Test my subnetting skills

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hi all

hope this is the correct place 4 this. We have just started subnetting at college and i wanna check im doing it right. Im NOT looking 4 answers, just a verification of my work

IP 192.168.0.3
Hosts required: 6
Subnets required: 4

subnet mask 255.255.255.248
Belongs To Subnet 192.168.0.0
Valid Addresses 192.168.0.1 To 192.168.0.6
Broadcast 192.168.0.7


IP 192.168.4.2
Hosts required: 60
Subnets required: 2

subnet mask 255.255.255.192
Belongs To Subnet 192.168.4.0
Valid Addresses 192.168.4.1 To 192.168.4.62
Broadcast 192.168.4.63

IP 192.168.34.56
Hosts required: 60
Subnets required: 2

subnet mask 255.255.255.192
Belongs To Subnet 192.168.34.0
Valid Addresses 192.168.34.1 To 192.168.34.62
Broadcast 192.168.34.63

 

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www.learntosubnet.com

 

thanx

 

datarunner,

Issues:

1. Subnet mask is too long.
2. There's something you forgot to account for with "Hosts required."

That's correct.

That's correct.

 

not sure wot u mean

 

ah typo 255.255.248.0 instead of 255.255.255.248 is that correct? Are subnet masks correct for the others or should they be 255.255.x.0

 

datarunner,

No, that typo would make it incorrect.

 

datarunner,

How many subnets would you get with a subnet mask of 255.255.255.128 ?

 

sorry buddy im lost now. im in class tomorrow so hopefully ill learn more. many thanks for your time.

 

datarunner,

Going back to the original problem...

If you use a subnet mask of 255.255.255.128 on that IP address, how many subnets would you get?

 

I know how to work out how hosts from the above (2x2x2x2x2x2x2x2=256) but dont know how to work out how many subnets that would give. Could you explain? Sorry to be a pain.

 

datarunner,

Sure I can explain. An IP address, just like any other network layer address such as IPX and Appletalk, has a network part and a host part. This is determined by applying the default mask to the IP address using the logical AND process. So in this case...

Code:

IP = IP address
DM = Default Mask
NA = Network Address

IP: 192.168.000.003
DM: 255.255.255.000
NA: 192.168.000.000

As you may or may not know, subnetting is the process of extending the network part of the IP address by borrowing bits from the host part of the IP address. For this case, the host part of the IP address of 192.168.0.3 is the last octet. So in order to subnet, you borrow bits from this last octet. So let's look at this last octet in binary.

Code:

SM = Subnet Mask

IP: 192.168.000.003
SM: 255.255.255.128

Convert last octet from decimal to binary

IP: 192.168.000.00000011
SM: 255.255.255.10000000

Now that results in 2 possible subnets, 192.168.0.0 and 192.168.0.128. However, in classful subnetting, you're not supposed to use the subnet with "all binary zeros" and "all binary ones" so that means you have 0 usable subnets.

Now, let's take this a step further and try using a subnet mask of 255.255.255.192 . How many usable subnets do we have now?

Code:

IP: 192.168.000.003
SM: 255.255.255.192

[Convert last octet to binary]

IP: 192.168.000.00000011
SM: 255.255.255.11000000

Possible subnets
[list=1]
[*]192.168.000.00000000 = 192.168.0.0
[*]192.168.000.01000000 = 192.168.0.64
[*]192.168.000.10000000 = 192.168.0.128
[*]192.168.0.0.11000000 = 192.168.0.192
[/list]

Since the "all zeros" and "all ones" subnets are not allowed, subnets 1 and 4 are unusable, therefore you have two usable subnets, 192.168.0.64 and 192.168.0.128.

So, let's see if you understand by figuring out how many usable subnets you get with a subnet mask of 255.255.255.224.

 

Hmmm, in my opinion there are four correct answers for the first problem. The given solution is correct, but there are other options.
SNM of 255.255.255.248 would give me 8 addresses in total which is sufficiant for 6 hosts, the network ID and the broadcast address.
There are 32 subnets here.

SNM 255.255.255.240 would give me 16 addresses (14 hosts) and 16 subnets. This would also work.

SNM 255.255.255.224 would give me 32 addresses (30 hosts) and 8 subnets. Also correct.

SNM 255.255.255.192 would give me 64 addresses (62 hosts) and 4 subnets. Also correct.

In the past is was common not to use the all-0-net and all-1-net, so 248, 240 and 224 would be correct answers.
Today the 192 would also work, but you might run into trouble with some equipment.

 

Hi,

You need to apply an easy way to subnet method to get a quick and accurate answer. Most Cisco college courses really do make this harder than they should.

In about 24 hours I will be updating www.subnetting-secrets.com with more explanations and some videos showing you how to do this.

It should be up on 4th September, at the moment it is just a one page site.

Paul
www.howtonetwork.net
www.subnetting-secrets.com
www.networksinc.co.uk

 

I was introduced to the lovely world of subnetting last week. out cisco tutor tried to teach us an 'easy' method to work out subnetting problems that none of us understood.

I spent all of saturday morning and most of Sunday night trying to teach myself subnetting, I think I have got it now but it it takes me ages to finish a subnetting problem.

Wish I'd read r.h lee's post sooner, its very clear and concise, and ciscoPaul.... looking forward to checking you site when its up and running.

Damn subnetting (shaking fist in angry manor)

 

VantageIsle,

You should be more proud of knowing how to properly and correctly work through a subnetting problem and get a correct answer but at a slow pace than to improperly work through a subnetting problem and quickly get the wrong answer. Subnetting becomes faster and faster with practice. Remember, only perfect practice makes perfect.

 

our tutor showed us a quick way without working out all the binarys.

say u wanted to subet 192.168.0.3 into 4 networks with 6 hosts in each. This gives u a subnet mask of 255.255.255.248

we take the 248 from 256 which = 8 or as we use 3 bits here its 2x2x2 = 8 possible 6 usable per network

so our networks number 0,8, 16, 24 etc (8 being the multiplier)

so 192.168.0.3 belongs in the 1st network (192168.0.0) with a broadcast address of 192.168.0.7

so really just deduct your submask figure from 256 and that gives u your network ranges, works for class b as well. hope it helps someone.

 

datarunner,

I am concerned that you are trying to avoid binary computations and remain in decimal because it will be very important later when you have to use Variable Length Subnet Masking or VLSM for short and decimal to binary to hexadecimal conversions.

See, this is also where binary would make things simpler than "memorizing the 8 times table all the way up to 256." For example, if you had the following binary number to convert to decimal, it's easier to just "read" the binary than sit there trying to figure out "multiples."

Code:

Binary: 10110000

Decimal
128
+32
===
160
+16
===
176

If you used some sort of complicated "multiples" system, then what you'd have to do is "figure out the multiples all the way up to the 128 decimal bit minus 64 decimal." This problem gets worse if the bits "move" such as binary 11010000. So, I think from a binary to decimal conversion perspective, it's easier to think in terms of what the binary _is_ (adding up the decimal numbers of the binary positions with a '1') instead of trying to apply some sort of multiples of 2, multiples of 4, multiples of 8, multiples of 16, multiples of 32, or multiples of 64.

 

i understand the decimal to binary conversons and the importance of them however for me, deducting the last octet of the subnet mask say 254 from 256 = 2 gives me a quick way of identifying which network an address should fall into

 

I learned both binary and non-binary methods, and in the end found using non-binary much faster (including doing VLSM stuff).

The Cisco Press CCNA exam guide and Todd Lammle's (Sybex) CCNA book both give lots of examples, using both types of methods.

Maybe it's something about how different people's brain's process things? Whichever works for you....