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Chapter 13: Problem 38

Suppose that a file initially contains \(r=120,000\) records of \(R=200\) bytes each in an unsorted (heap) file. The block size \(B=2400\) bytes, the average seek time \(s=\) \(16 \mathrm{ms}\), the average rotational latency \(r d=8.3 \mathrm{ms}\) and the block transfer time \(b t t=\) \(0.8 \mathrm{ms} .\) Assume that 1 record is deleted for every 2 records added until the total number of active records is 240,000 a. How many block transfers are needed to reorganize the file? b. How long does it take to find a record right before reorganization? c. How long does it take to find a record right after reorganization?

Short Answer

Expert verified
a. The number of block transfers needed to reorganize the file is 20,000. b. The time to find a record right before reorganization is 247 seconds. c. The time to find a record right after reorganization is 494 seconds.

Step by step solution

01

Calculate the total blocks

The total blocks can be calculated by dividing the total bytes by the block size. Given that each record is \(R=200\) bytes and initially we have \(r=120,000\) records, the total bytes will be \(r * R = 120,000 * 200 = 24,000,000\) bytes. And the block size \(B=2400\) bytes. Thus, the total blocks is \(24,000,000 / 2400 = 10,000\) blocks.
02

Find the total number of block transfers

Assuming that 1 record is deleted for every 2 records added until the total number of records is 240,000, that means there will be \(240,000 - 120,000 = 120,000\) records added. These additional records will require \(120,000 * R / B = 120,000 * 200 / 2400 = 10,000\) extra blocks. Hence, the total number of block transfers will be the sum of the initial blocks and the extra blocks needed for the added records, so, \(10,000 + 10,000 = 20,000\) block transfers.
03

Find the time to find a record before reorganization

Before reorganization, the time to find a record is the sum of the seek time \(s\), the rotational latency \(rd\), and the time to transfer a block. Assuming the file is unsorted and we need to read half the blocks on average, the total time will be \((s + rd + btt/2) * Total Blocks = (16 ms + 8.3 ms + 0.8 ms/2) * 10,000 = 247,000 ms or 247 seconds.
04

Find the time to find a record after reorganization

After reorganization, the time to find a record is the same as before but the total blocks are now \(20,000\). Therefore, the total time will be \((s + rd + btt/2) * Total Blocks = (16 ms + 8.3 ms + 0.8 ms/2) * 20,000 = 494,000 ms or 494 seconds.

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Most popular questions from this chapter

Discuss the techniques for allowing a hash file to expand and shrink dynamically. What are the advantages and disadvantages of each?

Load the records of Exercise 13.27 into an expandable hash file, using linear hash. ing. Start with a single disk block, using the hash function \(h_{0}=K \bmod 2^{0},\) and show how the file grows and how the hash functions change as the records are inserted. Assume that blocks are split whenever an overflow occurs, and show the value of \(n\) at each stage.

Write pseudocode for the insertion algorithms for linear hashing and for extendible hashing.

Write program code to access individual fields of records under each of the following circumstances. For each case, state the assumptions you make concerning pointers, separator characters, and so forth. Determine the type of information needed in the file header in order for your code to be general in each case. a. Fixed-length records with unspanned blocking. b. Fixed-length records with spanned blocking. c. Variable-length records with variable-length fields and spanned blocking. d. Variable-length records with repeating groups and spanned blocking. e. Variable-length records with optional fields and spanned blocking. f. Variable-length records that allow all three cases in parts \(c, d,\) and e.

How does disk mirroring help improve reliability? Give a quantitative example.
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