Modul 11-13 Flashcards
What is the prefix length notation for the subnet mask 255.255.255.224?
- /25
- /26
- /27
- /28
What is the prefix length notation for the subnet mask 255.255.255.224?
- /25
- /26
- /27
- /28
Explanation: The binary format for 255.255.255.224 is 11111111.11111111.11111111.11100000. The prefix length is the number of consecutive 1s in the subnet mask. Therefore, the prefix length is /27.
How many valid host addresses are available on an IPv4 subnet that is configured with a /26 mask?
- 254
- 190
- 192
- 62
- 64
How many valid host addresses are available on an IPv4 subnet that is configured with a /26 mask?
- 254
- 190
- 192
- 62
- 64
Which subnet mask would be used if 5 host bits are available?
- 255.255.255.0
- 255.255.255.128
- 255.255.255.224
- 255.255.255.240
Which subnet mask would be used if 5 host bits are available?
- 255.255.255.0
- 255.255.255.128
- 255.255.255.224
- 255.255.255.240
A network administrator subnets the 192.168.10.0/24 network into subnets with /26 masks. How many equal-sized subnets are created?
- 1
- 2
- 4
- 8
- 16
- 64
A network administrator subnets the 192.168.10.0/24 network into subnets with /26 masks. How many equal-sized subnets are created?
- 1
- 2
- 4
- 8
- 16
- 64
An administrator wants to create four subnetworks from the network address 192.168.1.0/24. What is the network address and subnet mask of the second useable subnet?
- subnetwork 192.168.1.64
subnet mask 255.255.255.192
- subnetwork 192.168.1.32
subnet mask 255.255.255.240
- subnetwork 192.168.1.64
subnet mask 255.255.255.240
- subnetwork 192.168.1.128
subnet mask 255.255.255.192
- subnetwork 192.168.1.8
subnet mask 255.255.255.224
An administrator wants to create four subnetworks from the network address 192.168.1.0/24. What is the network address and subnet mask of the second useable subnet?
- subnetwork 192.168.1.64
subnet mask 255.255.255.192
- subnetwork 192.168.1.32
subnet mask 255.255.255.240
- subnetwork 192.168.1.64
subnet mask 255.255.255.240
- subnetwork 192.168.1.128
subnet mask 255.255.255.192
- subnetwork 192.168.1.8
subnet mask 255.255.255.224
Explanation: The number of bits that are borrowed would be two, thus giving a total of 4 useable subnets:
192.168.1.0
192.168.1.64
192.168.1.128
192.168.1.192
Because 2 bits are borrowed, the new subnet mask would be /26 or 255.255.255.192
How many bits must be borrowed from the host portion of an address to accommodate a router with five connected networks?
- two
- three
- four
- five
How many bits must be borrowed from the host portion of an address to accommodate a router with five connected networks?
- two
- three
- four
- five
Explanation: Each network that is directly connected to an interface on a router requires its own subnet. The formula 2n, where n is the number of bits borrowed, is used to calculate the available number of subnets when borrowing a specific number of bits.
How many host addresses are available on the 192.168.10.128/26 network?
- 30
- 32
- 60
- 62
- 64
How many host addresses are available on the 192.168.10.128/26 network?
- 30
- 32
- 60
- 62
- 64
Explanation: A /26 prefix gives 6 host bits, which provides a total of 64 addresses, because 26 = 64. Subtracting the network and broadcast addresses leaves 62 usable host addresses.
How many host addresses are available on the network 172.16.128.0 with a subnet mask of 255.255.252.0?
- 510
- 512
- 1022
- 1024
- 2046
- 2048
How many host addresses are available on the network 172.16.128.0 with a subnet mask of 255.255.252.0?
- 510
- 512
- 1022
- 1024
- 2046
- 2048
Explanation: A mask of 255.255.252.0 is equal to a prefix of /22. A /22 prefix provides 22 bits for the network portion and leaves 10 bits for the host portion. The 10 bits in the host portion will provide 1022 usable IP addresses (210 – 2 = 1022).
What three blocks of addresses are defined by RFC 1918 for private network use? (Choose three.)
- 10.0.0.0/8
- 172.16.0.0/12
- 192.168.0.0/16
- 100.64.0.0/14
- 169.254.0.0/16
- 239.0.0.0/8
What three blocks of addresses are defined by RFC 1918 for private network use? (Choose three.)
- 10.0.0.0/8
- 172.16.0.0/12
- 192.168.0.0/16
- 100.64.0.0/14
- 169.254.0.0/16
- 239.0.0.0/8
Explanation: RFC 1918, Address Allocation for Private Internets, defines three blocks of IPv4 address for private networks that should not be routable on the public Internet.
10.0.0.0/8
172.16.0.0/12
192.168.0.0/16
A site administrator has been told that a particular network at the site must accommodate 126 hosts. Which subnet mask would be used that contains the required number of host bits?
- 255.255.255.0
- 255.255.255.128
- 255.255.255.224
- 255.255.255.240
A site administrator has been told that a particular network at the site must accommodate 126 hosts. Which subnet mask would be used that contains the required number of host bits?
