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Rewrote the parent selection to favour parents with a low ETX and a good rank. The old parent selection would always favour parents with a lower rank, regardless of their ETX, which in some cases lead to suboptimal parent selections. Also, the old code was buggy in that it would always pick the worst parent (the one with the highest ETX).

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This commit is contained in:
Adam Dunkels 2011-02-11 14:17:26 +01:00
parent a49c08caf4
commit ba6bf7dd6b
1 changed files with 26 additions and 6 deletions
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32
core/net/rpl/rpl-of0.c
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@ -46,6 +46,8 @@
#define DEBUG DEBUG_ANNOTATE
#include "net/uip-debug.h"
#include "net/neighbor-info.h"
static void reset(void *);
static rpl_parent_t *best_parent(rpl_parent_t *, rpl_parent_t *);
static rpl_rank_t calculate_rank(rpl_parent_t *, rpl_rank_t);
@ -82,6 +84,7 @@ calculate_rank(rpl_parent_t *p, rpl_rank_t base_rank)
return INFINITE_RANK;
}
return base_rank + DEFAULT_RANK_INCREMENT;
}
static rpl_parent_t *
@ -90,22 +93,39 @@ best_parent(rpl_parent_t *p1, rpl_parent_t *p2)
PRINTF("RPL: Comparing parent ");
PRINT6ADDR(&p1->addr);
PRINTF(" (confidence %d, rank %d) with parent ",
p1->local_confidence, p1->rank);
p1->etx, p1->rank);
PRINT6ADDR(&p2->addr);
PRINTF(" (confidence %d, rank %d)\n",
p2->local_confidence, p2->rank);
p2->etx, p2->rank);
/* Compare two parents by looking both and their rank and at the ETX
for that parent. We choose the parent that has the most
favourable combination. */
if(DAG_RANK(p1->rank, (rpl_dag_t *)p1->dag) * ETX_DIVISOR + p1->etx <
DAG_RANK(p2->rank, (rpl_dag_t *)p1->dag) * ETX_DIVISOR + p2->etx) {
return p1;
} else {
return p2;
}
/* This is the old code, which isn't used now, but left here in case
we would like to use it later (if the above code turns out to not
work as well as we expect it to. The old code first favoured the
parent with a lower rank, then used the ETX to compare two
parents with the same rank. This is not ideal since you may have
a parent with a low rank on the edge of your range that will have
a very bad ETX. But the code below would nevertheless pick that
one. */
if(p1->rank < p2->rank) {
return p1;
} else if(p2->rank < p1->rank) {
return p2;
}
if(p1->local_confidence > p2->local_confidence) {
if(p1->etx < p2->etx) {
return p1;
} else if(p2->local_confidence > p1->local_confidence) {
} else {
return p2;
}
return p2;
}