subnetting help

[viper92086]

hey guys so i'm studying for my exam and i figured out everything except for these two sample problems. i have the answers and SOME of the work involved but i cant figure out for the life of me how to get to that answer. Here are the questions:

A network of a site has network address 154.18.48.0. If its Direct Broadcast address is 154.18.51.255, then what network mask is used by this network?

154.18.48.0
154.18.00110000.00000000

154.18.51.255
154.18.00110011.11111111

Make net bits = 1

So,
255.255.252.0


* i cannot figure out how to get the "252" without the problem giving me the CIDR or number of hosts.



Here's the second question:
A site has been assigned the class B address 132.25.0.0. If the network administrator of this site wants to be able to introduce 29 subnets, then what mask (in dotted decimal notation) must he use? If the network addresses are assigned in an ascending order with first subnet the one that has all subnet bits equal to 0, then what is the network address and the direct broadcast address of the 21st network? You must use dotted decimal notation for all addresses.

2^5 = 32

132.25.0.0
132.25.10100|000.00000000

21st network, so find #20

Net: 132.25.160.0
Broadcast: 132.25.167.255


*so i know how to get the mask, (2^5 = 32 - 2, so it would be 11111111.1111111.11111000.00000000, or 255.255.248.0), but how do i come about the net and broadcast of the 21st network?



thanks guys.

 

[GrapeSoda]

A network of a site has network address 154.18.48.0. If its Direct Broadcast address is 154.18.51.255, then what network mask is used by this network?

154.18.48.0
154.18.00110000.00000000

154.18.51.255
154.18.00110011.11111111

Make net bits = 1

So,
255.255.252.0


* i cannot figure out how to get the "252" without the problem giving me the CIDR or number of hosts.

The broadcast address will have all of the host bits set to 1. So given:

154.18.00110000.00000000
154.18.00110011.11111111

We can see that the host portion consists of the last 2 bits in the third octet and the 4th octet. That leaves the first two octets and the first 6 bits in the third octet. So the subnet mask would be 255.255.11111100.00000000. 11111100 is 252 in decimal, so 255.255.252.0 is the subnet mask, or /22 in CIDR.

Here's the second question:
A site has been assigned the class B address 132.25.0.0. If the network administrator of this site wants to be able to introduce 29 subnets, then what mask (in dotted decimal notation) must he use? If the network addresses are assigned in an ascending order with first subnet the one that has all subnet bits equal to 0, then what is the network address and the direct broadcast address of the 21st network? You must use dotted decimal notation for all addresses.

2^5 = 32

132.25.0.0
132.25.10100|000.00000000

21st network, so find #20

Net: 132.25.160.0
Broadcast: 132.25.167.255


*so i know how to get the mask, (2^5 = 32 - 2, so it would be 11111111.1111111.11111000.00000000, or 255.255.248.0), but how do i come about the net and broadcast of the 21st network?
.

Just plug 20 in binary into the subnet portion of the IP address:

132.25.00000|000.00000000

add 20 (10100) to the subnet portion and get

132.25.10100000.00000000 (132.25.160.0) for the network ID
132.25/10100111.11111111 (132.25.167.255) for the broadcast