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!
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!
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.
Since 192.168.1.48-63 is one of the ranges, why would I choose option B and not C for the answer?
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.
Add 1 to the network address (first #) and subtract 1 from the broadcast (last #) to get the assignable address range.
/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.
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
Last edited by Bl8ckr0uter; 06-16-2009 at 04:22 PM.
That makes sense! Thank you!
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
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.
thank you for your replies to this answer, I couldn't figure this out until today
Thanks for posting the question Dynamik for the answer and explanation.
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