I have a class B network 132.245.0.0 each subnet requires 100 subnets and each subnet has 400 host addresses?

Question

bits borrowed,subnets created,hosts created subnet mask,host range

Answer 1

Subnet masks must be contiguous, which means block sizes must be even powers of 2, so you would need 128 subnets, each with [up to] 512 addresses. This is impossible within a class B. The maximum number of IPs possible in a single subnet class B is 65,533 ((256 * 256) - 3) (one for network address, one for gateway, one for broadcast.) Since each subnet incurs that same overhead of 3 unusable IPs...

Oh hmm, but you don't need all 128, nor all 512, so... I'm still not sure it's possible:

This is the table of legal masks from Class C to B:

255.255.254.0 = /23 - 2 Class C's
255.255.252.0 = /22 - 4 Class C's
255.255.248.0 = /21 - 8 Class C's
255.255.240.0 = /20 - 16 Class C's
255.255.224.0 = /19 - 32 Class C's
255.255.192.0 = /18 - 64 Class C's
255.255.128.0 = /17 - 128 Class C's
255.255.0.0 = /16 - 256 Class C's - Full Class B

So a mask of 255.255.254.0 should give you a subnet of 512 - 3 addresses, and the network addresses would be all the even numbers possible in quad 3, e.g.,

(first 2 quads are implicit, shown as *, for all but Network)
Network, Gateway, First, Last, Broadcast
-----------
132.245.0.0, *.0.1, *.0.2, *.1.254, *.1.255
132.245.2.0, *.2.1, *.2.2, *.3.254, *.3.255
132.245.4.0, *.4.1, *.4.2, *.5.254, *.5.255
132.245.6.0, *.6.1, *.6.2, *.7.254, *.7.255
etc...

That should meet your requirement, but there might be a flaw in it, because I can tell you for sure that in practice, this sort of subnetting is virtually never actually done. (Plus blocks greater than 256 blows out my subnet calculator.)

Good luck with your homework. :-)

Answer 2

"That should meet your requirement, but there might be a flaw in it, because I can tell you for sure that in practice, this sort of subnetting is virtually never actually done. (Plus blocks greater than 256 blows out my subnet calculator.)"

I'm not sure where that comment comes from.. In the 'real world' that is done all the time. Subnetting and Veriable Length Subnet Mask (VLSM) is the only reason we haven't completely ran out of IP addresses yet. It is called "Classeless" IP addressing...

The first answer is dead on.. You would want a /23 mask (255.255.252.0) that would give you 128 network addresses, each with 510 useable IP addresses.
the network address would be 0.0 - +1.255.
So, 132.245.0.0 is the network address, 132.245.1.255 would be the broadcast addres (all IPs between would be useable addresses. )
The second group would be; 132.254.2.0 as the network address, and 132.254.3.255. (all IPs between would be useable address)

Answer 3

100 subnets = 7 bits, 400 host = 9 bits because it would provide 512 hosts -2 from network and broadcast

Therefore, the subnet mask would be 255.255.254.0

Therefore, the first subnet would be 132.245.0.0 and broadcast is 132.245.1.255 with usable 132.245.0.1 - 132.245.1.254

the 2nd one network is 132.245.2.0 , broadcast 132.245.3.255 and the useable would be 132.245.2.1 - 132.245.3.254,

3nd network 132.245.4.0 , broadcast 132.245.5.255, useable 132.245.4.1 - 132.245.2.254

Once you read it, you'll notice the pattern

Answer 4

you would borrow 7 bits. That would give you 128 subnets and 510 address per subnet. subnet mask would be 255.255.254.0

1st subnet:
ID: 132.245.0.0
Range: 132.245.0.1 - 132.245.1.254
Broadcast: 132.245.1.255

2nd subnet
ID: 132.245.2.0
Range: 132.254.2.1-132.245.3.254
Broadcast: 132.245.3.255

3rd subnet
ID: 132.245.4.0
Range: 132.245.4.1-132.245.5.254
Broadcast: 132.245.5.255

and so on.........

Answer 5

Here are some places to practice subnetting:

http://www.easysubnet.com/
http://www.subnettingquestions.com/custom/bren/

The second one is completely free.

Answer 6

. Wrong forum G T F O .