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  1. Member
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    #1

    Default Subnetting Question what a nightmare

    Hello all,

    I am banging my head against the wall trying to figure out subnetting.

    Now if you convert 255.255.255.0 to binary it is 11111111 11111111 11111111 00000000

    I understand to create a subnet from an ip address say 192.168.1.0 this would allow 254 hosts (minus broad cast address and network id).

    Whats baffling me is how to create subnets from a single ip address and subnet mask. Can anyone please explain this in relation to the comptia network plus exam ?

    Thanks






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    #2
    Did you read this post?

    IPv4 Subnetting TechNotes
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    #3
    After reading the post try some practice questions here. You wont get it at first but being able to see the answer then try another one is very helpful.

    After about an hour you will probably start getting the answer right on your own.

    https://www.subnetting.net/Subnettin...?mode=practice
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    #4
    Not sure what you mean so if someone says 192.168.1.0/29 you have a hard time? explain please
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    #5
    Quote Originally Posted by Jon_Cisco View Post
    After reading the post try some practice questions here. You wont get it at first but being able to see the answer then try another one is very helpful.

    After about an hour you will probably start getting the answer right on your own.

    https://www.subnetting.net/Subnettin...?mode=practice

    Thankyou all for your input first question i tried to the following question

    What subnet mask would you use for the 172.17.0.0 network, such that you can get 40 subnets and 590 hosts per subnet?

    Now I understand please correct me if i am wrong. This is a class b network so if you want 40 subnets thats 20 bits taken from the subnet mask 255.255.0.0

    which is

    255.255.0.0

    Now where i fall apart is where i go from here my brain gets boggled

    128 64 32 16 8 4 2 1
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    #6
    Quote Originally Posted by TheNewITGuy View Post
    Not sure what you mean so if someone says 192.168.1.0/29 you have a hard time? explain please

    I think 192.168.1.0/29 means that if i convert this to binary then i am taking

    128 64 32 16 8 4 2 1
    0 0 0 1 1 1 0 0

    3 bits which means you can have 6 subnets of this. Is this correct ?

    Thanks
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    #7
    Quote Originally Posted by GHOSTRIDER2016 View Post
    Thankyou all for your input first question i tried to the following question

    What subnet mask would you use for the 172.17.0.0 network, such that you can get 40 subnets and 590 hosts per subnet?

    Now I understand please correct me if i am wrong. This is a class b network so if you want 40 subnets thats 20 bits taken from the subnet mask 255.255.0.0

    which is

    255.255.0.0

    Now where i fall apart is where i go from here my brain gets boggled

    128 64 32 16 8 4 2 1

    Yes, it's a Class B IP Address, and as such has a 16 bit Mask, giving one Subnet.

    The Mask 255.255.0.0 demonstrates this; two Octets used with all the bits 'set'
    1 1 1 1 1 1 1 1 | 1 1 1 1 1 1 1 1 | 0 0 0 0 0 0 0 0 | 0 0 0 0 0 0 0 0

    All the bits that are 'set' (or '1') map the Network IPs, and all the bits that are not 'set', map the Host IPs.

    Hosts = (2^16)-2 = 65534 [We have 16 Bits 'unset' in the Mask]

    Networks = 2^{the number of 'extra' bit used to map the Network}. To start with, that's zero. Any number to the power of zero = 1.

    The next bit is easier if we add the 2 host address back on to make 65536.

    Subnet have to go up in powers of two, while at the same time, we half the number of Hosts...

    2^1 = 2 | 65536/2 = 32768
    2^2 = 4 | 32768/2 = 16384
    2^3 = 8 | 16384/2 = 8192
    2^4 = 16 | 8192/2 = 4096
    2^5 = 32 | 4094/2 = 2048
    2^6 = 64 | 2048/2 = 1024

    That's as close to your 40 & 590 combination as you can get...

