mtu

[amms68958]

Hi ,

How will I calculate the MPLS mtu , to allow 5 labels.

1) 1518 + 5 * 4 bytes

or

2) 1518 + (20 bits * 5 ) + 12 bits .

If the answer is (1) ,please explain the

need of the same..?

Thanks in advance

Thomas.

[mheusinger]

Hi,

a label is 4 Bytes for mtu calculation. The "normal" MTU is 1500 bytes, which then gives 1500+5*4=1520 Bytes mpls mtu.

If you want to transport ethernet over MPLS it would calculate as 1518+2*4=1526 Bytes as a minimum.

There seems to be some confusion arising from the fact that both, the 4 Byte is a called a label and a field inside the label (the 20 Bits used for forwarding) is also called label.

So 20 Bit label + 3Bit exp + 1 Bit bottom-of-stack + 8 Bit TTL = 32 Bits = 4 Byte label.

Hope this helps! Please rate all posts.

Regards, Martin

[amms68958]

Hi Martin ,

Thanks for your reply.

I asked transport ethrnet over MPLs only.

but why every time ,we are adding the TTL value ,exp bits etc with each label.

Why cant we keep that in common and add the label sizes ie;20 bits with each label .

Or ,the standrad wont allow to do this ?

Thanks

[amms68958]

Hi,

Also ,in other words,the 4 bytes is the MPLS header size ,in which label size is only 20 bits .

Why with each addition of a label we are adding 4 bytes of the header ,rather than just 20 bits.

Hope my question is clear .

Thanks in advance

Thomas.

[mheusinger]

Hi,

4 Byte is the standard (RFC 3036 and the like) and it makes sense to have a fixed length. Assume you would have an IP header varying between 8 Byte (I need just source and destination IP) and 20 Bytes. How would the router find out from the bitpattern arriving, what belongs to IP header and what to payload?

Fixed length is nearly mandatory for hardware processing, which allows for Gigabit performance.

Regards, Martin