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

  1. non-uniform distribution of requests

  2. arm starting and stopping inertia

  3. higher capacity of tracks on the periphery of the platter

  4. use of unfair arm scheduling policies

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Correct option is (B). 

The positioning time, or random-access time, consists of two parts: the time necessary to move the disk arm to the desired cylinder, called the seek time, and the time necessary for the desired sector to rotate to the disk head, called the rotational latency.

the seek latency is not linearly proportional to the seek distance due to arm starting and stopping inertia.


I couldn’t find any satisfactory answer. I do not wish to challenge

other’s answer but here is what i think.

when the seek distance is less we need less time to travel,

when it is more we naturally need more time to travel.

The question asks why doesn’t the time change w.r.t. distance.


Inertia and all is good but thinking naturally the time should vary unless we

do some optimization like increasing our speed for travelling far and

decreasing our speed for travelling less, so that we can keep avg. time constant,

or maybe some special scheduling algorithm for special cases.

So i think d should be the answer. Please help and provide some source.

Is it Explanation :/

What is the meaning of the terms seek distance and seek latency? How is it different from seek time and rotational latency?


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4 Answers

31 votes
Best answer
The answer is B, because due to Inertia

Whenever your read-write head moves from $1$ track to another track, it has to face resistance due to change in state of motion including speed and direction, which is nothing but inertia. Hence the answer is B.
edited by


" Seek times are not linear compared with the seek distance traveled because of factors of acceleration and deceleration of the actuator arm. "

Seek time and seek distance travelled. What do you mean by those two terms?
Its from wikipedia.It means seek time is not directly dependent on the seek distance traveled , but also on the acceleration/deceleration (on inertia) of actuator arm.
19 votes
edited by


I also agree with this answer.
Have drawn inferences from the link above and this paper:

An excerpt:
"A seek is composed of
• a speedup, where the arm is accelerated until it reaches half of the seek distance or a fixed maximum

Though, it's a despair that the book that I encountered this question, marks the answer as C and the Gate solution key as D.

Have already wasted an hour on this question.. duh!

Why not use of unfair arm scheduling policies

What if SCAN or LOOK algorithms are used?

can u do something better than downvoting?

The answer cannot be B, latency used by itself is used to refer to rotational latency and not seek time. Even then I find no other option to be suitable.

1 vote
Answer is B

Because of the factors of acceleration and deceleration of actuator arm.
0 votes

Note that we are asked “why seek latency is not linearly proportional to the seek distance”, that is why “duration required to travel a distance is not linearly proportional to the distance”. This definitely has to do with either the road surface or road shape. The surface here is uniform magnetic, but road shape does change. Here we don’t take turns, but put break and go on reverse gear, again put break and drive straight. Stopping a moving car takes time, it doesn’t stop instantly. Similarly accelerating stopped car to full speed takes time.
Why not option A: We have not been asked anything related to going to different cities located far from each other.
Why not option B: Tracks might be related with rotational latency / timings but has nothing to do with seek time.
Why not option C: Question is simple. Its just asking why going from one position to another takes different time each time. Policy come in to play when we are asked to read different sectors to decide which sectors to read first (if I am correctly interpreting what is meant by policy).

1 comment

Seek Latency being linear with respect to seek distance is asked. Why are you talking about Rotational Latency? 


Looking at it from the perspective of physics, say the seek arm is a constant mover, with a velocity v. For a seek distance d, the time taken would be –

$t = \dfrac{d}{v}$

However, the actuator arm doesn’t move constantly – it does rest. Hence, there is an acceleration component and a deceleration component. Hence, the equations of motion would be –

$s_1 = ut_1 + \dfrac{1}{2}at_1^2$,    -----for the acceleration

$d = vt_2$,     -----when it attains max speed

$s_2 = ut_3 + \dfrac{1}{2}at_3^2$,    -----for the deceleration

Total seek distance = $s_1 + d + s_3$

As you can see, $s_1$ and $s_2$ have non-linear dynamics w.r.t time, option B is the answer.


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