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  1. Senior Member
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    #101
    same thing for this to

    You are designing a subnet mask for the 172.27.0.0 network. You want 30 subnets with up to 2000 hosts on each subnet. What subnet mask should you use?

    subnets you take the left bits and hosts are the right most bits

    11111111 1111111 0000000 0000000 etc

    you know if you want 30 subnets your closest option is the 32 bit as the next 1 up is 64 and that is to many.

    just write out the third octet of zeros and treat each zero as 2 and go
    0 x 0 = 4 x 0 = 8 x 0 = 16 x = 32

    how many bits is that ? 6 bits from the left which is

    255.255.248.0

    as you borrowed 6 to the power of 2 bits in the THIRD octet

    you will pick it up so quick in the vids trust me
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    #102
    Mattau - thanks for your replies and link - thanks for taking the time

    Jason
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  4. Audentis Fortuna Iuvat veritas_libertas's Avatar
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    #103
    I'm thankful for this thread. I've read many ways of doing subnetting and this the only one that makes it really simple for me to understand.
    Currently working on: Resting
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    #104
    Why isnt this thread a sticky?
    They need to make this a sticky.
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  6. Junior Member Registered Member
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    #105
    hi ...
    i am a CCNA beginner..
    can anybody explain me how to solve the below given question.
    Q) Find out the valid ip range.
    172.16.0.0/16 for 300 hosts.
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  7. Aperture Science Futura's Avatar
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    #106
    Quote Originally Posted by prsreek View Post
    hi ...
    i am a CCNA beginner..
    can anybody explain me how to solve the below given question.
    Q) Find out the valid ip range.
    172.16.0.0/16 for 300 hosts.

    write out 300 in binary, thats how many host bits u need, the rest can be for subnets.
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  8. Junior Member Registered Member
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    #107
    Quote Originally Posted by Futura View Post
    write out 300 in binary, thats how many host bits u need, the rest can be for subnets.
    i dont get it..
    plzz explain ..........
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  9. Member jdfriesen's Avatar
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    #108
    Quote Originally Posted by prsreek View Post
    hi ...
    i am a CCNA beginner..
    can anybody explain me how to solve the below given question.
    Q) Find out the valid ip range.
    172.16.0.0/16 for 300 hosts.
    First, you've got to know your powers of 2. 2^8 is 256, which won't be enough, so you've got to go to 2^9 = 512, which gives you room for 510 hosts.

    That means your network (assuming the zero subnet can be used, and that you're wanting the first valid address range) will be 172.16.0.0/23 (32-9 = 23).

    Next, to figure out your ip range, you've got to calculate the block size. There are a few ways to do that, but the one that I tend to use is subtract the last non-zero subnet mask value from 256. In this case, the subnet mask is 255.255.254.0 (7 leading 1's in the octet = 254). So the block size is 2 (256-254). That means that your first two subnets will be:
    • 172.16.0.0/23
    • 172.16.2.0/23
    Since the first and last address in the range are not valid to be assigned, your valid address range would be 172.16.0.1 - 172.16.1.254.
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  10. Member
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    #109
    Passed my ICND1 today and just wanted to say a massive thanks to Lord Flasheart (and all contributors to this thread) as this was the single biggest component of my finally having that 'Eureka' moment with subnetting!

    From really not understanding the process a short while ago, I scored 100% on the "implement and IP addressing scheme and IP services to meet network requirements for a small branch office" section of the exam

    Thanks loads everyone
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  11. Aspiring ArchitecT -DeXteR-'s Avatar
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    #110
    Quote Originally Posted by Jas21 View Post
    Passed my ICND1 today and just wanted to say a massive thanks to Lord Flasheart (and all contributors to this thread) as this was the single biggest component of my finally having that 'Eureka' moment with subnetting!

    From really not understanding the process a short while ago, I scored 100% on the "implement and IP addressing scheme and IP services to meet network requirements for a small branch office" section of the exam

    Thanks loads everyone
    Congrats on the Pass
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  12. Member
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    #111
    Thanks!!
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  13. Senior Member
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    #112
    I cant figure out WTH Im doing wrong... Sometimes it works and sometimes it doesnt...

    Heres one that it doesnt:

    Question: How many subnets and hosts per subnet can you get from the network 172.26.0.0/27?

    Answer: 2048 subnets and 30 hosts

    Really? I got nowhere near this.

    Can someone please explain?

    @Jas21, Congrats!
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  14. Member jdfriesen's Avatar
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    #113
    Quote Originally Posted by nimrod.sixty9 View Post
    I cant figure out WTH Im doing wrong... Sometimes it works and sometimes it doesnt...

