Database-Management-System
October 3, 2023Database-Management-System
October 4, 2023Database-Management-System
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Question 106
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The order of an internal node in a B+ tree index is the maximum number of children it can have. Suppose that a child pointer takes 6 bytes, the search field value takes 14 bytes, and the block size is 512 bytes. What is the order of the internal node?
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24
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25
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26
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27
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Question 106 Explanation:
Block size = (n-1) * key size + n * child pointer
Child pointer = 6 bytes
key size = 14 bytes
Block size = 512 bytes
⇒ 512 = (n-1)14 + n(6)
512 = 20n – 14
n = 512+14/20 = 526/20 = 26.3
∴ n = 26
Child pointer = 6 bytes
key size = 14 bytes
Block size = 512 bytes
⇒ 512 = (n-1)14 + n(6)
512 = 20n – 14
n = 512+14/20 = 526/20 = 26.3
∴ n = 26
Correct Answer: C
Question 106 Explanation:
Block size = (n-1) * key size + n * child pointer
Child pointer = 6 bytes
key size = 14 bytes
Block size = 512 bytes
⇒ 512 = (n-1)14 + n(6)
512 = 20n – 14
n = 512+14/20 = 526/20 = 26.3
∴ n = 26
Child pointer = 6 bytes
key size = 14 bytes
Block size = 512 bytes
⇒ 512 = (n-1)14 + n(6)
512 = 20n – 14
n = 512+14/20 = 526/20 = 26.3
∴ n = 26
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