想请教1个845的问题这题怎么算的？

jaska

Using the G.729 codec and the default cRTP configuration, what is the approximately Voip bandwidth savings across a Frame Relay link that adds a 6 byte frame header?
Select the best response
A.55%
B.58%
C.60%
D.63%

jaska

还有这题 我就是想不通 为什么deny的telnet可以访问control plane 而没有任何限制，相反其他所有的permit的都限制了，请大虾们指点一下 在下万分感谢！！！
The following example shows how to apply a QoS policy for aggregate CP services to Telnet traffic received on the control plane. Trusted hosts with source addresses 10.1.1.1 and 10.1.1.2 forward Telnet packets to the control plane without constraint, while allowing all remaining Telnet packets to be policed at the specified rate:
! Allow 10.1.1.1 trusted host traffic.
Router(config)# access-list 140 deny tcp host 10.1.1.1 any eq telnet
! Allow 10.1.1.2 trusted host traffic.
Router(config)# access-list 140 deny tcp host 10.1.1.2 any eq telnet
! Rate limit all other Telnet traffic.
Router(config)# access-list 140 permit tcp any any eq telnet
! Define class-map "telnet-class."
Router(config)# class-map telnet-class
Router(config-cmap)# match access-group 140
Router(config-cmap)# exit
Router(config)# policy-map control-plane-in
Router(config-pmap)# class telnet-class
Router(config-pmap-c)# police 80000 conform transmit exceed drop
Router(config-pmap-c)# exit
Router(config-pmap)# exit
! Define aggregate control plane service for the active Route Processor.
Router(config)# control-plane
Router(config-cp)# service-policy input control-plane-in
Router(config-cp)# exit

legend8826

NP的，不懂，待高手解决

jaska

     求助哈。。。。。。

leo

第一题选B 58%
考试以记住答案PASS为主，等你把所有题都想通了，题库早变N回了。
考完了有的是时间搞清楚这些问题，乖啊，赶紧背题去

legend8826

第一题答案是58%，请参考下面的外文笔记

Total bandwidth requirement=((Total packet size)x normal bandwidth requirment) / payload size
G.729 with Frame Relay and without cRTP: ((6+40+20bytes)x8kbps)/20bytes=26.4kbps
G.729 with Frame Relay and with cRTP: ((6+2+20bytes)x8kbps)/20bytes=11.2kbps
Total bandwidth requirement=(Total packet size x normal bandwidth requirment) / payload size
reduced from 26.4kbps to 11.2kbps 11.2/26.4=42%

The 42% you've managed to come up with your calculation is the VoIP bandwidth that will be consumed by the voice call, hence, the bandwidth "SAVINGS" across the link is 58%

ubuntu_boy1982

第二个题目考察的是你对ＡＣＬ和ＭＱＣ工具的理解
首先ＡＣＬ的ｄｅｎｙ和ｐｅｒｍｉｔ在这里的理解是“不匹配”和“匹配”
Router(config)# access-list 140 deny tcp host 10.1.1.1 any eq telnet
! Allow 10.1.1.2 trusted host traffic.
Router(config)# access-list 140 deny tcp host 10.1.1.2 any eq telnet
! Rate limit all other Telnet traffic.
Router(config)# access-list 140 permit tcp any any eq telnet
这里的意思是不匹配上面的两个地址的telnet的流量，匹配其他地址的telnet的流量。然后再后面的MQC工具中对匹配到的流量进行控制！

jaska

police 80000 conform transmit exceed drop 能解释一下吗？