UGC NET Computer Science December 2019 | Question 88

Question 88

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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

1. 1. 52 M - 32
2. 2. 36 M - 32
3. 3. 36 M - 36
4. 4. 32 M - 36

Answer : (2) 36 M - 32

Explanation

Total size of non-leaf block = total size of keys + total size of tree pointer
The size of non-leaf node in B-tree = m(P_b) + (m-1)(key+ P_r)
Here P_b is Block pointer and P_r is record pointer.

In question, its Given that
P_b = 4
key size = 32
since the size of P_r is not given in question consider it as zero.

Hence size of non-leaf node in B-tree
= m(4) + (m-1)(32)
= 36M - 32

So, option 2 is correct answer

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