B + tree

Question

Following parameters to construct B+ tree

search key - 15B

block size - 512B

record pointer - 9B

block pointer - 8B

Max num of keys that can be accomodated in each non leaf node of the tree is

Answer 1

Search key - 15B

Block pointer - 8B

Block size - 512B
Max num of keys that can be accomodated in each non leaf node of the tree is

(x+1)*8+x*15<=512
23x<=504

x=21

Comments

(x+1)*8+x*15<=512  - how did u arrive at this?

non leaf node (i.e) internal node

it is  (n * block pointer) +  (n-1) * key <= block size right?

(n * 8) + (n-1) *15 <=512

where did i go wrong

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1) You mentioned record pointer but took block pointer which is correct it should be block pointer in your representation of the formula.
2) Your approach is right
23n<=527
thus n=22 [which is total pointers]
for max keys : n-1=21

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got it :) i dint do n-1 to find the answer

max keys is always (n-1) incase of B and B+ tree?

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Yes ! In both B and B+ tree at non leaf node the number of keys will be 1 less than order of tree.