#dbms #file structure

Question

The order of a leaf node in a B+ tree is the maximum number of (value, data record pointer),
it can hold. Given that the block size is 5 Kbyte, field is 9 byte long and a block pointer is 6
byte long, what is the order of the leaf node?

Comments

question seems incomplete to me !

---

when nothing is given,then record pointer size is equals to block pointer size

---

blockpointer + n (key+recordPointer) <=5KB find the value of n.... is the ans...

---

yes question is incomplete .

Answer 1

Block Pointer = 6 bytes

Block size = 5 Kbytes = 5*1024 bytes = 5120 bytes

Search key field = 9 bytes

let us consider order of root node is 'o'

so , 

o*block pointer + (o-1)*search key field <= block size

o*6 + (o-1)*9 <= 5120

o <= 5129/15 

o=341 is the order of leaf node

Comments

formula you used is for internal node

Answer 2

Formula to calculate order of internal node is :

n * (data pointer + Search key) + block pointer<= block size

n *9 + 6 <= 5 *1024

n<= 5114/9

n= 568