GATE CSE 2015 | Set 2 | Question: 49

Question

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 _____

Comments

@sunil

You are correct. In official answer key range was $6.1 – 6.2$.

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whoever confused with controller here " in this ques controller is going to tansfer the data instead of disk whose transfer time we will calculate with the disk specification"

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Similar questions on total access time(seek time, avg latency, data transfer time) :

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Operating System: GATE IT 2007 | Question: 44, ISRO2015-34 (how to find avg seek time)

Answer 1

Average time to read/write $=$ Avg. seek time $+$ Avg. rotational delay $+$ Effective transfer time

Rotational delay $= \frac{60}{15}$ = $4$ ms

Avg. rotational delay $= \frac{1}{2} \times 4$ = $2$ ms

Avg. seek time $= 2 \times 2$ = $4$ ms

Disk transfer time $= \frac{512 \text{ Bytes}}{50*10^6 \text{ Bytes/sec}}$ = $0.0102$ ms

Effective transfer time $= 10 \times $  disk transfer time $=$ $0.102$ ms

So, avg. time to read/write $=$ $4 + 2 +0.0102 +0.102$ $= 6.11$ ms  $\bf{\approx \;\; 6.1}$ ms 

Reference: http://www.csc.villanova.edu/~japaridz/8400/sld012.htm

Comments

Disk transfer time is bits transfered per second.

There is no exact formula. you can just find how many bits are being transfered in 1 second. It's unit is bits/second

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hi what if it was asked only for the write operation on the disk, then is the Controller overhead counted?  It shouldn't because the controller transfer rate is faster than the disk write rate.

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thanks

Answer 2

Avg rotation latency :

$15,000\ rot\rightarrow60sec$

$1\ rot\rightarrow\ ?$

$1\ rot=4ms$

$\dfrac{1}{2}rot\rightarrow2ms$

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$Avg\ Seek\ time=2\times Avg\ rotation\ latency$

$Avg\ Seek\ time=4ms$

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$Data\ transfer\ time:$

$1\ sec\leftarrow 50\times 10^6B$

$?\leftarrow 512B$

$0.0102ms$

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$Controller's\ transfer\ time=10\times data\ transfer\ time$

$Controller's\ transfer\ time=0.102ms$

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Total$=6.1ms$

Comments

So why did you add Controller’s time? What role does it play?

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@Shubham_25

 L-6.2: Disk Access Time with Example | Seek Time, Rotational Time and Transfer Time - Bing video

 watch this video you will get it.

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The best solution

Answer 3

$\text{Key Concept :}$ 

Refer: Modern Operating Systems - Andrew S. Tanenbaum (Chapter 5, Section 5.1.4: Direct Memory Access), 
in that the disk read process is explained as follows:

    $\text{1.}$ The disk controller first reads the block from the disk into its internal buffer (page 344).
$$\text{Platter} \to \text{Disk Buffer}$$

    $\text{2.}$ Once valid data are in the controller’s buffer, DMA transfers the data to main memory (page 345).
$$\text{Disk Buffer} \to \text{Main Memory}$$

    

    $\text{3.}$ The disk first reads data into its internal buffer before DMA can start. (page 346–347).
 This confirms that the data path is strictly sequential (Platter -> Internal Buffer -> Main Memory) rather than concurrent.

    Therefore, the data path involved in a disk read is:

\begin{array}{c}
\text{Platter} \xrightarrow{\hspace{2.5cm}} \text{Disk Buffer} \xrightarrow{\hspace{1.5cm}} \text{Main Memory} \\[-1.5ex]
\hspace{0cm} \underbrace{\hspace{4.2cm}}_{\begin{smallmatrix} \text{Disk transfer} \\ \text{time} \end{smallmatrix}} \hspace{0.8cm} \underbrace{\hspace{4.2cm}}_{\begin{smallmatrix} \text{Controller} \\ \text{transfer time} \end{smallmatrix}}
\end{array}

 

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$\text{Given:}$

Rotation speed = $15000 \text{ RPM}$

Transfer rate = $50 \times 10^6 \text{ bytes/sec}$

Average seek time = $2 \times \text{average rotational latency}$

Sector size = $512 \text{ bytes}$

Controller transfer time = $10 \times \text{disk transfer time}$

 

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$\text{Step 1: Time for one rotation}$

15000 rotations occur in 60 seconds.

