Autonomous allocation of wireless body-sensor networks to frequency channels: modeling with repeated “balls-in-bins” experiments

Abstract

In this paper we model a situation where several wireless body sensor networks (WBSN) compete for occupation of a number of frequency channels. Each channel can host at most one WBSN with satisfactory performance and WBSNs have the ability to change their operating channel, subject to the constraint that they can only monitor the performance or occupancy of their current channel but not of any other channel. We consider a number of randomized schemes for changing the frequency channels and present and evaluate Markov chain models for these, building on a “balls-in-bins” approach.

Appendices

A Reducing system size in Eq. (5)

In this appendix we discuss how to reduce the size of the linear equation system (5) and (6), making it more accessible to numerical solution. The key roles are played by the notion of types introduced in Sect. 2 and the exploitation of symmetries.

To begin with, it is straightforward to see that we only need to consider terms for the colliding states in Eq. (6), as the hitting times in non-colliding states are always zero. Next, consider two colliding states ${\mathbf {s}}_a$ and ${\mathbf {s}}_b$ of the same type. Therefore, there exists a permutation $\sigma (\cdot )$ of the bins from 1 to N such that ${\mathbf {s}}_b = \sigma ({\mathbf {s}}_a)$. From the definition of the transition probabilities (Eqs. 3 and 4) it is straightforward to check that $p({\mathbf {t}}|{\mathbf {s}}) = p(\tau ({\mathbf {t}})|\tau ({\mathbf {s}}))$ holds for any permutation $\tau (\cdot )$, and therefore the vector $(p({\mathbf {t}}|{\mathbf {s}}_b) : {\mathbf {t}} \in {\mathcal {C}}_{N,K})$ of transition probabilities from state ${\mathbf {s}}_b$ to all colliding states is a permutation of the vector $(p({\mathbf {t}}|{\mathbf {s}}_a) : {\mathbf {t}} \in {\mathcal {C}}_{N,K})$, as applying the permutation $\sigma (\cdot )$ to all colliding states reproduces the entire set of colliding states.

Next, fix two different types ${\mathcal {T}}({\mathbf {s}})$ and ${\mathcal {T}}({\mathbf {t}})$. Since all states of the same type ${\mathcal {T}}({\mathbf {s}})$ are permutations of each other, a similar argument shows that the vector of transition probabilities $\left( p({\mathbf {t'}}|{\mathbf {s}}_a) : {\mathbf {t'}} \in {\mathcal {T}}({\mathbf {t}}) \right)$ from some state ${\mathbf {s}}_a \in {\mathcal {T}}({\mathbf {s}})$ to all states in ${\mathcal {T}}({\mathbf {t}})$ is a permutation of the transition probabilities $\left( p({\mathbf {t'}}|{\mathbf {s}}_b) : {\mathbf {t'}} \in {\mathcal {T}}({\mathbf {t}}) \right)$ from any other state ${\mathbf {s}}_b \in {\mathcal {T}}({\mathbf {s}})$ to all states in ${\mathcal {T}}({\mathbf {t}})$. If we now order the states according to their type (i.e. all states of the same type are grouped together) and leave out the non-colliding states, then the matrix of state transition probabilities for the remaining states can be written as a block matrix of the following form:

$$\begin{aligned} {\mathbf {P'}} = \left[ \begin{array}{llll} {\mathbf {A}}_{(K)}^{(K)} & {\mathbf {A}}_{(K-1,1)}^{(K)} & \ldots & {\mathbf {A}}_{(2,1,\ldots ,1)}^{(K)} \\ \ {\mathbf {A}}_{(K)}^{(K-1,1)} & {\mathbf {A}}_{(K-1,1)}^{(K-1,1)} & \ldots & {\mathbf {A}}_{(2,1,\ldots ,1)}^{(K-1,1)} \\ \ \ldots & \ldots & \ldots & \ldots \\ \ {\mathbf {A}}_{(K)}^{(2,1,\ldots ,1)} & {\mathbf {A}}_{(K-1,1)}^{(2,1,\ldots ,1)} & \ldots & {\mathbf {A}}_{(2,1,\ldots ,1)}^{(2,1,\ldots ,1)} \\ \end{array} \right] \end{aligned}$$

where the matrix ${\mathbf {A}}_{{\mathcal {T}}_2}^{{\mathcal {T}}_1}$ contains all transition probabilities from states ${\mathbf {s}} \in {\mathcal {T}}_1$ to states ${\mathbf {t}} \in {\mathcal {T}}_2$, and in such a matrix all the rows are permutations of each other. Considering only the colliding states, the Equations from (6) can be re-arranged as

$$\begin{aligned} \left( {\mathbf {P'}} - {\mathbf {I}}\right) \cdot {\mathbf {k}} = (-1, -1, \ldots , -1)^T \end{aligned}$$

