How Do the Multiple Tunable Receivers Per Node Affect the Efficiency of a WDM LAN? A Performance Analytical Study

Abstract

In this study, the performance of a WDM LAN of passive star topology is explored. An effective synchronous transmission WDMA protocol is proposed that employs a pre-transmission reservation scheme in order to improve the system output. Each station is equipped with a network interface with more than one tunable receivers in order to be able to receive more than one data packets during a time frame. As a results, the average probability of packet rejection at destination is significantly reduced as compared with the single tunable receiver per station case. The system performance is mathematically modelled based on Poisson statistics and finite station population. Also, an analytical framework is adopted to evaluate the effect of the number of tunable receivers per station on the total performance. Numerical results are comparatively studied for various numbers of data channels and stations. Finally, the propagation delay latency parameter is properly modelled while its impact on the system performance is analytically evaluated.

Appendix

We assume the model that consists of N data channels and M stations. We aim to analytically describe the distribution of the successfully transmitted data packets over the N data channels to the M stations. This model corresponds to the occupancy problem of the distribution of indistinguishable balls (data packets) to cells (destination stations), supposing that the arrangements should have equal probabilities. We consider indistinguishable packets transmitted to indistinguishable destination stations using Maxwell-Boltzman statistics [19].

We are interested in the probability $\Pr [A_{N} (s) = r]$ of r correctly received data packets at destination when s data packets have been successfully transmitted over the N-channel system, during a time frame, 1 ≤ s.

Let’s suppose that each station may transmit to any of the M stations (for the sake of generality we suppose that a station may send to and receive from itself). According to the Maxwell-Boltzman statistics, there are $M^{s}$ possible arrangements of the s successfully transmitted data packets to the M destination stations, each with equal and constant probability: $1/M^{s}$.

We consider that the distribution of s data packets to M stations provides the following result by the end of a frame:

• there are k ₀ of M destination stations, $k_{0} \in \left\{ {0,1,2, \ldots ,M} \right\}$: for each of them there is no successfully transmitted data packet destined to it,

• there are k ₁ of M destination stations, $k_{1} \in \left\{ {0,1,2, \ldots ,s} \right\}$: for each of them there is 1 successfully transmitted data packet destined to it, and so on. In general,

• there are k _i of M destination stations, $i,k_{i} \in \left\{ {1,2, \ldots ,s} \right\}$: for each of them there are i successfully transmitted data packets destined to it.

It is obvious that:

$$ \sum\limits_{i = 0}^{s} {k_{i} } = M $$

(14)

and:

$$ \sum\limits_{i = 0}^{s} {ik_{i} } = s $$

(15)

Since each destination station is capable of receiving up to x data packets per frame, it is:

$$ \sum\limits_{i = 0}^{x - 1} {ik_{i} } + \sum\limits_{{i = x,k_{i} \ne 0}}^{s} i x = r. $$

(16)

For each set of integers $\left\{ {k_{0} ,k_{1} ,k_{2} , \ldots ,k_{M} } \right\}$ that satisfy Eqs. (14), (15), and (16) and it is $k_{0} ,k_{1} , \ldots k_{M} \in \left\{ {0,1,2, \ldots ,M} \right\}$, the probability P _ki that: no data packet is destined to k ₀ stations, one data packet is destined to k ₁ stations, and so on; and generally, i data packets are destined to k _i stations is given in [20]:

$$ P_{ki} = \frac{M!s!}{{M^{s} \prod\limits_{i = 0}^{s} {k_{i} !} \prod\limits_{z = 1}^{s} {(z!)^{{k_{z} }} } }}. $$

(17)

Thus, the probability $\Pr [A_{N} (s) = r]$ is defined as the sum of the probabilities P _ki , for all possible sets of integers $\left\{ {k_{0} ,k_{1} ,k_{2} , \ldots ,k_{M} } \right\}$ that satisfy Eqs. (14), (15), and (16) and it is $k_{0} ,k_{1} , \ldots k_{M} \in \left\{ {0,1,2, \ldots ,M} \right\}$, i.e.:

$$ \Pr [A_{N} (s) = r] = \sum\limits_{{all{\text{ sets}}}} {\frac{M!s!}{{M^{s} \prod\limits_{i = 0}^{s} {k_{i} !} \prod\limits_{z = 1}^{s} {(z!)^{{k_{z} }} } }}} . $$

(18)