subnetting question

teaboy5 · 16 Jan 2007 at 19:52

Can someone help me out, i am not sure if i am doing this right at all

Example question is

Which address is a valid subnet if a 26 bit mask is used for subnetting?
172.16.43.16
172.16.128.32
172.16.243.64
172.16.157.96
172.16.47.224
172.16.192.252

So its using a 26 bit mask, which leaves 8 bits for the host part, so the mask should be 255.255.255.192

So i would say the answer to this question would be 172.16.192.252? Am i right here?

OzyOly · 16 Jan 2007 at 19:57

teaboy5 said:
Can someone help me out, i am not sure if i am doing this right at all

Example question is

Which address is a valid subnet if a 26 bit mask is used for subnetting?
172.16.43.16
172.16.128.32
172.16.243.64
172.16.157.96
172.16.47.224
172.16.192.252

So it has borrorwed 10 bits, with leaves me with a subnet mask of 255.255.252.0

So i would say the answer to this question would be 172.16.192.252? Am i right here?

You doing some test here and need the answer?

teaboy5 · 16 Jan 2007 at 19:59

if it where only that easy

I have some example questions and i am just going over them, and i am not sure if thats the one.

And if i was doing a test how would i have the questions already

OzyOly · 16 Jan 2007 at 20:04

teaboy5 said:
if it where only that easy

I have some example questions and i am just going over them, and i am not sure if thats the one.

And if i was doing a test how would i have the questions already

Cheating?

Anyway, on topic I don't know, sorry. Ask me in 4 years after I finish my masters degree in computer science, but for now, I ain't gotta clue.

Wait another 10mins and someone with the answer will tell you.

teaboy5 · 16 Jan 2007 at 20:07

OzyOly said:
Cheating?

Anyway, on topic I don't know, sorry. Ask me in 4 years after I finish my masters degree in computer science, but for now, I ain't gotta clue.

Wait another 10mins and someone with the answer will tell you.

Yeah i must be cheating, so must everyone else asking the same kind of question in this same section of the forums.

Dam me to hell for cheating!!!!! :rolleyes:

Any one with clue think i have done this right

A[L]C · 16 Jan 2007 at 20:13

shoudnt this be in networks?

teaboy5 · 16 Jan 2007 at 20:19

Should you know, never thought

mattbrown91 · 16 Jan 2007 at 20:47

I'll let you know when I've finished the CCNA

Chaos · 16 Jan 2007 at 21:13

http://www.learntosubnet.com might be of help.

OzyOly · 16 Jan 2007 at 21:17

teaboy5 said:
Yeah i must be cheating, so must everyone else asking the same kind of question in this same section of the forums.

Yes

AndrewP · 16 Jan 2007 at 21:37

/26 mask, so 6 bits (32-26) for the host part, 2^6 = 64 addresses per subnet so subnet IDs will be multiples of 64.

How did you come up with the answer in the OP?

teaboy5 · 16 Jan 2007 at 21:59

AndrewP said:
/26 mask, so 6 bits (32-26) for the host part, 2^6 = 64 addresses per subnet so subnet IDs will be multiples of 64.

How did you come up with the answer in the OP?

Lol i dont know, as i am not sure on how to work it out. Once i have seen what you have done above, i see it how it works now. So what would the answer be then?

Wait hang on

If its a 26 bit mask

it would be 255.255.255.192 leaving 8 bits left for the host is it not?

255 - 8 bits
255 - 8 bits
255 - 8 bits
192 - 2 bits using a total of 26 bits

AndrewP · 16 Jan 2007 at 22:14

teaboy5 said:
Lol i dont know, as i am not sure on how to work it out. Once i have seen what you have done above, i see it how it works now. So what would the answer be then?

Which of the listed subnets ends in a multiple of 64?

Wait hang on

If its a 26 bit mask

it would be 255.255.255.192 leaving 8 bits left for the host is it not?

The mask is right, but not the remaining bits, in binary the mask is 11111111.11111111.11111111.11000000 (26 1s) which leaves 6 bits (the 0s) for the host portion to use, giving 2^6=64 available addresses per subnet.

If you get stuck then going back to binary and working through might be an idea and whatever documentation you have would be better than my 'thought up as I go along' help

.

teaboy5 · 16 Jan 2007 at 22:17

AndrewP said:
Which of the listed subnets ends in a multiple of 64?

The mask is right, but not the remaining bits, in binary the mask is 11111111.11111111.11111111.11000000 (26 1s) which leaves 6 bits (the 0s) for the host portion to use, giving 2^6=64 available addresses per subnet.

If you get stuck then going back to binary and working through might be an idea and whatever documentation you have would be better than my 'thought up as I go along' help .

Sorry was about to post again, went over my notes there and found out the correct way of working it out. I would say the answer is the one ending with .64 and how i said there are 8 bits left i dont know. I got 10 in my head and couldn't get it out.

Thanks for answering