3003:4AB1:2C4E:8801::/54

[JoshGarmonCiscoCCNA]

I know how to get the number of total subnets from this...it would be 1,024

I just dont' know what those subnets ARE.

8801 = I000 I000 0000 000I

with the last 10 bits left over              00 0000 000I

88 00 0000 0000 = ______/64

88 00 0000 000I= ______/64

and so forth

how would i write that out in /64 terms?

the first two '00' are what is throwing me off.

[Seb Rupik · Accepted solution]

Hi there,

You first few subnets will be:

3003:4ab1:2c4e:8800::/64

3003:4ab1:2c4e:8801::/64

3003:4ab1:2c4e:8802::/64

By the time you have incremented through the subnets to reach the first two bits you will then have the following /64:s

01  --> 3003:4ab1:2c4e:8900::/64

10 --> 3003:4ab1:2c4e:8a00::/64

11 --> 3003:4ab1:2c4e:8b00::/64

...until the point where you have all network bits set to 1, you will have you last subnet:

3003:4ab1:2c4e:8bff::/64

Hope that answers your question.

cheers,

Seb.

[JoshGarmonCiscoCCNA]

Hey Seb!

Sorry for the delayed response.

So the first two '00' I don't worry about?

I just worry about the last '0000 0000'?

[Seb Rupik]

Since your network prefix is /54, this gives you ten bits to create your addressing plan with.

As shown in the reply the first two 00 will not come into play until your 256 allocation (assuming you were going sequentially), so your bit pattern will be 01 0000 0000 (3003:4ab1:2c4e:8900::/64).

By your 512 allocation you will have 10 0000 0000 (3003:4ab1:2c4e:8a00::/64)

768 allocation 11 0000 0000 (3003:4ab1:2c4e:8b00::/64)

1024 allocation 11 1111 1111 (3003:4ab1:2c4e:8bff::/64)

Hope that makes sense :)

cheers,

Seb.