+ Reply to Thread
Page 4 of 20 First 1234 567814 ... Last
Results 76 to 100 of 480
  1. Junior Member
    Join Date
    May 2009
    Posts
    3
    #76
    Quote Originally Posted by earweed View Post
    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.

    Thanks for getting back, much appriciated.

    I understand the logic but how would counting up in the block size of 4 achive 255 ? Are you able to explain further ?

    Are the steps I have used correct ?
    Reply With Quote Quote  

  2. SS -->
  3. Still a noob earweed's Avatar
    Join Date
    Mar 2010
    Location
    Mobile, Alabama
    Posts
    5,176

    Certifications
    BSIT, Proj+, A+, Net+, Sec+: MCTS: X5; MCITP:EA
    #77
    The block size you're using is for subnets and you could find the last valid subnet that way. The last valid host on your last valid subnet would be 10.57.255.254.
    Reply With Quote Quote  

  4. Junior Member
    Join Date
    May 2009
    Posts
    3
    #78
    Quote Originally Posted by earweed View Post
    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.
    OK I think I have got it...

    To calculate the block size would it be

    (A)
    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

    or (B)
    1. The above IP address is a call A (i.e.255.0.0.0)
    2 but the subnet mask which has been given is a 255.255.0.0
    3. So 255 (8 ones) +255(8 ones) +240 (4 onnes) = 20 ones
    4. The next boundry is 24, so 24-20 = 4
    5. Block size is 4
    6. 2 to the power of 4 = 16

    Based on the block size of 16

    172.57.0.0
    172.57.16.
    172.57.32
    172.57.48
    172.57.64
    miss a few....
    172.57.224
    172.57.240
    172.57.256

    Host range betweeen 172.57.240 and 172.57.256 is:
    241 -255 ?

    Though I thought you cannot use the last broadcast address which is 255 or does this only apply to the last octect ?

    Am I meant to subreact 4 from 16 or 24 in terms of the next boundry ? If the subnet mask is provided do I follow the subnet mask or the default subnet mask for the IP Address?

    Thanks
    Reply With Quote Quote  

  5. Junior Member Registered Member
    Join Date
    Sep 2010
    Posts
    2
    #79
    Hi all, I'm currently studying for CCNA. This post is in regards to the "broadcast subnet."

    For example, take the following:

    130.4.0.0 / 24 (255.255.255.0)

    130.4.0.0 --> Classful network/subnet zero
    130.4.1.0 --> First Nonzero subnet
    .
    .
    .
    130.4.255.0 --> which is the "broadcast subnet." Thus as are all addresses from 130.4.255.0 - 130.4.255.255 all "broadcast addresses"? or just 130.4.255.255 and 130.4.255.1 to 130.4.255.254 can be used as valid addresses? (This example follows the guidelines that the subnet number address and broadcast address are reserved.)

    Please let me know if any of the above info or my questions need clarification. Thank you.
    Last edited by lstudent; 09-13-2010 at 07:10 PM.
    Reply With Quote Quote  

  6. Senior Member chmorin's Avatar
    Join Date
    Feb 2010
    Location
    Texas
    Posts
    1,443

    Certifications
    CCNP:Voice, CCNA:V(IIUC), CCNA, CCENT, Security +, Network +, A+,CIW
    #80
    Quote Originally Posted by lstudent View Post
    Hi all, I'm currently studying for CCNA and working on finding all the give subnets given an address/mask.

    For example, take the following:

    130.4.0.0 / 24 (255.255.255.0)

    130.4.0.0 --> Classful network/subnet zero
    130.4.1.0 --> First Nonzero subnet
    .
    .
    .
    130.4.255.0 --> which is the "broadcast subnet." Thus as are all addresses from 130.4.255.0 - 130.4.255.255 all "broadcast addresses"? or just 130.4.255.255 and 130.4.255.1 to 130.4.255.254 can be used as valid addresses? (This example follows the guidelines that the subnet number address and broadcast address are reserved.)

    Please let me know if any of the above info or my questions need clarification. Thank you.
    Either I missed something in my CCNA classes or you are barking up the wrong tree.

