Subnet question, help plz

[tchang]

I have one question needs to be answerd:

If a school owns all the ips in 128.197.xxx.xxx
and I want to devide the network in to 2 segments,

1. what would be my subnet mask?

2. what would be the network addresses of the two segments?


I think the answer for the first one is 255.255.0.0 since its a class b network, am I right?

[jkrohn]

If you are dividing that network into two parts you will need to borrow two bits.
The subnet mask then becomes
255.255.192.0
The first network will be
128.197.64.1 = 127.197.010000.0000001
To
128.197.127.254 = 128.197.01111111.11111110
128.197.127.255 is your directed broadcast
128.197.64.0 is the wire

The second network will be
128.197.128.1 = 128.197.10000000.00000001
to
128.197.191.254 = 128.197.10111111.11111110
128.197.191.255 is your directed broadcast.
128.197.128.0 is the wire

128.197.255.255 is your standard broadcast

I did this all in my head so there oculd be some wrong numbers in there
Jkrohn

[tchang]

thank you so much for answering, so can I say this if my subnet mask is 255.255.240.0 and

my first device ip is 137.56.32.1 and
my last device ip should be 137.56.32.254

on the segment 137.56.32.0?

thank you