In FusionInsight HD, assuming a topology, spout concurrency is set to 3, bolt1 concurrency is 2, bolt2 concurrency is 3, and the number of workers is 2, then how are bolt1's executors distributed…

Question

Options

• AOne of the two workers is distributed to two
• BTwo workers each distribute one
• CThere is no fixed pattern
• DTwo workers are distributed

• A
  7% (2)
• B
  79% (23)
• C
  10% (3)
• D
  3% (1)

Explanation

Option B is correct because Apache Storm (the engine underlying FusionInsight HD) distributes executors across workers as evenly as possible using its default even-round-robin scheduler. With bolt1 having a concurrency of 2 executors and exactly 2 workers available, each worker receives exactly 1 executor - a clean 1-per-worker split.

Why the distractors are wrong:

• A ("one worker gets two") describes an uneven distribution, which only happens when executors outnumber workers (e.g., 3 executors across 2 workers), not in a 2-into-2 case.
• C ("no fixed pattern") is incorrect - Storm's scheduler is deterministic and follows a defined even-distribution algorithm, not random assignment.
• D ("two workers are distributed") is grammatically ambiguous but implies both executors go to one worker, which contradicts Storm's load-balancing behavior.

Memory tip: Think of Storm's scheduler as a card dealer - it deals executors to workers one at a time in round-robin order. So 2 executors + 2 workers = 1 card each, always evenly dealt. When executors divide evenly into workers, distribution is perfectly balanced; the uneven case (like 3 executors across 2 workers → 2+1) is the exception, not the rule.

No community discussion yet for this question.