PUERTA | GATE-IT-2004 | Pregunta 83

Un enlace satelital de 20 Kbps tiene un retardo de propagación de 400 ms. El transmisor emplea el esquema «regresar n ARQ» con n establecido en 10. Suponiendo que cada cuadro tiene 100 bytes de largo, ¿cuál es la tasa de datos máxima posible?
(A) 5 Kbps
(B) 10 Kbps
(C) 15 Kbps
(D) 20 Kbps

Respuesta: (B)
Explicación:

It uses the sliding window protocol for transmission of data. 
The question takes into consideration the variant of sliding window protocol 
namely GO BACK N ARQ. In this protocol the sender can have up to N packets 
unacknowledged that are still remaining in the pipeline. The receiver only 
sends cumulative acknowledgements. In case of encountering an error the sender 
has to resend all the data frames following the error.

According to the question:
The data rate of the link is 20 Kbps and the propagation delay = 400 ms
So, the time required to transmit 100 bytes long data will be given by

Transmission Time t = Number of bits to be transmitted / data rate of the link 
                    = (100* 8 bits) /20 Kbps =  40 ms

Now, the propagation delay is given as d = 400 ms

So the efficiency of the link is given by:

Efficiency E = N * t / ( t+ 2 * d )

Where N = window size

E = 10 * 40 / (40+2*400) = 0.476

So, the maximum data rate attainable = 0.476 * 20 Kbps = 9.52 Kbps

This is close to 10.
So, the answer will be 10Kbps.

Esta explicación ha sido aportada por Namita Singh.
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