GATE | GATE-IT-2004 |问题83

20 Kbps卫星链路的传播延迟为400 ms。发射机采用“返回n ARQ”方案，n设置为10。假设每个帧的长度为100字节，最大数据速率是多少？ （A） 5Kbps （B） 10Kbps （C） 15Kbps （D） 20Kbps 答复: （B） 说明：

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.

这一解释是由 纳米塔·辛格。 这个问题的小测验