Hi
I have a question regarding how to get valid IP ranges when subnetting Class A & B, even though I believe class A,b&C all use he same technique but for me class C is much easier because it only involves to work on one octet instead of two or three octets.
This first scenario I believe i is fully resolved, it is class C address with 16 subents and 14 hosts each.
1- 192.168.44.0/28
Subnets Valid IPs Broadcast
192.168.44.0 (1-14) 192.168.44.15
192.168.44.16 (17-30) 192.168.44.31
192.168.44.32 (33-46) 192.168.44.47
192.168.44.48 (49-62) 192.168.44.63
192.168.44.64 (65-7

192.168.44.79
192.168.44.80 (81-94) 192.168.44.95
192.168.44.96 (97-110) 192.168.44.111
192.168.44.112 (113-126) 192.168.44.127
192.168.44.128 (129-142) 192.168.44.143
192.168.44.144 (145-15

192.168.44.159
192.168.44.160 (161-174) 192.168.44.175
192.168.44.176 (177-190) 192.168.44.191
192.168.44.192 (193-206) 192.168.44.207
192.168.44.208 (209-222) 192.168.44.223
192.168.44.224 (225-23

192.168.44.239
192.168.44.240 (241-254) 192.168.44.255
The following two are also partially resolved and want some assitance ..
10.0.0.0/25
Subnets Valid IP Broadcast Address
10.0.0.0 [ ????] 10.0.127.255
10.0.0.128 [???? ] 10.0.?. ?
I am stuck here, Even though I belive there is 131072 subnets and 126 hosts per subnet, I dont know how to get that out so please help.
172.14.0.0/26
I know there is 1024 subnets, with 30 hosts each, but I didnt understand how to identify all those 1024 from network number, valid range to broadcast same as what I have shown above in the first scenario...
first four subnets are:
172.14.0.0
172.14.0.64
172.14.0.128
172.14.0.192
Beacuse of 192+64=255 , should I make it
172.14.1.0
172.14.1.64
172.14.1.128
172.14.1.192
172.14.2.0
172.14.2.64
172.14.2.128
172.14.2.192
Any suggestions and calarifications will be appreciated.
Cheers..........