I just typed ipconfig /all into the cmd.
The following was returned
Connection-specific DNS Suffix . : lan
Description . . . . . . . . . . . : Broadcom 802.11n Network Adapter
Physical Address. . . . . . . . . : 74-E5-43-7D-8B-65
DHCP Enabled. . . . . . . . . . . : Yes
Autoconfiguration Enabled . . . . : Yes
Link-local IPv6 Address . . . . . : fe80::39d0:ff36:ff66:2215%15(Preferred)
IPv4 Address. . . . . . . . . . . : 192.168.1.78(Preferred)
Subnet Mask . . . . . . . . . . . : 255.255.255.0
Lease Obtained. . . . . . . . . . : 26 January 2020 07:13:40
Lease Expires . . . . . . . . . . : 27 January 2020 07:13:40
Default Gateway . . . . . . . . . : 192.168.1.254
DHCP Server . . . . . . . . . . . : 192.168.1.254
DHCPv6 IAID . . . . . . . . . . . : 259319107
DHCPv6 Client DUID. . . . . . . . : 00-01-00-01-25-1D-01-86-D4-BE-D9-53-C6-37
DNS Servers . . . . . . . . . . . : 192.168.1.254
I know it's beyond the scope of this module, that is subnetting;
My IPv4 Address is 192.168.1.78
I have a subnet mask of 255.255.255.0 --> this is a network, network, network. host pattern
which means I'm on the 192.168.1.0 network.
now if I had a subnet mask of 255.255.255.240 what do you think this means?
Answer:
I think it means I've sub-netted my default subnet mask, I've taken or stolen binary bit's from the octet host portion of my IPv4 address.
This means more networks but at a cost of less hosts.
This means in effect we are creating smaller subnetworks out of the larger one, dividing it up .
But realising that although you will increase the amount of networks, each range will hold a smaller amount of hosts per subnetwork.
how many bits did I steal to do this?
Answer:
convert the custom subnet dotted decimal host portion to binary,
128 64 32 16 8 4 2 1 - binary values working
1 1 1 1 0 0 0 0 240 - 128 = 112
112 - 64 = 48
48 - 32 = 16
16 - 16 = 0
check: 128 + 64 + 32 + 16 = 240
111100002 = 24010 (denary value)
This means 4 bits where taken from the host octet portion of the IPv4 address.
next question :
How many usable new subnetworks does this give me?
to work this out :
use formula : Number of subnets = 2s
so 24 = 16 so 16 subnets.
How many hosts per subnet ?
use formula : Number of hosts per subnet = 2h -2
I originally had 8 bits in the host portion and 8 - 4 = 4
so 24 - 2 = 14
Total number of subnets (subnetworks) = 16
Total number of host address per subnetwork = 14
workout the difference in binary between the default network address binary number and the
custom subnetwork binary number: ( Note the count starts at 0 !)
Address Ranges per subnetwork
(0) 192.168.1. 0 0 0 0 . 0 0 0 0 192.168.1.0 - 192.168.1.15
(1) 192.168.1. 0 0 0 1 . 0 0 0 0 192.168.1.16 - 192.168.1.31
(2) 192.168.1. 0 0 1 0 . 0 0 0 0 192.168.1.32 - 192.168.1.47
(3) 192.168.1. 0 0 1 1 . 0 0 0 0 192.168.1.48 - 192.168.1.63
(4) 192.168.1. 0 1 0 0 . 0 0 0 0 192.168.1.64 192.168.1.79
(5) 192.168.1. 0 1 0 1 . 0 0 0 0 192.168.1.80 - 192.168.1.95
(6) 192.168.1. 0 1 1 0 . 0 0 0 0 192.168.1.96 - 192.168.1.111
(7) 192.168.1. 0 1 1 1 . 0 0 0 0 192.168.1.112 - 192.168.1.127
(8) 192.168.1. 1 0 0 0 . 0 0 0 0 192.168.1.128 - 192.168.1.143
(9) 192.168.1. 1 0 0 1 . 0 0 0 0 192.168.1.144 - 192.168.1.159
(10) 192.168.1. 1 0 1 0 . 0 0 0 0 192.168.1.160 - 192.168.1.175
(11) 192.168.1. 1 0 1 1 . 0 0 0 0 192.168.1.176 - 192.168.1.191
(12) 192.168.1 1 1 0 0 . 0 0 0 0 192.168.1.192 - 192.168.1.207
(13) 192.168.1 1 1 0 1 . 0 0 0 0 192.168.1.208 - 192.168.1.223
(14) 192.168.1 1 1 1 0 . 0 0 0 0 192.168.1.224 - 192.168.1.239
(15) 192.168.1 1 1 1 1 . 0 0 0 0 192.168.1.240 - 192.168.1.255
So to conclude there are a total of 16 subnets or 16 subnetworks, that counting from 0 to 15 which is like the range function in python index starting at 0 !
Within each subnetwork division there are 14 usable host addresses
this is because each sub-network's starting address
per pool of IPv4 addresses is the address of that subnetworks address itself ! ( the lowest IPv4 address)
and the highest value address per pool of subnet-work addresses is that subnetwork's broadcast address.
super-netting is the opposite more hosts but at cost of less networks.