GATE IT 2005 | Question: 72

Question

A channel has a bit rate of $4$ $kbps$ and one-way propagation delay of $20$ $ms$. The channel uses stop and wait protocol. The transmission time of the acknowledgment frame is negligible. To get a channel efficiency of at least $50$$\text{%}$, the minimum frame size should be

• A. $80$ $\text{bytes}$
• B. $80$ $\text{bits}$
• C. $160$ $\text{bytes}$
• D. $160$ $\text{bits}$

Comments

instead of Stop and Wait , if GBN  or selective repeat given,then how to solve?

Answer 1

For $50\%$ utilization with Stop-and-wait,

$\large\dfrac{t_t}{t_t+2t_p}\geq \dfrac{1}{2},$

where, $t_t-$ Transmission time, $t_p-$ propagation delay. Here, $t_t = \frac{L}{B},$ where $L$ is the frame length in bits and $B$ is the bitrate of the channel.

$\large2t_t\geq t_t+2t_p$

$\large t_t\geq 2t_p$

$\dfrac{L}{B}\large\geq 2\times t_p$

$L\geq 2\large\times t_p\times B$

$\implies L=2\times 20\times 10^{-3}\times 4\times 10^{3}=160\text{ bits}$

So, answer is D.

Answer 2

Answer: D

Transmission Time/(Transmission Time + 2*Propagation Time) = 0.5

=> Transmission Time/(Transmission Time + 2*20*10^-3) = 0.5

=> Transmission Time = 40*10^-3

Option D satisfies the Transmission Time as 160 bits/4 Kbps = 40*10^-3

Answer 3

Answer->D \

efficiency = transmision time of packet/(transmission time of packet+2xprop delay+tranmission time for ack(~0))