Subnetting Help! ....again

How would i do these
What is the broadcast address for a host address of 172.16.31.99/22?
172.16.255.255
172.16.31.255 (this is my answer, is it corect?)
172.16.31.103
172.16.63.255
172.16.24.111

You are given 172.16.112.0/19 as a host address.
What is the LAST address in the address space?


The first subnet for 172.16.0.0/25 is 172.16.0.128. What is the broadcast address for this first subnet?

Your largest network needs to support 100 hosts. You have been given a Class B network. What is the maximum number of bits you can borrow?

You are given 172.16.112.0/19 as a host address.
What is the FIRST address in the address space?

The first subnet for 172.16.0.0/25 is 172.16.0.128. What is the broadcast address for this first subnet?

Any way to do the rest of the questions?

How would i do these
What is the broadcast address for a host address of 172.16.31.99/22?
172.16.255.255
172.16.31.255 (this is my answer, is it corect?)
172.16.31.103
172.16.63.255
172.16.24.111

Yes Correct

You are given 172.16.112.0/19 as a host address.
What is the LAST address in the address space?

172.16.112.0/19 is within the network id of 172.16.96.0/19. This 19bits makes the 3rd octect go up in chuncks of 32 (i.e. 11111111 11111111 11100000, as you can see it leaves 5 0's which is 2^5 = 32). This gives network id's of 0-31 (not valid in some TCP/IP implementations),32-63,64-95,96-127,etc... therefore our LAST address is 172.16.127.254 (cannot have .255 as this is a broadcast address.

The first subnet for 172.16.0.0/25 is 172.16.0.128. What is the broadcast address for this first subnet?

Again, /25 is 11111111 11111111 11111111 10000000, as you can see, the last octect has 7 zero's, this gives us 2^7=128 which is our increment. Therefore we get the next subnet starting at 256. Therefore the bradcast address is 172.16.0.255

Your largest network needs to support 100 hosts. You have been given a Class B network. What is the maximum number of bits you can borrow?

2^7-2=126. Therefore 7 bits will give you 126 hosts. If thats what the question is asking, i'm not too clear what its actually asking

You are given 172.16.112.0/19 as a host address.
What is the FIRST address in the address space?

You should be able to work this out using example above

The first subnet for 172.16.0.0/25 is 172.16.0.128. What is the broadcast address for this first subnet?

Again, give this a go aswell, if you get stuck with the process post another question for guidence. Also, read the post in this forum about Subnetting and give it a go

Cheers

Ok got it thanks, i don't really use the binary method ie dividing the last octet in binary and putting 1 and 0s i usually get the magic number and work from there, i mostly know the answers but i m not confident with them for some reason. Thanks for your help

No problem, just post again if you have any further questions.

Wayne