Subnetting

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Hi think I have finally grasped subnetting. I've been given some home work but know answers are provided, just want to check I have got them right:


The IP address allocated by a network administrator is 150.16.0.0 255.255.192.0.
How many possible subnets and usable subnets are there? How many hosts and useable hosts in each subnet?

Answer?
4 subnets, 2 usable subnets
64 hosts, 62 useable

The IP address allocated by a network administrator is 198.220.8.0 255.255.255.240.
(i) What is the network address of the third (2) subnet?
(ii) What is the broadcast address of the tenth (9) subnet?
(iii) What is the address of the first useable host in the fourteenth (13) subnet?

Answer?
(i)198.220.8.32
(ii) 198.220.8.159
(iii) 198.220.8.209

 

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Your question needs more information:

The IP address allocated by a network administrator is 150.16.0.0 255.255.192.0.
How many possible subnets and usable subnets are there? How many hosts and useable hosts in each subnet?

How many subnets and hosts do you want? The 150.16.0.0 255.255.192.0 network can be split up many ways.

For your second question, you are also missing a crucial piece of information - what network is the sample network a part of?

Spice_Weasel

 

Also, all this stuff about "what is the network address of the third (2) subnet".

What is that ?

You either talking about the 2nd subnet or the 3rd subnet. I don't what it means when you say third (2) subnet.

An easier way to frame the question perhaps would be: what is the network address of the third subnet with either:

(a). ip subnet zero configured
or
(b). ip subnet zero not configured

 

Sorry we were told to use the default subnet masks.

150.16.0.0 255.255.192.0
Class B so network is 255.255.0.0
Then subnetworks 255.255.192.0

198.220.8.0 255.255.255.240
Class C so network is 255.255.255.0
Then subnetworks 255.255.255.240

Also we told to include the .0 subnets
So the 3rd would mean putting a value of 2 into the subnetwork bit.

 

Okay, let me give you a little tip, I don't know if anybody is gonna agree with me on this one not, but as far as subnetting goes, it doesn't really matter whether you're dealing with a class A or a class B or class C address, really. What matters is the subnet mask you get given with the ip address.

Say for instance if they give you an ip address with subnet mask of say:

10.1.1.65 255.255.255.192

Then all it means is that you have is an ip address with a 26 bit mask, that's all. The fact that it's a class A address doesn't really mean anything. By the same token, if they give an ip address and subnet mask of say:

192.168.63.250 255.225.224.0

Then all it means is that you have is an ip address with a 19 bit mask, that's all. The fact that it's a class C address doesn't really mean anything.

So don't let yourself get confused with the class thing.

 

In other words, you were told to use ip subnet zero ?

Okay. Well, the other part about putting a value of 2 into the subnetwork bit, I honestly don't know what that means.

 

Well i'm not exactly sure either lol.

Yes we were told to include subnet zero, and were told to use the default subnets just for this example! I think we we told that if we wanted to find the 3rd subnetwork we should but a value of to into the subnetwork bit eg.

Subnet mask 255.255.0.0 (11111111.11111111.0.0)

Subnetwork mask 255.255.192.0

To find 3rd subnet we put a binary value of 2 into the subnetwork bit (11111111.11111111.0100000.0) then AND it with the original ip.

 

Okay, I gotcha.

What you mean is that you were told to add 2 extra bits to the default mask, right ?

Anyways, 255.255.192.0, when converted into bits is

11111111.11111111.11000000.00000000

not

(11111111.11111111.0100000.0)

That means that the 255.255.192.0 mask translates into 18 network bits and fourteen host bits.

What was your question again ?

 

The IP address allocated by a network administrator is 150.16.0.0 255.255.192.0.
How many possible subnets and usable subnets are there? How many hosts and useable hosts in each subnet?

Answer?
4 subnets, 2 usable subnets
64 hosts, 62 useable

The IP address allocated by a network administrator is 198.220.8.0 255.255.255.240.
(i) What is the network address of the third (2) subnet?
(ii) What is the broadcast address of the tenth (9) subnet?
(iii) What is the address of the first useable host in the fourteenth (13) subnet?

Answer?
(i)198.220.8.32
(ii) 198.220.8.159
(iii) 198.220.8.209

Just want to check I have worked out the correct answers!

 

Sorry, I had to run a couple of errands. But I had a quick look at your questions and came up with the following answers:

Question 1:

There are 4 usable subnets for 150.16.0.0/18. These are:

150.16.0.0
150.16.64.0
150.16.128.0
150.16.192.0

The number of usable hosts for this network (bearing in mind that you can derive 14 host bits from an 18 bit mask) is:

2 raised to power 14 minus 2

16384 - 2

Answer: 16382 usable hosts

Question 2

The IP address allocated by a network administrator is 198.220.8.0 255.255.255.240

Using ip subnet zero:

i). The network address for the 3rd subnet is 198.220.8.32
ii). The broadcast address of the 10th subnet is 192.220.8.175
iii). The address of the first usable host in the fourteenth subnet is 198.220.8.225

 

O dam. Different answers

Going to have to take another look to see what I did wrong

 

You got the first one right anyway, so you're doing fine.

Just remove those items highlighted in red, because that's what's confusing you. If it's the tenth subnet, then it's the tenth subnet. The number 9 in brackets shouldn't be there. The same thing for the fourteenth subnet. The number 13 in brackets shouldn't be there.

Any subnetting questions you get in your exam wouldn't look like that anyway.