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OPERATIONS RESEARCHVol. 62, No. 5, September–October 2014, pp. 1179–1201ISSN 0030-364X (print) � ISSN 1526-5463 (online) http://dx.doi.org/10.1287/opre.2014.1300

© 2014 INFORMS

Double-Sided Batch Queues with Abandonment:Modeling Crossing Networks

Philipp Afèche, Adam Diamant, Joseph MilnerRotman School of Management, University of Toronto, Toronto, Ontario M5S 3E6, Canada

{afeche@rotman.utoronto.ca, adam.diamant09@rotman.utoronto.ca, jmilner@rotman.utoronto.ca}

We study a double-sided queue with batch arrivals and abandonment. There are two types of customers, patient ones whoqueue but may later abandon, and impatient ones who depart immediately if their order is not filled. The system matches unitsfrom opposite sides of the queue based on a first-come first-served policy. The model is particularly applicable to a class ofalternative trading systems called crossing networks that are increasingly important in the operation of modern financialmarkets. We characterize, in closed form, the steady-state queue length distribution and the system-level average system timeand fill rate. These appear to be the first closed-form results for a double-sided queuing model with batch arrivals andabandonment. For a customer who arrives to the system in steady state, we derive formulae for the expected fill rate andsystem time as a function of her order size and deadline. We compare these system- and customer-level results for ourmodel that captures abandonment in aggregate, to simulation results for a system in which customers abandon after somerandom deadline. We find close correspondence between the predicted performance based on our analytical results and theperformance observed in the simulation. Our model is particularly accurate in approximating the performance in systems withlow fill rates, which are representative of crossing networks.

Subject classifications : abandonment; double-sided queues; balking and reneging; batch arrivals; level crossing; crossingnetworks; queueing systems.

Area of review : Stochastic Models.History : Received December 2012; revisions received August 2013, April 2014; accepted May 2014. Published online in

Articles in Advance July 17, 2014.

1. IntroductionWe study a double-sided queue with batch arrivals andabandonment. Two types of customers, patient and impatient,arrive to each side of the queue. Only patient customers arewilling to queue. Each customer brings a random number ofunits, i.e., the order batch size. The arrival process for eachcustomer type and side is modeled as an independent com-pound Poisson process where the jump size, correspondingto the order batch size, is exponentially distributed. Unitsarriving to one side of the queue “serve” units queued onthe opposite side. That is, they are matched in a first-comefirst-served (FCFS), unit-for-unit policy and exit the systemimmediately. Units from impatient customers that are notmatched upon arrival abandon immediately. Units frompatient customers that are not served upon arrival join thequeue, but may later abandon. We model an aggregateabandonment process whereby patient units abandon thesystem at a constant, side-dependent rate. We provide thefollowing results.

1. We characterize in closed form the steady-state queuelength distribution and related system-level performancemetrics, i.e., the average queue length, system time, andsystem fill rate. Based on these results, we provide insightson how system performance depends on customer flowcharacteristics.

2. We derive formulae for customer-level performancemeasures, i.e., the expected fill rate of a customer with agiven service deadline and the expected system time ofpatient (nonabandoning) customers. These metrics are usefulin evaluating the experience of individual customers.

3. We compare the results for our model, which capturesabandonment in aggregate, to simulation results for a systemin which customers (not units) abandon after some randomdeadline. Such customer-level abandonment is commonlyassumed, and tractable analytically, in queueing models thatrestrict attention to single-unit arrivals, in contrast to oursetup. We find that there is a close correspondence betweenthe predicted performance based on our analytical results andthe performance observed in the simulation. Furthermore,we show that our model performs better than a model inwhich the abandonment rate is linear in the queue length.We conclude that our model is attractive, because of itstractability and accuracy in approximating the performanceof a system with customer-level deadlines.

To our knowledge, this paper is the first to establishclosed-form results for a double-sided queueing modelwith batch arrivals and abandonment. Previous work ondouble-sided queues has considered batch queues withoutabandonment (e.g., Kashyap 1966) and single-unit arrivalswith abandonment (e.g., Zenios 1999, Boxma et al. 2011).Further, we consider heterogeneous customers with type- and

1179

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Afèche, Diamant, and Milner: Double-Sided Batch Queues with Abandonment1180 Operations Research 62(5), pp. 1179–1201, © 2014 INFORMS

side dependent arrival rates, batch sizes, and abandonmentbehavior, previously not considered.

The model we study captures important features of theoperations of a crossing network (like Posit or Liquidnet),one of several variants of increasingly important financialmarkets. Crossing networks and other so-called “dark pools”act as alternative trading systems to the more familiartransparent markets such as the New York Stock Exchangeand NASDAQ. In contrast to transparent or “lit” exchanges,orders submitted to dark pools are completely hidden fromother market participants. At no point can parties observethe buy and sell queues in the system. Basic transactioninformation (security and number of shares) is only disclosedwell after a trade is carried out, whereas the identities ofthe trading parties are never disclosed. In 2010, 13.27% ofU.S. equities trading volume was transacted in dark pools,which represented a 30.7% increase over 2009 (Schack andGawronski 2011).

In crossing networks, the most common type of dark pool,traders (referred to below as customers) submit anonymousbuy or sell orders for a particular security along with theorder size and the maximum time to carry out the transaction,which we refer to as the deadline. Orders from counterpartiesare matched in FCFS order. Trades occur whenever there isliquidity on both sides of the market, and orders may bepartially filled. Trades are carried out at a price exogenouslygiven, for example, the midpoint of the bid/ask spread derivedfrom the national best bid and offer (NBBO). If an order isnot completely filled by its deadline, the remainder of theorder is canceled (the remaining shares may be submitted toa different exchange).

Many crossing networks accept market orders (an order tobuy or sell at the prevailing market price) and limit orders(an order to buy or sell at a specific price or better). We onlyconsider market orders as (1) in a number of crossingnetworks, the volume of limit orders at this time is smallcompared with market orders (ITG 2011) and (2) allowinglimit orders significantly changes the behavior of the queueunder consideration, and so is left for future work.

The remainder of the paper proceeds as follows. In §2,we review the related literature. In §3, we introduce thequeueing model. In §4, we derive closed-form expressionsfor the steady-state queue length distribution and the system’sexpected performance. We use these results to provideinsights into the effect of flow characteristics on systemperformance. In §5, we derive formulae for the expected fillrate and system time of a customer who arrives in steadystate with a specific order size and deadline. In §6, wecompare the results for our model to simulation results forsystems in which customers arrive with deadlines and/orhave nonexponential order size distributions. Our concludingremarks are in §7.

2. Literature ReviewIn this section, we review research on double-sided queue-ing models, dam and insurance models that are related

to our aggregate abandonment model, associated modelson perishable inventory, and recent research on financialtrading markets.

Kendall (1951) introduced the double-sided queueingmodel using the example of taxis and customers indepen-dently arriving to a queueing point. Solution methods forthe double-sided queueing model were introduced in theearly 1960s by Dobbie (1961). Additional research in thearea includes time-dependent arrival rates (Giveen 1963)and bulk service (Kashyap 1966). More recently, the modelhas been applied to assembly facilities producing multi-part components. Parts arrive independently according to arenewal process, but the component can only be assembledwhen all the parts are present. Papers that discuss theseassembly-like queues include Som et al. (1994) and Taka-hashi et al. (2000). Double-sided queueing models haveapplications in other areas, including parallel processing,database concurrency control, communication protocols, andinventory management.

Zenios (1999) and Boxma et al. (2011) consider double-sided queues with abandonment and apply the model to theorgan transplant waiting list in the United States. Organs andpatients arrive independently to a queue, but may abandonas organs can expire and patients can die. In contrast toour setup, these models focus on the arrival and departureof single units only and do not take into account batcharrivals. Kim et al. (2010) present a simulation model ofa double-sided, batch arrival, batch service queue withabandonment. They provide a numerical procedure to findthe state probabilities and various performance measures,but no analytical results.

Our research is related to dam and insurance models inwhich the state of the system changes either by a jump or acontinuous process (c.f. Asmussen 2003). Perry et al. (1999,2002) and Lee (2007) analyze dam processes whose systemstate can increase and decrease through jumps while alsodecreasing continuously. These papers focus on the transientproperties of the dam or insurance process such as the timeuntil the system is empty and the total amount of claimspaid prior to ruin. Further, such models are naturally definedonly on the half line in comparison to a double-sided queue.In comparison, we calculate metrics such as the number ofunits in the system, the fill rate, and the average amountof time a unit spends in the system. Further, our systemoperates perpetually and does not condition on a terminationevent (i.e., ruin or overflow), which is the foundation ofinsurance research and dam models.

Our research is also related to the application of queueingtheory to continuous review, perishable inventory systems asintroduced by Graves (1982) and Kaspi and Perry (1983).In such work impatient customers are either served frominventory or queue, while inventory units are supplied by acontinuous production process and expire after some fixedamount of time. This model has been modified to allow formultiple-unit demand and random customer abandonmenttimes (Kaspi and Perry 1984), items with random expiration

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Afèche, Diamant, and Milner: Double-Sided Batch Queues with AbandonmentOperations Research 62(5), pp. 1179–1201, © 2014 INFORMS 1181

times (Perry 1985), and state-dependent arrival and departurerates with finite queue lengths (Perry and Stadje 1999,Nahmias et al. 2004). See Karaesman et al. (2011) for acomprehensive survey of these and related studies. In contrastto our setup, models in this literature consider a supplyprocess that is constrained or controlled, which gives rise todifferent system dynamics.

The literature on crossing networks and other dark poolsis sparse. Hendershott and Mendelson (2002) and Ray (2010)study the conditions under which investors should use adark pool versus a traditional trading venue. Cebiroglu andHorst (2011) and Kratz and Schoeneborn (2011) investigateoptimal trading strategies when liquidating a portfolio withina finite time horizon in the presence of a dark pool and a litexchange. They examine the trade-off between executionuncertainty and price impact, incorporate adverse selectionand show that there are optimal execution strategies that useboth markets. Ready (2009) studies the constituency of thetraders in a dark pool and its relation to market liquidity.Buti et al. (2011) study the effects of dark pool trading onoverall trading volume and on the market quality and orderflow characteristics of the limit order book. Several papersfrom the industry, e.g., Sofianos (2007) and Mittal (2008),describe the various flavors of dark pools and their effect onthe trading system as a whole. All of these papers considerthe market-level effect of dark pools. They do not study theoperation of dark pools and how fill rates and system timesare affected by operational characteristics.

There has been considerable recent literature on theoperation of a limit order book in a visible exchange. Paperssuch as Parlour (1998), Foucault et al. (2005), Rosu (2009),and Maglaras and Moallemi (2011) study the strategicbehavior of traders given the amount of liquidity available ateach price. While limit orders are accepted at some crossingnetworks, their order books are not visible. Models forvisible exchanges are not directly applicable to crossingnetworks as they have different dynamics compared withtheir visible counterparts (c.f. Buti et al. 2011, Zhu 2012).

3. The ModelIn this section, we introduce the double-sided queueingmodel with batch arrivals and abandonment. Two types ofcustomers, patient and impatient, arrive on each side. Eachcustomer brings a number of units, her order size, to beserved by units from the opposite side. We refer to the sidesof the queue as a and b, denote the respective quantities bysubscripts a and b, and quantities pertaining to impatientcustomers by superscript I .

Units for each type and side arrive following a com-pound Poisson process with exponential jump (order) sizedistribution. The arrival processes and order sizes are mutu-ally independent. Let 8Na4t51 t ¾ 09 be the arrival processof side-a patient customers. Similarly define Nb4t5, N

Ia 4t5,

and N Ib 4t5. Let the arrival rates be �j , j ∈ 8a1 b9 for patient

and �Ij , j ∈ 8a1b9 for impatient customers. Let Xai be the

order size of the ith side-a patient customer arrival. Similarly,define Xbi, X

Iai, and XI

bi. Let the mean order size be �j 4�Ij 5,

j ∈ 8a1b9 for patient (impatient) customers. For notationalconvenience, let �j = 1/�j and �I

j = 1/�Ij , j ∈ 8a1 b9.

Orders arriving on one side “serve” (are matched with)units queued on the opposite side based on an FCFS, unit-for-unit policy. Each match involves an equal number ofunits from each side and matched units exit the systemimmediately. This means that orders may be partially filled,and at no point in time is there a queue on both sides. Unitsfrom impatient customers that are not served upon arrivalabandon immediately. Units from patient customers whoare not served upon arrival join the queue, but may laterabandon; in this regard, their patience is finite. We modelan aggregate side-dependent abandonment process. Wheny patient units are queued on side-a, the aggregate aban-donment rate is ra4y5¾ 0 (similarly rb4y5¾ 0). We assumeri4y5= ki for some constants ki ¾ 0, i ∈ 8a1 b9. We discussthis assumption below.

We define a stochastic process 8Y 4t51 t ¾ 09, whereY 4t5 ∈� is the queue length of patient units at time t ¾ 0.The sample paths of this process are right-continuous withleft limits; at any jump point s, we assume that Y 4s+5= Y 4s5.We map the queue length of patient side-a (side-b) units tothe negative (positive) half line, so that Y −4t5 is the numberof side-a, and Y +4t5 is the number of side-b units in queue.Define the abandonment rate function

r4y54

= −ra4�y�518y<09 + rb4y518y>091 for y ∈�1

where 1A is the indicator function of event A.The queue length process Y 4t5 satisfies the stochastic

differential equation

dY 4t5= −XaNa4t5dNa4t5+XbNb4t5

dNb4t5

− min6XIaN I

a 4t51 Y +4t−57dN I

a 4t5

+ min6XIbN I

b 4t51 Y −4t−57dN I

b 4t5− r4Y 4t55dt0

The first two terms give the jumps for the side-a and -bpatient arrivals, the third and fourth terms, the jumps for theimpatient arrivals, and the last term expresses the abandon-ment rate. Figure 1 shows a sample path of the Y 4t5 process.Although a unique solution to this stochastic differentialequation can be found by solving for Y 4t5 piecewise betweenthe jump times of Na1Nb1N

Ia , and N I

b , our analysis focusesexclusively on the steady-state properties of Y 4t5.

