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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 _____

edited | 5.7k views
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$6.11264$ is the exact answer, but how to know that up to how many bits we can round off the answer?

But I think, $6.11264$ is in the range of $6.1-6.2$ So, it should be considered as correct answer, can someone please confirm?
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Disk transfer time means data transfer time and usually, we transfer even a small amount of data by reading one entire sector in which it is stored, so consider one sector read time as the disk transfer time.

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

by Veteran (434k points)
edited
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avg. time to read/write = 4 + 2 + 0.0102  + 0.102 =6.112
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how it was 512/50*10^3 , as it was given 50*10^6 @Arjun
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@ Arjun sir,

I wrote 6.1024 as the answer and got it wrong. Is it not good practise to write ans upto 4 decimal digits. Kindly update me because it can make a difference.
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@Arjun sir

in the link controller overhead + transfer time is given....so ans should be 6.11 ??/?
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@amol No, floating point answers are in range- so you won't loose mark for extra precision in GATE.

@Lokesh Controller overhead same as controller transfer time?
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@Arjun. What is actually disk transfer time?

In some examples, they take it as time required to fill the controller buffer and some cases(as is here in this example), they take it time it takes for the disk controller to transfer it to memory.

I am confused.
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Is controller overhead different than controller transfer time?
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What is the formula to calculate  Avg Disk Transfer Time?
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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.
Ans 6.11
by Boss (13.7k points)
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Can someone please explain the procedure to calculate this?

Thanks!
+1 vote

by Junior (817 points)
+1 vote

Avg rotation latency :

$15,000\ rot\rightarrow60sec$

$1\ rot\rightarrow\ ?$

$1\ rot=4ms$

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

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

$Avg\ Seek\ time=4ms$

$Data\ transfer\ time:$

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

$?\leftarrow 512B$

$0.0102ms$

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

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

Total$=6.1ms$

by Loyal (5.5k points)
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Here controller transfer means what ???

does controller refers to R/W head ??

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Maybe. But I am not sure.