ISO-OSI-layers
March 3, 2024Question 8177 – SQL
March 4, 2024Database-Management-System
| Question 46 |
Consider the following relation schemas:
b-Schema = (b-name, b-city, assets)
a-Schema = (a-num, b-name, bal)
d-Schema = (c-name, a-number)
Let branch, account and depositor be respectively instances of the above schemas. Assume that account and depositor relations are much bigger than the branch relation.
Consider the following query:
Пc-name (σb-city = "Agra" ⋀ bal < 0 (branch ⋈ (account ⋈ depositor)
Which one of the following queries is the most efficient version of the above query?
| Пc-name (σbal < 0 (σb-city = “Agra” branch ⋈ account) ⋈ depositor) | |
| Пc-name (σb-city = “Agra”branch ⋈ (σbal < 0 account ⋈ depositor)) | |
| Пc-name (σb-city = “Agra” branch ⋈ σb-city = “Agra” ⋀ bal < 0 account) ⋈ depositor) | |
| Пc-name (σb-city = “Agra” ⋀ bal < 0 account ⋈ depositor)) |
Question 46 Explanation:
Answer should be (A) and not (B), because we are doing a join between two massive tables whereas in (A) we are doing join between relatively smaller table and larger one and the output that this inner table gives (which is smaller in comparison to join that we are doing in (B)) is used for join with depositor table with the selection condition.
Options (C) and (D) are invalid as there is no b-city column in a-schema.
Options (C) and (D) are invalid as there is no b-city column in a-schema.
Correct Answer: A
Question 46 Explanation:
Answer should be (A) and not (B), because we are doing a join between two massive tables whereas in (A) we are doing join between relatively smaller table and larger one and the output that this inner table gives (which is smaller in comparison to join that we are doing in (B)) is used for join with depositor table with the selection condition.
Options (C) and (D) are invalid as there is no b-city column in a-schema.
Options (C) and (D) are invalid as there is no b-city column in a-schema.
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