Discussion. The relationship between our system modeland the behavior of a crossing network rests on assumptionsregarding the arrival processes, the order size distributions,and the abandonment behavior.

Focusing on visible exchanges, Foucault et al. (2005) andRosu (2009) model the arrival process in financial marketsas Poisson, noting the acceptability of doing so under stable

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Afèche, Diamant, and Milner: Double-Sided Batch Queues with Abandonment1182 Operations Research 62(5), pp. 1179–1201, © 2014 INFORMS

Figure 1. Illustrative sample path of the queue lengthprocess 8Y 4t51 t ¾ 09 for ri4y5= ki, i ∈ 8a1 b9.

0 5 10 15 20 25–20

–10

0

10

20

t

Y(t

)

kb

ka

Queue of patient side-b units

Queue of patient side-a units

Note. The queue length of side-b (side-a) units is mapped to the positive(negative) half line.

market conditions. Gopikrishnan et al. (2000) and Maslovand Mills (2001) demonstrate that the order size distributionin transparent markets is well described by either a powerlaw or an exponential form. However, we are not aware ofstudies or publicly available data on the arrival processesand order size distributions for crossing networks, likely asa result of their recent advent and the anonymity they offertheir customers. In §6.2, we provide sensitivity analysis withrespect to the order size distribution.

We consider two abandonment processes. For impatientcustomers, we assume that they abandon immediately afterarriving to the system, which closely models a type of cus-tomers often seen within crossing networks who submit whatis known as “immediate-or-cancel” orders. Such customerstest the market by submitting orders (often of a relativelysmall size) to determine if a queue exists on one side orthe other, withdrawing their orders almost immediately ifnot filled. By doing so, they may gain information from thecrossing network that is unavailable otherwise, and exploitthis information by taking simultaneous positions in lit anddark markets.

For patient customers, we assume that they abandon afterqueueing for some time. Customers of this type provideliquidity in crossing networks and are necessary for theirfunctioning. Typically, such customers have made a commit-ment to buy or divest from a security. As such they willkeep their orders in the market for some period of timebefore withdrawing them. As discussed above, we modelthe aggregate abandonment of units from the queue ratherthan each individual’s abandonment to maintain tractability.In §6, we show that the results derived in our model are verysimilar to simulation results for a system with abandonmentmodeled at the customer level.

We assume that the abandonment function, ri4y5, isconstant for each i ∈ 8a1b9. Other research papers in thequeueing literature have made similar assumptions. In par-ticular, Adan et al. (2009) and Perry and Whitt (2011)assume constant abandonment rates when analyzing the

fluid and/or diffusion limit of their respective queueingmodels. An alternate assumption that is common in theliterature on single-sided, nonbatch queues is that the totalabandonment rate is linear in the queue length; Harris et al.(1987), Brandt and Brandt (2002) and Atar et al. (2014).In the online appendix (available as supplemental materialat http://dx.doi.org/10.1287/opre.2014.1300), we comparesystem performance measures of these two abandonmentmodels in a double-sided, batch arrival setting. This compari-son is made versus a simulation in which customers abandonafter some random deadline. In the linear abandonmentcase, the steady-state queue length distribution can onlybe computed numerically. We find that for systems withfill rates less than 50% (fill rates for crossing networks aretypically in the 1%–2% range), the model with a constantabandonment rate performs better.

Intuitively, one might expect that a linear abandonmentrate model would be better; the more units there are inqueue, the higher the cancellation rate. Such a model maybe appropriate when fill rates are high and queue lengths arelarge. However, in crossing networks, fill rates are typicallyvery low and the system is empty for a large fraction oftime, because the abandonment rate may be significant evenwhen the queue is short. The constant abandonment ratemodel captures this behavior. In contrast, under the linearmodel, the rate of abandonment becomes arbitrarily small asthe queue length approaches zero.

Furthermore, the variation in trader patience and ordersize typically seen in practice may result in a constantabandonment rate. In crossing networks, smaller orders tendto be submitted by impatient traders that leave immediatelyupon arrival, or shortly thereafter based on a predetermineddeadline. To the extent that there is a queue on the same sideupon arrival, these orders tend to leave unfilled. Therefore,their abandonment rate is similar to their constant arrivalrate. Conversely, patient traders tend to submit larger ordersand gradually withdraw units from these orders based ona timer. The timers are typically exogenous, i.e., queuelength independent, and may depend on the trading strategyor market-related factors. As a result, the abandonmentrate for a given patient trader may be constant over time.To the extent that few large orders exist simultaneously inthe market, which is typically the case, the aggregate rate ofabandonment would appear constant.

4. System-Level PerformanceIn this section, we characterize the stationary queue lengthdistribution using techniques from level crossing theory.We then present closed-form expressions for system-levelsteady-state performance metrics of interest, in particular,the average system time and the average system fill rate.We illustrate the use of these results to provide insights intothe effect of flow characteristics on system performance.

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Afèche, Diamant, and Milner: Double-Sided Batch Queues with AbandonmentOperations Research 62(5), pp. 1179–1201, © 2014 INFORMS 1183

4.1. Stationary Queue Length Distribution

Let the random variable Y denote the stationary limit ofthe process 8Y 4t51 t ¾ 09 if it exists; let F be its cumulativedistribution function, that is, limt→� P4Y 4t5¶ y5= F 4y5 forall y such that P4Y = y5= 0 if Y 4t5 converges in distribution.We will show that if the system is stable, there is a nonzeroprobability that the system is empty. Let �0 ≡ P4Y = 05represent this probability and let f 4y5 denote the densityof Y for y 6= 0.

Proposition 1. The density function f 4y5, solves the follow-ing differential equations:

For y > 0,

f ′′′4y5r4y5+f ′′4y543r ′4y5+4�b−�a−�Ia5r4y5

−4�a+�b+�Ia55+f ′4y543r ′′4y5+24�b−�a−�I

a5r′4y5

+4�Ia�a−�b�

Ia−�b�a5r4y55+f ′4y54�I

a4�a−�b5

+�b4�a+�Ia5+�a4�

Ia−�b55+f 4y54r ′′′4y5

+4�b−�a−�Ia5r

′′4y5+4�a�Ia−�b�

Ia−�b�a5r

′4y5

+�a�Ia�br4y55+f 4y54�a�b�

Ia+�I

a�b�a−�b�a�Ia5

=01 (1)

and for y < 0,

f ′′′4y5r4y5+f ′′4y543r ′4y5+4�Ib+�b−�a5r4y5

−4�a+�b+�Ib55+f ′4y543r ′′4y5+24�I

b+�b−�a5r′4y5

+4�Ib�b−�a�

Ib−�a�b5r4y55+f ′4y54�I

b4�a−�b5

−�b4�Ib−�a5−�a4�

Ib+�b55+f 4y54r ′′′4y5

+4�Ib+�b−�a5r

′′4y5+4�Ib�b−�a�

Ib−�a�b5r

′4y5

−�a�b�Ibr4y55+f 4y54�b�a�

Ib+�I

b�a�b−�a�b�Ib5

=00 (2)

(All proofs are in the appendix.)For r4y5 = −ka18y<09 + kb18y>09, the differential equa-

tions (1) and (2) are homogeneous with constant coefficientsover the half line. Three conditions determine their solution.The first is the normalization condition∫ �

−�

dF 4y5= 10 (3)

The second and third are flow balance conditions: In steadystate, the inflow rate of patient units into the system equalstheir outflow rate. The inflow rate is �a�a for patient side-aunits and �b�b for patient side-b units. To formalize theoutflow rates, let sa4y5 denote the expected number of unitsserved (matched) upon arrival of a patient side-a customergiven y > 0 units are in the queue. Then

sa4y5=∫ y

0�aze

−�azdz+∫ �

y�aye

−�azdz=1�a

41−e−�ay50

Let sIa4y5 denote the expected number of units servedupon arrival of an impatient side-a customer given y > 0units are in the queue. Then

sIa4y5=1�I

a

41 − e−�Iay50

Similarly,

sb4y5=1�b

41 − e�by5 and sIb4y5=1�I

b

41 − e�Iby51

are, respectively, the number of units served upon the arrivalof a patient and impatient side-b customer given y < 0 unitsare in queue.

The expected numbers of units served upon arrival ofpatient side-a and side-b customers are, respectively,

s̄a4

=

∫ �

0+

sa4y5f 4y5dy and s̄b4

=

∫ 0−

−�

sb4y5f 4y5dy0

For impatient side-a and side-b customers, the expectedvalues are, respectively,

s̄Ia4

=

∫ �

0+

sIa4y5f 4y5dy and s̄Ib4

=

∫ 0−

−�

sIb4y5f 4y5dy0

Recalling that an equal number of side-a and side-b unitsare served upon each arrival, the flow balance equations forpatient side-a and patient side-b units are, respectively,

�a�a = �as̄a +�b s̄b +�Ib s̄

Ib +

∫ 0−

−�

�r4y5�f 4y5dy1 (4)

�b�b = �as̄a +�b s̄b +�Ias̄

Ia +

∫ �

0+

r4y5f 4y5dy0 (5)

The left-hand sides of (4) and (5) are the inflow rates ofpatient units, the right-hand sides of (4) and (5) are therespective outflow rates. The outflow rate in each caserepresents the different ways that patient units can leavethe system. Consider the right-hand side of (4). The firstterm represents the rate of patient side-a units that arematched upon arrival with queued side-b units; s̄a is theaverage number of units matched per arrival. The second andthird terms represent, respectively, the rates of patient andimpatient side-b units that are matched upon arrival withqueued side-a units. The final term represents the averageabandonment rate of patient side-a units.

Theorem 1. For r4y5= −ka18y<09 + kb18y>09, the double-sided queue is stable if and only if

−ka −�Ib

�Ib

<�b

�b

−�a

�a

<�Ia

�Ia

+ kb0 (6)

In this case, the stationary distribution of the queue lengthsolves (1)–(5) and is given by

dF 4y5=

Uaeyäa dy1 y < 01

�01 y = 01

Ube−yäb dy1 y > 01

(7)

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where

�0 =1−Ua

äa

−Ub

äb

>01

Ua=

(

äa�a4äa+�b54äa+�Ib5

·(

�Ia�a�b+äb44�a+�I

a5�b−�b�a5

+4�a�b−�b�a5�Ia+kb�b4äb+�a54äb+�I

a5)

)

·4é5−1>01

Ub =

(

äb�b4äb+�a54äb+�Ia5

(

�Ib�a�b+äa

(

4�b+�Ib5�a−�a�b

)

+4�b�a−�a�b5�Ib+ka�a4äa+�b54äa+�I

b5)

)

·4é5−1>01

é =�a�b

(

äa

(

äb4�Ib�a+�I

a4�b+�Ib55

+�Ia4�b+�I

b5�a+�Ib�a�

Ia

)

+�Ia�

Ib�a�b+�I

b�a�Ia�b+�I

a�b�Ib�a

+ka(

äb4�a+�Ia5+�I

a�a+�a�Ia

+kb4äb+�a54äb+�Ia5)

·4äa+�b54äa+�Ib5+kb4äb+�a54äb+�I

a5

·(

äa4�b+�Ib5+�I

b�b+�b�Ib

)

+äb

(

4�a+�Ia5�

Ib�b+�I

a�b�Ib

)

)

1

and the constants äa > 0 and äb > 0 are given in the proofas explicit functions of the system parameters.

The stability condition (6) is intuitive. Recall �Ij = 1/�I

j

and �j = 1/�j . Let �i = ki +�Ij�

Ij for i1 j ∈ 8a1 b91 i 6= j and

interpret �i as the system’s excess side-i capacity, i.e., themaximum rate at which it can deplete patient units queuedon side-i through abandonment plus the flow of impatientside-j units. By (6), the system is stable if it has sufficientexcess capacity on whichever side it is required to processthe net inflow rate ��b�b −�a�a� into the queue.

To highlight the effect of impatience on system perfor-mance, consider a stable system in which patient customersabandon and there are impatient customers, i.e., ki > 0 and�Ii�

Ii > 0 for i ∈ 8a1 b9. We compare this original system to

two alternate systems, referred to as infinitely patient andabandonment-only systems. In an infinitely patient system,patient customers do not abandon, i.e., ka = kb = 0. In anabandonment-only system, there are no impatient customers,i.e., �I

a = �Ib = 0. Based on the proof of Theorem 1, we

make two remarks.

Remark 1. Consider the original system, an abandonment-only and an infinitely-patient system with the same parame-ters �i1�i1�

Ii , i ∈ 8a1 b9. Suppose that the abandonment-only

system has side-i abandonment rate �−i < �i, and the infinitely

patient system has side-j impatient unit arrival rate �+

i > �i,where for i1 j ∈ 8a1 b91 i 6= j;

�−

i = �i −�Ij�

Ij

äi�Ij

1 +äi�Ij

= ki +�Ij�

Ij

1 +äi�Ij

and

�+

i = �i + kiäi�Ij = ki41 +äi�

Ij5+�I

j�Ij 0

(8)

Then, all three systems have the same stationary queue lengthdistribution, even though the abandonment-only system haslower excess capacity, and the infinitely patient system hashigher excess capacity, compared to the original system.These differences are offset by differences in the variabilityof their patient unit outflow. Compared to the originalsystem, this variability is lower in the abandonment-onlysystem since the abandonment flow is constant, and higherin the infinitely patient system since the impatient flow isstochastic. Less variability implies less queueing. Thereforethe abandonment-only system can match the stationaryqueue length distribution of the original system with lessexcess capacity, whereas the infinitely patient system requiresmore excess capacity to do so. Note from (8) that �+

i /�−i =

1 +äi�Ij . The more fluid like the impatient unit arrivals,

i.e., the smaller �Ij for fixed �I

j�Ij , the more similar, and

the larger the decay rates äi of the distribution f , the moredissimilar the three systems.

