TCP/IP subnetting help plz 1

[1666]

Hi, I'm getting a bit confused with subnetting at the moment, just getting ready for my Win2k server MCP exam (yikes). I understand the way IP's are broken down into classes A,B,C, D etc and then into 4 octets and into binary values. I understand that the subnet mask separates the IPv4 into 2 sections the network part and host part. I had a question like this which I found difficult:

Your company has been assigned a network ID of 132.132.0.0. Your network has 5 subnets. The maximum number of hosts per subnet is 8000, and the minimum number of hosts per subnet is 6500. Which subnet mask should you use for the network?

A: 255255.192.0
B: 255.255.224.0
c: 255.255.240.0
D: 255.255.248.0

Answer is B

In "thick" explainations for me, how can this be split into 5 subnets, which is what they say can be? Apparently you can get 8190 hosts from this, how do I get this sum?

If this is all explained I think it will "click" for me...

Thanks

Andy

 

[Ntr0P]

Try this:

http://lantech.geekvenue.net/chucktips/jason/chuck/1022445898/index_html

It give a good and thorough explanation that saves me a lot of typing.

 

[hnd]

Only one thing: In IP-Adressing following Adresses are not allowed:
1. All bits are set to 1, because this would be the Broadcast address of the Subnet.
2. All host bits set 0, because this Adress is equal to the Netmask.

This disallows host-part 0 and 8191. Therefore you have 8190 addresses.

hnd
hasso55@yahoo.com

 

[GrnEyedLdy]

1666,

"In "thick" explainations for me, how can this be split into 5 subnets, which is what they say can be? Apparently you can get 8190 hosts from this, how do I get this sum?"

The way you determine how many host a given subnet can accommodate is by taking the number of bits left for HOSTS and raising it to the power of 2.

For example, if your subnet mask is 255.255.224.0, then you are using the first 19 bits from the left to designate the NETWORK id and the last 13 bits are left over for HOSTS. Now take 2 to the 13th and you will get the number of HOSTS available, (8192...however, since you cannot use all 1's or all 0's, you take 8192 and subtract 2 to come up with the available HOSTS which would be 8190).

So, in order to answer the question from above, you would have to first determine how many bits you would need in order to accommodate the 6500-8000 HOSTS on each network.

2^2 = 4
2^3 = 8
2^4 = 16
2^5 = 32
2^6 = 64
2^7 = 128
2^8 = 256
2^9 = 512
2^10 = 1024
2^11 = 2048
2^12 = 4096
2^13 = 8192

So you can see that in order to fulfill the requirements of the above question, you need to have 13 bits available to you for HOSTS.

255. 255. 224. 0 Decimal
11111111.11111111.11100000.00000000 Binary

With the leftmost 19 bits turned 'ON', (all 1's), that leaves you with 13 bits for HOSTS, (all 0's).

The additional 5 subnets come from the fact that you borrowed (turned ON) 3 bits from the third octect and 2^3 = 8 (remember to subtract 2), and you come up with the possibility of 6 new subnets. So by choosing the answer 255.255.224.0 you satisfy both the need for 5 new subnets and the need for 6500-8000 HOSTS on each new subnet.

Hope this makes sense

For more detail on subnetting, see the following thread, thread96-445600 entitled Subnetting and posted in the Windows 2000 Server forum.

Patty

 

[Marcs41]

Here is a free subnet calculator

http://www.radmin.com/products/famatech_nu/ipcalc.html

If the solution is here, let us know it was helpful so others can benefit from it as too