Determining IP Addresses

Johnny Johnson

Here is a practice question that I'm having trouble figuring out:

The correct answer is B, but I'm not sure how to go about getting the answer. Any help would be appreciated!

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dynamik

All you have to do is determine the range of IPs in that subnet.

/28 will give you the following as available options:
192.168.1.0-15
192.168.1.16-31
192.168.1.32-47
192.168.1.48-63
192.168.1.64-79
etc.

You can do a binary AND to arrive at the correct range as well.

Since you are given the router's IP, it's obvious which other IP falls within that range.

Johnny Johnson

Since 192.168.1.48-63 is one of the ranges, why would I choose option B and not C for the answer?

dynamik

That's the broadcast address for that subnet; you can't assign it to a device. You also can't assign the first IP in the range because that's the network address.

Stotic

Add 1 to the network address (first #) and subtract 1 from the broadcast (last #) to get the assignable address range.

Johnny Johnson

dynamik wrote: »

All you have to do is determine the range of IPs in that subnet.

/28 will give you the following as available options:
192.168.1.0-15
192.168.1.16-31
192.168.1.32-47
192.168.1.48-63
192.168.1.64-79
etc.

You can do a binary AND to arrive at the correct range as well.

Since you are given the router's IP, it's obvious which other IP falls within that range.

Could you be more specific as to how you go about finding the ranges/available options? (0-15, 16-31, etc.)

dynamik

/28 is 11111111.11111111.11111111.11110000, so the last four bits are for your hosts.

So you're going to have:

192.168.1.00000000-192.168.1.00001111 - 192.168.1.0 - 192.168.1.15
192.168.1.00010000-192.168.1.00011111 - 192.168.1.16 - 192.168.131
192.168.1.00100000-192.168.1.00101111 - 192.168.1.32 - 192.168.1.47
192.168.1.00110000-192.168.1.00111111 - 192.168.1.48 - 192.168.1.63
etc.

Since you know four bits can make 16 combinations, you can just keep incrementing the network number by 16. The last address is the one before the next network address (or the last address when you get to the last range). You really don't need to write it out in binary, but that's how it works.

Bl8ckr0uter

Johnny Johnson wrote: »

Could you be more specific as to how you go about finding the ranges/available options? (0-15, 16-31, etc.)

255.255.255.240 in dec

(11111111.11111111.11111111)= 255.255.255 (first three octets in subnet mask)

128 192 224 240
.1 1 1 1 0000 in binary (last octet in subnet mask)

They were finding the ranges by using the blocksizes (256-240=16)
Which means that you are counting by 16s while going through the subnets.

Which means that the ranges are 0-15, 16-31, etc

Johnny Johnson

That makes sense! Thank you!

Bl8ckr0uter

Johnny Johnson wrote: »

That makes sense! Thank you!

No thank-you. By telling you I have noticed the results of my training

lol.
Seriously I focused on blocksizes for a week and then it just came to me. Literally I had a neo at the end of matrix (1), I saw the code, I saw the 0's and 1's and it clicked.

techster79

192.168.1.62/28

A 192.168.1.47, broadcast address for 192.168.1.32/28
B 192.168.1.49, valid IP for subnet 192.168.1.48/28, 192.168.1.62 is also in this subnet
C 192.168.1.63, broadcast address for 192.168.1.48/28
D 192.168.1.68, valid IP for subnet 192.168.1.64/28, 192.168.1.62 is not in this subnet

Tryntotechit

My professor told us to look over this website when we were learning about subnetting. It is definitely worth it.
www.learntosubnet.com.

I had a problem using firefox with this site. I think I finally used netscape. Good luck and I hope this helps.

imafoxxx

[FONT=arial, sans-serif]thank you for your replies to this answer, I couldn't figure this out until today [/FONT]

mishy

Thanks for posting the question Dynamik for the answer and explanation.