Fast node cardinality estimation and cognitive MAC protocol design for heterogeneous machine-to-machine networks

Abstract

We design two estimation schemes, Method I and Method II, for rapidly obtaining separate estimates of the number of active nodes of each traffic type in a heterogeneous machine-to-machine (M2M) network with T types of nodes (e.g., those that send emergency, periodic, normal type data etc.), where $T\ge 2$ is an arbitrary integer. Method I is a simple scheme, and Method II is more sophisticated and outperforms Method I. Also, we design a medium access control (MAC) protocol that supports multi-channel operation for a heterogeneous M2M network with T types of nodes, operating as a secondary network using Cognitive Radio technology. In every time frame, our Cognitive MAC protocol uses the proposed estimation schemes to rapidly estimate the active node cardinality of each type, and uses these estimates to find the optimal contention probabilities to be used. We compute a closed form expression for the expected number of time slots required by Method I to execute, and a simple upper bound on it. Also, we analytically obtain expressions for the expected number of successful contentions per frame and the expected amount of energy consumed. Finally, we evaluate the performances of our proposed estimation schemes and Cognitive MAC protocol using simulations.

Appendices

Appendix 1: Proofs of the upper bounds obtained in Section 4.1.5

Proof of Theorem 1

Since $K_T = \sum _{i=0}^{t_T-1} I_{\{S_1^{(i)} = C, \ldots , S_{T-1}^{(i)} = C\}}$ (see Sect. 4.1.4), the expected number of slots required in the second phase is $E(K_T) = \sum _{i=0}^{t_T-1} P(C_i)$, where $C_i$ is the event that collisions occur in all the $T-1$ slots of block $B_i$. Now, $C_i = A_i \cup F_i \cup D_i$, where

$A_i$ : Event that at least two Type 1 nodes transmit in block $B_i$;

$F_i$ : Event that at least two each of Type 2, $\ldots$ , Type T nodes transmit in block $B_i$;

$D_i$ : Event that exactly one Type 1 node and one Type 2, $\ldots$ , one Type T node transmit in block $B_i$.

Using the union bound, upper bounds for $P(A_i)$, $P(F_i)$ and $P(D_i)$ are computed as,

$$\begin{aligned} P(A_i)\le & {} {{n_1} \atopwithdelims (){2}} p_i^2 \le \frac{n_1^2}{2}p_i^2; \end{aligned}$$

(11)

$$\begin{aligned} P(F_i)\le & {} {{n_2} \atopwithdelims (){2}} p_i^2 \ldots {{n_T} \atopwithdelims (){2}} p_i^2 \le (n_2 \ldots n_T)^2 \Big (\frac{p_i^2}{2}\Big )^{T-1}; \end{aligned}$$

(12)

$$\begin{aligned} P(D_i)\le & {} (n_1 p_i)\ldots (n_T p_i) = (n_1 \ldots n_T) p_i^T; \end{aligned}$$

(13)

where $p_i$ is given by (3).

Next, recall that $l_{n_r} = \lceil (\log _2n_r)\rceil$. Using the union bound, $P(C_i) \le P(A_i) + P(F_i) + P(D_i) \le ({n_1^2}/{2})p_i^2 + (n_2 \ldots n_T)^2 (p_i^2/{2})^{T-1} + n_1 \ldots n_T p_i^T$. Now,

$$\begin{aligned} E(K_T) = \sum _{i=0}^{t_T-1} P(C_i) = \sum _{i=0}^{l_{n_r} - 2} P(C_i) + \sum _{i= l_{n_r} - 1}^{t_T-1} P(C_i). \end{aligned}$$

(14)

Let us consider the first summation of (14),

$$\begin{aligned} \sum _{i=0}^{l_{n_r} - 2} P(C_i) \le \sum _{i=0}^{ l_{n_r} - 2} 1= l_{n_r} - 1. \end{aligned}$$

(15)