    There are really three types of addresses. Network, host, and broadcast. The first address in a subnet is the network address, and the last is the broadcast address. All inbetween is the host addreses.

    Using your example:
    130.4.0.0/24 is the network address, with 130.4.0.255 as the broadcast address.
    130.4.1.0/24 is the network address, with 130.4.1.255 as the broadcast address
    .
    .
    .
    130.4.255.0 is the network address, with 130.4.255.255 as the broadcast address.

    The broadcast address storms to all host addresses in the subnet, where as the network address is a unique identifier for the subnet.

    I have never heard of a broadcast subnet.

    To answer your question, 130.4.255/24 is a usable subnet for hosts, excluding of course the network and broadcast address.
    Last edited by chmorin; 09-13-2010 at 07:13 PM.
    Reply With Quote Quote  

  7. Junior Member Registered Member
    Join Date
    Sep 2010
    Posts
    2
    #81
    That answers my question, its probably just a general term for the last subnet in a list of subnets for a given address/mask. p.389 of CCENT/CCNA ICND1 uses the term "Broadcast Subnet" in chart 12-37 as the last subnet when listing all the subnets for a given address/mask.

    Thanks!
    Reply With Quote Quote  

  8. Senior Member chmorin's Avatar
    Join Date
    Feb 2010
    Location
    Texas
    Posts
    1,443

    Certifications
    CCNP:Voice, CCNA:V(IIUC), CCNA, CCENT, Security +, Network +, A+,CIW
    #82
    Quote Originally Posted by lstudent View Post
    That answers my question, its probably just a general term for the last subnet in a list of subnets for a given address/mask. p.389 of CCENT/CCNA ICND1 uses the term "Broadcast Subnet" in chart 12-37 as the last subnet when listing all the subnets for a given address/mask.

    Thanks!
    That is probably an accurate observation.
    Reply With Quote Quote  

  9. Junior Member Registered Member
    Join Date
    Sep 2010
    Posts
    1
    #83
    Quote Originally Posted by Morty3 View Post
    Very much alike Jeremy from CBT nuggets way. Find the last 1 in the mask, that one is your increment. If the last one is a 8, the nets also hop with and 8.
    Hi all
    I have a little problem with cbtnuggets method to calculate...for example
    10.0.0.0 255.0.0.0
    and I have to divide in 4 networks
    can someone try to calculate and give me a feedback how to do it (in cbtnugets method)...
    I need 4 networks:

    00000100 is 4 (So it takes 3 bits for number 4)

    so next step is to add those bits in subnet mask to get increment and new sub-net mask:

    11111111.11100000.00000000.00000000
    what makes
    255.224.0.0 and increment of 32(last bit in mask)

    but subnet calculator gives me this:


    Am I making a obvious mistake somewhere...please help..Im gona explode


    Sorry guys, found it at the end of lecture, I noticed before that 4 and 32 ect. dont give wright result...so
    2, 4, 8 , 16, 32 , 64, 128 give off calculations!!!!
    Those are exceptions!!!
    Last edited by Gargamel; 09-17-2010 at 05:54 PM. Reason: found the answer to my question :)
    Reply With Quote Quote  

  10. Cisco Moderator mikej412's Avatar
    Join Date
    May 2005
    Location
    Chicago
    Posts
    10,190

    Certifications
    CCNP CCIP CCSP CCVP CCDP CCDA CCNA CS-CIPSS CS-CIPTDS CS-CIPTOS CS-CIPCSS CS-CFWS CS-CVPNS CS-CISecS ISSP 4013 4011
    #84
    Quote Originally Posted by Gargamel View Post
    00000100 is 4 (So it takes 3 bits for number 4)
    ?

