Maybe it's me being stupid?

wedge1988

hey, it's happened before! ... not that i like to make a habit of this..

heres my question..

and i'm shortening it, but it should make sense..

In the boson test software for the ccna, a question asks about a network setup. the answers are:

1. can use any ip address between 192.168.11.193 and 192.168.11.223
2. can use any ip address between 192.168.11.193 and 192.168.11.222

now, the VLSM is /27 so to me, that makes the network break up into 32 host chunks..

so why am i asking?

am i going crazy? the answer was apparently number 2 above.

I'm sure it should really be number 1?

if it goes in 32 host chunks then it should be 192-224, where 192 and 224 are not usable but the other addresses are?

anybody else here think the software is wrong?

cheers!

earweed

32 host addresses would have you going from 192-223. It's a simple error.

capitanuionut

earweed wrote: »

32 host addresses would have you going from 192-223. It's a simple error.

Depends, maybe it reffers to host addresses and takes account about the broadcast address of the network, .223 will be the broadcast address of this network. But this is not a rule, more a recommandation to use the last useable address as the broadcast address.
It is possible to be an error.

EDIT:
Sorry about this...i am totally wrong...i was thinking at the gateway address that is a recomandation to use the last one. .223 is the broadcast address and this is a rule. .192 network address so the hosts lay between 193 - 222.

CodeBlox

earweed wrote: »

It's a simple error.

Wrong.

223 is the broadcast address

To find the number of host the method is 2^x - 2. So there are 30 usable IP addresses, 192.168.11.193 - 192.168.11.222 . It's not a "simple error" and choice 2 is correct.

Chris_

Quote:

""""""Depends, maybe it reffers to host addresses and takes account about the broadcast address of the network, .223 will be the broadcast address of this network. But this is not a rule, more a recommandation to use the last useable address as the broadcast address.
It is possible to be an error."""

I think you'll find that this is a rule!!!!

earweed

The simple error was that he had 32 addresses and said it went from 192-224 when that would be 33 addresses.

capitanuionut

Chris_ wrote: »

Quote:

""""""Depends, maybe it reffers to host addresses and takes account about the broadcast address of the network, .223 will be the broadcast address of this network. But this is not a rule, more a recommandation to use the last useable address as the broadcast address.
It is possible to be an error."""

I think you'll find that this is a rule!!!!

I was thinking of something else, and i also edited...sorry again, my bad!!!

CodeBlox

wedge1988 wrote: »

if it goes in 32 host chunks then it should be 192-224, where 192 and 224 are not usable but the other addresses are?

anybody else here think the software is wrong?

cheers!

Software is not wrong. 223 is broadcast so it's not usable. 224 is the start of the next subnet (subnet ID) and isn't usable for the next subnet.

chopsticks

wedge1988 wrote: »

hey, it's happened before! ... not that i like to make a habit of this..

heres my question..

and i'm shortening it, but it should make sense..

In the boson test software for the ccna, a question asks about a network setup. the answers are:

1. can use any ip address between 192.168.11.193 and 192.168.11.223
2. can use any ip address between 192.168.11.193 and 192.168.11.222

now, the VLSM is /27 so to me, that makes the network break up into 32 host chunks..

so why am i asking?

am i going crazy? the answer was apparently number 2 above.

I'm sure it should really be number 1?

if it goes in 32 host chunks then it should be 192-224, where 192 and 224 are not usable but the other addresses are?

anybody else here think the software is wrong?

cheers!

I find it interesting, and did some workings on it:



11000000 . 10101000 . 00001011 . 11000001 = 192.168.11.193
11111111 . 11111111 . 11111111 . 11100000 = 255.255.255.224 (/27)

---

AND them
11000000 . 10101000 . 00001011 . 11000000 = 192.168.11.192


So we know 192.168.11.192 is the Network ID now.

Hence, we will also know that the next network address from it will be the first valid host:

11000000 . 10101000 . 00001011 . 11000001 = 192.168.11.193
.
.
.
.
11000000 . 10101000 . 00001011 . 11011110 = 192.168.11.222
11000000 . 10101000 . 00001011 . 11011111 = 192.168.11.223

Therefore, we also get to know that 192.168.11.223 is the broadcast address for this range of IP addresses, and 192.168.222 will be the last valid host in this range.


So, I will choose (2) as my final answer.

wedge1988

haha, you guys have gone all out on this havn't you? I actually re-thought this 20 mins later and just got home so..

lol.

thanks for the help anyway ^_^

yeh i knew 192 is the network address
yeh i recalculated and did 223 as the broadcast.

this is where i went wrong:

1. didnt include 192 as the network mask.

2. did this: 192+8 = 200 + 24 = 224

3. DUH! 192 is an address too!

lol. i guess the lesson here is don't attempt to rush a test before you go out!

wedge1988

earweed wrote: »

32 host addresses would have you going from 192-223. It's a simple error.

and i agree, as stated above hahahaha.

IT WAS A SIMPLE ERROR "CodeBlox" ^.- lolz.