huyuanfeng2018 commented on code in PR #879:
URL: 
https://github.com/apache/flink-kubernetes-operator/pull/879#discussion_r1756220655


##########
flink-autoscaler/src/main/java/org/apache/flink/autoscaler/JobVertexScaler.java:
##########
@@ -378,28 +405,70 @@ protected static int scale(
 
         // Cap parallelism at either maxParallelism(number of key groups or 
source partitions) or
         // parallelism upper limit
-        final int upperBound = Math.min(maxParallelism, parallelismUpperLimit);
+        int upperBound = Math.min(maxParallelism, parallelismUpperLimit);
 
         // Apply min/max parallelism
         newParallelism = Math.min(Math.max(parallelismLowerLimit, 
newParallelism), upperBound);
 
         var adjustByMaxParallelism =
                 inputShipStrategies.isEmpty() || 
inputShipStrategies.contains(HASH);
         if (!adjustByMaxParallelism) {
-            return newParallelism;
+            return Tuple2.of(newParallelism, Optional.empty());
         }
 
-        // When the shuffle type of vertex inputs contains keyBy or vertex is 
a source, we try to
-        // adjust the parallelism such that it divides the maxParallelism 
without a remainder
-        // => data is evenly spread across subtasks
-        for (int p = newParallelism; p <= maxParallelism / 2 && p <= 
upperBound; p++) {
-            if (maxParallelism % p == 0) {
-                return p;
+        if (numPartitions <= 0) {
+            // When the shuffle type of vertex inputs contains keyBy or vertex 
is a source,
+            // we try to adjust the parallelism such that it divides the 
maxParallelism without a
+            // remainder => data is evenly spread across subtasks
+            for (int p = newParallelism; p <= maxParallelism / 2 && p <= 
upperBound; p++) {
+                if (maxParallelism % p == 0) {
+                    return Tuple2.of(p, Optional.empty());
+                }
+            }
+            // If parallelism adjustment fails, use originally computed 
parallelism
+            return Tuple2.of(newParallelism, Optional.empty());
+        } else {
+
+            // When we know the numPartitions at a vertex,
+            // adjust the parallelism such that it divides the numPartitions 
without a remainder
+            // => Data is evenly distributed among subtasks
+            for (int p = newParallelism; p <= upperBound && p <= 
numPartitions; p++) {
+                if (numPartitions % p == 0) {
+                    return Tuple2.of(p, Optional.empty());
+                }
             }
-        }
 
-        // If parallelism adjustment fails, use originally computed parallelism
-        return newParallelism;
+            // When the degree of parallelism after rounding up cannot be 
evenly divided by source
+            // PartitionCount, Try to find the smallest parallelism that can 
satisfy the current
+            // consumption rate.
+            for (int p = newParallelism; p > parallelismLowerLimit; p--) {
+                if (numPartitions / p > numPartitions / newParallelism) {
+                    if (numPartitions % p != 0) {
+                        p += 1;
+                    }

Review Comment:
   > Thanks for explaining in detail. I misread some of the code. It is correct 
that we need to add +1 when we have found a parallelism which yields a greater 
value for `num_partitions / p` than the initial `num_partitions / 
new_parallelism` because we have found the tipping point where we achieve the 
most utilization in terms of partitions per task.
   > 
   > I think we should return `new_parallelism` if all adaptation logic fails 
because using a potentially very small configured lower parallelism could make 
things a lot worse due to resource constraints.
   
   I want to explain the reason for using `parallelismLowerLimit`,  example:
   
   `numPartitions=35 ,newParallelism=24, upperBound = 30, parallelismLowerLimit 
= 19`
   
   Step1 cannot get a result, so it goes to step2, but step2 still cannot get a 
result because parallelismLowerLimit = 19 and the expected value of step2 is 
18, so it will eventually return 19
   
   



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