(13)

where ${\mathbf {k}}$ is the solution vector with the hitting times of all non-colliding states and ${\mathbf {I}}$ is the identity matrix of suitable dimension. Please note that in this equation the matrix $\left( {\mathbf {P'}} - {\mathbf {I}}\right)$ maintains the property of matrix ${\mathbf {P'}}$ of being a block matrix made up of matrices where all the rows are permutations of each other: in all the “diagonal matrices” ${\mathbf {A}}_{\mathcal {T}}^{\mathcal {T}}$ the respective diagonal elements are identical and remain so after subtracting the identity matrix. Furthermore, since we have omitted all transitions into non-colliding states (which occur with positive probability from every colliding state), the matrix ${\mathbf {P'}}$ is strictly sub-stochastic, which in turn makes the matrix $\left( {\mathbf {P'}} - {\mathbf {I}}\right)$ strictly diagonal-dominant and thus the equation system (13) is uniquely solvable. Note that its dimension is given by the size of the state space $|{\mathcal {S}}_{N,K}|$ with only the non-colliding states removed.

The key property which now allows to simplify the calculation of the hitting times is that all states of the same type have the same average hitting time. To see this, we argue as follows. If we denote by $\alpha _{{\mathcal {T}}_2}^{{\mathcal {T}}_1}$ the row sum of sub-matrix ${\mathbf {A}}_{{\mathcal {T}}_2}^{{\mathcal {T}}_1}$, then the system

$$\begin{aligned} \left( \left( \begin{array}{llll} \alpha _{(K)}^{(K)} & \alpha _{(K-1,1)}^{(K)} & \ldots & \alpha _{(2,1,\ldots ,1)}^{(K)} \\ \alpha _{(K)}^{(K-1,1)} & \alpha _{(K-1,1)}^{(K-1,1)} & \ldots & \alpha _{(2,1,\ldots ,1)}^{(K-1,1)} \\ \ldots & \ldots & \ldots & \ldots \\ \alpha _{(K)}^{(2,1,\ldots ,1)} & \alpha _{(K-1,1)}^{(2,1,\ldots ,1)} & \ldots & \alpha _{(2,1,\ldots ,1)}^{(2,1,\ldots ,1)} \\ \end{array} \right) - {\mathbf {I}} \right) \cdot {\mathbf {k'}} = (-1, -1, \ldots , -1)^T \end{aligned}$$

(14)

is again strictly diagonal-dominant and thus has a unique solution ${\mathbf {k'}} = (k_{(K)}, k_{(K-1,1)}, \ldots , k_{(2,1,\ldots ,1)})$. Furthermore, its dimension is given by the number of types $C_K$ existing for K balls [minus one, the non-colliding type $(1,1,\ldots ,1)$—compare Eq. (7)]. We construct a vector ${\mathbf {k}}$ from ${\mathbf {k'}}$ by starting with repeating value $k_{(K)}$ exactly $|{\mathcal {T}}(K)|$ times, followed by $|{\mathcal {T}}(K-1,1)|$ repetitions of $k_{(K-1,1)}$ and so forth. It is then clear that this vector ${\mathbf {k}}$ is a solution of the full equation system (13).

Let us consider the savings that can be obtained from using system (14) instead of (13). For $N=15$ and $K=10$ we have that $C_{10}=42$ and thus the system (14) has to be solved for 41 unknowns (after leaving out the single non-colliding type). As discussed before in Sect. 2.2, the full system (13) would have a dimension in the order of approximately 1.9 million unknowns.

