Question: A microprocessor clock rate is 10GHz. If 10 clock cycles are required to fetch and decode the instruction and 15 clock cycles are required to transfer each byte.

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Final Answer:         a. The length of the instruction cycle for a string of 64 bytes is 1150 clock cycles.         b. The worst-case delay for acknowledging an interrupt, assuming the instruction is noninterruptible, is 640 clock cycles.         c. The worst-case delay for acknowledging an interrupt, assuming the instruction can be interrupted after the instruction fetch, is 810 clock cycles.          Explanation:         a. To determine the length of the instruction cycle for a string of 64 bytes, we need to consider both the clock cycles for fetching and decoding the instruction and the clock cycles required to transfer each byte. The total number of clock cycles can be calculated as follows: (Clock Cycles = Fetch and Decode Cycles + (Bytes × Transfer Cycles per Byte)). Substituting the given values (10 clock cycles for fetch and decode, 15 clock cycles per byte, and 64 bytes) yields Clock Cycles = 10 + (64 × 15) = 1150.          b. In the worst-case scenario for acknowledging an interrupt with a noninterruptible instruction, the delay is equal to the time it takes to complete the current instruction cycle. Therefore, it's 1150 clock cycles.          c. If the instruction can be interrupted after the instruction fetch, the worst-case delay for acknowledging an interrupt is the time it takes to complete the current instruction cycle (1150 clock cycles) plus the time needed to fetch and decode the next instruction (10 clock cycles). This results in a total worst-case delay of 1160 clock cycles.          Learn more about Clock cycles(source)

Understanding Microprocessor Clock Cycles: Calculation and Analysis

Question 8 (15 marks)

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