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    #1

    Default Can anyone explain Subnet 255.255.254.0?

    PLEASE IGNORE THIS THREAD. I JUST NEED TO PRACTICE MORE

    I just finished watching the CBT nuggets videos of explaining subnetting. Its very informative by the way, it makes me wonder why there are people who are afraid of subnetting. But lets get back to the subject:
    How do I find the ranges of IP addresses for that subnet?
    I know 254 can subtract 128 so I know right away it takes all 8 bits for that address right? So that leaves the smallest increment of 1??????

    11111111.11111111.111111111.00000000


    I know thats not right because I've been doing practice excercises on subnettingquestions.com and I always get those 255.255.254.0 questions wrong. Can anyone please help


    PLEASE IGNORE THIS THREAD
    Last edited by VinUnleaded; 01-05-2011 at 11:10 PM.
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  3. Senior Member tomaifauchai's Avatar
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    #2
    255.255.255.128 = 25 bits mask = 126 hosts (128 -2)
    255.255.255.0 = 24 bits mask = 254 hosts (256 - 2)
    255.255.254.0 = 23 bits mask = 510 hosts (512 -2 )
    255.255.252.0 = 22 bits mask = 1022 hosts (1024 -2 )
    255.255.248.0 = 21 bits mask = 2046 hosts (2048 -2)

    You always multiply by 2 or divide by 2..and remove 2 hosts (broadcast and network)

    What i do to be fast is...
    255.255.254.0 = 1 bit less than 24 bits mask, so (256 x 2) - 2 hosts = 510


    And to know if an ip address is good or not regarding a certain subnet..i do the following:

    Is 192.168.1.64 - 255.255.255.192 a valid ip address?

    Substract the 192 from 255 and you get 63 which is the broadcast address of the ip above. So it isnt a valid IP, it is a network address.

    Is 192.168.1.64 - 255.255.255.128 a valid ip address?

    Substract the 128 from 255 and you get 127 which is the broadcast of the first subnet, so .64 is a valid address from the 192.168.1.0 subnet.
    Last edited by tomaifauchai; 01-05-2011 at 10:44 PM.
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    #3
    Quote Originally Posted by tomaifauchai View Post
    255.255.255.128 = 25 bits mask = 126 hosts (128 -2)
    255.255.255.0 = 24 bits mask = 254 hosts (256 - 2)
    255.255.254.0 = 23 bits mask = 510 hosts (512 -2 )
    255.255.252.0 = 22 bits mask = 1022 hosts (1024 -2 )
    255.255.248.0 = 21 bits mask = 2046 hosts (2048 -2)

    You always multiply by 2 or divide by 2..and remove 2 hosts (broadcast and network)

    What i do to be fast is...
    255.255.254.0 = 1 bit less than 24 bits mask, so (256 x 2) - 2 hosts = 510


    And to know if an ip address is good or not regarding a certain subnet..i do the following:

    Is 192.168.1.64 - 255.255.255.192 a valid ip address?

    Substract the 192 from 255 and you get 63 which is the broadcast address of the ip above. So it isnt a valid IP, it is a network address.

    Is 192.168.1.64 - 255.255.255.128 a valid ip address?

    Substract the 128 from 255 and you get 127 which is the broadcast of the first subnet, so .64 is a valid address from the 192.168.1.0 subnet.
    I have no idea what you are talking about LOL. Its not your fault, I will read the Sybex book once im done with the CBT series. Different people have different ways of subnetting. I was just wondering what went wrong with Jeremy's method for reverse engineering 255.255.254.0
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    #4
    255.255.254.0 = 11111111.11111111.11111110.00000000

    128+64+32+16+8+4+2 = 254

    I know 254 can subtract 128 so I know right away it takes all 8 bits for that address right? <-- is where you're going wrong, unless I don't understand what problem you're having.
    Last edited by Pratt2; 01-05-2011 at 11:29 PM.
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    #5
    Quote Originally Posted by Pratt2 View Post
    255.255.254.0 = 11111111.11111111.11111110.00000000

    128+64+32+16+8+4+2 = 254

    I know 254 can subtract 128 so I know right away it takes all 8 bits for that address right? <-- is where you're going wrong, unless I don't understand what problem you're having.
    That is exactly where I went wrong. I got the few methods mixed up. Thank you!
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    #6
    Quote Originally Posted by VinUnleaded View Post
    PLEASE IGNORE THIS THREAD.
    How do I find the ranges of IP addresses for that subnet?
    Imagine you have a subnet of 192.168.0.0 with a mask of 255.255.255.0

    Now you know that you have 254 hosts available. Why?

