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NEW QUESTION: 1
Which two options are valid for the number of bytes in a BGP AS number? (Choose two.)
A. 4 bytes
B. 8 bytes
C. 6 bytes
D. 2 bytes
E. 16 bytes
Answer: A,D
Explanation:
During the early time of BGP development and standardization, it was assumed that availability of a 16 bit binary
number to identify the Autonomous System (AS) within BGP would have been more than sufficient. The 16 bit AS
number, also known as the 2-byte AS number, provides a pool of 65536 unique Autonomous System numbers. The
IANA manages the available BGP Autonomous System Numbers (ASN) pool, with the assignments being carried out
by the Regional Registries.
The current consumption rate of the publicly available AS numbers suggests that the entire public 2-byte ASN pool
will be fully depleted. A solution to this depletion is the expansion of the existing 2-byte AS number to a 4-byte AS
number, which provides a theoretical 4,294,967,296 unique AS numbers. ARIN has made the following policy changes
in conjunction with the adoption of the solution.
The Cisco IOS BGP "4-byte ASN" feature allows BGP to carry a Autonomous System Number (ASN) encoded as a 4-
byte entity. The addition of this feature allows an operator to use an expanded 4-byte AS number granted by IANA.
NEW QUESTION: 2
Was ist eine Funktion des Wireless LAN Controllers?
A. Überwachen Sie die Aktivität in drahtlosen und kabelgebundenen LANs
B. Registrieren Sie sich bei einem einzelnen Zugriffspunkt, der den Verkehr zwischen kabelgebundenen und kabellosen Endpunkten steuert.
C. Verwenden Sie SSIDs, um zwischen drahtlosen Clients zu unterscheiden.
D. LWAPP-Pakete an Zugriffspunkte senden.
Answer: D
Explanation:
Lightweight APs (LAPs) is devices require no initial configuration. LAPs use the Lightweight Access Point Protocol (LWAPP) to communicate with a WLAN controller (WLC), as shown in the below figure. Controller-based APs are useful in situations where many APs are required in the network. As more APs are added, each AP is automatically configured and managed by the WLC.
NEW QUESTION: 3
Which command is used to examine detailed events processes by the router with Cisco UCCE?
A. showlogs rtr <options...>
B. dumplog router <options...>
C. Dumplog rtr <options...>
D. viewlogs rtr <options...>
Answer: C
NEW QUESTION: 4
You have enabled the Slow Query Log for a short period.
When you process the Slow Query Log, you receive the following snip of output:
Count: 100 Time=0 .22a (22s) Lock=0.00s (0s) Rows=0.0 (0), root[root] @localhost CREATE TABLE 't1' (id serial,id0 varchar(N) unique key,intcaoll INT (N)
,intco12 INT(N) ,intco13 INT(N) ,intco14 INT(N) ,intco15 INT(N)
,charcol1 VARVHAR(N) ,charcol2 VARCHAR(N) charcol3 VARCHAR (N)
,charcol4 VARVHAR(N) ,charcol5 VARCHAR(N) charcol6 VARCHAR (N)
,charcol7 VARVHAR(N) ,charcol8 VARCHAR(N) charcol9 VARCHAR (N) .charcol 10 VACHAR (N) ) Count: 64000 Time-0.02s (1213s) Lock=0.00s (6s) Rows=1.0 (64000), root [root]@ localhost SELECT intocl1, intco12, intco13, intco14, intco15, intco16,intco17, intco18
,intcol9, intcol10, charcol1, charcol2, charcol3, charcol4, charcol5, charcol6
,charcol7, charcol8, charcol9, charcol10 FROM t1 WHERE id = 's'
Count: 1 Time=0.02s (0s) Lock=0.00s (0s) Rows=1.0 (1) agent [agent] @localhost SELECT Select_priv, Repl_client_priv, Show_db_priv, Super_priv,
Process_priv FROM mysql.user WHERE CONCAT (user, 's', host) = CURRENT_USER () Count: 48000 Time=0.02s (778s) Lock=0.00 (3s) Rows=1.0 (48000), root[root]@localhost SELECT intocl1,intcol2,intcol3, intcol4, intcol5, charcol1, charcol2, charcol3
,charcol4, charcol5, charcol6, charcol7, charcol8, charcol9, charcol10 FROM t1 WHERE id = 's' You want to tune the query such that it provides the greatest overall time savings.
Which query will accomplish this?
A. SELECT intcol1, intcol2, intcol3, intcol4, intcol5, charcol1, charcol2, charcol3, charcol4, charcol5, charcol6, charcol7, charcol8, charcol9, charcol10
FROM t1
WHERE id = 's';
B. SELECT Select_priv, Repl_client_priv, Show_db_priv, Super_priv, Process_priv FROM mysql.user
WHERE CONCAT (user,'s', host) = CURRENT_USER();
C. CHEATE TABLE 't1' (id serial, id0 varchar (N) unique key, intcol1 INT (N)
,intcol2 INT (N), intcol3 INT(N) ,intcol4 INT(N), intcol5 INT(N), charool1 VARCHAR (N)
,charcol2 VARCHAR (N), charcol3 VARCHAR(N), charcol4 VARCHAR(N), charcol5 VARCHAR (N)
,charcol6 VARCHAR (N), charcol7 VARCHAR(N), charcol8 VARCHAR(N), charcol9 VARCHAR (N)
,charcol10 VARCHAR (N);
D. SELECT intcol1, intcol2, intcol3, intcol4, intcol5, intcol6, intcol7, intcol8, intcol9, Intcol10, intcol11, intcol12, intcol13, intcol14, intcol15, intcol16, intcol17, intcol18, intcol19, charcol10 FROM t1
WHERE id = 's';
Answer: C