Now consider the second summation of (14), in which $t_T$ can be written as $t_T = l_{n_r} + s$,

$$\begin{aligned}&\sum _{i={l_{n_r} - 1}}^{l_{n_r} + s - 1} P(C_i)\nonumber \\&\quad \le \sum _{i={l_{n_r} -1}}^{l_{n_r} + s -1} \left( \frac{n_1^2}{2}p_i^2 + \frac{(n_2 \ldots n_T)^2}{2^{T-1}} p_i^{2(T-1)} + {n_1 \ldots n_T} p_i^T\right) \nonumber \\&\quad = \frac{n_1^2}{2} \Bigg [\sum _{i={l_{n_r} -1}}^{l_{n_r} + s -2} \left( \frac{1}{4}\right) ^{i+1} + \left( \frac{1}{4}\right) ^{l_{n_r} + s-1} \Bigg ] \nonumber \\&\qquad + \frac{(n_2 \ldots n_T)^2}{2^{T-1}} \Bigg [\sum _{i={l_{n_r} -1}}^{l_{n_r} + s -2} \left( \frac{1}{4^{T-1}}\right) ^{i+1} + \left( \frac{1}{4^{T-1}}\right) ^{l_{n_r} + s-1}\Bigg ]\nonumber \\&\qquad + n_1 \ldots n_T \Bigg [\sum _{i={l_{n_r} -1}}^{l_{n_r} + s - 2} \left( \frac{1}{2^T}\right) ^{i+1} + \left( \frac{1}{2^T}\right) ^{l_{n_r} + s-1}\Bigg ] \ (\text{ by } (3)) \end{aligned}$$

(16)

Let us consider the series, $S(r,n) = 1 + r + r^2 + \ldots + r^{n-1} + r^{n-1}$, where n is an integer and $0< r < 1$. A simplified formula for the sum S(r, n) is as follows.

$$\begin{aligned} S(r,n) = \frac{1 - r^n}{1-r} + r^{n-1} = \frac{1-2r^n(1-1/2r)}{1-r}. \end{aligned}$$

(17)

Let us consider individual upper bounds for each quantity in (16). We get,²⁵²⁶²⁷

$$\begin{aligned}&\frac{n_1^2}{2} \Bigg [\sum _{i={l_{n_r} -1}}^{l_{n_r} + s -2} \left( \frac{1}{4}\right) ^{i+1} + \left( \frac{1}{4}\right) ^{l_{n_r} + s-1} \Bigg ] \nonumber \\&\quad \le \frac{n_1^2}{2} \left[ \frac{1}{n_r^2} + \frac{1}{4n_r^2} + \frac{1}{16n_r^2} + \ldots \ldots + \frac{1}{4^{s-2}n_r^2} + \frac{1}{4^{s-2}n_r^2 }\right] \nonumber \\&\quad = \frac{n_1^2}{2n_r^2} \Bigg [ \frac{1-2(1/4^s)(1-4/2)}{1-1/4}\Bigg ] \text { (by~(17))}\nonumber \\&\quad = \frac{2}{3} \frac{n_1^2}{n_r^2}\Big ( 1 + 2/4^s\Big ). \end{aligned}$$

(18)

$$\begin{aligned}&\frac{(n_2 \ldots n_T)^2}{2^{T-1}} \Bigg [\sum _{i={l_{n_r} -1}}^{l_{n_r} + s -2} \left( \frac{1}{4^{T-1}}\right) ^{i+1} + \left( \frac{1}{4^{T-1}}\right) ^{l_{n_r} + s-1}\Bigg ] \nonumber \\&\quad \le \frac{(n_2 \ldots n_T)^2}{2^{T-1}} \left( \frac{1}{4^{T-1}}\right) ^{\log _2n_r} \nonumber \\&\qquad \left[ 1 + \frac{1}{4^{(T-1)}} + \frac{1}{4^{2(T-1)}} + \dots +\frac{1}{4^{(s-1)(T-1)}} +\frac{1}{4^{(s-1)(T-1)}}\right] \nonumber \\&\quad = \Big (\frac{n_2 \ldots n_T}{n_r^{T-1}}\Big )^2 \frac{1}{2^{T-1} (1 - 4^{-(T-1)})} \nonumber \\&\qquad \bigg (1 - \frac{2}{4^{(T-1)s}} \Big (1 - \frac{4^{T-1}}{2}\Big )\bigg ) \text { (by~(17))}. \end{aligned}$$

(19)

$$\begin{aligned}&n_1 \ldots n_T \Bigg [\sum _{i={l_{n_r} -1}}^{l_{n_r} + s - 2} \left( \frac{1}{2^T}\right) ^{i+1} + \left( \frac{1}{2^T}\right) ^{l_{n_r} + s-1}\Bigg ] \nonumber \\&\quad \le n_1 \ldots n_T \left( \frac{1}{2^{T}}\right) ^{\log _2n_r} \nonumber \\&\qquad \left[ 1 + \frac{1}{2^T} + \frac{1}{2^{2T}} + \dots + \frac{1}{2^{(s-1)T}} + \frac{1}{2^{(s-1)T}}\right] \nonumber \\&\quad = \Big (\frac{n_1 \ldots n_T}{n_r^T}\Big ) \frac{1}{1 - 2^{-T}} \Big ( 1 - \frac{2}{2^{Ts}} (1 - 2^{T-1})\Big ) \text { (by~(17))}. \end{aligned}$$

