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
