Computer Performance

Last updated Nov 8, 2022 Edit Source

One indicator of performance is execution time of a program $$\text{Performance} = \frac{1}{\text{Execution Time}} = \frac{1}{time_{end} - time_{start}}$$ $$\text{Execution Time} = \text{Instruction Count} \times \text{Clocks Per Instruction} \times \text{Clock Period}$$

$$\text{Average CPI}=\text{Cycle Count}/\text{Instruction Count}$$

Clock period is the inverse of clock frequency Memory wall problem: a higher clock frequency may not result in better performance if the the time needed for memory access operations is slower than the CPU

Speed-up is the factor over the original machine of improved performance $$\text{Speedup} = \frac{Perf_a}{Perf_b}$$ We can define the execution time of an enhanced machine by the proportion of the program E that is improved and T the original time taken and S the enhancement factor $$T’ = (T\times (1-E))+\frac{T\times E}{S}$$

If the program is of a fixed workload: Let E be the fraction of program that is enhanced via parallelism, with maximum enhancement factor $S = \infty$, the maximum speedup is $$\text{Max Speedup} =lim_{s\rightarrow \infty}\frac{1}{1-E+\frac{E}{S}}=\frac{1}{1-E}$$ Let E be the fraction of program enhanced by Speedup $S_1$ and $1-E$ enhanced by Speedup $S_2$. $$Speedup=\frac{1}{\frac{1-E}{S_2}+\frac{E}{S_1}}$$

If the program is set to a fixed time period instead: do more parallel work in the same amount of time

$$ \begin{align} &IC=200+500+300=1000 \&T_c=100ns\\ &CPI_{average}=(200\times1+500\times2+300\times3)/1000=2.1\\ &T_{execution}=1000\times100\times2.1=210ms \end{align} $$ a. $$ \begin{align} &10=200\times10^6\times\frac{1}{200\times10^6}\times CPI_{avg}\\ &CPI_{avg}=10\\ &5=160\times10^6\times\frac{1}{300\times10^6}\times CPI_{avg}\\ &CPI_{avg}=9.375 \end{align} $$ b. $$ \begin{align} &IC_a=4\div(\frac{10}{200\times10^6})=80\times10^6\\ &IC_b=3\div(\frac{9.375}{300\times10^6})=96\times10^6 \end{align} $$ a. $$ \begin{align} &T_{M2}=(1+2+3+4)\times\frac{1}{500\times10^6}\times4\\ &T_{M3}=(2+2+4+4)\times\frac{1}{750\times10^6}\times4\\ &\text{Speedup}=1.25 \end{align} $$ b. $$ \begin{align} &T_{M1}=2\times\frac{1}{500\times10^6}\times1\\ &T_{M2}=1\times\frac{1}{500\times10^6}\times1\\ &T_{M3}=2\times\frac{1}{750\times10^6}\times1\\ &\text{Speedup M2 over M1}=2 &\text{Speedup M3 over M1}=1.5 \end{align} $$