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Keith_Richards [23]

3 years ago

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The answer to the question is what the

2 answers:

Brut [27]3 years ago

7 0

The correct answer is wrong with the correct spelling correct

11111nata11111 [884]3 years ago

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The answer to the question is what

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Assume a program requires the execution of 50 x 10^6 FP instructions, 110 x 10^6 INT instructions, 80 x 10^6 Load/Store (L/S) in

Svetradugi [14.3K]

Answer:

We can not improve CPI of FP instructions when we run the program two times faster because it would be negative.

Explanation:

Processor clock rate = 2 GHz

Execution Time = ∑ $(\frac{Clock cyles}{Clock rate})$

Clock cycles can be determined using following formula

Clock cycles = ( $CPI_{FP}$ x No. FP instructions )+ ( $CPI_{INT}$ x No. INT instructions) + ( $CPI_{L/S}$ x No. L/S instructions ) + ( $CPI_{branch}$ x No. branch instructions)

Clock cycles = ( 50 x $10^{6}$ x 1) + ( 110 x $10^{6}$ x 1) + ( 80 x $10^{6}$ x 4) + ( 16 x $10^{6}$ x 2)

Clock cycles = 512 x 10⁶

So,Initial Execution time for FP instructions is,

= $\frac{512(10^{6}) }{2(10^{9}) }$

Initial execution Time = 256 x 10⁻³

For 16 processors ,

clock cycle = 512 x 10⁶

Execution Time = 256 x 10⁻³

To run the program two times faster, half the number of clock cycles

( $\frac{Clockcycles}{2}$ )= ( $CPI_{FP}$ x No. FP instructions )+ ( $CPI_{INT}$ x No. INT instructions) + ( $CPI_{L/S}$ x No. L/S instructions ) + ( $CPI_{branch}$ x No. branch instructions)

$CPI_{FP improved}$ x No. FP instructions = ( $\frac{Clockcycles}{2}$ ) -[ ( $CPI_{INT}$ x No. INT instructions) + ( $CPI_{L/S}$ x No. L/S instructions ) + ( $CPI_{branch}$ x No. branch instructions)]

$CPI_{FP improved}$ x 50 x $10^{6}$ = ( $\frac{512(10)^{6} }{2}$ ) - [ ( 110 x $10^{6}$ x 1) + ( 80 x $10^{6}$ x 4) + ( 16 x $10^{6}$ x 2)]

$CPI_{FP improved}$ x 50 x $10^{6}$ = - 206 x $10^{6}$

$CPI_{FP improved}$ = - 206 x $10^{6}$ / 50 x $10^{6}$

$CPI_{FP improved}$ = - 4.12 < 0

3 0

4 years ago

What energy does a curcuit board run on

son4ous [18]

a curcuit board is powered by energy from the computers power soarce

6 0

3 years ago

Read 2 more answers

To provide some perspective on the dimensions of atomic defects, consider a metal specimen that has a dislocation density of 105

GenaCL600 [577]

Answer:

$62.14\ \text{miles}$

$6213727.37\ \text{miles}$

Explanation:

The distance of the chain would be the product of the dislocation density and the volume of the metal.

Dislocation density = $10^5\ \text{mm}^{-2}$

Volume of the metal = $1000\ \text{mm}^3$

$10^5\times 1000=10^8\ \text{mm}\\ =10^5\ \text{m}$

$1\ \text{mile}=1609.34\ \text{m}$

$\dfrac{10^5}{1609.34}=62.14\ \text{miles}$

The chain would extend $62.14\ \text{miles}$

Dislocation density = $10^{10}\ \text{mm}^{-2}$

Volume of the metal = $1000\ \text{mm}^3$

$10^{10}\times 1000=10^{13}\ \text{mm}\\ =10^{10}\ \text{m}$

$\dfrac{10^{10}}{1609.34}=6213727.37\ \text{miles}$

The chain would extend $6213727.37\ \text{miles}$

3 0

3 years ago

Determine the dimensions for W if W = P L^3 / (M V^2) where P is a pressure, L is a length, M is a mass, and V is a velocity.

Eva8 [605]

Correct answer is option E. No dimensions

As we know formula Pressure (P) is $\frac{F}{A}$

also,

Dimensional formula of <em>Pressure is </em> $M^{1}L^{-1}T^{-2}$
Dimensional formula of <em>length is L </em>
Dimensional formula of <em>mass is M</em>
Dimensional formula of <em>velocity is </em> $L^{1} T^{-1}$

So, as given W= $\frac{P*L^{3} }{M*V^{2} }$

Dimensional formula of W = $\frac{M^{1}L^{-1}T^{-2} L^{3} }{M^{1} L^{2}T^{-2} }$

since all terms get cancelled

Work is dimensionless i.e no dimensions

Learn more about dimensions here brainly.com/question/20351712

#SPJ10

6 0

2 years ago

How do Consumers sometimes interact with a producers?

Vika [28.1K]

I really don’t know good luck

7 0

3 years ago