On optimizing CSMA for wide area ad hoc networks

Abstract

The recent deployment of data-rich smart phones has led to a fresh impetus for understanding the performance of wide area ad hoc networks. The most popular medium access mechanism for such ad hoc networks is CSMA/CA with RTS/CTS. In CSMA-like mechanisms, spatial reuse is achieved by implementing energy-based guard zones. We consider the problem of simultaneously scheduling the maximum number of links that can achieve a given signal to interference ratio (SIR). In this paper, using tools from stochastic geometry, we study and maximize the medium access probability of a typical link. Our contributions are two-fold: (i) We show that a simple modification to the RTS/CTS mechanism, viz., changing the receiver yield decision from an energy-level guard zone to an SIR guard zone, leads to performance gains; and (ii) We show that this combined with a simple modification to the transmit power level—setting it inversely proportional to the square root of the link gain—leads to significant improvements in network throughput. Further, this simple power-level choice is no worse than a factor of two away from optimal over the class of all “local” power level selection strategies for fading channels, and further is optimal in the non-fading case. The analysis relies on an extension of the Matérn hard core point process which allows us to quantify both these SIR guard zones and this power control mechanism.

Appendix

1.1 A.1 Proofs of the theorems of Sect. 4

1.1.1 A.1.1 Proof of Lemma 1

In this section we prove Lemma 1 by studying the basic properties of the area of the union of two disks, which have the following relations: the centers of the two disks are at an arbitrary but fixed distance d and the radii of the two disks r ₁ and r ₂ satisfy the relation r ₁ r ₂=k. Note that without loss of generality, we can take k=1 and define the two radii to be hence x and $\frac{1}{x}$, as shown in Fig. 12. For any other value of k, we can always scale the problem by a factor of $\frac{1}{\sqrt{k}}$ and change d appropriately. We define U _d (x) to be the area of the union of the two disks defined above, with a distance d between the two centers and x being one of the radii.

Fig. 12

Union of two circles with centers of distance d away and having radius x and $\frac{1}{x}$

There is always a set of values X(d) for which the two boundary circles of respective radii x and 1/x intersect. Consider first that the set where 1 is an interior point of X(d). By elementary geometric arguments, is easy to see that for all x∈X(d),

$$U_d(x) = (\pi-\theta_1)x^2 + (\pi-\theta_2)\frac{1}{x^2} + \sin(\theta_3), $$

(35)

where θ ₁, θ ₂ and θ ₃ are defined as in Fig. 12 and where θ ₁, θ ₂ and θ ₃ are determined by the following equations:

(36)

(37)

(38)

Based on these expressions, we are now ready to prove Lemma 1.

Consider first the case where 1 is an interior point of X(d), which holds if and only if d<2. It is easy to see that it then suffices to prove the following conditions on the first derivative of U _d (x) to conclude the proof:

• (1) $U_{d}'(1)=0$;

• (2) $U_{d}'(x) > 0$ for any x>1, and $U_{d}'(x)<0$ for any x<1.

Indeed, in this case, if x∉X(d), then the smaller disk is included in the larger and simple continuity, monotonicity and symmetry arguments show that the minimum cannot be reached in this region.

In order to prove (1) and (2), we use (35) to obtain the following expression for $U_{d}'(x)$:

$$U_d'(x) = 2(\pi-\theta_1)x-2(\pi-\theta_2)\frac{1}{x^3} - \theta_{1x}'x^2-\theta_{2x}'\frac{1}{x^2}+\cos(\theta_3)\theta_{3x}',$$

where $\theta_{ix}'$ (i=1,2,3) denotes the first derivative of θ _i with respect to x. Now we show that the last four terms in the above expression cancel each other. Since

$$\theta_3=\pi-\theta_1-\theta_2,$$

we have

$$\theta_3'=-\theta_1'-\theta_2'.$$

Thus, we obtain

On the other hand, it is easy to check that

Since θ ₁ and θ ₂ also satisfies

$$\frac{\sin(\theta_1)}{\frac{1}{x}} = \frac{\sin(\theta_2)}{x},$$

we have

$$- \theta_{1x}'x^2-\theta_{2x}'\frac{1}{x^2}+\cos(\theta_3)\theta_{3x}' =0,$$

and thus

$$U_d'(x) = 2(\pi-\theta_1)x-2(\pi-\theta_2)\frac{1}{x^3}. $$

(39)

