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  1. Senior Member miller811's Avatar
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    #51
    here is my favorite site to test your skills

    IP Subnet Practice

    click the new problem button and then solve

    with practice, you will be able to subnet in your head
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    #52
    Why isnt this a sticky? Great thread!

    Jon
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    #53
    I believe LordFlasheart`s post deserves a sticky. That was the best post teaching how to do subnets that I have ever seen! Good job!
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    #54
    I have no words to describe how EASY you made for me subnetting! After using your method I am indeed laughing each time I have to subnet when I remind my self how easy you have made it for me! God bless you man
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    #55
    This thread was helpful to me, while I had 'got' subnetting I was bogged down in slow methods. This is damned fast, I think knowing the bog standard theory and then learning a quick technique which you know makes sense due to your grounding is great.

    Rep flashheart.
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    #56
    Quote Originally Posted by daveccna View Post
    This thread was helpful to me, while I had 'got' subnetting I was bogged down in slow methods. This is damned fast, I think knowing the bog standard theory and then learning a quick technique which you know makes sense due to your grounding is great.

    Rep flashheart.
    Seriously it's awesome, I can now do the questions on the subnetting questions internet site in less than 5 seconds in most cases.
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    #57
    I got stuck with these - Can someone give me a clear breakdown of how these answers were gotten pls
    What is the last valid host on the subnetwork 192.168.39.0/28?
    Answer: 192.168.39.14

    What is the last valid host on the subnetwork 172.19.176.0 255.255.252.0?
    Answer: 172.19.179.254

    What is the first valid host on the subnetwork that the node 10.5.178.10 255.255.240.0 belongs to?
    Answer:
    10.5.176.1
    Last edited by Nzastudios; 03-02-2010 at 01:23 PM.
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  9. Senior Member miller811's Avatar
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    #58
    Quote Originally Posted by Nzastudios View Post
    I got stuck with these - Can someone give me a clear breakdown of how these answers were gotten pls
    What is the last valid host on the subnetwork 192.168.39.0/28?
    Answer: 192.168.39.14

    What is the last valid host on the subnetwork 172.19.176.0 255.255.252.0?
    Answer: 172.19.179.254

    What is the first valid host on the subnetwork that the node 10.5.178.10 255.255.240.0 belongs to?
    Answer:
    10.5.176.1
    I]What is the last valid host on the subnetwork 192.168.39.0/28?
    [/I]Answer: 192.168.39.14

    since the mask is /28 the block size if 16 = 2 to the 4th or 2X2X2X2
    192.168.39.0 is the network address
    192.168.39.1 is the first valid host address
    192.168.39.15 would be the broadcast address
    which makes 192.168.39.14 as the last useable address in the subnet
    192.168.39.16 would start the next network segment with a /28 mask

    What is the last valid host on the subnetwork 172.19.176.0 255.255.252.0?
    Answer: 172.19.179.254
    block size is 4 =2X2 since 6 bits are used for subnetting
    172.19.176.0 is the network address
    172.19.179.255 would be the broadcast address
    172.19.179.254 would be the last address
    172.19.180.0 would be the next network address with a /252
    the key here is you are subnetting the 3rd octet and not the 4th

    What is the first valid host on the subnetwork that the node 10.5.178.10 255.255.240.0 belongs to?
    Answer:
    10.5.176.1[/QUOTE]

    240 mask means you are borrowing the first 4 bits of the third octet. Like the first example your block size is 16, Like the second question the subnetting is taking place in the third octet

    10.5.0.0 1st subnet
    10.5.16.0 2nd subnet
    10.5.32.0 3rd subnet
    10.5.48.0 4th subnet
    10.5.64.0 5th subnet
    10.5.80.0 6th subnet
    10.5.96.0 7th subnet
    10.5.112.0 8th subnet
    10.5.128.0 9th subnet
    10.5.144.0 10th subnet
    10.5.160.0 11th subnet
    10.5.176.0 12th subnet

    on the 12 subnet
    10.5.176.1 would be the first valid host address
    10.5.191.255 would be the broadcast address
    10.5.191.254 would be the last host address

    10.5.192.0 13th subnet
    etc

    Hope that helps

    The keys that help me figure this out is figure out the block size
    from the block size figure out your network address
    figure out the next network subnet from the block size
    one address above your network address is the first host
    one address below the next network segment is your broadcast address
    one address below your broadcast address is the last host address
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    #59
    I am actually starting to get this together in my head and its pretty motivating lol

