TCP/IP TABLES

These are the tables I memorized for the TCP/IP exam for the MCSE 3.51 track. 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. Of course now with the new Win2K exams you are expected to know TCP/IP as it is involved in almost all of the exams. Exam # 70 - 216 Administrating a Network Infrastructure is where your knowledge of TCP/IP is required the most.

			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

If you want a thorough detailed explanation of subnetting I suggest that you visit the following site:
www.mctgeek.com/students/tcpip.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.

Win2K Update

As you know with the new Win2K track tcp/ip is something that you need to know for all the exams. The following is useful information to know as it relates to the way the current exams look at tcp/ip.

As you probably already know the default subnet masks for the following classes can be represented as:

Class A 255.0.0.0 (11111111.00000000.00000000.00000000)
Class B 255.255.0.0 (11111111.11111111.00000000.00000000)
Class C 255.255.255.0 (11111111.11111111.11111111.00000000)

A new way of representing the mask is with the "/x" notation where x is the number of bits reserved for the network ID. Then the subnet masks above could then be represented by:

Class A /8 or 10.x.x.x /8
Class B /16 or 172.16.x.x./16
Class C /24 or 192.168.1.x /24

If we then start to apply subnetting to this notation then:

Let's borrow 4 bits from a class C address ==> 11111111.11111111.11111111.11110000
Since there are 3 octets with 8 bits and the final with 4 bits then the total is 28. The notation for the subnet mask is then "/28".
What is the subnet mask in the old convention. Refer to the following table:

Values of Binary Digits in an Octet

1	1	1	1	1	1	1	1
128	64	32	16	8	4	2	1

Since we borrowed 4 bits we start adding the values from left to right which = 128 + 64 + 32 + 16 ==> 240
So the mask is 255.255.255.240
So if you see an ip address 192.168.75.18 /28 it is 192.168.75.18 with a subnet mask of 255.255.255.240. The exams use this convention a lot so get to know it.

How many subnets does this mask represent?

You use the following formula:
  Number of subnets = 2^x - 2 where x is the number of bits borrowed from the host ID

Likewise
  Number of Hosts = 2^x - 2 where x is the number of unmasked host ID bits remaining.

Then with a mask of /28 or 255.255.255.240 (4 bits borrowed) the number of subnets = 2⁴ - 2 = 14
In this case the number of hosts would be the same calculation and the same value.
Of course the tables at the beginning of this web page will give you the same answer for a mask of 240

If I give you an ip address of 172.16.25.25 /21 What is the subnet class and mask? How many subnets are there? You can email me the answer and I will reply back. Easy enough? Well then you can score yourself bonus points if you can tell me in what subnet range of addresses does that address belong? I want to know 172.16.x.y - 172.16.x.y.

Hopefully that helps clarify the subnetting issue.

Note:

If all this calculating makes your brain wish for a rest I don't blame you. I found an IP Subnetting Calculator online. You can download the program at Download.Com. When you get to the site do a search for 'IP Subnet Calculator'.
 

Back to:

MCSE page

Top of this Page

Last updated on Mar. 9, 2001