Remark 2. Denote by double bar the abandonment-onlysystem with abandonment rates ¯̄ki = �i for i1 j ∈ 8a1 b91 i 6= j .Similarly, denote by double hat the infinitely patient system

with impatient unit arrival rates ˆ̂�Ij�

Ij = �i for i ∈ 8a1b9.

Let both systems have parameters �i1�i1�Ii , i ∈ 8a1b9, as

in the original system. Then they have the same excess

capacity as the original system. Letting ¯̄�0 and ˆ̂�0 denotethe steady-state probabilities that these systems are empty,we have

ˆ̂�0 <�0 <¯̄�01

which is consistent with Remark 1. Since the double-barsystem has lower outflow variability, it is more likely tobe empty than the original system, and vice versa for thedouble-hat system. Viewed differently, the abandonment ratesin the double-bar system exceed those of the abandonment-only system in Remark 1, i.e., �i > �−

i , and conversely, theimpatient flow rates in the double-hat system are lowerthan those of the infinitely patient system in Remark 1, i.e.,�i < �+

i .

4.2. Mean Queue Length, System Time, andSystem Fill Rate

In this subsection, we present system-level steady-stateperformance metrics of interest, in particular, the meanqueue length, average system time, and average system fillrate. These metrics are straightforward to compute in closedform based on (7).

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Queue Length. The expected queue length measures thebias of one side versus the other and is given by

Ɛ6Y 7=∫ �

−�

y dF 4y5=Ub

ä2b

−Ua

ä2a

0

For each side, the conditional mean queue length given thereis a queue is

Ɛ6�Y � � Y < 07=

∫ 0−

−��y�f 4y5dy

∫ 0−

−�f 4y5dy

=1äa

and

Ɛ6Y � Y > 07=1äb

0

The mean absolute queue length measures the expected totalexcess in the system and is given by

E6�Y �7=Ɛ6�Y � �Y <07·�4Y <05+Ɛ6Y �Y >07·�4Y >05

=Ua

ä2a

+Ub

ä2b

1

where �4Y < 05=Ua/äa and �4Y > 05=Ub/äb by (7).When considering the application of our model to a financialmarket, this metric would be an expression of the marketliquidity.

System Time. Let Wi, i ∈ 8a1b9 be the average systemtime of patient side-i units. By Little’s law,

Wa =E6�Y � · 18Y<097

�a�a

=E6�Y � � Y < 07�4Y < 05

�a�a

=Ua

�a�aä2a

and Wb =Ub

�b�bä2b

0 (9)

Impatient customers have a system time of zero.System Fill Rate. Units exit the system either by matching

with counterparties or by abandoning. Let the fill rate, �i,i ∈ 8a1 b9, be the average fraction of side-i units that match.

Proposition 2. The system fill rate of side-i units is given by

�i = 1 −ki4Ui/äi5+�I

i�Ii 41 −Uj/4äj41 +�I

iäj555

�i�i +�Ii�

Ii

for i1 j ∈ 8a1 b91 i 6= j0 (10)

The fill rate equals 1 minus the fraction of units that exitunserved. The denominator in (10) is the total arrival rate ofside-i units. The numerator is the expected rate of side-iunits that exit unserved. The first term is the abandonmentrate of patient side-i units, and the second is the arrival rateof impatient side-i units, multiplied by the average fractionof these units that exit unserved on arrival.

We also provide an expression for the fraction of patientunits that exit the system by matching with patient counter-parties. We refer to this metric as the patient-to-patient fillrate. Let �ij , i1 j ∈ 8a1b91 i 6= j be the average fraction ofpatient side-i units that match with patient side-j units.

Corollary 1. For patient side-i units served by patientside-j units, the fill rate is

�ij = 1 −ki4Ui/äi5+�I

j�Ij4Ui/4äi41 +�I

jäi555

�i�i

for i1 j ∈ 8a1 b91 i 6= j0 (11)

The patient-to-patient fill rate equals 1 minus the fractionof units that either exit unserved or are served by impatientcounterparty order flow.

4.3. Effect of Flow Characteristics onSystem Performance

Our closed-form expressions provide useful tools to studyhow the flow characteristics affect system performance.For illustration, we consider the performance impact of(1) asymmetric demand volume, (2) varying abandonment,and (3) asymmetric demand burstiness. For simplicity, weset �I

a = �Ib = 0.

Figure 2 shows the expected fill rates and system times forside-a and side-b units, and their weighted averages as thedemand mix varies. The arrival rates �a and �b vary between0.8 and 1.2, holding �a +�b = 2 with �a =�b = 100 and

Figure 2. Expected fill rates and system times for side-aand side-b units and their weighted averages,as functions of �a and �b , holding �a +�b = 2.

0.8 0.9 1.0 1.1 1.20.6

0.7

0.8

0.9

1.0� Fill rate

�Avg

�a�b

0.8 0.9 1.0 1.1 1.2�a

�a

0

5

10

15

W System time

Wa

(b)

(a)

Wb

WAvg

Note. �a =�b = 100, ka = kb = 3705.

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Table 1. Expected system fill rates and system times for side-a and side-b units comparing cases of low and highabandonment and symmetric and asymmetric demand characteristics.

Symmetric Asymmetric�a = �b = 1 �a =�b = 100 �a = 1, �b = 5 �a = 100, �b = 15

Low abandonment High abandonment Low abandonment High abandonmentSystem-level metric ka = kb = 3705 ka = kb = 11000 ka = kb = 3705 ka = kb = 11000

�a (%) 81087 9050 69003 7008�b (%) 81087 9050 92004 9044Wa 2067 0010 8009 0011Wb 2067 0010 0022 0002

ka = kb = 3705. We observe, as is intuitive, that the fill ratedecreases and the system time increases for side-a units astheir prevalence increases. Further, the weighted average fillrate given by �Avg = 4�a/4�a +�b55�a +4�b/4�a +�b55�b ismaximized at symmetry (�a = �b = 1), though it is somewhatinsensitive to changing parameter values. The weightedaverage system time is minimized at symmetry, and it issomewhat more sensitive to changing parameter values.The system behavior is nearly identical in the case where themean order sizes �a and �b vary from 80 to 120, holding�a +�b = 200 with �a = �b = 1 and ka = kb = 3705. Thisdemonstrates that the system behavior is closely related tothe overall arrival rate for each side, i.e., �i�i, i ∈ 8a1 b9.

Table 1 shows the effect of varying abandonment on the fillrates and system times of side-a and side-b units for severalcases. We observe that increasing the abandonment ratereduces the fill rates and system times. For the asymmetriccase, there are a total of �a�a = 100 side-a units arrivingper unit time while �b�b = 75. As in Figure 2, we observethe side with the higher overall arrival rate has lower fillrates and higher system times.

Next, we consider the trade-off between the fill rate andsystem time when customers on both sides bring the samenumber of units to the system per unit time, but arrivalson one side are more bursty, i.e., customers arrive lessfrequently with larger orders, whereas customers on the otherside arrive more frequently with smaller orders. Figure 3shows parametric plots of the fill rate versus the systemtime for side-a and side-b units as we vary �b ∈ 6000111007,fixing �b�b = 100, �a = 1, and �a = 100 (ka = kb = 3705).For both sides, the system time decreases in �b, implyingthat �b >�a for points to the left and �b <�a for points tothe right of the peak on the graph. The maximum fill rateoccurs at symmetry (�a = �b = 1). Noting that both sideshave the same fill rate since their inflow rates are the same,we observe that the side with more frequent, smaller ordersperforms better, i.e., lower system time for equal fill rate.For example, as shown in Figure 3, for �b = 002, both sideshave a fill rate of 8106%, but the expected system time ofside-b units is approximately double that of side-a units.

5. Customer-Level MeasuresIn this section, we consider the experience of an individualcustomer who brings a known number of units to the system

and a service deadline, i.e., a maximum time the customer iswilling to wait in queue. We assume this customer (referredto as the marked customer) randomly arrives to the systemin steady state. We characterize the expected system timeand fill rate for such a customer. Since these metrics dependon customer-specific attributes, they differ from the systemaverages given in §4.2.

We consider two types of marked customers, time-constrained and time-unconstrained. A time-constrainedcustomer arrives with a finite deadline T and abandons thequeue at that time, withdrawing all of his unserved units;none of his units abandon prior to this time. For such atime-constrained customer, we derive his expected fill rate.A time-unconstrained customer stays in the queue until herentire order is processed. That is, she is of infinite patience(T = �). By definition, her fill rate will be 100% and wederive her expected system time.

5.1. Clearing Time

We first define and compute the expected clearing time,which we use in §5.2 to compute the expected systemtime for a time-unconstrained customer. Suppose there areside-b units in queue, i.e., y > 0. Assume that none of theseunits abandon the queue, so that their departure is triggered

Figure 3. Parametric plots of expected system fill ratevs. expected system time for �b ∈ 6000111007,holding �b�b = 100, �a = 1, and �a = 100(ka = kb = 3705).

�b = 0.2�b = 1.0

�b > �a �b < �a

50 10 15 20

System time

81.0

81.2

81.4

81.6

81.8

82.0

Fill

rate

(%

)

Side-a units

Side-b units

Note. For each value of �b , we compute the system-level fill rate andsystem time for side-a and side-b units.

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exclusively by the arrival of patient and impatient side-acustomers. Define the clearing time �c

b 4y5 as the time forthese side-b units to leave the system by being served.

Let Z̄a4t5 be the number of side-a units that arrive to thesystem by time t, assuming Z̄a405= 0. That is,

Z̄a4t54

=

Na4t5∑

i=1

Xa1 i +

N Ia 4t5∑

i=1

XIa1 i

with 8Na4t51 t ¾ 09, 8N Ia 4t51 t ¾ 09, Xa1 i, and XI

a1 i as definedin §3. Let N̄a4t5=Na4t5+N I

a 4t5. Then 8N̄a4t51 t ¾ 09 isa Poisson process with rate �̄a = �a + �I

a. Let X̄a1 i be arandom variable with density ga4z5 and distribution functionGa4z5 given by

ga4z5=�a�ae

−�az +�Ia�

Iae

−�Iaz

�̄a

1

Ga4z5= 1 −�ae

−�az +�Iae

−�Iaz

�̄a

1

(12)

respectively. Then Z̄a4t5 is the compound Poisson process

Z̄a4t5=

N̄a4t5∑

i=1

X̄a1 i0

The clearing time for y side-b units is given by

�cb 4y5

4

= inf8t ¾ 02 Z̄a4t5¾ y9 for y > 00

Proposition 3. The expected value of �cb 4y5 for y > 0 is

Ɛ6�cb 4y57

=�a�

2a +�I

a4�Ia5

2

4�a�a +�Ia�

Ia5

2+

y

�a�a +�Ia�

Ia

−�a�

Ia4�a −�I

a52

4�a +�Ia54�a�a +�I

a�Ia5

2e−y44�a�a+�Ia�

Ia5/44�a+�Ia5�a�

Ia550

Note that limy↓0 E6�cb 4y57= 1/�̄a, the expected time until the

first arrival.A similar expression holds for the clearing time for the

side-a queue.

5.2. System Time for a Time-UnconstrainedCustomer

Consider a side-i, time-unconstrained marked customer whosubmits an order of size x and remains in the system untilher entire order is filled. Let T c

i 4x1 y5 be her conditionalsystem time given the queue length is y upon her arrival,let Ɛ6T c

i 4x1 y57 be its expectation, and let Ɛ6T ci 4x57 be her

unconditional expected system time. We assume any side-iunits in queue upon the arrival of the marked customerabandon the system at rate ki. Note this assumption isconsistent with a first come/first abandon assumption, whichwas not made in §3 but is needed here.

Suppose the marked customer is a side-b arrival and recallthat y < 0 indicates a queue of side-a units. There are threecases: (1) If y ¶ −x, all her units are served on arrivaland T c

b 4x1 y5= 0. (2) If −x < y¶ 0, then �y� of her unitsare served on arrival and the remaining x + y units jointhe queue, so that Ɛ6T c

b 4x1 y57= Ɛ6�cb 4x+ y57, the expected

time to clear these units. (3) If y > 0, she must queue andƐ6T c

b 4x1 y57 solves the following integro-differential equationobtained by conditioning on the first arrival of a side-acustomer, where z denotes that customer’s order size and tis the time elapsed since the arrival of the marked customer:

Ɛ6T cb 4x1y57

=

∫ �

y/kb

�̄ae−�̄at

(

t+∫ x

0ga4z5Ɛ6�

cb 4x−z57dz

)

dt

+

∫ y/kb

0�̄ae

−�̄at

(

t+∫ y−tkb

0ga4z5Ɛ6T

cb 4x1y−tkb−z57dz0

+

∫ x+y−tkb

y−tkb

ga4z5Ɛ6�cb 4x+y−tkb−z57dz

)

dt0 (13)

The time until the first side-a arrival is exponentially dis-tributed with mean 1/�̄a and her order size is distributed asga4z5. The first term in (13) expresses the case where thefirst side-a arrival occurs after all y units in queue abandon.The expected remaining system time of the marked customeris Ɛ6�c

b 4x− z57 if z < x, and zero otherwise. The secondterm expresses a renewal equation for the case where thefirst side-a arrival occurs before all y units have abandoned.In this case, if z < y − tkb, then y− tkb − z units remain inqueue in front of the marked customer. Her expected remain-ing system time is Ɛ6T c

b 4x1 y− tkb − z57. If z¾ y− tkb, thenher expected remaining system time is Ɛ6�c

b 4x+ y− tkb − z57if z < x+ y− tkb, and zero otherwise.

