=1             i=1
                                                                                                                         
                                             a max{Sv , Sw }                            if Sv ≤ kv and Sw ≤ kw ,         
                                          =   a max{kv , kv Sw /Sv }                     if Sw /Sv ≤ kw /kv and Sv ≥ kv ,   . (5)
                                                                                                                         
                                              a max{kw Sv /Sw , kw }                     if Sw /Sv ≥ kw /kv and Sw ≥ kw

The aggregate revenue depends on the total volume and weight of the loaded cargos, as such it differs from the

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detailed revenue. But these two revenues can be related to each other.


   Bounds on the Detailed Revenue: Our aggregate revenue is a lower bound for the detailed revenue
because
                                              N            N               N
                                  a max             vi ,         wi   ≤a         max{vi , wi }.            (6)
                                              i=1          i=1             i=1

The aggregate revenue would be equal to the detailed revenue when the density of all cargos is the same or
when the density of all cargos is below or above 1.
   We can also construct an upper bound for the detailed revenue. This is initiated with the next lemma.
First, let the interval [θ, θ] be the support of the density θ.

Lemma 1. i) If θ < 1 < θ, then for a cargo with given volume vi and density θi , we have

                             1−θ       θ−1
                                 θvi +     wi ≥ vi max{1, θi },                      where wi = vi θi .
                             θ−θ       θ−θ

ii) If θ ≤ 1 (θ ≥ 1), then the detailed revenue is equal to the aggregate revenue and both are based on volume
(weight).

Proof: i) We decompose the cargo (vi , wi ) in two cargos: one with the maximum density θ, the other with the
                                      1
minimum density θ; see Figure 2. Let vi be the volume of the cargo with the minimum density. After some
                    1
algebra, we obtain vi = (θvi − wi )/(θ − θ).

                         Weight


                             wi




                                                           θ
                                          θ
                                                           1
                                                  vi  viVolume
                    Figure 2: Decomposing a cargo into two cargos with extreme densities.


                                                                    1
   The revenue obtained from the minimum density cargo is based on vi while the revenue obtained from the
maximim density cargo is based on that cargo’s weight. The sum of these volume and weight is given by

                                              θvi − wi          θvi − wi
                                                       + wi − θ                          .
                                               θ−θ               θ−θ
We need to show that this sum is larger than max{vi , wi }, which determines the actual revenue. Equivalently,
we show that it is larger than both vi and wi = θi vi . By recalling θ < θi and θ < 1 < θ, we use algebra to

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derive the following:
               θvi − wi          θvi − wi                      θvi − wi          θvi − wi
                        + wi − θ              > vi and                  + wi − θ                  > wi .          (7)
                θ−θ               θ−θ                           θ−θ               θ−θ
To finish, note that
                         θvi − wi          θvi − wi                1−θ               θ−1
                                  + wi − θ                    =        θ vi +              wi .
                          θ−θ               θ−θ                    θ−θ               θ−θ
   ii) If θ ≤ 1, all the cargos are priced by their volume. The revenue made is a times the total volume of all
of the cargos. Hence, detailed and aggregate revenues are the same. The same conclusion can be made when
θ ≥ 1.


   Using Lemma 1 and summing up cargos one by one, we obtain
                                 N                                         N                        N
                                                               1−θ                        θ−1
         Detailed revenue = a         max{vi , wi } ≤ a            θ             vi + a                    wi .   (8)
                                i=1
                                                               θ−θ         i=1
                                                                                          θ−θ       i=1

This upper bound uses the aggregate volume and weight shown in the square brackets, so it can be constructed
without detailed data. Combining (6) and (8), we show that the detailed revenue is bounded respectively from
below and above by the aggregate and upper bound revenues.


   Offload Cost: Offloading the excess cargo leads to costs at the rate of b per chargeable weight. Since the
revenue is generated only from the loaded cargos, b < a is possible as well as b ≥ a in a well-posed problem.
   Since all the excess cargo is offloaded to a single airline at once, all of the excess cargo can be gathered
into a single cargo unit for cost computations. That is, each offloaded cargo affects the offloading cost only by
being a part of this single cargo unit. Thus, the offload cost does not depend on the individual cargos, but it
is based on the maximum of the total offloaded volume and weight.
   For expressing the offload costs, we still have the three cases in Figure 1. In the first case, there is no
offload since both showing up volume and weight are less than their respective capacities. In the second case
where the showing up volume is greater than the volume capacity, the cargo offloaded is simply based on the
difference between the showing up volume and the capacity. The same applies to the weight in the third case.
In summary, the offload cost is
                                               N               N
  Offload cost(Sv , Sw ; kv , kw ) = b max              vi ,            wi
                                             i=N +1          i=N +1
                                                                                                        
                                   0                                   if Sv ≤ kv and Sw ≤ kw ,         
                                =   b max{Sw − Sw kv /Sv , Sv − kv }    if Sw /Sv ≤ kw /kv and Sv ≥ kv ,
                                                                                                        
                                    b max{Sw − kw , Sv − Sv kw /Sw } if Sw /Sv ≥ kw /kv and Sw ≥ kw
                                                                                                  
                                   0                             if Sv ≤ kv and Sw ≤ kw ,         
                                =   b(1 − kv /Sv ) max{Sv , Sw }  if Sw /Sv ≤ kw /kv and Sv ≥ kv ,   .     (9)
                                                                                                  
                                    b(1 − kw /Sw ) max{Sv , Sw } if Sw /Sv ≥ kw /kv and Sw ≥ kw
   Profit: We denote by Π(r, θ) the profit based on the aggregate revenue if the density of the cargo is θ
and the magnitude of the overbooking limit at that density is r. Averaging the profit Π(r, θ) over all possible

                                                             10
physical densities, we obtain the profit Ψ(r(·)),
                                                                  ∞
                                      Ψ(r(.)) :=                      Π(r(θ), θ)dH(θ).                                        (10)
                                                              0

In (10), Π(·, ·) has two arguments which are both numbers, whereas Ψ(·) has a single argument which is a
function. The generic overbooking problem is

                                                        max Ψ(r(.)).                                                          (11)
                                                           r(.)

   In the sequel, we discuss aggregate revenue and detailed revenue under finite and infinite booking requests.
Therefore, we need profit notation for each version of the overbooking problem:
   • Πu (r, θ): Profit based on the upper bound revenue and finite booking requests.
   • Π∞ (r, θ) : Profit based on aggregate revenue and infinite booking requests.
   • Πu,∞ (r, θ) : Profit based on the upper bound revenue and infinite booking requests.
Π∞ and Πu,∞ are simply the profits Π and Πu under the limiting case of infinite booking requests. By putting
each version of the profit Π into (10) and (11), we obtain different versions of the overbooking problem.
   Let r∗ (θ) = argmaxr Π(r, θ). Then it is sufficient to find r∗ (θ) for each θ to solve the overbooking problem
in (11). This fact is established in the next theorem, which allows us to decompose the overbooking problem
over each density θ.

Theorem 1. The optimal overbooking curve is {r∗ (θ) : 0 ≤ θ < ∞}, i.e., it solves (11).

