Opportunistic Spectrum Access Through Cognitive Radio

Abstract

As mentioned above, CR is envisioned as a promising approach to deal with the spectrum scarcity in wireless communications, which enables unlicensed users to opportunistically exploit the spectrum owned by licensed users [1, 2]. In CR-VANETs, considering the highly dynamic mobility, VUs are expected to exploit more spatial and temporal spectrum opportunities along the road than stationary SUs.

Appendix

3.1.1 NE Condition for Uniform MAC

First, the situation for two channels is considered. Since C = 2 and $N \in \mathbb{Z}^{+}$, we have

$$\displaystyle\begin{array}{rcl} \frac{\varPsi _{1}} {n_{1}} \geq \frac{\varPsi _{2}} {n_{2} + 1}\ \text{and}\ \frac{\varPsi _{2}} {n_{2}} \geq \frac{\varPsi _{1}} {n_{1} + 1},& &{}\end{array}$$

(3.35)

which can be rewritten as follows:

$$\displaystyle\begin{array}{rcl} \frac{\varPsi _{1}} {\varPsi _{2}}n_{2} - 1 \leq n_{1} \leq \frac{\varPsi _{1}} {\varPsi _{2}}n_{2} + \frac{\varPsi _{1}} {\varPsi _{2}}.& &{}\end{array}$$

(3.36)

Substitute $n_{2} = N - n_{1}$ into (3.36), we obtain

$$\displaystyle\begin{array}{rcl} \frac{\varPsi _{1}N -\varPsi _{2}} {\varPsi _{1} +\varPsi _{2}} \leq n_{1} \leq \frac{\varPsi _{1}N +\varPsi _{1}} {\varPsi _{1} +\varPsi _{2}}.& &{}\end{array}$$

(3.37)

Since

$$\displaystyle\begin{array}{rcl} \frac{\varPsi _{1}N +\varPsi _{1}} {\varPsi _{1} +\varPsi _{2}} -\frac{\varPsi _{1}N -\varPsi _{2}} {\varPsi _{1} +\varPsi _{2}} = 1& &{}\end{array}$$

(3.38)

and

$$\displaystyle\begin{array}{rcl} -1 < \frac{\varPsi _{1}N -\varPsi _{2}} {\varPsi _{1} +\varPsi _{2}} < N& &{}\end{array}$$

(3.39)

Γ has at least one pure NE, in which

$$\displaystyle\begin{array}{rcl} n_{1} = \lceil \frac{\varPsi _{1}N -\varPsi _{2}} {\varPsi _{1} +\varPsi _{2}} \rceil \ \text{and}\ n_{2} = N - n_{1}.& &{}\end{array}$$

(3.40)

Next, we extend this conclusion to the situation where more than two channels are available, i.e., C > 2. When C > 2, any two arbitrary channels i and k, $i,k \in \mathcal{C}$ should satisfy (3.36) to constitute an NE. Thus,

$$\displaystyle\begin{array}{rcl} \frac{\varPsi _{k}} {\varPsi _{i}} n_{i} - 1 \leq n_{k} \leq \frac{\varPsi _{k}} {\varPsi _{i}} n_{i} + \frac{\varPsi _{k}} {\varPsi _{i}}.& &{}\end{array}$$

(3.41)

Define F _L, ki and F _U, ki as

$$\displaystyle\begin{array}{rcl} F_{L,ki} = \frac{\varPsi _{k}} {\varPsi _{i}} n_{i} - 1\ \text{and}\ F_{U,ki} = \frac{\varPsi _{k}} {\varPsi _{i}} n_{i} + \frac{\varPsi _{k}} {\varPsi _{i}}.& &{}\end{array}$$

(3.42)

Then, for channels i and {∀}k ≠ i, $i,k \in \mathcal{C}$, we have

$$\displaystyle\begin{array}{rcl} F_{L,ki} \leq n_{k} \leq F_{U,ki}.& &{}\end{array}$$

(3.43)

It holds that

$$\displaystyle\begin{array}{rcl} \sum _{k\neq i,k\in \mathcal{C}}F_{L,ki} \leq \sum _{k\neq i,k\in \mathcal{C}}n_{k} \leq \sum _{k\neq i,k\in \mathcal{C}}F_{U,ki}.& &{}\end{array}$$

(3.44)

By substituting $\sum _{k\neq i,k\in \mathcal{C}}n_{k} = N - n_{i}$ into (3.44), we have

$$\displaystyle\begin{array}{rcl} \frac{\varPsi _{i}N -\sum _{k\neq i,k\in \mathcal{C}}\varPsi _{k}} {\sum _{k\in \mathcal{C}}\varPsi _{k}} \leq n_{i} \leq \frac{\varPsi _{i}N +\varPsi _{i}(C - 1)} {\sum _{k\in \mathcal{C}}\varPsi _{k}}.& &{}\end{array}$$

