考虑选择性重复滑动窗口协议，该协议使用1 KB的帧大小在1.5 Mbps链路上以50毫秒的单向延迟发送数据。为了实现60％的链路利用率，表示序列号字段所需的最小位数为________。
(A) 3
(B) 4
(C) 5
(D) 6答案： (C)
解释：
Transmission delay = Frame Size/bandwidth
= (1*8*10^3)/(1.5 * 10^6)=5.33ms
Propagation delay = 50ms
Efficiency = Window Size/(1+2a) = .6
a = Propagation delay/Transmission delay
So, window size = 11.856(approx)
min sequence number = 2*window size = 23.712
bits required in Min sequence number = log2(23.712)
Answer is 4.56
Ceil(4.56) = 5 这个问题的测验