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NEW QUESTION: 1
Which two mechanisms can be used to eliminate Cisco Express Forwarding polarization? (Choose two.)
A. alternating cost links
B. the unique-ID/universal-ID algorithm
C. different hashing inputs at each layer of the network
D. Cisco Express Forwarding antipolarization
Answer: B,C
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/116376-technote-cef-00.html

NEW QUESTION: 2
You need to generate a report an all of the Exchange Server 2007 computers in your Exchange organization. The report must include:
What should you do?
A. Run the Get-OrganizationConfig cmdlet.
B. Run the Test-SystemHealth cmdlet.
C. Run the Health Check scan in the Exchange Best Practices Analyzer.
D. Run the Get-ExchangeServer I Format-List cmdlet.
Answer: C

NEW QUESTION: 3
Francis, the IBM WebSphere administrator has the option of installing IBM Connections 4.0 on either Linux or Windows operating system. Which statement is true about access requirements from an operating system perspective?
A. On Linux,a non-root user can be granted required access permissions to certain data directories thereby allowing non-root installs. OnWindows,only a local administrator can install Connections 4.0.
B. On Linux,a non-root user can be granted required access permissions to certain data directories thereby allowing non-root installs. OnWindows,especially in Windows 2008,the administrator account must be a domain administrator in order to install Connections 4.0.
C. Both on Windows and Linux,a non-administrator or non-root user can perform Connections installs,so long as proper access permissions aregranted to the required target directories.
D. On Windows,any user will be able to install Connections so long as the administrator has shared the target installation directories with FullControl permissions. On Linux,a non-root install is not possible.
Answer: A