find order

Question

What will be the order of B+ tree with a database of 5,00,000 records of 200 bytes each and the search key is 15 bytes?Assume tree and data pointers are are 5 bytes each and the index node is 1024 bytes?

Comments

As nothing is mention in the question so we can take internal node so answer here is 51

---

but what has to be done with the number of reccords given?

---

While Solving for internal nodes

15 * (n-1) should be = 15n - 15 but you took 15n - 5

Answer 1

the formula      nPb+(n-1)(k+Pd) <=  Block size

in given picture k+Pd= 15 put, how?

whereas Pb and Pd =5B

k+Pd =15+5 = 20 should there?

Comments

https://gateoverflow.in/232929/b-tree-b-tree?show=232938#a232938 u can see the same question. and i think u are confused between b and b+ tree and between leaf node and internal node of b+ tree.

Answer 2

for linternal node:-     P*(B.P.)+(P-1)K <=Block size

        5(P)+15(P-1)<=1024

  20P-15<=1024

P=51.9

P= ~51

for leaf node:    B.P.+(P-1)(K+R)<=1024

5+(P-1)(20)<=1024

20P-15<=1024

P=51.9

P=~51