GATE CSE 2004 | Question: 68

Question

A hard disk with a transfer rate of $10$ Mbytes/second is constantly transferring data to memory using DMA. The processor runs at $600$ MHz, and takes $300$ and $900$ clock cycles to initiate and complete DMA transfer respectively. If the size of the transfer is $20$ Kbytes, what is the percentage of processor time consumed for the transfer operation?

• A. $5.0 \%$
• B. $1.0\%$
• C. $0.5\%$
• D. $0.1\%$

Comments

Ans

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The first 300 cycles spent by the $CPU is for the DMA register initialization(source address, destination address, count register, and mode register).

The 900 cycles needed by the CPU at the end/completion of the data transfer are spent in servicing the ISR for the interrupt raised by the DMA after completion.

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@goku4199 in dma questions hard disk data transfer rate is given, so i have a doubt that , Is this the time for hard disk to I/O controller transfer or i/o controller to mm transfer

Answer 1

Clock cycle time = $\frac{1}{600 \times 10^6}$   [ Frequency = 1/Time]

 For DMA initiation and completion = $\frac{(900+300)}{600\times10^6}  = 2$ microsec .

 Disk Transfer rate $= 10$ Mbytes/sec

 $1$ byte $= \frac{1}{10^7}$ sec

 $20$ Kbytes $= 2$ milisec $= 2000$ micro sec

  Percentage $= \left (\frac{2}{2+2000} \right )\times100    =0.0999 ≃ 0.1\%$

 option (D)

 
$\%$ of CPU time consumed $=\frac{x}{x+y}$

If $x$ is the Data preparation time or Total cycle time used by CPU and $y$ is the Data transfer time, to calculate the fraction of CPU time to the data transfer time -  we use $\frac{x}{x+y}$ in burst mode.

Comments

When the data is being prepared in the burst mode, the CPU is blocked. So, how CPU is busy for the data preparation time x.

Someone please clear my doubt.

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@saurabh0709 CPU is involved in 2 stages here for Initialisation of the DMA and Handling interrupt after the Data transfer is complete. In between these 2 stages the CPU is idel and DMAC is in control of system bus. 

For more clarity please go through this lecture https://youtu.be/v3qnkhEu72s?si=alk-tkQH4wUgzVp-&t=5836

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thanks bro for youtube link !!

Answer 2

Clock cycle time = (1/600) x 10 ^6

 For DMA initiation and competion = (900+300)/(600 x 10 ^6) = 2 microsec .

Disk Transfer rate = 10 M bytes. 1 byte = 1/10^7 sec

20 K bytes = 2000 microsec Percentage = (2/2000)*100 =0.1%

so, ans D

.

Comments

How is it 2/2000 ??

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@Himanshu1

Because, "Hard disk is "constantly" transferring data to memory using DMA.

This means, Hard disk uses some sort of buffer.

The formula for this type of questions is x/y and not x/x+y.

Tag me, if you feel my perspective is wrong. 

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P0535_Yedidyah_Sagar

It is x/x+y here just because we are finding the CPU time taken which is
x = CPU time spent during DMA (Initialization and Final stage) / Total time spent during the complete DMA process (i.e y= Data transfer time for 20KB + x=CPU time)

i.e x/x+y *100 gives you the percentage of time spent by Processor during data transfer

 

Answer 3

Data transfer time= initialization time + DMA transfer time =300+900=1200 cycles

Data preparation time = time disk is taking to transfer 20KB to DMA

                                   = (20KB)/(10MBps) = 2msec or 1200000 cycles

thus CPU time consumed = (transfer time)/(prepartion time + transfer time)

                                        = [1200/(1200000+1200)] * 100

                                        =0.0999 = 0.1%

Answer 4

It is simple logical question. No need of so many calculations.
The answer is simply
 (20KB/10MB)x100
= (20KB/10x1024KB)x100
= 0.1%

The reason for this is whatever may be the processor speed and clock cycle . While taking percentage ratio they will get cancel led out.

Answer 5

Processor Speed = 600MHZ = $6*10^{8}$ cycles/second

Transfer time =   $\frac{20KB}{10MB}$ = 2ms = $12*10^{5}$ cycles

Total cycles consumed = 1200,000 + 300 + 900 = 120,1200 cycles
% of processor time consumed =  $\frac{120,1200*100}{6*10^{8}} =  \frac{12012}{60000}$= 0.2%

So, (D) can be chosen as option!

Comments

@manu00x

how you are getting 0.2?

i am getting 0.1022 %(approx)

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how?? what will be 12012/60000 = ?

Answer 6

Transfer rate=10 MB per second

Data=20 KB=20* 2 ¹⁰

So Time=(20 * 2 ¹⁰)/(10 * 2 ²⁰)= 2* 10^-3 =2 ms

Processor speed= 600 MHz=600 Cycles/sec

Cycles required by CPU=300+900 =1200

For DMA=1200

So time=1200/(600 *10 ⁶)=.002 ms

 In %=.002/2*100=.1%

So (D) is correct option

Comments

what kind of logic is this, first converted into millisecond and converted it into %??