GATE CSE 1996 | Question: 25

Question

A hard disk is connected to a $50$ MHz processor through a DMA controller.  Assume that the initial set-up of a DMA transfer takes $1000$ clock cycles for the processor, and assume that the handling of the interrupt at DMA completion requires $500$ clock cycles for the processor. The hard disk has a transfer rate of $2000$ Kbytes/sec and average block transferred is $4$ K bytes.  What fraction of the processor time is consumed by the disk, if the disk is actively transferring $100\%$ of the time?

Comments

X is not preperation time it's data transfer time . Note that CPU will be blocked only when it will not have any control of buses which will be during transfer of data

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Ans

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why every answer differ for this problem?

Answer 1

$2000$ $KB$ is transferred in $1$ second

 $4$ $KB$ transfer is $(4/2000 ) * 1000 \text{ ms} = 2 \text{ ms}$

Total cycle required for locking and handling of interrupts after DMA transfer control

$=(1000+500) \text{ clock cycle } = 1500 \text{ clock cycle }$

Now, $50$ $Mhz = 50 * 10^6 = 0.02 \text{ microsecond}$

So, $(1500  * 0.02 ) = 30 \text{ microsecond}$

$30 \mu s$ for initialization and termination and $\ 2 ms$ for data transfer.

The CPU time is consumed only for initialization and termination.

Fraction of CPU time consumed $=\dfrac{30\mu s}{(30\mu s+2\,ms)}=0.015$

Comments

@hetgate23

If (% time CPU is blocked) is asked , then in the numerator we will take the time till which CPU is idle, which is when system bus is with the DMA, and that time is nothing but the data transfer time(which is 2ms here).

is this correct?

and one more doubt :

the data prepared inside the IO is according to its speed(rate), so here 4KB data is prepared in 2ms right? and for the transfer time we need the memory access time.

please help in this!

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What would be cpu time consumed if data is transferred only during the idle cycle of the cpu.

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wrong ans sir

Answer 2

(30/30+2000)*100  = 1.4778%

Answer 3

using DMA for data transfer from a hard disk.
initial DMA setup = 1000 cycles
Interrupt handling on DMA completion = 500 cycles
Average size of data transfer = 4 KB
Each DMA transfer takes (4 KB)/(2000 KB/sec) = 2 X 10^-3 seconds
CPU cycles used for DMA transfer = 1000 + 500 = 1500
Total CPU cycles during DMA transfer = (50 X 10⁶) X (2 X 10^-3)
= 100 X 10³
Fraction of CPU cycles used for DMA = 1500 / 100 X 10³
= 1.5 %
 

Answer 4

According to me what i figure out is

No of cycle (ticks ) by a processor = 50MHhz= 50 Million = 50 *10^6 cycle in 1 second

The time taken by 1 cycle take 0.2 μsecond

Total cycle required for locking and hadnling of interrupt after DMA transfer control=(1000+500) clock cyle = 1500 clock cycle= (1500*0.2  μsecond) = 300 μsecond

Since 2000K bytes is send in 60*10^3 msec

therefore 4K bytes can be send in (4*60*10^3)/2000= 120msec

Now as i know in 10^3 m sec = 50*10^6 cycles happen

therfore in 120 msec  , no of cycle that will happen is 6*10^6 cycle.

Now since each cycle take 0.2  μsecond

therefore 6*10^6 cycle will take 1.2 second

therfore fraction of cpu time consumed by disk = 300  μsecond / (300  μsecond + 120msec)=2.49 msec

Comments

30 us for initialisation and termination and 2 ms for data transfer% of cpu time consumed =30us/(30us+2ms)*100=1.5%

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conver 2 milliseconds to microsecond dear  2ms=2000 us...so--> (30/2030)*100=1.47%

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@Arjun sir why isnt this the case –

2000MBps here M is 10^6

 

4KB here K should be 2^10?? Isnt it? Why here 10^3 taken for K?

Answer 5

Ans

Answer 6

The hard disk transfers data at 2000 KB/s, and the average block size is 4 KB.

Therefore, the number of block transfers per second is:

$$
\frac{2000}{4} = 500 \text{ blocks/sec}
$$

For each block transfer, the processor spends:

• 1000 clock cycles for DMA setup

• 500 clock cycles for interrupt handling

Hence, total processor overhead per block is:

$$
1000 + 500 = 1500 \text{ cycles}
$$

Since there are 500 block transfers per second, total processor cycles consumed per second are:

$$
500 \times 1500 = 750000 \text{ cycles/sec}
$$

The processor speed is:

$$
50 \text{ MHz} = 50 \times 10^6 \text{ cycles/sec}
$$

Therefore, the fraction of processor time consumed by the disk is:

$$
\frac{750000}{50 \times 10^6}
= 0.015
$$

$$
= 1.5%
$$

Hence, the fraction of processor time consumed is:

$$
\boxed{0.015 \text{ (or } 1.5\%\text{)}}
$$