+ Reply to Thread
Page 2 of 2 First 12
Results 26 to 37 of 37
  1. Member
    Join Date
    Mar 2007
    Posts
    40
    #26
    on subnettingquestions dot com im getting all of the questions right apart from
    the ones that ask what is the first availble host on a certain subnetwork, or the last
    available host on a specific subnetwork.

    All the examples are class C, I screw it up even when following your instructions.
    I have to do it in reverse, as you start with the subnetted IP.

    Question: What is the last valid host on the subnetwork 192.168.124.128 255.255.255.192?

    Answer: 192.168.124.190

    How do they get this? PLEASE!
    Reply With Quote Quote  

  2. SS
  3. Member
    Join Date
    Mar 2007
    Posts
    40
    #27
    Question: What is the last valid host on the subnetwork 192.168.124.128 255.255.255.192?


    Answer: 192.168.124.190

    The only thing left I can think off is

    Do we need to find the IP address the subnetwork 192.168.124.128 belongs too?
    How would we do that, as ANDing will just give the same answer as the subnetwork address. So we can only Anding process on a host ip to find its network id.


    192.168.124.128 is one of the two subnets on the network ip of unknown

    2 bits borrowed for hosting, 6 bits for hosting.

    I still dont understand the HOST ip range in this example ??

    63 is in the host, but I dont understand how it doesnt go up in increments of 63.
    Reply With Quote Quote  

  4. Senior Member
    Join Date
    Mar 2007
    Location
    Sunny Isles Beach, FL.
    Posts
    630

    Certifications
    Expired: CCNA, CCDA, CCNP. Working on re-certs starting with CCNA.
    #28
    It does.

    You have 4 subnets...

    192.168.124.0
    192.168.124.64
    192.168.124.128
    192.168.124.192

    So your subnets break down like this...

    192.168.124.0 (subnet)
    192.168.124.63 (broadcast)
    1-62 (host range)

    192.168.124.64 (subnet)
    192.168.124.127 (broadcast)
    65-126 (host range)

    192.168.124.128 (subnet)
    192.168.124.191 (broadcast)
    129-190 (host range)

    192.168.124.192 (subnet)
    192.168.124.255 (broadcast)
    193-254 (host range)

    Hope that helps.
    Reply With Quote Quote  

  5. Member
    Join Date
    Jan 2006
    Location
    N.Ireland
    Posts
    36

    Certifications
    A+ Net+ ccna, mcdst
    #29
    Hi again
    You must first work out the networks (what the network addresses are)
    Your question 192.168.124.128/26 what is the last address this is really pretty straight forward
    Let’s start with what are the subnets
    Well it’s a class c address (24 bits default in the mask) and they have borrowed 2 bits so the mask is 192 take this away from 256 = 64 are networks are starting at 0 and going up in increments of 64
    Ie…
    192.168.124.0
    192.168.124.64
    192.168.124.128
    192.168.124.192
    Now take the first network 192.168.124.0
    Well what is the first address going to be = 192.168.124.1
    The last address in that network will be = 192.168.124.62
    The broadcast address will be 192.168.124.63 (the broadcast address will always be the last address before the next network you must remember this always work out the broadcast address before your last address)

    Now take are second subnet 192.168.124.64
    The first address I can use is 192.168.124.65
    The last address I can use 192.168.124.126
    The broadcast address will be 192.168.124.127 (again the address before the next network)

    Now for your question what is the last valid host address 192.168.124.128/26
    Well going up in blocks of 64, 128 (aha were on the 128 network) now what is the next network up again add your block size 128+64 i=192 now what is the last address before this network 191(now remember this is 128s broadcast address )so take away another one =190
    So the last valid host on the 192.168.124.128/26 network is 192.168.124.190


    It takes a while but stick with it and you will begin to see patterns forming just try stick with class c networks first then class b will be even easier
    Reply With Quote Quote  

  6. Member
    Join Date
    Mar 2007
    Posts
    40
    #30
    you guys are great! Thanks for giving some random guy time to help him out.
    Ive 20 questions right in a row, so i've finally mastered it.
    I wouldnt have been able to do it without your illustrations!

    Now for the rest, I have 2 weeks to learn CCNA 3-4 and all the configuration codes etc from CCNA 2.

    I owe you guys some crack (coke if u prefer) and a stripper.

    NinjaMonkey
    Reply With Quote Quote  

  7. Member
    Join Date
    Jan 2006
    Location
    N.Ireland
    Posts
    36

    Certifications
    A+ Net+ ccna, mcdst
    #31
    No problem
    This is the best site I have come across for people willing to help you out, also check out the webmasters notes on this sits they are fantastic
    If you need anymore help just ask.

    stephen
    Reply With Quote Quote  

  8. Junior Member
    Join Date
    Apr 2005
    Posts
    2
    #32

    Default Easy way to figure out the broadcast number

    The above ways of inverting your bits to get the broadcast will work fine.

