In the experiments so far, we have run each heuristic for a total of 11 trials and found that when given an unbounded computational budget, DBHC outperformed SDHC on average for some instances, and vice versa for other instances.
When comparing the hill-climbing heuristics for solving MAX-SAT instance #11 for a total of 5 seconds , which allows enough time for both methods to reach local optima in all trials, we find that on average DBHC outperforms SDHC.
T ry running these experiments again, but this time increase the “TRIALS_PER_TEST” configuration variable in Lab2ExercisesTestFrameConfig to increase the sample size used in the experimentation. Which of the following observations can you now conclude?

Question

In the experiments so far, we have run each heuristic for a total of 11 trials and found that when given an unbounded computational budget, DBHC outperformed SDHC on average for some instances, and vice versa for other instances. 
 When comparing the hill-climbing heuristics for solving  MAX-SAT instance #11  for a total of  5 seconds , which allows enough time for both methods to reach local optima in all trials, we find that on average DBHC outperforms SDHC. 
 T ry running these experiments again, but this time increase the “TRIALS_PER_TEST” configuration variable in Lab2ExercisesTestFrameConfig to increase the sample size used in the experimentation. Which of the following observations can you now conclude?

Accepted Answer

SDHC outperforms DBHC.

Answer

Neither heuristic outperforms each other.

Answer

The performance of a heuristic depends on the number of trials.

Answer

DBHC outperforms SDHC.

Crowdly

In the experiments so far, we have run each heuristic for a total of 11 trials a...

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