Proposition 4. The conditional expected system time giventhat a time-unconstrained side-b customer submits an orderof size x to a queue that contains y units is

Ɛ6T cb 4x1 y57=

0 −�< y ¶−x1

Ɛ6�cb 4x+ y57 −x < y ¶ 01

Ɛ6T cb 4x1 y57 0 < y <�1

where

Ɛ6T cb 4x1y57

=C0 +C1y+12C1

K1 +√K2

K1

√K2 −K2

e−y4K1−√K25

−12C1

K1 −√K2

K1

√K2 +K2

e−y4K1+√K25+

x

�a�a+�Ia�

Ia

−�a�

Ia4�a−�I

a52

4�a+�Ia54�a�a+�I

a�Ia5

2e−x44�a�a+�Ia�

Ia5/44�a+�Ia5�a�

Ia551

and

K1 =4�a+�Ia+kb�a+kb�

Ia5/2kb1

K2 =(

�2a+4�I

a52+k2

b4�a−�Ia5

2−2kb�

Ia4�a+3�I

a5

+2�a4�Ia+kb4�a−�I

a55)

/4k2b1

C0 =(

4K21 −K254�

Ia4�

Ia5

2+�a�

2a5

·4kb+�a�a+�Ia�

Ia+2�I

a�a5−2K14�Ia�

Ia+�a�a5

2)

·(

4K21 −K254kb+�a�a+�I

a�Ia+2�I

a�a5

·4�Ia�

Ia+�a�a5

2)−1

1

C1 =4kb+�a�a+�Ia�

Ia+2�I

a�a5−10

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Observe that when there is a side-b queue, i.e., y > 0,the system time has a similar structure to the clearing time,having linear and exponential terms in x and y. The expectedsystem time for a time-unconstrained side-b individual withan order of size x is given by

Ɛ6T cb 4x57=

∫ �

−xƐ6T c

b 4x1 y57dF 4y50 (14)

Since the stationary queue length distribution given in (7) isof exponential type, Ɛ6T c

b 4x57 can be determined in closedform. Similar expressions hold for Ɛ6T c

a 4x1 y57 and Ɛ6T ca 4x57.

5.3. Fill Rate for a Time-Constrained Customer

Consider a side-i, time-constrained marked customer whosubmits an order of size x and remains in the system untilhis order is filled or his finite deadline, T time units, haspassed, whichever occurs first. We assume without loss ofgenerality that he arrives to the system in steady state attime 0. None of his units abandon until time T , when all hisremaining unserved units abandon the system en masse. Asin §5.2, we assume that units in queue in front of the markedcustomer abandon the system at rate ki. Let �c

i 4x1 T 5 be thecustomer’s expected fill rate, i.e., the expected fraction ofthe x units that are served by the deadline T .

Suppose the marked customer is a side-b arrival. First, ifT = 0, the customer is “impatient” as defined in §3, and wehave from (7) that

�cb4x105=

E6min4x1Y −57

x=

Ua41 − e−xäa5

ä2ax

0 (15)

For T > 0, we compute �cb4x1 T 5 by conditioning on Y , the

queue length at the arrival time of the marked customer, � ,the time when only units from the marked customer remainin queue, Z̄a4�5, the number of counterparty units that arriveby time � , and Z̄a4�1 T 5, the number of counterparty unitsthat subsequently arrive prior to T . We first characterizethe conditional fill rate given realizations of these randomvariables, and then derive the probability distribution foreach conditioning event to find �c

b4x1 T 5.Let y be the realization of Y . Consider the case where

y > 0, i.e., there are side-b units queued in front of themarked customer and these units abandon at rate kb. Then �is the time until the process Z̄a4t5 either hits or jumps overthe linearly decreasing boundary given by y− kbt. That is, �is the stopping time

� = inf8t ¾ 02 Z̄a4t5¾ y− kbt91

where � ∈ 401 y/kb7 for y > 00 (If y ¶ 0, then � = 0.) The fillrate depends on whether the last of the queued units infront of the marked customer abandons or is served, that is,whether Z̄a4t5, respectively, hits or jumps over the linearlydecreasing boundary y− kbt. Let Q = 4Z̄a4�5− 4y− kb�55

+

denote the overshoot for y > 0. If Z̄a4t5 hits the boundary,Q = 0. Otherwise, Q> 0 (almost surely). So given y and � ,

conditioning on Q is equivalent to conditioning on Z̄a4�5.For simplicity, we assume that patient and impatient side-aorder sizes are i.i.d., so that by (12), the overshoot givenQ> 0 is exponentially distributed according to ga4z5. Let qbe the realization of Q. Let z be the realization of Z̄a4�1 T 5,the number of side-a units that arrive between � and T .

The conditional fill rate is denoted by �cb4x1 T � y1 �1 q1 z5

and is given by

�cb4x1T �y1�1q1z5=

z−y

x∧11 if y¶01

z+q

x∧11 if y>01�¶T 1

01 if y>01� >T 0

(16)

If y¶ 0, then −y units are served immediately and theremainder can be served by the z arrivals over 601 T 7. Ifz−y ¾ x, then �c

b4x1 T � y1 �1 q1 z5= 1; otherwise �cb4x1 T � y1

�1 q1 z5= 4z− y5/x < 1. If y > 0 and � ¶ T , then q units areserved by the overshoot and the remainder can be servedby the z arrivals over 6�1 T 7. If y > 0 and � > T , then themarked customer leaves before the queue ahead of him isdepleted and �c

b4x1 T � y1 �1 q1 z5= 0.To compute the unconditional expectation �c

b4x1T 5, werequire the probability distributions for Z̄a4�1T 5 and Q.Because the arrivals are Poisson, Z̄a4�1 T 5∼ Z̄a4T − �5 for� ¶ T . Let �t4z5 be the distribution of Z̄a4t5. Recalling thejump size distribution ga4z5 associated with Z̄a4t5 definedin (12), we have

�t4z5=

�∑

n=1

e−�̄at4�̄at5

n

n!g∗4n5a 4z51 (17)

where g∗na 4z5 is the n-fold convolution of ga4z5. Example: If

�Ia = 0, then (17) becomes

�t4z5= e−�ate−�az�∑

n=1

4�at5n

n!

�naz

n−1

4n− 15!

= e−�ate−�az

√

�a�at

zI14z

√

�a�azt51

where I14 · 5 is the modified Bessel function of the first order.To determine the distribution of the overshoot Q, we need

the probability that it is zero or positive given y > 0. To doso, we define the defective densities for the time the processZ̄a4t5, respectively, hits 8h4t1 y59 or jumps over 8j4t1 y59 theboundary y − kbt, when y side-b units are in queue at t = 0

h4t1 y5=d

dt�4� ¶ t1 Z̄a4�5= y− kb�51

j4t1 y5=d

dt�4� ¶ t1 Z̄a4�5 > y− kb�50

Following Perry et al. (1999) and Zacks (2004),

h4t1 y5= e−�̄at�∑

n=1

4�̄at5n

n!g∗4n5a 4y− kbt51

j4t1 y5= �ae−4�̄at+�a4y−kb t55 +�I

ae−4�̄at+�I

a4y−kb t55 + e−�̄at

·

�∑

n=1

4�̄at5n

n!

(

G∗4n5a 4y− kbt5−G∗4n+15

a 4y− kbt5)

1

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where g∗4n5a 4y−kbt5 and G∗4n5

a 4y−kbt5 are the n-fold convolu-tions of ga4y−kbt5 and Ga4y−kbt5, respectively. Example. If�Ia = 0, then

h4t1y5=kb�ae−�ate−�a4y−kb t5

�∑

n=0

4�at5n+1

4n+15!�n

a4y−kbt5n

n!

=kbe−�ate−�a4y−kb t5

√

�a�at

y−kbtI1

(

2√

�a�at4y−kbt5)

1

and

j4t1y5=�ae−�ate−�a4y−kb t5

�∑

n=0

4�at5n

n!

�na4y−kbt5

n

n!

=�ae−�ate−�a4y−kb t5I0

(

2√

�a�at4y−kbt5)

1

where I04 · 5 and I14 · 5 are the modified Bessel functions ofzeroth and first order, respectively.

If t < y/kb, the hitting time has density h4t1 y5 whilethe jump time has density j4t1 y5. If � = y/kb, the process(almost surely) hits the boundary at Z̄a4y/kb5 = 0 (i.e.,there were no arrivals before time y/kb). This occurs withprobability �4� = y/kb5= exp4−y�̄a/kb5.

From (16) and taking the expectation over y, � , q, and z,the expected fill rate is

�cb4x1T 5

=

∫ −x

y=−�

dF 4y5+∫ 0

y=−x

∫ �

z=0

(

z−y

x∧1)

�T−t4z5dzdF 4y5

+

∫ �

y=0+

∫ min4T 1y/kb5

t=0

∫ �

z=0

(

z

x∧1)

�T−t4z5h4t1y5dzdtdF 4y5

+

∫ kbT

y=0+

∫ �

z=0

(

z

x∧1)

�T−y/kb4z5e−y4�̄a/kb5dzdF 4y5

+

∫ �

y=0+

∫ min4T 1y/kb5

t=0

∫ �

q=0

∫ �

z=0

(

z+q

x∧1)

·�T−t4z5ga4q5j4t1y5dzdqdtdF 4y50 (18)

A similar formula holds for a marked side-a customer.

6. Comparisons with Simulation Resultsfor Model with Customer Deadlines

In this section, we compare the results for our model, whichassumes a constant aggregate abandonment rate to simula-tion results for a system with customer-level abandonment.In the simulated system, unlike in the one modeled in §3,customers abandon the system at exponentially distributeddeadlines, and each abandoning customer withdraws all ofher unserved units at that time. We consider how closelythe key performance measures given by our model matchthose of this simulated system. In particular, we presentresults on the expected fill rate and the expected systemtime at the system and customer level. We address threequestions: (1) How do these metrics vary with changingparameter values? (2) How do these metrics, based on our

analytical results, compare with simulation results? (3) Howdoes our model perform when there are alternate order sizedistributions and customer trading preferences?

To summarize our results, we find that the measuresderived based on our model perform well in that they provideinsight into the system behavior and compare favorably tosimulation results. In particular, for system- and customer-level metrics there is a close correspondence between themodel and the simulated system. We conclude that our modelis attractive, because of its simplicity and tractability, andbecause of its accuracy in approximating the performance ofa simulated system with customer-level deadlines.

6.1. System-Level Performance Comparisons

In this section, we report system-level performance com-parisons for seven series of test cases (38 cases in total)without impatient customers, that is, �I

a = �Ib = 0. In each

case, we choose parameter values for �i, �i, and ti, i = a1 b,where ti denotes the mean deadline for side-i customers inthe simulation. For each test case, we simulate 5,000,000customer arrivals. To determine the expected system timeand fill rate using (9) and (10), we require the valuesof the abandonment rates ki, i = a1b. We set ki equalto the side-i abandonment rate observed in the simula-tion, given by the number of side-i units that abandonthe system divided by the time the system spends onside i.

In Table 2, we present the fill rate and system time forside-b that we observe in the simulation and the predictedvalues based on our model. We also show the relative differ-ence (observed-predicted)/predicted × 100%. The observedand the predicted performance exhibits the following behav-ior. The fill rate and system time are increasing in thecustomers’ patience (increasing ta = tb). The metrics aresensitive to changes in arrival rate (increasing �a = �b) andrelatively insensitive to changes in the order size (increasing�a =�b), when demand is balanced. However, increasingthe frequency of orders while decreasing the order size tohold the total demand constant (changing �a = �b while�a�a = �b�b = 100), increases the fill rate and decreases thesystem time. For asymmetric systems, we observe increaseddemand from the opposite side increases the fill rate andlowers the system time.

With respect to the model’s agreement with the simu-lation, we observe a very close correspondence betweenthe predicted and observed fill rates. This is to be expectedsince we are matching the abandonment rates in our modelto those observed in the simulation. However, because in thesimulation abandoning customers withdraw all their unservedunits at the same time, the observed fill rate can differ fromthe predicted one.

It is the difference between the predicted and observed sys-tem times that measures the discrepancy between our modeland the simulated one. We observe that our model systemati-cally overestimates the observed system time. The accuracyof our model in predicting the system time depends on

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Afèche, Diamant, and Milner: Double-Sided Batch Queues with Abandonment1190 Operations Research 62(5), pp. 1179–1201, © 2014 INFORMS

Table 2. Observed vs. predicted system-level average fill rate and average system time for side-b units.

