Kindlky correct my subnetting

satishtech

Subnetting Example1 :

172.16.192.0/19

50 lans 70 users required.

Next
5 lans and 10 users each is required.

1.
172.16.192.0 / 19

172.16.192.0 /19 =32
32 is Block Size

172.16.192- 224.0

192 to 224 is the network space available with block size 32
/19 = 2^3 = 8 Networks.

Network :
172.16.192.0 Network number
172.16.192.1 First Host
172.16.223.254 Last Host
172.16.223.255 Broadcast address


192 to 223 19th or 3rd octet varies by 32
0 to 128 25th octet varies
6 bits borrowed to /25 to get 50 LAN's

therefore we get
Network 1
1.172.16.192.0

Network2
2.172.16.192.128

Network 3
3.172.16.193.0

Network4
4.172.16.193.128

that is from 192 to 223

32 * 2 = 64 subnets

last subnet is
172.16.223.128 Network number
172.16.223.129 First Host
172.16.223.254 Last Host
172.16.223.255 Broadcast number

2.
Next for 5 Lans with 10 Users each.
we borrow 3 more bits after /25
2^3 = 8 total 8 subnets
4 host bits remain 2^4 =16 Hosts
Block size 16

first LAN is after choosing last subnet after /25 we use /28
172.16.223.128 /28 is network number
172.16.223.129 /28 is first host
172.16.223.142 /28 last host
172.16.223.143 /28 broadcast address

172.16.223.144 second network number
172.16.223.160 third...
172.16.223.176 fourth...

OctalDump

This looks ok, except I'm not sure why those two subnets overlap.

I think that you only need to 172.16.217.0 for the 50 /25s (I just did this by saying it's the same as 25 /24s, so 192 + 25). And they'd be as you describe:

Network 1
1.172.16.192.0

Network2
2.172.16.192.128

Network 3
3.172.16.193.0

etc to:

Network 50

172.16.216.128

So you can start at 172.16.217.0/28 for your /28s. which is easier since you only need 5:

172.16.217.0
172.16.217.16
172.16.217.32
172.16.217.48
172.16.217.64