Proof: Using (10), we write (11) as
                                                                              ∞
                                   max Ψ(r(.)) = max                              Π(r(θ), θ)dH(θ).
                                    r(.)                          r(.)    0

    ˜
Let r be the maximizer of the above expression, then
                                               ∞
                       max Ψ(r(.)) =                 r
                                                   Π(˜(θ), θ)dH(θ)
                       r(.)                0
                                               ∞                                                ∞
                                      ≤            max Π(r, θ)dH(θ) =                               Π(r∗ (θ), θ)dH(θ)
                                           0           r                                    0
                                      ≤ max Ψ(r(.)).                                                                          (12)
                                           r(.)

Hence, all the inequalities above are indeed equalities, which completes the proof.


   We now make the aggregate revenue formulation more explicit by writing Π(r, θ) formally:
                                                   N              N                                 N             N
                   Π(r, θ) = E a max                       vi ,          wi       − b max                vi ,            wi   (13)
                                                  i=1             i=1                           i=N +1          i=N +1

where vi and wi are random variables. Also, N is random as a consequence of vi and wi . By putting Π(r, θ)
into (10) and (11), we obtain the aggregate overbooking problem. Expression in (13) facilitates the comparison
of aggregate and upper bound overbooking problems. However, we often use a profit expression obtained via

                                                                         11
Π(r, θ) := E{Aggregate Revenue(Sv , Sw ) − Offload Cost(Sv , Sw )}, where Aggregate revenue and Offload cost
are plugged in respectively from (5) and (9). Details are provided in the next section.
     The detailed profit Πd (r, θ) is obtained by replacing the aggregate revenue in (13) with the detailed revenue,
so
                                         N                                 N               N
                       Πd (r, θ) = E a         max{vi , wi } − b max              vi ,            wi      .                     (14)
                                         i=1                             i=N +1          i=N +1

     To express the upper bound problem, we replace the revenue in (14) with the upper bound of the revenue
on the right-hand side of (8).

                                         N                         N                           N                N
                               1−θ                      θ−1
         Πu (r, θ) = E a           θ          vi + a                     wi − b max                    vi ,            wi   .
                               θ−θ      i=1
                                                        θ−θ        i=1                      i=N +1            i=N +1

We use (4) to rewrite the upper bound of the revenue:
                                                                                                                            
  1−θ
               N
                           θ−1
                                    N            a(αSv + βSw )                             if Sv ≤ kv and Sw ≤ kw ,         
a        θ        vi + a               wi   =      a(αkv + βkv Sw /Sv )                     if Sw /Sv ≤ kw /kv and Sv ≥ kv ,   ,
  θ−θ                      θ−θ                                                                                              
              i=1                  i=1             a(αk S /S + βk )w v    w         w       if Sw /Sv ≥ kw /kv and Sw ≥ kw

where α := (1 − θ)/(θ − θ) θ and β := (θ − 1)/(θ − θ). Deducting the offload cost in (9) from the upper
bound of the revenue, we obtain
                                                                                                               
               a(αSv + βSw )                                                  if Sv ≤ kv and Sw ≤ kw ,         
  Πu (r, θ) =   a(αkv + βkv Sw /Sv ) − b(1 − kv /Sv ) max{Sv , Sw }            if Sw /Sv ≤ kw /kv and Sv ≥ kv ,   . (15)
                                                                                                               
                a(αkw Sv /Sw + βkw ) − b(1 − kw /Sw ) max{Sv , Sw }            if Sw /Sv ≥ kw /kv and Sw ≥ kw

Once Πu (r, θ) is in hand, the upper bound overbooking problem is defined by (10) and (11).
     At this point, we have completed the formulation of the aggregate Π(r, θ), detailed Πd (r, θ), and upper
bound Πu (r, θ) profits; see (13), (14), and (15) respectively. In the next section, we rewrite the aggregate profit
without using the variables (vi , wi ), N or N . These variables do not appear in the upper bound profit either.
Indeed, (vi , wi ), N or N are not needed in the forthcoming sections. We use these variables in this section to
better explain the various profit functions and their relations with each other.
     In closing this section, we note that Π(r, θ) ≤ Πd (r, θ) ≤ Πu (r, θ) for every physical density θ. Thus, we
can solve the aggregate and upper bound overbooking problems to bound the optimal profit of the detailed
problem. Since both the aggregate and upper bound problems can be written without requiring individual
volume and weight (vi , wi ) data, these problems can be constructed with little effort by using the existing cargo
RM databases, which are notorious for not having the detailed data. Moreover, the actual revenue is indeed
           N          N
a max      i=1 vi ,   i=1 wi    when each of the N cargos come from the same shipper. In other words, to the
extent that the flight is serving a single shipper, the aggregate revenue is closer to the actual revenue. Lastly,
the aggregate overbooking problem can be solved easily as we discuss next.




                                                              12
4    Aggregate Overbooking Problem
    In this section, we solve the aggregate revenue problem. We first simplify the aggregate revenue and offload
cost given by (5) and (9). We then analyze this problem by considering infinite and finite booking requests.
Aggregate revenue in (5) can be simplified as follows.
                                                        
                                                         1                    if 1 ≤ kv /Sv and 1 ≤ kw /Sw ,
Aggregate Revenue(Sv , Sw ; kv , kw ) = a max{Sw , Sv }   kv /Sv               if kv /Sv ≤ kw /Sw and kv /Sv ≤ 1,
                                                        
                                                          kw /Sw               if kv /Sv ≥ kw /Sw and kw /Sw ≤ 1.
                                                                     1                       if 1 ≤ kv /Sv and 1 ≤ kw /Sw ,
                                              = a max{Sw , Sv }
                                                                     min{kv /Sv , kw /Sw }   if 1 ≥ min{kv /Sv , kw /Sw }.


                                              = a max{Sw , Sv } min{1, kv /Sv , kw /Sw }.

    For the offload cost in (9), we have
                                                 
                                                  0                        if 1 ≤ kv /Sv and 1 ≤ kw /Sw ,
Offload Cost(Sv , Sw , kv , kw ) = b max{Sw , Sv }   1 − kv /Sv               if kv /Sv ≤ kw /Sw and kv /Sv ≤ 1,
                                                 
                                                   1 − kw /Sw               if kv /Sv ≥ kw /Sw and kw /Sw ≤ 1
                                                             0                                 if 1 ≤ min{kv /Sv , kw /Sw },
                                        = b max{Sw , Sv }
                                                             max{1 − kv /Sv , 1 − kw /Sw }     if 1 ≥ min{kv /Sv , kw /Sw }
                                        = b max{Sw , Sv } [1 − min{1, kv /Sv , kw /Sw }] .

