请问RS题库中，这题的答案选哪个

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178题：
How is voice traffic entering this router on interface GigabitEthernetO/O being handled by the shown marking policy?
A. Any traffic matching access-list voice is trusted and marking is not changed.
B. All voice is being set to DSCP 0
C. All voice is being set to AF21
8 M- |0 U  J* KD. Any traffic matching access-list voice is set to EF
Answer: D
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1 U) V2 x7 `4 e3 A: j) m9 S: k, B
请问各位大神，此题答案是A还是D
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326Q中答案是A，227Q中答案是D，我个人倾向于选A
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! y2 @6 k. M$ T/ H/ U会的题不多，求正确答案

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RS 227Q题库中，此题答案是ACF，为啥不是ADF
& k. o1 ^1 C; `求各路大神解答，谢谢
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179. Which three pieces of information are carried on OSPF type 3 LSAs? (Choose three)
4 U2 F7 x4 V- E) D, m0 m: MA. metric
B. authentication type
' l$ Y* J8 B: O7 t9 J/ R# ^/ yC. link state
. T8 v, |; |* V6 T5 X; oD. IP subnet
E. external route tag
" j" e% i, J5 B2 F; `2 _1 dF. subnet mask
9 m5 F4 J4 v& uAnswer: ACF

shawwang1

没考到

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shawwang1 发表于 2016-7-2 15:54
没考到

- [# `9 W8 s" K& S$ P6 J) n你刚考过还是啥意思？

haiyu.dong

思科网站应该搜得到答案

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查了思科网站，显示如下
% s- I3 Z( n- u8 R" rSummary Links
# u+ m* u4 x! B7 p6 o( c8 K; rSummary Net Link States (Area 0)

1 K4 V0 w5 Y% V; }Link ID         ADV Router      Age    Seq#       Checksum
' a/ c& j) @2 U$ S203.250.15.0    203.250.15.67   792    0x80000002 0xAEBD
2 O: _8 }# Z$ T: ]$ `Area 0 has one summary link represented by the IP network address of the
$ U1 u) ^, \# V+ H: M' X% h: Qlink 203.250.15.0. This link was injected by the ABR RTC from area 1 into
area 0. A detailed view of this summary link follows, summary links for
0 i/ F& c# B0 o/ k5 B4 X1 [1 Warea 1 are not listed here:

/ p4 }1 i7 x# @9 Y1 [2 `$ ^; FRTC#show ip ospf database summary (area 1 is not listed)
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0 r: |+ n$ J1 i9 G& ]7 _$ q  x; W/ x              Summary Net Link States (Area 0)

+ d& s1 t( g3 T, b5 p+ N* z  LS age: 615
5 h/ w/ e: Z% E; ]/ |9 I: S/ O4 W  Options: (No TOS-capability)
( C- h/ I, O# t/ `' S9 _+ p  Q  LS Type: Summary Links(Network)
  Link State ID: 203.250.15.0 (summary Network Number)
  Advertising Router: 203.250.15.67
  LS Seq Number: 80000003
. ?0 x. a% Z% h4 ~% u  Checksum: 0xACBE
  Length: 28
' d3 t; ^- n! u3 x5 B- V  Network Mask: 255.255.255.192 TOS: 0  Metric: 64

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这样答案应该是ACF了

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大家都来看看一楼选哪个答案，谢谢了

shenka

选A。在policy-map中匹配但没有具体指定PHB行为的，默认不做改变，即marking is not changed

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shenka 发表于 2016-7-3 13:17
选A。在policy-map中匹配但没有具体指定PHB行为的，默认不做改变，即marking is not changed

收到，谢谢了

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shenka 发表于 2016-7-3 13:17
选A。在policy-map中匹配但没有具体指定PHB行为的，默认不做改变，即marking is not changed

* Z+ F( W4 M% F9 A收到，谢谢了

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在麻烦各位看227Q中的52题

52.Which two security risks can occur with PMTUD? (Choose two)
6 }' W$ O( w4 R. QA. An attacker can block valid Datagram Too small messages, to cause a DDos.
50.Refer to the exhibit.
Which kind of ICMPv6 packet is shown in the output?
B. An attacker can block valid Datagram Too small messages, to cause a Dos.
C. An attacker can slow data flow by sending false messages that indicate that the PMTU is significantly smaller than the true PMTU.
; L% U9 C8 Q" u4 K, c# C: D5 vD. An attacker can cause data lost by sending false messages that indicate that the PMTU is significantly smaller than the true PMTU.
5 F- h( P$ J, x9 D9 r1 QE. An attacker can cause data lost by sending false messages that indicate that the PMTU is significantly larger than the true PMTU.
F. An attacker can slow data flow by sending false messages that indicate that the PMTU is significantly larger than the true PMTU.
, b& p: }$ ]/ c* W  i! b9 _Answer: BD

  F4 j  n, W* ?8 R% q0 p* h我觉得答案应该选E，而非选D


6 ?" y" g. |, G( Y3 j, T' jPMTU（Path MTU，路径最大传输单元）是在源节点和目的节点之间的路径上的任一链路所能支持的链路MTU的最小值。
" B1 [9 Y+ |' ^: q, U6 |& Y$ l在IPv6网络中，分段不在中间路由器上进行。当需要传送的数据报文长度比链路MTU大时，只由源节点本身对数据报文进行分段，中间路由器不对数据报文进行再次分段。这就要求源节点在发送数据报文前能够发现整个发送路径上的所有链路的最小MTU，然后以该MTU值发送数据报文，这就是PMTU发现。

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shawwang1 发表于 2016-7-2 15:54
没考到

1 I( \4 B8 {# T' P4 W/ r0 E你刚考的吗

shawwang1

7.1考，没考到

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shawwang1 发表于 2016-7-3 17:27
* _3 d. \; w; [) }$ h: B7.1考，没考到

5 G1 A7 ]. p& ~' B恭喜恭喜

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根据以下配置信息
router eigrp 100
network 10.0.0.0
distance 80 10.1.12.0 0.0.0.255 10
access-list 10 permit 10.1.0.0 0.0.255.255
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Which prefixes will have their distance changed?
' j% x, p: R; OA. all prefixes in the range 10.1.12.0 - 10.1.12.255 learned from peers matching access-list 10
. x- H! f4 m: N* {0 V7 eB. all prefixes matching access-list 10 learned from peers in the range 10.1.12.0 - 10.1.12.255
C. all internal prefixes in the range 10.1.12.0- 10.1.12.255 learned from peers matching access-list 10
9 i1 D! t8 n- D) mD. all internak prefixes matching access-list 10 learned from peers in the range 10.1.12.0 - 10.1.12.255
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答案是选C，可我觉得应该选Ｂ，大家觉得呢