NET603: Business Subnetting with CIDR notation.
192.168.1.0/16
How many subnets: 1 network, no subnets
How many hosts: 65536
Net Mask: 255.255.0.0
Net Address: 192.168.0.0
Broadcast Address: 192.168.255.255
192.168.1.0/17
How many subnets: 2
How many hosts: 32768(-2) each
Net Mask: 255.255.128.0
Subnet 1:
Net Address: 192.160.0.0
Broadcast Address: 192.168.127.255
Subnet 2:
Net Address: 192.168.128.0
Broadcast Address: 192.168.255.255
192.168.1.0/18
How many subnets: 4
How many hosts: 16384(-2) each
Net Mask: 255.255.192.0
Subnet 1:
Net Address: 192.168.0.0
Broadcast Address: 192.168.63.255
Subnet 2:
Net Address: 192.168.64.0
Broadcast Address: 192.168.127.255
Subnet 3:
Net Address: 192.168.128.0
Broadcast Address: 192.168.191.255
Subnet 4:
Net Address: 192.168.192.0
Broadcast Address: 192.168.255.255
192.168.1.0/19
How many subnets: 8
How many hosts: 8192(-2) each
Netmask: 255.255.224.0
Net Addresses / Broadcast Addresses of each subnet:
1# 192.168.0.0–192.168.31.255
2# 192.168.32.0–192.168.63.255
3# 192.168.64.0–192.168.95.255
4# 192.168.96.0–192.168.127.255
5# 192.168.128.0–192.168.159.255
6# 192.168.160.0–192.168.191.255
7# 192.168.192.0–192.168.223.255
8# 192.168.224.0–192.168.255.255
192.168.1.0/20
How many subnets: 16
How many hosts: 4096(-2) each
Netmask: 255.255.240.0
Network Address — Broadcast Address.
Subnet 1# 192.168.0.0–192.168.15.255
Subnet 2# 192.168.16.0–192.168.31.255
Subnet 3# 192.168.32.0–192.168.47.255
Subnet 4# 192.168.48.0–192.168.63.255
Subnet 5# 192.168.64.0–192.168.79.255
Subnet 6# 192.168.80.0–192.168.95.255
Subnet 7# 192.168.96.0–192.168.111.255
Subnet 8# 192.168.112.0–192.168.127.255
Subnet 9# 192.168.128.0–192.168.143.255
Subnet 10# 192.168.144.0–192.168.159.255
Subnet 11# 192.168.160.0–192.168.175.255
Subnet 12# 192.168.176.0–192.168.191.255
Subnet 13# 192.168.192.0–192.168.207.255
Subnet 14# 192.168.208.0–192.168.223.255
Subnet 15# 192.168.224.0–192.168.239.255
Subnet 16# 192.168.240.0–192.168.255.255
192.168.1.0/21
How many subnets: 32
How many hosts: 2048(-2)
Net Mask: 255.255.255.248
#S — Net Address — Broadcast Address
#1: 192.168.0.0–192.168.7.255
#2: 192.168.8.0–192.168.15.255
#3: 192.168.16.0–192.168.23.255
#4: 192.168.24.0–192.168.31.255
#5: 192.168.32.0–192.168.39.255
#6: 192.168.40.0–192.168.47.255
#7: 192.168.48.0–192.168.55.255
#8: 192.168.56.0–192.168.63.255
To 32 etc.
The amount of subnets is figures out by ‘2 to the power of n bits’ in this case n=5
example: 2 to the power of 5 is 32
Each subnet increments by ‘8’, this is figured out by 256/32=8
The amount of hosts per subnet is calculated by ‘2 to the power of host bits’ in this case h=11 (8+3)
example: 2 to the power of 11 is 2048, due to the network and broadcast addresses of each subnet, this is reduced to 2046 for each.
Class Question:
A company has a 190.240.0.0/16 block. They now want to re-organise it into large blocks. They wish to have subnets that can handle at least 1000 hosts each.
Answer
190.240.0.0/22
Amount of subnets: 64 , (2⁶)
How many hosts: 1024(-2) , (2¹⁰)
Please calculate:
S = bits used for subnetting
H= bits used for the number of hosts in each subnet
Answer: nnnnnnnn.nnnnnnnn.sssssshh.hhhhhhhh
Also, please calculate the following for the second subnet:
subnet mask: 255.255.252.0
subnet address: 190.240.4.0
first usable address in that subnet: 190.240.4.1
last usable address in that subnet: 190.240.7.254
broadcast address of that subnet: 190.240.7.255