Then, we deduct the offload cost from the aggregate revenue to obtain the profit: max{Sw , Sv }[(a + b) min{1,
kv /Sv , kw /Sw }−b]. By taking the expected value of the profit, we arrive at the profit expression in the aggregate
overbooking problem:
                                                                                     kv       kw
                Π(r, θ) = E max{Sw (r, θ), Sv (r, θ)} (a + b) min 1,                       ,            −b    .          (16)
                                                                                  Sv (r, θ) Sw (r, θ)
    The aggregate overbooking problem can be written in more detail by inserting (3) into (16) and considering
two cases: the first (second) case occurs when the magnitude B(θ) of booking requests is smaller (larger) than
the overbooking limit r(θ). If the booking requests are larger than the overbooking limit, then the showing up
cargo depends only on the overbooking limit. Otherwise, it depends on the magnitude of the booking requests.
Formally,
                    1       r
                                                                             √           √
                                        Bξθ      Bξ                        kv 1 + θ 2 kw 1 + θ 2
 Π(r, θ) =                      max   √       ,√            (a + b) min 1,           ,            − b dFθ (B)dG(ξ)
                0       0              1+θ  2   1 + θ2                        ξB         ξθB
                    1       ∞
                                                                              √          √
                                        rξθ      ξr                         kv 1 + θ 2 kw 1 + θ 2
            +                   max   √       ,√             (a + b) min 1,            ,           − b dFθ (B)dG(ξ).
                0       r              1+θ  2   1 + θ2                         ξr          ξθr
                                                                                                                         (17)

The region defined by {(v, w) : v ≤ r(w/v)/ 1 + (w/v)2 , w ≤ (w/v)r(w/v)/ 1 + (w/v)2 } can be termed as
the acceptance region. On reading day 1, a booking request is accepted if the acceptance of this request keeps
the total accepted bookings within the acceptance region. Otherwise, the booking request is rejected. We call

                                                                13
this acceptance region policy. In a sense, we would like total bookings to be as close to r(·) as possible. This
can be achieved by repeating the acceptance region policy on the other reading days as well.


   Infinite booking requests: We now examine the profit when the magnitude of booking requests is
sufficiently large, e.g. B → ∞. “Sufficiently large” means B(θ) ≥ r(θ) for each θ. Modifying (17) for B → ∞,
the profit becomes
                         1
                                                                           √         √
         ∞                           rξθ      ξr                         kv 1 + θ2 kw 1 + θ2
        Π (r, θ) =           max   √       ,√             (a + b) min 1,          ,            − b dG(ξ).
                     0              1+θ  2   1 + θ2                         ξr        ξθr

   We remark that the optimal r∗ that maximizes Π(r, θ) (or Π∞ (r, θ)) can potentially depend on Fθ and G
(or only on G). The next subsection surprisingly establishes that the maximizer of Π(r, θ) and Π∞ (r, θ) is
actually the same, and hence it is independent of Fθ . That is, the distribution Fθ of booking requests, whose
estimation is complicated by data filtering, is not necessary to solve the aggregate problem.

4.1   Analysis with Finite and Infinite Booking Requests

   First, we show that the profit function with the infinite booking requests is concave in the overbooking
limit r. Then we establish that the partial derivatives with respect to r of the profits under both the finite and
infinite bookings are proportional. These arguments are presented in the next lemma and the ensuing theorem.

Lemma 2. i) Π∞ (r, θ) is concave in r.
ii) The partial derivatives of the expected profit with respect to the overbooking limit r when B is infinite and
B is finite are proportional:
                                                      ∂ ∞           ∂
                                            F θ (r)      Π (r, θ) =    Π(r, θ).
                                                      ∂r            ∂r
where F θ (r) = 1 − Fθ (r).
iii) Π(r, θ) is quasiconcave in r.
iv) Π(r, θ) and Π∞ (r, θ) have the same maximizer.

   Note that Π(r, θ) and Π∞ (r, θ) are two different profit functions. However, when maximized, both functions
yield the same optimal overbooking curve. This observation is formally stated in the next corollary.

Corollary 1. The aggregate overbooking problems with finite and infinite booking requests have the same optimal
overbooking curve solution.

Corollary 1 has an important practical implication: The estimation of the booking requests is not necessary to
construct and solve the aggregate overbooking problem. This is a nice feature of the aggregate formulation.
We reveal another nice feature of this formulation by further studying the shape of the optimal curve with the
next lemma.




                                                              14
Theorem 2. The optimal overbooking curve of the aggregate problem with finite and infinite booking requests
solves the necessary equations
                                    √
                                  kv 1+θ 2 /r(θ)                          1
                            a                      ξdG(ξ) − b           √           ξdG(ξ)               = 0 for θ ≤ kw /kv       (18)
                                 0                                    kv 1+θ2 /r(θ)
                                  √
                                kw 1+θ2 /θr(θ)                        1
                       a                         ξdG(ξ) − b            √             ξdG(ξ)              = 0 for θ ≥ kw /kv .     (19)
                            0                                        kw 1+θ 2 /θr(θ)

                                                    ∗        ∗                                        ∗      ∗
This curve is a box, whose acceptance region [0, lv ] × [0, lw ] is defined by the overbooking limits lv and lw such
                                         √
          ∗        ∗          ∗
that kw /lw = kv /lv , where lv = r∗ (θ)/ 1 + θ2 solves
                                                                                       ∗
                                                                                  kv /lv
                                                          b     1
                                                             =                             xdG(x).                                (20)
                                                         a+b   E(ξ)           0




   By Theorem 2, one only needs to solve the simple equation in (20) to find the optimal volume and weight
overbooking limits. Utilizing integration by parts, (20) can be written in a different form
                                                kv /lv                                                        kv /lv
                        b     1                                         1            kv        kv
                           =                             xdG(x) =                       G            −                 G(x)dx .   (21)
                       a+b   E(ξ)           0                          E(ξ)          lv        lv         0

This equation has a geometric interpretation when expressed as
                                                                              k /lv
                                  b       (kv /lv )G(kv /lv ) − 0 v                    G(x)dx                 Area B
                                    =                                                kv /lv
                                                                                                         =           .            (22)
                                  a   E(ξ) − [(kv /lv )G(kv /lv ) −                           G(x)dx]         Area A
                                                                                    0

Areas A and B are defined by G; see Figure 3. Note that sum of areas A and B is E(ξ).

                                   1
                                                                              G

                   Area A




                  b
      Area B =       E ( ξ)
                 a+b



                                  0
                                       0                  kv / l v                     1

                              Figure 3: Interpretation of the optimal volume overbooking limit.

   Next we illustrate the computation of r∗ (·) with three examples. The first example is very simple so it leads
to closed-form formulas. The second example is actually a counter-example, where we show that stochastically
larger show up rates do not necessarily imply lower overbooking limits. The last example is more realistic but
we can obtain the overbooking limits only numerically.

                                                                              15
Example 1: Suppose that the show up rate is uniformly distributed over [0, 1], so G(u) = u for 0 ≤ u ≤ 1.
By (22),
                                        b     Area B             2
                                                        1/2(kv /lv )
                                           =          =              .
                                       a+b   Area A+B      1/2
The acceptance region is indeed a box, whose sides are given by

                                    ∗        a+b      ∗                    a+b
                                   lv = kv       and lw = kw                   .      ♦
                                              b                             b
Example 2: We compare the overbooking limits under two show up rates ξ1 and ξ2 . Let ξ1 be uniform as in
Example 1 and let ξ2 have a cumulative density given by
                                                   2u2    if 0 ≤ u ≤ 1/2,
                                       G2 (u) =
                                                   u      if 1/2 ≤ u ≤ 1.

