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Answers and Explanations

  1. C. 172.16.0-3.0/22 breaks down into the following binary equivalents:

    172.16.0.0 = 10101100.00010000.00000000.00000000

    172.16.1.0 = 10101100.00010000.00000001.00000000

    172.16.2.0 = 10101100.00010000.00000010.00000000

    172.16.3.0 = 10101100.00010000.00000011.00000000

    Based on these binary values, you can see that the first 22 bits on each one of these IP addresses are the same. This makes the accurate summary address 172.16.0.0/22. All other answers are not valid.

  2. D. To find the valid number of hosts on a subnet, use the formula (2x) – 2, where "x" is the number of host bits in the subnet mask. A /21 (255.255.248.0) subnet mask has 11 host bits. (211) – 2 = 2046. All other answers are not valid.
  3. C, F. These styles of VLSM are common on the ICND2/CCNA exam. It's usually best to reverse-engineer the given subnets to see what ranges are in use and then compare the answers to fill in what's left over. In this case, the given ranges are:

    • 192.168.1.32-63/27 (Behind the Oracle router)
    • 192.168.1.64-95/27 (Behind the Morpheus router)
    • 192.168.1.96-111/28 (Behind the Trinity router)
    • 192.168.1.4-7/30 (Neo-Oracle WAN link)
    • 192.168.1.8-11/30 (Neo-Morpheus WAN link)

    With that given information, answer C (192.168.1.128-255/25) and answer F (192.168.1.0-3/30) best fit the scenario. Answers A, D, and E are incorrect because these are invalid subnet addresses. Answer B is incorrect because the /26 mask will only provide 64 IP addresses, which is not sufficient for the size of Network A. These addresses also overlap with the other office subnets.

  4. C, E. IPv6 uses three types of communication: Unicast (one-to-one), Multicast (one-to-many), and Anycast (one-to-closest). Answer C is incorrect because the concept of Broadcast messaging is tied with the IPv4 protocol and is no longer valid in IPv6. Answer E is incorrect because there is no such thing as Cryptocast messaging.
  5. A, B, D. An IPv6 address consists of eight octets that can be four hexadecimal characters each. Consecutive sets of zeros can be abbreviated with a double colon (::), but this can only be used once in each IP address. Leading zeros can also be dropped. Based on these rules, addresses from the question can be described as:

    • 2001:0db8:0000:0000:0000:0000:1428:57ab (Valid, eight octets)
    • 2001:0db8::1428:57ab (Valid, same address as above with abbreviation)
    • 2001::1685:2123::1428:57ab (Invalid use of double colon)
    • 2001:99:ab:1:99:2:1:9 (Valid, dropped leading zeros)
    • 2001:1428:57ab:1685:2123: 1428:57ab (Invalid, only seven octets)
  6. E. A dual-stack router can receive requests from both IPv4 and IPv6 clients on the same interface. This provides a smoother transition between the two protocols. Answers A, B, and D are incorrect because these all describe tunneling methods (IPv4 tunneled through an IPv6 network = 4to6). Answer C is incorrect because this describes a newer form of NAT that is able to translate between IPv4 and IPv6 addressing.
  7. E, I. 10.170.120-123.0/24 breaks into the following binary equivalents:

    • 10.170.120.0 = 00001010.10101010.01111000.00000000
    • 10.170.121.0 = 00001010.10101010.01111001.00000000
    • 10.170.122.0 = 00001010.10101010.01111010.00000000
    • 10.170.123.0 = 00001010.10101010.01111011.00000000

    Based on this, we can see that the first 22 bits of each subnet are the same, thus the summary address is 10.170.120.0/22 for Area 0 (answer E). 10.170.0-1.0/24 breaks into the following binary equivalents:

    • 10.170.0.0 = 00001010.10101010. 00000000.00000000
    • 10.170.1.0 = 00001010.10101010. 00000001.00000000

    Based on this, we can see that the first 23 bits of each subnet are the same, thus the summary address is 10.170.0.0/23 (answer I). All other answers are not valid.