    For myself, the quickest way I have found is the following..I'll use this paste from a subnet calculator to explain.

    The subnet is this example is based on the fourth octet...(The octet where the range is anything other than 255).

    In this case it's 240. So, with that in mind, subtract 240 from 255 which gives you 15.

    Add 15 to 32 (192.168.15.32)

    This gives you your broascast of 192.168.15.47

    To summarize, just subtract the last number in your mask from 255, and add it to the NETWORK number.

    Address: 192.168.15.32
    Netmask: 255.255.255.240 = 28
    Wildcard: 0.0.0.15
    Network: 192.168.15.32/28
    Broadcast: 192.168.15.47
    Reply With Quote Quote  

  9. Junior Member
    Join Date
    Apr 2005
    Posts
    2
    #33

    Default Subnet increments

    Another easy method for me in dealing with network increments, is very simple.

    Let's say you have an address of 192.168.1.17 with a mask of 255.255.255.224

    the mask in binary is 11111111.11111111.11111111.11100000

    128 64 32 16 8 4 2 1

    1 1 1 0 0 0 0 0

    Your network number increments are always based on the position of the last "1" in the mask.

    So with that in mind, 224 means that the last "1" is under the 32 postion. This means your network increments in 32.

    If it was 240, your network would increment by 16. It's just that simple.
    Reply With Quote Quote  

  10. Senior Member
    Join Date
    Mar 2007
    Location
    Sunny Isles Beach, FL.
    Posts
    630

    Certifications
    Expired: CCNA, CCDA, CCNP. Working on re-certs starting with CCNA.
    #34

    Default Re: Subnet increments

    Quote Originally Posted by remmy7593
    Another easy method for me in dealing with network increments, is very simple.

    Let's say you have an address of 192.168.1.17 with a mask of 255.255.255.224

    the mask in binary is 11111111.11111111.11111111.11100000

    128 64 32 16 8 4 2 1

    1 1 1 0 0 0 0 0

    Your network number increments are always based on the position of the last "1" in the mask.

    So with that in mind, 224 means that the last "1" is under the 32 postion. This means your network increments in 32.

    If it was 240, your network would increment by 16. It's just that simple.
    I like that. Good call.
    Reply With Quote Quote  

  11. Junior Member Registered Member
    Join Date
    Sep 2018
    Posts
    2
    #35
    the ip is classB and the defalt mask for it is 16 and the question show 192 mean /18 the networks can be 0 64 128 and 192 so how 151.100.255.1 is possible?
    Reply With Quote Quote  

  12. Junior Member Registered Member
    Join Date
    Sep 2018
    Posts
    2
    #36

    Question subnetting

    the ip is classB and the defalt mask for it is 16 and the question show 192 mean /18 the networks can be 0 64 128 and 192 so how 151.100.255.1 is possible?
    Reply With Quote Quote  

  13. Senior Member
    Join Date
    Aug 2016
    Location
    Richmond, VA
    Posts
    104

    Certifications
    CCNP R&S, CCNA Sec, A+, Net+, Sec+, Project+, Linux+/LPIC-1
    #37
    151.100.255.1 is just one of the many addresses in the 151.100.192.0/18 network. That network includes all IPs from 151.100.192.0 - 151.100.255.255. Even though the network number starts at 192, that doesn't mean it's only 192. The mask is 255.255.192.0. Meaning for each subnet, first, the 255s don't change, so it'll always be 151.100.x.x. The 192 means each subnet's third octet covers (256 - 192 = 64) 64 numbers. And then the 0 in the fourth octet is all 256.

    So the first IP in the 151.100.192.0/18 network is 151.100.192.1, then .2, .3, all the way to 151.100.192.255. And 151.100.192.255 isn't the broadcast IP, it's just another usable IP in the subnet. And the very next usable IP, which is a valid host IP, and still in the same 151.100.192.0/18 network, is 151.100.193.0. You could give a computer 151.100.193.0/18 and ping it. And then .1, .2 etc, and then you're at 151.100.193.255. The next ip...151.100.194.0. And you'll keep doing that, incrementing the third octet after the 4th octet rolls over, for the same 151.100.192.0/18 subnet until finally you're in the 151.100.255.X range. At this point, you'll go through the .1, .2, etc, and then at 151.100.255.254 is the last usable IP in the 151.100.192.0/18 network, and 151.100.255.255 is the broadcast.
    CCNP - Switch [X] Route [X] T-Shoot [X]
    Reply With Quote Quote  

+ Reply to Thread
Page 2 of 2 First 12

Social Networking & Bookmarks