Since G2 (u) ≤ G1 (u) for every u, ξ1 is stochastically smaller than ξ2 . Also suppose that b/(a + b) = 1/4. If the
                           ∗
show up rate is ξ1 , then lv = 2kv according to Example 1. If the show up rate is ξ2 , then u = kv /lv solves
                                                                      u
                                      E(ξ2 )/4 = uG2 (u) −                G2 (x)dx.                             (23)
                                                                  0

We can compute that E(ξ2 ) = 13/24, so the left-hand side of (23) is 13/96. However, the right-hand side of
(23) at u = 1/2 is 16/96. Thus, the left-hand side of (23) is smaller than its right-hand side when u = 1/2. To
                                                  ∗
achieve equality, we need u < 1/2, which implies lv > 2kv . Consequently, ξ2 yields a larger overbooking limit
than ξ1 does.    ♦
Example 3: Luo et al. (2005) model the cargo show up rate as a beta distribution with range [0, 1], which has
the p.d.f. of the form
                                      Γ(η)Γ(ζ) η−1
                                               x (1 − x)ζ−1           0 ≤ x ≤ 1,
                                      Γ(η + ζ)
where Γ is the gamma function. They use real life data to estimate parameters as η = 3.59 and ζ = 2.23. We
now use these estimated parameters within the beta distribution and set a=$5000. Since the value b is not
readily available, we consider three different values that b can take: $4000, equal to a=$5000 or $7000. Optimal
overbooking limits are computed respectively as kv /0.67, kv /0.69 and kv /0.72 times the capacities. That is,
the overbooking limits are decreasing in the offload cost.      ♦
    The aggregate formulation has two very desirable features: it does not need Fθ and its solution is a box.
Hovewer, these two by themselves do not entirely validate the aggregate formulation. What is left is to show
that the optimal aggregate profit does not differ much from the optimal detailed profit. This will be achived
by solving the upper bound overbooking problem and comparing it to the aggregate problem.


5    Upper Bound Overbooking Problem
    We now analyze the upper bound problem. We first solve for the optimal overbooking limit under infinite
and finite booking requests. Once the solution is obtained, we measure the gap between the optimal solutions
of the aggregate and upper bound problems.



                                                         16
   Infinite Booking Requests: We first write the profit as B → ∞. Suppose that θ ≤ kw /kv , so the third
case in Figure 1 does not happen. Inserting θ for Sw /Sv in (15) and Sv and Sw given by (3), we have
                                            √
                                       kv      1+θ 2
     u,∞
                                               r             rξ          rξθ
   Π       (r, θ)       =                               aα √      + aβ √       dG(ξ)
                                                            1+θ 2       1 + θ2
                                   0
                                       1
                                                                                                                       √
                                                                                     rξ       rξθ                    kv 1 + θ 2
                            +              √           aαkv + aβkv θ − b max       √       ,√                     1−                   dG(ξ)
                                   kv       1+θ 2                                   1 + θ2   1 + θ2                     rξ
                                            r
                                            √
                                       kv      1+θ 2
                                               r         arξ
                        =                              √       (α + βθ) dG(ξ)
                                   0                    1 + θ2
                                       1
                            +              √           akv (α + βθ) − b max {1, θ} rξ/ 1 + θ2 − kv                 dG(ξ).                  (24)
                                   kv       1+θ 2
                                            r


The profit expression for θ ≥ kw /kv is in the appendix. It is an algebraic exercise to show that the optimal
                   ∗
overbooking limit ru (θ) for the upper bound problem under the case of infinite booking requests satisfies the
following necessary conditions:
                                    √
                                  kv 1+θ2 /r                                       1
                    a(α + βθ)                                    max{1, θ}
                     √                                 ξdG(ξ) − b √              √               ξdG(ξ) = 0 for θ ≤ kw /kv ,               (25)
                      1 + θ2     0                                 1 + θ2      kv 1+θ 2 /r
                                  √
                                kw 1+θ 2 /rθ                                   1
              a(α + βθ)                                       max{1, θ}
               √                                    ξdG(ξ) − b √               √                 ξdG(ξ) = 0         for θ ≥ kw /kv .       (26)
                1 + θ2      0                                   1 + θ2       kw 1+θ2 /rθ

   We notice that the optimality equation of the upper bound problem under infinite bookings is very similar
to that of the aggregate problem. The only difference is in the terms multiplying the integrals. In view of
                                                        √                                            √
(18)-(19), we notice that the price a becomes a(α + βθ)/ 1 + θ2 and the cost b becomes b max {1, θ} / 1 + θ2 .
With this interpretation, the price and the cost are no longer constant, but they are now functions of the
density θ. Thus, the optimal overbooking curve for the upper bound problem is no longer a box. With the
next lemma, we compare the optimal overbooking limits of the aggregate and upper bound problems.

                                                                        ∗
Lemma 3. For every fixed density θ, Πu,∞ (r, θ) is concave and r∗ (θ) ≤ ru,∞ (θ).

Proof: We provide the proof only for θ ≤ kw /kv . The concavity follows from the fact that the left-hand side
of (25) is decreasing in r. We manipulate (25) to obtain
                                                                                             √
                                                                                           kv 1+θ2 /r
                                         b max {1, θ}          1
                                                            =                                           ξdG(ξ).                            (27)
                                   a(α + βθ) + b max {1, θ}   E(ξ)                     0

It suffices to compare the left-hand sides of (27) and (20), which respectively are

                                                             b max {1, θ}            b
                                                                                and     .
                                                        a(α + βθ) + b max(1, θ)     a+b

   For θ ≤ 1,
                                                          1−θ    θ−1    1−θ    θ−1
                                            α + βθ =          θ+     θ≥     θ+     θ=1
                                                          θ−θ    θ−θ    θ−θ    θ−θ


                                                                        17
Since max{1, θ} = 1, we have

                                              b max {1, θ}               b          b
                                                                 =               ≤     ,                                                           (28)
                                        a(α + βθ) + b max {1, θ}   a(α + βθ) + b   a+b

   For θ ≥ 1,
                                                 α      (1 − θ)θ θ − 1   1−θ θ−1
                                                   +β =         +      ≥    +    =1
                                                 θ      (θ − θ)θ θ − θ   θ−θ θ−θ
Since max{1, θ} = θ, we have

                                              b max {1, θ}               b           b
                                                                 =                ≤     ,                                                          (29)
                                        a(α + βθ) + b max({1, θ}   a(α/θ + β) + b   a+b
                                                                                            ∗
Thus, the left-hand side of (27) is less than that of (20). Consequently, we have r∗ (θ) ≤ ru,∞ (θ).


   This result is quite intuitive because the revenue function of the upper bound problem is larger than that
of the aggregate problem, while the offload costs are the same. Therefore, the upper bound profit emphasizes
the revenue. Higher revenue, though at the cost of higher offload cost, is obtained when the overbooking limits
are higher.


