On Sun, Jun 15, 2008 at 8:54 PM, jiwen zhang <[EMAIL PROTECTED]> wrote:
> Hello Omprakash :
> I have some questions about the interface CtpInfo .
> Firstly , i am puzzled to the meaning of some fields in routetable.
> in the structure routing_table_entry (TreeRouting.h), there are some
> information of the current node . the field of parent in route_info_t should
> be the parent of one neighbor in the routetable , and what does the ETX mean
> ? is it the the neighbor's parent's EXT or the neighbor's the ETX ?(the EXT
> here should be the path metric to the root??) .
>
> i have seen the implementation of CtpInfo in CtpRoutingEngineP, and i have
> interest to the command which get the neighbors' information.
> command uint16_t CtpInfo.getNeighborLinkQuality(uint8_t n) {
> return (n < routingTableActive)? call
> LinkEstimator.getLinkQuality(routingTable[n].neighbor):0xffff;
> }
> command uint16_t CtpInfo.getNeighborRouteQuality(uint8_t n) {
> return (n < routingTableActive)? call
> LinkEstimator.getLinkQuality(routingTable[n].neighbor) +
> routingTable[n].info.etx:0xfffff;
> }
> in my opinion , the command of getNeighborLinkQuality returns a value of the
> quality between the current node and this neighbor in the routetable .
> if the field of ETX in route_info_t is the neighbor's EXT to the root , i
> think the command getNeighborRouteQuality returns a value of path metric
> between the current node and the root node through this neighbor (which is
> the sum of this neighbor's ETX to the root and the quility between the
> current node and the neighbor node).
>
> the command Ctp.getEtx(uint16_t* etx) will return the value of the path
> metric between the current node and the root through the node's parent. i
> find there are some differencs with the way in
> CtpInfo.getNeighborRouteQuality.
> *etx = routeInfo.etx + evaluateEtx(call
> LinkEstimator.getLinkQuality(routeInfo.parent));
> and the function evaluateEtx() is
> uint16_t evaluateEtx(uint16_t quality) {
> return (quality + 10);
> }
>
> i don't know whether i have expressed the problem distinctly , i think the
> two ways have the difference of "10" . am i right ,Omprakash ? can you give
> me an explain?
Yes. The algorithms use the metric ETX*10 but the metric is encoded as
(ETX-1)*10 in the packets.
- om_p
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