Better time estimate for stress test tasks.
diff --git a/commons-rng-examples/examples-stress/src/main/java/org/apache/commons/rng/examples/stress/StressTestCommand.java b/commons-rng-examples/examples-stress/src/main/java/org/apache/commons/rng/examples/stress/StressTestCommand.java
index 8780094..5de6a32 100644
--- a/commons-rng-examples/examples-stress/src/main/java/org/apache/commons/rng/examples/stress/StressTestCommand.java
+++ b/commons-rng-examples/examples-stress/src/main/java/org/apache/commons/rng/examples/stress/StressTestCommand.java
@@ -683,7 +683,8 @@
final int nthOldest = Math.max(0, id - parallelTasks + remainder);
final long endTime = startTimes[nthOldest] + taskTime;
// Note: The current time is the most recent entry in the startTimes array.
- millis += endTime - startTimes[id];
+ // Ensure the addition is positive in the case where the estimate is too low.
+ millis += Math.max(0, endTime - startTimes[id]);
}
return millis;
@@ -695,10 +696,34 @@
* @return the estimated task time
*/
private long getEstimatedTaskTime() {
- // Use the median. This is less sensitive to outliers than the average.
- // For example PractRand may fail very fast for bad generators and this
- // will skew the average.
- return sortedDurations[completed / 2];
+ // Return median of small lists. If no tasks have finished this returns zero.
+ // as the durations is zero initialised.
+ if (completed < 4) {
+ return sortedDurations[completed / 2];
+ }
+
+ // Dieharder and BigCrush run in approximately constant time.
+ // Speed varies with the speed of the RNG by about 2-fold, and
+ // for Dieharder it may repeat suspicious tests.
+ // PractRand may fail very fast for bad generators which skews
+ // using the mean or even the median. So look at the longest
+ // running tests.
+
+ // Find long running tests (>50% of the max run-time)
+ int upper = completed - 1;
+ final long halfMax = sortedDurations[upper] / 2;
+ // Binary search for the approximate cut-off
+ int lower = 0;
+ while (lower + 1 < upper) {
+ int mid = (lower + upper) >>> 1;
+ if (sortedDurations[mid] < halfMax) {
+ lower = mid;
+ } else {
+ upper = mid;
+ }
+ }
+ // Use the median of all tasks within approximately 50% of the max.
+ return sortedDurations[(lower + completed - 1) / 2];
}
/**