   Finite Booking Requests: The show-up volume and weight now depend on both the overbooking limit
r and the magnitude of booking request B. We add two terms to Πu,∞ (r, θ) to handle the case B < r. The
expected profit when θ ≤ kw /kv is
                       √                                                                           √
                  kv   1+θ 2                                                                  kv   1+θ 2
                                   r                                                                           ∞
  u
                       r                aBξ                                                        r                 arξ
Π (r, θ) =                             √       (α + βθ) dFθ (B)dG(ξ) +                                             √       (α + βθ) dFθ (B)dG(ξ)
              0                0        1 + θ2                                            0                r        1 + θ2
                                             1                       r
                                        +         √                      akv (α + βθ) − b max {1, θ} Bξ/ 1 + θ2 − kv                dFθ (B)dG(ξ)
                                             kv      1+θ 2
                                                     r
                                                                 0
                                            1                    ∞
                                       +         √                       akv (α + βθ) − b max {1, θ} rξ/             1 + θ 2 − kv   dFθ (B)dG(ξ). (30)
                                            kv    1+θ 2
                                                  r
                                                             r

                                                             ∗
The profit expression for θ ≥ kw /kv is in the appendix. Let ru (θ) be a solution of the optimality equations, then
we have the next lemma. The proof of the lemma is in the appendix, which contains the optimality equations
under finite booking requests. Both the profit and the optimality conditions are more complicated under finite
booking requests.

                                    ∗          ∗
Lemma 4. For every fixed density θ, ru,∞ (θ) ≤ ru (θ).

   This lemma says that the optimal overbooking limits are tighter under infinite booking requests. Then,
the overbooking limits play a bigger role in the specification of the showing up cargo. Moreover, the showing
up cargo is larger for any fixed overbooking curve when the booking requests are infinite. Higher showing up
cargo leads to higher offload costs. To curtail the offload costs, the overbooking limits should be set lower with
infinite booking requests.



                                                                                   18
6    Performance of the Aggregate Overbooking Model
    We now compare the optimal aggregate profit with the optimal upper bound profit. If the gap between
the aggregate and upper bound optimal profits is small, we conclude that the aggregate model is a good
approximation of the detailed model so it can be used in practice to compute the optimal overbooking limits.
We recall that the detailed profit is bounded by the aggregate and upper bound, i.e.,

                 Π(r∗ (θ), θ)dH(θ) ≤ max          Πd (r(θ), θ)dH(θ) ≤           ∗
                                                                           Πu (ru (θ), θ)dH(θ).            (31)
                                         r(.)

We compute the percentage difference between the aggregate and the upper bound profits:
                                                     ∗
                                                Πu (ru (θ), θ)dH(θ) − Π(r∗ (θ), θ)dH(θ)
                        % Difference =                                                   .                  (32)
                                                                  ∗
                                                             Πu (ru (θ), θ)dH(θ)
The % difference is reported below for various problem parameters. When the % difference is small, the
aggregate problem approximates the detailed problem well.
    We next introduce the problem parameters. Show up rate distribution is taken as a beta distribution with
parameters (η = 3.59, ζ = 2.23). The weight and volume are respectively measured in ton and 5 m3 . We
arrange the volume to be in units of 5 m3 so that the standard density is simply 1. We set a=$5K and b =$7K.
The volume and weight capacities are 125 m3 and 10 tons, so kv = 25 and kw = 10.
    In our experiments, the booking requests are always normally distributed with standard deviation 16. We
try three different distributions whose means are 40, 50 and 60. To avoid trivial situations (such as smaller
booking requests than the available capacity), the means are set to be larger than the largest magnitude
√
 252 + 102 ≈ 27 of the cargo that can be fit into the aircraft. We also consider the case of infinite booking
requests. We have real-life data on the density of cargos carried by a leading airline. We fit a symmetric
triangular distribution to the data. The support for the density is [θ, θ] and the mean is 0.9 tons per 5 m3 .
Throughout our experimental studies, we keep the mean constant by fixing θ + θ = 1.8 and vary θ − θ.
    Table 2 contains the performance measure % difference under different expected booking requests (in
columns) and under increasing values of θ − θ (in rows). The % difference in the upper left side of the
table are smaller, which indicates that the aggregate model approximates the detailed model well when the
density does not vary much and the booking requests are smaller. In Table 2, we write the % differences less
(larger) than 5% (10%) in bold (italic) and draw a line between these and the rest of differences.
    Numbers in Table 2 must be interpreted with care. First of all, the aggregate model does not approximate
the detailed model when the densities are extremely variable and the booking requests are very large with
respect to the available cargo capacity. However, there are plenty of instances when the aggregate profit is
close to the upper bound profit, so it is even closer to the detailed (actual) profit. The interesting question is
what range of densities should we expect in practice. Consider the row of (θ = 0.5, θ = 1.3), which corresponds
to densities between 0.5/5=0.1 and 1.3/5=0.26 tons per m3 . These numbers are reasonable after noting that
air cargo is mostly composed of high-technology low density products such as consumer electronic products.
Moreover, protective packaging uses lightweight materials such as polystyrene, which lowers the density. For
(θ = 0.5, θ = 1.3), the % differences are between 3.7 and 5.3 for finite booking requests. Thus, the aggregate
model can justifably be used in this instance.

                                                        19
                             Support of the               % Difference
                            physical density   Expected booking requests
                                 (θ , θ)           40      50         60     B→∞
                              (0.70 , 1.10)       0.3     0.4        1.2       2.5
                              (0.65 , 1.15)       0.3     2.1        2.4       3.6
                              (0.60 , 1.20)       2.0     3.0        3.7       5.5
                              (0.55 , 1.25)       3.6     3.8        4.9       6.5
                              (0.50 , 1.30)       3.7     4.2        5.3       8.0
                              (0.45 , 1.35)       4.0     6.0        6.2       8.5
                              (0.40 , 1.40)       5.2     6.8        7.7       8.7
                              (0.35 , 1.45)       6.3     7.1        8.6       9.6
                              (0.30 , 1.50)       7.3     8.6        9.3      10.0
                              (0.25 , 1.55)       7.8     9.0       10.5      11.4
                              (0.20 , 1.60)       8.5     9.3       11.7      12.2
                              (0.00 , 1.80)      13.8    14.4       15.8      17.5

    Table 2: % Differences as the expected booking requests and the range of the physical densities vary.