B Sticky variant: derivation of transition probabilities

To derive explicit expressions for the transition probabilities of the sticky model we will use the framework of exponential generating functions (EGF) [14, 27, 38]. When applying EGF to balls-in-bin problems, a modified exponential power series

$$\begin{aligned} c(z) = \sum _{m=0}^\infty \frac{c_m}{m!} z^m \qquad (z \in \mathbb {C}) \end{aligned}$$

(15)

is associated to each bin, where the coefficients $c_m \in \left\{ 0,1 \right\}$ are chosen to express whether or not it is permissible to have m balls in the bin. The power series for several bins are then multiplied to get the overall EGF for the problem, and the m-th coefficient of the overall EGF gives the total number of allocations of balls to bins which satisfy all constraints simultaneously.

Exponential generating functions are one widely used class of generating functions, another class are ordinary generating functions. EGFs are appropriate in settings with labeled (i.e. distinct) entities. When given a general power series f(z), we denote by $\left\{ z^{n}\right\} f(z)$ the coefficient for the $z^n$ term of f(z), and for the particular case of an EGF c(z) formed according to Eq. (15), the coefficient $c_m$ is given by $c_m = m! \cdot \left\{ z^{m}\right\} c(z)$. Note that an EGF c(z) (taken as a power series around $z=0$) is an analytic / holomorphic function within its convergence radius, and it is a basic fact from complex analysis that we can recover the coefficients of a power series c(z) from the derivatives of c(z):

$$\begin{aligned} \frac{c_m}{m!} = \frac{(\partial _mc)(0)}{m!} \end{aligned}$$

(16)

where $\partial _m$ denotes the m-th complex derivative operator.

To transition from state $S \in {\mathcal {S}}$ to state $S+T \in {\mathcal {S}}$ we find the number of allocations of $K-S$ unsettled balls to bins such that

• Each of the S bins containing a settled ball can contain zero or more unsettled balls. To each such bin we assign the EGF $e^z$, in which all coefficients are one (compare Eq. 15).

• T of the bins that contained no settled ball before should now contain exactly one ball. To each of these bins we assign the EGF z (corresponding to $c_1=1$ and $c_m=0$ for $m \ne 1$).

• The remaining $N-S-T$ bins each contain either zero or at least two balls. To each of these bins we assign the EGF $(e^z-z)$, excluding the case of exactly one ball.

The EGF corresponding to these constraints is given by

$$f(z) = \left( e^z\right) ^S \cdot z^T \cdot \left( e^z - z\right) ^{N-(S+T)}$$

(17)

and the expression $(K-S)! \cdot \left\{ z^{K-S}\right\} f(z)$ gives the number of such allocations. This expression, however, refers to one particular choice of T bins out of the $N-S$ bins not containing a settled ball, and there are in total $\left(\begin{array}{c} {N-S} \\ {T} \end{array}\right)$ such choices. Therefore, the total number allocations of $K-S$ unsettled balls to bins satisfying the above constraints is given by

$$\begin{aligned} n(S,T) = \left(\begin{array}{c} {N-S} \\ {T} \end{array}\right) \cdot (K-S)! \cdot \left( \left\{ z^{K-S}\right\} f(z)\right) \end{aligned}$$

(18)

and since the total number of allocations of $K-S$ balls to N bins is given by $N^{K-S}$ the transition probability becomes

$$\begin{aligned} p_{S,S+T} = \frac{n(S,T)}{N^{K-S}} = \frac{ \left(\begin{array}{c} {N-S} \\ {T} \end{array}\right) \cdot (K-S)! \cdot \left( \left\{ z^{K-S}\right\} f(z)\right) }{N^{K-S}} \end{aligned}$$

(19)

To get explicit expressions, we expand the term $(e^z-z)^{N-(S+T)}$ of Eq. (17) using the binomial theorem, and simplify the resulting expression (exploiting $K\le N$) to arrive at:

$$\begin{aligned}&{n(S,T) = \left(\begin{array}{c} {N-S} \\ {T} \end{array}\right) \cdot (K-S)! \cdot } \\&\quad \left( \sum _{\nu = 0}^{K-(S+T)} (-1)^\nu \left(\begin{array}{c} {N-(S+T)} \\ {\nu } \end{array}\right) \frac{(N - T - \nu )^{K-(S+T)-\nu }}{(K-(S+T)-\nu )!} \right) \end{aligned}$$

