rocessor A has a CPI 1.2 and a clock rate of 1 GHz. Processor B has a CPI of 2.0 and a clock rate of 2 GHz. Processor C has a dock rate of 2.5 GHz (400 ps/cycle) and a CPI of 2.6. All three processors implement the same instruction set. Determine the time per instruction for each processor and write it in the correct spot. Which processor is faster? How many times faster is it than the slowest processor?

Explanation & answer:

To understand technical questions, it is necessary to understand all technical terms used.

Here,

CPI = cycles per instruction

clock rate = number of cycles the processor generates.

1 GHz = 1,000,000,000 cycles per second

= 1 nanosecond / cycle.

Processor A

CPI = 1.2 cycles per instruction

Frequency 1 GHz.

time per instruction,

Ta = (1.2 cycles/instr.) / 1 GHz

= (1.2 cycles / instr.) / (1 cycle / nanosecond)

= 1.2 nanoseconds / instruction

Processor B

CPI = 2.0 cycles per instruction

Frequency 2 GHz.

time per instruction,

Tb = (2.0 cycles/instr.) / 2 GHz

= (2 cycles / instr.) / (2 cycles / nanosecond)

= 1.0 nanosecond / instruction

Processor C

CPI = 2.6 cycles per instruction

Frequency 2.5 GHz.

time per instruction,

Tc = (2.6 cycles/instr.) / 2.5 GHz

= (2.6 cycles / instr.) / (2.5 cycles / nanosecond)

= 1.1 nanoseconds / instruction

The fastest is Tb = 1.0 nanosecond/instruction

The slowest is Ta = 1.2 nanoseconds/instruction

The ratio fastest/slowest = 1.2/1.0 = 1.2 times as fast

Note: cycles per instructions (CPI) are always in whole number cycles. CPI for different instructions may vary according to complexity of instruction. Decimal CPI usually indicates the value is a weighted mean of the set of instructions.