B+ tree Index

Question

Suppose size of disk block 1000 bytes and search key of 12 bytes, pointer size 8 bytes. How many minimum number of records in data file which leads 3 level dense B⁺ tree index?

My Approach:- 

Say P is the Order. and we know In dense index the number of records in DB file = No. of keys in B+tree index

P*8 + (P-1)*12 <= 1200

P = 50

Now

Level                    Min.Nodes                      Min_Child                     Min_Keys

  1                               1                                     2                                   1

  2                               2                                     2*25                             2*24

  3                               50                                    ---                                50*24

So total keys = 1 + 2*24 + 50*24 = 1249. 

Am i right?

Comments

In B+ tree all keys are at only last level then why are adding all levels keys?

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Ohk that's my mistake.One point of confirmation though the leaf nodes will always have all the keys right ?.So it means that 50*24 = 1200 should be the answer it internally contains all the keys am i right?

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See my solution ,in question given min no. Of keys so you have keep in every node min no. Of keys

Answer 1

Check my solution,

Comments

See my solution now ,for level 3 and min keys

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I got your answer brother the thing is u considered only the last level keys but in made easy solution they had taken 49 + 50*49 = 2499 and i don't know why they added 1 into it and gave the answer as 2450. Maybe they are wrong or there is something ambiguous. According to me only the last level keys should be considered.

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this ans should be 1250 instead 1200.. in b+ tree the min no of keys in leaf and non-leaf are not same.. in leaf it is ceil(order/2) but in non-leaf it's ceil(order/2)-1. hence in level 3 min keys would be 50*25=1250