[Intel-gfx] [PATCH v2 0/9] New DDB Algo and WM fixes
Kumar, Mahesh
mahesh1.kumar at intel.com
Thu Sep 8 11:26:25 UTC 2016
From: Mahesh Kumar <mahesh1.kumar at intel.com>
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
Changes since v1:
- Rebased the series on top of Paulo's patches
- Add changes to calculate WM in fixed point 16.16
- Address review comments for Transition WM
Mahesh Kumar (9):
drm/i915/skl: pass pipe_wm in skl_compute_(wm_level/plane_wm)
functions
drm/i915/skl+: use linetime latency instead of ddb size
drm/i915/skl: New ddb allocation algorithm
drm/i915: Decode system memory bandwidth
drm/i915/gen9: WM memory bandwidth related workaround
drm/i915/skl+: change WM calc to fixed point 16.16
drm/i915/bxt: Enable IPC support
drm/i915/bxt: set chicken bit as IPC y-tile WA
drm/i915/bxt: Implement Transition WM
drivers/gpu/drm/i915/i915_drv.c | 101 +++++++++
drivers/gpu/drm/i915/i915_drv.h | 29 +++
drivers/gpu/drm/i915/i915_params.c | 5 +
drivers/gpu/drm/i915/i915_params.h | 1 +
drivers/gpu/drm/i915/i915_reg.h | 29 +++
drivers/gpu/drm/i915/intel_display.c | 50 +++++
drivers/gpu/drm/i915/intel_drv.h | 12 +
drivers/gpu/drm/i915/intel_pm.c | 412 +++++++++++++++++++++++++++--------
8 files changed, 551 insertions(+), 88 deletions(-)
--
2.8.3
More information about the Intel-gfx
mailing list