All Exams Test series for 1 year @ ₹349 only
Question

For a magnetic disk with concentric circular tracks, the seek latency is not linearly proportional to the seek distance due to

The correct answer is

arm starting or stopping inertia

Understanding Magnetic Disk Seek Latency and Inertia

Magnetic disks store data on rotating platters in concentric circles called tracks. To read or write data, a read/write head attached to an arm must move to the correct track. The time it takes for the arm to move the head to the desired track is called seek latency or seek time.

The question asks why this seek latency is not linearly proportional to the seek distance. Linear proportionality would mean that if you double the distance the arm has to move (the number of tracks crossed), the time taken would also exactly double. However, this isn't typically the case in a real magnetic disk.

Why Seek Latency is Non-Linear: The Role of Arm Inertia

The main reason seek latency isn't linearly proportional to the seek distance is the physical mechanics of moving the disk arm. The arm has mass and therefore inertia. Inertia is the resistance of any physical object to any change in its state of motion. To move the arm, it must be accelerated from rest. Once it approaches the target track, it must be decelerated to a stop. These acceleration and deceleration phases take time and require forces.

The movement can be thought of in phases:

  1. Acceleration Phase: The arm starts moving and speeds up.
  2. Constant Velocity Phase: For longer seeks, the arm reaches a maximum speed and travels at that speed for a period. For very short seeks, this phase might be non-existent.
  3. Deceleration Phase: The arm slows down as it approaches the destination track.
  4. Settle Time: A small amount of time for the head to become perfectly stable over the track.

For short seek distances, the acceleration and deceleration phases (and settle time) constitute a larger proportion of the total seek time. The time taken during these phases does not scale linearly with the distance covered in these phases. For example, doubling a very short seek distance might not even allow the arm to reach its full speed, and the dominant factors are acceleration and deceleration, making the time relationship non-linear (closer to quadratic for constant acceleration, but complicated by real-world physics). As the seek distance increases, the constant velocity phase becomes more significant, making the relationship appear more linear, but the initial and final non-linear phases still prevent perfect linearity across all distances.

This arm starting or stopping inertia means that it takes a certain minimum time to start and stop the movement, regardless of how short the actual travel distance is. This fixed overhead contributes significantly to short seeks, breaking the linear relationship.

Analyzing Other Options

Let's look at why the other options do not explain the non-linear proportionality between seek latency and seek distance:

  • Non-Uniform distribution of requests: This affects the *average* seek time experienced by the system over a period, depending on where the requested data is located relative to the current head position. It influences the overall performance and average seek *distance*, but it doesn't change the fundamental relationship between time and distance for a single seek.
  • Higher capacity of tracks on the periphery of the plate: Outer tracks typically hold more sectors and thus more data because they are longer. This affects the data transfer rate once the head is positioned on the track, but it has no direct impact on the *time* taken to move the arm to that track (the seek latency).
  • Use of unfair arm scheduling policies: Disk scheduling algorithms (like FCFS, SSTF, SCAN, C-SCAN) determine the *order* in which pending read/write requests are serviced. Unfair policies might prioritize certain requests or areas of the disk, affecting which seeks are performed and when, but they do not alter the physics of the arm movement or the time-distance relationship for a given seek.

Therefore, the primary physical reason for the non-linear relationship between seek latency and seek distance is the arm starting or stopping inertia, which necessitates acceleration and deceleration phases that dominate short seeks and add non-linearity to longer ones.

Revision Table: Magnetic Disk Seek Components

Component of Disk Access Time Description Impact on Performance
Seek Time (Seek Latency) Time to move the read/write head to the correct track. Major contributor to access time, non-linear with distance.
Rotational Latency Time for the desired sector to rotate under the head. Average is half the time for one rotation.
Data Transfer Time Time to read or write the data block. Depends on data size and disk transfer rate.

Additional Information: Factors Affecting Disk Performance

Beyond seek latency, other factors significantly impact overall magnetic disk performance:

  • Rotational Speed (RPM): Higher RPM means lower rotational latency and potentially higher data transfer rates.
  • Track Density: More tracks per inch means higher capacity and potentially shorter maximum seek distances, but requires more precise arm movement.
  • Areal Density: More bits per square inch (combination of track density and bit density along the track) increases overall capacity and potential data transfer rates.
  • Cache/Buffer: Onboard disk cache can store frequently accessed data or buffer writes, reducing the need for physical disk access.
  • Disk Scheduling Algorithm: Optimizing the order of requests can minimize total seek time and improve throughput.

Understanding these components is crucial for optimizing storage system performance, but the non-linearity of seek time specifically relates to the physical mechanics of the disk arm and its inertia.

Was this answer helpful?

Important Questions from Secondary Memory

  1. Which of the following can be said about primary storage in comparison with secondary storage?

  2. __________ is the time taken to locate the disk arm to a specified track for data read/write.

  3. If one track of data can be transferred per revolution, then what is the data transfer rate ?

  4. Given below are two statements:

    Statement I: The disk has a total number of 2000 cylinders.

    Statement II: 51200 bytes is not a valid block size for the disk.

    In the light of the above statements, choose the correct answer from the options given below:

  5. If T is the capacity of a track in bytes, and S is the capacity of each surface in byte, then (T, S) = _____.

Need Expert Advice?

Start Your Preparation with Prepp Mobile App

Download the app from Google Play & App Store
Download the app from Google Play & App Store
Prepp Mobile App