An asymptotic approximation for TCP compound

Abstract

In this paper, we derive an approximation for throughput of TCP Compound connections under random losses. Throughput expressions for TCP Compound under a deterministic loss model exist in the literature. These are obtained assuming that the window sizes are continuous, i.e., a fluid behavior is assumed. We validate this model theoretically. We show that under the deterministic loss model, the TCP window evolution for TCP Compound is asymptotically periodic and is independent of the initial window size. We then consider the case when packets are lost randomly and independently of each other. We discuss Markov chain models to analyze performance of TCP in this scenario. We use insights from the deterministic loss model to get an appropriate scaling for the window size process and show that these scaled processes, indexed by p, the packet error rate, converge to a limit Markov chain process as p goes to 0. We show the existence and uniqueness of the stationary distribution for this limit process. Using the stationary distribution for the limit process, we obtain approximations for throughput, under random losses, for TCP Compound when packet error rates are small. We compare our results with ns2 simulations which show a good match and a better approximation than the fluid model at low p.

Appendix

Proposition 8

Suppose $\{X(p)\}$ is a family of positive random variables indexed by $p \in I$, for some $I \subset {\mathbb {R}}$ such that

$$\begin{aligned} {\mathbb {P}}(X(p) \ge y) \le c_1 \exp (-c_2 y), \end{aligned}$$

for all y, for some $c_1 > 0$ and $c_2 > 0$. Then there exists $t>0$ such that

$$\begin{aligned} \sup _{u \in (-t,t)} \sup _{p \in I} {\mathbb {E}}\left[ \mathrm{e}^{uX(p)}\right] < \infty . \end{aligned}$$

Proof

For any $u > 0$ and $p > 0$,

$$\begin{aligned} \begin{aligned} {\mathbb {E}}[\mathrm{e}^{uX(p)}]&= \int _{0}^{\infty } {\mathbb {P}}\left( \mathrm{e}^{uX(p)} \ge y\right) \mathrm{d}y \le 1 + \int _{1}^{\infty } {\mathbb {P}}(uX(p) \ge \log (y)) \mathrm{d}y \\&\le 1 + \int _{1}^{\infty } c_1 \exp \left( - \frac{c_2}{u} \log (y)\right) \mathrm{d}y = 1 + \int _{1}^{\infty } \frac{c_1}{y^{\frac{c_2}{u}}} \mathrm{d}y. \end{aligned} \end{aligned}$$

For $u < c_2$, $\int _{1}^{\infty } \frac{c_1}{y^{\frac{c_2}{u}}} < \infty$. Therefore $\sup _{u \in [0,c_2 - \epsilon ]} \sup _{p \in I} {\mathbb {E}}[\mathrm{e}^{uX(p)}] < \infty$, where $\epsilon$ is a positive constant less than $c_2$. For $u<0$, $uX(p) \le 0$ which implies $\sup _{p \in I} {\mathbb {E}}[\mathrm{e}^{uX(p)}] \le 1$, for $u < 0$. $\square$

Proposition 9

Let $\{X_p(x), x \in {\mathbb {R}}^+\}$, be a process, for $0< p < 1$, which converges to a limiting process X(x) uniformly in the sense that

$$\begin{aligned} \lim _{p \rightarrow 0} \sup _{x, y \le M} \Bigl | {\mathbb {P}}(X_p(x) \le y) - {\mathbb {P}}(X(x) \le y) \Bigr | = 0, \end{aligned}$$

(34)

for any finite M, and for each x, the limiting distribution, ${\mathbb {P}}(X(x) \le y)$ is continuous. Then,

$$\begin{aligned} \lim _{p \rightarrow 0} \sup _{x \le M} \Bigl | {\mathbb {E}}f(X_p(x)) - {\mathbb {E}}f(X(x)) \Bigr | = 0, \end{aligned}$$

for any $f:\mathbb {R^+}\rightarrow {\mathbb {R}}$ continuous with compact support.

Proof

Consider a continuous function f with compact support [0, K]. Such a function is uniformly continuous. Therefore, given any $\epsilon$, there exist m points $u_0 = 0< u_1< \cdots < u_m = K$, such that

$$\begin{aligned} \sup _{u_i< y< u_{i+1}} |f(y) - f(u_i)| < \epsilon , \end{aligned}$$

(35)

for all $i=1,2,\cdots , m$. We have

$$\begin{aligned} E[f(X_p(x)] = \int \limits _{0}^{K} f(u) {\mathbb {P}}(X_p(x) \in du). \end{aligned}$$

From (35),

$$\begin{aligned} \left| E[f(X_p(x)] - \sum \limits _{i=1}^{m-1} f(u_i) {\mathbb {P}}(X_p(x) \in (u_i,u_{i+1}]) \right| \le \epsilon . \end{aligned}$$

Similarly,

$$\begin{aligned} \begin{aligned} \left| E[f(X(x)] - \sum \limits _{i=1}^{m-1} f(u_i) {\mathbb {P}}(X(x) \in (u_i,u_{i+1}])\right| \le \epsilon . \end{aligned} \end{aligned}$$

Therefore,

$$\begin{aligned} \begin{aligned} \bigl |{\mathbb {E}}f(X_p(x) - {\mathbb {E}}f(X(x)) \bigr |&\le \sum \limits _{i=1}^{m} f(u_i) \bigl | {\mathbb {P}}(X_p(x) \in (u_i,u_{i+1}]) \\&\quad - {\mathbb {P}}(X(x) \in (u_i,u_{i+1}])\bigr | + 2 \epsilon \\&\le \sum \limits _{i=1}^{m} \parallel f \parallel _{\infty } \Bigl |{\mathbb {P}}(X_p(x) \in (u_i,u_{i+1}]) \\&\quad - {\mathbb {P}}(X(x) \in (u_i,u_{i+1}])\Bigl | + 2 \epsilon , \end{aligned} \end{aligned}$$

where $\parallel f \parallel _{\infty } = \sup \{f(x): x \in [0,K] \}$. Since f is continuous over a compact support, it is bounded and hence $\parallel f \parallel _{\infty } < \infty$. Therefore,

$$\begin{aligned} \begin{aligned} \lim _{p \rightarrow 0}&\sup _x \bigl |{\mathbb {E}}f(X_p(x) - {\mathbb {E}}f(X(x)) \bigr | \\&\le \lim _{p \rightarrow 0} \parallel f \parallel _{\infty } \sum \limits _{i=1}^{m} \sup _x \bigl | {\mathbb {P}}(X_p(x) \in (u_i,u_{i+1}]) - {\mathbb {P}}(X(x) \in (u_i,u_{i+1}]) \bigr | + 2 \epsilon , \\&= 2 \epsilon . \end{aligned} \end{aligned}$$

The second relation follows from the hypothesis (34). Since $\epsilon$ is arbitrary, we get the desired result. $\square$