another summerization question

ebykmebykm Member Posts: 29 ■□□□□□□□□□
networks

192.168.1. 0 /27
192.168.1.32 / 28
192.168.1.48 /28


the summarization would be 192.168.1.0 /26,

192.168.1.0 - 31
192.168.1.32 - 47
192.168.1.48 - 63

because all networks mentioned can be packed into x.x.1.0 - 64.

am i correct ?.

Thanks.

[edit] when using binary methord, should one use broadcast address ? [edit]

Comments

  • DirtySouthDirtySouth Member Posts: 314 ■□□□□□□□□□
    I believe you are correct. The way I figure it out is by writing out the last octet, of each of those subnets, in binary. Then by looking at them you can tell which bit will summarize all of them.
  • ebykmebykm Member Posts: 29 ■□□□□□□□□□
    DirtySouth wrote:
    I believe you are correct. The way I figure it out is by writing out the last octet, of each of those subnets, in binary. Then by looking at them you can tell which bit will summarize all of them.

    like this ?

    broadcast addr

    x.x.x.00011111 - 0 ~ 31

    x.x.x.00101111 - 32 ~ 47

    x.x.x.00111111 - 48 ~ 63

    so only 2 bits on the 4 th octet matches, what if the subnets starts at x.x.x.192. would summarry be x.x.x.192 /26 ?

    how about summerizing class A, B or C networks with default subnets,
    networks 172.0.0.0 /16 to 172.8.0.0 /16 ?

    simply take the block size ? ( 256 - 8 = 24icon_cool.gif so it becomes, 172.0.0.0 /21 ?

    thanks.
  • DirtySouthDirtySouth Member Posts: 314 ■□□□□□□□□□
    ebykm wrote:
    like this ?

    broadcast addr

    x.x.x.00011111 - 0 ~ 31

    x.x.x.00101111 - 32 ~ 47

    x.x.x.00111111 - 48 ~ 63

    so only 2 bits on the 4 th octet matches, what if the subnets starts at x.x.x.192. would summarry be x.x.x.192 /26 ?
    Yep...thats it. :)
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