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Anon25 [30]
3 years ago
13

A program executes 15 billion instructions. You are running this program in two systems: System A: has a processor with 3 GHz an

d an average CPI of 2.0 System B: has an average CPI of 1.2 and execution time is 1.5 X more than system A. What is the clock speed of processor in system B???????
Computers and Technology
1 answer:
katrin2010 [14]3 years ago
7 0

Answer:

CPU clock cycles = Instruction count x CPI.

CPU execution time =

= CPU clock cycles x Clock cycle.

= Instruction count x CPI x Clock cycle.

T =

I.

x CPI x C

Explanation:

i'm guessing

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Find the root using bisection method with initials 1 and 2 for function 0.005(e^(2x))cos(x) in matlab and error 1e-10?
frutty [35]

Answer:

The root is:

c=1.5708

Explanation:

Use this script in Matlab:

-------------------------------------------------------------------------------------

function  [c, err, yc] = bisect (f, a, b, delta)

% f the function introduce as n anonymous function

%       - a y b are the initial and the final value respectively

%       - delta is the tolerance or error.

%           - c is the root

%       - yc = f(c)

%        - err is the stimated error for  c

ya = feval(f, a);

yb = feval(f, b);

if  ya*yb > 0, return, end

max1 = 1 + round((log(b-a) - log(delta)) / log(2));

for  k = 1:max1

c = (a + b) / 2;

yc = feval(f, c);

if  yc == 0

 a = c;

 b = c;

elseif  yb*yc > 0

 b = c;

 yb = yc;

else

 a = c;

 ya = yc;

end

if  b-a < delta, break, end

end

c = (a + b) / 2;

err = abs(b - a);

yc = feval(f, c);

-------------------------------------------------------------------------------------

Enter the function in matlab like this:

f= @(x) 0.005*(exp(2*x)*cos(x))

You should get this result:

f =

 function_handle with value:

   @(x)0.005*(exp(2*x)*cos(x))

Now run the code like this:

[c, err, yc] = bisect (f, 1, 2, 1e-10)

You should get this result:

c =

   1.5708

err =

  5.8208e-11

yc =

 -3.0708e-12

In addition, you can use the plot function to verify your results:

fplot(f,[1,2])

grid on

5 0
3 years ago
One line of code is missing (marked in
Harman [31]

Answer: num1.plus(1);

Explanation:

4 0
3 years ago
For a new version of processor, suppose the capacitive load remains, how much more energy will the processor consume if we incre
erastovalidia [21]

Answer:

The answer is below

Explanation:

The amount of power dissipated by a processor is given by the formula:

P = fCV²

Where f = clock rate, C = capacitance and V = Voltage

For the old version of processor with a clock rate of f, capacitance C and voltage of V, the power dissipated is:

P(old) = fCV²

For the new version of processor with a clock rate of 20% increase = (100% + 20%)f = 1.2f, capacitance is the same = C and voltage of 20% increase = 1.2V, the power dissipated is:

P(new) = 1.2f × C × (1.2V)² = 1.2f × C × 1.44V² =1.728fCV² = 1.728 × Power dissipated by old processor

Hence, the new processor is 1.728 times (72.8% more) the power of the old processor

4 0
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Several people work with data at Erica’s office. She enters data. One of her coworkers enters new product numbers. Another cowor
Dvinal [7]

Answer:

a software program for storing, managing, and retrieving information

Explanation:

8 0
3 years ago
The county clerk's office is writing a program to search their database for citizen records based on information about that citi
Alborosie

Answer:

The correct pseudocode to find the records of all citizens over the age of 50 is IF(age > 50).

OR EACH item IN citzlist

{

WHILE(not end of citzlist)

{

IF(age > 50)

{

DISPLAY(name)

}

}

}

If this is run, it will bring out all the names of the citizen who are over the age of 50 in the list.

Explanation:

6 0
3 years ago
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