Official PYQ · 2015Set 2 · Q49I/O SchedulingNAT1 mark

GATE CS 2015 Set 2 Q49 — I/O Scheduling

Consider a typical disk that rotates at

15000

rotations per minute (RPM) and has a transfer rate of

50 \times 10^6

bytes/sec. If the average seek time of the disk is twice the average rotational delay and the controller's transfer time is

10

times the disk transfer time, the average time (in milliseconds) to read or write a

512

-byte sector of the disk is _____

View worked solution

Answer: 6.1 : 6.2

The average time to read or write a sector is the sum of the average seek time, the average rotational delay, the disk transfer time, and the controller's transfer time:

T_{\text{avg}} = T_{\text{seek}} + T_{\text{rot}} + T_{\text{trans}} + T_{\text{ctrl}}

Step 1: Calculate Average Rotational Delay ( $T_{\text{rot}}$ )

The disk rotates at $15000$ RPM (rotations per minute).

\text{Rotations per second} = \frac{15000}{60} = 250 \text{ rotations/sec}

\text{Time for one full rotation} = \frac{1}{250} \text{ seconds} = 4 \text{ ms}

The average rotational delay is half of the time for one full rotation:

T_{\text{rot}} = \frac{4 \text{ ms}}{2} = 2 \text{ ms}

Step 2: Calculate Average Seek Time ( $T_{\text{seek}}$ )

Given that the average seek time is twice the average rotational delay:

T_{\text{seek}} = 2 \times T_{\text{rot}} = 2 \times 2 \text{ ms} = 4 \text{ ms}

Step 3: Calculate Disk Transfer Time ( $T_{\text{trans}}$ )

The disk transfer rate is $50 \times 10^6$ bytes/sec, and the sector size is $512$ bytes.

T_{\text{trans}} = \frac{512 \text{ bytes}}{50 \times 10^6 \text{ bytes/sec}} = 10.24 \times 10^{-6} \text{ seconds} = 0.01024 \text{ ms}

Step 4: Calculate Controller's Transfer Time ( $T_{\text{ctrl}}$ )

Given that the controller's transfer time is $10$ times the disk transfer time:

T_{\text{ctrl}} = 10 \times T_{\text{trans}} = 10 \times 0.01024 \text{ ms} = 0.1024 \text{ ms}

Step 5: Calculate Total Average Access Time ( $T_{\text{avg}}$ )

T_{\text{avg}} = 4 \text{ ms} + 2 \text{ ms} + 0.01024 \text{ ms} + 0.1024 \text{ ms}

T_{\text{avg}} = 6.11264 \text{ ms}

Thus, the average time to read or write a sector is approximately 6.11 ms (which lies in the range 6.1 to 6.2).

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Published 10 Oct 2026

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More I/O Scheduling questions

GATE CS 2015 Set 2 Q49 — I/O Scheduling

Step 1: Calculate Average Rotational Delay (TrotT_{\text{rot}}Trot​)

Step 2: Calculate Average Seek Time (TseekT_{\text{seek}}Tseek​)

Step 3: Calculate Disk Transfer Time (TtransT_{\text{trans}}Ttrans​)

Step 4: Calculate Controller's Transfer Time (TctrlT_{\text{ctrl}}Tctrl​)

Step 5: Calculate Total Average Access Time (TavgT_{\text{avg}}Tavg​)

Related questions

Step 1: Calculate Average Rotational Delay ( $T_{\text{rot}}$ )

Step 2: Calculate Average Seek Time ( $T_{\text{seek}}$ )

Step 3: Calculate Disk Transfer Time ( $T_{\text{trans}}$ )

Step 4: Calculate Controller's Transfer Time ( $T_{\text{ctrl}}$ )

Step 5: Calculate Total Average Access Time ( $T_{\text{avg}}$ )