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