(20)

By (14)–(16) and (1819)–(20), we get:

$$\begin{aligned} E(K_T)&\le {} l_{n_r} - 1+ \frac{2}{3} \frac{n_1^2}{n_r^2}\Big (1+2/4^s\Big ) + \Big (\frac{n_2 \ldots n_T}{n_r^{T-1}}\Big )^2 \nonumber \\&\quad \times \frac{1}{2^{T-1} (1 - 4^{-(T-1)})} \bigg (1 - \frac{2}{4^{(T-1)s}} \Big (1 - \frac{4^{T-1}}{2}\Big )\bigg )\nonumber \\&\quad + \Big (\frac{n_1 \ldots n_T}{n_r^T}\Big ) \frac{1}{1 - 2^{-T}} \Big ( 1 - \frac{2}{2^{Ts}} (1 - 2^{T-1})\Big ). \end{aligned}$$

(21)

$\square$

Proof of Theorem 2

The expected number of slots required in the third phase is (see Sect. 4.1.4) $E(R_T) = \sum _{i=0}^{t_T-1} P(A_i)$, where the event $A_i$ is as defined in the proof of Theorem 1. Recall that $l_{n_r} = \lceil (\log _2n_r)\rceil$. From the inequalities shown in (111213), (15) and (1819), we get:

$$\begin{aligned} E(R_T) \le l_{n_r} - 1 + \frac{2}{3} \frac{n_1^2}{n_r^2}\Big (1+2/4^s\Big ). \end{aligned}$$

(22)

$\square$

Appendix 2: Method II in the special cases $T = 4$, $T = 5$, and $T = 6$

1.1 Scheme for $T = 4$

The first phase consists of $t_4$ blocks, each consisting of two slots. From Fig. 4, it follows that the symbol combinations given in the following table are used.

Type	Symbol Combination
1	$\alpha$ 0
2	$\alpha$$\alpha$
3	0 $\beta$
4	$\beta$$\beta$

1.1.1 First phase

In each slot of the first phase there are four possible outcomes: E, $\alpha$, $\beta$, and C. Hence, the total number of possible outcomes in the two slots of a block is $4^2 = 16$. It is easy to see that the bit patterns $B(b, i), b \in \{1, 2, 3, 4\}$, can be unambiguously found by the BS if no collision occurs in either of the slots of block $B_i$.²⁸ So next, we consider the cases in which at least one collision occurs per block. The number of cases in which a collision occurs in exactly one slot of a block is $\left( {\begin{array}{c}2\\ 1\end{array}}\right) \times 3 = 6$. The possible outcomes in a block in these 6 cases are shown in the first two columns of Table 2. The types of nodes listed under the ‘Active’ (respectively, ‘Inactive’) column are unambiguously identified by the BS as being active (respectively, inactive) at the end of the first phase; the types listed under the ‘Not Sure’ column are those for which ambiguity exists, and this ambiguity needs to be resolved in the second phase.²⁹

Table 2 This table shows the inferences drawn by the BS regarding the activity or inactivity of various types of nodes for each of the outcomes in which exactly one collision occurs per block

Table 3 shows the possible combinations of numbers of active nodes of different types for which the result CC may occur in a block of the first phase.³⁰ From this table, it can be seen that if the result CC occurs in a block, then ambiguity remains about the activity or inactivity of all four node types, and this ambiguity needs to be resolved in the second phase.

Table 3 This table shows the possible combinations of numbers of active nodes of different types for which the block result CC may occur

1.1.2 Broadcast packet (BP)

The list of type numbers for which ambiguity about activity or inactivity needs to be resolved in each block is informed to all nodes by the BS through a BP, which consists of a bit stream, just after the first phase (see Fig. 3). For a given T, let $S_{NS}(T)$ be the set of all possible subsets of types of nodes for which the BS may not be sure of their activity or inactivity for a block after the first phase; the case where the activity (or inactivity) of every type of node is known to the BS with certainty is included in the set $S_{NS}(T)$ as an empty set ($\emptyset$). From Tables 2 and 3, we can see that $S_{NS}(4) = \{ \{3, 4\}, \{1, 2\}, \{1, 2, 3, 4\}, \emptyset \}$. Suppose the bit strings 00, 01, 10 and 11 are used to represent the cases $\emptyset , \{3, 4\}, \{1, 2\}, \{1, 2, 3, 4\}$ respectively. The BS concatenates the bit strings corresponding to all the blocks of the first phase and sends the concatenated bit string in the BP; this concatenated list is received by all active nodes and they act accordingly during the second phase.³¹

