NMA01：网路命令综合分析.doc

网路诊断与分析实训岗位：网管( 7 ) 实训人员(1行1名字)请尽量独立自主完成，认真学习参考资料Part1文档和命令帮助（命令 /? ）为提高效率，建议开启多个命令窗口进行各项测试。完成后及时上交以便评审(请勿修改文件名)。课后由老师统一讲评。注意：命令记录要求记录的是文本内容(不是截图！)文本复制方法如下图所示：提示：可以多开几个命令视窗同时进行多个测试，提高效率！Ping 网路分析（1）分别 ping 和 命令记录网管7 C:ping Pinging 9 with 32 bytes of data:Request timed out.Reply from 9: bytes=32 time=6ms TTL=56Reply from 9: bytes=32 time=6ms TTL=56Request timed out.Ping statistics for 9: Packets: Sent = 4, Received = 2, Lost = 2 (50% loss),Approximate round trip times in milli-seconds: Minimum = 6ms, Maximum = 6ms, Average = 3ms网管7 C:ping Pinging 8 with 32 bytes of data:Request timed out.Request timed out.Request timed out.Reply from 8: bytes=32 time=26ms TTL=244Ping statistics for 8: Packets: Sent = 4, Received = 1, Lost = 3 (75% loss),Approximate round trip times in milli-seconds: Minimum = 26ms, Maximum = 26ms, Average = 6ms分析计算各自的网络距离，分析到哪个目标更近？结论计算过程及结论：到达目标更近（2）如何用 ping 获知 的主机名 ？命令记录网管7 C:ping -a Pinging with 32 bytes of data:Request timed out.Request timed out.Request timed out.Request timed out.Ping statistics for : Packets: Sent = 4, Received = 0, Lost = 4 (100% loss),Approximate round trip times in milli-seconds: Minimum = 0ms, Maximum = 0ms, Average = 0ms结论为本主机名（3）测试：ping 62 的 最大字节？（注意：只记录 最终的两次 （正负1个字节！）命令记录网管7 C:ping -l 1469 62Pinging 62 with 1469 bytes of data:Request timed out.Request timed out.Request timed out.Request timed out.Ping statistics for 62: Packets: Sent = 4, Received = 0, Lost = 4 (100% loss),Approximate round trip times in milli-seconds: Minimum = 0ms, Maximum = 0ms, Average = 0ms网管7 C:ping -l 1468 62Pinging 62 with 1468 bytes of data:Reply from 62: bytes=1468 time=52ms TTL=245Reply from 62: bytes=1468 time=58ms TTL=245Reply from 62: bytes=1468 time=52ms TTL=245Reply from 62: bytes=1468 time=49ms TTL=245Ping statistics for 62: Packets: Sent = 4, Received = 4, Lost = 0 (0% loss),Approximate round trip times in milli-seconds: Minimum = 49ms, Maximum = 58ms, Average = 52ms结论最大字节为1468（4）ping 命令记录网管7 C:ping Pinging with 32 bytes of data:Request timed out.Request timed out.Request timed out.Request timed out.Ping statistics for : Packets: Sent = 4, Received = 0, Lost = 4 (100% loss),Approximate round trip times in milli-seconds: Minimum = 0ms, Maximum = 0ms, Average = 0ms分析返回的ICMP差错报文表示什么问题？结论回答：超时路由分析 （1）tracert 分析路由过程命令记录网管7 C:tracert Tracing route to 30over a maximum of 30 hops: 1 1 ms tracert Tracing route to over a maximum of 30 hops 1 1 ms 1 ms tracert Tracing route to over a maximum of 30 hops 1 10 ms 1 ms pathping -q 255 62Tracing route to 62 over a maximum of 30 hops 0 nms07 4 1 2 3 4 5 6 7 2 8 1 9 1 10 65 11 13 12 7 13 62Computing statistics for 828 seconds.(2)Pathping 62 分析该路由通路的拥塞情况。(注意:时间需要几分钟！请保持切勿中断！每隔1分钟敲回车几下，以便数据显出。某些节点可能不回应ICMP(显示* lost 100%)，请忽略这些节点)命令记录网管7 C:pathping Tracing route to 62 over a maximum of 30 hops 0 nms07 4 1 2 3 4 5 6 7 2 8 1 9 1 10 65 11 13 12 7 13 62Computing statistics for 325 seconds. Source to Here This Node/LinkHop RTT Lost/Sent = Pct Lost/Sent = Pct Address 0 nms07 4 0/ 100 = 0% | 1 0ms 0/ 100 = 0% 0/ 100 = 0% 0/ 100 = 0% | 2 0ms 0/ 100 = 0% 0/ 100 = 0% 0/ 100 = 0% | 3 6ms 7/ 100 = 7% 7/ 100 = 7% 0/ 100 = 0% | 4 - 100/ 100 =100% 100/ 100 =100% 0/ 100 = 0% | 5 - 100/ 100 =100% 100/ 100 =100% 0/ 100 = 0% | 6 - 100/ 100 =100% 100/ 100 =100% 0/ 100 = 0% | 7 5ms 6/ 100 = 6% 6/ 100 = 6% 2 0/ 100 = 0% | 8 7ms 13/ 100 = 13% 13/ 100 = 13% 1 0/ 100 = 0% | 9 6ms 7/ 100 = 7% 7/ 100 = 7% 1 0/ 100 = 0% | 10 5ms 4/ 100 = 4% 4/ 100 = 4% 65 0/ 100 = 0% | 11 10ms 32/ 100 = 32% 32/ 100 = 32% 202.97.3