    It all came together for me (for those who are still baffled like I was for a longgggg time - From example below)

    What is the first valid host on the subnetwork that the node 10.5.178.10 255.255.240.0 belongs to?
    Answer: 10.5.176.1[/QUOTE]

    240 mask means you are borrowing the first 4 bits of the third octet. Like the first example your block size is 16, Like the second question the subnetting is taking place in the third octet

    10.5.0.0 1st subnet
    10.5.16.0 2nd subnet
    10.5.32.0 3rd subnet
    10.5.48.0 4th subnet
    10.5.64.0 5th subnet
    10.5.80.0 6th subnet
    10.5.96.0 7th subnet
    10.5.112.0 8th subnet
    10.5.128.0 9th subnet
    10.5.144.0 10th subnet
    10.5.160.0 11th subnet
    10.5.176.0 12th subnet

    on the 12 subnet
    10.5.176.1 would be the first valid host address
    10.5.191.255 would be the broadcast address
    10.5.191.254 would be the last host address

    10.5.192.0 13th subnet
    etc

    What is the last valid host on the subnetwork 172.19.176.0 255.255.252.0?
    Answer: 172.19.179.254
    block size is 4 =2X2 since 6 bits are used for subnetting

    Please confirm my thought process below is in the right track...

    The 6 bits
    128 64 32 16 8 4 2 1 =252 (128+64+32+16+8+4)
    1 1 1 1 1 1 0 0 = 6 Bits >> Now find block 8 - 6 = 2, then 2 x block diff(2)=4

    Now that I know its a block 4, I simply add the 4 to the 3rd octet in 172.19.176.0 (176+4=180)

    Therefore I focus on the range
    172.19.176.0
    172.19.180.0

    Then thats how you came about the answer? If my thinkin is correct I definitely finally understand Subnetting
    172.19.179.255 would be the broadcast address
    172.19.179.254 would be the last address

    figure out the next network subnet from the block size
    one address above your network address is the first host
    one address below the next network segment is your broadcast address
    one address below your broadcast address is the last host address


    -----------------------------------------------------------------------
    Quote Originally Posted by miller811 View Post
    I]What is the last valid host on the subnetwork 192.168.39.0/28?
    [/I]Answer: 192.168.39.14

    since the mask is /28 the block size if 16 = 2 to the 4th or 2X2X2X2
    192.168.39.0 is the network address
    192.168.39.1 is the first valid host address
    192.168.39.15 would be the broadcast address
    which makes 192.168.39.14 as the last useable address in the subnet
    192.168.39.16 would start the next network segment with a /28 mask

    What is the last valid host on the subnetwork 172.19.176.0 255.255.252.0?
    Answer: 172.19.179.254
    block size is 4 =2X2 since 6 bits are used for subnetting
    172.19.176.0 is the network address
    172.19.179.255 would be the broadcast address
    172.19.179.254 would be the last address
    172.19.180.0 would be the next network address with a /252
    the key here is you are subnetting the 3rd octet and not the 4th

    What is the first valid host on the subnetwork that the node 10.5.178.10 255.255.240.0 belongs to?
    Answer: 10.5.176.1
    240 mask means you are borrowing the first 4 bits of the third octet. Like the first example your block size is 16, Like the second question the subnetting is taking place in the third octet

    10.5.0.0 1st subnet
    10.5.16.0 2nd subnet
    10.5.32.0 3rd subnet
    10.5.48.0 4th subnet
    10.5.64.0 5th subnet
    10.5.80.0 6th subnet
    10.5.96.0 7th subnet
    10.5.112.0 8th subnet
    10.5.128.0 9th subnet
    10.5.144.0 10th subnet
    10.5.160.0 11th subnet
    10.5.176.0 12th subnet

    on the 12 subnet
    10.5.176.1 would be the first valid host address
    10.5.191.255 would be the broadcast address
    10.5.191.254 would be the last host address

    10.5.192.0 13th subnet
    etc

    Hope that helps

    Thank you so much Lordflashheart and Miller811 !!!

    If I get confused in the near future I will be back here again lol!

