| Freq | CPI | |
|---|---|---|
| ALU | 43% | 1 |
| LOAD | 21% | 4 |
| STORE | 12% | 4 |
| BRANCH | 12% | 2 |
| JUMP | 12% | 2 |
$N_{istr}=200$ $F=500MHz$
$CPI,T_{CPU}=?$
$CPI=0,431+0,214+0,124+0,122+0,12*2=2,23$
$T_{CPU}=\frac{CPIN_{istr}}F=\frac{2,23200}{50010^6}=\frac{2002,23}{0,510^9}=2002,23*\frac{1}{2^{-1}}\frac{1}{10^9}=2002,23210^{-9}\newline =2002,232ns=892ns$
oppure
$T_{CPU}=2002,23\frac{1}{500*10^6}=0,000000892s=892ns$
| N_istr | CPI | T_clock | |
|---|---|---|---|
| CPU_1 | 2N | 2 | T |
| CPU_2 | N | 1.5 | 2T |
$T_{CPU_1}=2N2T=4N*T$
$T_{CPU_2}=N1.52T=3N*T$
Quindi $T_{CPU_2}<T_{CPU_1}$
$MIPS=\frac {\frac{N_{istr}}{10^6}}{T_{CPU}}=\frac{N_{istr}}{10^6}*\frac{1}{T_{CPU}}=\frac{N_{istr}}{T_{CPU}*10^6}$
$MIPS_1=\frac {2N}{4NT10^6}=\frac{1}{2T10^6}=\frac{1}{2}\frac{1}{T10^6}$ $MIPS_2=\frac N{3NT10^6}=\frac{1}{3T10^6}=\frac{1}{3}\frac{1}{T10^6}$
$MIPS_1>MIPS_2$
Il MIPS può contraddire l’effettiva performance di una CPU
| CPI_1 | CPI_2 | |
|---|---|---|
| A | 1 | 2 |
| B | 2 | 2 |
| C | 3 | 4 |
| D | 4 | 4 |
La frequenza è equalmente distribuita.