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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

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UGC NET Computer Science (Paper 2) Dec 2019 Paper
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  1. 32 M - 32
  2. 36 M - 32
  3. 36 M - 36
  4. 32 M - 36

Answer (Detailed Solution Below)

Option 2 : 36 M - 32
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Detailed Solution

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Data:

Block size = 8192 bytes = 213 bytes

Key size = 32 byte

Block pointer = 4 byte

Order of the b-tree = M

Number of keys = M – 1 

Formula:

In a B-Tree, the size of non-leaf node = M × (Block pointer) + (M – 1) × (key + Record pointer)

Calculation:

Assuming Record pointer = 0  (since nothing is mentioned about it)

Total size = M(4) + (M - 1)(32) = 36 M - 32

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