Question

Suppose we have a sequential (ordered) file of 100,000 records where each record is 240 bytes. Assume that \(B=2400\) bytes, \(s=16 \mathrm{ms}, r d=8.3 \mathrm{ms},\) and \(b t t=0.8 \mathrm{ms}\) Suppose we want to make \(X\) independent random record reads from the file. We could make \(X\) random block reads or we could perform one exhaustive read of the entire file looking for those \(X\) records. The question is to decide when it would be more efficient to perform one exhaustive read of the entire file than to perform \(x\) individual random reads. That is, what is the value for \(X\) when an exhaustive read of the file is more efficient than random \(X\) reads? Develop this as a function of \(X\)

Short answer

The function involving \(X\) is hence \(X = \frac{s + rd + (\text{File Size}/B) \cdot btt}{s+rd+btt}\). On determining the value of \(X\), it can be used to make the decision as to which operation is more efficient, an exhaustive read or random reads.

Step-by-step solution

Compute the Time for One Exhaustive Read

First, it has to be determined how long it would take to perform one exhaustive read of the entire file. In such a task, the file's total size should be divided by the block size to identify the total number of blocks, which gives \(n\) and is computed as \(n = \frac{\mathrm{size~of~file}}{B}\). Once we have this, the time it takes to exhaustively read the file can be computed as \(T_{exh} = n \cdot btt\). Because the seek time and rotational delay would only apply to the first record in the block, it can be written as \(T_{exh} = s + rd + n \cdot btt\).

Compute the Time for Random Reads

Next, the time it would take to perform random reads needs to be calculated. In this scenario, each read includes a seek time, rotational delay, and block transfer time. Thus, the time for random reads is \(T_{rd} = X \cdot (s + rd + btt)\). As every record-read operation is independent, these parameters would apply to every read. They are then multiplied by \(X\) because there are \(X\) number of reads.

Formulate the Function of \(X\)

The value of \(X\) that will make exhaustive read time equal to the random read time is what we are searching. Therefore, set \(T_{exh} = T_{rd}\). By substituting from the formulas got in step 1 and 2, we get \(s + rd + (\frac{\mathrm{size~of~file}}{B} \cdot btt) = X \cdot (s + rd + btt)\). Find the value for \(X\) . This leads to the formulation of the function involving \(X\).

Key concepts

Sequential File Organization

When we talk about Sequential File Organization, we're referring to a system where data is organized in a specific, predictable sequence. This is typically based on a key field that arranges the records so they are in some kind of logical order – for example, alphabetical or numerical. Imagine this like a list of names in a phone book where each name's record follows the previous one alphabetically.

For the computer to read through a sequential file, it typically starts at the beginning and reads through each record, one by one, until it finds the desired data. This method is simple and reliable; however, when dealing with large files, as in our exercise, the efficiency can drastically vary depending on the tasks required. If you need to access most or all of the records, sequential access can be quite efficient – you just read from start to finish. But if you're only looking for one specific record, you might have to go through thousands of records first, which can be time-consuming.

Random Access

Random access is a way of reaching a particular record directly, without having to read through all the preceding records, unlike in a sequential file organization. This method is extremely useful when you need to access records intermittently or when the dataset is so large that sequential access would be unbearably slow.

Consider a jukebox as an analogy – you can directly pick the song (record) you want to play without going through every single song before it. In the context of our exercise, if you imagine having a file with 100,000 records and you're looking for just one specific record, using random access means that you can jump straight to it. However, random access typically requires additional mechanisms such as indexes to allow these direct reads, and it includes seek times and rotational delays that add to the overall access time.

Block Read Operations

With Block Read Operations, rather than accessing each record individually, data is read in blocks. This is an important strategy because computers actually read and write data in blocks to and from storage devices. A block is simply a group of records that the computer handles as a single unit, and block size is a critical factor in database system performance.

In database systems like the one mentioned in our exercise, reading a single block of data, even if it contains hundreds of records, can often be done with similar effort as reading a single record due to how storage devices operate. This is because the majority of the time taken in reading information is not in the actual transfer of data but in the seek time (the time it takes to move the read/write head of the disk to the required data location) and the rotational delay (the time waiting for the disk to rotate the required data under the read/write head).

File Access Efficiency

When evaluating File Access Efficiency, it involves understanding the quickest way to access the required data within a file. In the scenario presented in our exercise, we look at which method of reading – either an exhaustive read of the entire file or individual random reads – results in a lower total time when accessing multiple records.

To determine the most efficient method of file access, you must consider factors such as the total number of records needed (X), the size of each record, the block size, and the time consumed by the system for seeking, rotational delay, and block transfer. The efficiency depends on balancing these factors to minimize the time required to access the desired data. In some cases, such as when only a few records are needed, random reads may be more efficient. However, if many records are needed, it could be more time-efficient to perform an exhaustive read. The exercise provided guides us through calculating the break-even point to make such a decision.