Self-test 1.3 — suggested answer | Open Textbooks for Hong Kong

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1.	$\small Average\: CPI\small = \left ( 2\times 0.2 \right ) + \left ( 1\times 0.5 \right ) + \left ( 4 \times 0.3 \right ) = 2.1$

2.	$\small CPU\:time= \frac{IC \times CPI}{Clock rate} \small = \frac{150 millions \times 2.1}{200 MHz} = 1.575$

3.	$\small Since\: performance = \frac{1}{Execution time}, therefore:$
	$\small Performance\:improvement = \frac{\frac{Clock\: rate}{IC\times2}-\frac{Clock\:rate}{IC\times 4}}{\frac{Clock\:rate}{IC\times 4}} \times100\% = 100\%$

4.	$The\: average\: CPI\: of\: the\: new\: system$
	$= \left ( 2\times0.2 \right )\left ( 1\times0.5 \right )\left ( 4\times0.3\right )=1.5$
	$\small The\: CPU\: time\: is\: therefore = \frac{150 millions \times 1.5}{200 MHz} = 1.125\: seconds$
	$Thus,\: the\: performance\: improvement$
	$=\frac{1.575-1.125}{1.575} \times 100\% = 28.57\%$

5.	$Since\: CPU\: Time = \frac{IC \times CPI}{Clock\:rate}, therefore:$
	$\small \frac{150 million \times2.1}{Clock\:rate}=1.125\:seconds$
	$\small Clock\:rate=280MHz$

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