Let b(T) be the number of bits used to represent each element of the set $S_{NS}(T)$. Then the length of the BP (in terms of number of slots) for a given T is given by:

$$\begin{aligned} len_{BP}(T) = \lceil (b(T) \times t_T)/S_W\rceil . \end{aligned}$$

(23)

For $T=4$, $b(4) = \lceil \log _2 |S_{NS}(4)|\rceil = 2$. Hence, $len_{BP}(4) = \lceil (2 t_4)/S_W\rceil$.

1.1.3 Second phase

In this phase only active nodes of those types and hash values (blocks) participate, for which ambiguity exists after the first phase. The first phase block results $C \beta$ and $\alpha C$ (see Table 2) require one additional slot in the second phase to resolve the ambiguity. In case the result of a block is $C \beta$ (respectively, $\alpha C$), active Type 3 (respectively, Type 1) nodes corresponding to that block respond with symbol $\alpha$ and active Type 4 (respectively, Type 2) nodes corresponding to that block respond with symbol $\beta$ in their corresponding slot of the second phase. If the slot outcome is $\alpha$, then it implies that a node of Type 3 (respectively, Type 1) is active, else the slot outcome is $\beta$, and it implies that a node of Type 4 (respectively, Type 2) is active. Thus, the ambiguity for the block is resolved at the end of the second phase. In case the result CC occurs in a block of the first phase, then ambiguity exists regarding the activity or inactivity of nodes of all four types at the end of the first phase; then, for each of the groups $(\{\text {Type 1, Type 2}\}, \{\text {Type 3, Type 4}\})$, the estimation scheme for $T=2$ is used in the second phase to resolve this ambiguity. Note that Method II is recursive– if the result CC occurs in a block of the first phase while executing the scheme for $T = 4$, the scheme for $T = 2$ is used to resolve the ambiguity for that block in the second phase.

1.2 Scheme for $T = 5$

The first phase consists of $t_5$ blocks, each consisting of two slots. From Fig. 4, it follows that the symbol combinations given in the following table are used.

Type	Symbol Combination
1	$\alpha$ 0
2	$\alpha$$\alpha$
3	0 $\beta$
4	$\beta$$\beta$
5	$\beta$$\alpha$

1.2.1 First phase

The number of cases in which a collision occurs in exactly one slot of a block is $\left( {\begin{array}{c}2\\ 1\end{array}}\right) \times 3 = 6$. The possible outcomes in a block in these cases and the inferences drawn by the BS regarding the activity or inactivity of various types of nodes for each of these outcomes are shown in Table 4. From Table 4 it can be seen that the block results $C \alpha$, $C \beta$, $\alpha C$ and $\beta C$ require additional slots in the second phase to resolve ambiguity. Finally, it is easy to check that if the result CC occurs in a block, then ambiguity remains about the activity or inactivity of all five node types, and this ambiguity needs to be resolved in the second phase.

Table 4 This table shows the inferences drawn by the BS regarding the activity or inactivity of various types of nodes for each of the outcomes in which exactly one collision occurs per block

1.2.2 Broadcast packet (BP)

A BP similar to that described in Section B.1.2 is sent by the BS after the first phase. Using notation similar to that in that section, $S_{NS}(5) = \{ \{2, 5\}, \{3, 4\},$$\{1, 2\}, \{4, 5\}, \{1, 2, 3, 4, 5\},$$\emptyset \}$. So $b(5) = \lceil \log _2 |S_{NS}(5)|\rceil = 3$ and $len_{BP}(5)$ = $\lceil b(5) \times t_5/S_W\rceil = \lceil 3t_5/S_W\rceil$.

