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CIDR help

gurkamal01
Level 1
Level 1

Hi

Can anyone tell me why  /19 and smaller prefixes have less routing tables than /20, /21, .... for  classC

This is done with major Tier 1 ISP's

Will appreciate an response

Regards

1 Accepted Solution

Accepted Solutions

Jon Marshall
Hall of Fame
Hall of Fame

gurkamal01 wrote:

Hi

Can anyone tell me why  /19 and smaller prefixes have less routing tables than /20, /21, .... for  classC

This is done with major Tier 1 ISP's

Will appreciate an response

Regards

Not sure i fully understand your question so if i have misunderstood please clarify.

A /19 ie. 255.255.224.0 would give you 32 useable Class C subnets (if that is what you wanted to use)

A /20 ie. 255.255.240.0 would give you 16 useable Class Cs

A /21 ie. 255.255.248.0 would give you 8 useable Class Cs

so you can see that you would need 4 /21s  ie.  4 * 8 = 32 to represent one /19 and 2 /20s ie. 2 * 16 = 32 to represent one /19.

So to represent 32 class C subnets in the routing table you would need  4 entries using /21s, 2 entries using /20s and only one entry using a /19. This is how using /19s means fewer entries in the routing table.

Note this is assuming that the 32 class C subnets could indeed be summarised with a /19.

Jon

View solution in original post

3 Replies 3

Jon Marshall
Hall of Fame
Hall of Fame

gurkamal01 wrote:

Hi

Can anyone tell me why  /19 and smaller prefixes have less routing tables than /20, /21, .... for  classC

This is done with major Tier 1 ISP's

Will appreciate an response

Regards

Not sure i fully understand your question so if i have misunderstood please clarify.

A /19 ie. 255.255.224.0 would give you 32 useable Class C subnets (if that is what you wanted to use)

A /20 ie. 255.255.240.0 would give you 16 useable Class Cs

A /21 ie. 255.255.248.0 would give you 8 useable Class Cs

so you can see that you would need 4 /21s  ie.  4 * 8 = 32 to represent one /19 and 2 /20s ie. 2 * 16 = 32 to represent one /19.

So to represent 32 class C subnets in the routing table you would need  4 entries using /21s, 2 entries using /20s and only one entry using a /19. This is how using /19s means fewer entries in the routing table.

Note this is assuming that the 32 class C subnets could indeed be summarised with a /19.

Jon

Thank for the response

As per your reply you stated ....." 4 * 8= 32 to represent one /19 and 2 /20s ie. 2 * 16 = 32 to represent one /19".

Can you please tell me how we got the 8 and 16 for /21 and /20 respectively

gurkamal01 wrote:

Thank for the response

As per your reply you stated ....." 4 * 8= 32 to represent one /19 and 2 /20s ie. 2 * 16 = 32 to represent one /19".

Can you please tell me how we got the 8 and 16 for /21 and /20 respectively

Well, you could do it in binary and that would show you for sure. But a quick way to work these things out is to simply do the following -

/19 = 255.255.224.0   so take the first octet that is not 255, in this case it is the 3rd octet and it is 224. Then simply do 256 - 224 = 32 so you have 32 useable class Cs

/20 = 255.255.240.0  so 256 - 240 = 16 class Cs

/21 = 255.255.248.0  so 256 - 248 = 8 class Cs

Jon