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Consider a device of 1MBPS is operating in a cycle stealing mode of DMA .Whenever 16 B word is available it is transferred into memory in 4 microseconds. What is the % of time the processor is blocked due to DMA?

a) 10%

b) 20%

c) 80%

d) 90%
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you can try this approach also...

Ok let's do it.., we have given a device with speed 1MBps and it is operating in cycle stealing mode

As we know in DMA (transfer time + preparation time) both we need to consider and here preparation time for this device is

1sec --> 1MB

16Byte --> 16 μsec

Transfer time is given as 4 μsec

% of CPU blocked due to DMA =              ( Transfer time / (transfer time + preparation time) ) * 100

                                                   =               (4 / (4+16)) * 100

                                                    =              20

Hence option (B) is correct... 20%

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