Damn subnetting - Help!

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Giraffe }{unter
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Damn subnetting - Help!

Post by Giraffe }{unter »

never got my head around this...


From the IP addresses listed below, select the two that are on the same subnet. Assume that each address is using a subnet mask of 255.254.0.0

46.85.254.121
46.88.1.254
46.89.46.34
46.90.112.12

I'm looking for a way to answer the questing by understanding it.... not just giving me the answer. Anyone around to help?
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SOAPboy
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Post by SOAPboy »

[size=75][i]I once had a glass of milk.

It curdled, and then I couldn't drink it. So I mixed it with some water, and it was alright again.

I am now sick.
[/i][/size]
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SOAPboy
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Post by SOAPboy »

Holy fucking Quad post batman. i'm a putz :-\

edited by axbaby
[size=75][i]I once had a glass of milk.

It curdled, and then I couldn't drink it. So I mixed it with some water, and it was alright again.

I am now sick.
[/i][/size]
[img]http://img162.imageshack.us/img162/3631/171164665735hk8.png[/img]
Giraffe }{unter
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Post by Giraffe }{unter »

tx, reading now, anyone with a more narrow explanation?
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Scourge
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Post by Scourge »

SOAPboy wrote:Holy fucking Quad post batman. :-\
You could have deleted them one by one before someone else posted.
SOAPboy
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Post by SOAPboy »

Giraffe }{unter wrote:tx, reading now, anyone with a more narrow explanation?
Just read it imo. Its pretty simple if you can read through that. :)
Scourge wrote:
SOAPboy wrote:Holy fucking Quad post batman. :-\
You could have deleted them one by one before someone else posted.
LoL. yeah, i posted and went "doh, fuck it"
[size=75][i]I once had a glass of milk.

It curdled, and then I couldn't drink it. So I mixed it with some water, and it was alright again.

I am now sick.
[/i][/size]
[img]http://img162.imageshack.us/img162/3631/171164665735hk8.png[/img]
Giraffe }{unter
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Joined: Fri Mar 17, 2000 8:00 am

Post by Giraffe }{unter »

255.254.0.0 = 11111111.11111110.00000000.00000000
254 octet = 11111110

the unmasked area allows me 1 so


46.88.1.254
46.89.46.34
would be the correct answer
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Underpants?
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Post by Underpants? »

46.90.0.0 and 46.89.0.0 as well--two possible answers there, tubby.
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Foo
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Post by Foo »

Underpants? wrote:46.90.0.0 and 46.89.0.0 as well--two possible answers there, tubby.
Nope:
1011000 = 88
1011001 = 89
1011010 = 90
1111110 = Mask

So 90 violates the mask by altering the second bit.

FWIW though, I had to resort to paper to work out the binary values. My conversion ability is not what it was. At least I remembered the method ;()
Giraffe }{unter
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Post by Giraffe }{unter »

Foo wrote:
Underpants? wrote:46.90.0.0 and 46.89.0.0 as well--two possible answers there, tubby.
Nope:
1011000 = 88
1011001 = 89
1011010 = 90
1111110 = Mask

So 90 violates the mask by altering the second bit.

FWIW though, I had to resort to paper to work out the binary values. My conversion ability is not what it was. At least I remembered the method ;()
ahh so I was doing this wrong, but got the answer right I assumed the last bit was in the binary slot for 1 and you can alter the second octet by 1 increment. 88 + 1 = 89

so you are saying:

convert all second octet values to binary
all digits masked by ones in the subnet must be the same
the only digit in the binary that can change is the one not masked (0)?
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Dr_Watson
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Re: Damn subnetting - Help!

Post by Dr_Watson »

Giraffe }{unter wrote:never got my head around this...


From the IP addresses listed below, select the two that are on the same subnet. Assume that each address is using a subnet mask of 255.254.0.0

46.85.254.121
46.88.1.254
46.89.46.34
46.90.112.12

I'm looking for a way to answer the questing by understanding it.... not just giving me the answer. Anyone around to help?
46.x.x.x = class a
mask of 255.254.0.0 is borrowing 7 bits to subnet
producing 2^7 networks and 2^16 hosts
2^7=128
256/128 =2
so you have a network every 2 in the second octet of the decimal ip
so the .88 and .89 are on the same subnet. (since the subnets will start on evens and end on odds)
46.88.0.0 = network
46.89.255.255 = broadcast

if you need more explination about how to do the binary AND/XOR operations that explain the nuts and bolts i have some easy to understand slides laying around from a network theory class somewhere.
Giraffe }{unter
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Joined: Fri Mar 17, 2000 8:00 am

Re: Damn subnetting - Help!

Post by Giraffe }{unter »

Dr_Watson wrote:
Giraffe }{unter wrote:never got my head around this...


From the IP addresses listed below, select the two that are on the same subnet. Assume that each address is using a subnet mask of 255.254.0.0

46.85.254.121
46.88.1.254
46.89.46.34
46.90.112.12

I'm looking for a way to answer the questing by understanding it.... not just giving me the answer. Anyone around to help?
46.x.x.x = class a
mask of 255.254.0.0 is borrowing 7 bits to subnet
producing 2^7 networks and 2^16 hosts
2^7=128
256/128 =2
so you have a network every 2 in the second octet of the decimal ip
so the .88 and .89 are on the same subnet. (since the subnets will start on evens and end on odds)
46.88.0.0 = network
46.89.255.255 = broadcast

if you need more explination about how to do the binary AND/XOR operations that explain the nuts and bolts i have some easy to understand slides laying around from a network theory class somewhere.
When you have the time I would appreciate those slides. I believe my understanding of this is not 100% and would like to learn more :icon30: you can gmail them to me = giraffehunter
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Underpants?
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Post by Underpants? »

Foo wrote:
Underpants? wrote:46.90.0.0 and 46.89.0.0 as well--two possible answers there, tubby.
Nope:
1011000 = 88
1011001 = 89
1011010 = 90
1111110 = Mask

So 90 violates the mask by altering the second bit.

FWIW though, I had to resort to paper to work out the binary values. My conversion ability is not what it was. At least I remembered the method ;()
:( reading was not my strong point this day.
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