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NEW QUESTION: 1
Which two fundamental modifications, related to traffic forwarding, does MPLS introduce? (Choose two.)
A. For unicast routing, labels are assigned to FECs (in other words, IP prefixes).
B. For multicast routing, labels are assigned to IP multicast groups.
C. IP lookup is performed on every hop within the MPLS core.
D. IP destination routing is reduced to label lookup within the MPLS network.
Answer: A,D
Explanation:
MPLS works by tagging packets with an identifier (a label) to distinguish the LSPs. When a packet is received, the router uses this label (and sometimes also the link over which it was received) to identify the LSP. It then looks up the LSP in its own forwarding table to determine the best link over which to forward the packet, and the label to use on this next hop. A different label is used for each hop, and it is chosen by the router or switch performing the forwarding operation. This allows the use of very fast and simple forwarding engines, as the router can select the label to minimize processing. Ingress routers at the edge of the MPLS network use the packet's destination address to determine which LSP to use. Inside the network, the MPLS routers use only the LSP labels to forward the packet to the egress router.

In the diagram above, LSR (Label Switched Router) A uses the destination IP address on each packet to select the LSP, which determines the next hop and initial label for each packet (21 and 17). When LSR B receives the packets, it uses these labels to identify the LSPs, from which it determines the next hops (LSRs D and C) and labels (47 and 11). The egress routers (LSRs D and C) strip off the final label and route the packet out of the network. As MPLS uses only the label to forward packets, it is protocol-independent, hence the term "Multi-Protocol" in MPLS. Packet forwarding has been defined for all types of layer-2 link technologies, with a different label encoding used in each case.

NEW QUESTION: 2
コマンドラインで一般的なNetFlowデータを視覚化するコマンドは何ですか？
A. show mls sampling
B. show ip flow top-talkers
C. show ip cache flow
D. show mls netflow ip
E. show ip flow export
Answer: C
Explanation:
The following is an example of how to visualize the NetFlow data using the CLI. There are three methods to visualize the data depending on the version of Cisco IOS Software. The traditional show command for NetFlow is "show ip cache flow" also available are two forms of top talker commands. One of the top talkers commands uses a static configuration to view top talkers in the network and another command called dynamic top talkers allows real-time sorting and aggregation of NetFlow data. Also shown is a show MLS command to view the hardware cache on the Cisco Catalyst 6500 Series Switch.
The following is the original NetFlow show command used for many years in Cisco IOS Software.
Information provided includes packet size distribution; basic statistics about number of flows and export timer setting, a view of the protocol distribution statistics and the NetFlow cache.
R3#show ip cache flow
IP packet size distribution (469 total packets):
1-32 64 96 128 160 192 224 256 288 320 352 384 416 448
480 .000 .968 .000 .031 .000 .000 .000 .000 .000 .000 .000 .000 .000 .000 .000 512 544 576
1024 1536 2048 2560 3072 3584 4096 4608
.000 .000 .000 .000 .000 .000 .000 .000 .000 .000 .000 IP Flow Switching Cache, 278544 bytes 7 active, 4089 inactive, 261 added
1278 ager polls, 0 flow alloc failures Active flows timeout in 30 minutes Inactive flows timeout in
15 seconds IP Sub Flow Cache, 25736 bytes
1 active, 1023 inactive, 38 added, 38 added to flow 0 alloc failures, 0 force free 1 chunk, 1 chunk added last clearing of statistics never Protocol Total Flows Packets Bytes Packets Active(Sec) Idle(Sec) -------- Flows /Sec /Flow /Pkt
/Sec /Flow /Flow TCP-WWW 71 0.0 1 40 0.1 1.3 1.2 TCP-BGP 35 0.0 1 40 0.0 1.3 1.2 TCP-other
108 0.1 1 40 0.1 1.3 1.2 UDP-other 37 0.0 1 52 0.0 0.0 15.4 ICMP 3 0.0 5 100 0.0 0.0 15.3 Total:
254 0.2 1 42 0.4 1.1 3.5 (NetFlow cache below) SrcIf SrcIPaddress DstIf DstIPaddress Pr SrcP DstP Pkts Et1/0 172.16.7.2 Null 224.0.0.9 11 0208 0208 1 Et1/0 172.16.10.2 Et0/0 172.16.1.84
06 0087 0087 1
Et1/0 172.16.10.2 Et0/0 172.16.1.84 06 0050 0050 1
Et1/0 172.16.10.2 Et0/0 172.16.1.85 06 0089 0089 1
Et1/0 172.16.10.2 Et0/0 172.16.1.85 06 0050 0050 1
Et1/0 172.16.10.2 Et0/0 172.16.1.86 06 00B3 00B3 1
Et1/0 172.16.10.2 Et0/0 172.16.1.86 06 0185 0185 2
http://www.cisco.com/c/en/us/products/collateral/ios-nx-os-software/ios-netflow/prod_white_paper0900aecd80406232.html

NEW QUESTION: 3
What happens, by default, when one of the resources in a CloudFormation stack cannot be created?
A. Previously-created resources are kept but the stack creation terminates.
B. CloudFormation templates are parsed in advance so stack creation is guaranteed to succeed.
C. Previously-created resources are deleted and the stack creation terminates.
D. The stack creation continues, and the final results indicate which steps failed.
Answer: C
Explanation:
Explanation/Reference:

NEW QUESTION: 4
You have a database named DB1. You plan to create a stored procedure that will insert rows into three different tables. Each insert must use the same identifying value for each table, but the value must increase from one invocation of the stored procedure to the next. Occasionally, the identifying value must be reset to its initial value. You need to design a mechanism to hold the identifying values for the stored procedure to use. What should you do? More than one answer choice may achieve the goal. Select the BEST answer.
A. Create an identity column in each of the three tables. Use the same seed and the same increment for each table. Insert new rows into the tables by using the stored procedure. Use the DBCC CHECKIDENT command to reset the columns as needed.
B. Create a sequence object that holds the next value in the sequence. Retrieve the next value by using the stored procedure. Increment the sequence object to the next value by using an ALTER SEQUENCE statement. Reset the value as needed by using a different ALTER SEQUENCE statement.
C. Create a sequence object that holds the next value in the sequence. Retrieve the next value by using the stored procedure. Reset the value by using an ALTER SEQUENCE statement as needed.
D. Create a fourth table that holds the next value in the sequence. At the end each transaction, update the value by using the stored procedure. Reset the value as needed by using an UPDATE statement.
Answer: C
Explanation:
According to these references, the answer looks correct.
References: http://msdn.microsoft.com/en-us/library/ff878091.aspx http://msdn.microsoft.com/en-us/library/ms176057.aspx http://msdn.microsoft.com/en-us/library/ff878572.aspx http://msdn.microsoft.com/en-us/library/ff878058.aspx