1.2.3 Second phase

In this phase only active nodes of those types and hash values participate, for which ambiguity still exists after the first phase. To resolve the ambiguity in case of the block results $C \alpha$, $C \beta$, $\alpha C$, and $\beta C$, an approach similar to the one described in Section B.1.3 for the cases $C \beta$ and $\alpha C$ is used; in particular, one additional slot is required in the second phase. In case the block result CC occurs, the set of node types is divided into two groups: $(\{\text {Type 1, Type 2, Type 3}\}$, $\{\text {Type 4, Type 5}\})$. For the first (respectively, second) group, the scheme for $T = 3$ (respectively, $T=2$) is (recursively) used in the second phase to resolve the ambiguity.

1.3 Scheme for $T = 6$

The first phase consists of $t_6$ blocks, each consisting of 3 slots. From Fig. 4, it follows that the symbol combinations given in the following table are used.

Type	Symbol Combination
1	$\alpha$ 0 0
2	$\alpha$$\alpha$ 0
3	$\alpha$$\alpha$$\alpha$
4	0 0 $\beta$
5	0 $\beta$$\beta$
6	$\beta$$\beta$$\beta$

1.3.1 First phase

The number of cases in which a collision occurs in exactly one slot (respectively, two slots) of a block is $\left( {\begin{array}{c}3\\ 1\end{array}}\right) \times 3^2 = 27$ (respectively, $\left( {\begin{array}{c}3\\ 2\end{array}}\right) \times 3 = 9$). Out of these $27 + 9 = 36$ cases, in Table 5, we only show the cases where ambiguity about the activity or inactivity of some of the types of nodes exists and additional slots in the second phase are needed to resolve the ambiguity. Also, it is easy to check that if the result CCC occurs in a block of the first phase, then ambiguity remains about the activity or inactivity of all six node types, and this ambiguity needs to be resolved in the second phase.

Table 5 This table shows the inferences drawn by the BS regarding the activity or inactivity of various types of nodes for each of the outcomes in which one or two collisions occur per block and ambiguity about the activity or inactivity of some of the types of nodes exists

1.3.2 Broadcast packet (BP)

A BP similar to that described in Section B.1.2 is sent by the BS after the first phase. Using notation similar to that in that section, $S_{NS}(6) = \{ \{5, 6\},\{2, 3\}, \{1\}, \{1, 4, 5, 6\}, \{4\},$$\{1, 2, 3, 4\}, \{1, 2, 3, 4, 5, 6\}$, $\emptyset \}$. So $b(6) = \lceil \log _2 |S_{NS}(6)|\rceil = 3$ and $len_{BP}(6) = \lceil b(6) \times t_6/S_W\rceil = \lceil 3t_6/S_W\rceil$.

1.3.3 Second phase

In this phase only nodes of those types and hash values participate, for which ambiguity still exists after the first phase. To resolve the ambiguity in case of the block results listed in Table 5, an approach similar to those described in Sections B.1.3 and B.2.3 is used. For the block results $C C \beta$ and $\alpha C C$, two slots are required in the second phase to resolve the ambiguity, whereas for the other block results, one slot is sufficient. For example, in case the block result $CC\beta$ occurs, in the first of the two required slots, each active Type 1 node transmits symbol $\alpha$. If the slot result is empty, it implies that all Type 1 nodes are inactive, else at least one Type 1 node is active. In the second of the two required slots, each active Type 4 (respectively, Type 5, Type 6) node transmits symbol $\alpha$ (respectively, $\beta$, 0 (no transmission)). Only three outcomes are possible: if the outcome is $\alpha$ (respectively, $\beta$, E), it implies that a Type 4 (respectively, Type 5, Type 6) node is active and all nodes of the other two types are inactive. Finally, in case the result CCC occurs in a block of the first phase, then the set of all node types is divided into two groups of size 3: $\{1, 2, 3\}$ and $\{4, 5, 6\}$, and the scheme for $T = 3$ is (recursively) used twice in the second phase to resolve the ambiguity.

Appendix 3: Proofs of the analytical results provided in section 6

Proof of Proposition 1

There are $M=m$ successful contentions if and only if for some integers $k_1,\ldots , k_m$, the first $(k_1 - 1)$ contention attempts are unsuccessful with n contending nodes and the $k_1^{th}$ attempt is successful, $(k_1 + 1)^{th}$ to $(k_2 - 1)^{th}$ attempts are unsuccessful with $n- 1$ contending nodes and $k_2^{th}$ attempt is successful, …, and $(k_{m} + 1)^{th}$ to $W_m ^{th}$ attempts are unsuccessful with $n- m$ contending nodes. So we get the expression for $P(M=m)$ given in (7).