   In Table 3, we use the real-life data to list the number of overbooking instances whose density range θ − θ is
below the numbers in the first column of the table. Our data deals with 8 different flights and 7 reading days,
which gives us 56 populations of cargo density. We compute the range of the densities for each population and
place it in the table. A density population obtained on reading day 28 has a range wider than 1.8, that is why
the number in the eighth column and the last row is 7 as opposed to 8. 58.9% of the populations have density
smaller than 0.9 which corresponds to the row of (θ = 0.45, θ = 1.35) in Table 2. It is safe to conclude that %
difference would be 4-6.2 for about 60% of the instances. 21.5% of the instances have density range larger than
0.9 but smaller than 1.4. These correspond to rows of (θ = 0.45, θ = 1.35) and (θ = 0.2, θ = 1.6) in Table 2. In
these rows, the % difference ranges from 5.2 to 11.7, i.e., aggregate model is a reasonable approximation but
it is not very close to the detailed model. In summary, the aggregate model is a good approximate for 60% of
the real-life instances; it is reasonable for 20% of the instances; we do not suggest using the aggregate model in
the remaining 20% of the instances. We must also remark that the optimal solution of the aggregate problem
is a box while neither the detailed nor the upper bound problems have this property. Enforcing a box solution
on the detailed and the upper bound problems would decrease the % differences in all of our tables.
   Having justified the aggregate model for 60% of the real-life instances, we check if this justification holds
when the show up rate variance varies. In Table 4, we increase the variance of the show up rate and report the
change in the % difference. Since the show up rate has a beta distribution with parameters (η, ζ), its expected
value and variance are η/(η + ζ) and ηζ/((η + ζ)2 (η + ζ + 1)). Varying the expected value while the variance
is constant should have a similar effect as manipulating the booking request, which is already investigated in
Table 2. That is why we keep the expected value constant and vary the variance. Note that the expected
value is unaltered when η and ζ are multiplied by the same constant. When the constant is larger than 1,
the variance decreases; otherwise it increases. The current values of the parameters are multiplied by 4, 2,
1, 1/1.5, 1/1.8 to obtain the rows in Table 4. Doing so keeps the expected value at 0.61 while the variance
ranges from 0.98/100 to 5.59/100. In these experiments, the physical density has a triangular distribution with

                                                       20
                                        Number of instances with      Percentage
                                       density range < first column    of instances
                              First       Reading Days (RDs)              over
                             column    2 5 7 10 14 21 28                all RDs
                               0.5     3 2 0        2   3     3  3         28.6%
                               0.7     4 3 3        3   4     6  3         46.4%
                               0.9     4 3 4        6   5     7  4         58.9%
                               1.0     4 5 4        6   6     7  5          66.1%
                               1.1     5 5 5        6   6     7  5          69.6%
                               1.2     5 5 6        6   6     7  6          73.2%
                               1.3     5 6 6        6   6     7  6          75.0%
                               1.4     5 6 7        7   7     7  6          80.4%
                               1.6     7 7 8        7   7     7  6          87.5%
                               1.8     8 8 8        8   8     8  7          98.2%

            Table 3: Number of instances whose density range is smaller than a particular number.

(θ = 0.4, θ = 1.4) and the booking requests have a normal density with mean 40 and variance 16. As expected,
the aggregate profits are decreasing in the variance. More importantly, the gap between the aggregate and the
upper bound profits is closing with higher variances. Consequently, a detailed formulation should not be built
when the show up rates are extremely variable. Then, it makes more sense to use an aggregate model while
working to reduce the show up rate uncertainty.

                                 Show up rate                    Aggregate
                 Parameters (η, ζ)             100 · Variance   Profit in K $      % Difference
                 (14.3 , 8.92)=4(3.59,2.23)         0.98            48.7                  6.9
                 (7.18 , 4.46)=2(3.59,2.23)         1.87            47.7                  6.4
                 (3.59 , 2.23)=Base case            3.46            41.8                  5.2
                 (2.39 , 1.48)=(3.59,2.23)/1.5      4.85            40.0                  1.6
                 (1.99 , 1.23)=(3.59,2.23)/1.8      5.59            37.0                  0.7

                         Table 4: % Difference as the show up rate uncertainty varies.



7    Summary and Conclusions
    We formulate an aggregate cargo overbooking problem which is appropriate for B2B transactions. The
solution to this problem is a curve parameterized by the cargo density. We also provide a detailed and an
upper bound formulation for the same problem. The gap between the optimal profits of upper bound and the
aggregate problems is studied to show that the aggregate formulation is justifiable in 60% of the cases found
in our real-life data.
    We provide easy-to-solve equations to find the optimal overbooking curve, which turns out to be a box.
We establish that this solution is insensitive to the booking requests and it can be derived only from the show
up rate distribution. Thus, not only that the solution is easy to obtain, but also it does not require the type
of data which is difficult to come by in the cargo industry. We believe that these advantages would make the

                                                      21
solution implementable and popular in practice.
   As we discuss earlier, our aggregate profit does not approximate the detailed profit well when the density is
very variable. This is easy to identify in practice and if it turns out to be the case, a detailed formulation must
be pursued. Such a formulation can be sought as a future research topic. However, analytic results may be
difficult to obtain and simulation-based optimization can be a viable approach. Another avenue to extend the
ideas in the current paper is pursuing a dynamic, multi-period formulation. In these future research endeavors,
the parametrization of the overbooking curve by the density can be a useful device to decompose the problem
over densities. This paper also has the potential to galvanize the research in multi-dimensional overbooking
in the traditional application areas of RM such as Railways, Maritime shipping, Hotel management, Show,
concert and sport event planning.


Appendix
Proof of Lemma 2: i) We consider the derivative of Π∞ (r, θ) and simplify it for θ ≤ kw /kv starting from the
second equality below.
                                     1
                                                                                                √         √
   ∂ ∞                                   ∂          ξθ       ξ                                kv 1 + θ2 kw 1 + θ2
      Π (r, θ) =                            max   √      ,√                    (a + b) min r,          ,                                 − br dG(ξ)
   ∂r                            0       ∂r        1+θ 2   1 + θ2                                 ξ        ξθ
                                                                                                                                                 (33)
                                     1
                                                                                           √
                                                 θ          1          ∂                 kv 1 + θ 2
                        =                max  √        ,√            ξ    (a + b) min r,                                    − br dG(ξ)
                                 0              1+θ  2    1 + θ2       ∂r                    ξ
                                                                              √                                             
                                                                         kv   1+θ 2                                         
                                                                                                          1
                                              θ         1                      r
                        = max              √       ,√            a                     ξdG(ξ) − b              √       ξdG(ξ) .                  (34)
                                            1 + θ2    1 + θ2         0                                   kv   1+θ 2
                                                                                                               r
                                                                                                                             

Similarly, for the case when θ ≥ kw /kv we have
                                                                                            √                                             
                                                                                       kw    1+θ 2                                        
                                                                                                                        1
               ∂ ∞                                   θ        1                              rθ
                  Π (r, θ) = max                  √       ,√                    a                     ξdG(ξ) − b            √        ξdG(ξ) .    (35)
               ∂r                                  1 + θ2   1 + θ2                 0                                  kw    1+θ 2         
                                                                                                                            rθ

                                                                                                        √
Clearly, the expressions inside the curly brackets in (34) and (35) are nonincreasing in r. Since max{θ/ 1 + θ2 ,
  √
1/ 1 + θ2 } is nonnegative, it follows that the derivative of Π∞ (r, θ) is nonincreasing in r. This completes the
proof of i).
   ii) In the remaining arguments, θ is fixed, so we drop it from B(θ). We consider the partial derivative of
Π(r, θ), which is composed of two terms; see (17). The derivative of the first term in (17) is

               1            r
                                                                                √           √
                   ∂                        Bθξ       Bξ                      kv 1 + θ 2 kw 1 + θ 2
                                max       √        ,√          (a + b) min 1,           ,              − b dFθ (B) dG(ξ)
          0        ∂r   0                  1+θ  21   1 + θ2                      Bξ          Bθξ
               1
                                                                    √          √
                      rθξ      rξ                                 kv 1 + θ2 kw 1 + θ2
     =              √       ,√                     (a + b) min 1,          ,              − b fθ (r) dG(ξ).           (36)
          0          1+θ  2   1 + θ2                                  rξ        rθξ