    The keys that help me figure this out is figure out the block size
    from the block size figure out your network address
    figure out the next network subnet from the block size
    one address above your network address is the first host
    one address below the next network segment is your broadcast address
    one address below your broadcast address is the last host address[/QUOTE]
    Last edited by Nzastudios; 03-08-2010 at 02:03 PM.
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  11. Censorship is Un-American JockVSJock's Avatar
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    #60
    This thread is golden!!! Should be a sticky.
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    #61
    Quote Originally Posted by LordFlasheart View Post
    Hi all,

    I've received an email from one of your members asking me to post up my technique for subnetting as links to external blogs are not allowed due to forum rules. I know that he benefited from it and he wishes to help out others so here goes:

    First of all I need you to get rid of all of the negative thoughts surrounding subnetting. Put down all of the books that you have read about the subject and navigate away from other sites claiming to provide an easy way to subnet. This technique requires no charts, just simply the know-how to work with the powers of 2.

    We need to start with the fundamentals of IP addressing. An IP address is made up of 32 bits, split into 4 octets (oct = 8, yes?). Some bits are reserved for identifying the network and the other bits are left to identify the host.

    There are 3 main classes of IP address that we are concerned with.

    Class A Range 0 - 127 in the first octet (0 and 127 are reserved)
    Class B Range 128 - 191 in the first octet
    Class C Range 192 - 223 in the first octet

    Below shows you how, for each class, the address is split in terms of network (N) and host (H) portions.

    NNNNNNNN . HHHHHHHH . HHHHHHHH . HHHHHHHH Class A
    NNNNNNNN . NNNNNNNN . HHHHHHHH . HHHHHHHH Class B
    NNNNNNNN . NNNNNNNN . NNNNNNNN . HHHHHHHH Class C

    At each dot I like to think that there is a boundary, therefore there are boundaries after bits 8, 16, 24, and 32. This is an important concept to remember.

    We will now look at typical questions that you may see on subnetting. More often than not they ask what a host range is for a specific address or which subnet a certain address is located on. I shall run through examples of each, for each class of IP address.

    What subnet does 192.168.12.78/29 belong to?

    You may wonder where to begin. Well to start with let's find the next boundary of this address.

    Our mask is a /29. The next boundary is 32. So 32 - 29 = 3. Now 2^3 = 8 which gives us our block size i.e. 2 to the power of 3 equals 8.

    We have borrowed from the last octet as the 29th bit is in the last octet. We start from zero and count up in our block size. Therefore it follows that the subnets are:-

    192.168.12.0
    192.168.12.8
    192.168.12.16
    192.168.12.24
    192.168.12.32
    192.168.12.40
    192.168.12.48
    192.168.12.56
    192.168.12.64
    192.168.12.72
    192.168.12.80
    .............etc

    Our address is 192.168.12.78 so it must sit on the 192.168.12.72 subnet.

    What subnet does 172.16.116.4/19 sit on?

    Our mask is /19 and our next boundary is 24. Therefore 24 - 19 = 5. The block size is 2^5 = 32.

    We have borrowed into the third octet as bit 19 is in the third octet so we count up our block size in that octet. The subnets are:-

    172.16.0.0
    172.16.32.0
    172.16.64.0
    172.16.96.0
    172.16.128.0
    172.16.160.0
    .............etc

    Our address is 172.16.116.4 so it must sit on the 172.16.96.0 subnet. Easy eh?

    What subnet does 10.34.67.234/12 sit on?

    Our mask is 12. Our next boundary is 16. Therefore 16 - 12 = 4. 2^4 = 16 which gives us our block size.

    We have borrowed from the second octet as bit 12 sits in the second octet so we count up the block size in that octet. The subnets are:-

    10.0.0.0
    10.16.0.0
    10.32.0.0
    10.48.0.0
    .............etc

    Our address is 10.34.67.234 which must sit on the 10.32.0.0 subnet.

    Hopefully the penny is starting to drop and you are slapping the side of your head realising that you were a fool to think that subnetting was hard. We will now change the type of question so that we have to give a particular host range of a subnet.

    What is the valid host range of of the 4th subnet of 192.168.10.0/28?

    Easy as pie! The block size is 16 since 32 - 28 = 4 and 2^4 = 16. We need to count up in the block size in the last octet as bit 28 is in the last octet.

    192.168.10.0
    192.168.10.16
    192.168.10.32
    192.168.10.48
    192.168.10.64
    .................etc

    Therefore the 4th subnet is 192.168.10.48 and the host range must be 192.168.10.49 to 192.168.10.62, remembering that the subnet and broadcast address cannot be used.

    What is the valid host range of the 1st subnet of 172.16.0.0/17?

