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A device with data transfer rate of 8 KBps is connected to a CPU. Data is transferred bytewise. Let interrupt overhead be 2 μsec. The byte transfer time between the device interface register and CPU or memory is negligible. The minimum performance gain of operating the device under interrupt mode over operating it under program-controlled mode is __________ . (Upto 1 decimal places)
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In Programmed I/O, the CPU issues a command and waits for I/O operations to complete.

So here, CPU will wait for 1 sec to transfer 8 KB of data.

overhead in programmed I/O =1 sec

In Interrupt mode , data is transferred word by word (here word size is 1 byte as mentioned in question
"Data is transferred byte-wise").
So to transfer 1 byte of data overhead is 2×10^−6 sec
Thus to transfer 8 KB of data overhead is= 2×10^−6×8×10^3  sec

Performance gain  =1/ 2×10^−6×8 ×10^3=62.5
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In Interrupt I/O formula is : x/y 

Programmed control I/O : x/x+y

Here x=time taken by cpu in interrupt processing and y = time taken by i/o device to transfer data.

Data transfer rate of 8 KBPS 

so,  8 * 210 B transfer in 1 seconds 

1 B transfer in  (1/8192 ) * 106  μsec = 0.000122 * 106  μsec = 122.07   μsec = 122  μsec

hence y = 122 and x = 2 

for Interrupt I/O processor time consumed is : (2/122) * 100 = 1.63 %

for Programmed I/O mode , processor time consumption is : ( 2/2+122) * 100 = 1.61 %

The minimum performance gain of operating the device under interrupt mode over operating it under program-controlled mode is:

( 1.63-1.61/1.61 )  = 0.63

hence performance gain is 0.6 

Same type of question :

https://gateoverflow.in/114797/interrupt-i-o

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In programed I/O, CPU does continuous polling,

8*103 byte …………….. 1 sec

1 byte …………………….. 1/ 8* 103 = 125 μsec

 In interrupt mode CPU is interrupted on completion of i\o,
To transfer 1B CPU does 2 μsec of processing(since transfer time between other components is negligible).
Gain = 125 / 2= 62.5
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in programmed control it will continuously poll the device so for 1 byte to get ready is wasted so 1B/8KBps so 125micro sec

in interrupt mode it will be 2micro sec so performance increase will be 125/2 = 62.5
Answer:

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