This is a rather simple expression which has a simple geometrical interpretation. As shown in Fig. 13, (39) simply indicates that the first derivative of the union of the two disks is determined by the difference of the length of the two non-intersecting arcs, since perturbing x causes one disk to expand its area while the other disk reduces. The increase and the decreased are mainly determined by the lengths of the arcs while the intersecting part of the arcs does not contribute to the first order derivative. From here, we can easily see that

$$U_d'(1) = 0$$

since θ ₁=θ ₂ in that case. When x>1, we can bound $U_{d}'(x)$ as in the following:

$$U_d'(x) > 2(\pi-\theta_1)-2(\pi-\theta_2) = 2(\theta_2-\theta_1) > 0.$$

Similarly, we can conclude $U_{d}'(x)<0$ for x<1, which completes the proof in this case.

Fig. 13

First derivative of the area of the union of two disks

If d>2, then, for x in a neighborhood 1, the area of the union is the sum of the areas and the minimum in this region is reached for x=1. Outside this region, simple continuity and monotonicity arguments based on the evaluation of $U_{d}'(x)$ for x∈X(d) allow one to show that the minimum cannot be found there. Similarly, it cannot be found in the region where one disk is included in the other and this concludes the proof.

1.1.2 A.1.2 Proof of Theorem 2

For notational brevity, we denote by $P_{Z}^{*}=f^{*}(H_{Z})$ the optimizing transmit power law for (16). To obtain the MAP of a link with link length d we can again place the transmitter of the link at the origin with a link vector D ₀ with length d and with an arbitrary orientation. Under fading the MAP is further averaged over the fading state of the link F ₀, i.e.

$$\mbox{MAP}(d) = \int\mbox{MAP}(d,x)\, dP(F_0\le x),$$

with MAP(d,x) the MAP of a link with length |D ₀|=d and fading F ₀=x. So, to obtain (16), it suffices to show that for any fading state x

$$\mbox{MAP}_{\mathrm{ISR}}(d,x) \ge\frac{1}{2}\sup_{f:\mathbb {R}^+\rightarrow\mathbb{R}^+}\mbox{MAP}_{P_Z=f(H_Z)}(d,x).$$

Given the link vector d and fading state x the number of neighbors of link l ₀ is a Poisson random variable and, by arguments similar to those used in the proof of (7), we get

$$\mbox{MAP}_{\mathrm{ISR}}(d,x) = \frac{1-e^{-\bar{N}(d,x)}}{\bar{N}(d,x)},$$

where $\bar{N}(d,x)$ is the mean number of neighbors of a typical link with length d and fading x. From here, we notice that it suffices to show, for any given d and x,

$$\bar{N}_{P_Z^*}(d,x) \le\bar{N}_{\mathrm{ISR}}(d,x) \le2\bar {N}_{P_Z^*}(d,x), $$

(40)

since we can bound MAP_ISR(d,x) as follows:

$$\mbox{MAP}_{\mathrm{ISR}}(d,x) = \frac{1-e^{-\bar{N}_{\mathrm {ISR}}(d,x)}}{\bar{N}_{\mathrm{ISR}}(d,x)} \ge\frac{1-e^{-\bar {N}_{P_Z^*}(d,x)}}{2\bar{N}_{P_Z^*}(d,x)}.$$

We now derive an expression for $\bar{N} (d,x)$. For notational brevity, we remove the (d,x) argument in $\bar{N}$, MAP and other related quantities. We have

