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  1. Senior Member thenjduke's Avatar
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    #26
    Wow this concept is so much easier then any video or book I have read. I actually like this.
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    #27
    Hi folks,

    I have two questions :


    1) How can I count the next boundary?


    Quote:
    Originally Posted by kriscamaro68
    Ok so I got this question on subnettingquestions.com and didn't see a way to answer it with your method:
    Question: You are designing a subnet mask for the 10.0.0.0 network. You want 3800 subnets with up to 3800 hosts on each subnet. What subnet mask should you use?

    Answer: 255.255.240.0


    For your first question, how many bits do you need to borrow to accommodate 3800 subnets? The answer would be 12 bits as 2 ^ 12 = 4096. Your network address is a /8 by default as it is a Class A address so 8 + 12 = /20 mask.
    !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!?
    2) may you explain It more by detail?


    My regards
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  4. Senior Member miller811's Avatar
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    #28
    Quote Originally Posted by gozila View Post
    Hi folks,

    I have two questions :


    1) How can I count the next boundary?




    !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!?
    2) may you explain It more by detail?


    My regards
    Read the very post in the string... should be all you need
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  5. Member LAN_Man's Avatar
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    #29
    Thanks everyone for the great post. It helps alot.
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  6. Junior Member
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    #30
    Thanks for the post, it helped out a lot. Though I still find myself struggling on class a & b. Guess more practice will help.
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  7. CCNA in progress Satcom's Avatar
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    #31
    thanks for the write up I hate to even think about sub-netting when I think about study. I would rather learn how routing protocols work and how routers talk to each other than subnet. Back to the books

    bookmarked this thread for future reference going to get heavy on subnetting!
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  8. Left for Android! =)
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    #32
    Quote Originally Posted by Satcom View Post
    thanks for the write up I hate to even think about sub-netting when I think about study. I would rather learn how routing protocols work and how routers talk to each other than subnet. Back to the books

    bookmarked this thread for future reference going to get heavy on subnetting!
    You have to learn to crawl before you can run. I still credit this post with helping me get over the subnet "hump".
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  9. Senior Member
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    #33
    is there an easy to work this one out

    class a network 10.0.0.0 create a subnet mask for the 600 subnets. THen Identify the 100th subnet

    I can work out doing the subnet easy but is there an easier way to finding out the 100th subnet rather than writing each subnet down like this

    10.0.0.0
    10.0.64.0
    10.0.128.0
    10.0.192.0

    It would take forever
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    #34
    I'm kinda confused. How do you know which octet to subtract from?

    Like this:

    "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."

    How do you know which boundary to use, 8, 16, 24, 32 to subtract?
    Last edited by cleanwithit; 10-11-2009 at 10:06 PM.
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  11. Senior Member billscott92787's Avatar
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    #35
    Quote Originally Posted by cleanwithit View Post
    I'm kinda confused. How do you know which octet to subtract from?

    Like this:

    "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."

    How do you know which boundary to use, 8, 16, 24, 32 to subtract?




    Look at it this way. You have a /29 mask. This means that there are 29 "on" bits in the subnet mask.

    11111111.11111111.11111111.11111 | 000

    So, since you have a /29 you see the split in the fourth octect. That's where you know how to determine the subnet blocked and the next available block.

    Subnet block: 256 - 248 = 8, this means that each subnet block size is 8

    192.168.12.0 /29
    192.168.12.8 /29
    192.168.12.16 / 29
    192.168.12.24 /29
    192.168.12.32 /29
    192.168.12.40 /29
    192.168.12.48 /29
    192.168.12.56 /29
    192.168.12.64 /29
    192.168.12.72 /29
    192.168.12.80 /29
    192.168.12.88 /29
    192.168.12.96 /29
    192.168.12.104 /29
    and so on.
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  12. Senior Member miller811's Avatar
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    #36
    Quote Originally Posted by cleanwithit View Post
    I'm kinda confused. How do you know which octet to subtract from?