    That is the binary number 4 -- but 3 bits gives you 8 possible bit choices/options/networks -- 000 001 010 011 100 101 110 111 (0 through 7)

    1 bit gives you 2 choices (or 2 networks) -- 0 or 1 (0 through 1)

    2 bits gives you 4 choices (or 4 networks) -- 00 01 10 11 (0 through 3)
    Reply With Quote Quote  

  11. Junior Member Registered Member
    Join Date
    Sep 2010
    Posts
    1
    #85

    Unhappy 23 bit mask

    Just like to say thanks to lordflasheart

    could any help to me undetstand 23 bit mask
    I have been using lordflasheart process but still cant get 23bit mask

    the bit which is confusing is where you draw the boundry 24 - 23 =1 this is very a get confused

    please help

    Thanks storma
    Reply With Quote Quote  

  12. Senior Member
    Join Date
    Mar 2010
    Posts
    129

    Certifications
    A+, Network+, Security+
    #86

    Default My subnetting howto

    Hosts howto:

    max = /32
    Determining number of host’s:

    2^(number of zeroes in the subnet mask)-2

    if /26, 32-26=6

    2^6-2=62 total hosts

    ----------------------------------------
    8, 16, 24, 32 boundries

    Subnet howto:

    1) determine octet position from netmask
    2) Take netmask subtract from boundry position
    3) do 2^boundry difference = x, block size
    4) count up by block size value
    5) stop when block size value is close to ip address.

    -------------------------------------------------------
    What subnet does 192.168.12.78/29 belong to?

    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.

    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

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

  13. Junior Member
    Join Date
    Jun 2010
    Posts
    14

    Certifications
    A+
    #87
    Glad I found this post, it really helped me understand.
    Reply With Quote Quote  

  14. Senile old fart laidbackfreak's Avatar
    Join Date
    Oct 2007
    Location
    wandering t'internet
    Posts
    991

    Certifications
    CISSP, CCVP, CCNAV, CCNAS, CCNA
    #88
    This should be made a sticky.
    Reply With Quote Quote  

  15. Member
    Join Date
    Jan 2006
    Location
    Cape Town, South Africa
    Posts
    33

    Certifications
    MCP, MCSA +Messaging, MCSE, CNA, Network+, I-Net+, CIW-A, Linux LPI 101 & 102, CCNA, CCNP
    #89
    I've just gone through the first page of this post now and already understand subnetting a lot better than from going through a lot of other books and tutorials. Thanks a lot!
    Reply With Quote Quote  

  16. Member
    Join Date
    Jan 2006
    Location
    Cape Town, South Africa
    Posts
    33

    Certifications
    MCP, MCSA +Messaging, MCSE, CNA, Network+, I-Net+, CIW-A, Linux LPI 101 & 102, CCNA, CCNP
    #90
    Ok, I'm still having problems with these type of questions, where they give an ip and subnet mask, and ask for the host etc. It's been answered before, but I still don't get it.

    Quote Originally Posted by Nzastudios View Post
    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.
    How does 240 mean that you're borrowing the first 4 bits of the third octet???
    Reply With Quote Quote  

  17. Member gandalph's Avatar
    Join Date
    Apr 2006
    Posts
    33
    #91
    Third octect would look like this

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


    11110000 = 240
    Reply With Quote Quote  

  18. Member
    Join Date
    Jan 2006
    Location
    Cape Town, South Africa
    Posts
    33

    Certifications
    MCP, MCSA +Messaging, MCSE, CNA, Network+, I-Net+, CIW-A, Linux LPI 101 & 102, CCNA, CCNP
    #92
    Quote Originally Posted by gandalph View Post
    Third octect would look like this

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


    11110000 = 240
    So how do you get to that?
    Reply With Quote Quote  

  19. Senior Member miller811's Avatar
    Join Date
    Oct 2007
    Location
    Nashville
    Posts
    896

    Certifications
    CCNP, CCDP, MCSA, Security +. Network +, A+
    #93
    Quote Originally Posted by Spoonroom View Post
    Ok, I'm still having problems with these type of questions, where they give an ip and subnet mask, and ask for the host etc. It's been answered before, but I still don't get it.

