all 5 comments

[–]zanfarCCENT 0 points1 point  (4 children)

Can you give a page reference? It's hard to understand your question without context.

[–]mingoleg[S] 0 points1 point  (3 children)

Page is 329 in the ICND1 book.

[–]zanfarCCENT 2 points3 points  (2 children)

Gotcha. Thanks for the image, but most of us have at least one of Odom or Lammle, so we can look up page numbers pretty quickly.

Don't worry about the math, I have no idea why he added it to the table. The classfull ranges are defined, not calculated.

Class A is 27-2 because bit 0 must be 0b0 so you only have 7 optional bits in the first 8. The -2 is because the and networks are reserved

Class B is 214 because bits 0-1 must be 0b10, so you only have 14 optional bits in the first 16.

Class C is 221 because bits 0-2 must be 0b110 so you only have 21 optional bits in the first 24.

Mostly, ignore this. Classfull addressing is not really useful beyond getting your cert. Just memorize the class letter, first-octet ranges, and default subnets for each class.

Knowing the other classes (D, E) and reserved ranges (0, 127) is also helpful.

[–]mingoleg[S] 0 points1 point  (1 child)

Thanks for looking it up! I have my classless subnetting down pretty well, so this was throwing me off. I think I'm understanding though.

1-126 are always going to have bit 0 used, 128-191 always need 0-1 used, and 192-223 are always going to have bits 1-2 used.

When solving for # host bits in subnetting, those octets are already determined so they aren't counted anyway.

[–]zanfarCCENT 2 points3 points  (0 children)

It sounds like you got it. The math is weird because he is trying to apply subnetting rules to classful ranges. The subnetting problems work out to:

how many /8 networks can you subnet into,
not counting and

28-1 - 1 - 1 = 126

how many /16 networks can you subnet into

216-2 = 24 × 210 ≈ 16K

how many /24 networks can you subnet into

224-3 = 21 × 220 ≈ 2M