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Subnet Question

aaron0011aaron0011 Member Posts: 330
in CCNA & CCENT
Can someone answer and explain this question for me? I am studying for the CCNA and subnetting just isn't sticking for me. I'd go take the exam tomorrow if I knew there would be no subnetting questions on the test.

Question: How many subnets and hosts per subnet can you get from the network 172.17.0.0/26?

Subnettingquestions.com says 1024 networks and 62 hosts. I get the hosts amount with 6 bits used for hosts, 2^6-2=62. How can you have 1024 networks from a 255.255.255.192 mask?
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    oli356oli356 Member Posts: 364
    /26 = 11111111.11111111.11111111.11000000

    How many 0's are there? 6. The formula for hosts is 2^x-2=
    How many 1's are there? 10. 2 from the last octet and 8 from the 3rd. The formula for subnets is 2^x=
    Lab:
    Combination of GNS3 and Cisco equipment if required.
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    aaron0011aaron0011 Member Posts: 330
    Why are you counting the eight 1s from the third octet? Because it's a class B address?
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    MickQMickQ Member Posts: 628 ■■■■□□□□□□
    Exactly
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    aaron0011aaron0011 Member Posts: 330
    Got it, thanks.
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    aaron0011aaron0011 Member Posts: 330
    Question: What is the first valid host on the subnetwork that the node 10.251.252.186 255.255.240.0 belongs to?

    Ok, so I know 10.0.0.0/20 allows for 4096 networks with 4094 hosts per network. What is the shortcut for identifying ranges in such a large subnet?
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    Ltat42aLtat42a Member Posts: 587 ■■■□□□□□□□
    Use this chart to help with subnetting & block sizes.

    hth
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    MosGuyMosGuy Member Posts: 195
    aaron0011 wrote: »
    Question: What is the first valid host on the subnetwork that the node 10.251.252.186 255.255.240.0 belongs to?

    Ok, so I know 10.0.0.0/20 allows for 4096 networks with 4094 hosts per network. What is the shortcut for identifying ranges in such a large subnet?

    Remember every octet eventually loops back around. Note the "interesting" (non 255) octet. For example: start from 10.251.0.0, find the increment for the 3rd octet & calculate from there.
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    atorvenatorven Member Posts: 319
    Memorize that chart that Ltat42a posted, once you have that in your head everything will be easier.

    As MosGuy said, start from 10.251.0.0 and find the increment, which in this case is 256-240=16, if you keep incrementing you will get to the 10.251.240.0, the next increment will make it 10.251.255.0 so now you know that your 10.251.252.186 falls within the 10.251.240.0 network, the valid hosts start from 10.251.240.1 as 10.251.240.0 is the subnet ID.
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    MickQMickQ Member Posts: 628 ■■■■□□□□□□
    You can memorise a chart, or you can learn how the machine does it.
    Personally, I can jot up the powers of 2 in my head and it becomes second nature. After a while, you'll find that you can just glance at a subnet expressed in 255.255.240.0 or /20 notation and be able to convert between the two, know how many subnets you have, and also if it's valid!
    192.168.10.0 /20 is not valid; 172.16.10.0 /20 is valid ;)

    1. As atorven said, spot the octet where the action is happening.
    2. See if it's vaild (like my class C/B example).
    3. Find your increment (/20): 24 - 20 = 4...... 2^4(remember it's powers!) = 16.
    256 (2^8 ) - 16 (2^4) = 240. Now we know that /20 = 255.255.240.0
    We also have the increment for the different subnets - 16.
    What's our address? Let's use 172.20.0.0 and find the first four valid ranges.

    Subnet............valid host ranges............................broadcast address
    172.20.0.0......172.20.0.1 to 172.20.15.254...........172.20.15.255
    172.20.16.0....172.20.16.1 to 172.20.31.254..........172.20.31.255
    172.20.32.0....172.20.32.1 to 172.20.47.254..........172.20.47.255
    172.20.48.0....172.20.48.1 to 172.20.63.254..........172.20.63.255

    Notice where the 16 comes in and where it's added.

    And a final note of warning - make sure you are careful about which octet things are happening in!
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