                                                                           22
The derivative of the second term is
           1          ∞
                                                                  √          √
               ∂                            rθξ      rξ         kv 1 + θ2 kw 1 + θ2
                      max                 √       ,√
                                                 (a + b) min 1,            ,            − b dFθ (B) dG(ξ)
          0    ∂r  r                       1 + θ2   1 + θ2          rξ        rθξ
            1    ∞
                                                                  √          √
                    ∂         rθξ       rξ                      kv 1 + θ2 kw 1 + θ2
    =                 max √          ,√          (a + b) min 1,            ,            − b dFθ (B) dG(ξ)
          0    r   ∂r         1 + θ2   1 + θ2                       rξ        rθξ
            1
                                                           √           √
                       rθξ       rξ                      kv 1 + θ2 kw 1 + θ2
    −         max √          ,√          (a + b) min 1,           ,             − b fθ (r) dG(ξ).       (37)
          0           1 + θ2    1 + θ2                      rξ          rθξ

   When summing up (36) and (37), the second term on the right-hand side of (37) cancels (36). Thus, by
changing the order of integration in the first term on the right-hand side of (37), we have
                              ∞
    ∂
       Π(r, θ) =                  dFθ (B) ×
    ∂r                    r
                              1
                                                                                          √          √
                                  ∂                rξθ      ξr                          kv 1 + θ 2 kw 1 + θ 2
                                     max         √       ,√              (a + b) min 1,           ,             − b dG(ξ)
                          0       ∂r              1+θ  2   1 + θ2                          ξr         ξθr
                                      ∂ ∞
                    = F θ (r)            Π (r, θ),                                                                          (38)
                                      ∂r
where we use (33) to obtain the last equality.
   iii) Quasiconcavity of Π(r, θ) is due to the concavity of Π∞ (r, θ) and (38).
   iv) Setting (38) equal to zero, we obtain candidate solutions r1 and r2 , which respectively satisfy

                                                   ∂ ∞
                                                      Π (r, θ)          = 0 and F θ (r) = 0.
                                                   ∂r            r=r1

Since Π(r = 0, θ) = 0 and Π(r, θ) is a continuous and quasiconcave function, it must have one of the shapes
depicted in Figure 4. With the curve on the left, r1 maximizes Π(r, θ). With the curve on the right, both r1
and r2 maximize Π(r, θ) as Π(r, θ) is constant for r ≥ r2 . Then, we can arbitrarily pick r1 as the maximizer.
Consequently, r1 maximizes both Π(r, θ) and Π∞ (r, θ).

 ∏(r,θ)                                              ∏(r,θ)




           r1                     r          r                            r       r1   r
                                      2                                       2

                          Figure 4: For fixed θ, possible shapes of Π(r, θ) and its maximizers.

                                                                                     √           √
Proof of Theorem 2: We first set the derivative in (34) equal to zero. The term max{θ/ 1 + θ2 , 1/ 1 + θ2 }
is positive for 0 < θ < ∞. Moreover, it approaches 1 as θ approaches 0 or ∞. Thus, this term can be dropped


                                                                         23
from the equality to obtain the necessary condition in (18) for θ ≤ kw /kv . The condition in (19) for θ ≥ kw /kv
is obtained in a similar fashion after starting with (35).
                                 √                         √
    After letting lv (θ) = r(θ)/ 1 + θ2 and lw (θ) = r(θ)θ/ 1 + θ2 , the necessary conditions above can be
written as
                                                         kv
                                                        lv (θ)
                                                                                           1
                                            a                      ξg(ξ)dξ − b                     ξg(ξ)dξ = 0 for θ ≤ kw /kv
                                                                                           kv
                                                    0                                     lv (θ)
                                                      kw
                                                     lw (θ)
                                                                                          1
                                            a                      ξg(ξ)dξ − b                     ξg(ξ)dξ = 0              for θ ≥ kw /kv .
                                                                                          kw
                                                 0                                       lw (θ)


These two equations yielding the optimal lv (θ) and lw (θ) do not depend on θ. Both lv (θ) and lw (θ) are constant
in θ, which can be eliminated from lv (θ) and lw (θ) notations. Thus, the acceptance region is a box given by
                                                                  ∗      ∗                            ∗        ∗
{(v, w) : v ≤ lv , w ≤ lw }. In view of the equations above that lv and lw must satisfy, we have kw /lw = kv /lv .
Manipulating the optimality equation corresponding to the case when θ ≤ kw /kv , we obtain (20).


   Upper bound profits for θ ≥ kw /kv are as follows.
Infinite Booking Requests:
                                                                       √
                                                               kw           1+θ 2
                                                                           rθ         arξ
         Πu,∞ (r, θ)                    =                                           √       (α + βθ) dG(ξ)
                                                           0                         1 + θ2
                                                               1
                                                +                  √                akw (α/θ + β) − b max {1, θ} rξ/                   1 + θ2 − kw /θ     dG(ξ).
                                                           kw           1+θ 2
                                                                       rθ


Finite Booking Requests:
                       √                                                                                           √
                  kw    1+θ 2                                                                                 kw    1+θ 2
                                        r                                                                                        ∞
   u
                       rθ                    aBξ                                                                   rθ                  arξ
 Π (r, θ) =                                 √       (α + βθ)dFθ (B)dG(ξ) +                                                           √       (α + βθ)dFθ (B)dG(ξ)
              0                     0        1 + θ2                                                       0                  r        1 + θ2
                                            1                              r
                                +                √                             akw (α/θ + β) − b max{1, θ} Bξ/ 1 + θ2 − kw /θ                           dFθ (B)dG(ξ)
                                            kw       1+θ 2
                                                    rθ
                                                                       0
                                            1                          ∞
                                +               √                              akw (α/θ + β) − b max{1, θ} rξ/ 1 + θ2 − kw /θ                       dFθ (B)dG(ξ).
                                        kw        1+θ 2
                                                 rθ
                                                                   r

Proof of Lemma 4: We obtain the optimality condition only for θ ≤ kw /kv . Define A(r, θ) as
                                      √
                                    kv 1+θ2 /r                     r
                                                                       Bξ           Bξθ
       A(r, θ) :=                                                 aα √       + aβ √        dFθ (B)dG(ξ)
                                                                      1+θ 2         1 + θ2
                                0                            0
                                    1                        r
                                                                                                           √
                                                                                  bBξ                    kv 1 + θ2
                       +          √                            [aαkv + aβkv θ − √        max {1, θ} (1 −           )]dFθ (B)dG(ξ).
                                kv 1+θ 2 /r                0                      1 + θ2                    Bξ

And let B(r, θ) := Πu (r, θ) − A(r, θ). A(r, θ) and B(r, θ) respectively correspond to the parts of the profit (30)




