Re: [RFC Patch net-next 3/3] net: phy: lan87xx: added ethtool SQI support

[Andrew Lunn]

On Thu, Mar 24, 2022 at 03:48:57PM +0000, Arun.Ramadoss@xxxxxxxxxxxxx wrote:
> Hi Andrew,
> 
> Thanks for the review.
> 
> On Mon, 2022-03-21 at 19:36 +0100, Andrew Lunn wrote:
> > EXTERNAL EMAIL: Do not click links or open attachments unless you
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> > 
> > > +#define T1_DCQ_SQI_MSK                       GENMASK(3, 1)
> > > +static int lan87xx_get_sqi(struct phy_device *phydev)
> > > +{
> > > +     u16 sqi_value[LAN87XX_SQI_ENTRY];
> > > +     for (i = 0; i < LAN87XX_SQI_ENTRY; i++) {
> > > +
> > > +             sqi_value[i] = FIELD_GET(T1_DCQ_SQI_MSK, rc);
> > > +
> > > +     /* Sorting SQI values */
> > > +     sort(sqi_value, LAN87XX_SQI_ENTRY, sizeof(u16),
> > > lan87xx_sqi_cmp, NULL);
> > 
> > Sort is quite heavyweight. Your SQI values are in the range 0-7
> > right?
> > So rather than have an array of LAN87XX_SQI_ENTRY entries, why not
> > create a histogram? You then just need to keep 8 uints. There is no
> > need to perform a sort to discard the outliers, simply remove from
> > the
> > outer histogram buckets. And then you can calculate the average.
> > 
> > That should be faster and use less memory.
> > 
> >      Andrew
> 
> I could get the algorithm for replacing array of LAN87XX_SQI_ENTRY(200)
> to array of 8 (sqi values 0 to 7) and increment the array[sqi_value]
> for every reading. And calculate the Average = ( 1 * array[1] + 2 *
> array[2] ... + 7 * array[7])/LAN87XX_SQI_ENTRY. By this way we get the
> average for 200 entries.
> But I couldn't get the algorithm on how to discard the outliers from
> the buckets. our main aim is to average from array[40] to arrary[160]
> value. Can you bit elaborate on how to remove the outer histogram
> buckets.

So your raw results look something like

array[0] = 10
array[1] = 10
array[2] = 25
array[3] = 100
array[4] = 50
array[5] = 1
array[6] = 4
array[7] = 0

To discard the lower outliers, take 40 away from the array[0],
array[1], array[2], etc. To discard the upper outliers, take 40 away
from array[7], array[6], array[5], etc. So you should end up with:

array[0] = 0
array[1] = 0
array[2] = 5
array[3] = 100
array[4] = 15
array[5] = 0
array[6] = 0
array[7] = 0

and then calculate the average: (2*5 + 3*100 + 4*15) / 120 = 3.

    Andrew