    /17 tells us that the block size is 2^(24-17) = 2^7 = 128. We are borrowing in the 3rd octet as bit 17 is in the 3rd octet. Our subnets are:-

    172.16.0.0
    172.16.128.0

    The first subnet is 172.16.0.0 and the valid host range is 172.16.0.1 to 172.16.127.254. You must remember not to include the subnet address (172.16.0.0) and the broadcast address (172.16.127.255).

    What is the valid host range of the 7th subnet of address 10.0.0.0/14?

    The block size is 4, from 16 - 14 = 2 then 22 = 4. We are borrowing in the second octet so count in the block size from 0 seven times to get the seventh subnet.

    The seventh subnet is 10.24.0.0. Our valid host range must be 10.24.0.1 to 10.27.255.254 again remembering not to include our subnet (10.24.0.0) and the broadcast address (10.27.255.255)
    .

    What if they give me the subnet mask in dotted decimal?

    If you're lucky and they give you a mask in dotted decimal format then you should have an even easier time. All you need again is your block size.

    Let's say they have given a mask of 255.255.255.248 and you wish to know the block size. Here's the technique:

    1. Starting from the left of the mask find which is the first octet to NOT have 255 in it.
    2. Subtract the number in that octet from 256 to get your block size e.g. above it is 256 - 248 = block size of 8.
    3. Count up from zero in your block size in the octet identified in step 1 as you have learned above (the example above would be in the last octet).

    Another example is a mask of 255.255.192.0 - you would simply count up in 256 - 192 = 64 in the third octet.

    One more example is 255.224.0.0 - block size is 256 - 224 = 32 in the second octet.

    What now?

    Now it's time to go and pick up those books again and go straight to the practice questions, completely by-passing any of their techniques. Use my method and you will be laughing!

    Happy subnetting!
    can you explain me both for example

    What is the valid host range of the 7th subnet of address 10.0.0.0/14?

    i use the method so 16-14 = 2
    now 2^2= 4

    so size of the blocks is 4 for so

    10.0.0.0
    10.4.0.0
    10.8.0.0
    10.12.0.0
    10.16.0.0
    10.20.0.0
    10.24.0.0
    10.28.0.0

    so i thought the valid range was 10.25.0.0 to 10.26.0.0 it works for class C
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    #62
    Wow. I signed up just to say Thank You.

    This process allows me to subnet in my head within 5-10 seconds on most questions. I have some trouble with how many hosts can be created, and how many subnets can be created questions ... is that just pure memorization of tables?

    I'll post some questions if I still can't get a method around them.
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    #63
    Quote Originally Posted by gouki2005 View Post
    can you explain me both for example

    What is the valid host range of the 7th subnet of address 10.0.0.0/14?

    i use the method so 16-14 = 2
    now 2^2= 4

    so size of the blocks is 4 for so

    10.0.0.0
    10.4.0.0
    10.8.0.0
    10.12.0.0
    10.16.0.0
    10.20.0.0
    10.24.0.0
    10.28.0.0

    so i thought the valid range was 10.25.0.0 to 10.26.0.0 it works for class C
    You should be right, unless the question specifically states you can not use IP Subnet-Zero.
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  15. Still a noob earweed's Avatar
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    #64
    People keep asking why this isn't a sticky. It's probably because it get's referenced by so many people here at TE that it gets posted to at least weekly and maintains a high spot in the forum on its own.
    No longer work in IT. Play around with stuff sometimes still and fix stuff for friends and relatives.
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  16. Senior Member alan2308's Avatar
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    #65
    Quote Originally Posted by earweed View Post
    People keep asking why this isn't a sticky. It's probably because it get's referenced by so many people here at TE that it gets posted to at least weekly and maintains a high spot in the forum on its own.
    It is referenced in the CCNA FAQ sticky. Which everyone should read through.

    And besides that, there's way too many epic threads here to sticky them all. The first 5 pages would all be stickies if we did that.
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    #66
    Okay ... I'm having trouble answering questions like this:

    You have a Class B network address and need 500 subnets with about 100 hosts per subnet? What subnet mask should be used?

    Is this going to be pure memorization of a table listing how many subnets and hosts per mask, or is there a mathematical way to figure it out for each problem? I'd prefer the latter.
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  18. Senior Member miller811's Avatar
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    #67
    Quote Originally Posted by Local Native View Post
    Okay ... I'm having trouble answering questions like this:

    You have a Class B network address and need 500 subnets with about 100 hosts per subnet? What subnet mask should be used?

