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  1. Member amart83's Avatar
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    #1

    Default Max subnets per subnet mask

    What is the best way to determine the maximum amount of subnets a default subnet mask can produce?
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    #2
    What do you mean by "default subnet mask?"

    The number is going to depend on your host requirements. You could use all the available bits to obtain the maximum, but then you wouldn't be able to have any hosts
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  4. New Member royal's Avatar
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    #3
    1111111.0000000.00000000.00000000 = 255.0.0.0 (Class A Default Subnet Mask) means that you can borrow 22 bits for subnets since you have to leave 2 over at minimum for clients. This means 2^22 for subnets which means 4,194,304 subnets

    1111111.1111111.00000000.00000000 = 255.255.0.0 (Class B Default Subnet Mask) means that you can borrow 14 bits for subnets since you have to leave 2 over at minimum for clients. This means 2^14 for subnets which means 16,384 subnets

    1111111.1111111.11111111.00000000 = 255.255.255.0 (Class C Default Subnet Mask) means that you can borrow 6 bits for subnets since you have to leave 2 over at minimum for clients. This means 2^6 for subnets which means 16 subnets

    Because we're always leaving 2 bits over for clients, for each scenario you'll have (2^2)-2 = 2 clients. Not a whole lot! But you definitely won't have to worry about broadcast traffic hitting every client!

    If you want more clients, take 1 bit away from the subnet and add it to the clients. For example, for our Class C scenario, we can instead use 3 bits for the subnet which leaves 5 bits for the clients. Because of this, we'll have 2^3 for subnets which means 8 possible subnets and (2^5)-2 for clients which means 30 clients on each subnet.
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  5. Member amart83's Avatar
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    #4
    Sorry, I should have stated my question correctly.

    What is the formula to determine how many networks a subnet mask can produce?

    Do I use 2^X (where X is the left most bits?)

    If that's the case, then:

    255.0.0.0: 2^8 = 256 networks?
    255.255.0.0: 2^16 = 65,536 networks?
    255.255.255.0: 2^24 = 16,777,216 networks?
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  6. New Member royal's Avatar
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    #5
    I already answered your question by showing you the calculations in my post...
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  7. Member amart83's Avatar
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    #6
    Quote Originally Posted by royal
    I already answered your question by showing you the calculations in my post...
    I appreciate your help royal but my first question was worded incorrectly. Please refer to my second question. I've re-read your post but it doesn't answer the question I originally intended to ask.
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  8. New Member royal's Avatar
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    #7
    Quote Originally Posted by amart83
    Quote Originally Posted by royal
    I already answered your question by showing you the calculations in my post...
    I appreciate your help royal but my first question was worded incorrectly. Please refer to my second question.
    Yes, 2^x where x is the left most bits. And again, I show all this in my calculations. You look at the amount of bits I have in the left most and you can see I use the 2^X calculations. What do you find confusing about in my original post where I show the calculations and how things are done?
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    #8
    You need to practice using the above formula then it will answer all your questions, you might also need to check this site I used to study subnetting it's a good primer if your confused.

    http://www.semsim.com/ccna/tutorial/...ubnetting.html
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  10. Member amart83's Avatar
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    #9
    Quote Originally Posted by royal
    Quote Originally Posted by amart83
    Quote Originally Posted by royal
    I already answered your question by showing you the calculations in my post...
    I appreciate your help royal but my first question was worded incorrectly. Please refer to my second question.
    Yes, 2^x where x is the left most bits. And again, I show all this in my calculations. You look at the amount of bits I have in the left most and you can see I use the 2^X calculations. What do you find confusing about in my original post where I show the calculations and how things are done?
    I'm looking at your example and your not using 2^X where X is all the left most bits.

    Your example is 2^X where X is the bits you borrowed (ie, in your example, X disregards the default network bits for the given class)
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  11. Senior Member meadIT's Avatar
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    #10
    Quote Originally Posted by amart83

    Your example is 2^X where X is the bits you borrowed (ie, in your example, X disregards the default network bits for the given class).

    That's exactly what a subnet is. It's when you borrow bits to subnet. If you're asking how many networks (not sub-networks) are in each class, then your 2^8, 2^16, and 2^24 calculations are correct. I think your questions may be unclear.
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  12. New Member royal's Avatar
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    #11
    Quote Originally Posted by amart83
    Your example is 2^X where X is the bits you borrowed (ie, in your example, X disregards the default network bits for the given class)
    That's in my last example. In the first 3 examples I do it all based on the default subnet mask. Hence I'm using all left-most bits instead of borrowed bits.
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  13. Member amart83's Avatar
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    #12
    A little late to reply but...

    I was just looking at things the wrong way.

    Thanks to everyone who helped and have a happy turkey day!
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