Determining Address Block Size

having a little trouble with this portion of the MSPress Book 70-291

here is the question and answer please help me understand how they got b=6

what is the address block size i would need for 60 subnets and 120 hosts a piece?

equation n =32-(b+h)

answer

32-(6+7)=19

I know how they got 7 but how did they get 6 from 60 subnets
On the road to MCITP......

Comments

  • dynamikdynamik Banned Posts: 12,314 ■■■■■■■■□□
    I think the formulas just confuse things.

    You have 32 bits that are used for networks and hosts. You borrow bits from the host bits in order to create subnets. IP addresses consist of four octets (8 bits), each of which have values of 0-255.

    To find the number of hosts or subnets, just set things up in binary.

    [erroneous info removed]

    For these types of questions, it really helps to just memorize what bit position gives you which value.

    bits--- 1-2-3--4---5--6---7---8----9----10---11
    12--etc
    value- 2-4-8-16-32-64-128-256-512-1024-2048-4096-etc

    (even if you can't remember a value, just remember it's twice the value before it and half the value after it)

    As you can see, 7 bits meets the requirement of 120 (remember to subtract 2, so you will get 126 usable addresses), and 6 bits meets the requirements of 60 (you typically do not need to subtract 2 here as modern equipment supports subnets of all 0s or all 1s, be sure to read the question carefully).
  • PashPash Member Posts: 1,601 ■■■■■□□□□□
    having a little trouble with this portion of the MSPress Book 70-291

    here is the question and answer please help me understand how they got b=6

    what is the address block size i would need for 60 subnets and 120 hosts a piece?

    equation n =32-(b+h)

    answer

    32-(6+7)=19

    I know how they got 7 but how did they get 6 from 60 subnets

    Two to the power of 6 is 64. Your in the clear.

    I am not sure how MSPress teach subnetting because I never read those chapters but search on these forums for quality subnetting links and questions.
    DevOps Engineer and Security Champion. https://blog.pash.by - I am trying to find my writing style, so please bear with me.
  • PashPash Member Posts: 1,601 ■■■■■□□□□□
    dynamik wrote:
    If you need 120 hosts, you will need 7 bits since 1+2+4+8+16+32+64 = 127 (remember to subtract 2, so you will get 125 usable addresses)

    You need 6 bits to obtain 60 subnets because 1+2+4+8+16+32 = 63 (you typically do not need to subtract 2 here as modern equipment supports subnets of all 0s or all 1s, be sure to read the question carefully).

    2^7 = 127?

    2^6 = 63?

    Not sure about that mate :P

    Remember the block numbers 2 4 8 16 32 64 128 = 254

    Computing is all about these wonderful numbers.
    DevOps Engineer and Security Champion. https://blog.pash.by - I am trying to find my writing style, so please bear with me.
  • dynamikdynamik Banned Posts: 12,314 ■■■■■■■■□□
    Yea, I did it right on the bottom lol. I went too fast and mixed up a couple of concepts, and you had already responded by the time I noticed.
  • PashPash Member Posts: 1,601 ■■■■■□□□□□
    No worries like. Best way to learn is think of real world questions, like what network a host is in using the ip and subnet mask. Designing a network, subnet and host requirements.

    And yeh your right its a bad day to learn subnetting from us two because of our lack of maths skills. LOL
    DevOps Engineer and Security Champion. https://blog.pash.by - I am trying to find my writing style, so please bear with me.
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