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Question 11747 – ISO-OSI-layers
March 3, 2024
Question 8177 – SQL
March 4, 2024
Question 11747 – ISO-OSI-layers
March 3, 2024
Question 8177 – SQL
March 4, 2024

Question 10783 – Relational-Algebra

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-nameb-city = "Agra" ⋀ bal < 0 (branch ⋈ (account ⋈ depositor)  

Which one of the following queries is the most efficient version of the above query?

Correct Answer: A

Question 16 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.
A
Пc-namebal < 0b-city = “Agra” branch ⋈ account) ⋈ depositor)
B
Пc-nameb-city = “Agra”branch ⋈ (σbal < 0 account ⋈ depositor))
C
Пc-nameb-city = “Agra” branch ⋈ σb-city = “Agra” ⋀ bal < 0 account) ⋈ depositor)
D
Пc-nameb-city = “Agra” ⋀ bal < 0 account ⋈ depositor))
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