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I have the next issue : A company has the network ID:192.168.10.0/24 , the company is composed of 10 department. What is the maximum number of the users in each department that can be covered by this address space ?

i need to know the way of finding the maximum users step by step in order to finding the solution fot other issues

Omar
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  • You need to clarify your question: it is that space only for users??, you don' t need to put on this network services like servers, printers?, it is this network connected directly into a router or L3 switch?, do your network need to reach the internet or is completely isolated? – Orlando Gaetano Jan 22 '16 at 20:32
  • to be honest ... it's an exam question ... it is as like this and nothing more I can tell it depends on subneting ... I don't know what to do !! only a company with 10 departments !! – Omar Jan 22 '16 at 20:43
  • Your question is out of topic here i guess, no matter of that i will give you an answer – Orlando Gaetano Jan 22 '16 at 20:52
  • This post should help you out. Hint: you cannot divide a /24 in to 10 equal sized subnets. Expect some wasted space. – IsAGuest Jan 22 '16 at 20:57
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    Unfortunately, all "education, certification, or homework" questions are explicitly off-topic here. – Ron Maupin Jan 22 '16 at 21:03
  • it's question year ago exam ... i'm not trying to cheat !! and it may issue in real work ! – Omar Jan 22 '16 at 21:10

1 Answers1

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You have 192.168.10.0/24 and 10 departments:

Firts, i will suppose that you don' t have any requirement about your address space, because i will give you the basic answer.

What you need to do its subnetting looking 10 different networks>

You need to remember that there is a rule for that: (2^n) where n is the number of bits If you take 3 bits: 8 networks, NOT If you take 4 bits: 16 networks, YES

192.168.10.0000 0000 the last octet it is in bit notation

192.168.10.0000 0000

192.168.10.0001 0000

192.168.10.0010 0000

192.168.10.0011 0000

192.168.10.0100 0000

192.168.10.0101 0000

192.168.10.0110 0000

192.168.10.0111 0000

192.168.10.1000 0000

192.168.10.1001 0000

192.168.10.1010 0000

192.168.10.1011 0000

192.168.10.1100 0000

192.168.10.1101 0000

192.168.10.1110 0000

192.168.10.1111 0000

The part of the octet that i has write in bold form, that bits now will be part of network, and ther your mask will move to 28.

On that way, you have your 10 required networks and 6 networks more.

Then your network will have 2^n -2 host, cause the first address is your networks address and the last is your broadcast address. Every network will have then 14 hosts.

Orlando Gaetano
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  • thank you ... so there is no need for the address to be addressed exactly 10 partitions ...? – Omar Jan 22 '16 at 21:22
  • Oh no, for do that, you need to do VLSM, where ones department will be more address like other, but you will use more wisely your address space. That is not a easy method and you need to practice a lot – Orlando Gaetano Jan 22 '16 at 21:25