    Like this:

    "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."

    How do you know which boundary to use, 8, 16, 24, 32 to subtract?
    you are almost there
    since the mask is 29 = 3 the block size is 8 as you have stated.
    the first subnet is 192.168.12.0 hosts 192.168.12.1 - 192.168.12.6 broacast 192.168.12.7
    2nd subnet 192.168.12.8 hosts 192.168.12.9 - 192.168.12.14 broadcast 192.168.12.15
    etc....
    the address you were given is 198.168.12.78 /29

    multiplying by 8 you would realize 80 is subnet, and one less is 72....
    192.168.12.72 subnet
    hosts 192.168.12.73-192.168.12.78
    broadcast 192.168.12.79

    hope that helps
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    #37

    Default Subnetting

    Hi guys. New to subnetting, and having a few problems. I found out how to get the network number through this great method, but I'm still having trouble.

    I need to figure out :

    If the IP (Host) valid?

    What is the Network (Subnet) Number? ( I think this is covered by using the method above)

    What is the Broadcast Number?

    What is the valid IP (Host) Range?

    What is the subnet mask? If given dotted decimal convert to the prefix length and vice versa?

    The first one is :

    192.168.10.23 / 25


    The second one is :

    192.168.1.253
    255.255.255.240


    I'm just looking for a little help.
    I've gotten the basic concept, but that's about it.


    Thanks guys.
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  14. Senior Member miller811's Avatar
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    #38
    Quote Originally Posted by ukman2003 View Post
    Hi guys. New to subnetting, and having a few problems. I found out how to get the network number through this great method, but I'm still having trouble.

    I need to figure out :

    If the IP (Host) valid?

    What is the Network (Subnet) Number? ( I think this is covered by using the method above)

    What is the Broadcast Number?

    What is the valid IP (Host) Range?

    What is the subnet mask? If given dotted decimal convert to the prefix length and vice versa?

    The first one is :

    192.168.10.23 / 25


    The second one is :

    192.168.1.253
    255.255.255.240


    I'm just looking for a little help.
    I've gotten the basic concept, but that's about it.


    Thanks guys.
    practice.......you can figure out all of the answers by working through the methods listed in the thread

    192.168.10.23 / 25 = 255.255.255.128 1 bit is borrowed...
    128
    64
    32
    16
    8
    4
    2
    1

    that determines the subnets possible
    256/128 = 2 subnets .0 and .128
    256/64 = 4 subnets .0 and .64 and .128 and .192
    256/32 = 8 subnets .0 and .32 and .64 and .96 and .128 etc
    256/16 = 16 subnets .0 and .16 and .32 and .48 and .64 etc
    256 /8 = 32 subnets .0 and .8 and .16 and .24 and .32 etc
    256/4 - 64 subnets .0 and .4 and .8 and .12 and .16 etc


    192.168.10.0 is the subnet
    valid host range is 192.168.10.1 - 192.168.10.126
    broadcast address 192.168.10.127
    so your address falls into the range so yes it is valid
    the second subnet starts at 192.168.10.128
    this allowed them to take a network and divide it... subnetting


    192.168.1.253 255.255.255.240 = /28 = 16 bit address blocks
    subnet 192.168.1.240
    valid host range 192.168.1.241 -192.168.1.254
    broadcast 192.168.1.255
    so yes it is valid host because it is in the valid range

    in this case 192.168.1.0 was subnetted as follows
    1 subnet 192.168.1.0 /28
    2 subnet 192.168.1.16 /28
    3 subnet 192.168.1.32 /28
    4 subnet 192.168.1.48 /28
    5 subnet 192.168.1.64 /28
    6 subnet 192.168.1.80 /28
    etc
    your example was the last subnet available with the /28
    find the subnet size by examining the mask
    one ip above the subnet it the first available host address
    one ip address below the next subnet is the broadcast
    one ip address below the broadcast is the last host address
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    #39
    Really good guide.

