Modern network traffic modeling based on binomial multiplicative cascades

Abstract

In this paper we present a new multifractal approach for modern network traffic modeling. The proposed method is based on a novel construction scheme of conservative multiplicative cascades. We show that the proposed model can faithfully capture some main characteristics (scaling function and moment factor) of multifractal processes. For this new network traffic model, we also explicitly derive analytical expressions for the mean and variance of the corresponding network traffic process and show that its autocorrelation function exhibits long-range dependent characteristics. Finally, we evaluate the performance of our model by testing both real wired and wireless traffic traces, comparing the obtained results with those provided by other well-known traffic models reported in the literature. We found that the proposed model is simple and capable of accurately representing network traffic traces with multifractal characteristics.

Appendix

1.1 Appendix A: Proof of Lemma 1

Let $X( {{\Delta }t_0 })=2^n.{Y}.{R}( {b_1 }){R}( {b_1 b_2 })\ldots {R}(b_1 \ldots b_k )$ for $N\gg 1$. Thus, the mean of process ${X}(\Delta {t}_0 )$ is given by:

$$\begin{aligned} {E}[{X}( {{\Delta t}_0 })]={E}[2^{n}.{Y}.{R}( {{b}_1 }){R}( {{b}_1 {b}_2 })\ldots {R}({b}_1 \ldots {b}_{k} )]. \end{aligned}$$

(39)

Using the fact that ${E}[ {X( {{\Delta }t_{N} })} ]=2^{-N}$, we have:

$$\begin{aligned} {E}[ {X( {{\Delta }t_0 })} ]={E}[{Y}]. \end{aligned}$$

(40)

Assuming $Y$ have lognormal distribution:

$$\begin{aligned} {E}[ {{X}( {{\Delta t}_0 })} ]={E}[ {Y} ]={e}^{{m}+{v}^2/2}. \end{aligned}$$

(41)

1.2 Appendix B: Proof of Lemma 2

Using variance definition

$$\begin{aligned} \hbox {Var}[ {{X}( {{\Delta t}_0 })} ]=E[ {{X}( {{\Delta t}_0 })^2} ]-E[{X}( {{\Delta t}_0 })]^2. \end{aligned}$$

(42)

Then

$$\begin{aligned} \hbox {Var}[ {{X}( {{\Delta t}_0 })} ]={E}[ {{Y}^2} ]2^{2{N}}{E}[ {{R}^2} ]^{N}-{E}[{X}( {{\Delta t}_0 })]^2 \end{aligned}$$

(43)

or

$$\begin{aligned} \hbox {Var}[ {{X}( {{\Delta t}_0 })} ]={E}[ {{Y}^2} ]2^{2{N}}{E}[ {{R}^2} ]^{N}-{E}[{Y}]^2. \end{aligned}$$

(44)

Therefore

$$\begin{aligned} \hbox {Var}[ {{X}( {{\Delta t}_0 })} ]=e^{2m+2v^2}.\,2^{2N}.\,\left( {\frac{( {\alpha +\beta })(\alpha +\beta +1)}{(\alpha +1)\alpha }}\right) ^N-e^{2m+v^2} \end{aligned}$$

(45)

1.3 Appendix C: Proof of Lemma 4

Let ${m}=2^{k}$ and ${x}_{( {{N}-{k}})} =\sum _{( {{k}-1}){m}+1}^{{km}} {X}( {k})$, the aggregate process ${X}^{m}$ at stage $( {{N}-{k}})$ can be represented as

$$\begin{aligned} {X}^{m}=2^{-{k}}{x}_{( {{N}-{k}})} \end{aligned}$$

(46)

The variance of aggregate process ${X}^{m}$ can be written as

$$\begin{aligned} \hbox {Var}[{X}^{m}]={E}[ {( {2^{-{k}}{x}_{({N}-{k})} })^2} ]-{E}^2[ {( {2^{-{k}}{x}_{({N}-{k})} })} ] \end{aligned}$$

(47)

Using the fact that the process ${X}$ can be viewed as the product of a lognormal random variable ${Y}$ and a multiplicative cascade ${\mu }$, the following relation is valid:

$$\begin{aligned} {x}_{({N}-{k})} =\mathop \sum \limits _{( {{k}-1}){m}+1}^{{km}} {{Y}\mu }(\Delta {t}_{k} ). \end{aligned}$$

(48)

Using Eq. (46) and moments of the aggregate process of the multiplicative cascade given by:

$$\begin{aligned} \frac{1}{{m}}\mathop \sum \limits _{{i}=1}^{m} {X}( {i})^{q}=\frac{1}{{m}}\mathop \sum \limits _{{i}=1}^{m} ({Y}_{i} {\mu }( {\Delta {t}_{i} }))^{q} \end{aligned}$$

(49)

we have

$$\begin{aligned} \hbox {Var}[{X}^{m}]&= {E}\left[ \left( {2^{-{k}}\frac{1}{2^{k}}\mathop \sum \limits _{( {{k}-1}){n}+1}^{{kn}} {{Y} \mu }(\Delta {t}_{i})^2}\right) \right] \nonumber \\&\quad -{E}^2\left[ {\left( {2^{-{k}}\frac{1}{2^{k}}\mathop \sum \limits _{( {{k}-1}){n}+1}^{{kn}} {{Y}\mu }(\Delta {t}_{i} )}\right) } \right] \end{aligned}$$

(50)

$$\begin{aligned} \hbox {Var}[{X}^{m}]&= {E}\left[ {{Y}^2} \right] 2^{2{N}}{E}\left[ {{R}^2} \right] ^{{N}-{k}}-{E}[{Y}]^2\left( {\frac{1}{2}}\right) ^{2({N}-{k})}\end{aligned}$$

(51)

$$\begin{aligned} \hbox {Var}[{X}^{m}]&= {e}^{2{m}+2{v}^2}\left( {\frac{{\alpha }({\alpha }+1)}{( {{\alpha }+{\beta }})({\alpha }+{\beta }+1)}}\right) ^{{N}-{k}}-{e}^{2{m}+{v}^2}2^{2{k}-2{N}}. \end{aligned}$$

(52)