A database file has 6,000,000 records, which occupy disk storage at a density of

Question

A database file has 6,000,000 records, which occupy disk storage at a density of 12 records per block. A daily update modifies 4.5 percent of the file and we assume that the updates are distributed uniformly across the records. Weekly the system must rewrite any block, which has an update to any of the records In the block. In the lines provided. (Do your work on scratch paper first.) The probability that an Individual block will be rewritten p = ___ Show work: ___ Let X denote the Binomial Distribution associated with all the blocks in the file. The parameters of this distribution, n, p, are: n = ____, p = ___ Show work: ___ The average number of blocks rewritten mu_x = ___ Show work: ____

Explanation / Answer

Solution:

(1) Given, total number of records = 6,000,000

    Each block has 12 records, so, total number of blocks = 6,000,000/12 = 500000

    Each update modifies 4.5% of the file, so, (4.5/100) × 6,000,000 = 270000 records

    Therefore, each update modifies, 270000/12 = 22500 blocks

   The probability that an individual block will be rewritten = 22500 / 500000 = 0.045

(2) n = Number of blocks = 6000000/12 = 500000

    p = 4.5% = 0.045

These would be the values of parameters.

(3) Total number of records on the database file = 6,000,000

12 records per block are stored.

Hence, number of blocks = 6,000,000/12 = 500000

Daily update modifies 4.5% of the file.

Therefore, number of records modified per week = 6,000,000 × 0.045 × 7 = 1890000

The average number of blocks rewritten = x = 1890000/12 = 157500

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