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# subnetting

what is subnetting?how is it done?suppose 198.53.202.0 is a network address and we want 4 subnets.find the following

1)number of bits required for subnetting
4)starting host id
5)last host id for each subnet.

i have my network exam coming up so please explain me the solutions i m currently refering furozan and tannenbaum
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shilpi84
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7 Solutions

We cannot do homework for you.  If you want, post what YOU think the answers are and we can tell you if you are right.
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Author Commented:
i dont know what are standard and custom subnet masks!!not written in any book so what do i do?
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Commented:
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Commented:
Better yet, search on "subnetting" on this web site and you will see tons of topics with very useful information.
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Commented:
All classes of IP networks can be divided into smaller networks called subnetworks or subnets. And the process is called subnetting.

Well before going to the answers, first you should know the IP Address Classes. There are 3 IP classes. Class A, Class B & Class C. These are also called commercial classes.

In Class A address, the first octet is the network portion, so the Class A address of 10.1.25.1, has a major network address of 10. Octets 2,3 & 4 or say next 24 bits are the hosts. Class A addresses can be used for more than 65536 hosts.

In Class B, the first two octet is network portion. Now, 172.16.122.204 is a Class B IP address where 172 & 16 are the major network address & octets 3 & 4 are hosts.
Range between 256 to 65536 hosts.

In Class C, the first 3 octets are network portion & the last one is hosts. So 193.18.9.40 here 193.18.9 is for network portion & 40 is host. In Class C there can hosts less than 254.

Now, to solve ur problem, the first duty is to classify in which class you IP is, from First Octet Rule, as follows...

ADDRESS CLASS          FIRST OCTET in DECIMAL             HIGH-ORDER BITS
--------------------         ------------------------------             ---------------------
Class A                       1-126                                           0
Class B                       128-191                                       10
Class C                       192-223                                       110

So, form the above chart we can say that the IP 198.53.202.0 is a Class C address, as we can see the first octet lies between 192-223.

So, in Class C address the last octet is for host. So 198.53.202.1, 198.53.202.2, 198.53.202.3, 198.53.202.4 can be said subnets of 198.53.202.0.

1) 64
2) 255.255.255.0 [as this is the default subnet musk for Class C address] #
3) 255.255.255.192 *
4) Starting host ID is "0" followed by 64, 128 & 192 *
5) 192 *

*Explained:~ Using a subnet mask of 255.255.255.192, your 198.53.202.0 network then becomes the four networks 198.53.202.0, 198.53.202.64, 198.53.202.128 and 198.53.202.192. These four networks would have as valid host addresses, as required in your question.

# Standard Subnet Musks are as follows...

Class A - 255.0.0.0 [As last 3 octets are hosts]
Class B - 255.255.0.0 [As last 2 octets are hosts]
Class C - 255.255.255.0 [As last octet is for hosts]

Hope now u can understand it well...
Milan
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Commented:
Milan: class based configuration is obsolete.  CIDR is the new standard.  If you are going to do the kid's homework, at least get it right :-)

17:57:06 tigger [Linux] user=ntop pwd=~ \$ whois -h whois.radb.net 198.53.202.0
route:              198.53.0.0/16
descr:              NETBLK-ISTAR0004
origin:             AS852
remarks:            Proxy route entry added on behalf of
remarks:            iSTAR internet Inc.
remarks:            originating in AS852
mnt-by:             MAINT-AS852
changed:            abuse@telus.com 20020115

You will notice that the given address is part of a larger block assigned to a specific ISP, Telus.com - so the given address could be ANYTHING subnetted from that /16 assignment...

-----Burton
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Commented:
Hello Burton,

>> "class based configuration is obsolete.  CIDR is the new standard.  If you are going to do the kid's homework, at least get it right :-)"

Don't know whether it is obsolete or not. Anyways still in Technical Institution & Colleges this concept is alive & this kind of problems/ assignments are done with the class concept. So, ... put ur solution if any, instead doing comments.

Thanks
MilanKM
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Commented:
Any school still teaching class-based addressing is one I would RUN away from.

The first RFC for CIDR (1519) is dated September 1993.

-----Burton
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Commented:

You can not determine this with the information given, because the network mask for the base address is missing.

-----Burton
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Commented:
bstrauss3,
Actually  because it seems to be homework, which is much like a cert test, you have to assume the basics. that it is a class C address and there fore the solution would be simple.

secondly CIDR and class-based are both valid, and both widley used, mostly in conjunction. Everyone should be familiar with both.
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Commented:
IMHO, Understanding the classes is fundamental to understanding the subnet mask.  I don't know how its taught now, but I learned that a class A is 1-126 .  Maybe they teach it now that it's everything under the left most bit is a natural /8 and everything under the two left most bits is a /16.

