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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

  1. $32 M -32$
  2. $36 M – 32$
  3. $36 M – 36$
  4. $32 M – 36$
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$\underline{\textbf{Answer:}\Rightarrow}$

$\underline{\textbf{Explanation:}\Rightarrow}$

The size of non-leaf node $=\mathbf{m(P_b) + (m-1)(key+P_r)}$

where $\mathbf{P_b}$ is the Block pointer and $\mathbf{P_r} $ is the record pointer.

Now, given $\mathbf{P_b = 4}$ and size of key $=32$ but size of $\mathbf{P_r}$ is not given in the question.

$\therefore$ Assume it $0$.

$\therefore \text{Total size} = \mathbf{m(4) + (m-1)(32)=36M-32}$
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Since B tree has M-1 keys in its block, therefore, the size of keys= (M-1)x32

b tree also has M tree pointer in its block so the total size of tree pointer =Mx4

therefor total size of block =total size of keys + total size of tree pointer

                                           =(M-1)x32+Mx4

                                             =36M-32
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In a B-tree of order M,

There are M block pointers and (M-1) keys.

So total size of a node is=M* size of a block pointer+(M-1)* size of a key

$=M*4+(M-1)*32$

$=36M-32$

Hence option 2 is the correct answer.

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