问题描述
为什么在TCP的Go-Back-N算法中，窗口大小(N)必须小于序列号空间(S):S> N?我试图自己弄清楚它，但不要安静地得到它
Why in TCP's Go-Back-N Algorithm window size(N) has to be smaller than the sequence number space(S): S>N? I tried figuring it out myself but don't quiet get it
推荐答案
假定序列空间为四个(序列号0、1、2、3).假设窗口大小也是4.发件人发送了4个数据包，其序号为(0,1,2,3).接收器接收所有四个数据包.因此它发送4个确认(0、1、2、3).现在假设所有确认都丢失了.发件人将重新发送所有四个数据包，但接收者将假定它们是新的数据包.为了避免由于丢失确认而引起的混乱，我们保持n<
Assume that sequence space was four (sequence numbers 0,1,2,3). Lets say window size was also 4. Sender sends 4 packets with sequence numbers (0,1,2,3). Receiver receives all four packets. So it sends 4 acknowledgements(0,1,2,3). Now assume all acknowledgements are lost. Sender will resend all four packets but receiver will assume they're the new ones. To avoid confusions arising from lost acknowledgements, we keep n < s
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