假设在一条链接上使用停止等待协议，其比特率为每秒64千比特，传播延迟为20毫秒。假设确认的传输时间和节点处的处理时间可以忽略不计。然后，达到至少50％的链接利用率的最小帧大小(以字节为单位)为_________。
(A) 160
(B) 320
(C) 640
(D) 220答案： (B)
解释：
Transmission or Link speed = 64 kb per sec
Propagation Delay = 20 milisec
Since stop and wait is used, a packet is sent only
when previous one is acknowledged.
Let x be size of packet, transmission time = x / 64 milisec
Since utilization is at least 50%, minimum possible total time
for one packet is twice of transmission delay, which means
x/64 * 2 = x/32
x/32 > x/64 + 2*20
x/64 > 40
x > 2560 bits = 320 bytes GATE键中的答案表示160个字节，但是答案键似乎不正确。请参阅此处的问题36。
这个问题的测验