Help with broadcast question

lorddalronlorddalron Registered Users Posts: 8 ■□□□□□□□□□
Just when you think you've got a hang of it, you go and get stumped.

Question:

What is the broadcast address for the subnetwork on which host A resides?

Host A= 10.149.177.147/12

How do i get the answer 10.159.255.255, or more to the point, the .159?

Any help much appreciated

Comments

  • HackedAliasHackedAlias Member Posts: 34 ■■□□□□□□□□
    You have an address of 10.149.177.147 /12 which would mean you're 'borrowing' 12 bits. Visually looks like this:

    11111111.11110000.00000000.00000000

    So the first octet is unchanged, now you need to find the interval for the rest. Now I'll move to the second octet.

    11110000

    Each digit from left to right represents what the interval will be by the 1's (Network bits) 0's for the host bits.
    2^4(4 1's) = 16 Network Bits
    2^4 (4 0's) = 16 Host bits

    What helps some people is to count down from 128 from left to right. So each bit would be:
    128, 64, 32, 16 etc..

    So, your second octet will be increasing by intervals of 16.

    Doing simple math we find that Host A with an address of 10.149.177.147 /12 resides on the network of:
    10.144.0.0 (I simply added 16 until I was as close to 149 as possible.)

    Now to find the broadcast you would add 16 to 144. Which would be 160(the next sub network) and then subtract 1 since we want the broadcast address which is the last address in a subnet. So going backwards from 10.160.0.0 the next immediate address is:

    10.159.255.255

    The broadcast address for your network.
  • CoolhandlukeCoolhandluke Member Posts: 118
    Can't really explain it any simpler than that.
    [CCENT]->[CCNA]->[CCNP-ROUTE]->COLOR=#0000ff]CCNP SWITCH[/COLOR->[CCNP-TSHOOT]
  • lorddalronlorddalron Registered Users Posts: 8 ■□□□□□□□□□
    Great. I understand that and its been refreshed. Thankyou

    Network address of 10.144.0.0
    Broadcast address of 10.159.255.255


    So why was i doing it like this:

    10.149.177.147/12

    12=00001100 with the increment being 4 (Lowest bit value)

    Then going up in 4's.

    Ive changed my subnetting style a few times and after watching CBT Nuggets i decided to go with his way because it seemed pretty simple but maybe I was getting my methods mixed up (Well clearly I was) but was was i thinking of?
  • oli356oli356 Member Posts: 364
    The way hacked demonstrated is how I do it and how cbtnuggets do it.

    i can't quite understand what you were doing. Keep practising on the random subnetting question site and your master it.

    Also make sure you put the bit notation at the end not beginning.. I guess it doesn't really matter (well it makes it look confusing in my opinion) but this is how it's done :)
    not: /24 192.168.0.1
    do: 192.168.0.1 /24
    Lab:
    Combination of GNS3 and Cisco equipment if required.
  • kurosaki00kurosaki00 Member Posts: 973
    /12 = 255.240.0.0

    Now we get the magic number
    How many addresses you can get in a network? 256 (2^8 = 256)
    Now we 256 - 240 = 16
    16 is your magic number, is the number of range of hosts per subnet.

    Meaning:
    10.0.0.0 (10.0.0.1 - 10.15.255.255) 10.15.255.255
    10.16.0.0 (10.16.0.1 - 10.31.255.254 ) 10.31.255.255
    and so on

    Your subnets will be 16 on 16, first address is subnet ID, last address is broadcast of that subnet
    everything in between are the usable address
    First address is subnet ID, last address is broadcast ID

    Now you can do this the hard way and do all the subnets till you get your wanted one, or some basic math
    or we can do some ANDing


    We convert your IP to binary (below your subnetmask in binary)

    00001010.10010101.10010101.10010011
    11111111.11110000.00000000.00000000

    00001010.10010000.00000000.00000000
    =10.144.0.0

    So hes in the 10.144.0.0
    we add 16 and we have the next subnet ID
    10.144.0.0
    10. 160.0.0
    so the address before that is the broadcast address, the range would be
    10.144.0.0 (10.144.0.1 - 10.159.255.254 ) 10.159.255.255

    Clearly your IP is in this subnet, so thats the broadcast address

    This may have seemed like a lot, but essentially if they give me the subnet mask, the exercise is pretty solved itself.
    You got subnet mask, means you got "magic number" (range), and you got subnet mask you use it with ANDing, you got youself the specific IP range for the address
    and thats it.

    If this was an exam question I would have wrote
    magic number
    subnet and ip in binary
    anding result
    thats the range
    thats it
    meh
  • lorddalronlorddalron Registered Users Posts: 8 ■□□□□□□□□□
    Thanks guys. I have pretty much mastered it but i think in my haste i get carried away and get things mixed up.


    Ive just literally booked my exam. 29th July. Cant wait to get it to move on to my next certs but scared at the same time as i dont want to fail to get help back.

    Im going to do it though and hang the cert with pride in my kitchen icon_cheers.gif


    Anyway......icon_study.gif
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