(41)

Since F _0,Z and F _Z,0 are i.i.d. random variables, we can further simplify (41) as

(42)

which can be interpreted as the area of union of two disks, averaged over four independent random variables D _Z ,F _Z ,F _0,Z and F _Z,0. We first average over F _0,Z and $F_{Z_{0}}$ and keep D _Z and F _Z fixed for now. With D _Z and F _Z fixed, the radii of the two disks have the similar property as in the non-fading case, except that they are further perturbed by two i.i.d. random variables F _0,Z and F _Z,0. For an arbitrary realization (F _0,Z ,F _Z,0)=(f ₁,f ₂), (f ₁≠f ₂), the area of the union of the disks is not going to be minimized by ISR, which fails to create the same radius due to the asymmetric perturbations. However, it is easy to see that (F _0,Z ,F _Z,0)=(f ₁,f ₂) and (F _0,Z ,F _Z,0)=(f ₂,f ₁) have equal probability due to the i.i.d. property. We thus consider the average of these two fading states and hope that ISR minimizes the average area of the unions of disks of the two fading states.

Thus, we start from the expression in (42) and calculate the average area of union of two disks in the two fading states when (F _Z,0,F _0,Z )=(f ₁,f ₂) and (F _Z,0,F _0,Z )=(f ₂,f ₁)

(43)

(44)

where U _d (x) is defined as in (35) and c,d,t,x are determined as follows:

ISR enforces x=1, which is not necessarily the optimizing x for the sum of the areas in (43). Next, we show that for any choice of d>0 and t>0, the following relationship is true:

$$U_d(t)+U_d\biggl(\frac{1}{t}\biggr) \le2\inf_{x>0} U_d(tx)+U_d\biggl(\frac{x}{t}\biggr).$$

To see this, we define

$$V(x) = \pi x^2 + \pi\frac{1}{x^2},$$

which is the sum of the area of two disks with radii x and $\frac{1}{x}$. So for any d>0, we have

$$U_d(x) \le V(x) \le2U_d(x),$$

where the inequality on the right comes from the fact that

$$U_d(x)\ge\pi x^2 \quad\mathrm{and}\quad U_d(x) \ge\pi\frac{1}{x^2},$$

which leads to

$$V(x)=\pi x^2+\pi\frac{1}{x^2}\le2U_d(x). $$

(45)

Thus, we have

$$V(tx)+V\biggl(\frac{x}{t}\biggr) \le2\biggl(U_d(tx)+U_d\biggl(\frac{x}{t}\biggr)\biggr), \quad\forall x>0,$$

which leads to

(46)

(47)

(48)

Here the equality in (47) comes from the fact that V(x) is convex and it is easy to check $V'(tx)+V'(\frac{x}{t})|_{x=1}=0$. Thus, x=1 is the minimizing x for $V(tx)+V(\frac{x}{t})$.

Thus, the integrand in (44) at most loses a fraction of $\frac{1}{2}$ when we use ISR, or equivalently, letting x=1, as compared to any other choices of x. It is easy to see that given this, the relationship in (40) is indeed true, which completes the proof.

1.2 A.2 Proofs of the theorems of Sect. 5

1.2.1 A.2.1 Proof of Theorem 3

The links emanating from the points Z and 0 are not neighbors iff

$$|Z+D_Z|^2 > \varepsilon_T^{-\frac{2}{\alpha}}\quad\mbox{and}\quad |Z-D_0|^2 > \varepsilon_T^{-\frac{2}{\alpha}}.$$

The probability that the points Z and 0 are not neighbors is

(49)

