subnetting/VLSM

Could someone assist me with the following. Given IP address 192.168.24.0/22. A network requires 400 hosts. Thus I came with the 192.168.24.0/23 to 192.168.25.255 IP range to meet the number of hosts.

Question: is above one big network, thus one broadcast address or do class C networks broadcast at .255 address regardles of size/#hosts?

Thanks.

Avctually, the 192.168.24.0/22 is given for the requirement of following network
RouterA requires 50 hosts
RouterB 400 hosts
RouterC 50 hosts
RouterD 200 hosts

It all depends on how you look at your IP address when it comes to finding your broadcast address. If you look at your whole IP space, 192.168.24.0/22, then the broadcast address would be 192.168.27.255. Now, if you begin to subnet this network, each subnet will be treated separately when it comes to their broadcast address, min and max host, etc. 192.168.24.0/23 broadcast address would be 192.168.25.255, for example.

Question: is above one big network, thus one broadcast address or do class C networks broadcast at .255 address regardles of size/#hosts?

Also, the broadcast address is not always a .255 address, especially when you are using variable length subnet masks. Here's an example for your network, you have a requirement for two subnets w/50 hosts. You could subnet part of your network as follows:

192.168.26.0/25* for one of the 50 host nets
192.168.26.1 is host min., 192.168.26.126 is host max, 192.168.26.127 is broadcast address.

192.168.26.128/25* for the other.
192.168.26.129 is host min., 192.168.26.254 is host max, 192.168.26.255 is broadcast address.

So as can be seen, the broadcast address is not necessarily a .255 address. If you have any questions about the arithmetic involved or any further questions, please ask.
*You could use a /26 mask if you want to use less address space. I just used /25 to demonstrate the broadcast address as I did.

JWJ,
Thanks for your reply. Could you explain how drived the broadcast address for 192.168.24.0/22, would be 192.168.27.255.

Thanks.
Said

Ok, so your network is 192.168.24.0/22.

In binary, the network 192.168.24.0:
11000000.10101000.00011000.00000000

and the subnet mask 255.255.252.0:
11111111.11111111.11111100.00000000

The bolded area is the host port of this network as is applied with this subnet mask. To find your broadcast IP address, change all the host bits from 0 to 1.

Broadcast address:
11000000.10101000.00011011.11111111

If you convert this to decimal you get 192.168.27.255