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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 1 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. 3 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 8 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 9 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). 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