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I have a question, and I have some confusion.

I have to create subnets for six different teams. All IP addresses should start with 172.168.1.x only.

Team 1 -> 5 Hosts
Team 2 -> 16 Hosts
Team 3 -> 13 Hosts
Team 4 -> 37 Hosts
Team 5 -> 65 Hosts
Team 6 -> 12 Hosts

It looks like it is not possible. I think I have to reduce the host count from some team.

Can anyone can help?

Ron Maupin
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karthikeayan
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    Classful networking is dead. And even with the old definition of Classful an address starting with 178 can never ever be a Class C address. –  Mar 26 '16 at 08:31

1 Answers1

6

In other words you have a /24 network that you want to divide in 6 networks.

Let's reorder the teams by size:

Team 5 : 65 hosts  
Team 4 : 32 hosts  
Team 2 : 16 hosts  
Team 3 : 13 hosts  
Team 6 : 12 hosts  
Team 1 : 5 hosts

and compare them with subnet sizes:

A /24 network comprise 256 addresses from which you must remove the network, broadcast and gateway addresses leaving 253 addresses for hosts.

Similarly:

/25 : 125 hosts  
/26 :  61 hosts  
/27 :  29 hosts  
/28 :  13 hosts  
/29 :   5 hosts 

So we see that we need at a minimum:

Team 5 : 65 hosts -> /25  
Team 4 : 32 hosts -> /26  
Team 2 : 16 hosts -> /27  
Team 3 : 13 hosts -> /28  
Team 6 : 12 hosts -> /28  
Team 1 :  5 hosts -> /29  

Can we divide a /24 network to fit all those networks in ?

A way to look at it would be:

  • Divide the /24 in two /25 networks.
    The first /25 is assigned to Team5, the second /25 is further sub-netted.

  • Divide the second /25 in two /26 networks.
    The first /26 is assigned to Team4, the second is further sub-netted.

  • Divide the second /26 in two /27 networks.
    The first /27 is assigned to Team2, the second is further sub-netted.

  • Divide the second /27 in two /28 networks.
    the two /28 are assigned to Team3 and Team6.

Nothing left for Team1... ...too bad.

You could for example:

  • Group Team 1 with another team (so both team will use the same network).
  • User for another team 2 networks, i.e.:
    Separate the the first /25 that was assigned to Team 5 in two /26 networks.
    Assign the first one to the first 63 hosts of Team5.
    Subnet the second one in four /28 networks
    Assign the first /28 to the other members of Team 5, the second one to team 1 and you have two /28 available.
JFL
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