(20)

which for $T=0$ simplifies to

$$n(S,0) = (K-S)! \cdot \left( \sum _{\nu =0}^{K-S} (-1)^\nu \left(\begin{array}{c} {N-S} \\ {\nu } \end{array}\right) \frac{(N-\nu )^{K-S-\nu }}{(K-S-\nu )!} \right)$$

(21)

Note that a straightforward calculation yields that in particular we have

$$p_{K,K} = 1$$

(22)

$$p_{K-1,K-1}= \frac{K-1}{N}$$

(23)

$$p_{K-2,K-2}= \frac{N + K^2 - 5 K + 6}{N^2}$$

(24)

C Sticky variant: monotonicity of self-transition probabilities

In this Appendix we show that the self-transition probabilities $p_{S,S}$ in the Markov model of the sticky variant (compare Eqs. (20) and in particular (9)) are strictly monotonically increasing in S for $K \ge 2$, i.e. that for $S \in \left\{ 0, 1, \ldots , K-1 \right\}$ we have $p_{S,S} < p_{S+1,S+1}$. Note that the self-transition probabilities are also all positive, which can be established probabilistically from the fact that colliding balls can jump with positive probability back to exactly where they currently are, so that they remain in the colliding state.

The EGF corresponding to the diagonal entries can be specialized from Eq. (17) for fixed state S to become:

$$f_S(z) = e^{zS} \cdot \left( e^z - z\right) ^{N-S}$$

(25)

by setting $T=0$, and valid for all $0 \le S \le K \le N$. When taken as a power series, note that $f_S(z)$ has non-negative coefficients, since $g_1(z)=e^{z}=1+\frac{z}{1!} + \frac{z^2}{2!} + +\frac{z^3}{3!}+\cdots$ and $g_2(z)=e^z-z = 1 + \frac{z^2}{2!} + \frac{z^3}{3!}+\cdots$ have only non-negative coefficients and $f_S(z)$ is formed out of $g_1(z)$ and $g_2(z)$ by multiplications. Note that the power series $g_3(z)=z$ also has only non-negative coefficients.

We can say more: for $S\ne 0$ a factor $g_1(z)$ is present which has strictly positive coefficients throughout, and all other factors of $f_S(\cdot )$ contribute a 1, so in $f_S(\cdot )$ there is at least one term of any order and all the coefficients are truly positive. For $S=0$ all coefficients of $f_S(\cdot )$ of order zero or of order two or more are strictly positive. The coefficient of order one (for the term z) is zero, but this is relevant only for $K=1$. So for all $K \ge 2$ and all S the coefficients of $f_S(\cdot )$ are strictly positive.

From the general expression for the transition probabilities (Eq. 8) and by taking into account that extracting coefficients is equivalent to taking derivatives (Eq. 16) we get

$$\begin{aligned} p_{S,S}= & \frac{(K-S)! \cdot \left( \left\{ z^{K-S}\right\} f_S(z)\right) }{N^{K-S}}\\ = & \frac{(K-S)! \cdot \left[ \frac{\partial _{K-S}f_S(0)}{(K-S)!}\right] }{N^{K-S}} \\= & \frac{\partial _{K-S}f_S(0)}{N^{K-S}} \end{aligned}$$

(26)

With this, the condition $p_{S,S} < p_{S+1,S+1}$ is equivalent to the condition

$$\begin{aligned}&\partial _{K-S}\left[ e^{zS}\left( e^z-z\right) ^{N-S}\right] (0) \\&\quad < N \cdot \left[ \partial _{K-S-1}\left[ e^{z(S+1)}\left( e^z-z\right) ^{N-S-1}\right] (0)\right] \end{aligned}$$

(27)

For $S=0$ the left-hand side of this relation becomes

$$\begin{aligned} \partial _{K-1}\left[ \partial _1\left( e^z-z\right) ^N\right] (0) = N \cdot \partial _{K-1}\left[ \left( e^z-z\right) ^{N-1} \cdot \left( e^z-1\right) \right] (0) \end{aligned}$$

and the right-hand side will be

$$N \cdot \left[ \partial _{K-1}\left[ e^z\left( e^z-z\right) ^{N-1}\right] (0)\right]$$

