Chapter 28 - Fundamentals of IP Version 6 Flashcards
Which of the following was a short-term solution to the IPv4 address exhaustion problem?
a. IP version 6
b. IP version 5
c. NAT/PAT
d. ARP
C.
NAT, specifically the PAT feature that allows many hosts to use private IPv4 addresses while being supported by a single public IPv4 address, was one short-term solution to the IPv4 address exhaustion problem. IP version 5 existed briefly as an experimental protocol and had nothing to do with IPv4 address exhaustion. IPv6 directly addresses the IPv4 address exhaustion problem, but it is a long-term solution. ARP has no impact on the number of IPv4 addresses used.
A router receives an Ethernet frame that holds an IPv6 packet. The router then makes a decision to route the packet out a serial link. Which of the following statements is true about how a router forwards an IPv6 packet?
a. The router discards the Ethernet data-link header and trailer of the received frame.
b. The router makes the forwarding decision based on the packet’s source IPv6 address.
c. The router keeps the Ethernet header, encapsulating the entire frame inside a new IPv6 packet before sending it over the serial link.
d. The router uses the IPv4 routing table when choosing where to forward the packet.
A.
Routers use the same process steps when routing IPv6 packets as they do when routing IPv4 packets. Routers route IPv6 packets based on the IPv6 addresses, listed inside the IPv6 header in the IPv6 packets, by comparing the destination IPv6 address to the router’s IPv6 routing table. As a result, the router discards the incoming frame’s data-link header and trailer, leaving an IPv6 packet. The router compares the destination (not source) IPv6 address in the header to the router’s IPv6 (not IPv4) routing table and then forwards the packet based on the matched route.
Which of the following is the shortest valid abbreviation for
FE80:0000:0000:0100:0000:0000:0000:0123?
a. FE80::100::123
b. FE8::1::123
c. FE80::100:0:0:0:123:4567
d. FE80:0:0:100::123
D.
If following the steps in the book, the first step removes up to three leading 0s in each quartet, leaving FE80:0:0:100:0:0:0:123. This leaves two strings of consecutive all-0 quartets; by changing the longest string of all 0s to ::, the address is FE80:0:0:100::123.
Which of the following is the shortest valid abbreviation for
2000:0300:0040:0005:6000:0700:0080:0009?
a. 2:3:4:5:6:7:8:9
b. 2000:300:40:5:6000:700:80:9
c. 2000:300:4:5:6000:700:8:9
d. 2000:3:4:5:6:7:8:9
B.
This question has many quartets that make it easy to make a common mistake: removing trailing 0s in a quartet of hex digits. To abbreviate IPv6 addresses, only leading 0s in a quartet should be removed. Many of the quartets have trailing 0s (0s on the right side of the quartet), so make sure to not remove those 0s.
Which of the following is the unabbreviated version of IPv6 address 2001:DB8::200:28?
a. 2001:0DB8:0000:0000:0000:0000:0200:0028
b. 2001:0DB8::0200:0028
c. 2001:0DB8:0:0:0:0:0200:0028
d. 2001:0DB8:0000:0000:0000:0000:200:0028
A.
The unabbreviated version of an IPv6 address must have 32 digits, and only one answer has 32 hex digits. In this case, the original number shows four quartets and a ::. So, the :: was replaced with four quartets of 0000, making the number have eight quartets. Then, for each quartet with fewer than four digits, leading 0s were added so that each quartet has four hex digits.
Which of the following is the prefix for address 2000:0000:0000:0005:6000:0700:0080:0009, assuming a mask of /64?
a. 2000::5::/64
b. 2000::5:0:0:0:0/64
c. 2000:0:0:5::/64
d. 2000:0:0:5:0:0:0:0/64
C.
The /64 prefix length means that the last 64 bits, or last 16 digits, of the address should be changed to all 0s. That process leaves the unabbreviated prefix as 2000:0000:0000:0005:0000:0000:0000:0000. The last four quartets are all 0s, making that string of all 0s be the longest and best string of 0s to replace with ::. After removing the leading 0s in other quartets, the answer is 2000:0:0:5::/64.