- 255.255.255.0
- 255.255.255.128
- 255.255.255.224
- 255.255.255.240
Explanation: The subnet mask of 255.255.255.0 has 8 host bits. The mask of 255.255.255.128 results in 7 host bits. The mask of 255.255.255.224 has 5 host bits. Finally, 255.255.255.240 represents 4 host bits.
What is the usable number of host IP addresses on a network that has a /26 mask?
- 256
- 254
- 64
- 62
- 32
- 16
What is the usable number of host IP addresses on a network that has a /26 mask?
- 256
- 254
- 64
- 62
- 32
- 16
Explanation: A /26 mask is the same as 255.255.255.192. The mask leaves 6 host bits. With 6 host bits, 64 IP addresses are possible. One address represents the subnet number and one address represents the broadcast address, which means that 62 addresses can then be used to assign to network devices.
Which address prefix range is reserved for IPv4 multicast?
- 240.0.0.0 – 254.255.255.255
- 224.0.0.0 – 239.255.255.255
- 169.254.0.0 – 169.254.255.255
- 127.0.0.0 – 127.255.255.255
Which address prefix range is reserved for IPv4 multicast?
- 240.0.0.0 – 254.255.255.255
- 224.0.0.0 – 239.255.255.255
- 169.254.0.0 – 169.254.255.255
- 127.0.0.0 – 127.255.255.255
Explanation: Multicast IPv4 addresses use the reserved class D address range of 224.0.0.0 to 239.255.255.255.
A high school in New York (school A) is using videoconferencing technology to establish student interactions with another high school (school B) in Russia. The videoconferencing is conducted between two end devices through the Internet. The network administrator of school A configures the end device with the IP address 209.165.201.10. The administrator sends a request for the IP address for the end device in school B and the response is 192.168.25.10. Neither school is using a VPN. The administrator knows immediately that this IP will not work. Why?
- This is a loopback address.
- This is a link-local address.
- This is a private IP address.
- There is an IP address conflict.
A high school in New York (school A) is using videoconferencing technology to establish student interactions with another high school (school B) in Russia. The videoconferencing is conducted between two end devices through the Internet. The network administrator of school A configures the end device with the IP address 209.165.201.10. The administrator sends a request for the IP address for the end device in school B and the response is 192.168.25.10. Neither school is using a VPN. The administrator knows immediately that this IP will not work. Why?
- This is a loopback address.
- This is a link-local address.
- This is a private IP address.
- There is an IP address conflict.
Which three addresses are valid public addresses? (Choose three.)
- 198.133.219.17
- 192.168.1.245
- 10.15.250.5
- 128.107.12.117
- 172.31.1.25
- 64.104.78.227
Which three addresses are valid public addresses? (Choose three.)
- 198.133.219.17
- 192.168.1.245
- 10.15.250.5
- 128.107.12.117
- 172.31.1.25
- 64.104.78.227
Explanation: The ranges of private IPv4 addresses are as folllows:
10.0.0.0 – 10.255.255.255
172.16.0.0 – 172.31.255.255
192.168.0.0 – 192.168.255.255
A message is sent to all hosts on a remote network. Which type of message is it?
- limited broadcast
- multicast
- directed broadcast
- unicast
A message is sent to all hosts on a remote network. Which type of message is it?
- limited broadcast
- multicast
- directed broadcast
- unicast
Explanation: A directed broadcast is a message sent to all hosts on a specific network. It is useful for sending a broadcast to all hosts on a nonlocal network. A multicast message is a message sent to a selected group of hosts that are part of a subscribing multicast group. A limited broadcast is used for a communication that is limited to the hosts on the local network. A unicast message is a message sent from one host to another.
A company has a network address of 192.168.1.64 with a subnet mask of 255.255.255.192. The company wants to create two subnetworks that would contain 10 hosts and 18 hosts respectively. Which two networks would achieve that? (Choose two.)
- 192.168.1.16/28
- 192.168.1.64/27
- 192.168.1.128/27
- 192.168.1.96/28
- 192.168.1.192/28
A company has a network address of 192.168.1.64 with a subnet mask of 255.255.255.192. The company wants to create two subnetworks that would contain 10 hosts and 18 hosts respectively. Which two networks would achieve that? (Choose two.)
- 192.168.1.16/28
- 192.168.1.64/27
- 192.168.1.128/27
- 192.168.1.96/28
- 192.168.1.192/28
Explanation: Subnet 192.168.1.64 /27 has 5 bits that are allocated for host addresses and therefore will be able to support 32 addresses, but only 30 valid host IP addresses. Subnet 192.168.1.96/28 has 4 bits for host addresses and will be able to support 16 addresses, but only 14 valid host IP addresses.
Which address is a valid IPv6 link-local unicast address?
- FEC8:1::FFFF
- FD80::1:1234
- FE80::1:4545:6578:ABC1
- FE0A::100:7788:998F
- FC90:5678:4251:FFFF
Which address is a valid IPv6 link-local unicast address?
