Subnetting question

satishtechsatishtech Member Posts: 243
10.0.0.0/17

is my subnetting correct ?

/17 = 255.255.128.0

Network bits = 8
Subnet bits = 9
Subnets = 2^9 =512
Host bits = 17

First subnet = 10.0.0.0
First host = 10.0.0.1
Last Host = 10.0.0.254
Broadcast address =10.0.0.255

Next subnet is 10.0.1.0 and next is 10.0.2.0 uptil 10.0.127.0
this gives a total of 128 subnets from 0 to 127

but where are the 512 subnets ?
kindly correct me.

Comments

  • clarsonclarson Member Posts: 903 ■■■■□□□□□□
    going to need to do better than that

    you are right up to host bits
    32-17=15 host bits
    broadcast address is the network address with all host bits set to one
    nnnn nnnn.ssss ssss.s111 1111.1111 1111
    network is 10.0.0.0 so nnnn nnnn = 10 first subnet ssss ssss.s are all zeros
    10.0.127.255
    last host is broadcast minus one
    10.0.127.254
    next subnet is all zeros plus one 10.0.1hhh hhhh.hhhh hhhh with all the hosts set to zero
    10.0.128.0
    next subnet 10.1.0.0, then 10.1.128.0, then 10.2.0.0
  • hodgey87hodgey87 Member Posts: 232
    Quickly working this out

    Broadcast Address: 10.0.127.255
    First host: 10.0.0.1
    Last host: 10.0.127.254
    Total host count: 32766
  • PristonPriston Member Posts: 999 ■■■■□□□□□□
    Given that you didn't post the subnetting question, I assume you started with 10.0.0.0/8 and subnetted into 10.0.0.0/17
    Looking at the binary can help sometimes

    nnnn nnnn . ssss ssss . s hhh hhhh . hhhh hhhh
    0000 1010 . 0000 0000 . 0 000 0000 . 0000 0000 = 10.0.0.0 First subnet
    0000 1010 . 0000 0000 . 1 000 0000 . 0000 0000 = 10.0.128.0 Second subnet
    0000 1010 . 0000 0001 . 0 000 0000 . 0000 0000 = 10.1.0.0 third subnet
    0000 1010 . 0000 0001 . 1 000 0000 . 0000 0000 = 10.1.128.0 fourth subnet

    nnnn nnnn . ssss ssss . s hhh hhhh . hhhh hhhh
    0000 1010 . 0000 0000 . 0 1111 1111 . 1111 1111 = 10.0.127.255 First subnet broadcast
    0000 1010 . 0000 0000 . 1 1111 1111 . 1111 1111 = 10.0.255.255 Second subnet broadcast
    0000 1010 . 0000 0001 . 0 1111 1111 . 1111 1111 = 10.1.127.255 third subnet broadcast
    0000 1010 . 0000 0001 . 1 1111 1111 . 1111 1111 = 10.1.255.255 fourth subnet broadcast
    A.A.S. in Networking Technologies
    A+, Network+, CCNA
  • satishtechsatishtech Member Posts: 243
    so the last subnets will be 10.255.0.0 and 10.255.128.0
    2nd octet block size is 1
    3rd octet block size is 128

    but if the IP/network was 10.8.9.0/17
    will the same above rules apply like 2nd and 3rd octet varies
    or is it that now only 3rd octet varies with a block size of 128 ?
  • clarsonclarson Member Posts: 903 ■■■■□□□□□□
    the block size is 128 in the 3rd octet.

    you are only incrementing the 3rd octet. which when going from 128 to 256 causes a carry to the 2nd octet to happen.

    10.8.9.0/17 is a host address. it's network address is 10.8.0.0/17
    the block size is the same 128 so next network is
    10.8.128.0/17 and the next is
    10.9.0.0/17
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