| Class B | Class C |

Class B Borrowed-Bits Summary
Borrowed_Bits	Subnet Mask	Subnets	Hosts	Range
2	255.255.192.0	2	16382	16384
3	255.255.224.0	6	8190	8192
4	255.255.240.0	14	4094	4096
5	255.255.248.0	30	2046	2048
6	255.255.252.0	62	1022	1024
7	255.255.254.0	126	510	512
8	255.255.255.0	254	254	256
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	16382	2	4

Question: If you have a Class B IP network with a 10-bit subnet mask, how many subnets and hosts are usable? Answer: 1022 subnets, each with 62 hosts. Class B (16 bits for networks by default) 10 additional borrowed bits: Subnets: 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 [10] 4 8 16 32 64 128 256 512 1024 [-2] = 1022 subnets Hosts: Bits Remaining: 16 (default for Class B) + 10 Borrowed Bits = 26 32 - 26 = [ 8 Bits Remaining - 2 ] = 6 2 x 2 x 2 x 2 x 2 x 2 4 8 16 32 64 [-2] = 62 [ range is .0 .1 - .62 .63 ] Another way: Class B with 10-bit subnet mask: [ 16 (default for Class B) + 10 = 26 ] /26 26 - 16 = 10 10 = 8 + 2 8 bits = 255 or [ 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255 ] 2 bits = 192 or [ 128 + 64 = 192 ] Default subnet mask for a class B: 255.255.0.0 New subnet mask with 10-bit subnet mask: 255.255.255.192 Magic Number: 256 - 192 = 64 Subnet IP Range Broadcast address .0 .1 - .62 .63 .64 .65 - .126 .127 .128 .129 - .190 .191 .192 .193 - .254 .255 Question: You have a subnet mask of 255.255.255.248 in a Class B network. How many usable subnets and hosts are available? Answer: 8190 subnets, each with 6 hosts. Magic Number: 256 - 248 = 8 Default subnet mask for a Class B: 255.255.0.0 (16 bits) We can use up 14 borrowed bits: Using: 8 bits = 255 or [ 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255 ] Using: 5 bits = 248 or [ 128 + 64 + 32 + 16 + 8 = 248 ] Total of 13 bits: 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 [ 8192 - 2 = 8190 subnets] 2 Subnet Host IP Range Broadcast address .0 .1 - .6 .7 Hosts = 6 .8 .9 - .14 .15 .16 .16 - .22 .23 .24 .25 - .30 .31 --> Question: Given the IP address of 162.53.21.12 -- plan for 126 hosts on the subnet that includes this address. What subnet mask would you use? Answer: 255.255.255.128 Default subnet mask for Class B: 255.255.0.0 [ Default 16-bits used for networks ] Hosts: 2 x 2 x 2 x 2 x 2 x 2 x 2 [ 6 bits used for hosts] 4 8 16 32 64 128 [ 128 - 2 = 126 hosts] New Subnet mask: We have already used up 7 bits for hosts. 16 - 7 = 9 BB bits for networks Using: 8 bits = 255 or [ 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255 ] Using: 1 bit = 128 or [ 128 = 128 ] Total of 9 bits: 255.255.255.128 Magic Number: 256 - 128 = 128 Subnet Host IP Range Broadcast address 255.255.0 .0 .1 - .126 .127 Hosts = 126 255.255.0 .128 .129 - .254 .255 255.255.1 .0 .1 - .126 .127 255.255.1 .128 .129 - .254 .255 255.255.2 .0 .1 - .126 .127 255.255.2 .128 .129 - .254 .255 --> Note: Memorize the table above or understand the "anding" math operation. Either will stand you in good stead.

BSIG:

Subnetting - Class B Address