Assume that a machine includes a floating point (FP) unit, which contains an FP square root...

Assume that a machine includes a floating point (FP) unit, which contains an FP square root

(FPSQR) instruction. Suppose we have the following measurements. Frequency of FP operations = 20%, Frequency of FPSQR = 4% Average CPI of FP operations = 4.0, CPI of FPSQR = 20

Average CPI of other instructions (non-FP instructions) = 1.4

What is the average CPI of the machine?
If the average CPI of FP is decreased to 3, what is the average CPI of the new machine?
If the CPI of FPSQR is decreased to 2, what is the average CPI of the new machine?
Which design improvement is better of b and c?

Expert Solution

SOLUTION-

Given,
Frequency of FP operations =20%,
Frequency of FPSQR = 4%
Average CPI of FP operations = 4.0,
CPI of FPSQR=20
Average CPI of other instructions(no-FP instructions) = 1.4
100 - 20 (Freq _FP ) = 80 (Freq _non-FP)

(a) What is the average CPI of the machine?
Answer:
CPI _machine= (CPI _FP ∗ Freq _FP) + (CPI _non−FP ∗ Freq _non−FP)
= (4.0∗20%) + (1.4∗80)
= (4.0∗0.2) + (1.4∗0.8)
= 0.8 + 1.92
= 2.72 (Average CPI original)

(b) If the average CPI of FP is decreased to 3, what is the average CPI of the new machine?Answer:
CPI _new = CPI _avg − Freq _FP ∗ (CPI _FP − CPI _FPnew)
= 2.72 − 20 ∗ (4−3)
= 2.72 − 0.2 ∗ (4−3)
= 2.72 - 0.2 * 1
= 2.52 (CPI new average)

(c) If the CPI of FPSQR is decreased to 2, what is the average CPI of the new machine?Answer:
CPI _new=CPI _avg−Freq _FPSQR∗(CPI _FPSQR−CPI _FPSQRnew)
= 2.72 − 4 ∗ (20−2)
= 2.72 − 0.04 (18)
= 2.72 − 0.72
= 2 (CPI new average)

(d) Which design improvement is better of b and c?
Answer:
Speed _FP = CPI _avg / CPI _new
= 2.72 / 2.52
= 1.07

Speed _FP = CPI _avg / CPI _new
= 2.72 / 2
= 1.36

So we can say that B is a better improvement.

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venereology answered 1 year ago

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