3 ms·
> For large transfers The implicit assumption is that a fragmented packet will be slightly more than MTU. So for a large transfer - say N packets, of packet si
by majke 9y ago
> For large transfers
The implicit assumption is that a fragmented packet will be slightly more than MTU. So for a large transfer - say N packets, of packet size MTU + 1 - you will see N "first fragments" fragmented to the size of MTU, and another N last fragments of size 1. (actually IP header + 1)
In other words 50% of the packets flowing will have size of 1 which is super bad use of router switching resources.
- gsich 9y agoBut wouldn't that be multiple transfers?
- majke 9y agoBy "transfer" here I mean a thing like a curl on a large file.
- tyingq 9y agoIt would even apply to a single socket send() of say, a 4k buffer, which is pretty common. Broken MTU negotiation could easily result in twice as many packets as needed in that scenario.
- haimez 9y agoTwice as many packets per reduced MTU hop. The worst case is pretty bad and far north of 2x the number of packets sent from origin.
- im3w1l 9y agoThe worst case is that the routers support N, N-1, N-2, N-3, which means one additional packet per hap. I know what you are thinking: "well what if it's N/2, N/4, N/8 etc". And the answer is that in such a network you are screwed anyway. Lots of small packets would be unavoidable.