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The relative performance of the IBM 360 Model 75 is 50 times that of the 360 Model 30, yet the instruction cycle time is only 5 times as fast. How do you account for this discrepancy?

User Iroyo
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Final answer:

The relative performance enhancement of the IBM 360 Model 75 over the Model 30 is attributed to advancements in integrated circuits, efficiency of parallel processing, pipeline architecture, and instruction sets, not just cycle time speed. This reflects Moore's Law and the rapid evolution of computing technology.

Step-by-step explanation:

The question deals with the discrepancy between the IBM 360 Model 75's relative performance, which is 50 times that of the 360 Model 30, and its instruction cycle time, which is only 5 times as fast. To account for this discrepancy, we must understand that computer performance is not solely determined by instruction cycle time. Multiple factors such as instruction set efficiency, pipeline architecture, data processing capabilities, and parallelism contribute to the overall performance of a computer system. For instance, integrated circuits and Moore's Law, which describes the rapid increase in computing power, lead to significant improvements in performance beyond just clock speed.

The IBM 360 Model 75 likely incorporated more advanced technology, including better use of parallel processing, sophisticated pipelines, and more efficient use of instruction sets, which would allow it to execute more instructions concurrently relative to cycle time. Therefore, while its instruction cycle time was only 5 times faster, these other improvements in technology and architecture resulted in a performance that was 10 times that multiplier, yielding a 50 times overall performance improvement.

User Bluelovers
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