I think it's a little easier learning a Class A goes to 126, a B to 191, and a C from 192 to 224 and that the masks are /8, /16, and /24.
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Commented:
Also, as lrmoore has often pointed out, some older equipment and software is not compatible with non-class based masks. I recently downloaded a current little utility that wouldn't accept other than classful masks.
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Commented:
Usually I respect your opinions PC, but here you're wrong - dead wrong.

For example, a huge portion of the US gets their service via cable internet - 24.0.0.0/8.  But that block was delgated by IANA to ARIN in 2001.

That's been chunked up in /12s through /19s... almost at random...

24.0.0.0/12 is Comcast, as is 24.16.0.0/13
24.24.0.0/14 is RoadRunner as is 24.28.0.0/15
etc.

The same is true of many of the original Class A's - check out http://www.iana.org/assignments/ipv4-address-space.

See all the blocks allocated to other registries?  Those are subdivided up!  Plus all the reserved blocks and Military assignements, etc.

So instead of 1..126 being /8s, only about half of 1..56 are really at the /8 level.

Any tool or hardware that won't support CIDR is obsolete and has been for many years...  You wouldn't teach AutoShop on a model T, would you?

-----Burton

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Commented:
...

>>24.0.0.0/12 is Comcast, as is 24.16.0.0/13
>>24.24.0.0/14 is RoadRunner as is 24.28.0.0/15

... but, these are SUBNETS of the "natural" 24.0.0.0/8 network.  Ok, so ARIN subnetted their original block - just like the rest of us subnet within our own networks.  So what?

Someone new to networking still has to have a starting point when given a typical CCNA level subnetting question - figuring out host portion and network portion of a subnet.
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Commented:
you know most network equipment you have to know classfull snm in order to configure acls/interfaces etc right? some support both snm and cidr but most do not.
so you have to know both.
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Commented:
pc - The point is that if I give you an address in that 24 block, or in 41 or any of the other divided blocks, you can't derive ANY useful information from that address as to the proper subnet mask.  Teaching somebody that the can is only going to cause problems to them and to others later on.

jabiii - no.  To configure any modern equipment you need to understand CIDR.  Period.

The OP gave us a specific address.  MilanKM jumped in assuming that since 198 > 192 it was a /24.

In fact, it's an unknown subneted portion of a /22, subnetted from Telus's /16!:

09:35:55 tigger [Linux] user=ntop pwd=~ \$ whois -h whois.arin.net 198.53.202.0
[Querying whois.arin.net]
[Redirected to rwhois.telus.net:4321]
[Querying rwhois.telus.net]
[rwhois.telus.net]
%rwhois V-1.5:003ab7:00 rwhois.telus.net (by Network Solutions, Inc. V-1.5.7.3)
network:Class-Name:network
network:ID:123.198.53.0.0/16
network:Auth-Area:198.53.0.0/16
network:Network-Name:EDTNABXBAR03
network:IP-Network:198.53.200.0/22
network:Org-Name:TELUS-HSIA-EDTNABXB
network:City:Edmonton
network:State-Province:AB
network:Country-Code:CA
network:Postal-Code:T6C 3C4
network:Updated:2005-05-11 (21:45:08)
network:Created:2004-02-09 (21:17:22)
network:Abuse-Contact:abuse@telus.com (1-604-444-5791)
network:Tech-Contact:swip@swip.ca.telus.com

%ok

Now do you see why teaching classes is dangerous?

-----Burton
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Commented:
>>suppose 198.53.202.0 is a network address and we want 4 subnets.find the following

In the interest of time, I'm cheating and using SolarWinds Subnet Calculator.

Since 198.53.202.0 is specified to be a network address - the LARGEST network it could be would be a /23.

So, beginning with the assumption that the mask is 255.255.254.0 and there are 510 hosts available,

Details are as follows:

Subnet Bits      : 25
Host Bits        : 7
Possible Number of Subnets : 4
Hosts per Subnet : 126

198.53.202.0      255.255.255.128      126      198.53.202.1  to  198.53.202.126                      198.53.202.127
198.53.202.128      255.255.255.128      126      198.53.202.129  to  198.53.202.254      198.53.202.255
198.53.203.0      255.255.255.128      126      198.53.203.1  to  198.53.203.126                       198.53.203.127
198.53.203.128      255.255.255.128      126      198.53.203.129  to  198.53.203.254      198.53.203.255
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Commented:
Thanks shilpi84,
--Rob
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