                                                                                                    24
where B ≤ r and B ≥ r. Then, we have
                                    √                √
                                                       1+θ 2
                                                                      √
                                                                        1+θ2
                                                                                           √
    ∂                      r
                                  kv 1 + θ 2     B kv r          Bθ kv r                 kv 1 + θ 2
       A(r, θ) =               (−            )[aα √          + aβ √          ]fθ (B)dBg(            )
    ∂r                 0             r2             1 + θ2          1 + θ2                   r
                             √
                           kv 1+θ 2 /r
                                                  rξ           rξθ
               +                             aα √       + aβ √       g(ξ)dξfθ (r)
                       0                         1 + θ2       1 + θ2
                                       √                                        √
                                                                                    2
                                                                                                  √
                                                                                                      2            √
                           r
                                     kv 1 + θ 2                              B kv r1+θ
                                                                                        Bθ kv 1+θ                kv 1 + θ 2
               −                   −                      akv (α + βθ) − b   √         , √ r              (1 −     √       ) ×
                       0                r2                                      1 + θ2     1 + θ2             B kv r1+θ 2
                                                                                                               √
                                                                                                             kv 1 + θ 2
                                                                                                   dFθ (B)g(             )
                                                                                                                  r
                           1
                                                                                                   √
                                                                     rθξ      rξ                 kv 1 + θ 2
               +               √           akv (α + βθ) − b        √       ,√                 1−               g(ξ)dξfθ (r).
                       kv 1+θ2 /r                                   1 + θ2   1 + θ2                 rξ


                                           √                       √
                                                                        2
                                                                                    √
                                                                                          2                 √
      ∂                            ∞
                                         kv 1 + θ2             r kv 1+θ      rθ kv 1+θ                    kv 1 + θ 2
         B(r, θ) =                     −                     aα √ r      + aβ √ r               dFθ (B)g(            )
      ∂r                       r            r2                    1 + θ2         1 + θ2                       r
                                     √
                                   kv 1+θ2 /r
                                                      rξ          rξθ
                   −                             aα √      + aβ √       g(ξ)dξfθ (r)
                                                     1+θ 2       1 + θ2
                               0
                                     √
                                   kv 1+θ2 /r        ∞
                                                              ξ           ξθ
                   +                                     aα √      + aβ √       dFθ (B)g(ξ)dξ
                                                             1+θ 2       1 + θ2
                               0                 r
                                           √                                  √
                                                                               1+θ2
                                                                                                         √
                                   ∞
                                         kv 1 + θ2                        r kv r                      kv 1 + θ 2
                   −                   −                  akv (α + βθ) − b √         max {1, θ} (1 −       √       ) ×
                               r            r2                               1 + θ2                    r kv r1+θ2
                                                                                                               √
                                                                                                             kv 1 + θ2
                                                                                                 dFθ (B)g(             )
                                                                                                                  r
                                   1
                                                                                            √
                                                                    rξ                   kv 1 + θ 2
                   −             √              akv (α + βθ) − b √       max {1, θ} (1 −               g(ξ)dξfθ (r)
                               kv 1+θ 2 /r                        1 + θ2                      rξ
                                   1              ∞
                                                                                       √
                                                       ∂      rξ                     kv 1 + θ 2
                   +             √                  −b     √        max {1, θ} 1 −                  dFθ (B)g(ξ)dξ.
                               kv 1+θ 2 /r      r      ∂r    1 + θ2                      rξ

   We add the partial derivatives of A(r, θ) and B(r, θ), cancel some of the terms and take constants out of




                                                                       25
integrals to obtain
                            √                             √                                             r
    ∂ u                   kv 1 + θ2 akv (α + βθ)       kv 1 + θ2
       Π (r, θ) =       −                          g                                                        BdFθ (B)
    ∂r                        r2            r               r                                       0
                          √                           √
                        kv 1 + θ 2                 kv 1 + θ 2
                    +               akv (α + βθ)g(             )Fθ (r)
                            r2                          r
                                          √                √                                                r
                                       kv 1 + θ 2        kv 1 + θ2                                                  B
                    + bkv max {1, θ} −               g                                                          (     − 1)]dFθ (B)
                                            r2                r                                         0           r
                                                      √
                                                    kv 1+θ2 /r                          1
                           a(α + bθ)                               b max {1, θ}
                    +       √        Fθ (r)            ξg(ξ)dξ − √              Fθ (r)   √        ξg(ξ)dξ                                                    (39)
                              1 + θ2         0                         1 + θ2          kv 1+θ2 /r
                                 √                   √             r
                               kv 1 + θ2 kv        kv 1 + θ2
                    =       −                  g                     BdFθ (B){a(α + βθ) + b max {1, θ}}
                                   r2        r         r         0
                                √                  √
                             kv 1 + θ 2          kv 1 + θ 2
                    +                     kv g               Fθ (r) {a(α + βθ) + b max{1, θ}}
                                 r2                  r
                                                      √
                                                    kv 1+θ2 /r                 1
                           a(α + bθ)                      b max {1, θ}
                    +       √        Fθ (r)                 √       ξg(ξ)dξ −
                                                                       Fθ (r)   √        ξg(ξ)dξ          (40)
                             1 + θ2   0                       1 + θ2          kv 1+θ2 /r
                                                      √              √
                                                    2
                                                   kv 1 + θ 2      kv 1 + θ 2     1 r
                    = (a(α + βθ) + b max {1, θ}) −             g                        BdFθ (B) − Fθ (r)
                                                       r2               r         r 0
                                                                √
                                                              kv 1+θ 2 /r                                                1
                            F (r)
                    +      √ θ             a(α + bθ)                        ξg(ξ)dξ − b max {1, θ}                        √              ξg(ξ)dξ   .         (41)
                             1 + θ2                       0                                                             kv 1+θ2 /r

The second inequality is due to distributing the third term to the first and the second in (39). The third equality
is obtained by combining the first and the second terms in (40). The optimality condition for θ ≤ kw /kv follows
from setting (41) equal to zero. The optimality condition for θ ≥ kw /kv can be similarly obtained to be

                                2
                                  √                                   √                             r
                               kw 1 + θ 2                           kw 1 + θ 2             1
  (a(α + βθ) + b max {1, θ}) −                                  g                                       BdFθ (B) − Fθ (r)
                                  r2 θ2                                rθ                  r    0
                                            √
                                          kw 1+θ 2 /rθ                                      1
       Fθ (r)
     +√                 a (α + bθ)                       ξdG(ξ) − b max {1, θ}               √                        ξdG(ξ)           =0    for θ ≥ kw /kv . (42)
       1 + θ2                         0                                                    kw 1+θ2 /rθ

                       ∗          ∗
   Next, we show that ru,∞ (θ) ≤ ru (θ). To compare the optimal solutions, we study the sign of the terms in
the square brackets in (41). We have
                    r                                                       r                                                      r
            1                                      1                                                                      1
                        BdFθ (B) − Fθ (r) =              rFθ (r) −              Fθ (B)dB        − Fθ (r) = −                           Fθ (B)dB ≤ 0,
            r   0                                  r                    0                                                 r    0

where the first inequality is due to the integration by parts. Hence, (41) is larger than the left-hand side of
                                                                     ∗
(25). Since the left-hand side of (25) is greater than zero for r ≤ ru,∞ (θ), (41) is also greater than zero over
                                ∗                                  ∗
the same range. Thus, the root ru (θ) of (41) must be larger than ru,∞ (θ).




                                                                            26
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