我正在修改网络考试,我不确定以下答案:
Consider the effect of using slow-start on a link with a 10ms round-trip time
and no congestion. The receive window is 24KB and the maximum segment
size is 2KB. How long does it take before the first full window can be sent
in one transmission round?
每次确认后,最大段大小增加1还是加倍?如果它加倍,答案是50ms,因为2KB ^ 5 = 32KB所以在5次跳闸之后MSS将等于32KB并且由于10ms往返时间它将是10x5 = 50ms?
答案 0 :(得分:0)
我在这里找到了你问题的解决方案(问题4):
http://web.eecs.utk.edu/~qi/teaching/ece453f06/hw/hw7_sol.htm
您或多或少是正确的,除非您忽略最后一个RTT(因为窗口大小超过24KB)。所以你的答案是:
4 x 10 = 40ms。