I Introduction
Consider a wireless system with a single access point that has identical channels, where is a positive integer. Each channel can support one file transmission, so that up to files can be transmitted simultaneously. Different types of users want to upload files to the access point. To do this, they first need to find an idle channel. At the start of every upload attempt, a user randomly scans a subset of the channels in hopes of finding at least one channel that is idle. Let be the size of the subset that is scanned and assume that . Example numbers are and , so that every user randomly scans 5 of the 25 channels. If an idle channel is found, the user sends a single file over that channel (if multiple idle channels are found, the choice of which one to use is made arbitrarily). If no idle channel is found, the users react differently depending on their type: Persistent users try again later, while nonpersistent users leave and do not return.
An example of this situation is when the access point is in a fixed location, such as in a coffee shop. Persistent users are customers who find a table at the coffee shop, stay for an extended period of time, and use their wireless devices during their stay. Nonpersistent users either walk past the coffee shop without entering, or enter only for a short time (perhaps to place a takeout order). The goal of this paper is to establish a Markov chain model for this system and to analyze the model to obtain steady state behavior for throughput and blocking probability.
The system operates in continuous time over the timeline . A channel is said to be busy at time if it is currently being used, that is, if there is a user that is transmitting a file over that channel. Let be the total number of channels that are busy at time . Suppose a user who is not currently transmitting attempts to access the network at a time for which . This user scans a random subset of channels to determine which (if any) are not being used, with all subsets equally likely. Let denote the conditional probability that a user successfully finds an idle channel, given that . For example, if then for all . If then
(1) 
An interesting feature of this model is that the probability of successfully finding an idle channel at time depends only on , not on which types of users are using each channel. However, is not enough to describe the state of the system: The full system state is described by a multidimensional Continuous Time Markov Chain (CTMC).
First, this paper considers the case when all users are nonpersistent. There are different classes of nonpersistent users. Users for each class arrive according to independent Poisson processes with rates
. File sizes are independent and exponentially distributed with rate
for each class . The system state is , where is the number of class users currently transmitting over the network, and . The state space grows exponentially with . Fortunately, the system exhibits a latent reversible property. A general theory of reversibility in Markov chains is described in [1]. Our CTMC is similar to (but not the same as) the open migration processes described in [1]. This similarity motivates us to guess a particular productform steady state distribution. The guess is validated by showing that it satisfies the detailed balance equations. This leads to a closed form expression for the steady state mass function in terms of the and values. This yields several insightful results, including an expression for the long term access success probability that depends only on the number of channels , the number of channels scanned , and a single loading parameter .Next, this paper considers the scenario with both persistent and nonpersistent users. As before,
is the vector that specifies the number of nonpersistent user of each type that are currently transmitting. In addition, there are
persistent users, where is a positive integer. Each persistent user can be in one of three activity states: Idle, Waiting, Transmitting. Let denote the current activity state of persistent user . The dynamics of are described by the 3state diagram of Fig. 1:
Idle (): Persistent user is not using its wireless device and thus has no files to send. This user stays in the idle state for an independent and exponentially distributed time with parameter .

Waiting (): Persistent user is waiting to either attempt transmission of a new file, or go to the idle state, according to independent racing exponential variables of parameters and . If the user attempts a transmission but fails, it remains in the waiting state.

