题目详情

某卫星信道数据速率为64Kb/s,帧长4000比特,采用停等ARQ协议,则信道的最大利用率为()。

  • A.0.480
  • B.0.125
  • C.0.104
  • D.0.010

正确答案及解析

正确答案
C
解析

发送一个帧的时间=发送时间+传播时间+【应答帧的发送时间+返回时间】,由于应答帧很小,题干没有给出,可以忽。因此总的时间=4000/64000+270+270=602.5ms而这段时间里,真正用于数据帧发送的时间=4000/64000=0.0625S,相当于62.5ms,因此最大利用率=62.5/602.5=0.1037,因此选C

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