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5 Written questions

5 Matching questions

  1. 192.168.100.17/29
  2. How many subnets?
  3. Given 172.16.0.0/18, How many subnets?
  4. Why subnet?
  5. Classful Routing
  1. a 4
  2. b Facilitated spanning of large geographical distances
  3. c /29 is 255.255.255.248. The fourth octet is a block size of 8. 0, 8, 16, 24. The host is in the 16 subnet, broadcast of 23. Valid host 17-22
  4. d all interfaces within the classful address space have the same subnet mask
  5. e 2^x where x is the number of masked bits (or 1's)

5 Multiple choice questions

  1. 32-bit value that allows the recipient of IP packets to distinguish the network ID portion of the IP address from the host ID portion of the IP address
  2. the 10.32 subnet; The broadcast is 10.63
  3. A /25 mask is 255.255.255.128. Used with a Class B network, the third and fourth octets are used for subnetting with a total of 9 subnet bits, 8 bits in the third octet and 1 bit in the fourth octet. Since there is only 1 bit in the fourth octet, the bit is either off or on—which is a value of 0 or 128. The host in the question is in the 0 subnet, which has a broadcast address of 127 since 128 is the next subnet.
  4. A Class B network ID with a /22 mask is 255.255.252.0, with a block size of 4 in the third octet. The network address in the question is in subnet 172.16.16.0 with a broadcast address of 172.16.19.255. Only option E even has the correct subnet mask listed, and 172.16.18.255 is a valid host.
  5. .0.1 - .63.254, .64.1 - .127.254, .128.1 - .191.254, .192.1 - .255.254

5 True/False questions

  1. 192.168.100.66/27/27 is 255.255.255.224. The fourth octet is a block size of 32. Count by 32s until you pass the host address of 66. 0, 32, 64. The host is in the 32 subnet, broadcast address of 63. Valid host range of 33-62.

          

  2. You have a network that needs 29 subnets while maximizing the number of host addresses available on each subnet. How many bits must you borrow from the host field to provide the correct subnet mask?A Class B network ID with a /22 mask is 255.255.252.0, with a block size of 4 in the third octet. The network address in the question is in subnet 172.16.16.0 with a broadcast address of 172.16.19.255. Only option E even has the correct subnet mask listed, and 172.16.18.255 is a valid host.

          

  3. Variable Length Subnet Masks (VLSMs)/25 (with ip subnet-zero), /26, /27, /28, /29, /30

          

  4. A router receives a packet on an interface with a destination address of 172.16.46.191/26. What will the router do with this packet?A /21 is 255.255.248.0, which means we have a block size of 8 in the third octet, so we just count by 8 until we reach 66. The subnet in this question is 64.0. The next subnet is 72.0, so the broadcast address of the 64 subnet is 71.255.

          

  5. Given 192.168.10.0/28, What are the valid subnets?0 & 128

          

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