Measuring performance

"This processor runs at 4 GHz" says how often its clock ticks, not how fast your program runs. The time a program takes on the CPU is:

CPU time = instruction count × CPI × clock period = instruction count × CPI ÷ clock rate

programs = {"A at 3 GHz, CPI 1.0": (2e9, 1.0, 3e9), "B at 4 GHz, CPI 2.0": (2e9, 2.0, 4e9)}

for name, (count, cpi, clock) in programs.items():
    seconds = count * cpi / clock
    print(name, "->", round(seconds, 3), "s")

CPI from an instruction mix

Different instructions take different numbers of cycles. CPI is their weighted average:

mix = {"ALU": (0.50, 1), "load": (0.20, 5), "store": (0.10, 3), "branch": (0.20, 2)}   # share, cycles

cpi = 0
for kind, (share, cycles) in mix.items():
    cpi += share * cycles
    print(kind, ":", share, "x", cycles, "cycles")
print("average CPI:", round(cpi, 2))

Comparing two machines by clock rate alone misleads: a 3 GHz processor with CPI 1 beats a 4 GHz one with CPI 2 on the same program. MIPS (millions of instructions per second = clock rate ÷ CPI ÷ 10⁶) misleads too, because a machine with simpler instructions needs more of them. The only fair test is to run real programs: benchmarks such as SPEC CPU.

Amdahl's law

Speeding up one part helps only as much as that part matters. If a fraction p of the time can be made s times faster:

overall speedup = 1 ÷ ((1 − p) + p ÷ s)

Make 80% of a program 10 times faster and the whole is only 1 ÷ (0.2 + 0.08) ≈ 3.6 times faster. Even with an infinitely fast part, it can never beat 1 ÷ 0.2 = 5. The lesson: speed up the common case first.

Mark lesson complete