Daily Archives: February 20, 2013

3.2.a Troubleshoot reverse path forwarding

they got this right…

An RPF check examines the source address of an incoming packet and checks it against the router’s unicast routing table to see what interface should be used to get back to the source network. If the incoming multicast packet is using that interface, the RPF check passes, and the packet is forwarded. If the multicast packet comes in a different interface, the RPF check fails, and the packet is discarded.

 

6.2.a CoS and DSCP mapping

0   0   0   0   0   0 = 0 CS0/Default

32  16  8   4   2   1

1   1   1   1   1   1 = 63

IP Precedence follows CS and the AF’s within the CS

CS increments by 8

CS1 = 8  decimal and 001 000 binary

CS2 = 16 decimal and 010 000 binary

CS3 = 24 decimal and 011 000 binary

CS4 = 32 decimal and 100 000 binary

CS5 = 40 decimal and 101 000 binary

CS6 = 48 decimal and 110 000 binary

CS7 = 56 decimal and 111 000 binary

notice without the lagging zero’s the leftmost binary bits = 1 through 7 or the CS designations, with the rightmost  bits they equal the decimal value of the CS..

ie, 32 + 16 + 8 + 0 + 0 + 0 = 56 or CS 7 or 111 000

or if taken without the rightmost bits 111 simply equals 7… the logic is pure…

there is no mystery in this…

There are 3 AF values within each class selector. The IP precedence equals the class selector in every case. The first value of the AF equals the CS

CS1 AF1

CS1 equals 8 equals 001 000

CS1 has 3 AF vlaues AF1 AF1 AF1

each AF vlaue’s second number increments by 1

AF11 AF12 AF13 whose ip precedence always equals 1 because it’s CS1

it is not difficult to understand that AF31 has to reside in CS3 and has ipp of 3 for every AF

CS3 = 24

AF31 = 26

AF32 = 28

AF33 = 30

because CS4 = 32 or the CS value times 8

and it’s first AF41 = 34

it’s second AF has to be 42 which equals 36 decimal because the decimal values increment by two for each AF

the binary values simply match the decimal value