nptel

Question

Consider a B-tree with the following specifications: Page size c = 2 kB; Size of key γ = 32 bytes; Size of pointer η = 8 bytes. What is the order m of the B+-tree? How many records can a tree of height 3 store?

doubt:

x->pointer to block

8*x +(8+32)*(x-1)<=2048

x<=43.5

but given solution is

2*1024>=32*m+8(m+1)+8*m=42.2=42

please guide where i am wrong

Comments

2*1024>=32*m+8(m+1)+8*m=42.2=42

This is the structure of B tree not B+ tree.

Above can also be written as

(32+8)*m + 8(m+1)

where m is the order of tree, which is the maximum number of ( Key + Record pointer ) Pair.

which is exactly same as (key + Record pointer) Order + Block pointer * (order + 1) <= Block size

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yes solution is according to B tree .

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I need to apologize for the above,sorry

but

order= no of child pointers in a node

order*(block pointer)+ order-1( key value+record pointer)<=block size

by this i am getting 43.5