    Is this going to be pure memorization of a table listing how many subnets and hosts per mask, or is there a mathematical way to figure it out for each problem? I'd prefer the latter.
    it is simple math and the more you practice the easier it becomes.

    a class b address is 255.255.0.0 or a /16 with 16 bits for the network and 16 bits for hosts

    solve the first part of the question and it will lead you to the second part of the question.

    since you need 500 subnets, first determine how many bits are required to reach 500 at a minimum

    2/8 = 256 = /24
    2/9 = 512 = /25

    so your class B address would be given a mask of 255.255.255.128 or /25

    this leave 128 bits-2 or 2/7 left for hosts....on each of the 512 subnets
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  19. Senior Member Somnipotent's Avatar
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    #68
    this thread really nailed down my subnetting confidence. thanks!!
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  20. Senior Member Somnipotent's Avatar
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    #69
    Quote Originally Posted by miller811 View Post
    here is my favorite site to test your skills

    IP Subnet Practice

    click the new problem button and then solve

    with practice, you will be able to subnet in your head
    great site miller! subnettingquestions.com was getting kinda boring
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    #70
    Edit: Nevermind
    Last edited by ScytheX10; 06-18-2010 at 11:12 AM.
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  22. Member anobomski's Avatar
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    #71

    Angry Subnetting when given a required number of clients

    Hi,

    could someone explain how to use this method method when you are only given a required number of host and clients but no mask.

    For example:
    Needed usable subnets: 250
    Network address: 109.0.0.0
    figure out the Custom Subnet Mask

    and:
    A service provider has given you the Class C network range 209.50.1.0. break it into as many subnets as possible as long as there are at least 50 clients per network.

    miller811 posted an explanation to a similar question (below)but i dont get it.

    Quote:
    Originally Posted by Local Native
    Okay ... I'm having trouble answering questions like this:

    You have a Class B network address and need 500 subnets with about 100 hosts per subnet? What subnet mask should be used?

    ANS:
    it is simple math and the more you practice the easier it becomes.

    a class b address is 255.255.0.0 or a /16 with 16 bits for the network and 16 bits for hosts

    solve the first part of the question and it will lead you to the second part of the question.

    since you need 500 subnets, first determine how many bits are required to reach 500 at a minimum

    2/8 = 256 = /24
    2/9 = 512 = /25

    so your class B address would be given a mask of 255.255.255.128 or /25

    this leave 128 bits-2 or 2/7 left for hosts....on each of the 512 subnets
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  23. Senior Member bermovick's Avatar
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    #72
    It's a LOT easier when you're working to match the number of hosts. 100 hosts per subnet. The lowest bit that can encompass 100 hosts is 128: 10000000 (starting in the 4th octet and working left). That gives 8+8+8+1 network bits and 7 host bits = a mask of 255.255.255.128 or /25 in CIDR.
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  24. Senior Member miller811's Avatar
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    #73
    Quote Originally Posted by anobomski View Post
    Hi,

    could someone explain how to use this method method when you are only given a required number of host and clients but no mask.

    For example:
    Needed usable subnets: 250
    Network address: 109.0.0.0
    figure out the Custom Subnet Mask

    and:
    A service provider has given you the Class C network range 209.50.1.0. break it into as many subnets as possible as long as there are at least 50 clients per network.

    miller811 posted an explanation to a similar question (below)but i dont get it.

    Quote:
    Originally Posted by Local Native
    Okay ... I'm having trouble answering questions like this:

    You have a Class B network address and need 500 subnets with about 100 hosts per subnet? What subnet mask should be used?

    ANS:
    it is simple math and the more you practice the easier it becomes.

    a class b address is 255.255.0.0 or a /16 with 16 bits for the network and 16 bits for hosts

    solve the first part of the question and it will lead you to the second part of the question.

    since you need 500 subnets, first determine how many bits are required to reach 500 at a minimum

    2/8 = 256 = /24
    2/9 = 512 = /25

    so your class B address would be given a mask of 255.255.255.128 or /25

    this leave 128 bits-2 or 2/7 left for hosts....on each of the 512 subnets
    always start with this chart to determine the class of address, and then determine the standard mask based on the address

    There are 3 main classes of IP address that we are concerned with.

    Class A Range 0 - 127 in the first octet (0 and 127 are reserved)
    Class B Range 128 - 191 in the first octet
    Class C Range 192 - 223 in the first octet

    Below shows you how, for each class, the address is split in terms of network (N) and host (H) portions.

