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Question

In a B-Tree, each node represents a disk block. Suppose one block holds 8192 bytes. Each key uses 32 bytes. In a B-tree of order M there are M – 1 keys. Since each branch is on another disk block. We assume a branch is of 4 bytes. The total memory requirement for a non-leaf node is

The correct answer is

36 M - 32

Understanding B-Tree Non-Leaf Node Memory Requirements

In a B-Tree data structure, information is stored in nodes, which typically correspond to disk blocks to minimize disk I/O. The structure of a node, especially a non-leaf node, is crucial for understanding memory usage and performance. A non-leaf node in a B-Tree of order \(M\) contains keys and pointers (or branches) to child nodes.

Let's break down the components of a non-leaf node and calculate its memory footprint based on the given information:

  • The B-Tree is of order \(M\).
  • A non-leaf node in a B-Tree of order \(M\) contains \(M-1\) keys.
  • A non-leaf node in a B-Tree of order \(M\) contains \(M\) branches (pointers) to its child nodes.
  • Each key occupies 32 bytes.
  • Each branch (pointer) occupies 4 bytes.

The total memory requirement for a non-leaf node is the sum of the memory used by the keys and the memory used by the branches.

Memory used by keys = (Number of keys) \(\times\) (Size of each key)

Memory used by keys = \((M-1) \times 32\) bytes

Memory used by branches = (Number of branches) \(\times\) (Size of each branch)

Memory used by branches = \(M \times 4\) bytes

Total memory requirement for a non-leaf node = Memory used by keys + Memory used by branches

Total memory requirement = \((M-1) \times 32 + M \times 4\)

Now, let's simplify this expression:

Total memory requirement = \(32M - 32 + 4M\)

Total memory requirement = \((32M + 4M) - 32\)

Total memory requirement = \(36M - 32\)

Therefore, the total memory requirement for a non-leaf node in this B-Tree of order \(M\) is \(36M - 32\) bytes.

This calculation shows how the order of the B-Tree (\(M\)) and the sizes of keys and branches directly influence the memory needed for each non-leaf node. The block size (8192 bytes) mentioned in the question is the maximum capacity of a disk block, which would be the upper limit for \(36M - 32\) in a practical implementation to ensure the node fits within a single block.

Revision Table: B-Tree Node Structure

Review the components contributing to the memory size of a non-leaf B-Tree node:

  • Order (M): Defines the capacity of the node.
  • Number of Keys: \(M-1\) keys in a non-leaf node of order \(M\).
  • Number of Branches: \(M\) branches in a non-leaf node of order \(M\).
  • Key Size: 32 bytes per key in this specific case.
  • Branch Size: 4 bytes per branch (pointer) in this specific case.

Additional Information: B-Tree Properties and Order

The order \(M\) of a B-Tree is a critical parameter that defines the structure and performance characteristics of the tree. Understanding its properties helps in analyzing the memory requirements and efficiency.

  • In a B-Tree of order \(M\), every node (except the root) must have at least \(\lceil M/2 \rceil\) children (and thus at least \(\lceil M/2 \rceil - 1\) keys). The root must have at least 2 children (unless it is a leaf).
  • Every non-leaf node has exactly \(M\) children and \(M-1\) keys.
  • All leaf nodes are at the same level.
  • The size of a node is often chosen to fit within a single disk block to minimize disk I/O operations, which are much slower than main memory operations. The calculated memory requirement \(36M - 32\) must be less than or equal to the block size (8192 bytes) for a valid B-Tree configuration with this block size. This constraint is used to determine the maximum possible order \(M\) for a given block size and key/pointer sizes.
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Important Questions from Indexing - Teaching

  1. Consider a B-tree of height h, minimum degree t ≥ 2 that contains any n-key, where n ≥ 1. Which of the following is correct?

  2. Suppose you have a Linux file system where the block size is 2K bytes, a disk address is 32 bits, and an i-node contains the disk addresses of the first 12 direct blocks of file, a single indirect block, and a double indirect block. Approximately, what is the largest file that can be represented by an i-node?  

  3. The order of a leaf node in a B+ tree is the maximum number of (value, data record pointer) pairs it can hold. Given that the block size is 1K bytes, data record pointer is 7 bytes long, the value field is 9 bytes long and a block pointer is 6 bytes long, what is the order of the leaf node?

  4. Suppose a B +tree is used for indexing a database file. Consider the following information :

    size of the search key field = 10 bytes, block size = 1024 bytes, size of the record pointer = 9 bytes, size of the block pointer = 8 bytes.

    Let K be the order of internal node and L be the order of leaf node of B +tree, then (K, L) = ______.

  5. The clustering index is defined on the fields which are of type

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