Rate of abandonment Fill rate (%) System time

Varying parameter Value ka kb Obs. Pred. Rel. diff. (%) Obs. Pred. Rel. diff. (%)

ta = tb 0001 10,062 10,062 0098 0099 −0071 0001 0001 −00000005 2,053 2,053 4077 4082 −1006 0005 0005 −20420010 1,052 1,052 8088 9006 −2006 0009 0010 −40560050 248.30 248.30 30090 32050 −4064 0035 0040 −140201000 145.90 145.90 45010 47030 −4070 0055 0069 −190805000 54.30 54.30 73060 74060 −1030 1032 1084 −28030

10000 37.50 37.50 81040 81090 −0050 1085 2067 −3006020000 26.30 26.30 87000 87010 −0010 2061 3080 −31040

100000 11.60 11.60 94020 94030 −0001 5068 8062 −34020

�a = �b 0010 1,005 1,005 0099 0099 −0020 0010 0010 −00510050 1,025 1,025 4071 4076 −1015 0010 0010 −20375000 1,240 1,240 30090 32050 −4086 0007 0008 −14030

10000 1,458 1,458 45010 47030 −4078 0005 0007 −19090

�a =�b 10000 105.00 105.00 8090 9008 −1089 0009 0010 −403550000 524.60 524.60 8089 9008 −2004 0009 0010 −4038

200000 2,100 2,100 8089 9007 −2000 0009 0010 −403211000000 10,502 10,502 8091 9007 −1084 0009 0010 −4044

�a = �b 0010 10,046 10,046 0098 0099 −0073 0010 0010 −00374�a�a = �b�b = 1005 0020 5,054 5,054 1095 1096 −0046 0010 0010 −0094

0050 2,050 2,050 4071 4076 −1011 0010 0010 −20312000 549.10 549.10 16000 16060 −3030 0008 0009 −70785000 248.10 248.10 30090 32050 −4085 0007 0008 −14030

10000 145.90 145.90 45010 47030 −4074 0005 0007 −20000100000 37.54 37.54 81040 81090 −0056 0002 0003 −30000

tb 0001 1,050 10,052 5009 5020 −2013 0001 0001 −00420005 1,050 2,050 6082 6097 −2003 0005 0005 −20260050 1,049 247.70 21050 22050 −4071 0039 0046 −150301000 1,049 145.60 31050 33030 −5048 0069 0087 −210105000 1,050 54.34 59010 60030 −2002 2004 2088 −29000

20000 1,049 26.13 77020 77050 −0033 4056 6071 −32010

�b 0010 1,050 1,004 9039 9048 −0098 0009 0009 −00400050 1,050 1,025 9013 9029 −1072 0009 0009 −20092000 1,049 1,100 8042 8066 −2076 0009 0010 −8069

10000 1,042 1,567 5079 6019 −6060 0009 0015 −38060

�a 0010 10,078 1,050 1078 1082 −2009 0010 0010 −40834�a�a = 1005 0050 2,049 1,050 6015 6027 −1092 0009 0010 −4056

2000 550.00 1,048 11040 11080 −3029 0009 0009 −400210000 155.20 1,048 14080 17040 −14070 0009 0009 −1045

Notes. Abandonment rates ka and kb are set to the observed values in the simulation. Other parameter values are �a = �b = 100, �a =�b = 10000,ta = tb = 0010, except as noted in the first two columns of the table.

the mean patience of the simulated customers relative totheir mean interarrival time. Specifically, in cases wherethe mean deadline is larger than the mean interarrival time(ta = tb ¾ 1/�a = 1/�b), implying significant fill rates, weobserve a relative difference of approximately 20%–30%between the observed and predicted system times. How-ever, in cases where customers are relatively impatient(ta = tb < 1/�a = 1/�b), the fill rates are lower and thesystem time predictions of our model are much more accu-rate. It is important to point out that these latter cases arerepresentative of operating regimes in crossing networks,where relatively low fill rates, between 1%–2%, are the norm.

6.2. Sensitivity Analysis of System-LevelPerformance Measures

In this section, we study the effects of impatient customers,nonexponential order size distributions, and patient cus-tomers with counterparty preferences, on the system-levelperformance metrics.

Impatient Customers. Table 3 summarizes the effects ofimpatient customers on the system-level metrics. We observethe fill rate of patient customers increases and their sys-tem time decreases as the overall arrival rate of impatientunits increases. That is, impatient customers improve theaverage performance of patient customers. As expected, the

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Table 3. Effect of impatient customers on the observed and predicted system-level average fill rates and averagesystem times.

Fill rate (%)Rate of System time

Overall Patient Impatient (patient)abandonment

Rel. Rel. Rel. Rel.�I

a =�Ib �I

a = �Ib ka = kb Obs. Pred. diff. (%) Obs. Pred. diff. (%) Obs. Pred. diff. (%) Obs. Pred. diff. (%)

100.0 000 11052 8088 9006 −2006 8088 9006 −2006 N/A N/A N/A 00091 00095 −4056001 11050 8085 9003 −2006 9032 9049 −1080 4045 4052 −1054 00091 00095 −4047005 11049 8070 8087 −1087 10087 11009 −2001 4034 4044 −2022 00089 00093 −4012100 11049 8053 8067 −1064 12075 13001 −2003 4028 4033 −1016 00087 00090 −3079200 11046 8018 8032 −1068 16035 16062 −1060 4010 4016 −1027 00084 00087 −3049

1000 11039 6016 6022 −1000 37002 37034 −0086 3010 3011 −0045 00062 00063 −10985000 11021 2077 2077 −0020 72016 72018 −0003 1038 1039 −0030 00028 00028 −0027

10000 11013 1064 1064 −0012 83059 83059 0000 0082 0082 −0009 00017 00017 −001610.0 001 11050 8096 9013 −1089 8097 9014 −1091 8000 7093 0085 00091 00095 −4039

005 11049 9018 9036 −1093 9025 9043 −1094 7089 7091 −0028 00091 00095 −4037100 11049 9044 9062 −1083 9060 9079 −1094 7085 7088 −0033 00090 00095 −4025200 11048 9088 10005 −1074 10029 10050 −1099 7080 7083 −0044 00090 00094 −4029

1000 11045 11052 11057 −0041 15063 15075 −0076 7040 7039 0019 00084 00088 −30895000 11038 10059 10060 −0012 34095 35013 −0051 5072 5070 0041 00065 00067 −2061

10000 11034 8057 8055 0022 49032 49049 −0035 4046 4046 0007 00051 00051 −00742.0 001 11049 8092 9009 −1087 8092 9009 −1088 8044 8050 −0070 00091 00095 −4048

005 11049 8097 9015 −1095 8097 9015 −2002 8051 8050 0008 00091 00095 −4033100 11050 9004 9021 −1090 9006 9023 −1083 8050 8049 0007 00091 00095 −4032200 11050 9023 9035 −1025 9026 9038 −1030 8052 8048 0048 00091 00095 −4034

1000 11047 10006 10024 −1076 10040 10061 −2001 8037 8037 −0002 00090 00094 −40235000 11045 12013 12005 0065 16041 16027 0088 7086 7084 0030 00084 00088 −4074

10000 11047 12035 12027 0066 22040 22029 0049 7034 7026 1012 00078 00080 −3033

Notes. Impatient customer parameters are shown in the first two columns of the table. Patient customer parameters: �a = �b = 100, �a =�b = 10000,and ta = tb = 001; abandonment rates ka and kb are set to the observed values in the simulation.

fill rate of impatient customers decreases as their arrivalrate increases. We also see good correspondence betweenobserved and predicted values of the fill rate and the systemtime. Finally, the relative difference between the observedand predicted system time decreases as the proportion ofimpatient customers in the system increases.

Nonexponential Order Size Distributions. Next, we studythe effect of order size variability on system-level perfor-mance under constant abandonment rates. Specifically, weconsider a system that is identical to the model introducedin §3, except that order sizes are sampled from a uniform ora lognormal distribution. We calculate system-level averagefill rates and system times using our results for exponentialorder size distributions and simulation for uniform andlognormal distributions. Table 4 shows representative resultsfor four of the 38 parameter settings (without impatientcustomers) in Table 2. Similar results hold for systems withpatient and impatient customer types.

We find that the average system times mainly depend onthe order size distribution through its coefficient of variation(CV), and that the average fill rates are insensitive to theorder size distribution. For distributions with a CV near 1.0,the average system times correspond well with those forexponential order size distributions, whereas distributionswith lower (higher) CV result in shorter (longer) average

system times. These observations are consistent with theo-retical results for single-sided queues with nonexponentialservice times. Although not shown in Table 4, we observe aclose correspondence in average fill rates for all cases; fornonexponential order size distributions fill rates are within3%, and in most cases, within 1% of those for exponentialdistributions.

Next, we compare the average fill rates and system timesin our model with the values observed in a simulatedsystem where patient customers have hyperexponential ordersize distributions and abandon at exponentially distributeddeadlines (impatient customers’ order sizes are exponentiallydistributed). That is, we consider two patient customer types,each with an exponential order size distribution, one withthe same mean as impatient orders, the other with a multipleM > 1 thereof. The mean order size of patient customersequals �I 441 −p5+pM5, where p ∈ 40115 is the fraction ofcustomers with the larger mean order.

We present results in Table 5. We find the observed patientcustomer fill rate decreases and their system time increasesin the CV of the order size. That is, increasing the variabilityof patient customer order sizes (by introducing two customertypes) hurts their performance in the system. In contrast, thefill rate of the impatient customers increases in the overallarrival rate of patient customer units, regardless of theirorder size CV.

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Afèche, Diamant, and Milner: Double-Sided Batch Queues with Abandonment1192 Operations Research 62(5), pp. 1179–1201, © 2014 INFORMS

Table 4. Effect of order size variability on system-level performance under constant aggregate abandonment rates: averagefill rates and system times for an exponential order size distribution, and relative differences in system timescompared to simulations with uniform and lognormal order size distributions.

Relative difference in system time (%)Rate of

abandonment Exponential distribution Uniform distribution Lognormal distribution

ti ki Fill rate (%) System time CV = 0029 CV = 0058 CV = 005 CV = 100 CV = 200

0.01 10,062 0099 0001 −4607 −3500 −3706 −2040 146050.10 1,052 9006 0010 −4505 −3402 −3704 −1090 151051.00 145090 47030 0069 −45090 −3303 −3707 −0036 1470310.00 37050 81090 2067 −4508 −3304 −3703 1060 15009

Notes. Patient customer parameters: �a = �b = 100, �a =�b = 10000; the abandonment rates ka = kb are chosen to match the values in four of thetest cases in Table 2, and the first column below indicates the corresponding values of ta = tb in the model with customer deadlines. No impatientcustomers.

As shown in the previous examples, our model approxi-mates the fill rates in systems with different abandonmentbehavior or nonexponential order size distributions fairlyclosely. Table 5 shows that the fill rate approximation errormay be significant for systems with different abandon-ment behavior and nonexponential order size distributions.We observe that the absolute error in the fill rate approxima-tions grows with the implied CV.

Recall from §6.1 that our model overestimates the systemtime for systems with order-level abandonment and exponen-tially distributed order sizes. We observe in Table 5 that forCV values larger than 1.5, our model underestimates theobserved system time. This is consistent with the systemtime increasing in the order size CV.

Patient Counterparty Preference. Impatient customers aresometimes viewed as acting in a predatory manner. A naturalresponse by patient customers may be to request to tradeonly with patient counterparties. We study the effect ofsuch counterparty preferences on system performance via

Table 5. Predicted average fill rates and system times for exponential order size distributions and constant aggregateabandonment rates, and observed values in simulated system where patient customers have hyperexponential ordersize distributions and abandon at exponentially distributed deadlines.

Fill rate (%)System time

Overall Patient Impatient (patient)Patient order size Rate ofdistribution abandonment

Rel. Rel. Rel. Rel.M p Mean CV ka = kb Obs. Pred. diff. (%) Obs. Pred. diff. (%) Obs. Pred. diff. (%) Obs. Pred. diff. (%)

10 0000 100 1000 11049 8053 8067 −1064 12075 13001 −2003 4028 4033 −1016 00087 00090 −30790025 325 1097 31608 6038 9041 −32020 6074 10037 −34099 5024 6030 −16081 00093 00088 50690050 550 1053 61026 6097 9043 −26006 7012 9086 −27078 6013 7005 −13010 00093 00090 30500075 775 1023 81308 8013 9051 −14054 8027 9077 −15034 7001 7053 −6091 00092 00092 −00501000 11000 1000 101486 9043 9062 −1099 9058 9080 −2020 7089 7088 0007 00090 00095 −4036

50 0000 100 1000 11049 8053 8067 −1064 12075 13001 −2003 4028 4033 −1016 00087 00090 −30790025 11325 2046 151181 3066 8076 −58020 3053 8086 −60014 5041 7043 −27019 00096 00087 110240050 21550 1069 281422 5031 8082 −39080 5026 8086 −40061 6053 7090 −17038 00095 00089 60030075 31775 1028 401713 7019 9001 −20019 7018 9003 −20049 7050 8021 −8070 00093 00093 00311000 51000 1000 521525 9000 9021 −2032 9001 9023 −2036 8055 8050 0058 00091 00095 −4032

Note. Parameters: �a = �b = �Ia = �I

b = 100; �a =�b = 10000641 −p5+Mp7 for the values of M and p indicated in the first two columns of the table;�I

a =�Ib = 10000; ta = tb = 001; and the abandonment rates ka and kb are set to the observed values in the simulation.

simulation. Let CP denote the fraction of patient customersthat request to trade only with patient counterparties. IfCP = 0, then patient customers have no matching preference.As CP increases, a larger fraction of patient customers willrequest to trade exclusively with other patient customers.When CP = 1, all patient customers will only trade withother patient customers.

In Table 6, we separately compare the fill rates and systemtimes for impatient and patient customers with and withoutpreferences. As the CP value increases, the overall systemfill rate decreases as patient customers are more likely totrade exclusively with other patient customers. For eachclass of patient customer, the fill rates and system times arerelatively insensitive to changes in the CP value. Patientcustomers with counterparty preference have a lower fillrate and longer system time compared to those without.Increasing the order flow from impatient customers amplifiesthis performance difference between patient customers withand without a counterparty preference. We conclude that

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Table 6. Effect of counterparty preference: Simulated system-level average fill rates and system times.

Fill rate (%) System time

�Ia = �I

b CP Overall Impatient Patient No pref. Pref. Patient No pref. Pref.

