Assuming that you already know all the basics and fundamentals of subnetting.
I am going to show you how you can solve questions about subnetting very easily with the help of the chart below:
Lets see how this chart help us to easy subnetting by solving questions. . .
Question # 1:
Given the network ID: 192.168.123.0. What subnet mask will you need for 10 subnet and also write down the first valid subnet???
Answer:
SUBNETMASK:
Because the IP belong to class C so we don't have to worry about first three octates they are default for class C i.e.
SUBNETMASK:
Because the IP belong to class C so we don't have to worry about first three octates they are default for class C i.e.
255.255.255.-----
so we only concerned with last octate. . .
so we only concerned with last octate. . .
A quick look at the chart indicates that 14 subnets yields 240 subnet mask which is our required subnet. so the answer is 255.255.255.240.
First Subnet:
Look above the subnet mask which is in this case is 240,and you see 16.This is your first subnet address i.e.
192.168.123.16
Now,if you want all 10 subnet addresses simply add the range to previous subnet address, for second subnet address add the range i.e 16 in this case to first subnet address i.e
192.168.123.32
Look above the subnet mask which is in this case is 240,and you see 16.This is your first subnet address i.e.
192.168.123.16
Now,if you want all 10 subnet addresses simply add the range to previous subnet address, for second subnet address add the range i.e 16 in this case to first subnet address i.e
192.168.123.32
Question # 2:
Given the network ID: 172.190.0.0. What subnet mask will you need for 10 subnet???
Given the network ID: 172.190.0.0. What subnet mask will you need for 10 subnet???
Answer:
Because the IP belong to class B , the first two octate will remain default i.e
255.255.----.------
we have to concerned with last two octates.Now look at the chart 14 subnet yields 240 subnet mask so the subnet mask for the given IP is:255.255.240.0
Question # 3:
What is the broadcast address of the network 192.168.102.192/28??
Answer:
28 followed by slash indicates that there is 4bits borrowed for subnetting.
Now look at the chart from left to right the 4th bit yields 240 subnet mask.Now subtract this subnet mask from 255 i.e
255-240=15.
add 15 to last octate of given IP i.e
192.168.102.192 +15 i.e.
192.168.102.207 is the broadcast address of given address.
I hope it will help you. . .
