Question 7680 – UGC-NET DEC-2019 Part-2
November 27, 2023UGC-NET DEC-2019 Part-2
November 27, 2023Database-Management-System
| Question 447 |
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
| 32 M – 32 | |
| 36 M – 32 | |
| 36 M – 36 | |
| 32 M – 36 |
Question 447 Explanation:
The size of non-leaf node in B-tree = m(Pb) + (m-1)(key+ Pr)
Where Pb is Block pointer and Pr is record pointer.
In question,Pb = 4 and key size = 32 and since the size of Pr 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
Where Pb is Block pointer and Pr is record pointer.
In question,Pb = 4 and key size = 32 and since the size of Pr 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
Correct Answer: B
Question 447 Explanation:
The size of non-leaf node in B-tree = m(Pb) + (m-1)(key+ Pr)
Where Pb is Block pointer and Pr is record pointer.
In question,Pb = 4 and key size = 32 and since the size of Pr 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
Where Pb is Block pointer and Pr is record pointer.
In question,Pb = 4 and key size = 32 and since the size of Pr 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