1.0 0000 8052 4027 12076 12076 N/A 00087 00087 N/A0025 7052 3020 11083 12087 8071 00088 00087 000910050 6050 2014 10086 12092 8080 00089 00087 000910075 5048 1007 9088 12092 8086 00090 00087 000911000 4045 0000 8091 N/A 8091 00091 N/A 00091

10.0 0000 6020 3010 37016 37016 N/A 00063 00063 N/A0025 4082 2031 29087 37024 7082 00070 00062 000920050 3048 1054 22095 37071 8024 00077 00062 000920075 2016 0078 15094 38010 8057 00085 00062 000921000 0081 0000 8089 N/A 8089 00091 N/A 00091

Notes. CP is the fraction of patient customers who exclusively trade with other patient customers. Pref. refers to such customers, No Pref. topatient customers who trade with all counterparties. Parameters: �i = 100, �i =�I

i = 10000 and ti = 001 for i = 8a1b9, �Ia = �I

b and CP as noted inthe first two columns of the table.

allowing patient customers to have a counterparty preferencedoes not meaningfully change the performance for patientcustomers who do not have such a preference. The cost tothe system is born through the reduced fill rate. However,allowing such customers may reduce the demand fromimpatient customers because of their decreased fill rate. Thismay be seen by some patient traders as a benefit.

Representative Examples with Imbalanced AggregateOrder Flows. We next consider representative examples ofsystems often seen in practice, with unbalanced aggregateorder flows between the two sides, and with the following mixof customer types: (1) small, frequently arriving impatientcustomers, (2) small, frequent patient customers, and (3) large,infrequent very patient customers. We investigate how thefill rates perform, especially those of the patient customers.We also study how well our model captures the systemperformance with this customer mix.

The numerical examples are constructed under the follow-ing assumptions. All arrivals are Poisson, simulated ordersizes are sampled from a shifted exponential distributionwith minimum order size of 100, and abandonment timesare exponentially distributed. The impatient customers andfrequent patient customers have a mean order size of 1,000units, with the patient customers abandoning after 2 minuteson average. The large, infrequent very patient customersarrive only on side-a. They have an average arrival rate of 1every 2 hours, a mean order size of 50,000 units, and anaverage abandonment time of 40 minutes. We fix the arrivalrate of frequent patient side-a customers at 100 per hour.We vary the arrival rate of the impatient customers (on bothsides of the queue) and the arrival rate of the patient side-bcustomers. The results for side-a of the queue are presentedin Table 7.

As discussed in §3, overall fill rates for crossing networksare typically small, between 1%–2%. Further, in manycrossing networks, patient-to-patient fill rates are typicallywell below 1%. In our representative examples, we observelow overall fill rates and low patient-to-patient fill rates occurin two scenarios. Either there must be significant imbalance

between the side-a and side-b total order flow with far greaternumbers of arrivals on side-a (�a�a +�I

a�Ia � �b�b +�I

b�Ib)

or there must be a very large excess of impatient versuspatient order flow (�I

i�Ii � �i�i for i ∈ 8a1 b9). These are the

cases typically seen in crossing networks. Furthermore, weobserve that the absolute difference between the theoreticaland simulated fill rate and system time values are small.These examples therefore demonstrate that our simplifiedmodel captures well the performance of typical crossingnetworks with unbalanced aggregate order flow and severaldemand types.

6.3. Customer-Level Performance Comparisons

In this section, we focus on customer-level metrics insteadof system-level averages. We consider the customer-levelfill rates and system times derived in our model (§5) andcompare them to the observed performance of an individualcustomer in the simulated system with customer deadlines.We simulate the system for the same test cases as in §6.1 butrandomly insert “marked” side-b customers with order size xand service deadline T . For each test case, we simulate10,000,000 customer arrivals with 100,000 inserted markedcustomers. This translates to 100 nonmarked customersbetween consecutive marked customer arrivals. We testedthe robustness of our results by rerunning the simulationsand varying the number of nonmarked customer arrivalsbetween marked customers from 50 to 1,000. We found novariation in the reported metrics.

Table 8 summarizes for each case the parameter values andresults. In each case, we set the marked customer’s order sizeto the average order size of side-b customers, i.e., x =�b.For time-constrained customers, we set T = 0 or T = tb,the mean deadline for a side-b customer. We compare theaverage fill rate observed in the simulation with its predictedvalue computed using (15) for T = 0 and (18) for T = tb . Fora time-unconstrained marked customer, T = � by definition,and we compare the average system time observed in thesimulation with its predicted value computed based on (14).

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Table 7. Predicted and simulated side-a average fill rates and system times for systems with imbalance in aggregate orderflows.

Fill rate (%)Rate of System time

Varying arrival rates abandonment Overall (side-a) Patient to patient (side-a) (patient side-a)

�Ia �I

b �b ka kb Obs. Pred. Abs. diff. Obs. Pred. Abs. diff. Obs. Pred. Abs. diff.

100.0 10000 10000 146179108 97103508 430077 440292 10215 360461 380807 20346 00050 00009 00041100.0 1000 1000 177167207 64166508 60261 60602 00341 50299 50637 00338 00102 00043 00059100.0 500 500 180111703 62155204 30212 30414 00202 20719 20912 00193 00110 00054 00057100.0 200 200 184105106 63129308 10292 10377 00085 10095 10175 00079 00114 00057 00057100.0 100 100 183191600 64179709 00647 00695 00049 00548 00593 00045 00112 00062 000501,000.0 10000 10000 143118506 72181509 110477 120237 00760 290118 300480 10362 00054 00011 000441,000.0 1000 1000 176109705 65172104 10425 10558 00133 40772 50200 00428 00104 00045 000581,000.0 500 500 180127908 58126609 00729 00795 00065 20480 20695 00216 00110 00053 000561,000.0 200 200 184193304 56134208 00288 00319 00031 00965 10087 00121 00112 00055 000571,000.0 100 100 183127107 73152403 00146 00161 00015 00500 00557 00057 00113 00063 000501,000.0 10000 10000 143142804 71141008 110427 120244 00817 280984 300465 10481 00049 00011 000391,000.0 10000 1000 155152508 62194006 50145 50479 00334 30636 30913 00277 00071 00018 000521,000.0 10000 500 154123209 63192304 40780 50166 00386 10901 20006 00105 00067 00020 000471,000.0 10000 200 155147405 63101309 40585 40933 00348 00746 00808 00061 00072 00020 000531,000.0 10000 100 154147305 63121609 40484 40889 00405 00383 00407 00024 00070 00020 000501,000.0 1100000 10000 106183701 72150300 170598 180837 10240 120850 140057 10208 00013 00002 000101,000.0 1100000 1000 109138703 64183804 110777 120543 00766 10369 10501 00132 00015 00003 000121,000.0 1100000 500 107157100 58157306 110474 120229 00756 00681 00757 00076 00016 00003 000131,000.0 1100000 200 107145402 61127903 110149 120022 00873 00278 00303 00025 00015 00003 000121,000.0 1100000 100 110161703 66179906 110003 110904 00901 00141 00151 00010 00014 00003 0001210,000.0 10100000 10000 60111400 67139702 20382 20601 00219 10788 10858 00071 00002 00000 0000210,000.0 10100000 1000 64135503 72101904 10524 10663 00139 00138 00187 00049 00002 00000 0000110,000.0 10100000 500 102138005 53186101 10408 10602 00195 00078 00093 00015 00001 00000 0000110,000.0 10100000 200 76156502 N/A 10432 10578 00146 00028 00037 00010 00002 00000 0000110,000.0 10100000 100 137103708 N/A 10352 10552 00200 00016 00019 00003 00002 00000 00001

Notes. Arrivals are Poisson, simulated order sizes are sampled from a shifted exponential distribution with minimum order size of 100, andabandonment times are exponentially distributed. There are three customer types, impatient, frequent patient, and infrequent very patient.The impatient customers and the frequent patient customers have a mean order size of 1,000 units, with the patient customers abandoning after2 minutes on average. The arrival rates of the impatient customers and the frequent patient side-b customers are varied as shown in the firstthree columns of the table. The frequent patient side-a customers have an arrival rate of 100 per hour. The infrequent very patient customers onlyarrive on side-a. They have an arrival rate of 1 every 2 hours, a mean order size of 50,000 units, and an average abandonment time of40 minutes. Note that cells with N/A represent simulations where the side-b queue is always empty, so abandonment rate kb is not well defined.

Two key observations stand out from our results in Table 8.First, the performance predictions based on the analyticalresults for our model provide accurate approximations ofobserved performance. The predicted fill rate for the time-constrained customers closely matches the observed fillrate. When T = 0, the predicted fill rate is within 4% ofthe observed fill rate for all but three of the test cases.When T > 0, the predicted fill rate is always lower thanthe observed fill rate. The percentage error is largest whenthe absolute error is small. Similarly, we observe that thepredicted system time approximates the observed systemtime well. Further, this approximation is very accurate whenthe fill rate is low, which is typically the case in crossingnetworks.

Second, the results quantify the trade-offs, which anindividual customer faces between fill rate and system timein a double-sided queue. Specifically, we observe that thefill rate for T = 0 is approximately one-half that of the valueat T = tb (except in the cases where ta 6= tb). Comparing the

case of T = tb to T = �, we observe that the time requiredto fill an order completely may be many times longer thanthe time required to achieve a small percentage fill. We alsoobserve that the expected time to achieve a 100% fill rate isinversely related to the arrival rate.

7. Concluding RemarksWe study a double-sided queue with batch arrivals andabandonment. We characterize in closed form the steady-statequeue length distribution and system-level performancemetrics, i.e., the system time and fill rate. We provideinsights on how system performance depends on customerflow characteristics such as asymmetries in demand volumeand variability, and the share of patient versus impatientcustomers. We also provide the expected fill rate of atime-constrained customer and the expected system time ofa time-unconstrained customer. These metrics are usefulin evaluating the experience of individual customers with

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Table 8. Observed vs. predicted customer-level average fill rate for time constrained and average system time fortime-unconstrained marked side-b customers with order size x =�b.

Time-constrained Time-constrained Time-unconstrainedfill rate (%), T = 0 fill rate (%), T = tb system time (T = �)

Varying parameter Value Obs. Pred. Rel. diff. (%) Obs. Pred. Rel. diff. (%) Obs. Pred. Rel. diff. (%)

ta = tb 0001 0062 0062 0000 1024 0099 25000 1098 1098 00000005 2093 2099 −2000 5096 4081 23090 1096 1096 00000010 5054 5068 −2050 11030 9026 22000 1091 1091 −00000050 19030 19080 −2060 39040 34070 13060 1074 1074 −00001000 27060 28010 −1060 57010 51090 10010 1071 1072 −00505000 41090 41030 1050 90000 86010 4052 2005 2032 −11060

10000 44080 44020 1040 97020 93010 4044 2052 3000 −1601020000 46040 46010 0060 99070 97010 2072 3026 4004 −19030

100000 47060 48040 −1080 100000 99080 0018 7015 8072 −18000�a = �b 0010 0063 0063 0079 1024 0099 25030 19090 19090 0004

0050 3001 3000 0060 6003 4082 24090 3092 3091 00315000 19000 19080 −3098 39030 34090 12040 0035 0035 0021

10000 27050 28010 −2009 56060 52030 8021 0017 0017 −1028�a =�b 10000 5061 5069 −1037 11010 9027 20030 1092 1091 0046

50000 5060 5069 −1067 11010 9027 20020 1091 1091 −0018200000 5065 5068 −0054 11010 9027 20010 1091 1091 0008

11000000 5064 5069 −0082 11010 9027 20020 1092 1091 0037�a = �b 0010 0063 0063 0020 1029 0099 29090 19090 19090 00134�a�a = �b�b = 1005 0020 1024 1024 0035 2046 1097 25000 9087 9090 −0031

0050 2094 3000 −1069 6002 4081 25010 3093 3091 00602000 9097 10030 −3028 20000 17020 16070 0093 0092 00535000 19010 19080 −3024 39020 34050 13080 0035 0035 −0003

10000 27040 28010 −2047 57010 50080 12030 0017 0017 −0088100000 44090 44020 1047 97010 88060 9064 0003 0003 −2086

tb 0001 5078 5093 −2060 6035 4025 49040 1091 1090 00280005 5077 5082 −0090 8063 6056 31060 1091 1091 00110050 4075 4085 −1092 28030 25060 10040 1099 2001 −00871000 4022 4017 1000 42080 38070 10060 2015 2020 −20395000 2052 2049 1047 84030 78060 7031 3023 3067 −12000

20000 1036 1041 −3090 99040 94070 4097 5065 7016 −21010�b 0010 6007 6020 −2022 12000 9080 22050 1091 1090 0027

0050 5092 5097 −0082 11060 9056 21080 1090 1091 −00102000 5003 5017 −2057 10060 8071 22000 1093 1093 0025

10000 2054 2037 6099 7025 5024 38030 2000 2003 −1035�a 0010 0092 0093 −0037 1086 1081 2079 10090 10090 −00444�a�a = 1005 0050 3056 3064 −2029 7027 6049 12000 2090 2091 −0050

2000 7037 7082 −5077 14090 11040 31040 1042 1041 003810000 8054 12030 −30071 17090 17030 3056 1002 0098 4003

Notes. Abandonment rates ka and kb are set to the observed values in the simulation. Other parameter values are �a = �b = 100, �a =�b = 10000,ta = tb = 0010, except as noted in the first two columns of the table.

specific attributes as opposed to system averages. We comparethese system- and customer-level results based on our model,which captures abandonment in aggregate, to simulationresults for a system in which customers abandon aftersome random deadline. We show that the performancemeasures predicted based on our model approximate thesimulated values well, particularly in cases where the fillrate is relatively low, which is typically the case in crossingnetworks. Given these results and because of its tractability,we conclude that our model is quite attractive.

Our model captures the main operational features ofcrossing networks. Our results have several implications forthe behavior of these systems. We show that the side (buy or

sell) that brings more demand has a lower fill rate and highersystem time. Customers generally prefer high fill rates andlow system times. If both sides are looking to achieve this,then we would expect balanced systems to result. Whendemand is balanced, we observe that the side with morefrequent, smaller orders achieves a lower system time and,by necessity, the same fill rate. Further, we find that highervariability in the order sizes of the patient customers leadsto lower fill rates, while system times remain relativelyconstant. This indicates that in balanced systems, there isincentive for traders to place small orders.

