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
Calculate the difference of maximum order of leaf node of B-tree (P) and maximum order of leaf node of B+ tree (Q)?
(A) 210
(B) 195
(C) -175
(D) None of these.
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, option (C) is correct.
Quiz of this Question
Share your thoughts in the comments
Please Login to comment...