SHA256
5.4 KiB
5.4 KiB
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?
Answer: ______
2. You normally write nine separate 0.7 KB items every second. How many WCUs are required?
Answer: ______
3. You normally write four separate 3.01 KB items every second. How many WCUs are required?
Answer: ______
4. You transactionally write three separate 2.01 KB items every second. How many WCUs are required?
Answer: ______
5. You transactionally write six separate 1 KB items every second. How many WCUs are required?
Answer: ______
6. You perform strongly consistent reads of five separate 5 KB items every second. How many RCUs of throughput are required?
Answer: ______
7. You perform strongly consistent reads of three separate 12 KB items every second. How many RCUs of throughput are required?
Answer: ______
8. You perform eventually consistent reads of nine separate 0.8 KB items every second. How many RCUs of throughput are required?
Answer: ______
9. You perform eventually consistent reads of five separate 8.01 KB items every second. How many RCUs of throughput are required?
Answer: ______
10. You perform eventually consistent reads of seven separate 8 KB items every second. How many RCUs of throughput are required?
Answer: ______
11. You transactionally read five separate 4.01 KB items every second. How many RCUs of throughput are required?
Answer: ______
12. You transactionally read three separate 11 KB items every second. How many RCUs of throughput are required?
Answer: ______
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?
Answer: ______
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?
Answer: ______
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?
Answer: ______
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?
Answer: ______
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?
Answer: ______
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?
Answer: ______
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?
Answer: ______
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?
Answer: ______
Answer key
Show answers and calculations
ceil means round up to the next whole number. Sizes are divided by the 1 KB write or 4 KB read increment before rounding.
- 12 WCUs —
ceil(1.8 / 1) × 6 = 2 × 6. - 9 WCUs —
ceil(0.7 / 1) × 9 = 1 × 9. - 16 WCUs —
ceil(3.01 / 1) × 4 = 4 × 4. - 18 WCUs —
ceil(2.01 / 1) × 3 × 2 = 3 × 3 × 2. - 12 WCUs —
ceil(1 / 1) × 6 × 2 = 1 × 6 × 2. - 10 RCUs —
ceil(5 / 4) × 5 = 2 × 5. - 9 RCUs —
ceil(12 / 4) × 3 = 3 × 3. - 4.5 RCUs —
ceil(0.8 / 4) × 9 × 0.5 = 1 × 9 × 0.5. - 7.5 RCUs —
ceil(8.01 / 4) × 5 × 0.5 = 3 × 5 × 0.5. - 7 RCUs —
ceil(8 / 4) × 7 × 0.5 = 2 × 7 × 0.5. - 20 RCUs —
ceil(4.01 / 4) × 5 × 2 = 2 × 5 × 2. - 18 RCUs —
ceil(11 / 4) × 3 × 2 = 3 × 3 × 2. - 11 WCUs —
(ceil(0.6 / 1) × 2) + (ceil(2.4 / 1) × 3) = 2 + 9. - 5.5 RCUs —
(ceil(5 / 4) × 2) + (ceil(1 / 4) × 3 × 0.5) = 4 + 1.5. - 7 writes/s —
28 / ceil(3.2 / 1) = 28 / 4. - 5 transactional item writes/s —
30 / (ceil(2.2 / 1) × 2) = 30 / 6. - 10 reads/s —
15 / (ceil(9 / 4) × 0.5) = 15 / 1.5. - 8 transactional item reads/s —
32 / (ceil(5 / 4) × 2) = 32 / 4. - 6 provisioned RCUs —
ceil(ceil(3 / 4) × 11 × 0.5) = ceil(5.5). - 10 provisioned RCUs —
ceil(ceil(12.01 / 4) × 5 × 0.5) = ceil(10).
For questions 1–14, RCUs describe exact throughput requirements. Provisioned settings must be whole numbers, as explicitly requested in questions 19–20.