考虑连接两个相距 8000 公里的系统的网络。网络带宽为每秒 500 × 10 6位。媒体的传播速度为每秒 4 × 10 6米。需要为此网络设计一个Go-Back-N滑动窗口协议。平均数据包大小为 10 7位。网络将被充分利用。假设节点处的处理延迟可以忽略不计。那么，序列号字段的最小比特大小必须是________。
(一) 2
(乙) 4
(三) 8
(四) 16答案： (C)
解释：
Propagation time = (8000 * 1000)/ (4 * 10^6)
= 2 seconds
Total round trip propagation time = 4 seconds
Transmission time for one packet = (packet size) / (bandwidth)
= (10^7) / (500 * 10^6)
= 0.02 seconds
Total number of packets that can be transferred before an
acknowledgement comes back = 4 / 0.02 = 200
Maximum possible window size is 200.
In Go-Back-N, maximum sequence number should be one more than
window size.
So total 201 sequence numbers are needed. 201 different sequence
numbers can be represented using 8 bits.这个问题的测验