    Heres one that it doesnt:

    Question: How many subnets and hosts per subnet can you get from the network 172.26.0.0/27?

    Answer: 2048 subnets and 30 hosts

    Really? I got nowhere near this.

    Can someone please explain?

    @Jas21, Congrats!
    The way I calculate the number of available subnets, is to subtract the CiDR value of the network class, from the CiDR value of the subnets, and then take 2 to the power of whatever the result is. In this case, it's a class B network, so the CiDR value is 16. 27 - 16 = 11, so to calculate the number of subnets, you can do 2^11 = 2048.

    To calculate the number of hosts per network, calculate the block size, and subtract 2. In this case, the block size is 32 (/27 = 255.255.255.224, 256 - 224 = 32), so 32 - 2 = 30 hosts per subnet.
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  15. Aspiring ArchitecT -DeXteR-'s Avatar
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    #114
    Quote Originally Posted by nimrod.sixty9 View Post
    I cant figure out WTH Im doing wrong... Sometimes it works and sometimes it doesnt...

    Heres one that it doesnt:

    Question: How many subnets and hosts per subnet can you get from the network 172.26.0.0/27?

    Answer: 2048 subnets and 30 hosts

    Really? I got nowhere near this.

    Can someone please explain?

    @Jas21, Congrats!
    since it is a class B address it has a default subnet mask of 255.255.0.0
    In the given question /27 means 255.255.255.224 ,you have exactly 5 host bits and 11 subnet bits here
    so 2^5 -2 = 30 hosts
    similarly 2^11=2048 subnets
    Hope that helped
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  16. Senior Member
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    #115
    Im confused, again. This says 8 16 24 and 32 are boundries. "The next boundry is..". So my math was done the same way. 27, the next boundry is 32. So I suppose thats where I messed up. So now I really dont know which one Im supposed to subtract.

    How did you figure "5 host bits and 11 subnet bits"? Would either of you like to break down the problem for me?

    Thanks for the reponses!
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  17. Member jdfriesen's Avatar
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    #116
    Quote Originally Posted by nimrod.sixty9 View Post
    Im confused, again. This says 8 16 24 and 32 are boundries. "The next boundry is..". So my math was done the same way. 27, the next boundry is 32. So I suppose thats where I messed up. So now I really dont know which one Im supposed to subtract.

    How did you figure "5 host bits and 11 subnet bits"? Would either of you like to break down the problem for me?

    Thanks for the reponses!
    There are 32 bits total, and 27 are excluded from the host portion, leaving 32-27=5 bits for the host section. Given that it is a class B network, which is /16, there are 27-16=11 bits for the subnet portion.
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  18. Senior Member
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    #117
    Quote Originally Posted by nimrod.sixty9 View Post
    Question: How many subnets and hosts per subnet can you get from the network 172.26.0.0/27?

    Answer: 2048 subnets and 30 hosts

    Really? I got nowhere near this.

    Can someone please explain?
    It might help if you diagram it out. (N=Network, S=Subnet, H=Host)

    172.26.0.0/27

    The 27 indicates the network and subnet. Everything left over are hosts.

    It's a class B address so you know that the first 16 bits are for the network and cannot be touched so it will look like the following:

    NNNNNNNN.NNNNNNNN.--------.--------

    Now consider the /27. 16 out of the 27 are network which leaves 11 as the subnet number.

    NNNNNNNN.NNNNNNNN.SSSSSSSS.SSS-----

    Fill in the last remaining because they are the hosts

    NNNNNNNN.NNNNNNNN.SSSSSSSS.SSSHHHHH

    Now do the math
    S^11 ( 2^11 ) = 2048 Subnets
    H^5-2 ( 2^ 5 - 2 ) = 30 Hosts

    To figure out the subnet if you want to convert from / notation to decimal remember to add from left to right (or memorize the following)

    128, 192, 224, 240, 248, 252, 254, 255

    In this case it's a /27 so there are 3 S bits assigned so it's a 224 subnet or
    255.255.255.224

    I was having a heck of a time until I read through this thread, did both binary and decimal methods and learned some shortcuts. Writing it out in binary really helped me because then I could start visualizing it.

