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Question 103 - DBS-C01 discussion

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An electric utility company wants to store power plant sensor data in an Amazon DynamoDB table.

The utility company has over 100 power plants and each power plant has over 200 sensors that send data every 2 seconds. The sensor data includes time with milliseconds precision, a value, and a fault attribute if the sensor is malfunctioning. Power plants are identified by a globally unique identifier.

Sensors are identified by a unique identifier within each power plant. A database specialist needs to design the table to support an efficient method of finding all faulty sensors within a given power plant.

Which schema should the database specialist use when creating the DynamoDB table to achieve the fastest query time when looking for faulty sensors?

A.
Use the plant identifier as the partition key and the measurement time as the sort key. Create a global secondary index (GSI) with the plant identifier as the partition key and the fault attribute as the sort key.
Answers
A.
Use the plant identifier as the partition key and the measurement time as the sort key. Create a global secondary index (GSI) with the plant identifier as the partition key and the fault attribute as the sort key.
B.
Create a composite of the plant identifier and sensor identifier as the partition key. Use the measurement time as the sort key. Create a local secondary index (LSI) on the fault attribute.
Answers
B.
Create a composite of the plant identifier and sensor identifier as the partition key. Use the measurement time as the sort key. Create a local secondary index (LSI) on the fault attribute.
C.
Create a composite of the plant identifier and sensor identifier as the partition key. Use the measurement time as the sort key. Create a global secondary index (GSI) with the plant identifier as the partition key and the fault attribute as the sort key.
Answers
C.
Create a composite of the plant identifier and sensor identifier as the partition key. Use the measurement time as the sort key. Create a global secondary index (GSI) with the plant identifier as the partition key and the fault attribute as the sort key.
D.
Use the plant identifier as the partition key and the sensor identifier as the sort key. Create a local secondary index (LSI) on the fault attribute.
Answers
D.
Use the plant identifier as the partition key and the sensor identifier as the sort key. Create a local secondary index (LSI) on the fault attribute.
Suggested answer: D

Explanation:


Plant id as partition key and Sensor id as a sort key. Fault can be identified quickly using the local secondary index and associated plant and sensor can be identified easily.

asked 16/09/2024
Derrick Dave T Alvarez
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