  8. D. This subnet mask gives you four additional subnets using VLSM, with up to 14 hosts per subnetwork. Answer A is incorrect, as it is a higher subnet mask than your original /26, which is actually called supernetting. Answer B is incorrect because it is your original subnet mask. Answer C is incorrect, as it does not give you enough subnets. Answer E is incorrect, as it gives you enough subnets (six), but you would have only six hosts.
  9. A, C, and D. A subnet mask of 255.255.255.240 divides the fourth octet into subnet parts: the highest four bits and a host port (the lowest four bits). You simply check the fourth octet to ensure that all subnet and host parts are okay. The host bit portion cannot be 0000 or 1111. Answers A, C, and D are correct because 33 in decimal is 00100001, 119 in decimal is 01110111, and 126 in decimal is 1111110. Answer B is incorrect, as 112 in decimal is 1110000 in binary. This is not a valid host address in this network. All its host bits are zero. Answer E is incorrect, as 175 in decimal is 10101111 in binary. All host bits are ones. This is the local broadcast address and cannot be used as a host address. Answer F is incorrect, as 208 in decimal is 11010000 in binary. This is not a valid host address in this network, and all its host bits are zero.
  10. B, D, F. The departments will use the following subnet masks:

    Department

    Number of Users

    Subnet Mask

    Corporate

    117

    255.255.255.128 (126 hosts)

    Customer Support

    15

    255.255.255.224 (30 hosts)

    Financial

    25

    255.255.255.224 (30 hosts)

    HR

    5

    255.255.255.248 (6 hosts)

    Engineering

    5

    255.255.255.248 (6 hosts)

    All other answers are invalid.

  11. C. It is not a valid host address; 192.168.5.95/27 is a directed broadcast address for the 192.168.5.64 network. Answer A is incorrect, as you can certainly assign Class C addresses to any type of interface. Answer B is incorrect, as the /27 mask is the 255.255.255.224 subnet mask, which is perfectly valid. Answer D is incorrect because 192.168.5.95/27 represents a broadcast address rather than a network address. Answer E is incorrect because it is a private IP address. Answer F is incorrect, as the fact that it is a private IP address will not cause it to be refused by an interface.
  12. B. One of the IPv6 transition schemes includes an IPv6 to IPv4 (6to4) tunneling method. This allows you to tunnel your IPv6 networks through an IPv4 network. Answer A is incorrect because the VPN is actually using the IPv4 Internet as its connection point. Answer C is incorrect because 4to6 tunnels involve IPv4 networks tunneling through an IPv6 network. Answer D is incorrect because NAT-PT is a form of NAT that can handle translations between protocol suites. It is not related to VPN technology.
  13. B, C. 172.16.112.0/24 and 172.16.113.0/24 can be summarized into the single entry 172.16.112.0/23. Answers A, D, and E represent addresses that cannot be summarized into a single routing entry that only encompasses two of the /24 network ranges.
  14. Figure 8.10

    Figure 8.10 Network diagram.

  15. B, C, E. RIPv2, EIGRP, and OSPF all support Variable Length Subnet Mask (VLSM) capabilities. Answers A and D are incorrect because RIPv1 and IGRP are classful routing protocols and do not support VLSM.
  16. A. When auto-summarization is enabled, EIGRP will summarize networks back to their classful boundary anytime a discontiguous network is reached (such as the transition from 10.0.0.0/8 networks to 172.16.0.0/16 networks). In this case, R1 will summarize the corporate network back to 10.0.0.0/8 as it passes the routing update to R2. It will also summarize all the 172.16.x.x networks back to 172.16.0.0/16 as it passes the routing update into the corporate network. Answers B, C, and D are incorrect because auto-summarization only summarizes back to a classful boundary rather than to more specific subnet masks. Answer E is incorrect because EIGRP can handle VLSM with auto-summarization enabled.
  17. D. IPv6 addresses have been expanded to 128-bit addressing from the 32-bit addressing of IPv4. This provides a virtually inexhaustible number of addresses (although, I'm sure many thought the same of the IPv4 address space). All other answers do not apply.
  18. E. The Internet-valid addresses are considered "global" addresses in IPv6. They are specified to begin with 2000::/3. Answer A is incorrect because private addresses are for use in a private network, as it currently happens in IPv4 addressing. Answer B is incorrect because global addresses have replaced public addresses. Answers C and D do not apply directly to IPv6 addressing.
  19. C. Link-local addressing is a new concept when moving from IPv4 to IPv6. Link-local addresses are used to communicate directly on a link. This is used for communication such as establishing OSPF neighbor relationships or sending RIP routes. Answer A is incorrect because global addresses can access the Internet directly. Answer B is incorrect because private IPv6 addresses can route through an organization. The addresses shown in answers D and E do not exist in the IPv6 environment.
  20. F. IPv6 addresses are assigned using the ipv6 address command. In IPv6, there is no decimal version of the subnet mask; all subnet masks are written in bit-notation. Answers A, C, and E are incorrect because they use the decimal version of the subnet mask. Answers B and D are incorrect because they use the incorrect command.
  21. B. The exact syntax to enable the RIPng (RIP for IPv6) routing protocol is ipv6 router rip <tag>. The tag can be anything from a number to a name; in this question, the tag was "RIPng". This tag must be used when enabling RIP on an interface-by-interface basis. Answers A, C, and D will produce invalid syntax messages.
  22. B, D, F, H. To perform the most efficient VLSM, always begin with the biggest subnet first. The ranges that will properly address the network in Figure 8.8 are as follows:

