B+ Tree _ 40q

Question

Calculate the order of B+ tree internal node and leaf node respectively if the disk block size is 512 Bytes. Assume that block pointer, record pointer and key field are of 7B, 8B, and 10B respectively

(A) 30,29

(B) 31,30

(C) 29,28

(D) 32,31

Comments

I got 30 and 28 :P

n(10+8) + 7 <=512 ==> n =28(internal)

n*7+(n-1)*10 <= 522 ==> n=30(Leaves)

Edit: mistakenly written internal and leaves, it should be

n(10+8) + 7 <=512 ==> n =28(leaves)

n*7+(n-1)*10 <= 522 ==> n=30(internal )

 

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answer is (A)... @Hemanth_13

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getting 31 and 30. is it correct?

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what is order here?

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For internal node(Non-leaf)$ = 30$ and for leaf $=28$

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As nothing is specified we take the general definition

Order for

Internal nodes: max No of record pointers.

Leaf nodes: max No of index record pointers in a block

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Sorry Sir @Utkarsh Joshi in answer key Option is (A) 30,29

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@BOB

i'm not getting $29?$(leaf)

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@BOB can you please provide their solution.. need to look where I'm missing with leaf node

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I am getting $\text{30 and 28}.$
$\text{Internal node}\rightarrow 7P + 10(P-1) \leq 512 \Rightarrow P =30$
$\text{Leaf}\rightarrow Q(8+10) + 7 \leq 512 \Rightarrow Q =28$

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sorry, @Hemanth_13 sir for so last replying

in the image below that was not the gape... it's just blank...

 

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For leaf nodes it should be p not p-1 if any change to standard definition it should be specified in the question