Yet another subnetting Question....

[Chr1sUK]

I have managed to get to grips with most of this subnetting business but there are still a few things that stump me when I am questioned on them.

I was doing some revision earlier today, and this question came up on a practice test:

"Given the IP Address 172.156.100.100, and mask 255.255.255.224, choose the appropriate broadcast address?"

- 172.156.100.127
- 172.156.100.255
- 172.156.100.31
- 172.156.100.128

I thought it could be .255 - but it isnt!

If somebody could tell me the answer and explain why it is the answer I would greatly appreciate it.

Thanks,

Chris

 

[Parcival21]

Hi Chris,

We are used to have the .255 as the broadcast address. The reason for that that most admins use class A, B, or C in their network. A class A net always starts at th address xxx.xxx.xxx.0 which is always the network address and ends with the address xxx.xxx.xxx.255.
The first address of a network is always reserved for the network address and the last address of the subnet is always reserved for the broadcast address. So if the last address of your subnet is not 255 this is not your broadcast address.
Now that was the way it should be done I think. Now you have to find the last address of your subnet. this will be the broadcast address.

bye
busche

 

[Parcival21]

Sorry,
should have finished that one. There might me a better way to solve this question.
now that you know that the first three digits of you last octet are part of the network address you have to do the following:
100 is 0 times 128(2 to the power of 7)
is 1 times 64 (2 to the power of 6)
is 1 times 32 (2 ................5)
So your network address is:
xxxx xxxx.xxxx xxxx.xxxx xxxx.0110 0000
and your broadcast address is:
..............................0111 1111
Which is 2to the power of 6 + 2 to the power of 5+....
The final outcom should be 127 if I didn't calculate wrong.

 

[Microbyte]

since you got the basic ip subnetting, now you have to throw away your classful ip address state of mind, and enter into the world of classless ip addressing.

with the given ip address 172.156.100.100 and mask 255.255.255.224
you should focus subnet mask.

since first 3 octets are all "one"s in subnet mask, let's focus onto 4th octets.

ip address =100
128 064 032 016 008 004 002 001
-------------------------------
0 1 1 0 0 1 0 0

but the mask is = 224

128 064 032 016 008 004 002 001
-------------------------------
n n n x x x x x

n = subnet-work = 128 + 64 + 32 = 224
x = host

since broadcast address has all "one"s in HOST bit of the ip address.

128 064 032 016 008 004 002 001
-------------------------------
0 1 1 0 0 1 0 0
becomes
128 064 032 016 008 004 002 001
-------------------------------
0 1 1 1 1 1 1 1

equals
127...right?
so, broadcast address for this specific address 172.156.100.100 (or specific network = 172.156.100.96) is
172.156.100.127

hope this helps

Microbyte

 

[Chr1sUK]

Thanks for your help there folks, I understand that.

Although, according to the Cisco CDROM, the answer is 172.156.100.128!

 

[Chr1sUK]

I notice you said its all ones for the broadcast address, but why have you left one as 0?

 

[Microbyte]

brodcast address being 172.156.100.128 is dead wrong. Only way that address would be a broadcast address is....hmm...never. however, it can be a network address.

and, i presume your last threat is for me, and specifically...

"since broadcast address has all "one"s in HOST bit of the ip address"

this comment..right?

the "0" you are referring to is, the (sub)network bit, not the host bit.

Microbyte

 

[Microbyte]

correction...i mistakenly typed the word "threat" instead of "thread"...wow, what a difference one single letter makes? isn't it awfully wonderful?

Microbyte

 

[Parcival21]

just one more thing.
As you can see in Microbyte's post:
128 064 032 016 008 004 002 001
-------------------------------
n n n x x x x x
Always start with dividing your ip address into network part(being the first part) and host-part(x). if you have done that, write down the network address (the host part should be all zeros. Then write down the broadcast address where the host part should be all ones. The ones and zeros in the network bits have nothing to do with an address being a broadcast or a network address.