* Scalability
* VLSM support
* Minimal overhead
* Support for connecting networks using routers of multiple vendors
Which of the following routing protocol would best serve your needs?
A. VTP
B. RIP version 1
C. EIGRP
D. OSPF
E. IGRP
F. CDP
Answer: D
Explanation:
Since one of the requirements is that the routing protocol must support other vendors, our
only choices are RIP and OSPF. Since RIP version 1 does not support VLSM, OSPF is
the only choice.
Incorrect Answers:
A. VTP is the VLAN Trunking Protocol. This is not a routing protocol.
B. RIP version one does not support VLSM. Note that RIPv2 does support VLSM, and
would be a valid choice.
C, E. EIGRP and IGRP are Cisco proprietary routing protocols, and are not supported by
other vendors.
F. CDP is the Cisco Discovery Protocol, which is used to exchange information between
Cisco devices. It can only be used between Cisco routers and switches, and it is not a
routing protocol.
When designing OSPF networks; what is the purpose of using a hierarchical design? (Select all choices that apply)
When designing OSPF networks; what is the purpose of using a hierarchical design? (Select all choices that apply)
A. To reduce the complexity of router configuration
B. To speed up convergence
C. To confine network instability to single areas of the network
D. To reduce routing overhead
E. To lower costs by replacing routers
F. To decrease latency
Answer: B, C, D
Explanation:
An OSPF network designed in a hierarchical fashion with different areas is used because
a small change in the topology of a single area won't force every router to run the SPF
algorithm. Changes in one area are limited to that area only, not to every router within the
entire network. Confining the topology changes to one area reduces the overhead and
speeds the convergence of the network.
A. To reduce the complexity of router configuration
B. To speed up convergence
C. To confine network instability to single areas of the network
D. To reduce routing overhead
E. To lower costs by replacing routers
F. To decrease latency
Answer: B, C, D
Explanation:
An OSPF network designed in a hierarchical fashion with different areas is used because
a small change in the topology of a single area won't force every router to run the SPF
algorithm. Changes in one area are limited to that area only, not to every router within the
entire network. Confining the topology changes to one area reduces the overhead and
speeds the convergence of the network.
In network that support VLSM, which network mask should be used for point-to-point WAN links in order to reduce waste of IP addresses?
In network that support VLSM, which network mask should be used for
point-to-point WAN links in order to reduce waste of IP addresses?
A. /24
B. /30
C. /27
D. /26
E. /32
Answer: B
Explanation:
A 30-bit mask is used to create subnets with two valid host addresses. This is the exact number needed for a point-to-point connection.
point-to-point WAN links in order to reduce waste of IP addresses?
A. /24
B. /30
C. /27
D. /26
E. /32
Answer: B
Explanation:
A 30-bit mask is used to create subnets with two valid host addresses. This is the exact number needed for a point-to-point connection.
If an Ethernet port on router was assigned an IP address of 172.16.112.1/20, what is the maximum number of hosts allowed on this subnet?
If an Ethernet port on router was assigned an IP address of 172.16.112.1/20, what is the maximum number of hosts allowed on this subnet?
A. 1024
B. 2046
C. 4094
D. 4096
E. 8190
Answer: C
Explanation:
Given IP address of 172.16.112.1 / 20,
subnet mask: 255.255.240.0
max. num of hosts =(( 2^12) -2 ) = 4096-2 = 4094
A. 1024
B. 2046
C. 4094
D. 4096
E. 8190
Answer: C
Explanation:
Given IP address of 172.16.112.1 / 20,
subnet mask: 255.255.240.0
max. num of hosts =(( 2^12) -2 ) = 4096-2 = 4094
Which of the following IP addresses fall into the CIDR block of 115.54.4.0/22? Select three
A. 115.54.8.32
B. 115.54.7.64
C. 115.54.6.255
D. 115.54.3.32
E. 115.54.5.128
F. 115.54.12.128
Answer: B, C, E
Explanation:
Given CIDR block of 115.54.4.0 /22:
subnet mask : 255.255.252.0
theIP address range would be 115.54.4.1 to 115.54.7.254.
Therefore, 115.54.5.128 (E),115.54.6.255 (C) and 115.54.7.64 (B) are correct.