- FEC8:1::FFFF
- FD80::1:1234
- FE80::1:4545:6578:ABC1
- FE0A::100:7788:998F
- FC90:5678:4251:FFFF
Explanation: IPv6 LLAs are in the fe80::/10 range. The /10 indicates that the first 10 bits are 1111 1110 10xx xxxx. The first hextet has a range of 1111 1110 1000 0000 (fe80) to 1111 1110 1011 1111 (febf).
Which of these addresses is the shortest abbreviation for the IP address: 3FFE:1044:0000:0000:00AB:0000:0000:0057?
- 3FFE:1044::AB::57
- 3FFE:1044::00AB::0057
- 3FFE:1044:0:0:AB::57
- 3FFE:1044:0:0:00AB::0057
- 3FFE:1044:0000:0000:00AB::57
- 3FFE:1044:0000:0000:00AB::0057
Which of these addresses is the shortest abbreviation for the IP address: 3FFE:1044:0000:0000:00AB:0000:0000:0057?
- 3FFE:1044::AB::57
- 3FFE:1044::00AB::0057
- 3FFE:1044:0:0:AB::57
- 3FFE:1044:0:0:00AB::0057
- 3FFE:1044:0000:0000:00AB::57
- 3FFE:1044:0000:0000:00AB::0057
Explanation: The rules for reducing the notation of IPv6 addresses are:
1. Omit any leading 0s (zeros) in any hextet.
2. Replace any single, contiguous string of one or more 16-bit hextets consisting of all zeros with a double colon (::) .
3. The double colon (::) can only be used once within an address.
A network administrator has received the IPv6 prefix 2001:DB8::/48 for subnetting. Assuming the administrator does not subnet into the interface ID portion of the address space, how many subnets can the administrator create from the /48 prefix?
- 16
- 256
- 4096
- 65536
A network administrator has received the IPv6 prefix 2001:DB8::/48 for subnetting. Assuming the administrator does not subnet into the interface ID portion of the address space, how many subnets can the administrator create from the /48 prefix?
- 16
- 256
- 4096
- 65536
Explanation: With a network prefix of 48, there will be 16 bits available for subnetting because the interface ID starts at bit 64. Sixteen bits will yield 65536 subnets.
Given IPv6 address prefix 2001:db8::/48, what will be the last subnet that is created if the subnet prefix is changed to /52?
- 2001:db8:0:f00::/52
- 2001:db8:0:8000::/52
- 2001:db8:0:f::/52
- 2001:db8:0:f000::/52
Given IPv6 address prefix 2001:db8::/48, what will be the last subnet that is created if the subnet prefix is changed to /52?
- 2001:db8:0:f00::/52
- 2001:db8:0:8000::/52
- 2001:db8:0:f::/52
- 2001:db8:0:f000::/52
Explanation: Prefix 2001:db8::/48 has 48 network bits. If we subnet to a /52, we are moving the network boundary four bits to the right and creating 16 subnets. The first subnet is 2001:db8::/52 the last subnet is 2001:db8:0:f000::/52.
Consider the following range of addresses:
- 2001:0DB8:BC15:00A0:0000::
- 2001:0DB8:BC15:00A1:0000::
- 2001:0DB8:BC15:00A2:0000::
- …
- 2001:0DB8:BC15:00AF:0000::
The prefix-length for the range of addresses is ???
Consider the following range of addresses:
- 2001:0DB8:BC15:00A0:0000::
- 2001:0DB8:BC15:00A1:0000::
- 2001:0DB8:BC15:00A2:0000::
- …
- 2001:0DB8:BC15:00AF:0000::
The prefix-length for the range of addresses is /60 .
Explanation: All the addresses have the part 2001:0DB8:BC15:00A in common. Each number or letter in the address represents 4 bits, so the prefix-length is /60.
What type of IPv6 address is FE80::1?
- loopback
- link-local
- multicast
- global unicast
What type of IPv6 address is FE80::1?
- loopback
- link-local
- multicast
- global unicast
Explanation: Link-local IPv6 addresses start with FE80::/10, which is any address from FE80:: to FEBF::. Link-local addresses are used extensively in IPv6 and allow directly connected devices to communicate with each other on the link they share.
What is used in the EUI-64 process to create an IPv6 interface ID on an IPv6 enabled interface?
- the MAC address of the IPv6 enabled interface
- a randomly generated 64-bit hexadecimal address
- an IPv6 address that is provided by a DHCPv6 server
- an IPv4 address that is configured on the interface
What is used in the EUI-64 process to create an IPv6 interface ID on an IPv6 enabled interface?
- the MAC address of the IPv6 enabled interface
- a randomly generated 64-bit hexadecimal address
- an IPv6 address that is provided by a DHCPv6 server
- an IPv4 address that is configured on the interface
Explanation: The EUI-64 process uses the MAC address of an interface to construct an interface ID (IID). Because the MAC address is only 48 bits in length, 16 additional bits (FF:FE) must be added to the MAC address to create the full 64-bit interface ID.