    Originally Posted by Nzastudios
    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


    How does 240 mean that you're borrowing the first 4 bits of the third octet???
    the easiest way to figure those out is to do the following


    determine the interesting octet. (the non 255. one)
    in this case it is the third octet, which is 240
    each octet totals 256 (0-255)
    so 256-240 = 16
    your block size is 16.

    so with the address 10.5.178.10 you would start with the non interesting octets (255.255.0.0)
    10.5.x.x
    since your block size is 16 you would apply it to the base
    10.5.0.0
    10.5.16.0
    10.5.32.0
    10.5.48.0
    10.5.64.0
    10.5.80.0
    10.5.96.0
    10.5.112.0
    10.5.128.0
    10.5.144.0
    10.5.160.0
    10.5.176.0 (winner winner chicken dinner)
    10.5.192.0

    now you have determined the subnet it belongs in. the next ip address 10.5.176.1 is the first available host

    the broadcast address is easily determined by taking the next subnet (10.5.192.0) and subtracting 1 = 10.5.191.255

    the last available host would be one address below the broadcast
    10.5.191.255 - 1 = 10.5.191.254

    hope that helps
    Reply With Quote Quote  

  20. Member
    Join Date
    Jan 2006
    Location
    Cape Town, South Africa
    Posts
    33

    Certifications
    MCP, MCSA +Messaging, MCSE, CNA, Network+, I-Net+, CIW-A, Linux LPI 101 & 102, CCNA, CCNP
    #94
    Quote Originally Posted by miller811 View Post
    the easiest way to figure those out is to do the following


    determine the interesting octet. (the non 255. one)
    in this case it is the third octet, which is 240
    each octet totals 256 (0-255)
    so 256-240 = 16
    your block size is 16.

    so with the address 10.5.178.10 you would start with the non interesting octets (255.255.0.0)
    10.5.x.x
    since your block size is 16 you would apply it to the base
    10.5.0.0
    10.5.16.0
    10.5.32.0
    10.5.48.0
    10.5.64.0
    10.5.80.0
    10.5.96.0
    10.5.112.0
    10.5.128.0
    10.5.144.0
    10.5.160.0
    10.5.176.0 (winner winner chicken dinner)
    10.5.192.0

    now you have determined the subnet it belongs in. the next ip address 10.5.176.1 is the first available host

    the broadcast address is easily determined by taking the next subnet (10.5.192.0) and subtracting 1 = 10.5.191.255

    the last available host would be one address below the broadcast
    10.5.191.255 - 1 = 10.5.191.254

    hope that helps
    This made absolutely no sense when I read through it the first time, but then I tried another example, and it worked!

    This is so cool!

    Thx for the help, lemme go try some more examples...
    Reply With Quote Quote  

  21. Burn Baby Burn! Cisco Inferno's Avatar
    Join Date
    Oct 2010
    Location
    Denver, CO
    Posts
    945

    Certifications
    CCNA:R/S, MCSA:2012, MS Specialist: Server Virtualization, MCDST, A+, N+, S+, A.A:CIS
    #95
    Could anyone seem to help me with these type of problems? i have no clue how to start. every other question takes me 4 seconds.
    heres an example.
    thanks


    Question: How many subnets and hosts per subnet can you get from the network 172.19.0.0/28?
    Answer: 4096 subnets and 14 hosts
    Reply With Quote Quote  

  22. Senior Member miller811's Avatar
    Join Date
    Oct 2007
    Location
    Nashville
    Posts
    896

    Certifications
    CCNP, CCDP, MCSA, Security +. Network +, A+
    #96
    Quote Originally Posted by Cisco Inferno View Post
    Could anyone seem to help me with these type of problems? i have no clue how to start. every other question takes me 4 seconds.
    heres an example.
    thanks


    Question: How many subnets and hosts per subnet can you get from the network 172.19.0.0/28?
    Answer: 4096 subnets and 14 hosts
    this is fairly easy to do. It is all based on the powers of 2....
    see the chart below.

    1 - 2
    2 - 4
    3 - 8
    4 - 16
    5 - 32
    6 - 64
    7 - 128
    8 - 256
    9 - 512
    10 - 1024
    11 - 2048
    12 - 4096
    13 - 8192
    14 - 16384
    15 - 32768
    16 - 65536

    So in your example the address given is a class B address. So under normal circumstances it is using a 16 bit mask. 255.255.0.0
    In this example it is borrowing 12 of the possible bits for the subnets, and 4 bits for hosts.

    so 2 to the 12 = 4096 = to the number of subnets
    and then 4 bits are used for the number of hosts 2 to the 4 or 16... but you need to subtract 2 bits, one for the network address and one for the broadcast on each of the 4096 subnets......