    • Server Farm: 150.60.130.0-127/25 (answer H)
    • College of Education: 150.60.130.128-159/27 (answer B)
    • College of Business: 150.60.130.160-191/27 (answer F)
    • Administration: 150.60.130.192-207/28 (answer D)

    Answers A, C, E, and G are incorrect because they would each address a portion of the network but would not function correctly with the other given subnets.

  23. The binary equivalents of the shown addresses are as follows, reflected in answer E:

    192.168.112.0 = 11000000.10101000.01110000.00000000

    192.168.113.0 = 11000000.10101000.01110001.00000000

    192.168.114.0 = 11000000.10101000.01110010.00000000

    192.168.115.0 = 11000000.10101000.01110011.00000000

    192.168.116.0 = 11000000.10101000.01110100.00000000

    192.168.117.0 = 11000000.10101000.01110101.00000000

    192.168.118.0 = 11000000.10101000.01110110.00000000

    192.168.119.0 = 11000000.10101000.01110111.00000000

    This shows that the first 21 bits of all these addresses are the same making the summary address 192.168.112.0/21. Answers A, B, and C are incorrect because 192.168.110.0 is an inaccurate starting point for the subnets. Answers D and F assume the wrong subnet mask.

  24. The binary equivalents of the shown addresses are as follows, reflected in answer D:

    172.16.4.0 =

    10101100.00010000.00000100.00000000

    172.16.5.0 =

    10101100.00010000.00000101.00000000

    172.16.6.0 =

    10101100.00010000.00000110.00000000

    172.16.128.0 =

    10101100.00010000.10000000.00000000

    Because of the last subnet (172.16.128.0/24), a good summarization is not possible with these subnets. We must drop back to the classful summarization of 172.16.0.0/16. This helps demonstrate why discontiguous network addressing can destroy your network summarization efficiency. Answers A, B, and C are incorrect because the summarization addresses fail to encompass the 172.16.128.0/24 subnet.

  25. D. The IP address 172.20.2.255/23 comes from the range of 172.20.2.0 through 172.30.3.255. The network ID 172.20.2.0 cannot be assigned to a host. The broadcast ID 172.20.3.255 cannot be assigned to a host. Everything in the middle of the range will function just fine. The other answers do not apply.
  26. A, B, C. Given the 10.5.12.0/22 subnet, we can find the range of addresses to be 10.5.12.0 to 10.5.15.255. This makes Answers A, B, and C correct. Answers D and E are incorrect because they belong to the next subnet.
  27. B. There are two ways of shortening an IPv6 address: removing a single group of consecutive zeros by using the double colon (::) and removing leading zeros from an octet. Answer B (2001:ab9:0:0:3::59ff:1ac5) shortens the octets by removing leading zeros and abbreviates the second group of consecutive zeros by using the ::. Answer A incorrectly uses a :: twice in the IPv6 address. Answer C has too many characters in one of the octets. Answer D uses the underscore character, which is invalid.
  28. Figure 8.11

    Figure 8.11 Network diagram.

  29. B. To find the number of valid host IP addresses, use the formula (2x) – 2, where "x" represents the number of host bits. In this case, the FastEthernet interface has the subnet mask 255.255.252.0, which uses 10 hosts bits. (210) – 2 = 1022. The other answers are not valid.
  30. A. To enable the IPv6 protocol, use the command ipv6 unicast-routing from global configuration mode. All other answers produce an invalid syntax or incomplete command message.
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