Question

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

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

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.

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.

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.

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.

Key concepts

Block Transfers

In the context of database management, block transfers refer to the movement of blocks of data between disk storage and memory. A block is a continuous section of memory that holds multiple records. For a given problem, we start with a set of records, each with a defined size. To compute the number of blocks, we divide the total size of the records by the block size. Once we know the number of blocks, we can determine the number of block transfers required for different operations, such as reorganization.
Reorganization often involves adding or deleting records. This example illustrates calculating block transfers by adding records until they reach a certain total number. We compute the required blocks for these records and determine how many block transfers are needed. The total block transfers are an important measure of the efficiency and cost of database operations, impacting both performance and resource utilization.

Record Reorganization

Record reorganization involves updating and arranging records in a file to improve access and efficiency. In unsorted or heap files, records are stored in no particular order, affecting retrieval time since many blocks may need to be accessed to locate a specific record. With reorganization, records can be sorted or structured differently to optimize search and update efficiency.
Reorganization can add substantial overhead in block transfers and processing time. In this particular scenario, reorganization involves increasing the number of active records up to a certain value, requiring accounting for new blocks and recalculating necessary block transfers. By reorganizing, it is intended to make future data access quicker and more efficient, although it initially increases processing time due to extra operations involved in arranging the data.

Seek Time

Seek time is the duration it takes for a disk drive's read/write head to locate the position of a data block on the disk. This operation is a component of the time needed to access data from storage. In your database, seek time is significant because it contributes to the overall latency of data retrieval.
Given parameters, seek time can be constant while accessing different records, influencing the calculation of total access time before and after reorganization. Organizations and individuals aim to minimize seek times by optimizing data storage formats and structures, such as reorganizing data to reduce the average distance the read/write head must move, consequently enhancing database performance.

Rotational Latency

Rotational latency is the delay waiting for the desired disk sector to rotate under the read/write head after a seek operation. It is an essential part of the total time to access a record from disk storage. Disk platters spin at a consistent speed, so rotational latency depends on how quickly a required data block rotates to the read/write head following the seek operation.
In our exercise, given as 8.3 ms, rotational latency is added to seek time and other factors to compute the time required to find a record. Improvements in rotational latency can be achieved through better disk technology, but reorganizing data to reduce unnecessary full rotations can also contribute to reduced latency across the entire data retrieval process. It's important for efficiency in scenarios involving frequent read/write operations.