(3.45)

Similar to (3.38) and (3.39), it can be proved that

$$\displaystyle\begin{array}{rcl} \frac{\varPsi _{i}N +\varPsi _{i}(C - 1)} {\sum _{k\in \mathcal{C}}\varPsi _{k}} -\frac{\varPsi _{i}N -\sum _{k\neq i,k\in \mathcal{C}}\varPsi _{k}} {\sum _{k\in \mathcal{C}}\varPsi _{k}} > 1& &{}\end{array}$$

(3.46)

and

$$\displaystyle\begin{array}{rcl} -1 < \frac{\varPsi _{i}N -\sum _{k\neq i,k\in \mathcal{C}}\varPsi _{k}} {\sum _{k\in \mathcal{C}}\varPsi _{k}} < N.& &{}\end{array}$$

(3.47)

Then, for any $\mathcal{C}$ and $\mathcal{N}$, (3.18) has at least one solution, which is

$$\displaystyle\begin{array}{rcl} n_{i} = \lceil \frac{\varPsi _{i}N -\sum _{k\neq i,k\in \mathcal{C}}\varPsi _{k}} {\sum _{k\in \mathcal{C}}\varPsi _{k}} \rceil + W_{0},& &{}\end{array}$$

(3.48)

where $W_{0} \in \{ 0,1,2,\ldots,\lceil \frac{\varPsi _{i}N+\varPsi _{i}(C-1)} {\sum _{k\in \mathcal{C}}\varPsi _{k}} \rceil -\lceil \frac{\varPsi _{i}N-\sum _{k\neq i,k\in \mathcal{C}}\varPsi _{k}} {\sum _{k\in \mathcal{C}}\varPsi _{k}} \rceil - 1\}$. With $\sum _{i\in \mathcal{C}}n_{i} = N$, we have (3.24). Thus, the game Γ has at least one pure NE (3.24) is called NE condition of the spectrum access game Γ when uniform MAC is used.

3.1.2 Proof of Proposition 2

Assume that for a given round R _t , the congestion vector $\mathbf{n}(S_{t}) =\{ n_{1},n_{2},\ldots,n_{C}\}$ composes a pure NE. According to (3.18), for each channel $i \in \mathcal{C}$,

$$\displaystyle\begin{array}{rcl} \varPsi _{i}r(n_{i}) \geq \varPsi _{k}r(n_{k} + 1),\ \forall k \in \mathcal{C},k\neq i.& &{}\end{array}$$

(3.49)

Then for a new round R _t+1, a new vehicle joins the game and chooses its best response according to the existing strategy profile, i.e., n(S _t ). Consider its best response is channel m, and thus the new congestion vector is $\mathbf{n}(S_{t+1}) =\{ n_{1},\ldots,n_{m} + 1,\ldots,n_{C}\}$. For the new congestion vector, we have the following observations:

• 1. For each channel $i \in \mathcal{C},i\neq m$, $\varPsi _{i}r(n_{i}) \geq \varPsi _{k}r(n_{k} + 1),\ \forall k \in \mathcal{C}\setminus \{i,m\}$ holds because the number of vehicles that choose the channels other than channel m does not change, and r(n _i ), i ≠ m remains unchanged.

• 2. $\varPsi _{m}r(n_{m} + 1) \geq \varPsi _{k}r(n_{k} + 1),\forall k \in \mathcal{C},k\neq m$. This statement holds due to that channel m is the best response for the new vehicle.

• 3. $\varPsi _{k}r(n_{k}) \geq \varPsi _{m}r(n_{m} + 1 + 1),\forall k \in \mathcal{C},k\neq m$. Remember in round t, Ψ _k r(n _k ) ≥ Ψ _m r(n _m + 1). r(n) is a non-increasing function, and thus $r(n_{m} + 1) \geq r(n_{m} + 1 + 1)$.

Therefore, n(S _t+1) also constitutes a pure NE. For a specific game, the first vehicle chooses the channel with largest ECA and of course composes a pure NE. Then, for each round, the strategies of vehicles which have participated in the game constitute a new pure NE, until all vehicles have chosen their strategies.

3.1.3 Proof of Corollary 3.3.1

For any round in Proposition 2, assume that the congestion vector $\mathbf{n}(S) =\{ n_{1},n_{2},\ldots,n_{C}\}$ constitutes a pure NE and Ψ _i is sorted so that $\varPsi _{1} \geq \varPsi _{2} \geq \ldots \geq \varPsi _{C}$. The efficiency of the NE is

$$\displaystyle\begin{array}{rcl} \mathcal{E}_{S} =\sum _{ i=1}^{C}f(n_{ i}).& &{}\end{array}$$

(3.50)

Remember that in slotted ALOHA, $f(n) = (1 - \frac{1} {n})^{n-1}$. A new vehicle comes and finds there are more than one best response (BR).