    Thank you, will defiantly pass this on.
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    #40

    Post What is the broadcast address of the network 10.4.64.0 255.255.240.0?

    well this was a great tut to learn subnetting.

    i have a problem

    Question :-
    What is the broadcast address of the network 10.4.64.0 255.255.240.0?
    ANSWER : 10.4.79.255

    but the problem is how to find out broadcast address when there are 4096 subnets and 4094 host. because in exam you have to do it quickly without wasting time. is there a trick to find out quickly first valid host, last valid host and broadcast address.
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  17. Senior Member miller811's Avatar
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    #41
    Quote Originally Posted by Assassin2005 View Post
    well this was a great tut to learn subnetting.

    i have a problem

    Question :-
    What is the broadcast address of the network 10.4.64.0 255.255.240.0?
    ANSWER : 10.4.79.255

    but the problem is how to find out broadcast address when there are 4096 subnets and 4094 host. because in exam you have to do it quickly without wasting time. is there a trick to find out quickly first valid host, last valid host and broadcast address.
    1. find the subnet size by examining the mask
    2. one ip address above your network address is the first available host address
    3. one ip address below the next subnet is the broadcast (use the subnet size to find the boundary)
    4. one ip address below the broadcast is the last host address
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  18. Small World isn't it Firemarshalbill.com's Avatar
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    #42
    Try my whitepaper on subnetting, I tried to make it as easy and understandable as possible

    http://www.linkedin.com/in/firemarshalbill

    and if you are on linkedin send me a link request
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  19. Senior Member georgemc's Avatar
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    #43
    Quote Originally Posted by thomas130 View Post
    is there an easy to work this one out

    class a network 10.0.0.0 create a subnet mask for the 600 subnets. THen Identify the 100th subnet

    I can work out doing the subnet easy but is there an easier way to finding out the 100th subnet rather than writing each subnet down like this

    10.0.0.0
    10.0.64.0
    10.0.128.0
    10.0.192.0

    It would take forever

    You've almost answered it yourself...

    For every 4 increments in your third octet you 2nd octet will increase by 1.

    so....100 / 4 = 25 increments of your second octet will get you to 100


    10.0.0.0 - 10.24.255.255 is your first 100 subnets (2nd octet 0 thru 24)

    The last subnet in that range (the 100th) should be 10.24.192.0


    It's very late and I'm completely out of it right now...so if I'm completely wrong and off-base with this, forgive me

    I hope this helps...


    George
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  20. Junior Member
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    #44

    Default Still confused and getting a bit ticked......

    Everytime I think I finally think I get subnetting I try a few questions and get them wrong. My blood pressure cant take much more of this. Here are a few of my main concerns:

    1. What if my mask matches the next boundary? For instance, 172.21.197.0/24 well 24-24 = 0 and 2 to the power of zero .... well I think you can see that this is not going to work.

    2. I am still baffled by which octet you are supposed to be working with and when. I followed the tutorial here to a tee when doing questions on subnettingquestions.com and ALWAYS turned out wrong. For instance it would seem I would use the 3rd octect since that is where the 24th bit is but I get the answer wrong all the time with this logic.

    3. I am still 100% confusedon how to answer questions when given the mask in decimal form.

    I very much appreciate all of your great information but I cant help but get frustrated when I constantly get this wrong over and over and over again.
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  21. Small World isn't it Firemarshalbill.com's Avatar
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    #45
    OK as you can see everyone on the posting does it a little different, I think it depends on how your brain is wired.

