The above IP address
would translate into
11000001 = 128 + 64 + 1
= 193
00001010 = 8 + 2 = 10
00011110 = 16 + 8 + 4 +
2 = 30
00000010 = 2
So in decimal form, the
IP address is:
193 . 10 . 30 . 2
The network ID in the
address consists of a certain number of bits starting from the left.
The host ID consists of
the remaining bits.
---------------------------------------------------------------------------------------------------------
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)
--------------------------------------------------------------------------------------------------
Creating
Your TCP/IP Chart
To get your answers
quickly during the exam, you should use a chart that you've written down
during the astronomy tutorial.
A 1 - 126, N = 8
bits
B 128 - 191, N = 16
bits
C 192 - 223, N = 24
bits, where N = how many bits in Network ID
N + S + H = 32, where S
= # of bits in Subnet ID, H = # of bits in Host ID