    Hope that helps.
    Reply With Quote Quote  

  23. Burn Baby Burn! Cisco Inferno's Avatar
    Join Date
    Oct 2010
    Location
    Denver, CO
    Posts
    945

    Certifications
    CCNA:R/S, MCSA:2012, MS Specialist: Server Virtualization, MCDST, A+, N+, S+, A.A:CIS
    #97
    Quote Originally Posted by miller811 View Post
    this is fairly easy to do. It is all based on the powers of 2....
    see the chart below.

    1 - 2
    2 - 4
    3 - 8
    4 - 16
    5 - 32
    6 - 64
    7 - 128
    8 - 256
    9 - 512
    10 - 1024
    11 - 2048
    12 - 4096
    13 - 8192
    14 - 16384
    15 - 32768
    16 - 65536

    So in your example the address given is a class B address. So under normal circumstances it is using a 16 bit mask. 255.255.0.0
    In this example it is borrowing 12 of the possible bits for the subnets, and 4 bits for hosts.

    so 2 to the 12 = 4096 = to the number of subnets
    and then 4 bits are used for the number of hosts 2 to the 4 or 16... but you need to subtract 2 bits, one for the network address and one for the broadcast on each of the 4096 subnets......

    Hope that helps.
    aha the key is to use the additional bits on top of already using 255.255.0.0!

    thanks. basically lammle's method learned in 30 seconds of reading this post instead of 3 days on a page
    2017 Goals
    [x] MCSA: Server 2012 [X]70-410 [X]70-411 [x]74-409

    Reply With Quote Quote  

  24. Senior Member sthompson86's Avatar
    Join Date
    Apr 2010
    Location
    Pearl, Ms
    Posts
    370

    Certifications
    A.A.S Computer Technology, CCNA, CCENT, A+, Network+, Security+
    #98
    Thanks for this post. I agree with the poster above, out of all the ways I have learned sub netting this week, this way is by far the fastest.

    I am horrible at math well not horrible, I just do not trust my self doing it in my head, so the less processes I have to do for sub netting the better, for I always write everything out.
    Reply With Quote Quote  

  25. Member
    Join Date
    Jan 2011
    Location
    UK
    Posts
    51

    Certifications
    ITIL v3 Network + CCENT CCNA 70-685 Win 7 MEDST
    #99
    Hi

    I joined to say thanks for this method - it has started to come together for me after struggling with subnetting using other methods for ages - (even Jeremy's CBT Nuggets method)

    However, I am still struggling with these types of questions:

    Question: 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?

    Answer: 255.255.248.0


    Is there an easy way of quickly knowing what mask(s) give you number of networks and hosts?

    I also have problems subnetting in the 3rd octet when it's a class B address - this gets me every time - class C I have no problems with! For example:

    Question: What is the last valid host on the subnetwork 10.194.176.0 255.255.240.0?

    Answer: 10.194.191.254


    Thanks again - great method and resource - Jason

    EDIT - Actually I worked both these out from post#6 on this thread! I am buzzing!!
    Last edited by Jas21; 01-16-2011 at 05:34 PM.
    Reply With Quote Quote  

  26. Senior Member
    Join Date
    Dec 2010
    Posts
    218

    Certifications
    CCENT, CCNA, CCNP R&S
    #100
    hi Jas. This guy helped me alot with how he does it. he uses a system called the magic number.

    basically all you have to do is find the magic number and its all easy from there.
    you should watch a have of his subnetting vids

    YouTube - danscourses's Channel

    What is the last valid host on the subnetwork 10.194.176.0 255.255.240.0?

    the "important octet is the third octet as it is something other than 255.

    what number bit is 240 ? it is the number 16 bit thats your magic number!

    so youre only interested in counting up in 16's in the THIRD octet.

    in this case you can just say to yourself. 10 x 16 is 160. 160 + 16 is 176 + another 16 = 192

    the subnet range is therefor 176 - 192

    host range 177 - 191.254 with the broadcast being 191.255
    Reply With Quote Quote  

+ Reply to Thread
Page 4 of 20 First 1234 567814 ... Last

Social Networking & Bookmarks