Quake3World

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?

http://www.cisco.com/warp/public/701/3.html

There ya go.

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

edited by axbaby

tx, reading now, anyone with a more narrow explanation?

SOAPboy wrote:Holy fucking Quad post batman. :-\

You could have deleted them one by one before someone else posted.

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"

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

46.90.0.0 and 46.89.0.0 as well--two possible answers there, tubby.

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 ;()

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)?

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.

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

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.