Plugging this back into condition (27) after canceling N, subtracting the left-hand side from the right-hand-side and using the linearity of the derivative we get the condition

$$\begin{aligned} 0&< \partial _{K-S}\left[ e^z \left( e^z-z\right) ^{N-1} - \left( e^z-z\right) ^{N-1} \cdot \left( e^z-1\right) \right] (0)\\&= \partial _{K-S}\left[ \left( e^z-z\right) ^{N-1}(e^z - \left( e^z-1\right) \right] (0)\\&= \partial _{K-S}\left[ \left( e^z-z\right) ^{N-1}\right] (0) \end{aligned}$$

which is just extracting the $K-S$ coefficient out of a power series (an integer power of $g_2(z)$) with coefficients that are non-negative in general and strictly positive from the $z^2$ coefficient onwards. Hence, for $N \ge K\ge 2$ and $S=0$ the claim is true. For $0 < S \le K-1$ we get for the left-hand side of condition (27):

$$\begin{aligned}&\partial _{K-S-1}\left[ \partial _1\left[ e^{zS}\left( e^z - z\right) ^{N-S} \right] \right] (0) \\&\quad = \partial _{K-S-1}\left[ Se^{zS}\left( e^z - z\right) ^{N-S} + e^{zS}(N-S)\left( e^z - z\right) ^{N-S-1}\left( e^z-1\right) \right] (0) \end{aligned}$$

With this the condition to check becomes

$$\begin{aligned} 0&< \partial _{K-S-1} \left[ N e^{z(S+1)}\left( e^z-z\right) ^{N-S-1} - Se^{zS}\left( e^z - z\right) ^{N-S} - e^{zS}(N-S)\left( e^z - z\right) ^{N-S-1}\left( e^z-1\right) \right] (0)\\&= \partial _{K-S-1} \left[ e^{zS}\left( e^z-z\right) ^{N-S-1} \left( N e^z - S\left( e^z-z\right) - (N-S)\left( e^z-1\right) \right) \right] (0)\\&= \partial _{K-S-1} \left[ e^{zS}\left( e^z-z\right) ^{N-S-1} \left( Sz + (N-S) \right) \right] (0)\\ \end{aligned}$$

which again is built from (a sum of) products of $g_1(z)$, $g_2(z)$ and $g_3(z)$ and has strictly positive coefficients, verifying the claim.

D Sticky variant: strict positivity of transition probabilities

In this Appendix we argue that the transition probabilities above the diagonal $p_{S,S+T}$ (with $T > 0$) in the Markov model of the sticky variant (compare Eqs. (20) and in particular (9)) are strictly positive, with one exception.

The starting point is again the EGF f(z) from Eq. (17), which is a product of integer multiples of the three elementary EGFs $g_1(z)=e^{z}$, $g_2(z)=e^z-z$ and $g_3(z)=z$.

The case of a state $0< S < K$ is straightforward: the EGF f(z) contains the factor $z^T$ and the two other factors $(g_1(z))^S$ and $(g_2(z))^{N-(S+T)}$ both contribute a $z^0=1$ term, so the $z^T$ term in f(z) is strictly positive.

For the state $S = 0$ we distinguish three cases:

• With $T=K$ we are asking about state transition probability $p_{0,K}$ which from elementary combinatorial considerations is given by $N \cdot (N-1) \cdot \cdots \cdot (N-K+1) > 0$.

• With $T=K-1$ we are asking that we are jumping out of a state in which no ball is settled into a state where every ball but one is settled—but to make this happen the first $N-1$ balls must jump alone into an empty bin (otherwise they would not settle) and the remaining ball cannot jump into the same bin as any of the others (otherwise none of the two would settle), so the last ball must jump into another empty bin, where it settles. More formally, for $S=0$ and $T=K-1$ we get the generating function $f(z)=z^{K-1}\cdot \left( e^z-z\right) ^{N-K+1}$ which for $K \ge 3$ does not contain a term in z, so the coefficient for z (and thus the transition probability into this state) are 0.

• For $1 \le T \le K-2$ the argument is very similar to the argument for $S > 0$.