We also observe that there is little incentive for customersto be patient. We find that if one side is less patient than the

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other, there is a small decrease in the fill rate compared to alarge decrease in the system time. Similarly, increasing thenumber of impatient customers benefits the patient customersby increasing their fill rate and lowering their system time.The implication for crossing networks is that we can expectsystems with impatient customers who bring small orders.Because traders are exposed to price risk while waiting, onewould expect that they will be of limited patience, choosingto have only a small percentage of their orders filled. Onealternate case we explore is that some patient customersmay respond not by lowering their patience, but by tradingexclusively with patient counterparties. We observe thatdoing so reduces their fill rate and increases their systemtime—again a decrease in their performance.

The difficulty with these outcomes is that low customerpatience combined with small frequent orders results inlow trading volume, a central concern to crossing networkoperators. To alleviate this problem, crossing networks couldintroduce incentives for customers to increase their patience.As is currently the case, anonymity in the marketplace doesprovide some incentive to be patient as it allows largertraders to hide the true size of their position. In doing so,they can keep their order in the market longer. To furtherencourage patience, fees may be imposed on traders whocancel their orders. Many crossing networks currently employthis strategy as it limits the number of impatient traders thatsubmit many small orders over a short time span. Additionalmechanisms include the allowance of limit orders, whichprotect patient customers from price swings. As we note,there has been some research on the use of limit orders invisible markets. However, their use in dark markets has notbeen investigated and future research should address this.

Supplemental Material

Supplemental material to this paper is available at http://dx.doi.org/10.1287/opre.2014.1300.

Acknowledgments

The authors would like to thank the area editor, associate editor,and the referees for their many helpful suggestions. Support for thisresearch was provided by the Natural Sciences and EngineeringResearch Council of Canada [Grant 261986].

Appendix. Proofs

Proof of Proposition 1. Level crossing theory states that at anygiven y, in steady state, the rate of upcrossings must equal the rateof downcrossings, see Brill (2008). This implies the followinglevel crossing equation for f 4y5:

r4y5f 4y5

+�a

[

∫ �

yexp8−�a4z− y59f 4z5dz+�0 exp8�ay918y¶09

]

+

[

�Ia

∫ �

yexp8−�I

a4z− y59f 4z5dz

]

18y>09

= �b

[

∫ y

−�

exp8�b4z− y59f 4z5dz+�0 exp8−�by918y¾09

]

+

[

�Ib

∫ y

−�

exp8�Ib4z− y59f 4z5dz

]

18y<090

The r4y5f 4y5 term represents the steady-state abandonment rateof patient units from the system at level y 6= 0. The bracketedterms on the left-hand side of this level crossing equation representthe steady-state rates of downcrossings of level y due to arrivalsof patient and impatient side-a units, respectively; these arrivalstrigger down jumps of the queue length process. Similarly, thebracketed terms on the right-hand side of the level crossing equationrepresent the steady-state rates of upcrossings of level y due toarrivals of patient and impatient side-b units, respectively.

Writing the level crossing equation more compactly by takingthe integral over the atom at 0,

r4y5f 4y5+�a

∫ �

yexp8−�a4z− y59dF 4z5

+

[

�Ia

∫ �

yexp8−�I

a4z− y59dF 4z5

]

18y>09

= �b

∫ y

−�

exp8�b4z− y59dF 4z5

+

[

�Ib

∫ y

−�

exp8�Ib4z− y59dF 4z5

]

18y<090 (19)

The third-order differential Equations (1) and (2) are obtainedby differentiating (19) thrice and eliminating the integrals throughsubstitution. �Proof of Theorem 1. Note that all parameter values are positive.

For i 6= j ∈ 8a1 b9, let

Ai =ki

�i�j�Ij

1 (20)

Bi = ki

(

1

�j�Ij

−1

�i�Ij

−1

�j�i

)

−�i +�j +�I

j

�i�j�Ij

1 (21)

Ci = ki

(

1�i

−1�j

−1

�Ij

)

+�i

(

�j +�Ij

�i�j�Ij

)

+�j

(

�Ij −�i

�i�j�Ij

)

+�Ij

(

�j −�i

�i�j�Ij

)

1 (22)

Di = ki +�Ij

�Ij

+�j

�j

−�i

�i

0 (23)

For r4y5= kb18y>09 −ka18y<09, the characteristic equation corre-sponding to (1) is

Abä3+Bbä

2+Cbä+Db = 0 4for y > 051 (24)

where ä4= f ′4y5/f 4y50 Similarly, the characteristic equation corre-

sponding to (2) is

Aaä3−Baä

2+Caä−Da = 0 4for y < 050 (25)

Let äb1, äb2

, and äb3, be the roots of (24) and äa1

, äa2, and

äa3be the roots of (25). Given the cubic form of (24)–(25), we

know from the method of characteristic polynomials that dF 4y5 hasthe following form:

dF 4y5

=

4Ua1eyäa1 +Ua2

eyäa2 +Ua3eyäa3 5dy1 y < 01

�01 y = 01

4Ub1eyäb1 +Ub2

eyäb2 +Ub3eyäb3 5dy1 y > 00

(26)

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The proof proceeds in the following four steps:Step 1. We show that F 4y5 can be a probability distribution if

and only if the stability condition (6) holds, in which case dF 4y5has the form given in (7) with äa > 0 and äb > 0.

Step 2. We show how to determine the constants äa, äb , Ua,and Ub from (20)–(23) and (3)–(5).

Step 3. We show that Ua1Ub > 0.Step 4. We show that �0 > 0.Step 1. We show that F 4y5 can be a probability distribution if

and only if the stability condition (6) holds, in which case dF 4y5has the form given in (7) with äa > 0 and äb > 0.

Lemma 1. Equation (24) has one negative and two positive realroots iff kb +�I

a/�Ia >�b/�b −�a/�a holds. Similarly, (25) has one

positive and two negative real roots iff ka+�Ib/�

Ib >�a/�a−�b/�b .

The proofs of Lemmas 1–6 are relegated to the online appendix.Let äb2

, äb3be the positive roots of (24), and äa2

, äa3be

the negative roots of (25). As a consequence of Lemma 1 and(26), for f 4y5 to be well defined in the limit as y → �, it mustbe that Ub2

=Ub3= 0. Similarly, for f 4y5 to be well defined as

y → −�, it must be that Ua2=Ua3

= 0. Thus, letting äa4=äa1

> 0,äb

4= −äb1

> 0, Ua4=Ua1

, and Ub4=Ub1

, the expression for dF 4y5in (26) simplifies to the form in (7), i.e.,

dF 4y5=

Uaeyäady y < 01

�0 y = 01

Ube−yäbdy y > 00

(27)

Step 2. We show how to determine the constants äa, äb , Ua,and Ub . First, note that äa and äb can be expressed in closed form.Let

Pi =B2i − 3AiCi

9A2i

1 Ri =2B3

i − 9AiBiCi + 27A2iDi

54A3i

1

�i = cos−1

(

Ri√

P 3i

)

1 for i ∈ 8a1 b91

(28)

where 4Ai1Bi1Ci1Di5 for i ∈ 8a1b9 are given in (20)–(23). Then,from e.g., Pachner (1983),

äa = −2√

Pa cos(

�a + 2�3

)

−Ba

3Aa

1

äb = 2√

Pb cos(

�b3

)

+Bb

3Ab

0

(29)

Using (27), the flow balance equations (4)–(5) can be reduced to

�a

�a

=�aUb

ä2b +äb�a

+�bUa

ä2a +äa�b

+�IbUa

ä2a +äa�

Ib

+kaUa

äa

1

�b

�b

=�aUb

ä2b +äb�a

+�bUa

ä2a +äa�b

+�IaUb

ä2b +äb�

Ia

+kbUb

äb

0

Solving these two equations for Ua and Ub provides the constantsgiven in the theorem. Then the normalization condition (3) can bereduced to

�0 = 1 −Ua

äa

−Ub

äb

0

Step 3. Next, we show that Ua, Ub > 0 for any system thatsatisfies (6). Let

�a =�Ib

äa +�Ib

and �b =�Ia

äb +�Ia

0 (30)

We use Lemma 2 to prove Lemma 3.

Lemma 2. If kb +�Ia/�

Ia >�b/�b −�a/�a, then kb +�b >�b/�b −

�a/�a. Similarly, if ka +�Ib/�

Ib >�a/�a −�b/�b , then ka +�a >

�a/�a −�b/�b .

Lemma 3. If −ka −�Ib/�

Ib <�b/�b −�a/�a <�I

a/�Ia + kb , then

Ua1Ub > 0 and �0 < 1.

Step 4. It remains to show that �0 > 0. We proceed in threesteps.

Step 4.1. Lemma 4 shows for any system without impatientcustomers that (6) implies �0 > 0.

Step 4.2. Corresponding to any original system, we define thefollowing “double-hat” system and denote its stationary distribution

function by ˆ̂F . In the double-hat system, patient customers do notabandon, and the arrival rates of impatient customers

ˆ̂�Ii

4= �I

i + kj�Ii 1 for i 6= j ∈ 8a1 b90 (31)

The other parameters of the original and the double-hat system

agree. Lemma 5 shows that (6) implies ˆ̂�0 > 0.Step 4.3. Corresponding to any original system, we define

the following “hat” system and denote its stationary distribution byF̂ . In the hat system, patient customers do not abandon, and thearrival rates of impatient customers

�̂Ii

4= �I

i + kj4�Ii +äj51 for i 6= j ∈ 8a1 b90 (32)

The other parameters of the original and the hat system agree.Lemma 6 shows that the hat system and the original system havethe same stationary queue length distribution, in particular, �0 = �̂0.

It follows from Steps 4.2 and 4.3 that �0 > 0 for any system thatsatisfies (6): The hat and double-hat systems are identical, exceptthat the arrival rates of impatient customers in the hat system

(weakly) exceed those in the double-hat system, i.e., �̂Ii ¾

ˆ̂�Ii for

i ∈ 8a1 b9 by (31)–(32). The hat system is therefore more likely to

be empty, and we conclude by Lemmas 5–6 that �0 = �̂0 ¾ˆ̂�0 > 0.

Step 4.1. Consider an arbitrary system without impatient cus-tomers (�I

a = �Ib =�I

a =�Ib = 0) and denote its stationary distribu-

tion by F̃ . Then, executing the steps in the proof of Proposition 1on the level crossing Equation (19) with r4y5= kb18y>09 − ka18y<09

and �Ia = �I

b = 0, one can show that (1) and (2) reduce to

f̃ ′′4y5+ f̃ ′4y5

(

kb4�b −�a5− 4�a +�b5

kb

)

+ f̃ 4y5

(

�b�a −�a�b −�b�akbkb

)

= 01 if y > 01

f̃ ′′4y5+ f̃ ′4y5

(

ka4�b −�a5+ 4�a +�b5

ka

)

+ f̃ 4y5

(

�a�b −�b�a −�b�akaka

)

= 01 if y < 00

(33)

Executing Steps 1–3 with appropriate modifications, it follows thatthere is a solution f̃ to (33) and (3)–(5), which is unique, if and onlyif the stability condition (6) holds, i.e., −ka <�b/�b −�a/�a < kb .In this case, dF̃ has the same form as in (7), that is,

dF̃ 4y5=

˜Uaey˜äa dy1 y < 01

˜�01 y = 01˜Ube

−yä̃b dy1 y > 01

with ˜�0 = 1 −˜Ua

˜äa

−˜Ub

˜äb

1

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and where

˜äa = �̃a − �̃a > 01 ˜äb = �̃b − �̃b > 01

�̃a =�a +�b

2ka+

�b −�a

21

�̃a =

√

�̃2a +

�b�a −�a�b

ka+�a�b1

�̃b =�a +�b

2kb+

�a −�b

21

�̃b =

√

�̃2b +

�a�b −�b�a

kb+�a�b1

and

˜Ua =�a˜äa4˜äa+�b54�a�b−�b�a+kb�b4˜äb+�a55

�a�b4kb�b4˜äb+�a5+ka4�a+kb4˜äb+�a554˜äa+�b55>01

˜Ub =�b˜äb4˜äb+�a54�b�a−�a�b+ka�a4˜äa+�b55

�a�b4kb�b4˜äb+�a5+ka4�a+kb4˜äb+�a554˜äa+�b55>00

Lemma 4. In a system without impatient customers if −ka <

�b/�b −�a/�a < kb , then ˜�0 > 0.