    Because the HOST portion of the subnet in bold 192.168.0.0 equal the formula 2^8 - 2 = 254
    So your valid hosts would be 192.168.0.1 to 192.168.0.254

    The network address is: 192.168.0.0
    The broadcast address: 192.168.0.255
    1. 2 being the base of binary (Base 2)
    2. 8 being the number of bits for the hosts
    3. -2 being the broadcast address and the network address removed.
    So, if you have a mask of 255.255.254.0, you add a bit for the hosts bringing it to (2^9) - 2 = 510
    So your valid hosts would be 192.168.0.1 to 192.168.1.254 and then 192.168.1.0 would be a valid host since it is included in the subnet.

    The network address is: 192.168.0.0
    The broadcast address: 192.168.1.255

    Hope it helps
    Tom
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    #7
    Thanks that clear up alot
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    #8
    I just started seriously learning to subnet today. After reading different methods on this forum, Odom's book, and other sources I had no idea what any of it really meant until I sat down with paper and pencil and bumbled through it. Now this thread makes sense, so I suppose that's a good sign.
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    #9

    Default dsas

    Quote Originally Posted by ehnde View Post
    I just started seriously learning to subnet today. After reading different methods on this forum, Odom's book, and other sources I had no idea what any of it really meant until I sat down with paper and pencil and bumbled through it. Now this thread makes sense, so I suppose that's a good sign.
    It is one of the most complicated part of the ccna to get and i thank my teacher because he passed about 20 hours explaining it to us.

    But at the end, you will find yourself the BEST method of calculating it after comparing them all. I have nothing against the Jeremy's method of writing 11111111.11111111.11111111.00000000, but when you'll get some experience with it, you won't use it anymore.

    Learn subnet masks on your finger tips

    • 0 <---- /8 - /16 - /24
    • 128 <- /9 - /17 - /25
    • 192 <- /10 - /18 - /26
    • 224 <- /11 - /19 - /27
    • 240 <-/12 - /20 - /28 <--- MIDDLE
    • 248 <- /13 - /21 - /29
    • 252 <- /14 - /22 - /30
    • 254 <- /15 - /23 - /31
    • 255 <- /16 - /24 - /32

    And here's a very nice way of dumping Binary/Hexa/Decimal values for your exams in 1:30. I just made that video

    Screenjelly - Screen recording by Toma Hack
    Last edited by tomaifauchai; 01-06-2011 at 03:40 AM.
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    #10
    Quote Originally Posted by tomaifauchai View Post
    It is one of the most complicated part of the ccna to get and i thank my teacher because he passed about 20 hours explaining it to us.

    But at the end, you will find yourself the BEST method of calculating it after comparing them all. I have nothing against the Jeremy's method of writing 11111111.11111111.11111111.00000000, but when you'll get some experience with it, you won't use it anymore.

    Learn subnet masks on your finger tips

    • 0 <---- /8 - /16 - /24
    • 128 <- /9 - /17 - /25
    • 192 <- /10 - /18 - /26
    • 224 <- /11 - /19 - /27
    • 240 <-/12 - /20 - /28 <--- MIDDLE
    • 248 <- /13 - /21 - /29
    • 252 <- /14 - /22 - /30
    • 254 <- /15 - /23 - /31
    • 255 <- /16 - /24 - /32

    And here's a very nice way of dumping Binary/Hexa/Decimal values for your exams in 1:30. I just made that video

    Screenjelly - Screen recording by Toma Hack
    How do you learn subnet masks on your finger tips? I understand the table, just not the memorization technique. Could you please elaborate? I'm going to commit this table to memory.
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    #11
    Quote Originally Posted by ehnde View Post
    How do you learn subnet masks on your finger tips? I understand the table, just not the memorization technique. Could you please elaborate? I'm going to commit this table to memory.
    Okay, the first subnets you should memorize are the /24 to /30 portion.
    /31 and /32 being useless for the ccna i would say.

    So first step, put this in your mind.
    • 255.255.255.000 = /24
    • 255.255.255.128 = /25
    • 255.255.255.192 = /26
    • 255.255.255.224 = /27
    • 255.255.255.240 = /28 <- MIDDLE of them
    • 255.255.255.248 = /29
    • 255.255.255.252 = /30
    • 255.255.255.254 = /31
    • 255.255.255.255 = /32
    By having a middle reference, it's easier to figure where you are in the range for faster reverse engineering.

    So basicly, if you know that a /28 is .240, you know that a /29 is .248 etc...
    Plus the .240 (/28 mask is 16 subnets and 16 hosts ( 14 valid hosts ).


    Second good hint

    If you get a subnet like 192.168.0.0 with a mask 255.255.255.192, without hesitation you must calculate
    256 - 192 = 64 hosts per subnets ( subtract 2 for bcast and network)

    255.255.255.128 = 256 - 128 = 128 - 2
    255.255.255.192 = 256 - 192 = 64 - 2
    255.255.255.224 = 256 - 224 = 32 - 2
    255.255.255.240 = 256 - 240 = 16 - 2

    and so on...

