Given Data

Clock $f = 10$ GHz $\Rightarrow T_c = 1/f = 0.1$ ns $= 10^{-10}$ s

Type	CPI$_i$ ($c_i$)	$I_i$
Control	5	5000
Data transfer	4	4000
Floating	2	2000
Integer	1	1000

Total instructions $I = \sum I_i = 5000+4000+2000+1000 = 12000$

Total cycles $C = \sum c_i I_i$

$C = 5000\times5 + 4000\times4 + 2000\times2 + 1000\times1$
$C = 25000 + 16000 + 4000 + 1000 = 46000$ cycles

(a) CPI (Average Clock Cycles per Instruction)

$CPI = \dfrac{C}{I} = \dfrac{46000}{12000} = 3.833... \approx 3.83$ cycles/instr

If weighted: $CPI = \sum (I_i/I)\times c_i = 0.4167\times5 + 0.3333\times4 +0.1667\times2+0.0833\times1 =3.83$

(b) Execution Time

$T_{cpu}= C \times T_c = \dfrac{C}{f}$

$T_{cpu}= \dfrac{46000}{10\times10^9}= 4.6\times10^{-6}$ s $= 4.6 \mu s$

Alternatively $T_{cpu}= I \times CPI \times T_c =12000\times3.833\times0.1ns =4600 ns$

$T_{cpu}=4.6 \mu s$ (if $f=1GHz$, $46 \mu s$)

(c) MIPS Rate

$MIPS = \dfrac{I}{T_{cpu}\times10^6} = \dfrac{f}{CPI\times10^6}$

$MIPS = \dfrac{10\times10^9}{3.833\times10^6}= 2608.7$

$\approx 2609$ MIPS $\approx 2.61\times10^3$ MIPS $=2.61$ BIPS

Check: $MIPS =12000/(4.6\times10^{-6}\times10^6)=12000/4.6=2608.7$

If using MIPS = I/(T in $\mu s$) same result.

Summary: CPI $\approx 3.83$, $T \approx 4.6 \mu s$ at 10GHz ($46 \mu s$ at 1GHz), ~2609 MIPS.