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NEW QUESTION: 1
Which is the default SRST Certificate Provider port?
A. 0
B. 1
C. 2
D. 3
Answer: D
Explanation:
Explanation/Reference:
Reference: https://www.cisco.com/c/en/us/td/docs/voice_ip_comm/cucm/admin/8_6_1/ccmcfg/bccm-861- cm/b02srst.pdf

NEW QUESTION: 2
All of the following are inventory carrying costs, except:
A. Insurance.
B. Opportunity cost on inventory investment.
C. Obsolescence and spoilage.
D. Inspections.
Answer: D
Explanation:
Choice "d" is correct. Inspections. Inspections are part of order costs, not carrying costs.
Choices "a" and "c" are incorrect. Inventory carrying costs include all costs associated with warehousing
(storing) inventory (e.g., storage, insurance, obsolescence, and spoilage associated with holding
inventory).
Choice "b" is incorrect. The economic cost of holding inventory includes the implicit (opportunity) cost of
foregoing a return on the money invested in inventory.

NEW QUESTION: 3
Which two mechanisms can be used to eliminate Cisco Express Forwarding polarization? (Choose two.)
A. different hashing inputs at each layer of the network
B. Cisco Express Forwarding antipolarization
C. the unique-ID/universal-ID algorithm
D. alternating cost links
Answer: A,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/116376technote-cef-00.html

NEW QUESTION: 4
You executed the following command on a Linux machine:
sed s/hello/fun/g file1.txt | tr A-z -Z> file2.txt
What three things would this command do? (Choose three.)
A. Save all changes to file2.txt.
B. Translate all uppercase characters to lowercase.
C. Replace all occurrences of the entry hello with the entry fun.
D. Translate all lowercase characters to uppercase.
E. Modifies file1.txt and creates file2.txt.
F. Replace all occurrences of the entry fun with the entry hello.
Answer: A,C,D