[PATCH 0/8] Implement DDB algorithm and WM cleanup

Mahesh Kumar mahesh1.kumar at intel.com
Mon Feb 27 09:07:51 UTC 2017


This series implements new DDB allocation algorithm to solve the cases,
where we have sufficient DDB available to enable multiple planes, But
due to the current algorithm not dividing it properly among planes, we
end-up failing the flip.
It also takes care of enabling same watermark level for each
plane, for efficient power saving.
This series also implements Transition Watermarks and Gen-9 related
arbitrated display bandwidth Workarounds.

There are two steps in current WM programming.

1. Calculate minimum number of blocks required  for a WM level to be
enabled. For 1440x2560 panel we need 41 blocks as minimum number of
blocks to enable WM0. This is the step which doesn't use vertical size.
It only depends on Pipe drain rate and plane horizontal size as per the
current Bspec algorithm.
So all the plane below have minimum  number of blocks required to enable
WM0 as 41
    Plane 1  - 1440x2560        -    Min blocks to enable WM0 = 41
    Plane 2  - 1440x2560        -    Min blocks to enable WM0 = 41
    Plane 3  - 1440x48          -    Min blocks to enable WM0 = 41
    Plane 4  - 1440x96          -    Min blocks to enable WM0 = 41

2. Number of blocks allotted by the driver
    Driver allocates  12 for Plane 3   &  16 for plane 4

    Total Dbuf Available = 508
    Dbuf Available after 32 blocks for cursor = 508 - (32)  = 476
    allocate minimum blocks for each plane 8 * 4 = 32
    remaining blocks = 476 - 32 = 444
    Relative Data Rate for Planes
       Plane 1  =  1440 * 2560 * 3  =  11059200
       Plane 2  =  1440 * 2560 * 3  =  11059200
       Plane 3  =  1440 * 48   * 3  =  207360
       Plane 4  =  1440 * 96   * 3  =  414720
       Total Relative BW            =  22740480

-   Allocate Buffer
    buffer allocation = (Plane relative data rate / total data rate)
		    * total remaming DDB + minimum plane DDB
     Plane 1  buffer allocation = (11059200 / 22740480) * 444 + 8 = 223
     Plane 2  buffer allocation = (11059200 / 22740480) * 444 + 8 = 223
     Plane 3  buffer allocation = (207360   / 22740480) * 444 + 8 = 12
     Plane 4  buffer allocation = (414720   / 22740480) * 444 + 8 = 16

In this case it forced driver to disable Plane 3 & 4. Driver need to use
more efficient way to allocate buffer that is optimum for power.

New Algorithm suggested by HW team is:

1. Calculate minimum buffer allocations for each plane and for each
    watermark level

2. Add minimum buffer allocations required for enabling WM7
    for all the planes

Level 0 =  41 + 41 + 41 + 41  = 164
Level 1 =  42 + 42 + 42 + 42  = 168
Level 2 =  42 + 42 + 42 + 42  = 168
Level 3 =  94 + 94 + 94 + 94 =  376
Level 4 =  94 + 94 + 94 + 94 =  376
Level 5 =  94 + 94 + 94 + 94 =  376
Level 6 =  94 + 94 + 94 + 94 =  376
Level 7 =  94 + 94 + 94 + 94 =  376

3. Check to see how many buffer allocation are left and enable
the best case. In this case since we have 476 blocks we can enable
WM0-7 on all 4 planes.
Let's say if we have only 200 block available then the best cases
allocation is to enable Level2 which requires 168 blocks

Mahesh Kumar (8):
  drm/i915/skl+: calculate pixel_rate & relative_data_rate in fixed
    point
  drm/i915/skl+: use linetime latency if ddb size is not available
  drm/i915/skl: Fail the flip if no FB for WM calculation
  drm/i915/skl+: no need to memset again
  drm/i915/skl+: ddb min requirement may exceed allocation
  drm/i915/skl+: Watermark calculation cleanup
  drm/i915/skl: New ddb allocation algorithm
  drm/i915/skl+: consider max supported plane pixel rate while scaling

 drivers/gpu/drm/i915/i915_drv.h      |  52 +++-
 drivers/gpu/drm/i915/intel_display.c |   3 +
 drivers/gpu/drm/i915/intel_drv.h     |   2 +
 drivers/gpu/drm/i915/intel_pm.c      | 511 +++++++++++++++++++++++------------
 4 files changed, 384 insertions(+), 184 deletions(-)

-- 
2.11.0



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