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Consider a relation R(A B C) with attribute size of A as 8 bytes. Disk block size is 512 bytes and block pointer is 8 bytes. The best choice for degree (maximum value) for B+ tree, if B+ tree was used for creating indexing on R(A B C) is _________.
Correct answer is '32'. Can you explain this answer?
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Consider a relation R(A B C) with attribute size of A as 8 bytes. Dis...
Let p be degree of B + tree internal node
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(p - 1) keys + p Block pointers should fit in a block ie (p-1) keys + p Block pointers size ⇐ 512
(2p - 1) × 8 ⇐ 512
p ⇐ 65/2
p = 32
If you take p = 33 node size becomes 520 bytes so not possible to fit in a block hence the correct answer is 32.
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Consider a relation R(A B C) with attribute size of A as 8 bytes. Dis...
Explanation:
To determine the best choice for the degree (maximum value) for a B-tree, we need to consider the disk block size, the size of the attributes, and the size of the block pointers.
Disk Block Size:
The disk block size is given as 512 bytes. This means that each disk block can store a maximum of 512 bytes of data.
Attribute Size:
The size of attribute A is given as 8 bytes.
Block Pointer Size:
The size of the block pointer is given as 8 bytes.
Indexing on R(A B C):
When creating indexing on relation R(A B C), we need to consider the size of the attributes and the block pointers.
B-tree:
In a B-tree, each node can have a maximum of (degree - 1) keys and (degree) child pointers. The degree represents the maximum number of keys that can be stored in a node.
To determine the best choice for the degree, we need to consider the size of the attributes and the block pointers.
Calculating the Degree:
To calculate the degree for the B-tree, we need to consider the space required for each key and each child pointer.
Space required for each key = size of attribute A = 8 bytes
Space required for each child pointer = size of block pointer = 8 bytes
Total space required for each key and child pointer = 8 + 8 = 16 bytes
The maximum number of keys and child pointers that can be stored in a single disk block can be calculated as follows:
Maximum number of keys and child pointers per block = Disk block size / Total space required for each key and child pointer
Maximum number of keys and child pointers per block = 512 / 16 = 32
Therefore, the best choice for the degree (maximum value) for the B-tree is 32.
To determine the best choice for the degree (maximum value) for a B-tree, we need to consider the disk block size, the size of the attributes, and the size of the block pointers.
Disk Block Size:
The disk block size is given as 512 bytes. This means that each disk block can store a maximum of 512 bytes of data.
Attribute Size:
The size of attribute A is given as 8 bytes.
Block Pointer Size:
The size of the block pointer is given as 8 bytes.
Indexing on R(A B C):
When creating indexing on relation R(A B C), we need to consider the size of the attributes and the block pointers.
B-tree:
In a B-tree, each node can have a maximum of (degree - 1) keys and (degree) child pointers. The degree represents the maximum number of keys that can be stored in a node.
To determine the best choice for the degree, we need to consider the size of the attributes and the block pointers.
Calculating the Degree:
To calculate the degree for the B-tree, we need to consider the space required for each key and each child pointer.
Space required for each key = size of attribute A = 8 bytes
Space required for each child pointer = size of block pointer = 8 bytes
Total space required for each key and child pointer = 8 + 8 = 16 bytes
The maximum number of keys and child pointers that can be stored in a single disk block can be calculated as follows:
Maximum number of keys and child pointers per block = Disk block size / Total space required for each key and child pointer
Maximum number of keys and child pointers per block = 512 / 16 = 32
Therefore, the best choice for the degree (maximum value) for the B-tree is 32.
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Consider a relation R(A B C) with attribute size of A as 8 bytes. Disk block size is 512 bytes and block pointer is 8 bytes. The best choice for degree (maximum value) for B+ tree, if B+ tree was used for creating indexing on R(A B C) is _________.Correct answer is '32'. Can you explain this answer?
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Consider a relation R(A B C) with attribute size of A as 8 bytes. Disk block size is 512 bytes and block pointer is 8 bytes. The best choice for degree (maximum value) for B+ tree, if B+ tree was used for creating indexing on R(A B C) is _________.Correct answer is '32'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about Consider a relation R(A B C) with attribute size of A as 8 bytes. Disk block size is 512 bytes and block pointer is 8 bytes. The best choice for degree (maximum value) for B+ tree, if B+ tree was used for creating indexing on R(A B C) is _________.Correct answer is '32'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider a relation R(A B C) with attribute size of A as 8 bytes. Disk block size is 512 bytes and block pointer is 8 bytes. The best choice for degree (maximum value) for B+ tree, if B+ tree was used for creating indexing on R(A B C) is _________.Correct answer is '32'. Can you explain this answer?.
Consider a relation R(A B C) with attribute size of A as 8 bytes. Disk block size is 512 bytes and block pointer is 8 bytes. The best choice for degree (maximum value) for B+ tree, if B+ tree was used for creating indexing on R(A B C) is _________.Correct answer is '32'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about Consider a relation R(A B C) with attribute size of A as 8 bytes. Disk block size is 512 bytes and block pointer is 8 bytes. The best choice for degree (maximum value) for B+ tree, if B+ tree was used for creating indexing on R(A B C) is _________.Correct answer is '32'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider a relation R(A B C) with attribute size of A as 8 bytes. Disk block size is 512 bytes and block pointer is 8 bytes. The best choice for degree (maximum value) for B+ tree, if B+ tree was used for creating indexing on R(A B C) is _________.Correct answer is '32'. Can you explain this answer?.
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Consider a relation R(A B C) with attribute size of A as 8 bytes. Disk block size is 512 bytes and block pointer is 8 bytes. The best choice for degree (maximum value) for B+ tree, if B+ tree was used for creating indexing on R(A B C) is _________.Correct answer is '32'. Can you explain this answer?, a detailed solution for Consider a relation R(A B C) with attribute size of A as 8 bytes. Disk block size is 512 bytes and block pointer is 8 bytes. The best choice for degree (maximum value) for B+ tree, if B+ tree was used for creating indexing on R(A B C) is _________.Correct answer is '32'. Can you explain this answer? has been provided alongside types of Consider a relation R(A B C) with attribute size of A as 8 bytes. Disk block size is 512 bytes and block pointer is 8 bytes. The best choice for degree (maximum value) for B+ tree, if B+ tree was used for creating indexing on R(A B C) is _________.Correct answer is '32'. Can you explain this answer? theory, EduRev gives you an
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