$\text{Time for one rotation } = \frac{60}{15000} \text{ sec} = 0.004 \text{ sec} = 4 \text{ ms}$

$\text{Average rotational latency } = \frac{\text{Time for one rotation}}{2} = 2 \text{ ms}$

$\therefore$ $\boxed{\text{Average seek time } = 2 \times \text{rotational latency} = 2 \times 2 \text{ ms} = 4 \text{ ms}}$

 

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$\text{Step 2: Disk transfer time for one sector}$

$\text{Transfer rate } = 50 \times 10^6 \text{ bytes/sec}$

$\text{Time to transfer 512 bytes } = \frac{512}{50 \times 10^6} \text{ sec} = 1.024 \times 10^{-5} \text{ sec} = 0.01024 \text{ ms}$

$\therefore$ $\boxed{\text{Disk transfer time } = 0.01024 \text{ ms}}$
 

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$\text{Step 3: Controller transfer time}$

$\text{Controller transfer time } = 10 \times \text{disk transfer time}$

$\therefore$ $\boxed{\text{Controller transfer time} = 10 \times 0.01024 \text{ ms } = 0.1024 \text{ ms}}$

 

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$\text{Step 4: Total average access time}$

$\text{Average access time } = \text{Seek time} + \text{Rotational delay} + \text{Disk transfer time} + \text{Controller transfer time}$

                                            $= 4 + 2 + 0.01024 + 0.1024$

                                            $= 6.11264 \text{ ms}$

$\therefore$ $\boxed{\text{Average access time } \approx 6.11 \text{ ms}}$

Answer 4

Ans 6.11

Comments

Can someone please explain the procedure to calculate this?

Thanks!

Answer 5

Answer is 6.1

Answer 6

the absolute total time required to service a sector access includes the average seek time, the average rotational latency, the physical disk data transfer time, and the controller's processing overhead.

$$\text{Total Time} = \text{Seek Time} + \text{Rotational Delay} + \text{Disk Transfer Time} + \text{Controller Time}$$

To calculate the average rotational delay, we first find the total duration of one single complete 360-degree rotation using the given speed of 15000 RPM:

$$\text{Time for 1 rotation} = \frac{60 \text{ seconds}}{15000} = 0.004 \text{ seconds} = 4 \text{ ms}$$

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

$$\text{Rotational Delay} = \frac{4 \text{ ms}}{2} = 2 \text{ ms}$$

The problem states that the average seek time is exactly twice this calculated average rotational delay value:

$$\text{Seek Time} = 2 \times 2 \text{ ms} = 4 \text{ ms}$$

The physical disk transfer time is the time taken to read or write a 512-byte sector given the raw hardware transfer rate of $50 \times 10^6 \text{ bytes/sec}$:

$$\text{Disk Transfer Time} = \frac{512 \text{ bytes}}{50 \times 10^6 \text{ bytes/sec}} = 10.24 \times 10^{-6} \text{ seconds} = 0.01024 \text{ ms}$$

The controller's total data handling overhead is explicitly given as 10 times this physical disk transfer time layer:

$$\text{Controller Time} = 10 \times 0.01024 \text{ ms} = 0.1024 \text{ ms}$$

Summing all four individual time metrics gives the average time needed to read or write the 512-byte sector:

$$\text{Total Time} = 4 \text{ ms} + 2 \text{ ms} + 0.01024 \text{ ms} + 0.1024 \text{ ms} = 6.11264 \text{ ms}$$

$$\text{Average Time} \approx 6.11 \text{ ms}$$