• 1. If BR₁ corresponds to a free channel i when BR₂ corresponds to channel j that has been selected by at least one vehicle, then BR₁ leads to a NE with efficiency:

  $$\displaystyle\begin{array}{rcl} \mathcal{E}_{S1} = \mathcal{E}_{S} +\varPsi _{i} > \mathcal{E}_{S}.& & {}\end{array}$$

  (3.51)

  BR₂ leads to a NE with efficiency:

  $$\displaystyle\begin{array}{rcl} \mathcal{E}_{S2} = \mathcal{E}_{S}-\varDelta < \mathcal{E}_{S}.& & {}\end{array}$$

  (3.52)

  where Δ is the loss of f(n _j ) since f(n) decreases with n. Obviously, $\mathcal{E}_{S1} > \mathcal{E}_{S2}$.

• 2. Consider that BR₁ and BR₂ correspond to channel i and j with n _i  ≥ 1 and n _j  ≥ 1, respectively. Without loss of generality, consider Ψ _i  > Ψ _j . Under this condition, it is clear that n _i  > n _j , or else channel i and j cannot be the best response simultaneously. Consider only the total utility of users choosing channel i and j since other channels are not affected in this round. BR₁ will lead to an NE with utility $\mathcal{E}_{1} =\varPsi _{i}f(n_{i} + 1) +\varPsi _{j}f(n_{j})$, while BE₂ will lead to an NE with utility $\mathcal{E}_{2} =\varPsi _{i}f(n_{i}) +\varPsi _{j}f(n_{j} + 1)$. Using the property of the pure NE, we have $\varPsi _{i}r(n_{i} + 1) \geq \varPsi _{j}r(n_{j} + 1)$ and $\varPsi _{j}r(n_{j} + 1) \geq \varPsi _{i}r(n_{i} + 1)$, and thus $\varPsi _{i}r(n_{i} + 1) =\varPsi _{j}r(n_{j} + 1)$, i.e., $\varPsi _{i}\frac{f(n_{i}+1)} {n_{i}+1} =\varPsi _{j}\frac{f(n_{j}+1)} {n_{j}+1}$. Let

  $$\displaystyle\begin{array}{rcl} \varPsi _{i} = \frac{\frac{f(n_{j}+1)} {n_{j}+1} } {\frac{f(n_{i}+1)} {n_{i}+1} } \varPsi _{j} =\alpha \varPsi _{j}.& & {}\end{array}$$

  (3.53)

To prove

$$\displaystyle\begin{array}{rcl} \begin{array}{ll} \mathcal{E}_{1} -\mathcal{E}_{2} & =\varPsi _{i}f(n_{i} + 1) +\varPsi _{j}f(n_{j}) - (\varPsi _{i}f(n_{i}) +\varPsi _{j}f(n_{j} + 1)) \\ & =\varPsi _{j}[\alpha (f(n_{i} + 1) - f(n_{i})) + f(n_{j}) - f(n_{j} + 1)] > 0,\end{array} & &{}\end{array}$$

(3.54)

is equivalent to prove

$$\displaystyle\begin{array}{rcl} \alpha = \frac{\frac{f(n_{j}+1)} {n_{j}+1} } {\frac{f(n_{i}+1)} {n_{i}+1} } < \frac{f(n_{j}) - f(n_{j} + 1)} {f(n_{i}) - f(n_{i} + 1)},& &{}\end{array}$$

(3.55)

since $f(n) - f(n + 1) > 0$.

$$\displaystyle\begin{array}{rcl} & & \frac{\frac{f(n_{j}+1)} {n_{j}+1} } {\frac{f(n_{i}+1)} {n_{i}+1} } < \frac{f(n_{j}) - f(n_{j} + 1)} {f(n_{i}) - f(n_{i} + 1)} \\ & & \qquad \Leftrightarrow \frac{\frac{f(n_{j}+1)} {n_{j}+1} } {f(n_{j}) - f(n_{j} + 1)} < \frac{\frac{f(n_{i}+1)} {n_{i}+1} } {f(n_{i}) - f(n_{i} + 1)} \\ & & \qquad \Leftrightarrow g(n) = \frac{\frac{f(n+1)} {n+1} } {f(n) - f(n + 1)}\text{ increasing with }n \geq 1 \\ & & \qquad \Leftrightarrow g'(n) > 0,\text{ when }n \geq 1. {}\end{array}$$

(3.56)

We skip the tedious proof of (3.56) to simplify the exposition. Then, we have $\mathcal{E}_{1} > \mathcal{E}_{2}$.