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In a database file structure, the search key field is 9 bytes long, the block size is 512 bytes, a record pointer is 7 bytes and a block pointer is 6 bytes. The largest possible order of a leaf node in a B+ tree implementing this file structure is
  • a)
    23
  • b)
    24
  • c)
    34
  • d)
    44
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
In a database file structure, the search key field is 9 bytes long, th...
From the structure of B+ tree we can get the following equation:

In B+ tree, non leaf node has no record pointer: (by putting q = 0) 

Largest possible order of a non-leaf node = 34.
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Most Upvoted Answer
In a database file structure, the search key field is 9 bytes long, th...
Explanation:

B-tree is a data structure that is used to store large amounts of data in a database. It is a balanced tree structure that allows for efficient insertion, deletion, and searching of data. The order of a B-tree is the maximum number of children a node can have. The order of a leaf node is the maximum number of records that can be stored in a leaf node.

Given:

- Search key field size = 9 bytes
- Block size = 512 bytes
- Record pointer size = 7 bytes
- Block pointer size = 6 bytes

To find:

The largest possible order of a leaf node in a B-tree.

Solution:

First, we need to calculate the size of a record in the leaf node. It is the sum of the search key field size and the record pointer size.

Record size = Search key field size + Record pointer size
Record size = 9 + 7
Record size = 16 bytes

Next, we need to calculate the maximum number of records that can be stored in a leaf node. It is the maximum number of records that can fit in a block of size 512 bytes.

Max records per block = Block size / Record size
Max records per block = 512 / 16
Max records per block = 32 records

Now, we need to calculate the size of a block pointer. It is the size of a pointer to a child node.

Block pointer size = 6 bytes

Finally, we can calculate the largest possible order of a leaf node using the formula:

Order = (Block size - Block pointer size) / (Record pointer size + Search key field size)
Order = (512 - 6) / (7 + 9)
Order = 34

Therefore, the largest possible order of a leaf node in a B-tree is 34, which is option C.
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In a database file structure, the search key field is 9 bytes long, the block size is 512 bytes, a record pointer is 7 bytes and a block pointer is6 bytes. The largest possible order of a leaf node in a B+ tree implementing this file structure isa)23b)24c)34d)44Correct answer is option 'C'. Can you explain this answer?
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