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NEW QUESTION: 1
Which two mechanisms can be used to eliminate Cisco Express Forwarding polarization? (Choose two.)
A. the unique-ID/universal-ID algorithm
B. alternating cost links
C. Cisco Express Forwarding antipolarization
D. different hashing inputs at each layer of the network
Answer: A,D
Explanation:
This document describes how Cisco Express Forwarding (CEF) polarization can cause suboptimal use of redundant paths to a destination network. CEF polarization is the effect when a hash algorithm chooses a particular path and the redundant paths remain completely unused.
How to Avoid CEF Polarization
- Alternate between default (SIP and DIP) and full (SIP + DIP + Layer4 ports) hashing inputs configuration at each layer of the network.
- Alternate between an even and odd number of ECMP links at each layer of the network.The CEF load-balancing does not depend on how the protocol routes are inserted in the routing table. Therefore, the OSPF routes exhibit the same behavior as EIGRP. In a hierarchical network where there are several routers that perform load-sharing in a row, they all use same algorithm to load-share.
The hash algorithm load-balances this way by default:
1: 1
2: 7-8
3: 1-1-1
4: 1-1-1-2
5: 1-1-1-1-1
6: 1-2-2-2-2-2
7: 1-1-1-1-1-1-1
8: 1-1-1-2-2-2-2-2
The number before the colon represents the number of equal-cost paths. The number after the colon represents the proportion of traffic which is forwarded per path.
This means that:
- For two equal cost paths, load-sharing is 46.666%-53.333%, not 50%-50%. - For three equal cost paths, load-sharing is 33.33%-33.33%-33.33% (as expected). - For four equal cost paths, load-sharing is 20%-20%-20%-40% and not 25%-25%
25%-25%.
This illustrates that, when there is even number of ECMP links, the traffic is not load-
balanced. - Cisco IOS introduced a concept called unique-ID/universal-ID which helps avoid CEF polarization. This algorithm, called the universal algorithm (the default in current Cisco IOS versions), adds a 32-bit router-specific value to the hash function (called the universal ID - this is a randomly generated value at the time of the switch boot up that can can be manually controlled). This seeds the hash function on each router with a unique ID, which ensures that the same source/destination pair hash into a different value on different routers along the path. This process provides a better network-wide load-sharing and circumvents the polarization issue. This unique -ID concept does not work for an even number of equal-cost paths due to a hardware limitation, but it works perfectly for an odd number of equal-cost paths. In order to overcome this problem, Cisco IOS adds one link to the hardware adjacency table when there is an even number of equal-cost paths in order to make the system believe that there is an odd number of equal-cost links.
Reference: http://www.cisco.com/c/en/us/support/docs/ip/express-forwarding-cef/116376technote-cef-00.html

NEW QUESTION: 2
手動またはポリシーベースのブロックボリュームバックアップからのブロックボリュームの復元について正しい説明はどれですか。
A. 元のブロックボリュームのバックアップが存在する同じ可用性ドメインに新しいボリュームとして復元する必要があります。
B. 異なるリージョンの任意の可用性ドメインに新しいボリュームとして復元できます。
C. 同じリージョン内の任意の可用性ドメインに新しいボリュームとして復元できます。
D. バックアップとは異なるサイズの新しいボリュームとして復元できます。
Answer: C
Explanation:
References:

NEW QUESTION: 3
WebサーバーはPHPを実行しており、侵入テスターはLFIを使用して、URLを介してパラメーターを渡すことでコマンドを実行しています。これが可能なのは、サーバーログがPHPシステム（）関数を実行するためにポイズニングされたためです。次のうちどれがpasswdファイルの内容を取得しますか？
A. ''&CMD=cat ../../../../etc/passwd7id=34'
B. ''&CMD=cat / etc/passwd%&id= 34''
C. ''&system(CMD) ''cat /etc/passed&id=34''
D. ''&CMD_cat /etc/passwd--&id-34''
Answer: D

NEW QUESTION: 4
Click the Exhibit button.

Which two statements are true regarding the output shown in the exhibit? (Choose two.)
A. Both ESIs use the same VNI.
B. The ESI 00:24:24:24:24:24:24:24:24:24 is an auto-generated ESI.
C. Both ESIs are generated from the router ID.
D. The ESI 05:00:00:fe:4d:00:00:56:0e:00 is an auto-generated ESI.
Answer: A,D