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New Member

Ip subnetting

Hi all,

I am currently trying to create 3 subnets with a class A ip address (10.14.227.0 /24).Unfortunately, my knowledge for subnetting is close to non existence and i would appreciate if someone could help me with it thanks.

1 ACCEPTED SOLUTION

Accepted Solutions

Re: Ip subnetting

Hi,

I think you are trying to equally subnet 10.14.227.0/24 to 3 subnets. That won't happen.

However, you can create four equal subnets out of it;

10.14.227.0/26 that is 10.14.227.1 to 10.14.227.62 (max 62 hosts)

10.14.227.64/26 that is 10.14.227.65 to 10.14.227.126 (max 62 hosts)

10.14.227.128/26 that is 10.14.227.129 to 10.14.227.190 (max 62 hosts)

10.14.227.192/26 that is 10.14.227.193 to 10.14.227.254 (max 62 hosts)

or three subnets (two equal, one bigger)

10.14.227.0/25 that is 10.14.227.1 to 10.14.227.126 (max 126 hosts)

10.14.227.128/26 that is 10.14.227.129 to 10.14.227.190 (max 62 hosts)

10.14.227.192/26 that is 10.14.227.193 to 10.14.227.254 (max 62 hosts)

/25 subnet mask is 255.255.255.128

/26 subnet mask is 255.255.255.192

Step-by-step tutorial from http://www.ralphb.net/IPSubnet/ just select "Next" to go to next page.

Subnet calculator from Boson and other utilities http://www.boson.com/FreeUtilities.html

7 REPLIES
Bronze

Re: Ip subnetting

New Member

Re: Ip subnetting

I have tried the subnetting calcualtor but its seems not to help, please can u help me subnet the ip address 10.14.227.0/24 and give my the host range in each of the 3 subnets .thanks

Re: Ip subnetting

Adedayo,

I don't know if your question was intended for learning subnetting or just looking for a solution. Assuming you are looking for a solution you can break the /24 into 3 subnets the following way. Instead, if you are trying to learn subnetting there are plenty of free tools available online or let us know.

10.14.227.0/26 - 10.14.227.1 - 10.14.227.62 are usable addresses. .63 is the broadcast address for this subnet.

10.14.227.64/26 - 10.14.227.65 - 10.14.227.126 are usable addresses .127 is the broadcast address.

10.14.227.128/25 - 10.14.227.129 - 10.14.227.254 are usable addresses. .255 is the broadcast address.

HTH

Sundar

Re: Ip subnetting

Hi,

I think you are trying to equally subnet 10.14.227.0/24 to 3 subnets. That won't happen.

However, you can create four equal subnets out of it;

10.14.227.0/26 that is 10.14.227.1 to 10.14.227.62 (max 62 hosts)

10.14.227.64/26 that is 10.14.227.65 to 10.14.227.126 (max 62 hosts)

10.14.227.128/26 that is 10.14.227.129 to 10.14.227.190 (max 62 hosts)

10.14.227.192/26 that is 10.14.227.193 to 10.14.227.254 (max 62 hosts)

or three subnets (two equal, one bigger)

10.14.227.0/25 that is 10.14.227.1 to 10.14.227.126 (max 126 hosts)

10.14.227.128/26 that is 10.14.227.129 to 10.14.227.190 (max 62 hosts)

10.14.227.192/26 that is 10.14.227.193 to 10.14.227.254 (max 62 hosts)

/25 subnet mask is 255.255.255.128

/26 subnet mask is 255.255.255.192

Step-by-step tutorial from http://www.ralphb.net/IPSubnet/ just select "Next" to go to next page.

Subnet calculator from Boson and other utilities http://www.boson.com/FreeUtilities.html

New Member

Re: Ip subnetting

If you use the last subnet with a broadcast address of 10.14.227.255, and a broadcast is sent, would this broadcast not get dilevered to ALL hosts on all 4 subnets of the 10.14.227.0 network? Could this not create unnecessary congestion for hosts on the first 3 subnets?

I'm also curious as to why the private 10.0.0.0 network was carved into 10.14.227.0. This leaves only 6 bits for hosts and 18 bits for subnets. Unless it is for practice with the equivalent of a class C network, what is the reason?

New Member

Re: Ip subnetting

"If you use the last subnet with a broadcast address of 10.14.227.255, and a broadcast is sent, would this broadcast not get dilevered to ALL hosts on all 4 subnets of the 10.14.227.0 network?"

Hopefully for security reasons you have IP directed broadcasts disabled so this won't be an issue.

Silver

Re: Ip subnetting

Hi!

Good Day. Please see my summary about subnetting. Hope this helps!

Regards,

Albert

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