    FYI: This is based off what I read in the Odom ICND1 book. I don't want to be accused of plagiarism.
    Last edited by j-man; 03-05-2011 at 12:23 AM. Reason: The red. I had host instead of network. This might have confused people. My bad
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  19. Senior Member geek4god's Avatar
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    #118
    Quote Originally Posted by j-man View Post
    It might help if you diagram it out. (N=Network, S=Subnet, H=Host)

    Now do the math
    S^11 ( 2^11 ) = 2048 Subnets
    H^5-2 ( 2^ 5 - 2 ) = 30 Hosts
    Any tips on doing the 2 to the power of X math? Or just memorize them? I got 2^9 down do I just need to memorize the rest and if so what is the max power I need?
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  20. Burn Baby Burn! Cisco Inferno's Avatar
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    #119
    Quote Originally Posted by geek4god View Post
    Any tips on doing the 2 to the power of X math? Or just memorize them? I got 2^9 down do I just need to memorize the rest and if so what is the max power I need?
    just keep doubling.
    2,4,8,16,32,64,128.... youll memorize it soon.
    2017 Goals
    [x] MCSA: Server 2012 [X]70-410 [X]70-411 [x]74-409

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  21. Senior Member
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    #120
    Quote Originally Posted by geek4god View Post
    Any tips on doing the 2 to the power of X math? Or just memorize them? I got 2^9 down do I just need to memorize the rest and if so what is the max power I need?
    Nope. Exactly what Cisco Inferno said.

    IMHO you should at least have 2^13 or 2^14 ready at will.

    This is one of the things I always jot down before any practice exams. I'll start at ^13 and go to at least ^22
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  22. Aspiring ArchitecT -DeXteR-'s Avatar
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    #121
    Quote Originally Posted by nimrod.sixty9 View Post
    Im confused, again. This says 8 16 24 and 32 are boundries. "The next boundry is..". So my math was done the same way. 27, the next boundry is 32. So I suppose thats where I messed up. So now I really dont know which one Im supposed to subtract.

    How did you figure "5 host bits and 11 subnet bits"? Would either of you like to break down the problem for me?

    Thanks for the reponses!
    Think the above posts will clear it for you hopefully ! Btw which book you referring to .Imo todd lammle is good for subnetting for starters
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  23. Senior Member
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    #122
    That cleared it up very well, thank you all. I didnt realize subtract from class for subnet bits. I can do the math and the numbers, but this one confused me:

    Question: What valid host range is the IP address 172.17.247.48/23 a part of?

    Answer: 172.17.246.1 through to 172.17.247.254

    I got 2 Subnets and 510 Hosts...
    With that how do I figure out the above .246 and .247?

    Also, how heavy in subnetting in the Net+ exam?
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  24. Member
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    #123
    There isn't any subnetting in the Network + exam (certainly wasn't when I took it a few years back!)
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  25. Member jdfriesen's Avatar
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    #124
    Quote Originally Posted by nimrod.sixty9 View Post
    I can do the math and the numbers, but this one confused me:

    Question: What valid host range is the IP address 172.17.247.48/23 a part of?

    Answer: 172.17.246.1 through to 172.17.247.254

    I got 2 Subnets and 510 Hosts...
    With that how do I figure out the above .246 and .247?
    In this question you've got to figure out the block size. There are a couple ways to do that. The way I do it is 256 - the mask of the last non-zero octet. In that case, /23 = 255.255.254.0, so they last non zero octet is 254, so 256-254=2 as the block size.

    With the block size known, you've got to then find the start of the current subnet. With a block size of 2, it's pretty easy, it's going to be 246, so the first valid address is 172.17.246.1 (.0 isn't allowed as that's the network address). The next valid subnet given a block size of 2 would be 172.17.248.0/23, so that means the last allowable address is going to be 172.17.248.254 (.255 is the broadcast address).

    To subnet quickly, you've got to be able to quickly count in multiples of 2, 4, 8, 16, 32, 64 and 128, which enables you to find the start of each subnet quickly.
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  26. Senior Member
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    #125
    Quote Originally Posted by jdfriesen View Post
    In this question you've got to figure out the block size. There are a couple ways to do that. The way I do it is 256 - the mask of the last non-zero octet. In that case, /23 = 255.255.254.0, so they last non zero octet is 254, so 256-254=2 as the block size.
    Forget the subtraction stuff.

    128 192 224 240 248 252 254 255 - last non-zero of SM

    128 64 32 16 8 4 2 1 - block size

    That's the block size based on the final non zero octet of the subnet mask.

    While it's important to be able to come up with the numbers quickly, it's also extremely important to know why and how those numbers are found.

    Learn both the binary and decimal ways to manually calculate the subnet, first host, last host and broadcast and it will start to make sense.

    *edit* jdfriesen - the last bit wasn't directed at you. It's a general statement to anyone that wants to learn to subnet.
    Last edited by j-man; 03-05-2011 at 12:47 AM.
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