|
| |
IP Subnetting –
Solution
For the solution, focus in on the interesting octet, the
octet in the mask that is not all one’s or all zero’s.
Convert the interesting octet into binary for both the address and mask
and do a logical AND ( 1 + 1 = 1; 0 + anything = 0 ), this is the subnet or wire
address. To help keep the host bits
verses the network/subnet bits straight, draw a line after the last one in the
mask. For the broadcast of the
subnet, change all the host bits (bits to the right of the line) to ones.
To come up with the range of valid addresses (first and last host
addresses), set all bits, except the least significant bit to zero. The least significant bit should be set to a one, this is the
first address in the subnet. Reverse
those bits, in other words, change all the host bits to ones, except the least
significant…that one should be set to a zero.
That is the last address in the subnet.
For the number of host addresses in a subnet, use the formula 2n
– 2 (where n = number of host bits).
Example:
|
Address
|
172
|
16
|
37
|
186
|
|
Mask
|
255
|
255
|
255
|
192
|
|
Address in
Binary
|
|
|
|
10
|
11
1010 |
|
Mask in
Binary
|
|
|
|
11
|
00
0000 |
|
Subnet in
Binary
|
|
|
|
10
|
00
0000 |
|
Broadcast
in Binary
|
|
|
|
10
|
11
1111 |
|
First
Address in Binary
|
|
|
|
10
|
00
0001 |
|
Last
Address in Binary
|
|
|
|
10
|
11
1110 |
|
Subnet
|
172
|
16
|
37
|
128
|
|
Broadcast
|
172
|
16
|
37
|
191
|
|
First
Address
|
172
|
16
|
37
|
129
|
|
Last
Address
|
172
|
16
|
37
|
190
|
Number of hosts =
26 – 2 = 64 – 2 = 62
|
172.16.37.186/26
|
|
|
|
|
Subnet
|
172.16.37.128
|
|
|
Broadcast
|
172.16.37.191
|
|
|
Number of Hosts
|
62
|
|
|
First Host Address
|
172.16.37.129
|
|
|
Last Host Address
|
172.16.37.190
|
1.
Question
|
Address
|
10
|
0
|
199
|
237
|
|
Mask
|
255
|
255
|
252
|
0
|
|
Address in
Binary
|
|
|
1100 01
|
11 |
|
|
Mask in
Binary
|
|
|
1111 11
|
00 |
|
|
Subnet in
Binary
|
|
|
1100 01
|
00 |
|
|
Broadcast
in Binary
|
|
|
1100 01
|
11 |
255 |
|
First
Address in Binary
|
|
|
1100 01
|
00 |
1
|
|
Last
Address in Binary
|
|
|
1100 01
|
11 |
254
|
|
Subnet
|
10
|
0
|
196
|
0
|
|
Broadcast
|
10
|
0
|
199
|
255
|
|
First
Address
|
10
|
0
|
196
|
1
|
|
Last
Address
|
10
|
0
|
199
|
254
|
Number of hosts = 210 – 2 = 1024 – 2 = 1022
|
10.0.199.237/22
|
|
|
|
|
Subnet
|
10.0.196.0
|
|
|
Broadcast
|
10.0.199.255
|
|
|
Number of Hosts
|
1022
|
|
|
First Host Address
|
10.0.196.1
|
|
|
Last Host Address
|
10.0.199.254
|
2.
Question
|
Address
|
172
|
16
|
37
|
186
|
|
Mask
|
255
|
255
|
255
|
224
|
|
Address in
Binary
|
|
|
|
101
|
1
1010 |
|
Mask in
Binary
|
|
|
|
111
|
0
0000 |
|
Subnet in
Binary
|
|
|
|
101
|
0
0000 |
|
Broadcast
in Binary
|
|
|
|
101
|
1
1111 |
|
First
Address in Binary
|
|
|
|
101
|
0
0001 |
|
Last
Address in Binary
|
|
|
|
101
|
1
1110 |
|
Subnet
|
172
|
16
|
37
|
160
|
|
Broadcast
|
172
|
16
|
37
|
191
|
|
First
Address
|
172
|
16
|
37
|
161
|
|
Last
Address
|
172
|
16
|
37
|
190
|
Number of hosts = 25 – 2 = 32 – 2 = 30
|
172.16.37.186/27
|
|
|
|
|
Subnet
|
172.16.37.160
|
|
|
Broadcast
|
172.16.37.191
|
|
|
Number of Hosts
|
30
|
|
|
First Host Address
|
172.16.37.161
|
|
|
Last Host Address
|
172.16.37.190
|
3.
Question
|
Address
|
192
|
168
|
14
|
87
|
|
Mask
|
255
|
255
|
255
|
192
|
|
Address in
Binary
|
|
|
|
01
|
01
0111 |
|
Mask in
Binary
|
|
|
|
11
|
|
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