Illustrated TCP/IP
by Matthew G. Naugle Wiley Computer Publishing, John Wiley & Sons, Inc. ISBN: 0471196568 Pub Date: 11/01/98 |
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A customer has the base network address of 150.1.0.0 with a subnet mask of 255.255.0.0, or /16 prefix.
This time we are not interested in a requirement of subnets. All we know is that we must be able to have 100 hosts on each subnet. Each subnet will not have that many, but the largest one will, and without multinetting, we must use a mask small enough to accommodate that number. In order to support 100 hosts, 7 bits are needed, which allows for 126 addresses (2n7 2). This will allow for future growth. The next-lowest mask yields 62 addresses (2n6 2), so we must allow for 7 bits. Always assign a mask that allows for future growth.
Next we must determine the subnet mask for the network number. Since we will be reserving 7 bits for host assignment, this will leave 25 bits left for the network mask (32 bits 7 bits = 25 bits). This gives a subnet mask of 255.255.255.128, or /25 prefix. The natural mask for Class B is 255.255.0.0. This mask is 255.255.255.128, which allows for 9 bits to be assigned to the subnet mask, thereby allowing for 512 subnets to be defined. The subnet numbers range from 0 to 521. This gives the range of subnets of 150.1.0.0 (providing for the zero subnet) through 150.1.255.128 (using all 9 bits including the all-1s subnet).
Now that we have separated the subnets from the hosts, we should list them:
Subnets | Host Range |
---|---|
150.1.0.0 through 150.1.255.128 | 1 through 125 (2n7 2) |
150.1.1.0 (x host reserved bits) | Host 1 (x = network/subnet reserved bits) |
10010110 . 00000001 . 00000001 . 0xxxxxxx | xxxxxxxx.xxxxxxxx.xxxxxxxx.x0000001 |
150.1.1.0 | Host 127 |
10010110 . 00000001 . 00000001 . 0xxxxxxx | xxxxxxxx.xxxxxxxx.xxxxxxxx.x1111111 |
Another Try
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