21
http:\\ciscocertifications.info
6
255.255.252.0
62
1022
4
7
255.255.254.0
126
510
2
8
255.255.255.0
254
254
1
9
255.255.255.128
510
126
128
10
255.255.255.192
1022
62
64
11
255.255.255.224
2046
30
32
12
255.255.255.240
4094
14
16
13
255.255.255.248
8190
6
8
14
255.255.255.252
16,382
2
4
Class C Addresses# of bitsSubnet maskSubnetsHostsRange
2
255.255.255.192
2
62
64
3
255.255.255.224
6
30
32
4
255.255.255.240
14
14
16
5
255.255.255.248
30
6
8
6
255.255.255.252
62
2
4
Here's how it works.
QUESTION: If you have a class B IP network with a 10-bit subnet mask, how many subnets and hosts
can you have?
ANSWER: 1022 subnets with 62 hosts (just look on the table for this answer)
QUESTION: You have an IP address of 172.16.13.5 with a subnet mask of 255.255.255.128. What is
your network ID and what range is the range of addresses in this subnet.
ANSWER: Network ID is 172.16.13.0, range is 172.16.13.1 - 172.16.13.126
(Since you are subnetting all 8-bits in the 3rd octet, the number in the 3rd octet becomes part of yournetwork ID. By looking at the table you see you have 126 hosts in each subnet. You also see the ad-dress range for each subnet is 128. Since the 0 is you network address and 127 is your broadcast ad-dress, the valid range of hosts addresses in this subnet is 172.16.13.1 - 172.16.13.126 = 126).
QUESTION: You have a subnet mask of 255.255.255.248 in a class B network. How many subnets
and hosts do you have?
ANSWER: 8190 subnets, each with 6 hosts.
QUESTION: If you have a Class C network with a 6-bit subnet mask, how many subnets and hosts do
you have?
ANSWER: 62 subnets, each with 2 hosts.
QUESTION: You have an IP address of 172.16.3.57 with an 11-bit subnet mask. What is the Network
ID, range of subnet addresses, and Broadcast address for this subnet?
ANSWER: Network ID = 172.16.3.32
= 1
Host Ids = 172.16.3.33 - 172.16.3.62
= 30
Broadcast Address = 172.16.3.63
= 1
32
By looking at the table above, you can see that a class B address with an 11 bit subnet mask has aRANGE of 32 with 30 HOSTS. Since this is a class B address we know that the first 2 octets are theoriginal Network ID (172.16.0.0). Since we are subnetting all 8-bits of the 3rd octet, then the 3rd octet