# DynamoDB Read/Write Units — Numeric Practice, Set 3

20 questions. Give a numeric answer for each, using `1 = <number>`, `2 = <number>`, and so on.

## Assumptions

- Assume provisioned capacity and steady workloads.
- All item sizes are exact DynamoDB item sizes, including attribute names and values.
- Each read accesses an individual item, not a Query or Scan. Transactional operations count individual items, not transaction API calls.
- Normal writes are non-transactional. All writes create new items.
- Ignore indexes, replication, retries, failed requests, burst capacity and partition limits.
- Questions 1–14 ask for exact throughput requirements: fractional RCU answers are valid; do not round the final result to a whole number.
- Questions 15–18 ask for the maximum number of individual item operations per second.
- Questions 19–20 ask for whole-number provisioned capacity settings.

## Questions
### 1. You normally write six separate 1.8 KB items every second. How many WCUs are required?
Writing to DynamoDB splits writes into 1KB blocks. 
$2KB \times 6Hz = 12WCU$

Answer: 12 WCUs

### 2. You normally write nine separate 0.7 KB items every second. How many WCUs are required?
$1KB \times 9Hz = 9WCU$

Answer: 9 WCUs

### 3. You normally write four separate 3.01 KB items every second. How many WCUs are required?
$4KB \times 4Hz = 16WCU$

Answer: 16 WCUs

### 4. You transactionally write three separate 2.01 KB items every second. How many WCUs are required?
Transactional writes have $\times 2$ multiplier.
$3KB \times 3Hz \times 2 (transaction) = 18WCU$

Answer: 18WCUs

### 5. You transactionally write six separate 1 KB items every second. How many WCUs are required?
$1KB \times 6Hz \times 2(transaction) = 12WCU$

Answer: 12WCUs

### 6. You perform strongly consistent reads of five separate 5 KB items every second. How many RCUs of throughput are required?
Reading from DynamoDB splits reads into 4KB blocks.
$ceil(5KB/4KB) \times 5Hz = 2 \times 5Hz = 10RCU$

Answer: 10RCUs

### 7. You perform strongly consistent reads of three separate 12 KB items every second. How many RCUs of throughput are required?
$ceil(12KB/4KB) \times 3Hz = 3 \times 3Hz = 9RCU$

Answer: 9RCUs

### 8. You perform eventually consistent reads of nine separate 0.8 KB items every second. How many RCUs of throughput are required?
Eventually consistent reads have $\times 0.5$ multiplier.
$ceil(1KB/4KB) \times 9Hz \times 0.5(eventual) = 1 \times 9Hz \times 0.5 = 4.5RCU$

Answer: 4.5RCUs

### 9. You perform eventually consistent reads of five separate 8.01 KB items every second. How many RCUs of throughput are required?
$ceil(8.01KB/4KB) \times 5Hz \times 0.5(eventual)$
$= 3 \times 5Hz \times 0.5$
$= 15 \times 0.5$
$\therefore\ = 7.5 RCU$

Answer: 7.5RCUs

### 10. You perform eventually consistent reads of seven separate 8 KB items every second. How many RCUs of throughput are required?
$ceil(8KB/4KB) \times 7Hz \times 0.5(eventual) = 2 \times 7Hz \times 0.5 = 7RCU$

Answer: 7RCUs

### 11. You transactionally read five separate 4.01 KB items every second. How many RCUs of throughput are required?
Transactional reads have $\times 2$ multiplier.
$ceil(4.01KB/4KB) \times 5Hz \times 2(transaction) = 2 \times 5Hz \times 2 = 20 RCU$

Answer: 20RCUs

### 12. You transactionally read three separate 11 KB items every second. How many RCUs of throughput are required?
$ceil(11KB/4KB) \times 3Hz \times 2(transaction) =$
$=3 \times 3Hz \times 2$
$=18 RCU$

Answer: 18RCUs

### 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?
$ceil(0.6KB/1KB) \times 2Hz + ceil(2.4KB/1KB) \times 3Hz$
$= 1 \times 2Hz + 3 \times 3Hz = 11WCU$

Answer: 11WCUs

### 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?
$ceil(5KB/4KB) \times 2Hz + ceil(1KB/4KB) \times 3Hz \times 0.5(eventual)$
$= 2 \times 2 + 1 \times 3 \times 0.5 = 4 + 1.5 = 5.5 RCU$

Answer: 5.5RCU

### 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?
$floor(28 WCU / ceil(3.2KB/1KB))$ 
$= floor(28 WCU / 4 WCU) = 7Hz$
$\therefore$ 7 writes a second of 3.2KB items (up to 4KB @ 7Hz).

Answer: 7 writes a second.

### 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?
$floor(30WCU/ceil(2.2KB/1KB)/2(transaction))$
$=30WCU/3/2$
$=30WCU/6$
$=5Hz$
$\therefore$ 5 writes a second of 2.2KB items (up to 3KB @ 5Hz).

Answer: 5 writes a second.

### 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?
$floor(15RCU/ceil(9KB/4KB)/0.5(eventual))$
$=floor(15RCU/3/0.5)$
$=floor(5/0.5)$
$=floor(10)$
$=10Hz$
$\therefore$ 10 reads a second of 9KB items (up to 12KB @ 10Hz).

Answer: 10 reads a second

### 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?
$floor(32RCU/ceil(5KB/4KB)/2(transaction))$
$=floor(32/2/2)$
$=32 / 4$
$= 8Hz$

Answer: 8 reads a second

### 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?
$ceil(3KB/4KB) \times 11Hz \times 0.5(eventual)$
$= 1 \times 11 \times 0.5 = 5.5RCU$

Answer: 6 RCUs

### 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?
$ceil(12.01KB/4KB) \times 5Hz \times 0.5(eventual)$
$= 4 \times 2.5$
$=10 RCU$

Answer: 10 RCU