Subnetting questions.

CaptainCharismaCaptainCharisma Member Posts: 36 ■■□□□□□□□□
Alright I can subnet pretty well and try to do most of it in my head, I practice at work when I've got nothing to do (which is a good chunck of my day), but I do not understand some of the answers to some of the questions on subnettingquestions.com. Something like:

"How many subnets and hosts per subnet can you get from the network 10.0.0.0 255.255.240.0?"

Answer (according to them): 4096 subnets and 4094 hosts

The hosts part is correct, but shouldn't their answer for the subnets be 16, because the 255.255.240.0 subnet mask limits you to dealing only with the 3rd octet for subnets.

Will somebody please clarify wether I'm right or wrong on this, because this is driving me crazy.

Comments

  • gojericho0gojericho0 Member Posts: 1,059 ■■■□□□□□□□
    You also have all the subnet bits in the second and third octet since this is a class A address
  • NPA24NPA24 Member Posts: 588 ■■□□□□□□□□
    Let me see if I can try to explain this one. I've been visiting this website daily for the past 2 weeks in preparation for the 640-822 exam.

    I see that in your post you understand how they got the host. I'm going to explain both just for anyone else curious in how the answer was achieved.

    Question: "How many subnets and hosts per subnet can you get from the network 10.0.0.0 255.255.240.0?"

    First, you have to know which class this network is on. It starts with 10.x.x.x so we know that this is a class A network. This is very important because I think this is where you are getting confused.

    Second, you have to break down the subnet into binary.
    11111111.11111111.11110000.00000000 = 255.255.240.0

    Third, since this is a class A network, you want to focus getting your answers on just the 2nd, 3rd, and 4th octet.

    Fourth, to find the subnets you will take all the 1's from the 2nd, 3rd, and 4th octet add it up which amounts to 12.
    You then do 2 to the 12th power = 4096.

    Fifth, to find the hosts you will take the 0's from the 2nd, 3rd, and 4th octet add it up which amounts to 12.
    You then do 2 to the 12th power subtract 2 = 4094.


    I hope this makes sense. But most people I've talked to say that subnetting Class A is usually NOT on the test. But I think if you are going to become a network engineer in the future, this information is very important to learn.

    I hope this helps you as well as others.

    PM if you need anymore help.
  • CaptainCharismaCaptainCharisma Member Posts: 36 ■■□□□□□□□□
    NPA24 wrote:
    Let me see if I can try to explain this one. I've been visiting this website daily for the past 2 weeks in preparation for the 640-822 exam.

    I see that in your post you understand how they got the host. I'm going to explain both just for anyone else curious in how the answer was achieved.

    Question: "How many subnets and hosts per subnet can you get from the network 10.0.0.0 255.255.240.0?"

    First, you have to know which class this network is on. It starts with 10.x.x.x so we know that this is a class A network. This is very important because I think this is where you are getting confused.

    Second, you have to break down the subnet into binary.
    11111111.11111111.11110000.00000000 = 255.255.240.0

    Third, since this is a class A network, you want to focus getting your answers on just the 2nd, 3rd, and 4th octet.

    Fourth, to find the subnets you will take all the 1's from the 2nd, 3rd, and 4th octet add it up which amounts to 12.
    You then do 2 to the 12th power = 4096.

    Fifth, to find the hosts you will take the 0's from the 2nd, 3rd, and 4th octet add it up which amounts to 12.
    You then do 2 to the 12th power subtract 2 = 4094.


    I hope this makes sense. But most people I've talked to say that subnetting Class A is usually NOT on the test. But I think if you are going to become a network engineer in the future, this information is very important to learn.

    I hope this helps you as well as others.

    PM if you need anymore help.

    Wow, that clears it up for me. Thanks.
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