B. 115.54.7.64
C. 115.54.6.255
D. 115.54.3.32
E. 115.54.5.128
F. 115.54.12.128
Answer: B, C, E
Explanation:
Given CIDR block of 115.54.4.0 /22:
subnet mask : 255.255.252.0
theIP address range would be 115.54.4.1 to 115.54.7.254.
Therefore, 115.54.5.128 (E),115.54.6.255 (C) and 115.54.7.64 (B) are correct.
What is the network address for a host with the IP address 201.100.5.68/28?What is the network address for a host with the IP address 201.100.5.68/28?
A. 201.100.5.0
B. 201.100.5.32
C. 201.100.5.64
D. 201.100.5.65
E. 201.100.5.31
F. 201.100.5.1
Answer: C
Explanation:
This is a C ip with a subnet mask of 255.255.255.240
the host 201.100.5.68/28 belong to the second subnet which is 201.100.5.64
this is determined by doing the following:
subnets?2^4-2=14
hosts?2^4-2=14
valid subnet range?256-240=16
16+16=32,16+32=48,16+48=64,64+16=80 and so as you can see the ip
201.100.5.68 belongs to the second subnet which is.64
B. 201.100.5.32
C. 201.100.5.64
D. 201.100.5.65
E. 201.100.5.31
F. 201.100.5.1
Answer: C
Explanation:
This is a C ip with a subnet mask of 255.255.255.240
the host 201.100.5.68/28 belong to the second subnet which is 201.100.5.64
this is determined by doing the following:
subnets?2^4-2=14
hosts?2^4-2=14
valid subnet range?256-240=16
16+16=32,16+32=48,16+48=64,64+16=80 and so as you can see the ip
201.100.5.68 belongs to the second subnet which is.64
Which of the following IP addresses is a private IP address? Select all that apply.
A. 12.0.0.1
B. 168.172.19.39
C. 172.20.14.36
D. 172.33.194.30
E. 192.168.42.34
Answer: C, E
Explanation:
RFC 1918 Private Address Space:
Range of IP Addresses Class of Networks Number of Network
10.0.0.0 to
10.255.255.255.255
A 1
172.16.0.0 to
172.31.255.255
B 16
192.168.0.0 to
192.168.255.255
C 256
B. 168.172.19.39
C. 172.20.14.36
D. 172.33.194.30
E. 192.168.42.34
Answer: C, E
Explanation:
RFC 1918 Private Address Space:
Range of IP Addresses Class of Networks Number of Network
10.0.0.0 to
10.255.255.255.255
A 1
172.16.0.0 to
172.31.255.255
B 16
192.168.0.0 to
192.168.255.255
C 256
Which of the following are true regarding a network using a subnet mask of 255.255.248.0?
A. It corresponds to a Class A address with 13 bits borrowed.
B. It corresponds to a Class B address with 4 bits borrowed.
C. The network address of the last subnet will have 248 in the 3rd octet.
D. The first 21 bits make the host portion of the address.
E. This subnet mask allows for 16 total subnets to be created.
F. The subnetwork numbers will be in multiples of 8.
Answer: A, C, F
Explanation:
This subnet mask includes the first 5 bits within the third octet, so for a class A address
13 bits will be used for the mask (8 bits in the second octet plus 5 in the third).
Since the first 5 bits are used in this octet, that means that remaining 3 bits in this octet
will be available for hosts, so each network will be a factor of 8, making the last available
subnet with a .248 in the third octet.
B. It corresponds to a Class B address with 4 bits borrowed.
C. The network address of the last subnet will have 248 in the 3rd octet.
D. The first 21 bits make the host portion of the address.
E. This subnet mask allows for 16 total subnets to be created.
F. The subnetwork numbers will be in multiples of 8.
Answer: A, C, F
Explanation:
This subnet mask includes the first 5 bits within the third octet, so for a class A address
13 bits will be used for the mask (8 bits in the second octet plus 5 in the third).
Since the first 5 bits are used in this octet, that means that remaining 3 bits in this octet
will be available for hosts, so each network will be a factor of 8, making the last available
subnet with a .248 in the third octet.
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