Transmitting (): Persistent user is currently transmitting a file over a channel. This user stays in the transmitting state for an independent and exponentially distributed time with parameter .
The system state is . The total number of busy channels is
where is an indicator function that is if and else. A key aspect of Fig. 1 is that the transitions for persistent user occur with transition rate , which multiplies the access attempt rate by the current success probability . In particular, this transition rate depends on the current number of busy channels, which depends on the history of events associated with all users. Thus, the 3state “minichain” of Fig. 1 for each persistent user is only a partial view of the larger CTMC: These
different “minichains” are coupled in a nontrivial way. Using insight obtained from the case when all users are nonpersistent, we guess that the system is reversible, guess a particular steady state structure, and verify these guesses by showing that the detailed balance equations hold. The resulting steady state probabilities for this case have a simple product form solution. Unfortunately, there are exponentially many states (the number of states can easily be larger than the current estimate on the number of atoms in the universe) and it is not obvious how to sum the joint probabilities to obtain individual user performance. We provide a polynomialtime method for computing the exact sums by using a discrete Fourier transform.
Ia Related work
Our system model is similar to recent work in [2] that also treats multichannel systems that scan a subset of the available channels before transmission. The work in [2] also assumes that users are in one of three states (idle, probing, transmitting), which is similar to our 3state persistent user structure. The work in [2] does not solve the resulting steady state probabilities, rather, they develop meanfield results that are asymptotically accurate when all users have identical parameters and when the network size scales to infinity. In contrast, our work provides the exact steady state values for the continuous time Markov chain for any system size and for heterogeneous user parameters. It also treats the case when both persistent and nonpersistent users are present. It should be emphasized that our work exploits a latent reversibility property that does not exist in the model of [2]. In particular the 3state user dynamics of [2] can roughly be viewed as similar to those of Fig. 1 with the exception that there is no transition, and the transition is replaced with a transition. Intuitively it is clear that if it is possible to have a transition but impossible to have a transition in the opposite direction, then reversibility fails. It is not clear if exact steady state behavior can be obtained when reversibility fails; that remains an important open question and mean field analysis is an important technique for those situations.
Our model of persistent users accounts for heterogeneous human user activity, where users can be in various states depending on their activity patterns. The topic of mathematical models for humanbased activity patterns for wireless communication is of recent interest. For example, related Markovbased models of human user activity and human response times are treated in [3] for wireless scheduling; related 2state user activity models are used in [4] to treat file downloading as a constrained restless bandit problem.
Ii Nonpersistent users
This section considers the case where all users are nonpersistent. Each user arrives once and makes one attempt to access a channel. If the access is successful then the user transmits its file, else it leaves and does not return. This is a useful model for highly mobile wireless systems that pass by an access point for a short time. In the coffee shop example, these are users that either walk past the coffee shop but do not enter, or enter the shop and stand in line for a takeout order but do not stay for long. If they cannot obtain access after one attempt, they do not try again.
Assume there are classes of such users, where is a positive integer. Users from each class arrive according to independent Poisson processes with rates . Each user has one file to send. File service times are independent. Files from class users have service times that are exponentially distributed with parameter . Assume that and for all .
Iia Markov chain model
The system can be modeled as a continuous time Markov chain (CTMC) with vector state , where is the number of type files currently transmitting at time . The state space is given by the set of all vectors that have nonnegative integer components such that , where is the number of channels (assume is a positive integer). Let be the number of busy channels. A user that arrives to the system scans a subset of the channels to find one that is idle. For each define as the conditional probability that a newly arriving user finds an available channel, given that . The value of associated with finding at least one idle channel in a system with busy channels and total channels is given in (1). We shall call the conditional success probability function. Our mathematical analysis does not require to have the form (1) and allows for more general success probability functions. We assume only that satisfies the following basic properties:
(2)  
(3)  
(4) 
Requirement (2) ensures is a valid probability for each ; requirement (3) ensures that it is possible to utilize all channels simultaneously (for example, if this were violated by having but , then a system that is initially empty could never have more than 5 active channels, which underutilizes the existing 10 channels); requirement (4) enforces the physical constraint that the system cannot support more than active channels simultaneously. The particular success probability function in (1) indeed satisfies (2)(4).
To completely describe the Markov chain structure of this system, it remains to specify the transition rates. The transition rates between two states and are as follows: Fix an integer and define as the vector that is in entry and zero in all other entries. Let and be two states in the state space . Then

Transition rate is given by
This is the product of the arrival rate with the success probability given that the new user scans when the system state is .

