CCNP Route FAQ: IP Version 6 Addressing
Figure: IPv4 Classless and Classful Addressing, IPv6 Addressing
Q1. Which of the following is the shortest valid abbreviation for FE80:0000:0000:0000:0010:0000:0000:0123?
a. FE80::10::123
b. FE8::1::123
c. FE80:0:0:0:10::123
d. FE80::10:0:0:123
Answer: D. Inside a quartet, any leading 0s can be omitted, and one sequence of one or more quartets of all 0s can be replaced with “::”. The correct answer replaces the longer three-quartet sequence of 0s with ::.
Q2. An ISP has assigned prefix 3000:1234:5678::/48 to Company1. Which of the following terms would typically be used to describe this type of public IPv6 prefix?
a. Subnet prefix
b. ISP prefix
c. Global routing prefix
d. Registry prefix
Answer: C. The name of the prefix generally represents the group to which the prefix is given, with the exception of the term global routing. IANA assigns a prefix to a registry (registry prefix). The registry may assign a subset of that range as a prefix to an ISP (ISP prefix). That ISP then subdivides that range of addresses into prefixes and assigns a prefix to one of its customers (site prefix, also called global routing prefix). The Enterprise network engineers then further subdivides the range, often with prefix length 64, into subnet prefixes.
Q3. Which of the following answers lists either a protocol or function that can be used by a host to dynamically learn its own IPv6 address? (Choose two.)
a. Stateful DHCP
b. Stateless DHCP
c. Stateless autoconfiguration
d. Neighbor Discovery Protocol
Answer: A and C. IPv6 supports stateful DHCP, which works similarly to IPv4’s DHCP protocol to dynamically assign the entire IP address. Stateless autoconfiguration also allows for the assignment by finding the prefix from some nearby router and calculating the Interface ID using EUI-64 format. Stateless DHCP simply supplies the DNS server IP addresses, and NDP supplies Layer 2 mapping information.
Q4. Which of the following is helpful to allow an IPv6 host to learn the IP address of a default gateway on its subnet?
a. Stateful DHCP
b. Stateless RS
c. Stateless autoconfiguration
d. Neighbor Discovery Protocol
Answer: D. Stateless autoconfiguration only helps a host learn and form its own IP address, but it does not help the host learn a default gateway. Stateless RS is not a valid term or feature. Neighbor Discovery Protocol (NDP) is used for several purposes, including the same purpose as ARP in IPv4, plus to learn configuration parameters such as a default gateway IP address.
Q5. Which of the following answers lists a multicast IPv6 address?
a. 2000::1:1234:5678:9ABC
b. FD80::1:1234:5678:9ABC
c. FE80::1:1234:5678:9ABC
d. FF80::1:1234:5678:9ABC
Answer: D. Global unicast addresses begin with 2000::/3, meaning the first three bits match the value in hex 2000. Similarly, unique local addresses match FD00::/8, and link local addresses match FE80::/10 (values that begin with FE8, FE9, FEA, and FEB hex). Multicast IPv6 addresses begin FF00::/8, meaning the first two hex digits are F.
Q6. Router R1 has two LAN interfaces and three serial interfaces enabled for IPv6. All the interfaces use link local addresses automatically generated by the router. Which of the following could be the link local address of R1’s interface S0/0?
a. FEA0::200:FF:FE11:0
b. FE80::200:FF:FE11:1111
c. FE80::0213:19FF:FE7B:0:1
d. FEB0::211:11FF:FE11:1111
Answer: B. When created automatically, link local addresses begin FE80::/64, because after the prefix of FE80::/10, the device builds the next 54 bits as binary 0s. Statically assigned link local addresses simply need to confirm to the FE80::/10 prefix. As a result, only two answers are candidates with a beginning quartet of FE80. Of these, only one has only hex 0s in the second, third, and fourth quartets, making answer B the only valid answer.
Q7. Router R1 has the following configuration. Assuming R1’s F0/0 interface has a MAC address of 0200.0011.1111, what IPv6 addresses will R1 list for interface F0/0 in the output of the show ipv6 interface brief command?
interface f0/0
ipv6 address 2345:0:0:8::1/64
a. 2345:0:0:8::1
b. 2345:0:0:8:0:FF:FE11:1111
c. FE80::8:0:FF:FE11:1111
d. FE80:0:0:8::1
Answer: A and C. The ipv6 address command does not list an eui-64 parameter, so R1 does not form its global unicast address using the EUI-64 format. However, it does form its link local address using EUI-64. The show ipv6 interface brief command lists both the global unicast and link local addresses in its output.
Q8. Router R1 lists the following output from a show command. Which of the following is true about R1?
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R1# show ipv6 interface f0/0 FastEthernet0/0 is up, line protocol is up IPv6 is enabled, link-local address is FE80::213:19FF:FE12:3456 No Virtual link-local address(es): Global unicast address(es): 2000::4:213:19FF:FE12:3456, subnet is 2000:0:0:4::/64 [EUI] Joined group address(es): FF02::1 FF02::2 FF02::1:FF:12:3456 |
a. R1’s solicited node multicast address is FF02::1:FF:12:3456.
b. R1’s 2000::4:213:19FF:FE12:3456 address is a global unicast with all 128 bits statically configured.
c. Address FF02::2 is R1’s solicited node multicast.
d. R1’s solicited node multicast, not listed in this output, would be FF02::213:19FF:FE12:3456.
Answer: A. The group addresses listed in the output are the all IPv6 hosts address (FF02::1), the all IPv6 routers address (FF02::2), and the solicited node address that is based on R1’s global unicast address (FF02::1:FF:12:3456). Also, R1’s global unicast address is listed correctly in answer B, but the “[EUI]” notation implies that R1 derived the interface ID portion using EUI-64 conventions.
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