The random vector D _Z +Z is Gaussian with independent coordinates. If Z=(x,y), the first coordinate of D _Z +Z has mean x and variance σ ². Its second coordinate has mean y and variance σ ². Therefore, |D _Z +Z|²/σ ² follows a non centered χ ² distribution with 2 degrees of freedom and with parameter

$$\theta_{|Z|}= \frac{|Z|^2}{\sigma^2},$$

that is, with density

$$f_{|Z|}(s)= \sum_{i=0}^{\infty}e^{-\theta_{|Z|}/2} \frac{(\theta_{|Z|}/2)^i}{i!} g_{2(i+1)}(s)$$

on ℝ⁺, where g _q (s) is the density of the centered χ ² distribution with q degrees of freedom. The tail distribution function of the density f _|Z| is

$$\overline{F}_{|Z|} (s)= \sum_{i=0}^{\infty}e^{-\theta_{|Z|}/2} \frac {(\theta_{|Z|}/2)^i}{i!} \frac{\varGamma(s/2,i+1)}{\varGamma(i+1)},$$

and the expression for C(|Z|) follows.

The conditional probability that the points Z and 0 are not neighbors is

(50)

Let Z=(x,y)=(Rcos(θ),Rsin(θ)), −D ₀=(rcos(ψ),rsin(ψ)). Given Z and given that |D ₀|=r, the condition $|Z-D_{0}|^{2} > \varepsilon_{T}^{-\frac{2}{\alpha}}$ can be rewritten as

$$R^2 +r^2 +2 r R \cos(\theta-\psi) >\varepsilon_T^{-\frac{2}{\alpha}}$$

or equivalently as

Hence, since ψ is uniform on [0,2π], we get

$$P\bigl(|Z-D_0|^2 > \varepsilon_T^{-\frac{2}{\alpha}}\mid|D_0|\bigr)= \frac{1}{\pi}\arccos \biggl(\frac{\varepsilon_T^{-\frac{2}{\alpha }}-r^2-R^2}{2rR}\biggr)$$

and the formula of C(R,r) follows.

1.2.2 A.2.2 Proof of Theorem 4

In the SIR guard zones case without fading and without power control, the links emanating from the points Z and 0 are not neighbors iff

$$|Z+D_Z|^2 > a |D_Z|^2 \quad \mbox{and}\quad |Z-D_0|^2 > a |D_0|^2,$$

(51)

with $a=\gamma_{T}^{2/\alpha}$. Let Z=(x,y) and D _Z =(u,v). The condition |Z+D _Z |²>a|D _Z |² can be rewritten as

$$u^2(1-a) +2xu + x^2 +v^2(1-a) +2yv +v^2 >0.$$

If a<1, this is equivalent to

$$ \biggl(u + \frac{x}{1-a} \biggr)^2 + \biggl(v +\frac{y}{1-a} \biggr)^2 > \bigl(x^2+y^2\bigr)\frac{a}{(1-a)^2}.$$

(52)

If a>1, this is equivalent to

$$ \biggl(u + \frac{x}{1-a} \biggr)^2 + \biggl(v +\frac{y}{1-a} \biggr)^2 < \bigl(x^2+y^2\bigr)\frac{a}{(1-a)^2}.$$

(53)

Finally, if a=1, this is equivalent to

$$ 2xu+2yv+x^2+y^2 > 0.$$

(54)

Hence, the probability that the points Z and 0 are not neighbors is

(55)

If a<1,

(56)

where the first relation follows from the independence assumptions and the second from (52).

Similarly, if a>1,

(57)

whereas if a=1

(58)

with a⋅b denoting the scalar product of the vectors a and b in ℝ².

In the case a≠1, the random vector D _Z +1/(1−a)Z is Gaussian with independent coordinates. If Z=(x,y), its first coordinate has mean x/(1−a) and variance σ ². Its second coordinate has mean y/(1−a) and variance σ ². Therefore, |D _Z +1/(1−a)Z|²/σ ² follows a non-centered χ ² distribution with 2 degrees of freedom and with parameter

$$\widehat{\theta}_{|Z|}= \frac{|Z|^2}{(1-a)^2 \sigma^2},$$

that is, with density $\widehat{f_{|Z|}}(s)$ and tail distribution function $\overline{\widehat{F_{|Z|}}} (s)$ as given above. The expressions for C(|Z|) when a≠1 then follow.

