|
Home > Archive > CCNA > May 2002 > Variable Length Subnet Masking/Classless
You are viewing an archived Text-only version of the thread.
To view this thread in it's original format and/or if you want to reply to
this thread please [click here]
| Author |
Variable Length Subnet Masking/Classless
|
|
|
| Hi everyone! After a year of Cisco class I don't recall learning how to work with Variable Length Subnet Masking/Classless. Subnetting is easy to understand but this stuff really is confusing.
I figure if anyone one would know it would be someone in Cisco. I posted this question on two other sites but no one seems to be able to help.
Can someone help explain or better yet direct me to site that will help understand how to do this:
Assume that an ISP owns the address block 200.25.0.0/ 16. This block represents 65, 536 (2^ 16) IP addresses From the 200.25.0.0/ 16 block it wants to allocate the 200.25.16.0/ 20 address block . This smaller block represents 4,096 (2^ 12) IP addresses (or 16 /24s).
In a classless environment, the ISP is free to cut up the pie any way it wants. It could slice up the original pie into 2 pieces (each 1/ 2 of the address space) and assign one portion to Organization A, then cut the other half into 2 pieces (each 1/ 4 of the address space) and assign one piece to Organization B, and finally slice the remaining fourth into 2 pieces (each 1/ 8 of the address space) and assign it to Organization C and Organization D.
Step #1: Divide the address block 200.25.16.0/ 20 into two equal size slices. Each block represents one- half of the address space or 2,048 (211) IP addresses. ISP's Block 11001000.00011001.0001 0000.00000000 200.25.16.0/ 20
Org A: 11001000.00011001.0001 0 000.00000000 200.25.16.0/ 21
Reserved: 11001000.00011001.0001 1 000.00000000 200.25.24.0/ 21
Step #2: Divide the reserved block (200.25.24.0/ 21) into two equal size slices. Each block represents one- fourth of the address space or 1,024 (210) IP addresses.
Reserved 11001000.00011001.00011 000.00000000 200.25.24.0/ 21
Org B: 11001000.00011001.00011 0 00.00000000 200.25.24.0/ 22
Reserved 11001000.00011001.00011 1 00.00000000 200.25.28.0/ 22
Step #3: Divide the reserved address block (200.25.28.0/ 22) into two equal size blocks. Each block represents one- eight of the address space or 512 (29) IP addresses.
Reserved 11001000.00011001.000111 00.00000000 200.25.28.0/ 22
Org C: 11001000.00011001.000111 0 0.00000000 200.25.28.0/ 23
Org D: 11001000.00011001.000111 1 0.00000000 200.25.30.0/ 23
I wish this was enough to help me understand, but I still don't completely get it. I need a bigger picture with more details
Can anyone help? | |
| mikop 2002-05-17, 12:21 pm |
| Try working from a class C rather than a B, the numbers are easier to handle.
htere is a link to older post
http://www.examnotes.com/forums/sea...rder=descending
look over them and see if it help.
just remember, its all counting bits, move one here and one there. technically, ther eis nothing new, its just that before, subnetting involve cutting pies of equal size. so your little kid sister got as large a pie as you and can't finish, where you didn't get enuff and still hungry, but can't touch hers. | |
|
| Thanks the link on the link that you gave me really helped. I appreciate it. Ive been trying to crack this one for 3 days and your input helped |
|
|
|
|