5 5 votes Consider a $32$ bit, $10$ MIPS processor with an interrupt driven interface. Suppose a hard disk has a $16$ bit data bus and is connected to the processor and its transfer rate is $50 \:KB$ per second. Calculate the Processor time required to service hard disk when it is active? (Assume interrupt overhead is $20$ instructions)__________ (instructions per second) $256000$ instructions per second $512000$ instructions per second $1024000$ instructions per second None of the above CO & Architecture applied-course-2019-mock1 io-handling co-and-architecture interrupts + – Applied Course 1.5k views answer comment Share Follow Print See 1 comment 1 1 comment reply mehul vaidya commented Jun 4, 2019 reply Follow flag This is Good question , Assuming that this question is correct. 0 0 replyShare Please log in or register to add a comment.
2 2 votes Since the hard disk has a $16$-bit data bus, it can transfer two bytes at one time. Thus its transfer rate is $50/2 = 25 \: K$ half-words ($16$-bits each) per second. This corresponds to an overhead of $20$ instructions or $25 \: K \times 20 =25 \times 2^{10} \times 20 = 512000$ instruction per second. Applied Course answered Jan 16, 2019 Applied Course comment Share Follow 0 reply Please log in or register to add a comment.