In the case a=1, the random variable 2Z⋅D _Z ∈ℝ is Gaussian, centered and of variance 4(x ²+y ²)σ ². Hence, when denoting by A a $\mathcal{N}(0,1)$ random variable,

$$\phi\bigl({|Z|}\bigr)=P\bigl(2 \sigma|Z| A> -|Z|^{2}\bigr)=Q \biggl(-\frac{|Z|}{2\sigma } \biggr).$$

The conditional probability that the points Z and 0 are not neighbors is

(59)

Let Z=(x,y)=(Rcos(θ),Rsin(θ)), −D ₀=(rcos(ψ),rsin(ψ)). The condition |Z−D ₀|²>a|D ₀|² can be rewritten as

$$R^2 +r^2 +2 r R \cos(\theta-\psi) > ar^2$$

or equivalently as

The formula for C(R,r) follows.

1.2.3 A.2.3 Proof of Theorem 5

The links emanating from the points Z and 0 are not neighbors iff

$$|Z+D_Z|^2 > a |D_Z| |D_0|\quad\mbox{and}\quad |Z-D_0|^2 > a |D_0||D_Z|,$$

with $a=\gamma_{T}^{2/\alpha}$. Let Z=(x,y)=(Rcos(θ),Rsin(θ)), D _Z =(scos(ϕ),ssin(ϕ)) and D ₀=(u,v)=(rcos(ψ),rsin(ψ)). The above condition can be rewritten as

or equivalently as

and

The expression for C(|Z|) follows.

The conditional probability that the points Z and 0 are not neighbors is

(60)

Let Z=(x,y)=(Rcos(θ),Rsin(θ)), −D ₀=(rcos(ψ),rsin(ψ)). The condition |Z−D ₀|²>a|D ₀|² can be rewritten as

$$R^2 +r^2 +2 r R \cos(\theta-\psi) > ar^2$$

or equivalently as

The expression for C(|Z|,r) follows.

1.2.4 A.2.4 Proof of Theorem 6

The links emanating from the points Z and 0 are not neighbors iff

$$|Z+D_Z|^2 > \biggl(\frac{F_{0,Z}}{ \varepsilon_T}\biggr)^{\frac{2}{\alpha}} \quad\mbox {and}\quad |Z-D_0|^2 >\biggl(\frac{F_{Z,0}}{ \varepsilon_T} \biggr)^{\frac{2}{\alpha}}.$$

Hence, the probability that the points Z and 0 are not neighbors is

(61)

The random variable |D _Z +Z|² follows a non centered χ ² with two degrees of freedom and with parameter

$$\theta_{|Z|}= \frac{|Z|^2}{\sigma^2}.$$

The random variable F _0,Z is exponential with parameter 1. Hence the probability of collisions of the two links is

$$ C\bigl(|Z|\bigr)= 1 - \phi\bigl({|Z|}\bigr)^2$$

(62)

with

1.2.5 A.2.5 Proof of Theorem 7

The links emanating from the points Z and 0 are not neighbors iff

Hence, the probability that the points Z and 0 are not neighbors is

(63)

By the same arguments as above,

Hence, if |Z|=R,

with ϕ given as in the theorem.

1.2.6 A.2.6 Proof of Theorem 8

The links emanating from the points Z and 0 are not neighbors iff

Hence, the probability that the points Z and 0 are not neighbors is

(64)

By the same arguments as above, and using the fact that the ratio of two independent exponential random variables of parameter 1 has the density $\frac{1}{(1+x)^{2}}$, we see that the last expression can be evaluated as

Using the same approach as in (29), we get

with ψ(R,a,b) as given in the theorem.