Database-Management-System
October 3, 2023Database-Management-System
October 4, 2023Database-Management-System
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Question 94
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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?
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63
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64
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67
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68
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Question 94 Explanation:
Disk Block size = 1 KBytes = 210 Bytes = 1024 Bytes
Data recorder pointer size, r = 7 bytes
Value size, v = 9 bytes
Disk Block pointer, P = 6 bytes
⇒ Order of leaf be m, then
r*m+v*m+p <= 1024
7m+9m+6 <= 1024
16m <= 1024 – 6
m <= 63
Data recorder pointer size, r = 7 bytes
Value size, v = 9 bytes
Disk Block pointer, P = 6 bytes
⇒ Order of leaf be m, then
r*m+v*m+p <= 1024
7m+9m+6 <= 1024
16m <= 1024 – 6
m <= 63
Correct Answer: A
Question 94 Explanation:
Disk Block size = 1 KBytes = 210 Bytes = 1024 Bytes
Data recorder pointer size, r = 7 bytes
Value size, v = 9 bytes
Disk Block pointer, P = 6 bytes
⇒ Order of leaf be m, then
r*m+v*m+p <= 1024
7m+9m+6 <= 1024
16m <= 1024 – 6
m <= 63
Data recorder pointer size, r = 7 bytes
Value size, v = 9 bytes
Disk Block pointer, P = 6 bytes
⇒ Order of leaf be m, then
r*m+v*m+p <= 1024
7m+9m+6 <= 1024
16m <= 1024 – 6
m <= 63
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