E Sticky variant: convergence speed towards absorbing state

The claim of a geometric convergence speed of the Markov chain for the sticky variant towards the absorbing state 0 is in itself not a surprise, as it is similar to well-known results from the literature (compare for example [5, Chap. 6]), but the proofs of these frequently rely on the Perron-Frobenius theorem, which is only valid for irreducible transition matrices, which our upper-triagonal matrix ${\mathbf {P}}$ with the entries derived in “Appendix B” is not. We therefore proceed to verify this directly. Since ${\mathbf {P}}$ is diagonalizable (compare Sect. 4.2), we can express ${\mathbf {P}}$ as ${\mathbf {P}} = {\mathbf {U}} \cdot {\mathbf {D}} \cdot {\mathbf {U}}^{-1} =: {\mathbf {U}} \cdot {\mathbf {D}} \cdot {\mathbf {V}}$, where ${\mathbf {D}} = {{\mathrm{diag}}}(\lambda _0, \lambda _1, \ldots , \lambda _K)$ is a diagonal matrix made up of the eigenvalues / diagonal elements of ${\mathbf {P}}$ and the columns of ${\mathbf {U}}$ are given by the (linearly independent) right eigenvectors ${\mathbf {x_i}}$ for eigenvalues $\lambda _i$.

Let $M_j$ be a non-negative, integer-valued random variable denoting the number of steps it takes to get from starting state $j \in \left\{ 0,\ldots ,K-1 \right\}$ to the absorbing state K. The average hitting time is then the expectation $E\left[ M_j\right]$. For this expectation we can use the well-known so-called survivor representation [27]:

$$\begin{aligned} E\left[ M_j\right] = \sum _{n=0}^\infty \Pr \left[ M_j>n\right] \end{aligned}$$

If we denote by $\left[ {\mathbf {v}}\right] _{i}$ the i-th component of a vector ${\mathbf {v}}$ then we clearly have from the setup of our system that

$$\begin{aligned} \Pr \left[ M_j > n\right] = \left[ {\mathbf {e_K}} - {\mathbf {e_j}} \cdot {\mathbf {P^n}}\right] _{K} = 1 - \left[ {\mathbf {e_j}} \cdot {\mathbf {P^n}} \right] _{K} \end{aligned}$$

holds, where ${\mathbf {e_k}}$ is the k-th unit vector in ${\mathbb {R}}^{K+1}$ (taken as a row vector here). Recalling $\lambda _K=1$ we get

$$\begin{aligned} {\left[ {\mathbf {e_j}} \cdot {\mathbf {P^n}}\right] _{K}}&= \left[ {\mathbf {e_j}} \cdot {\mathbf {U}} \cdot {\mathbf {D}}^n\cdot {\mathbf {V}}\right] _{K} \\&= \lambda _0^n u_{j,0} v_{0,K} + \lambda _1^n u_{j,1} v_{1,K} \ldots + \lambda _{K-1}^n u_{j,K-1} v_{K-1,K} + u_{j,K} v_{K,K} \end{aligned}$$

Since convergence to the absorbing state happens with probability one and recalling $\lambda _i < 1$ for $i \in \left\{ 0,1,\ldots ,K-1 \right\}$ we see

$$\begin{aligned} 0= & \lim _{n \rightarrow \infty } \left[ {\mathbf {e_K}} - {\mathbf {e_j}} \cdot {\mathbf {P^n}}\right] _{K} \\= & 1 - u_{j,K} v_{K,K} \end{aligned}$$

(28)

and therefore

$$\begin{aligned} { \left[ {\mathbf {e_K}} - {\mathbf {e_j}} \cdot {\mathbf {P^n}}\right] _{K}}&= -\left( \lambda _0^n u_{j,0} v_{0,K} + \lambda _1^n u_{j,1} v_{1,K} \ldots \right.\\&\left.\quad +\, \lambda _{K-1}^n u_{j,K-1} v_{K-1,K}\right) \\&= -\, \lambda _{K-1}^n \left( \left( \frac{\lambda _0}{\lambda _{K-1}}\right) ^n u_{j,0} v_{0,K} + \cdots \right.\\&\left.\quad+\, \left( \frac{\lambda _{K-1}}{\lambda _{K-1}}\right) ^n u_{j,K-1} v_{K-1,K}\right) \\&\le \lambda _{K-1}^n \cdot \left\| {\mathbf {u_j}}\right\| _2\cdot \left\| {\mathbf {v}}\right\| _2 \end{aligned}$$

(29)

(using the abbreviations ${\mathbf {u_j}}=(u_{j,0},u_{j,1},\ldots ,u_{j,K-1})$ and ${\mathbf {v}}=(v_{0,K},v_{1,K},\ldots ,v_{K-1,K})$). This verifies the claim.