Computer Organization and Design MIPS Edition, Fifth Edition: The Hardware/Software Interface (The Morgan Kaufmann Series in Computer Architecture and Design)
Expert Solution & Answer
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Chapter 1, Problem 1.8.3E

Explanation of Solution

We have,

The reduction in total dissipated power is 10%

Then,

(Snew+Dnew)(Sold+Dold)=0.90 (1)

Dnew=C×V2new×F (2)

Sold=Vold×I (3)

Snew=Vnew×I (4)

From the equation (2)

Vnew=[DnewC×F]1/2 (5)

From the equation (1)

Dnew=0.90×(Sold+Dold)-Snew (6)

From the equation (3) and (4)

SnewVnew=SoldVold

Snew=Vnew×SoldVold (7)

Substitute, “10” for “Sold” and “1.25” for “Vold” in the equation (7)

Snew=Vnew×101.25

Snew=Vnew×8

Substitute, “Vnew×8” for “Snew”, “10” for “Sold” and “90” for “Dold” in the equation(6)

Dnew=0.90×(10+90)-(Vnew×8)=(0.90×100)-(Vnew×8)=90-(Vnew×8)

Substitute “90-(Vnew×8)” for “Dnew”, “3

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