CCNA English-06: Subnetting in class
Subnetting in class
A buddy of mine obtained a job in a multinational organisation earlier than ending his studies. Then of direction no one else amongst our pals joined the job. And this pal of mine would get paid on the twenty fifth or 26th. So at the give up of the month, when we have been in a tug of war, this pal was once buying and shopping. Even then we used to have a lot of exciting with him due to the fact he talked very little and usually tried to be alone. One day I requested this pal if you have joined IT then you can do some IT work! Oto acquired indignant and stated if he can not work, did he see my face and provide me a job! I notion to myself that some thing is going to be recognized now. So I without delay stated sorry, so if you do not be aware of the job, how did the job occur and I straight away requested what category IP in your workplace the use of the category right away said. I requested again, why in class? He stated with a little bit of wisdom, "Hey, you do not know. The motive of the use of us in our category is to have much less community however greater hosts." From this we apprehend that if the quantity of community IDs is much less and the range of host IDs is greater then we will pick its IP in class. Also,
The first eight bits are used in NetAir in Class A and the subsequent 24 bits are used as host bits. So it is understood that when chosen in the class, nearly three instances the community tackle can be used as the host address. And the default masks of the category is 255.0.0.0.
Let's subnet in classification A with a (1) bit on.
10.0.0.0/9
255.128.0.0
Network wide variety = 2^1 = 2
Number of hosts = 2^23-2 = 8338606
Subnet ID = 256-128 = 128
Network-1 = 10.0.0.0
First host = 10.0.0.1
10.0.0.1
10.0.0.3
Last host = 10.127.255.254
Broadcast tackle = 10.127.255.255
–
Network-2 = 10.128.0.0
First host = 10.128.0.1
10.128.0.2
10.128.0.3
Last host = 10.255.255.254
Broadcast tackle = 10.255.255.255
===================================
In the equal way, in case of type B, we subnet via turning on two (2) bits.
10.0.0.0/10
255.192.0.0
Network range = 2^2 = 4
Number of hosts = 2^22-2 = 4194302
Subnet ID = 256-192 = 64
–
Network-1 = 10.0.0.0
First host = 10.0.0.1
10.0.0.1
10.0.0.3
Last host = 10.63.255.254
Broadcast tackle = 10.63.255.255
–
Network-2 = 10.64.0.0
First host = 10.64.0.1
10.64.0.2
10.64.0.3
Last host = 10.127.255.254
Broadcast tackle = 10.127.255.255
–
Network-3 = 10.128.0.0
First host = 10.128.0.1
10.128.0.2
10.128.0.3
Last host = 10.191.255.254
Broadcast tackle = 10.191.255.255
Network-4- = 10.192.0.0
First Host = 10.192.0.1
10.192.0.2
10.192.0.3
Last host = 10.255.255.254
Broadcast tackle = 10.255.255.255
===================================
The kind of questions that the check contains.
10.0.0.0/6 Each subnet will have at least 5000 hosts.
Q: How many bits do you want to flip on to create 5,000 hosts?
Answer:
2^12 = 4096-2 = 4094
2 ^ 13 = 8192-2 = 8190
In different words, if we flip on thirteen bits, our want will be fulfilled.
Q: So what will be the new subnet?
Answer:
Since the default subnet of the type has been grew to become on 255.0.0.0 and thirteen extra bits then (32-13) = 19 bits. If you flip on 19 bits, then 255 for the first 6 bits. 255 for the subsequent eight bits, then the relaxation is three bits. The fee of three bits is 224. So the subnet mask
255.255.224.0
We received this facts from the area we discovered earlier
128-192-224-240-248-252-254-255
1--2--3--4--5--6--7--8
Answer: What is the block size?
256-224
= 32
Q: What is the fifth network?
10.0.0.0
10.0.32.0
10.0.64.0
10.0.96.0
10.0.128.0
I am ending right here like today. Our subsequent tutorial is VLSM. And to see the preceding lectures, go to the predominant web page of the route by using following the hyperlink below.
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