Question

c) Assume a block size of 4096 bytes; one node per block; disk references of 8 bytes each; and a record size of 160 bytes, with 10 bytes for the key: i) How many records will be stored in each leaf node? i In any index node? i) For the M-ary tree, what is the value of M?

Answer 1

Indices are usually implemented using B+ trees. In B+ trees, only the leaf nodes have data/records and not the internal/non-leaf nodes. Here is a diagram of an index B+ tree:

INTERNAL NODE P, Ki Ki- P Ki Ko Pe Ki,XKi LEA F NODE: Ke Pe Pnet K, D, KD P node pointer Ki key Di: data pointer Prext pointer value to the next leaf node

Now let us try to calculate the number of record references will be stored in the leaf nodes. Note that records are not stored in the nodes, just their references are stored. Now for simplicity let us assume that the disk reference is of the same size as the page reference (8 bytes). Let the number of keys in the leaf node be c.

Blocksize > C* (key diskreference)pointer

4096 c (10 8)8

4088 -227 18 C

i) Hence, 227 record references will be stored in the leaf nodes.

As we see from the image, no nodes apart from the leaf nodes store any records. Hence,

ii) No record reference will be stored in other index nodes.

Let us try to find the maximum pointers that can be stored in the internal nodes similar to the first part. Let's say there are c pointers, therefore according to the image, there will be c-1 keys.

Blocksizec * pointer(c- 1)* key

4096 c 8(c- 1) * 10

4106 18c

4106 228 C 18

Hence, there are 228 pointers in each non-leaf node which is what M denotes in an M-ary tree.

iii) M = 228

Note that in real world implementations, the disk references and page pointers are of different sizes. Also there are more things stored in the nodes for practical purposes. Therefore, these numbers are only indicative and not exact!

Feel free to ask any doubts and I'll be happy to help!