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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)

1. $256000$ instructions per second
2. $512000$ instructions per second
3. $1024000$ instructions per second
4. None of the above

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This is Good question , Assuming that this question is correct.

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.