20 Kbps 卫星链路的传播延迟为 400 毫秒。发送器采用“go back n ARQ”方案，其中 n 设置为 10。假设每帧长 100 字节，可能的最大数据速率是多少?
(A) 5Kbps
(B) 10Kbps
(C) 15Kbps
(D) 20Kbps答案：(乙)
解释：
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.
此解释由Namita Singh 提供。
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