Question

A 25 Kbps satellite link has a propagation delay of 400 ms. The transmitter employs the "go back n ARQ" scheme with n set to 10. Assuming that each frame is 100 bytes long, what is the maximum data rate possible?

A) 5 Kbps
B) 10 Kbps
C) 15 Kbps
D) 20 Kbps

Answer 1

Answer on Question#39316 - Math - Other

A 25 Kbps satellite link has a propagation delay of 400 ms. The transmitter employs the "go back n ARQ" scheme with n set to 10. Assuming that each frame is 100 bytes long, what is the maximum data rate possible?

A) 5 Kbps

B) 10 Kbps

C) 15 Kbps

D) 20 Kbps

Solution

The transmission delay $t_{tr}$ for sending the packet is equal to

$t_{tr} = \frac{L}{C} = \frac{100 \cdot 8}{25000} = 0.032$ seconds and $t_{ack} = 0.4$ seconds. Let $t_{ack}$ be the time until satellite receives an acknowledgment for a sent packet. In the time interval $T = t_{tr} + t_{ack} = 0.032 + 0.4 = 0.432$ we have that satellite sends a packet of length $L$. The average (transmission) rate (in bits per seconds) with which satellite sends data to endpoint is a function of $n$.

Let

$$N_{0} = \frac{t_{ack}}{t_{tr}} = \frac{0.4}{0.032} = 12.5 \approx 13$$

be the maximum number of packets satellite can send while waiting for an ACK. The rate $x(n)$ at which satellite sends packets to endpoint as a function of $n$ is given by

$$x(n) = \left\{ \begin{array}{c} \frac{nL}{T} = \frac{10 \cdot 800}{0.432} \approx 18519 \text{ bps}, \quad n \leqslant N_{0} + 1 \\ C = 25000 \text{ bps}, \quad n > N_{0} + 1 \end{array} \right. . \text{Thus, } x(10) \approx 20 \text{ Kbps}.$$

Answer

D)