Answers to Review Questions

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# Answers to Review Questions

1.

C. A CIDR address of /22 is 255.255.252.0, which is a block size of 4 in the third octet. We would keep counting by four until we passed the host address of 210.0. Counting by fours, we would get our subnet number as 208.0 with a broadcast address of 211.255.

2.

B, C, E. This is a Class A address with a 255.255.252.0 mask. The third octet has a block size of 4, and the host address in the question is in subnet 4, so the valid hosts are 4.1 through 7.254.

3.

C. This is a pretty simple question. A /28 is 255.255.255.240, which means that our block size is 16 in the fourth octet. 0, 16, 32, 48, 64, 80, etc. The host is in the 64 subnet.

4.

F. A CIDR address of /19 is 255.255.224.0. This is a Class B address, so that is only 3 subnet bits, but 13 host bits, or 8 subnets, each with 8190 hosts.

5.

B. If you use the mask 255.255.255.0, that only gives you 8 subnet bits, or 256 subnets. You are going to have to use 1 subnet bit from the fourth octet, or 255.255.255.128. This is 9 subnet bits (2

^{9 }= 512).6.

A, D. To answer this, you must first know that a /20 is 255.255.240.0, which means that there is a block size of 16 in the third octet. The summary network in the third octet is 144 and the next summary network is 160, so the valid range is 144.1 through 159.254.

7.

D. A point-to-point link uses only two hosts. A /30, or 255.255.255.252, mask provides two hosts per subnet.

8.

C. 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.

9.

B. This one takes some thought. 255.255.255.0 would give you 256 subnets, each with 254 hosts—doesn’t work for this question. 255.255.254.0 would provide 128 subnets, each with 510 hosts; the second option looks good. 255.255.252.0 is 64 subnets, each with 1022 hosts. So 255.255.254.0 is the best answer.

10.

C. A /29 is 255.255.255.248, which is a block size of 8 in the fourth octet. The subnets are 0, 8, 16, 24, 32, 40, etc. 192.168.19.24 is the 24 subnet and since 32 is the next subnet, the broad- cast address for the 24 subnet is 31. 192.168.19.26 is the only correct answer.

11.

B, C, E. A Class A network address with a /18 is 255.255.192.0. The subnets in the third octet are 0, 64, 128, 192. The network address in the question is 110.64.0.0, with a broadcast of

110.64.127.255

, since the next subnet is 110.64.128.0. Answers B, C, and E are correct host IDs.

12.

B. You need 5 subnets, each with at least 16 hosts. The mask 255.255.255.240 provides 16 sub- nets with 14 hosts—this will not work. The mask 255.255.255.224 provides 8 subnets, each with 30 hosts. This is the best answer.

13.

B, E. A mask of 255.255.255.0 will provide 256 subnets, each with 254 hosts. This will not work. We need more subnets, so we’ll move to the right and take some bits from the host bits in the fourth octet. 255.255.255.128 (/25) provides 512 subnets, each with 126 hosts. 255.255.255.192 (/26) provides 1024 subnets, each with 62 hosts.