$\square$

Proof of Proposition 2

Note that there are a total of $W_M$ uplink slots in the frame. In each of these slots, some of the active nodes are in transmission state and the rest are in idle state. So $\mathscr {E}_{UL} = \sum _{i=1}^{W_M} \big (L_i \gamma _T + (n-L_i)\gamma _I\big )$. Taking expectations and conditioning on the values taken by M, we get the expression given in (8).

$\square$

Computation of An Expression for$E(L_i/M=m)$:

$E(L_i/M=m)$ can be calculated as:

$$\begin{aligned}&E(L_i/M=m)\nonumber \\&\quad = \sum _{j=0}^m \left( \sum _{l_i = 0}^{n-j} l_i P(L_i = l_i/M=m, N_i = n-j)\right) \nonumber \\&\qquad P(N_i = n-j/M=m) \end{aligned}$$

(24)

$P(N_i = n-j/M=m)$ for all i and j can be found using the fact that $N_1 = n$ with probability 1 and the following recursion:

$$\begin{aligned}&P(N_i = n-j/M=m) \nonumber \\&\quad = P(N_{i-1} = n-j/M=m)\nonumber \\&\qquad \Big (1 - P_{i-1}(S_{n-j}/M=m)\Big )\nonumber \\&\qquad + P(N_{i-1} = n-j+1/M=m)\nonumber \\&\qquad \Big (P_{i-1}(S_{n-j+1}/M=m)\Big ) \end{aligned}$$

(25)

where $P_i(S_{x}/M=m)$ is the probability of success when x nodes contend in UL slot i given $M=m$ and $j \in \{0, 1, \dots ,i-1\}$. Now,

$$\begin{aligned} P_i(S_{n-j}/M=m) = \frac{P_i(M=m/S_{n-j}) P_i(S_{n-j})}{P(M=m)}, \end{aligned}$$

(26)

where $P_i(S_{x}) = x p_x (1 - p_x)^{x-1}$ is the probability of success when x nodes contend in UL slot i and $P_i(M=m/S_{x})$ is the probability that $M = m$ given that a success occurs when x nodes contend in UL slot i. Next, similar to (7), we get:

$$\begin{aligned}&P_i(M=m/S_{n-j}) \nonumber \\&\quad = \sum _{k_1=i+1}^{W_m - m + 1} \sum _{k_2=k_1 + 1}^{W_m - m + 2} \dots \sum _{k_{m}=k_{m-1} + 1}^{W_m} \nonumber \\&\quad (1 - r_{n-j-1})^{k_1 - i - 1}r_{n-j-1} (1 - r_{n-j-2})^{k_2 - k_1 - 1}r_{n-j-2} \nonumber \\&\quad \dots (1 - r_{n-m})^{W_m - k_m}. \end{aligned}$$

(27)

for $m \ge j+1$ and $P_i(M=m/S_{n-j}) = 0$ for $m < j+1$. Next:

$$\begin{aligned} P(L_i= & {} l_i | M = m, N_i = n-j) \nonumber \\= & {} \frac{P(L_i = l_i, M = m| N_i = n-j)}{P(M=m|N_i = n-j)} \nonumber \\= & {} \frac{P(M=m|L_i = l_i, N_i = n-j)P(L_i = l_i| N_i = n-j)}{P(M=m|N_i = n-j)} \end{aligned}$$

(28)

Note that closed form expressions for $P(M=m|L_i = l_i, N_i = n-j)$ and $P(M=m|N_i = n-j)$ can be computed similar to the computation of $P_i(M=m/S_{n-j})$ in (27). Also, clearly:

$$\begin{aligned} P(L_i = l_i| N_i = n-j) = \left( \begin{array}{c} n-j \\ l_i \\ \end{array} \right) p_{n-j}^{l_i} (1 - p_{n-j})^{n-j-l_i}. \end{aligned}$$

(29)

Finally, by using (7), (24)−(29), we get an expression for $E(L_i/M=m)$.

Proof of Proposition 3

In DL slots, contending nodes are in reception state and the rest are in idle state. So $\mathscr {E}_{DL} = \sum _{i=1}^{W_M} \big (N_i \gamma _R + (n-N_i)\gamma _I\big )$. Taking expectations and conditioning on the values taken by M, we get the expression given in (9).

$\square$

Computation of An Expression for$E(N_i/M=m)$:

By definition of conditional expectation, $E(N_i/M=m) = \sum _{j=0}^m (n-j) P(N_i = n-j/M=m)$. Also, $P(N_i = n-j/M=m)$ is given by (25).