process(bytes);
BitUtils.process(bytes);
app.BitUtils.process(bytes);
util.BitUtils.process(bytes);
import util.BitUtils.*; process(bytes);
SomeApp cannot use the process method in BitUtils.
第1题:
A. the K1 SONET byte; the K2 SONET byte
B. the K2 SONET byte;the K1 SONET byte
C. transmitter switching;receiver detection
D. Line Overhead bytes;Path Overhead bytes
E. Path Overhead bytes;Line Overhead bytes
第2题:
A.TheMPLSheaderis2bytes,andtheEXPfieldis3bitslong.
B.TheMPLSheaderis1byte,andtheEXPfieldis3bitslong.
C.TheMPLSheaderis4bytes,andtheEXPfieldis3bitslong.
D.TheMPLSheaderis3bytes,andtheEXPfieldis3bitslong.
第3题:
Which two statements are true when the extended system ID feature is enabled? ()
第4题:
UPSR protection switching is initiated by (), while BLSR is initiated by ().
第5题:
Given classes defined in two different files: 1.package util; 2.public class BitUtils{ 3.private static void process(byte[]b){} 4.} 1.package app; 2.public class SomeApp{ 3.public static void main(String[]args){ 4.byte[]bytes=newbyte[256]; 5.//insert code here 6.} 7.} What is required at line 5 in class SomeApp to use the process method of BitUtils?()
第6题:
What are the header sizes for point-to-point and multi-point GRE(also known asmGRE) with tunnel key?()
第7题:
Given classes defined in two different files: 1. package util; 2. public class BitUtils { 3. public static void process(byte[]) { /* more code here */ } 4. } 1. package app; 2. public class SomeApp { 3. public static void main(String[] args) { 4. byte[] bytes = new byte[256]; 5. // insert code here 6. } 7. } What is required at line 5 in class SomeApp to use the process method of BitUtils?()
第8题:
process(bytes);
BitUtils.process(bytes);
util.BitUtils.process(bytes);
SomeApp cannot use methods in BitUtils.
import util.BitUtils.*; process(bytes);
第9题:
forwards the packet
fragments the packet
drops the packet silently
drops the packet and sends an ICMP message
第10题:
process(bytes);
BitUtils.process(bytes);
app.BitUtils.process(bytes);
util.BitUtils.process(bytes);
import util.BitUtils. *; process(bytes);
SomeApp cannot use the process method in BitUtils.
第11题:
process(bytes);
BitUtils.process(bytes);
app.BitUtils.process(bytes);
util.BitUtils.process(bytes);
importutil.BitUtils.*;process(bytes);
SomeApp cannot use the process method in BitUtils.
第12题:
The BID is made up of the bridge priority value(two bytes) and bridge MAC address (six bytes).
The BID is made up of the bridge priority (four bits), the system ID (12 bits), and a bridge MAC address (48 bits).
The BID is made up of the system ID (six bytes) and bridge priority value (two bytes).
The system ID value is the VLAN ID (VID).
The system ID value is a unique MAC address allocated from a pool of MAC addresses assigned to the switch or module.
The system ID value is a hex number used to measure the preference of a bridge in the spanning-tree algorithm.
第13题:
Usersofasite-to-siteVPNarereportingperformanceproblems.TheVPNconnectionemploysIPSecandGREandtraversesseveralEthernetsegments.TheVPNpacketsarebeing fragmentedastheytraversethelinks.Whatwouldbetwomethodstoovercomethisproblem?()
A.EmploypathMTUdiscovery.
B.SettheMTUhigherthan1500bytes.
C.Turnoffpre-fragmentationforIPSec.
D.SettheMTUvalueto1400bytes.
第14题:
A.Themaximumnumberofbytesthatcantraversethisinterfacepersecondis1500.
B.Theminimumsegmentsizethatcantraversethisinterfaceis1500bytes.
C.Theminimumpacketsizethatcantraversethisinterfaceis1500bytes.
D.Themaximumpacketsizethatcantraversethisinterfaceis1500bytes.
E.Themaximumframesizethatcantraversethisinterfaceis1500bytes.
第15题:
What are disadvantages of Network Address Translation Transparency?()
第16题:
A router receives an IPv6 packet which is 2000 bytes in length. The MTU of the outgoing interface is 1500 bytes. What action will the router take?()
第17题:
IS-IS的NET地址至少需要()bytes,其中,()bytes的(),()bytes的()和()bytes的(),最多为()bytes.
第18题:
Given classes defined in two different files: 1. package util; 2. public class BitUtils { 3. private static void process(byte[] b) { } 4. } 1. package app; 2. public class SomeApp { 3. public static void main(String[] args) { 4. byte[] bytes = new byte[256]; 5. // insert code here 6. } 7. } What is required at line 5 in class SomeApp to use the process method of BitUtils?()
第19题:
Themaximumnumberofbytesthatcantraversethisinterfacepersecondis1500.
Theminimumsegmentsizethatcantraversethisinterfaceis1500bytes.
Theminimumpacketsizethatcantraversethisinterfaceis1500bytes.
Themaximumpacketsizethatcantraversethisinterfaceis1500bytes.
Themaximumframesizethatcantraversethisinterfaceis1500bytes.
第20题:
the K1 SONET byte; the K2 SONET byte
the K2 SONET byte;the K1 SONET byte
transmitter switching;receiver detection
Line Overhead bytes;Path Overhead bytes
Path Overhead bytes;Line Overhead bytes
第21题:
process(bytes);
BitUtils.process(bytes);
app.BitUtils.process(bytes);
util.BitUtils.process(bytes);
import util.BitUtils.*; process(bytes);
SomeApp cannot use the process method in BitUtils.
第22题:
process(bytes);
BitUtils.process(bytes);
util.BitUtils.process(bytes);
SomeApp cannot use methods in BitUtils.
import util.BitUtils.*; process(bytes);
第23题:
4 bytes for both
4 bytes, and 8 bytes respectively
8 bytes for both
24 bytes for both
第24题:
There is no disadvantage.
It will add 16 bytes per packet of overhead(8 bytes UDP header, and 8 bytes of NON-IKE Marker).
It will add 36 bytes per packet of overhead (20 bytes of IP header, 8 bytes of UDP header, 8 bytes of NON-IKE Marker).
It will require that both crypto peers support this feature in the IOS version. Both crypto peers must also have this feature enabled.