Re: [net-next PATCH 28/29] fm10k: Add support for ptp to hw specific files

[Alexander Duyck <alexander.h.duyck@intel.com>]

On 09/19/2014 12:38 AM, Richard Cochran wrote:
> On Thu, Sep 18, 2014 at 06:40:30PM -0400, Alexander Duyck wrote:
> 
>> +static s32 fm10k_adjust_systime_pf(struct fm10k_hw *hw, s32 ppb)
>> +{
>> +	u64 systime_adjust;
>> +
>> +	/* if sw_addr is not set we don't have switch register access */
>> +	if (!hw->sw_addr)
>> +		return ppb ? FM10K_ERR_PARAM : 0;
>> +
>> +	/* we must convert the value from parts per billion to parts per
>> +	 * 2^48 cycles.  In addition we can only use the upper 30 bits of
>> +	 * the value when making the change so that restricts us futher.
>> +	 * The math for this is equivilent to ABS(pps) * 2^40 / 10 ^ 9,
> 
> Huh? Converting ppb to parts per 2^48 should be (ppb * 2^48 / 10^9),
> shouldn't it?
> 
> Did you mean, "we must convert ... to parts per 2^40 cycles" ?
> 
> Also, the comment about using the upper 30 bits is not clear. Do you
> mean that the hardware ignores the two least significant bits?

So to break it all down.

1.  The hardware provides a value that can be specified in parts per
2^48, however that value spans across 2 registers starting at bit 24 in
the first register.  You can actually make out a bit more if you look at
the end of the patch.  The layout for the registers look something like
this:

 /* Registers contained in BAR 4 for Switch management */
 #define FM10K_SW_SYSTIME_CFG	0x0224C
 #define FM10K_SW_SYSTIME_CFG_ADJUST_MASK	0xFF000000
 #define FM10K_SW_SYSTIME_ADJUST	0x0224D
 #define FM10K_SW_SYSTIME_ADJUST_MASK		0x3FFFFFFF
 #define FM10K_SW_SYSTIME_ADJUST_DIR_NEGATIVE	0x80000000

2.  The value is in 2 ^ 48 units, and the OS wants to give us a value in
parts per billion, not parts per 256 * 2 ^ 40.  So in order to simplify
things I dropped the lower 8 bits and only update
FM10K_SW_SYSTIME_ADJUST register giving me an adjust value multiplied by
256.   So the adjustment value I am using now is:
	(256 / (256 * (2 ^ 40)))
		this simplifies down to
	1 / (2 ^ 40)

The value 1 / (2 ^ 40) implies that we are still looking at something
close to parts per trillion, however since I am now only adjusting one
register instead of 2 it meets my needs since those lower 8 bits are an
even smaller part of the whole.

3.  The last bit in the adjustment is to deal with the the fact that we
have a power of 2, not a power of 10 and the adjustments are in parts
per billion.  So we would need to convert by 1024 ^ 3 / 1000 ^ 3.  So
the whole thing broken down would be:
	value / (10 ^ 9) == adj_val / (2 ^ 40)
		so if I solve for adj_val
	value * (2 ^ 40) / (10 ^ 9) == adj_val
		then reduce it by dropping the shared values of 2
	value * (2 ^ 31) / (5 ^ 9) == adj_val

Hopefully that explains how I came to that value.

Thanks,

Alex