subnet mask question

Given a subnet mask of 255.255.255.224. which of the following addresses can be assigned to network hosts
the options given are (choose three)
1. 15.234.118.63
2. 92.11.178.93
3. 134.178.18.56
4. 192.168.16.87
5. 201.45.116.159
6. 217.63.12.192
Plz explain how to identify the correct answers.

thanks.

I'm not going to just hand out the answer, but I'll give some big clues:

The "block size" for a 255.255.255.224 mask is 32. This means that the network numbers are:

x.y.z.0
x.y.z.32
x.y.z.64
x.y.z.96
x.y.z.128
etc.

which provide 32 IP addresses per network. The first address is the network address and the last one is the broadcast address, hence there are 30 useable host addresses per network.

The easy way to determine the block size is to subtract the non-255 octet in the subnet mask from 256.

From these network numbers, it's easy enough to look at the IP addresses that you listed and decide which are valid host addresses rather than network or broadcast addresses.

If you're not comfortable using such a short cut, there are several very useful posts on the forum which deal with the basics of converting everything to binary and working it out from first principles.

Hi steveP,

I got the answer, thanks for your quick reply.

Interesting Steve P. Trying to figure out the answers. Why are u stingy with answer (lol) those are CCNA questions i guess so may be the dude is reading up and needs quick help?

SteveP can u elaborate some more? I'd be taking my exams soon and i really need to get sticky and refreshed.


SteveP Wrote:

From these network numbers, it's easy enough to look at the IP addresses that you listed and decide which are valid host addresses rather than network or broadcast addresses.

Did not get that paragraph.


Thanks

From these network numbers, it's easy enough to look at the IP addresses that you listed and decide which are valid host addresses rather than network or broadcast addresses.

Did not get that paragraph.


Thanks

I'll answer that specific question.

255.255.255.224 has 27 network bits, remaining with 5 host bits.
This translates to 2^5 = 32 hosts per subnet.

hence;
x.y.z.0
x.y.z.32
x.y.z.64
x.y.z.96
x.y.z.128
x.y.z.160
x.y.z.192
x.y.z.224


Each subnet has a total of 32 ip addresses but only 30 ip addresses are usable. This is because the 1st address of each subnet is the subnet I.D and the last address of each subnet is the broadcast.

In subnet 1: x.y.z.0
x.y.z.0 is the subnet I.D and x.y.z.33 is the broadcast address

In subnet 2: x.y.z.32
x.y.z.32 is the subnet I.D and x.y.z.63 is the broadcast address

This goes the same for all the other subnets.

trying to patch tru with ur explanation losh but relating it to lav_plsb1's initial question, if i had to choose an answer i will choose 1 and 6 using ur explanation of id and broadcast please correct me if i am wrong?

thanks man