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TCP/IP TABLES
 
These are the tables I memorized for the TCP/IP exam. Before starting the exam I took a moment and wrote out the following tables up to the subnetting section below on the piece of paper provided. With these tables I did not have to use the calculator once during the exam which saved a lot of time.
 
Hosts
Bits
Subnet Mask
Bit Values
Class A
Class B
Class C
No. of Subnets
1
 -------
128
-------
-------
--------
-------
2
192
64
4,194,302
16,382
62
2
3
224
32
2,097,150
8,190
30
6
4
240
16
1,048,574
4,094
14
14
5
248
8
524,286
2,046
6
30
6
252
4
262,142
1,022
2
62
7
254
2
131,070
510
0
126
8
255
1
65,534
254
0
254
Note: These values in these tables are the accurate values. You may want to round off the big numbers to make it easy to memorize.
Class
No. of Networks
Hosts per Network
Range of Host ID's (1st Octet)
A
126
16,777,214
1 - 126
B
16,384
65,534
128 - 191
C
2,097,152
254
192 - 223
 
TCP/IP Address Resolution
NetBios
Hosts
Cache
Local
WINS
Hosts
Broadcast
DNS
LM Hosts
Cache
Hosts
WINS
DNS
Broadcast
1
LM Hosts
Having trouble with Subnetting ?

If you want a thorough detailed explanation of subnetting I suggest that you visit the following site:
http://www.hardcoremcse.com/ctributor/tcpip3.htm

 Here is a simplified explanation of how to get the subnet address ranges from various subnet masks:

YOU HAVE A CLASS B ADDRESS THUS: 131.107.X.Y
SUBNET MASK OF: 255.255.224.0

 Q1/. How do I calculate the THIRD & FOURTH OCTETS of the IP ADDRESS?

Look at my Tcp/ip table at the top of this page. With a subnet mask of 224 on a class B address you will see that there are 6 subnets and the bit value is 32. The starting X will be multiples of the bit value 32 with the starting Y being equal to 1 and the finishing X value will be 1 less than the next multiple of 32 and the Y value will be 254.

On a class C address such as 131.107.100.X things change a little. The X starting value is multiples of the bit value plus 1. The X ending value becomes the normal value that you would use above minus 1. The following table depicts this:
Subnet Mask: Class B = 255.255.224.0 Class C = 255.255.255.224
 
Subnet
Host ID Range - Class B
Host ID Range - Class C
Beginning
End
Beginning
End
1
131.107.32.1 131.107.63.254 131.107.100.33 131.107.100.62
2
131.107.64.1 131.107.95.254 131.107.100.65 131.107.100.94
3
131.107.96.1 131.107.127.254 131.107.100.97 131.107.100.126
4
131.107.128.1 131.107.159.254 131.107.100.129 131.107.100.158
5
131.107.160.1 131.107.191.254 131.107.100.161 131.107.100.190
6
131.107.192.1 131.107.223.254 131.107.100.193 131.107.100.222

YOU HAVE A CLASS A ADDRESS THUS: 18.X.Y.Z
SUBNET MASK OF: 255.224.0.0

 Q2/. How do I calculate the 2ND, 3RD & 4TH octets?

Now the Class A will follow the same idea. Again there are 6 subnets. The bit value is 32. With the ip address of 18.x.y.z., the starting x will be multiples of the bit value 32, y will be 1 and z will be 1. The ending x will be one less then the next multiple, y will be 255 and z will be 254.

Subnet
Host ID Range - Class A
Beginning
End
1
18.32.1.1
18.63.255.254
2
18.64.1.1
18.95.255.254
3
18.96.1.1
18.127.255.254
4
18.128.1.1
18.159.255.254
5
18.160.1.1
18.191.255.254
6
18.192.1.1
18.223.255.254
Hope that the above helps clarify the subnetting issue.