    NNNNNNNN . HHHHHHHH . HHHHHHHH . HHHHHHHH Class A
    NNNNNNNN . NNNNNNNN . HHHHHHHH . HHHHHHHH Class B
    NNNNNNNN . NNNNNNNN . NNNNNNNN . HHHHHHHH Class C


    For example:
    Needed usable subnets: 250
    Network address: 109.0.0.0
    figure out the Custom Subnet Mask

    currently the mask for this subnet would be 255.0.0.0 since it is a class A address

    creating subnets from the current address you would start here.

    255.128.0.0 would = 2 subnets 109.0.0.0, 109.128.0.0
    255.192.0.0 would = 4 subnets 109.0.0.0, 109.64.0.0 x.128.0.0, x.192.0.0
    255.224.0.0 would = 8 subnets 109.0.0.0 109.32.0.0 x.64.0.0 ... x.224.0.0
    255.240.0.0 would = 16 subnets 109.0.0.0 - 109.240.0.0
    255.248.0.0 would = 32 subnets 109.0.0.0 - 109.248.0.0
    255.252.0.0 would =l 64 subnets 109.0.0.0 - 109.252.0.0
    255.254.0.0 would =l 128 subnets 109.0.0.0 - 109.254.0.0
    255.255.0.0 would =256 subnets 109.0.0.0 - 109.255.0.0

    so the custom mask of 255.255.0.0 would meet the requirements of the 250 subnets.
    ----------------------------------------------------------------

    and:
    A service provider has given you the Class C network range 209.50.1.0. break it into as many subnets as possible as long as there are at least 50 clients per network.

    so you know it is a class C address address which means the mask is 255.255.255.0

    this allows for up to 254 hosts on with the stand mask. 256-2 = 254

    so if you borrow the first bit the new mask would be
    255.255.255.128 this would allow 2 subnets from the current address
    209.50.1.0, 209.50.1.128 but this borrowing of 1 bit would equal 126 hosts per subnet, 128-2 = 126

    255.255.255.192 would borrow 2 bits from the standard mask and would give you 62 hosts per subnet 64-2 = 62

    255.255.255.224 would borrow 3 bits from the standard mask and would give you 30 hosts per subnets 32 -2 = 30

    so the mask of 255.255.255.192 would meet the requirement

    209.50.1.0 = first subnets with hosts 209.50.1.1 - 209.50.1.62
    209.50.1.64 = 2nd subnet with hosts 209.50.1.65 - 209.50.1.126
    209.50.1.128 = 3rd subnet with hosts 209.50.1.129 - 209.50.1.190
    209.50.1.192 = 4th subnet with hosts 209.50.1.193 - 209.50.1.254

    Hope that helps
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  25. Junior Member
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    #74

    Default help - question

    Question: What is the last valid host on the subnetwork 10.57.240.0 255.255.240.0?
    Answer: 10.57.255.254


    Hi,

    Some help needed please... I am unable to get my head around the above question. Could someone please explain how the answer is 10.57.255.254 ?

    I'm am calculating the above as follows:

    1. The above IP address is a class A ( i.e. 255.0.0.0)
    2. 240 is equal to four ones ( 128+64+32+16)
    3. So 255(8 ones) +240 (4 ones) = 12 ones
    4. The next boundry is 16, so 16-12 = 4
    5. block size is 4
    6 so, 10.57.0.0
    10.57.4.0
    10.57.8.0

    7. Counting up in the block size of 4 does not give 10.57.255.254

    Thanks,
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  26. Still a noob earweed's Avatar
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    #75
    Quote Originally Posted by aekash View Post
    Question: What is the last valid host on the subnetwork 10.57.240.0 255.255.240.0?
    Answer: 10.57.255.254


    Hi,

    Some help needed please... I am unable to get my head around the above question. Could someone please explain how the answer is 10.57.255.254 ?

    I'm am calculating the above as follows:

    1. The above IP address is a class A ( i.e. 255.0.0.0)
    2. 240 is equal to four ones ( 128+64+32+16)
    3. So 255(8 ones) +240 (4 ones) = 12 ones
    4. The next boundry is 16, so 16-12 = 4
    5. block size is 4
    6 so, 10.57.0.0
    10.57.4.0
    10.57.8.0

    7. Counting up in the block size of 4 does not give 10.57.255.254

    Thanks,
    When you said LAST valid host you automatically answered your question without requiring any work. Since 10.57.255.255 is the broadcast address the 10.57.255.254 is the LAST valid host.
    No longer work in IT. Play around with stuff sometimes still and fix stuff for friends and relatives.
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