局域网上相距2km的两个站点，采用同步传输方式以10Mb/s的速率发送150000字节大小的IP报文。假定数据帧长为1518字节，其中首部为18字节；应答帧为64字节。若在收到对方的应答帧后立即发送下一帧，则传送该文件花费的总时间为（ ）ms（传播速率为200m/μs），线路有效速率为（ ）Mb/s。
问题1选项
A.1.78
B.12.86
C.17.8
D.128.6
问题2选项
A.6.78
B.7.86
C.8.9
D.9.33
正确答案及解析
正确答案
解析
答案： D、D
总时间的=传输时延+传播时延
先算总的传输时延：
一共150000×8=1200000位，数据帧是（1518-18）×8=12000位，那么一共1200000/（1518-18）×8=100帧。一帧的传输时延是1518×8/10000000，那么总的传输时延是100×1518×8/10000000=0.12144s，而由于是必须等待应答信号，所以应答信号的总传输时延是100×64×8/10000000=0.00512s。总传输时延是0.12144+0.00512=0.1265。
再算总的传播时延：2×100×2000/200000000=0.002s。所以总时延等于128.6ms。
1200000/X=0.1286，那么有效速率是9.33Mbps。
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