    I assume when you say decimal format you mean something like 255.255.255.240. Remember that when we are talking mask we are starting at the left with 1's. each group between the periods represents 8 bits hence 255 = 11111111 because all 8 bits are 1's. When you see that value change as in my example here 240 that is the field we are going ot work in. the way to add up the value to determine the 1's is like this

    on a scrap piece of paper at the top right this (these are the 2 raised to the powers)

    128 64 32 16 8 4 2 1

    then you are going to subtract these values from the 240 until you reach 0 and put a 1 in the column if you can subtract.

    so you get

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

    because
    240 - 128 = 112

    112 - 64 = 48

    48 - 31 = 16

    16 - 16 = 0

    so we now see that the first 4 bits are ones and the last 4 are 0's.
    to give us the / (cidr) value we add up the one's. so for 255.255.255.240 we get 8+8+8+4 = 28 or as we normally write it /28

    Good luck I hope this helps
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    #46
    Quote Originally Posted by subnetaggravation View Post
    Everytime I think I finally think I get subnetting I try a few questions and get them wrong. My blood pressure cant take much more of this. Here are a few of my main concerns:

    1. What if my mask matches the next boundary? For instance, 172.21.197.0/24 well 24-24 = 0 and 2 to the power of zero .... well I think you can see that this is not going to work.

    2. I am still baffled by which octet you are supposed to be working with and when. I followed the tutorial here to a tee when doing questions on subnettingquestions.com and ALWAYS turned out wrong. For instance it would seem I would use the 3rd octect since that is where the 24th bit is but I get the answer wrong all the time with this logic.

    3. I am still 100% confusedon how to answer questions when given the mask in decimal form.

    I very much appreciate all of your great information but I cant help but get frustrated when I constantly get this wrong over and over and over again.
    If im right this is one of the most easyiest subnet. 172.21.0.0 is the network address. 255.255.255.0 is the subnet mask.

    Subnet = 256

    Host = 254

    Valid subnet - 1.0 all the way to 255.

    Remember that this is a class B address by looking at the first octect. therefore the first two octect dont change at all. the third is the subnet which in this case is all 1's = 256 the host is all 0's = 256 - 2 = 254
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    #47
    Quote Originally Posted by Picker View Post
    If im right this is one of the most easyiest subnet. 172.21.0.0 is the network address. 255.255.255.0 is the subnet mask.

    Subnet = 256

    Host = 254

    Valid subnet - 1.0 all the way to 255.

    Remember that this is a class B address by looking at the first octect. therefore the first two octect dont change at all. the third is the subnet which in this case is all 1's = 256 the host is all 0's = 256 - 2 = 254
    Ok I think you answered my question about which octet to manipulate, so a class A address we would manipulate the second octect, class B the 3rd and class C the 4th?
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    #48

    Default Here is part of my problem...

    lol well I couldve swore Cisco standard practice did not allow subnet zero but I guess I was wrong, no wonder I was off most of the time.
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  25. Senior Member Geetar28's Avatar
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    #49

    Default A Huge Thank You!!!!

    Dude, I signed up for the forum just to THANK YOU for posting this!!!!

    I was able to understand subnetting from the long route used in the Bryant method...but I knew that it took too long to answer the questions. It took me a bit to streamline things but now I'm able to answer EVERY question at subnettingquestions.com with ease.

    I write out the powers of 2: 2 4 8 16 32 64 128 256 etc...up to 65,536 and I write out the possible masks i.e. 128 192 224 240 248 252 254 255. This allows me to answer most every question within just a few seconds (many I can even do in my head now!!)

    My only regret is not finding this easy method before drudgging through all the other garbage out there.
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  26. Senior Member
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    #50
    I cannot thank you enough for this! I thought i would never understand sub netting and then found this! I must admit it took me a whole day before it started to click. As time went by it started making a little bit more sense and then a little more and then BAM!!!

    I can now go onto subnettingquestions.com - Free Subnetting Questions and Answers Randomly Generated Online and answer all the questions in my head. Phew!!

    Now that part is over i guess its time to get WAN's down pat

    But thank you again for introducing me to this concept. I LOVE IT!! I even just showed my manager a few questions and answered them all in about 8 seconds and he was utterly impressed
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