On Wed, Oct 11, 2023 at 08:00:22PM +0800, Abel Wu wrote:
On 5/31/23 7:58 PM, Peter Zijlstra Wrote:
/*
+ * If we want to place a task and preserve lag, we have to
+ * consider the effect of the new entity on the weighted
+ * average and compensate for this, otherwise lag can quickly
+ * evaporate.
+ *
+ * Lag is defined as:
+ *
+ * lag_i = S - s_i = w_i * (V - v_i)
+ *
+ * To avoid the 'w_i' term all over the place, we only track
+ * the virtual lag:
+ *
+ * vl_i = V - v_i <=> v_i = V - vl_i
+ *
+ * And we take V to be the weighted average of all v:
+ *
+ * V = (\Sum w_j*v_j) / W
+ *
+ * Where W is: \Sum w_j
+ *
+ * Then, the weighted average after adding an entity with lag
+ * vl_i is given by:
+ *
+ * V' = (\Sum w_j*v_j + w_i*v_i) / (W + w_i)
+ * = (W*V + w_i*(V - vl_i)) / (W + w_i)
+ * = (W*V + w_i*V - w_i*vl_i) / (W + w_i)
+ * = (V*(W + w_i) - w_i*l) / (W + w_i)
+ * = V - w_i*vl_i / (W + w_i)
+ *
+ * And the actual lag after adding an entity with vl_i is:
+ *
+ * vl'_i = V' - v_i
+ * = V - w_i*vl_i / (W + w_i) - (V - vl_i)
+ * = vl_i - w_i*vl_i / (W + w_i)
+ *
+ * Which is strictly less than vl_i. So in order to preserve lag
Maybe a stupid question, but why vl'_i < vl_i? Since vl_i can be negative.
So the below doesn't care about the sign, it simply inverts this
relation to express vl_i in vl'_i:
+ * we should inflate the lag before placement such that the
+ * effective lag after placement comes out right.
+ *
+ * As such, invert the above relation for vl'_i to get the vl_i
+ * we need to use such that the lag after placement is the lag
+ * we computed before dequeue.
+ *
+ * vl'_i = vl_i - w_i*vl_i / (W + w_i)
+ * = ((W + w_i)*vl_i - w_i*vl_i) / (W + w_i)
+ *
+ * (W + w_i)*vl'_i = (W + w_i)*vl_i - w_i*vl_i
+ * = W*vl_i
+ *
+ * vl_i = (W + w_i)*vl'_i / W
And then we obtain the scale factor: (W + w_i)/W, which is >1, right?
As such, that means that vl'_i must be smaller than vl_i in the absolute
sense, irrespective of sign.