From c05aa606e0196eee6afb1e14a88c5306ea14fd15449c67bc0cbb4e219bae9151 Mon Sep 17 00:00:00 2001 From: Baldyga Date: Mon, 2 Nov 2026 15:39:36 +0000 Subject: [PATCH] did test --- .../Practice/Read and Write Units - Set 3.md | 121 ++++++++++-------- 1 file changed, 69 insertions(+), 52 deletions(-) diff --git a/DynamoDB/Practice/Read and Write Units - Set 3.md b/DynamoDB/Practice/Read and Write Units - Set 3.md index 57d423a..17232ea 100644 --- a/DynamoDB/Practice/Read and Write Units - Set 3.md +++ b/DynamoDB/Practice/Read and Write Units - Set 3.md @@ -14,115 +14,132 @@ - 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. +$1KB \times 6Hz = 12WCU$ -Answer: ______ +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: ______ +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: ______ +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: ______ +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: ______ +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: ______ +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: ______ +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: ______ +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: ______ +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: ______ +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: ______ +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: ______ +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: ______ +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: ______ +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: ______ +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: ______ +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: ______ +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: ______ +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: ______ +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: ______ +Answer: 10 RCU -## 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. - -1. **12 WCUs** — `ceil(1.8 / 1) × 6 = 2 × 6`. -2. **9 WCUs** — `ceil(0.7 / 1) × 9 = 1 × 9`. -3. **16 WCUs** — `ceil(3.01 / 1) × 4 = 4 × 4`. -4. **18 WCUs** — `ceil(2.01 / 1) × 3 × 2 = 3 × 3 × 2`. -5. **12 WCUs** — `ceil(1 / 1) × 6 × 2 = 1 × 6 × 2`. -6. **10 RCUs** — `ceil(5 / 4) × 5 = 2 × 5`. -7. **9 RCUs** — `ceil(12 / 4) × 3 = 3 × 3`. -8. **4.5 RCUs** — `ceil(0.8 / 4) × 9 × 0.5 = 1 × 9 × 0.5`. -9. **7.5 RCUs** — `ceil(8.01 / 4) × 5 × 0.5 = 3 × 5 × 0.5`. -10. **7 RCUs** — `ceil(8 / 4) × 7 × 0.5 = 2 × 7 × 0.5`. -11. **20 RCUs** — `ceil(4.01 / 4) × 5 × 2 = 2 × 5 × 2`. -12. **18 RCUs** — `ceil(11 / 4) × 3 × 2 = 3 × 3 × 2`. -13. **11 WCUs** — `(ceil(0.6 / 1) × 2) + (ceil(2.4 / 1) × 3) = 2 + 9`. -14. **5.5 RCUs** — `(ceil(5 / 4) × 2) + (ceil(1 / 4) × 3 × 0.5) = 4 + 1.5`. -15. **7 writes/s** — `28 / ceil(3.2 / 1) = 28 / 4`. -16. **5 transactional item writes/s** — `30 / (ceil(2.2 / 1) × 2) = 30 / 6`. -17. **10 reads/s** — `15 / (ceil(9 / 4) × 0.5) = 15 / 1.5`. -18. **8 transactional item reads/s** — `32 / (ceil(5 / 4) × 2) = 32 / 4`. -19. **6 provisioned RCUs** — `ceil(ceil(3 / 4) × 11 × 0.5) = ceil(5.5)`. -20. **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. - -