BLOCKING / INCREMENTING SHORTCUT

In the previous example, we had the following situation in decimal form:

Network 131.107. y. z (Class B address)
Subnet mask: 255.255.224. 0

We found previously that the possible range of one of the subnets is: 131.107.32.1 through 131.107.63.254

SHORTCUT METHOD

To arrive at that result, we worked with binary bits. Next, we will use a shortcut to arrive at the same answer using only decimal numbers. This shortcut avoids the use of binary numbers.

Step 1

Take the rightmost non-zero number of the subnet mask (224). Subtract this number from 256. (256 - 224 = 32) This is the increment.

Step 2

Append the increment to your network address. (131.107. 32. z)

Step 3

Repeatedly add the increment (32) to the result in step 2 until you reach the original rightmost non-zero number of the subnet mask (224).

131.107. 64.z = 131.107. ( 32 + 32) .z
131.107. 96.z = 131.107. ( 64 + 32) .z
131.107.128.z = 131.107. ( 96 + 32) .z
131.107.160.z = 131.107. (128 + 32) .z
131.107.192.z = 131.107. (160 + 32) .z

131.107.224.z but we do not accept this one because 224 corresponds to the case of a subnet ID with all 1's. (224 = 11100000)

Step 4

Write out the smallest and largest IP addresses for your subnets. For our first subnet, we have 131.107.32.z. The smallest possible value is 131.107.32.1 because our host ID cannot have all 0's. The largest value on this subnet must be smaller than the start of the next subnet, which is 131.107.64.1. In addition, our host ID cannot have all 1's. So we eliminate 131.107.63.255 as a valid IP address. The largest value for an IP address on this subnet would be 131.107.63.254.**

** Note:

11111111 . 11111111 . 11100000 . 00000000 (subnet mask 255.255.224.0)
10000011 . 01101011 . 001 11111 . 11111110 (131.107.63.254,accepted)
10000011 . 01101011 . 001 11111 . 11111111 (131.107.63.255, largest binary value but not accepted because of all 1's in host ID)
10000011 . 01101011 . 010 00000 . 00000001 (131.107.64.1, smallest value for next subnet)

Using the above logic, we find these to be range of values for all 6 subnets:

Subnet 1: 131.107. 32. 1 through 131.107. 63.254
Subnet 2: 131.107. 64. 1 through 131.107. 95.254
Subnet 3: 131.107. 96. 1 through 131.107.127.254
Subnet 4: 131.107. 128.1 through 131.107.159.254
Subnet 5: 131.107. 160.1 through 131.107.191.254
Subnet 6: 131.107. 192.1 through 131.107.223.254

Example 2 (Class A IP addresses)

Network : 10. X. y. z
Subnet Mask 255.240.0.0

What are the ranges of IP addresses for all subnets? 240 is the rightmost nonzero number in the subnet mask. 256 - 240 = 16 (increment) Append the increment to the network address until you reach 240.

10. 16.0.z
10. 32.0.z = 10. ( 16 + 16) .0 .z
10. 48.0.z = 10. ( 32 + 16) .0 .z
10. 64.0.z = 10. ( 48 + 16) .0 .z
10. 80.0.z and so on
10. 96.0.z
10.112.0.z
10.128.0.z
10.144.0.z
10.160.0.z
10.176.0.z
10.192.0.z
10.208.0.z
10.224.0.z
10.240.0.z but we do not accept this one because this corresponds to all 1's in the subnet ID.

Subnet 1: 10. 16. 0. 1 through 10.31.255.254 **
Subnet 2: 10. 32. 0. 1 through 10.47.255.254
Subnet 3: 10. 48. 0. 1 through 10.63.255.254
Subnet 4: 10. 64. 0. 1 through 10.79.255.254
and so forth

** Note:

11111111 . 11110000 . 00000000 . 00000000 (subnet mask 255.240.0.0)
00001010 . 00011111 . 11111111 . 11111110 (10.31.255.254)

is the largest possible acceptable IP address on the same subnet that is less than:

00001010 . 00100000 . 00000000 . 00000001 (10.32. 0. 1)