03-30-2022 11:26 AM
I'm trying to interpret the ACL commands for extended ACL applied to BGP, which is slightly different than applied ti IGP
I need help with the subnet mask part
first entry 255.255.0.0 0.0.0.0
is /16 prefix as 255.255.0.0 resolves to /16 in standard (non-wildcard) prefix definition
the 2nd line
255.255.255.0 0.0.0.0
resovles to /24 prefix and the wildcard portion is 0 (bold) so there is no range
the third line 255.255.255.0 0.0.0.255
resolves to /24 for first part and /32 for bolded wildcard portion
so a /24 to /32 prefix length
the last box makes no sense to me
255.255.255.128 resolves to /25, as 128 takes it from 24 to 25 bits
this is where I get confused
0.0.0.127 resolves to /32?
The last row in the image aboce says "a /25 to /32" prefix length
I'm, interpreting the extended ACLs applied to BGP as standard mask and wildcard mask, and to me I do not see how /127 resolves to /32.
Solved! Go to Solution.
03-30-2022 12:48 PM - edited 03-30-2022 12:48 PM
You really need to write it out in binary to make sense ie. an octet has 8 bits with values -
128 64 32 16 8 4 2 1
for the 3rd line the subnet mask is 255.255.255.0 ie. /24 and then you have the whole of the last octet to use which gives you the values 25 - 32.
If you add up 128 + 64 + 32 + 16 etc. you get 255 which is where the .255 comes in for the last octet of the wildcard mask.
For the 4th line your subnet mask is 255.255.255.128 ie. you have used the first bit of the last octet for the subnet so you only have the last 7 bits left which covers values 26 - 32.
If you add up 64 + 32 + 16 etc. in the last octet you get 127 which is where the .127 value comes in for the last octet of the wildcard mask.
Does that make sense ?
Jon
03-30-2022 11:54 AM
There´s no image in your post.
03-30-2022 12:39 PM
image posted
03-30-2022 12:48 PM - edited 03-30-2022 12:48 PM
You really need to write it out in binary to make sense ie. an octet has 8 bits with values -
128 64 32 16 8 4 2 1
for the 3rd line the subnet mask is 255.255.255.0 ie. /24 and then you have the whole of the last octet to use which gives you the values 25 - 32.
If you add up 128 + 64 + 32 + 16 etc. you get 255 which is where the .255 comes in for the last octet of the wildcard mask.
For the 4th line your subnet mask is 255.255.255.128 ie. you have used the first bit of the last octet for the subnet so you only have the last 7 bits left which covers values 26 - 32.
If you add up 64 + 32 + 16 etc. in the last octet you get 127 which is where the .127 value comes in for the last octet of the wildcard mask.
Does that make sense ?
Jon
03-30-2022 01:28 PM
yes
128 64 32 16 8 4 2 1
the 2nd line
255.255.255.0 0.0.0.0
gives /24 exact
the third line the command is 255.255.255.0 0.0.0.255
so 1+ 2 + 4 + 8 + 16 + 32 + 64 + 128
= 255
as user defines the 0.0.0.255 wildcard to allow for this range
255.255.255.128
is /25 prefix
we use 0.0.0.127 as 7 digits to play around with
starting from 0 in wildcard
0 1 (1digit) 3(2 digit) 7 (3 digit) 15(4digit) 31(5digit) 63(6digit) 127 (7 digit) 255(8digit)
which makes it a range /25 - /32
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