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Question
It points out a mistake in using a part of the performance equation as a measure of performance. For example, examine these two processors. P1 operates at a clock frequency of 4 GHz, has an average CPI of 0.9, and needs to execute 5.0E9 instructions. P2 runs at 3 GHz, with an average CPI of 0.75, needing to execute 1.0E9 instructions. 1. A common misunderstanding is assuming that the processor with the highest clock rate has the best performance. Determine if this holds true for P1 and P2. 2. Another misconception is that the processor with the greater number of executed instructions necessarily has a longer CPU time. If processor P1 processes 1.0E9 instructions and both processors have unchanged CPI values, calculate how many instructions P2 can complete in the same duration that P1 uses to execute 1.0E9 instructions. 3. A frequent error is to use MIPS (millions of instructions per second) to evaluate the performance of different processors, believing that the one with the highest MIPS is the best. Verify whether this applies to P1 and P2. 4. MFLOPS (millions of floating-point operations per second) is another common metric, defined as MFLOPS = No. FP operations / (execution time x 1E6), but it suffers from the same issues as MIPS. Assuming 40% of the instructions executed on both P1 and P2 are floating-point instructions, calculate the MFLOPS values for the programs.
Answer:
(1) We apply the following formula:
CPU time = number of instructions x CPI / Clock rate
By substituting 1 GHz = 10⁹ Hz, we find:
CPU time₁ = 5 x 10⁹ x 0.9 / 4 GHz
= 4.5 x 10⁹ / 4 x 10⁹ Hz = 1.125 s
and,
CPU time₂ = 1 x 10⁹ x 0.75 / 3 GHz
= 0.75 x 10⁹ / 3 x 10⁹ Hz = 0.25 s
This shows that P2 is significantly faster than P1 because CPU₂ is shorter than CPU₁
(2)
Determine the CPU time of P1 using (*)
CPU time₁ = 1 x 10⁹ x 0.9 / 4 GHz
= 0.9 x 10⁹ / 4 x 10⁹ Hz = 0.225 s
Next, we need the count of instructions₂ so that CPU time₂ = 0.225 s, applying (*) with clock rate₂ = 3 GHz and CPI₂ = 0.75
Thus, instruction count₂ x 0.75 / 3 GHz = 0.225 s
Consequently, instruction count₂ = 0.225 x 3 x 10⁹ / 0.75 = 9 x 10⁸
Thus, P1 can handle more instructions than P2 within the same time frame.
(3)
We remember that:
MIPS = Clock rate / CPI x 10⁶
So, MIPS₁ = 4 GHz / 0.9 x 10⁶ = 4 x 10⁹ Hz / 0.9 x 10⁶ = 4444
MIPS₂ = 3 GHz / 0.75 x 10⁶ = 3 x 10⁹ / 0.75 x 10⁶ = 4000
This indicates that P1 has a higher MIPS
(4)
Now we note that:
MFLOPS = FLOPS Instructions / time x 10⁶
= 0.4 x instructions / time x 10⁶ = 0.4 MIPS
Accordingly,
MFLOPS₁ = 1777.6
MFLOPS₂ = 1600
Again, P1 boasts a greater MFLOPS
