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In a B-Tree, each node represents a disk block.
Suppose one block holds 8192 bytes. Each key uses 32 bytes. In a B-tree of order M there are M – 1 keys. Since each branch is on another disk block. We assume a branch is of 4 bytes. The total memory requirement for a non-leaf node is
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Eduncle Best Answer
The size of non-leaf node in B-tree = m(Pb) + (m – 1)(key + Pr)
Here Pb is Block pointer and Pr is record pointer.
Given That
Pb = 4 and key size = 32 and since the size of Pr is not given in question consider it as zero.
Hence size of non-leaf node in B-tree = m(4) + (m – 1)(32) = 36M – 32