Step 4.2. Now, consider the “double-hat” system, which has thesame parameters �i1�i1�

Ii , i ∈ 8a1b9 as the original system, but

patient customers do not abandon ( ˆ̂ka =ˆ̂kb = 0), and ˆ̂

�Ii = �I

i +kj�Ii

for i 6= j ∈ 8a1 b9 by (31). (The two systems are identical if and onlyif ka = kb = 0.) Then, with r4y5= 0, (1) and (2) in Proposition 1reduce to two homogeneous, constant-coefficient second-orderdifferential equations,

ˆ̂f ′′4y5+

ˆ̂f ′4y5

·

( ˆ̂�Ia4�b−�a5+�a4�b−�I

a5−�b4�a+�Ia5

�a+�b+ˆ̂�Ia

)

+ˆ̂f 4y5

·

(

−ˆ̂�Ia�a�b−�a�

Ia�b+�b�a�

Ia

�a+�b+ˆ̂�Ia

)

=01 if y>01

ˆ̂f ′′4y5+

ˆ̂f ′4y5

·

( ˆ̂�Ib4�b−�a5+�b4�

Ib−�a5+�a4�b+�I

b5

�a+�b+ˆ̂�Ib

)

+ˆ̂f 4y5

·

(

�a�b�Ib−

ˆ̂�Ib�a�b−�b�a�

Ib

�a+�b+ˆ̂�Ib

)

=01 if y<00

(34)

Executing Steps 1–3 with appropriate modifications, it follows that

there is a solution ˆ̂f to (34) and (3)–(5), which is unique, if and

only if the stability condition (6) holds, i.e., −ˆ̂�Ib/�

Ib <�b/�b −

�a/�a <ˆ̂�Ia/�

Ia, or equivalently, −ka −�I

b/�Ib <�b/�b −�a/�a <

kb +�Ia/�

Ia. In this case, d ˆ̂F has the same form as (7), that is,

d ˆ̂F 4y5=

ˆ̂Uaey

ˆ̂äa dy1 y < 01

ˆ̂�01 y = 01ˆ̂Ube

−yˆ̂äb dy1 y > 0

with ˆ̂�0 = 1 −

ˆ̂Ua

ˆ̂äa

−

ˆ̂Ub

ˆ̂äb

1

(35)

and where

ˆ̂äa =

ˆ̂�a − ˆ̂�a > 01 ˆ̂

äb =ˆ̂�b − ˆ̂�b > 01 (36)

ˆ̂�a =

ˆ̂�Ib4�b −�a5+�b4�

Ib −�a5+�a4�b +�I

b5

24�a +�b +ˆ̂�Ib5

1

ˆ̂�a =

√

√

√

√ ˆ̂�2a +

�a�b�Ib

�a +�b +ˆ̂�Ib

(

�b

�b

+

ˆ̂�Ib

�Ib

−�a

�a

)

1

ˆ̂�b =

ˆ̂�Ia4�a −�b5+�a4�

Ia −�b5+�b4�a +�I

a5

24�a +ˆ̂�Ia +�b5

1

ˆ̂�b =

√

√

√

√ ˆ̂�2b +

�a�b�Ia

ˆ̂�Ia +�a +�b

(

�a

�a

+

ˆ̂�Ia

�Ia

−�b

�b

)

1

(37)

and

ˆ̂Ua =

(

�a

�a

ˆ̂äa4

ˆ̂äa+�b54

ˆ̂äa+�I

b5

·

(

ˆ̂�Ia�a+�a�

Ia−

�b

�b

�a�Ia+

ˆ̂äb

(

�a+ˆ̂�Ia−

�b

�b

�a

)))

·( ˆ̂äa

ˆ̂äb

(

�aˆ̂�Ib+

ˆ̂�Ia4�b+

ˆ̂�Ib5)

+ˆ̂äa

( ˆ̂�Ia4�b+

ˆ̂�Ib5�a+�a

ˆ̂�Ib�

Ia

)

+ˆ̂äb

(

4�a+ˆ̂�Ia5

ˆ̂�Ib�b+

ˆ̂�Ia�b�

Ib

)

+�aˆ̂�Ib�

Ia�b+

ˆ̂�Ia�a4

ˆ̂�Ib�b+�b�

Ib5)−1

>01

ˆ̂Ub =

(

�b

�b

ˆ̂äb 4

ˆ̂äb+�a54

ˆ̂äb+�I

a5

·

(

ˆ̂�Ib�b+�b�

Ib−

�a

�a

�b�Ib+

ˆ̂äa

(

�b+ˆ̂�Ib−

�a

�a

�b

)))

·( ˆ̂äa

ˆ̂äb

(

�aˆ̂�Ib+

ˆ̂�Ia4�b+

ˆ̂�Ib5)

+ˆ̂äa

( ˆ̂�Ia4�b+

ˆ̂�Ib5�a+�a

ˆ̂�Ib�

Ia

)

+ˆ̂äb

(

4d�a+ˆ̂�Ia5

ˆ̂�Ib�b+

ˆ̂�Ia�b�

Ib5+�a

ˆ̂�Ib�

Ia�b

+ˆ̂�Ia�a4

ˆ̂�Ib�b+�b�

Ib5)−1

>00

Lemma 5. If −ka −�Ib/�

Ib <�b/�b −�a/�a <�I

a/�Ia + kb , then

ˆ̂�0 > 0.

Step 4.3. Now, consider the “hat” system, which has the sameparameters �i1�i1�

Ii , i = 8a1 b9 as the original system, but patient

customers do not abandon (k̂a = k̂b = 0), and �̂Ii = �I

i +kj4�Ii +äj5

for i 6= j ∈ 8a1b9 by (32). (The two systems are identical if andonly if ka = kb = 0.)

Lemma 6. The hat system and the original system have the samestationary queue length distribution: F̂ = F .

Thus we have shown that the double-sided queue is stable ifand only if (6) holds, in which case the stationary queue lengthdistribution has the properties specified in the theorem. �

Proof of Proposition 2. Consider the system fill rate for side-aunits. Let La denote the average number of units that enter thesystem per unit time:

Laã=

�a

�a

+�Ia

�Ia

0

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Side-a units can leave the system unserved in one of the followingthree ways:

1. When there is a queue of patient side-a units, they abandonat rate ka. Let O1

a represent the rate side-a units exit the system inthis way.

O1a

ã= ka

∫ 0−

−�

f 4y5dy =kaUa

äa

2. When an impatient side-a order arrives to an empty systemor a queue of patient side-a units, the entire order abandons uponarrival without any units being served. Let O2

a represent the exitrate in this way.

O2a

ã=

�Ia

�Ia

(

�0 +

∫ 0−

−�

f 4y5dy

)

=�Ia

�Ia

(

1 −Ub

äb

)

3. When an impatient side-a order arrives to a system of queuedpatient side-b units and only part of the order is matched, theremaining units abandon the system without being served. Let O3

a

represent the exit rate in this way.

O3a

ã= �I

a

∫ �

0+

(

�Ia

∫ �

y4z− y5e−�I

az dz

)

f 4y5dy

=�Ia

�Ia

∫ �

0+

e−�Iayf 4y5dy =

�Ia

�Ia

Ub

äb +�Ia

Thus the fill rate for side-a units is

�a = 1 −

∑3i=1 Oi

a

La

= 1 −

(

kaUa

äa

+�Ia

�Ia

(

1 −Ub

äb

)

+�Ia

�Ia

(

Ub

äb +�Ia

))

·

(

�Ia

�Ia

+�a

�a

)−1

= 1 −

(

kaUa

äa

+�Ia

�Ia

(

1 −Ub

äb

(

11 +äb/�

Ia

)))

·

(

�Ia

�Ia

+�a

�a

)−1

0

Substituting �a = 1/�a and �Ia = 1/�I

a gives (10) for side-a units.The derivation of the side-b fill rate is identical. �

Proof of Proposition 3. Let M̄a4y5= min8n2∑n

i=1 X̄a1 i ¾ y9.Observe M̄a4y5 is a stopping time. Further, if V̄a111 V̄a121 0 0 0 are theinterarrival times of the process 8N̄a4t51 t ¾ 09, then

�cb 4y5=

M̄a4y5∑

i=1

V̄a1 i0

Therefore by Wald’s equation,

Ɛ6�cb 4y57= Ɛ

[M̄a4y5∑

i=1

V̄a1 i

]

= Ɛ6V̄a1 i7Ɛ6M̄a4y570

Let m4y5 = Ɛ6M̄a4y57. Observe Ɛ6V̄a1 i7 = 1/�̄a. From basicrenewal theory (Karlin and Taylor 1975),

m4y5= 1 +

∫ y

0m4y− s5ga4s5ds0 (38)

Substituting ga4s5 into (38) and letting z= y− s,

m4y5= 1 +

∫ y

0m4z5

(

�a

�̄a

�ae−�a4y−z5

+�Ia

�̄a

�Iae

−�Ia4y−z5

)

dz0

Solving this integral equation gives

m4y5=C2 −C1�a +�I

a

�Ia�a +�a�

Ia

e−4y4�Ia�a+�a�Ia55/4�a+�Ia5

+4�a +�I

a5�a�Ia

�Ia�a +�a�

Ia

y1

where C1 and C2 are constants of integration solved for by using theboundary conditions. Trivially m405= 1, and by taking the derivativeof (38) and letting y → 0, we get m′405 = 4�a�a + �I

a�Ia5/�̄a.

Solving, we find

C1 =�a�

Ia4�a−�I

a52

4�a+�Ia54�

Ia�a+�a�

Ia5

and C2 =�a�

Ia4�a−�I

a52

4�Ia�a+�a�

Ia5

20

Consequently,

Ɛ6�cb 4y57=

1

�̄a

m4y5

or

Ɛ6�cb 4y57

=4�I

a52�2

a+�2a4�

Ia5

2 +�a�Ia4�

2a+4�I

a525

4�a+�Ia54�

Ia�a+�a�

Ia5

2+

�a�Ia

�Ia�a+�a�

Ia

y

−�a�

Ia4�a−�I

a52

4�a+�Ia54�

Ia�a+�a�

Ia5

2e−y44�Ia�a+�a�

Ia5/4�a+�Ia550

Substituting �a = 1/�a, �Ia = 1/�I

a and simplifying gives theresult. �Proof of Proposition 4. Letting w = y− kbt, we can simplify(13) as

Ɛ6T cb 4x1 y57

=1

�a +�Ia

+H4x1 y5+1kb

∫ y

0

∫ w

0e−44�a+�Ia5/kb54y−w5

· 4�a�ae−�ax +�I

a�Iae

−�Iax5Ɛ6T c

b 4x1w− z57dzdw1 (39)

where

H4x1y5

=1

4�a+�Ia54�

Ia�a+�a�

Ia5

2

·e−4�2ay+�Ia4�

Iay+kb�ax5+�a42�

Iay+kb�

Iax55/4kb4�a+�Ia55

·(

−�a�Ia4�a−�I

a52+e44�

Ia�a+�a�

Ia5x5/4�a+�Ia5

·(

4�Ia5

2�2a�

Iax+�2

a4�Ia5

2�ax+�a�Ia44�

Ia5

2

+�a�Ia4−2+�I

ax5+�2a41+�I

ax55))

+1

4�Ia�a+�a�

Ia5

2

(

�a�a

�a+�Ia−kb�a

41−e−4y4�a+�Ia−kb�a55/kb 5

·(

�a4�Ia5

2xe−y�a −�Ia�

Iae

−y�a

+�Ia�

Iae

−44y+x5�Ia�a+�a4y�a+x�Ia55/4�a+�Ia5+�I

a�ae−y�a

−�Ia�ae

−44y+x5�Ia�a+�a4y�a+x�Ia55/4�a+�Ia5+�I

a�a�Iaxe

−y�a)

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+�Ia�

Ia

�a+�Ia−kb�

Ia

(

1−e−4y4�a+�Ia−kb�Ia55/kb

)

·(

�Ia�

2axe

−y�Ia +�a�

Iae

−y�Ia

−�a�Iae

−44y+x5�a�Ia+�Ia4x�a+y�I

a55/4�a+�Ia5

−�a�ae−y�I

a +�a�ae−44y+x5�a�

Ia+�Ia4x�a+y�I

a55/4�a+�Ia5

+�a�ae−y�I

ax�Ia

)

)

0

From (39), we obtain the third-order differential equation

¡3Ɛ6T cb 4x1 y57

¡y3+

�a +�Ia + kb4�a +�I

a5

kb

¡2Ɛ6T cb 4x1 y57

¡y2

+4�a + kb�a5�

Ia +�I

a4�a + 2�Ia5

kb

¡Ɛ6T cb 4x1 y57

¡y

−�a�

Ia

kb= 00

The solution of this equation satisfies

Ɛ6T cb 4x1 y57=C0 +C1y−C2

e−y4K1−√K25

K1 −√K2

−C3e−y4K1+

√K25

K1 +√K2

+C44x51 (40)

where C1, K1, and K2 are given in the statement of Proposition 4.The terms, C0, C2, C3, and C44x5, are independent of y and areobtained by solving three boundary conditions. Note that K1 ±

√K2

is positive for all nonnegative parameter values.The first boundary condition specifies that when a submitted

order for x units finds the queue empty (i.e., y = 0), then theexpected system time equals the clearing time. By Proposition 3,

Ɛ6T cb 4x1057

=�a�

2a +�I

a4�Ia5

2

4�a�a +�Ia�

Ia5

2+

x

�a�a +�Ia�

Ia

−�a�

Ia4�a −�I

a52

4�a +�Ia54�a�a +�I

a�Ia5

2e−x44�a�a+�Ia�

Ia5/44�a+�Ia5�a�

Ia550

The other conditions follow by observing that because the arrivalsare compound Poisson, the marginal change in the expected systemtime is zero as the number of queued customers get small:

limy→0

¡ Ɛ6T cb 4x1 y57

¡y= 0 and lim

y→0

¡2 Ɛ6T cb 4x1 y57

¡2y= 00

With these boundary conditions, (40) is solved with C0 as specifiedin the statement of Proposition 4, and with

C2 = −12C1

K1 +√K2

√K2

1 C3 =12C1

K1 −√K2

√K2

1

C44x5=x

�a�a +�Ia�

Ia

−�a�

Ia4�a −�I

a52

4�a +�Ia54�a�a +�I

a�Ia5

2

· e−x44�a�a+�Ia�Ia5/44�a+�Ia5�a�

Ia550 �

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Philipp Afèche is an associate professor of operations manage-ment at the University of Toronto’s Rotman School of Management.His research interests include service operations and supply chainmanagement problems, focusing on pricing, service design, perfor-mance analysis, and response time management.

Adam Diamant is a Ph.D. candidate at the University ofToronto’s Rotman School of Management. His research interestsinclude stochastic modeling and optimization with applications tofinancial markets and healthcare systems.

Joseph Milner is an associate professor of operations manage-ment at the University of Toronto’s Rotman School of Management.His research interests include service operations management,revenue management, and healthcare applications.