    64 Subnets and (1024-2) 1022 Hosts, and we have used an 'extra' 6 Mask Bits to achieve it.

    So, adding the 'extra' Mask Bits to the Mask, gives us...
    1 1 1 1 1 1 1 1 | 1 1 1 1 1 1 1 1 | 1 1 1 1 1 1 0 0 | 0 0 0 0 0 0 0 0 = 255.255.252.0


    In CIDR notation, that would be written 172.17.0.0 /22 [the '22' = 16+6; sixteen bits from the standard mask plus six 'extra' bits we used to create the additional 63 Subnets]

    Is that any help?
    Last edited by rob42; 12-25-2016 at 08:51 PM.
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    #8
    Hi Rob,

    Going to digest your post thanks for chiming in. I am going through the ip address stuff again. From my understanding when you see a /24. Since there are 32 bits in a V4 address. Then the number of ip addresses would be 32-24 = 8 ip addresses for a subnet.
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    #9
    Hi NewITGuy

    I have also worked out that /29 means 32-29 = 3 ip addresss on that subnet.

    Guys going to go to bed and get up early morning and study ip addressing again. Its slowly sinking in jezuzzzz man i need to figure this out asap.

    Thanks all
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    #10
    Quote Originally Posted by GHOSTRIDER2016 View Post
    Hi Rob, Going to digest your post thanks for chiming in. I am going through the ip address stuff again. From my understanding when you see a /24. Since there are 32 bits in a V4 address. Then the number of ip addresses would be 32-24 = 8 ip addresses for a subnet.
    Yes. A /24 means '24 Bits used to Map the Network, leaving 8 Bits for the Host IPs, but it's ( 2 ^ 8 ) - 2 = 254 Host IP Addresses. It's hard to start with, but you'll get your head around it. I did, and I'm no Einstein
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    #11
    Thanks Rob, Jon and IT New Guy. I am going to be studying for the next 15 hours as i have an exam tomorrow. I have a hot cup of coffee, modifinal and some bourbon biscuits to keep me going.
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  13. Senior Member dontstop's Avatar
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    #12
    Quote Originally Posted by GHOSTRIDER2016 View Post
    Thankyou all for your input first question i tried to the following question

    What subnet mask would you use for the 172.17.0.0 network, such that you can get 40 subnets and 590 hosts per subnet?

    Now I understand please correct me if i am wrong. This is a class b network so if you want 40 subnets thats 20 bits taken from the subnet mask 255.255.0.0

    which is

    255.255.0.0

    Now where i fall apart is where i go from here my brain gets boggled

    128 64 32 16 8 4 2 1
    See if this helps, it's a quick worked solution to the problem:

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    #13
    Quote Originally Posted by GHOSTRIDER2016 View Post
    Thanks Rob, Jon and IT New Guy. I am going to be studying for the next 15 hours as i have an exam tomorrow. I have a hot cup of coffee, modifinal and some bourbon biscuits to keep me going.
    You're very welcome. All the best with your exam. Let us know how you get on. Good luck with it.

    Cheers.
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    #14
    Not quite but close

    so 32 bits - 29 bits is indeed 3, but we need to do a 2^3 -2 or you can do 256-248 since /29 = 255.255.255.248 and 0 - 255 is 256 possible values


    This tell us our block size

    so if we had to look at it both ways: 256-248 = 8-2 = 6 usable IP addresses in our blocks and 256/8 = 32 -2 = 30 usable subnets




    Quote Originally Posted by GHOSTRIDER2016 View Post
    Hi NewITGuy

    I have also worked out that /29 means 32-29 = 3 ip addresss on that subnet.

    Guys going to go to bed and get up early morning and study ip addressing again. Its slowly sinking in jezuzzzz man i need to figure this out asap.

    Thanks all
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  16. Junior Member n4pst3r's Avatar
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    #15
    >> Just sharing <<

    The best VLSM and CIDR lesson I had came from...
    CCNA Routing and Switching Study Guide: Exams 100-101, 200-101, and 200-120 by Todd Lamle

    Teaches you subnetting with different address class and subnet mask, at the end of it...I could subnet offhead in my brain
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