Transition rate is given by
This is because there are currently jobs of type that are actively using channels, and each has an exponential service rate equal to .
Since the system state can change by at most one at any instant of time, there are no other types of transitions and so for states that do not have the above form. It is not difficult to see that the Markov chain is irreducible, so that it is possible to get from any state of the state space to any other state in (the requirement (3) and the fact that for all ensure this).
IiB Basic Markov chain theory
This subsection recalls basic Markov chain theory (see, for example, [1][5][6]). Consider a continuous time Markov chain (CTMC) with a finite or countably infinite state space and transition rates for all . Assume that for all . The states of can be viewed as nodes of a graph; the links of the graph are defined by statepairs such that ; the CTMC is said to be irreducible if this graph has a path from every node to every other node. A probability mass function over the state space is a vector that satisfies for all and . The goal is to find a mass function that satisfies the following global balance equations:
(5) 
It is well known that if the CTMC is irreducible, then there is at most one probability mass function that solves (5). If such a mass function exists, then it is the unique steady state mass function for the CTMC. If the CTMC is irreducible and has a finite state space, then such a steady state solution always exists.
An irreducible CTMC is said to be reversible if there exists a probability mass function that satisfies the following detailed balance equations:
(6) 
It is well known that if a probability mass function solves the detailed balance equations, then it also satisfies the global balance equations and hence is the unique state state. Indeed, if (6) holds then for each we can sum (6) over all to obtain:
and thus (5) holds. However, not all CTMCs are reversible. That is, not all CTMCs have steady states that satisfy (6).
IiC Steady state analysis for nonpersistent users
It is not obvious whether or not the Markov chain for our system of nonpersistent users is reversible. Fortunately, the system is similar to an open migration process with reversibility properties as described in [1]. An open migration process is a system with colonies that can be described by a Markov chain of the type , where is the current population of colony , transitions between states occur when a single member of colony moves to colony , and transition rates for such events depend only on the current population . Such a migration process can almost be used to model the multiaccess system of interest, where the number of type jobs currently using channels can intuitively be viewed as the population of “colony .” However, the multiaccess system is not a migration system because transition structure is different and transition rates depend on the sum population . Nevertheless, reversibility properties of the current system can be established. To this end, define
The technique behind the next theorem is to guess a probability mass function and then show the guess satisfies (6). The structure of the guess is not obvious. However, to gain intuition, note that we constructed our guess for the probability of state by observing the “birthdeathlike” structure of the system in Fig. 2 and guessing that steady state is a product of terms that include factors of the type (which are also factors in the steady state mass function of a 1dimensional queue) as well as factors that multiply the chain of success probabilities over all . Once a good guess is made, it is not difficult to verify the guess satisfies the detailed balance equations.
Theorem 1
Under this nonpersistent user model with any success probability function that satisfies (2)(4) we have
a) The CTMC is reversible and the unique steady state distribution is
(7) 
where is the positive constant that makes the probabilities sum to 1, and we use the convention that .
b) The steady state probability that there are channels in use is
where represents a random state vector with distribution equal to the steady state distribution.
c) The constant is equal to
(8) 
Define the mass function according to (7). It suffices to show that this mass function satisfies (6). Since there are only two types of possible transitions, it suffices to show that
It is easy to verify that this equation holds for as given in the statement of the theorem. This proves part (a).
To prove (b), we have
where we have used the multinomial expansion:
This proves part (b). Part (c) immediately follows from part (b).
The success probability of each newly arriving job depends on the current state of the system and not on the class of that job. Since all jobs arrive as Poisson arrivals, and Poisson arrivals see time averages (“PASTA,” see, for example, [6]), it follows that jobs of all classes have the same long term success probability for finding an available channel. Define as this long term success probability. Specifically, if represents a random vector with distribution given by the steady state distribution in Theorem 1, then is defined
With this definition of the success probability , the long term rate of accepted jobs of type is jobs/time, and the long term rate of dropped jobs of type is jobs/time. Remarkably, the value of depends only on , not on the individual values, as shown in the following corollary.
Corollary 1
The long term success probability is given by
and the result follows by substituting the expression for from part (b) of Theorem 1.
IiD Plots for example cases
Corollary 1 shows that the success probability depends only on the conditional success probabilities and on the loading parameter . This is an insightful result: We can understand the success probability through the single parameter , regardless of the number of classes of nonpersistent users and regardless of the specific and parameters for each class . Notice that, by Little’s theorem, is equal to the steady state average number of actively transmitting users there would be in a virtual system with infinite resources: The virtual system has an infinite number of servers, each new file of the virtual system receives its own server with probability 1, and no files are dropped.
Fig. 3 plots the success probability versus (the number of channels that each user scans) for the case of channels and using the probabilities given in (1). The values are shown. The case is when the average number of active users in a virtual system with infinite resources is equal to 10, the number of channels in the actual system. This can be viewed as a threshold case: When exceeds (as plotted for the case in Fig. 3) then success probability is necessarily strictly less than even when the number of channels sensed is equal to . On the other hand, by choosing we obtain a success probability above when or .
Better performance is obtained when the number of channels is increased while the values are increased by the same factor: Figs. 4 and 5 show performance for the case channels and channels, respectively, with corresponding values that maintain the same ratio of as in the first figure. It can be seen that success probability increases to near 1 when the loading is small (). In all of the plots of Figs. 35 it can be seen that success probability is relatively flat for large values of : A considerable amount of energy can be saved by just scanning a small subset of the total number of channels.
Iii Persistent and nonpersistent users
Fix as a positive integer and suppose that, in addition to the classes of nonpersistent users, there are individual persistent users with activity states and behavior parameters , as shown in Fig. 1. The classes of nonpersistent users have parameters and for all . The values of all parameters , for and are assumed to be positive.
Recall that is the current number of nonpersistent users of type transmitting, for . The system state is . The total number of busy channels is
where is an indicator function that is 1 if persistent user is transmitting at time , and else. Let be a success probability function defined for that satisfies (2)(4) (an example function is in (1)). As before, if any user attempts access at a time such that , its conditional success probability is . Notice from Fig. 1 that the transition rates for the transitions of each persistent user depend on the current value of .
Iiia Markov chain model
Let be the state space of the system: This is the set of all vectors , where and , such that for all , for all , and
For simplicity of notation, for each define as the number of busy channels associated with state :
To completely describe the transition rates of this CTMC, let and be two distinct states in . There are three types of transitions that can occur between states and :

