发送方使用“停止并等待” ARQ协议进行帧的可靠传输。帧大小为1000字节，发送方的传输速率为80 Kbps(1Kbps = 1000位/秒)。确认的大小为100字节，接收器的传输速率为8 Kbps。单向传播延迟为100毫秒。假设没有帧丢失，则发送方吞吐量为__________字节/秒。
(A) 2500
(B) 2000
(C) 1500
(D) 500答案： (A)
解释：
Total time = Transmission-Time + 2* Propagation-Delay + Ack-Time.
Trans. time = (1000*8)/80*1000 = 0.1 sec
2*Prop-Delay = 2*100ms = 0.2 sec
Ack time = 100*8/8*1000 = 0.1 sec.
Total Time = 0.1 + 0.2 + 0.1 = 0.4 sec.
Throughput = ((L/B)/Total time) * B,
L = data packet to be sent and
B = BW of sender.
Throughput = L/Total Time
= 1000/0.4
= 2500 bytes/sec.
这个问题的测验