Consider the following data for B-tree and B+ tree,
Block size is 8 KB, Data pointer is 10 B, Block pointer is of 15 B and key size is 10 B
Which of the following option is correct ?
Note – This question is multiple select questions (MSQ).
(A) Maximum order of leaf node of B-tree is 230
(B) Maximum order of leaf node of B+ tree (Q) is 405
(C) Difference of maximum order of leaf node of B-tree (P) and maximum order of leaf node of B+ tree (Q) -175
(D) All of these are correct.
Answer: (C)
Explanation:
Given,
Block size = 8 KB, Data pointer or record pointer = Rp = 10 byte Block pointer or child pointer = Bp = 15 byte Key size/pointer = Kp = 10 byte
Since, maximum order (P) of leaf node of B tree is given by formula:
P(Bp) + (P-1)*(Kp + Rp) ≤ Block size P(15) + (P-1)*(10+10) ≤ 8 KB 15P + 20P - 20 ≤ 8*1024 35P ≤ 8192+20 35P ≤ 8212 35P ≤ 8212/35 P ≤ 234.63 P = 234
Take floor value for maximum.
And, maximum order (Q) of leaf node of B+ tree is given by formula:
(Q-1)*(Kp + Rp) + (Bp) ≤ Block size (Q-1)*(10+10) + 15 ≤ 8 KB 20Q - 20 + 15 ≤ 8*1024 20Q - 5 ≤ 8192 20Q ≤ 8192+5 Q ≤ 8197/20 Q ≤ 409.85 Q = 409
Take floor value for maximum.
Therefore P – Q = 234 – 409 = -175
So, only option (C) is correct.
Quiz of this Question
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