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Consider a system in which DMA technique is used to transfer 16 MB of data from an I/O device into memory. The bandwidth of I/O device is 128 KB/s. The clock cycle of the CPU is 10 microseconds and word length of CPU is 32 bits. What percentage of time is the CPU in busy mode (upto 2 decimal place) ?
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Bandwidth of I/O is mentioned =128KB/s

1 sec -- 128 Kbyte.

and we have to transfer of data size = 16MB

16MB how much time ?

1 byte = 1/128K sec

therefore 16MB -- ---> 16M/128k = (2^4*2^20)/(2^7*2^10) = 2^7 sec

CPU cycle time is 10 microsecond  

% CPU busy = 10*10^-6/(10*10^-6 + (128 *10^6)/10^6 micro sec)

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