Nonpersistent user (): Recall that is a vector of size with a in component and zeros in all other components.

Transitions have rate:

Transitions have rate


Persistent user ():

Transitions have rate:

Transitions have rate:


Persistent user ():

Transitions have rate:

Transitions have rate:

It is not difficult to show that the CTMC is irreducible. Indeed, every state can reach the state from a sequence of transitions that includes no new arrivals, has each transmitting user finish, and has all persistent users eventually move to the Idle state. Likewise, the state can reach every state in .
IiiB Steady state probabilities
Motivated by the “birthdeathlike” structure of the persistent user dynamics shown in Fig. 1 and by the structure of the steady state probabilities for the nonpersistent user case, we make the following guess about steady state: With we suggest
(10) 
where for all and is a constant that makes all probabilities sum to 1.
Theorem 2
The CTMC for this system with persistent and nonpersistent users is reversible and the steady state distribution is given by (10).
It suffices to show that defined by (10) satisfies the detailed balance equations. Consider states and in . We consider the three possible transition types:

Nonpersistent users (): For simplicity of notation we consider these transitions for nonpersistent class (the result is similar for a general nonpersistent class ). Fix and , both being states in . Then

Persistent user (): For simplicity of notation we consider these transitions for persistent user (the result is the same for a general persistent user ). Fix and , both being states in . Then

Persistent user (): For simplicity of notation we consider these transitions for persistent user (the result is the same for a general persistent user ). Fix and , both being states in . Then
The steady state probabilities in the above theorem can be simplified by aggregating all nonpersistent users. Consider a state . Define as the number of nonpersistent users associated with this state. Define as the steady state probability that the total number of nonpersistent users is and the state of the persistent users is . Define and define
where the subscript emphasizes that counts the number of busy persistent users from the vector . In particular, the total number of busy channels for a vector is . A vector is said to be a legitimate vector if .
Corollary 2
For this system with persistent and nonpersistent users we have for all legitimate vectors :
(11) 
where is the same constant used in Theorem 2, , and for . Further, in the special case when there are no nonpersistent users (so that ) we have
IiiC Solution complexity
The formulas (10) and (11) establish steady state probabilities for a very large number of system states. The number of states grows exponentially in the problem size. For example, just considering the 3 possibilities for each persistent user, we find the number of states is at least . If then , meaning that the number of states is larger than the current estimate for the number of atoms in the universe. Thus, it is not immediately clear how to compute the constant , and how to use the formulas (10) and (11) to calculate things such as the marginal fraction of time that persistent user 1 is busy, the throughput and success probability of persistent user 1, and the throughput and success probabilities of the different classes of nonpersistent users. For some problems that involve reversible networks, such as the admission control problems in [6] that are solved by truncation of queues, it can be shown that even calculating the proportionality constant to within a reasonable approximation is NPhard [7] (see also [8] for factor graph approximation methods). Fortunately, our problem has enough structure to allow efficient computation of all of these things via a discrete Fourier transform. That is developed next.
IiiD Calculating
Define
We can sum the probabilities in (11) by grouping states into those that have persistent users that are busy, for , and nonpersistent users:
(12) 
where we define by
(13) 
Notice that these values are defined for all , even if the number of persistent users is larger than the number of channels (so that only are used in (12)).
It is difficult to obtain the value of by a direct summation in (13) because there are so many terms. However, we can define a related polynomial function defined for all complex numbers :
For any given , the value can be easily computed as a product of (complexvalued) terms. We observe that
This motivates a discrete Fourier transform approach: Define and define
The sequence is the discrete Fourier transform of . The inverse transform gives
Of course, these values of only need to be computed for
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