    The networks would be
    • 192.168.0.0
    • 192.168.0.64
    • 192.168.0.128
    • 192.168.0.192
    The broadcasts addresses would be
    • 192.168.0.63
    • 192.168.0.127
    • 192.168.0.191
    • 192.168.0.255


    For subnets under /24 it is slightly different because you borrowing bits on the mask.

    Let say you have 192.168.0.0 with a mask of 255.255.192.0

    You also do 256 - 192 = 64 but this isnt your available hosts, this is your CIDR Step for subnets subdivision.
    So every 64 you have a new network or subnet (whatever you call it)

    Your networks (subnets) would be:
    • 192.168.0.0
    • 192.168.64.0
    • 192.168.128.0
    • 192.168.192.0
    Your broadcasts would be:
    • 192.168.63.255
    • 192.168.127.255
    • 192.168.191.255
    • 192.168.255.255
    The total of valid hosts per subnet would be (2^14) - 2
    • 2 being the base 2 ( 0 or 1 )
    • 14 being 11111111.11111111.11000000.00000000 (255.255.192.0)
    • -2 being broadcast and network removed.
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    #12
    Quote Originally Posted by tomaifauchai View Post
    Okay, the first subnets you should memorize are the /24 to /30 portion.
    /31 and /32 being useless for the ccna i would say.

    So first step, put this in your mind.
    • 255.255.255.000 = /24
    • 255.255.255.128 = /25
    • 255.255.255.192 = /26
    • 255.255.255.224 = /27
    • 255.255.255.240 = /28 <- MIDDLE of them
    • 255.255.255.248 = /29
    • 255.255.255.252 = /30
    • 255.255.255.254 = /31
    • 255.255.255.255 = /32
    By having a middle reference, it's easier to figure where you are in the range for faster reverse engineering.

    So basicly, if you know that a /28 is .240, you know that a /29 is .248 etc...
    Plus the .240 (/28 mask is 16 subnets and 16 hosts ( 14 valid hosts ).


    Second good hint

    If you get a subnet like 192.168.0.0 with a mask 255.255.255.192, without hesitation you must calculate
    256 - 192 = 64 hosts per subnets ( subtract 2 for bcast and network)

    255.255.255.128 = 256 - 128 = 128 - 2
    255.255.255.192 = 256 - 192 = 64 - 2
    255.255.255.224 = 256 - 224 = 32 - 2
    255.255.255.240 = 256 - 240 = 16 - 2

    and so on...

    The networks would be
    • 192.168.0.0
    • 192.168.0.64
    • 192.168.0.128
    • 192.168.0.192
    The broadcasts addresses would be
    • 192.168.0.63
    • 192.168.0.127
    • 192.168.0.191
    • 192.168.0.255


    For subnets under /24 it is slightly different because you borrowing bits on the mask.

    Let say you have 192.168.0.0 with a mask of 255.255.192.0

    You also do 256 - 192 = 64 but this isnt your available hosts, this is your CIDR Step for subnets subdivision.
    So every 64 you have a new network or subnet (whatever you call it)

    Your networks (subnets) would be:
    • 192.168.0.0
    • 192.168.64.0
    • 192.168.128.0
    • 192.168.192.0
    Your broadcasts would be:
    • 192.168.63.255
    • 192.168.127.255
    • 192.168.191.255
    • 192.168.255.255
    The total of valid hosts per subnet would be (2^14) - 2
    • 2 being the base 2 ( 0 or 1 )
    • 14 being 11111111.11111111.11000000.00000000 (255.255.192.0)
    • -2 being broadcast and network removed.

    Wow that is awesome! I was able to subnet without writing much on paper but it took a while though. Gonna defintely improve by using these hints for sure
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    #13
    Yes, these methods only applies for speed purposes. You must understand the process of binary etc.. it's important! Because in IPV6, there's no more decimals so you must understand hexa and binary.

    But the Subnet Step trick is particularly interesting to know when you'll face a silly Cisco question like...

    What is the subnet of the ip address 192.168.39.124 255.255.240.0 ?

    256 - 240 = 16

    192.168.0.0
    192.168.16.0
    192.168.32.0
    192.168.48.0
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    #14
    Quote Originally Posted by ehnde View Post
    I just started seriously learning to subnet today. After reading different methods on this forum, Odom's book, and other sources I had no idea what any of it really meant until I sat down with paper and pencil and bumbled through it. Now this thread makes sense, so I suppose that's a good sign.
    All the quick methods they teach you are meaningless if you don't understand the underlying binary and how it functions. Eventually you should be at the point where you don't need any spiffy "methods," you will just understand it in your head and be able to do it that way.

    At least thats how i do it. I was honestly surprised when a bunch of other people in my department started an email chain about their favorite subnet calculators...I replied, "My head."
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