On 11/04/17 14:59, George Amanakis wrote:
Hi Jonathan,
I have some questions regarding the algorithm to choose a tin to dequeue
from in sch_cake.c:
-----------8<------------
int oi, best_tin=0;
s64 best_time = 0xFFFFFFFFFFFFUL;
for(oi=0; oi < q->tin_cnt; oi++) {
int tin = q->tin_order[oi];
b = q->tins + tin;
if((b->sparse_flow_count + b->bulk_flow_count) > 0) {
s64 tdiff = b->tin_time_next_packet - now;
if(tdiff <= 0 || tdiff <= best_time) {
best_time = tdiff;
best_tin = tin;
}
}
}
-----------8<------------
1) best_time is defined as a positive signed integer, this equals to 78
hours if I did the calculations right. Why did you choose this? Did you
mean to define it as "-1"?
2) If you meant to define it as "-1", the condition "tdiff <= best_time"
would not matter. I can see no case where "tdiff > 0" and "tdiff <=
best_time".
Could you shed some light into this?
Thank you,
George
I'll try to answer this based on a vague bit of understanding...and then
Jonathan can shoot me down as well :-)
So the 'best_time' is signed because we can have a packet in a tin that
is not yet due to be sent (a positive result...ie. we have got here
early) and/or we can have a tin that is due now/overdue (a zero/negative
result, we've got here late)
The complication is that we can have multiple tins overdue, so we want
the highest priority *and* least overdue (least late) tin - this is the
reason for searching in tin_order[oi] and as a result tdiff can be <=0
and bigger than best_time.
best_time is initialised to the 'most early' time possible.
Kevin (awaits to be shot at :-) )
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