prototype: supply 1.25 credits for tiering merges

This is necessary to accomodate 5-way merges due to holding back runs
that are too small.

Author: mheinzel

prototypes/ScheduledMerges.hs

@@ -466,20 +466,24 @@ bufferToRun = id
 supplyCredits :: Credit -> Levels s -> ST s ()
 supplyCredits n ls =
   sequence_
-    [ supplyMergeCredits (n * creditsForMerge mr) mr | Level mr _rs <- ls ]
+    [ supplyMergeCredits (ceiling (fromIntegral n * creditsForMerge mr)) mr
+    | Level mr _rs <- ls
+    ]
 
 -- | The general case (and thus worst case) of how many merge credits we need
 -- for a level. This is based on the merging policy at the level.
 --
-creditsForMerge :: MergingRun s -> Credit
+creditsForMerge :: MergingRun s -> Float
 creditsForMerge SingleRun{}                           = 0
 
--- A levelling merge is 5x the cost of a tiering merge.
--- That's because for levelling one of the runs as an input to the merge
--- is the one levelling run which is (up to) 4x bigger than the others put
--- together, so it's 1 + 4.
-creditsForMerge (MergingRun MergePolicyLevelling _ _) = 5
-creditsForMerge (MergingRun MergePolicyTiering   _ _) = 1
+-- A levelling merge has 1 input run and one resident run, which is (up to) 4x
+-- bigger than the others.
+-- It needs to be completed before another run comes in.
+creditsForMerge (MergingRun MergePolicyLevelling _ _) = (1 + 4) / 1
+
+-- A tiering merge has 5 runs at most (once could be held back to merged again)
+-- and must be completed before the level is full (once 4 more runs come in).
+creditsForMerge (MergingRun MergePolicyTiering   _ _) = 5 / 4
 
 type Event = EventAt EventDetail
 data EventAt e = EventAt {