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# DynamoDB Read/Write Units — Numeric Practice, Set 3
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20 questions. Give a numeric answer for each, using `1 = <number>`, `2 = <number>`, and so on.
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## Assumptions
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- Assume provisioned capacity and steady workloads.
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- All item sizes are exact DynamoDB item sizes, including attribute names and values.
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- Each read accesses an individual item, not a Query or Scan. Transactional operations count individual items, not transaction API calls.
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- Normal writes are non-transactional. All writes create new items.
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- Ignore indexes, replication, retries, failed requests, burst capacity and partition limits.
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- Questions 1–14 ask for exact throughput requirements: fractional RCU answers are valid; do not round the final result to a whole number.
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- Questions 15–18 ask for the maximum number of individual item operations per second.
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- Questions 19–20 ask for whole-number provisioned capacity settings.
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## Questions
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### 1. You normally write six separate 1.8 KB items every second. How many WCUs are required?
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Writing to DynamoDB splits writes into 1KB blocks.
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$2KB \times 6Hz = 12WCU$
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Answer: 12 WCUs
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### 2. You normally write nine separate 0.7 KB items every second. How many WCUs are required?
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$1KB \times 9Hz = 9WCU$
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Answer: 9 WCUs
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### 3. You normally write four separate 3.01 KB items every second. How many WCUs are required?
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$4KB \times 4Hz = 16WCU$
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Answer: 16 WCUs
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### 4. You transactionally write three separate 2.01 KB items every second. How many WCUs are required?
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Transactional writes have $\times 2$ multiplier.
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$3KB \times 3Hz \times 2 (transaction) = 18WCU$
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Answer: 18WCUs
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### 5. You transactionally write six separate 1 KB items every second. How many WCUs are required?
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$1KB \times 6Hz \times 2(transaction) = 12WCU$
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Answer: 12WCUs
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### 6. You perform strongly consistent reads of five separate 5 KB items every second. How many RCUs of throughput are required?
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Reading from DynamoDB splits reads into 4KB blocks.
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$ceil(5KB/4KB) \times 5Hz = 2 \times 5Hz = 10RCU$
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Answer: 10RCUs
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### 7. You perform strongly consistent reads of three separate 12 KB items every second. How many RCUs of throughput are required?
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$ceil(12KB/4KB) \times 3Hz = 3 \times 3Hz = 9RCU$
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Answer: 9RCUs
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### 8. You perform eventually consistent reads of nine separate 0.8 KB items every second. How many RCUs of throughput are required?
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Eventually consistent reads have $\times 0.5$ multiplier.
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$ceil(1KB/4KB) \times 9Hz \times 0.5(eventual) = 1 \times 9Hz \times 0.5 = 4.5RCU$
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Answer: 4.5RCUs
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### 9. You perform eventually consistent reads of five separate 8.01 KB items every second. How many RCUs of throughput are required?
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$ceil(8.01KB/4KB) \times 5Hz \times 0.5(eventual)$
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$= 3 \times 5Hz \times 0.5$
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$= 15 \times 0.5$
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$\therefore\ = 7.5 RCU$
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Answer: 7.5RCUs
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### 10. You perform eventually consistent reads of seven separate 8 KB items every second. How many RCUs of throughput are required?
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$ceil(8KB/4KB) \times 7Hz \times 0.5(eventual) = 2 \times 7Hz \times 0.5 = 7RCU$
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Answer: 7RCUs
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### 11. You transactionally read five separate 4.01 KB items every second. How many RCUs of throughput are required?
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Transactional reads have $\times 2$ multiplier.
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$ceil(4.01KB/4KB) \times 5Hz \times 2(transaction) = 2 \times 5Hz \times 2 = 20 RCU$
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Answer: 20RCUs
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### 12. You transactionally read three separate 11 KB items every second. How many RCUs of throughput are required?
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$ceil(11KB/4KB) \times 3Hz \times 2(transaction) =$
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$=3 \times 3Hz \times 2$
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$=18 RCU$
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Answer: 18RCUs
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### 13. Each second, you normally write two 0.6 KB items and three 2.4 KB items to the same table. How many WCUs are required in total?
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$ceil(0.6KB/1KB) \times 2Hz + ceil(2.4KB/1KB) \times 3Hz$
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$= 1 \times 2Hz + 3 \times 3Hz = 11WCU$
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Answer: 11WCUs
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### 14. Each second, you perform strongly consistent reads of two 5 KB items and eventually consistent reads of three 1 KB items from the same table. How many RCUs of throughput are required in total?
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$ceil(5KB/4KB) \times 2Hz + ceil(1KB/4KB) \times 3Hz \times 0.5(eventual)$
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$= 2 \times 2 + 1 \times 3 \times 0.5 = 4 + 1.5 = 5.5 RCU$
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Answer: 5.5RCU
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### 15. You have 28 WCUs available. What is the maximum number of normal item writes per second you can sustain if each item is 3.2 KB?
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$floor(28 WCU / ceil(3.2KB/1KB))$
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$= floor(28 WCU / 4 WCU) = 7Hz$
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$\therefore$ 7 writes a second of 3.2KB items (up to 4KB @ 7Hz).
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Answer: 7 writes a second.
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### 16. You have 30 WCUs available. What is the maximum number of transactional item writes per second you can sustain if each item is 2.2 KB?
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$floor(30WCU/ceil(2.2KB/1KB)/2(transaction))$
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$=30WCU/3/2$
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$=30WCU/6$
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$=5Hz$
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$\therefore$ 5 writes a second of 2.2KB items (up to 3KB @ 5Hz).
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Answer: 5 writes a second.
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### 17. You have 15 RCUs available. What is the maximum number of eventually consistent item reads per second you can sustain if each item is 9 KB?
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$floor(15RCU/ceil(9KB/4KB)/0.5(eventual))$
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$=floor(15RCU/3/0.5)$
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$=floor(5/0.5)$
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$=floor(10)$
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$=10Hz$
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$\therefore$ 10 reads a second of 9KB items (up to 12KB @ 10Hz).
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Answer: 10 reads a second
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### 18. You have 32 RCUs available. What is the maximum number of transactional item reads per second you can sustain if each item is 5 KB?
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$floor(32RCU/ceil(5KB/4KB)/2(transaction))$
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$=floor(32/2/2)$
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$=32 / 4$
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$= 8Hz$
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Answer: 8 reads a second
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### 19. You perform eventually consistent reads of eleven separate 3 KB items every second. What is the minimum WHOLE NUMBER of RCUs you must provision?
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$ceil(3KB/4KB) \times 11Hz \times 0.5(eventual)$
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$= 1 \times 11 \times 0.5 = 5.5RCU$
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Answer: 6 RCUs
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### 20. You perform eventually consistent reads of five separate 12.01 KB items every second. What is the minimum WHOLE NUMBER of RCUs you must provision?
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$ceil(12.01KB/4KB) \times 5Hz \times 0.5(eventual)$
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$= 4 \times 2.5$
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$=10 RCU$
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Answer: 10 RCU
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