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Assume a system has a TLB hit ratio of 90%. It requires 15 nanoseconds to access the TLB, and 85 nanoseconds to access main memo

andriy [413]

Answer:

310 ns

Explanation:

Given that

TLB hit ratio = 90%

TLB hit ratio = 90/100

TLB hit ratio = 0.9

Time needed to access TLB = 15 ns

Time needed to access main Memory = 85 ns

Effective memory access time = ?.

The formula for finding the effective memory access time is given by

The effective memory access time = [TLB Hit ratio (main memory access time + required time to access TLB) + [2 * (main memory access time + required time to access TLB)] * (2 - TLB hit ratio)]

On substituting the values given in the question to the equation, we have

The effective memory access time = [0.9 (85 + 15) + [2 * (85 + 15)] * (2 - 0.90)]

The effective memory access time =

[(0.9 * 100) + (2 * 100) * 1.1]

The effective memory access time =

(90 + (200 * 1.1))

The effective memory access time =

90 + 220

The effective memory access time =

310 ns

3 0

3 years ago

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Samantha was calculating a mathematical formula on an electronic spreadsheet. She used multiple values to recalculate the formul

Bezzdna [24]

<span>Random access memory. This problem requires you to know what the different types of memory are and their relative advantages and disadvantages. Let's look at them and see why 3 are wrong and one is correct. read-only memory: Otherwise known as ROM, this type of memory stores code that can't be over written. Used frequently for constant lookup values and boot code. Since it can't be written to by normal programs, it can't hold temporary values for Samantha. So this is the wrong choice. random-access memory: Otherwise known as RAM, this type of memory is used to store temporary values and program code. It is quite fast to access and most the immediately required variables and program code is stored here. It can both be written to and read from. This is the correct answer. hard disk: This is permanent long term readable and writable memory. It will retain its contents even while powered off. But accessing it is slow. Where the contents of RAM can be accessed in nanoseconds, hard disk takes milliseconds to seconds to access (millions to billions of times slower than RAM). Because it's slow, this is not the correct answer. But it's likely that Samantha will save her spreadsheet to hard disk when she's finished working with it so she can retrieve the spreadsheet later to work on again. compact disk: This is sort of the ROM equivalent to the hard disk. The data stored on a compact disk can not be over written. One way of describing the storage on a compact disk is "Write Once, Read many times". In most cases it's even slower than the hard disk. But can be useful for archiving information or making backups of the data on your computer.</span>

5 0

4 years ago

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HELP ASAP!! WILL MARK BRAINLIEST

lesya692 [45]

Answer:20*i

Explanation:

Because the size will expand as its being written out to then have i

4 0

3 years ago

List at least 5 features that can be used to format a report in word 2013

Masteriza [31]

Answer:

Following are the features for formatting a report in word 2013

Design Tab
Convenient Layout options and alignment
Resume work option
Managing longer documents in a better way
Enhanced table features

Explanation:

In word 2013, many of the new features were introduced:

The design tab was given a new look, including templates options and much more.
The layout option was also made much convenient by putting all necessary option under the right-click option.
Word started to present an option for resuming the work were you left last time. This made the working much efficient.
By dealing with headings the word 2013 version made it easier to compact the document by collapsing heading and focus on the specific part you want.
Many of the options were added to tables formatting such as border, colors, line weight and much more.

i hope it will help you!

8 0

3 years ago

g % Create a program to plot the motion of the ping pong ball with drag. % The program will take inputs for velocity in m/s and

Gala2k [10]

Answer:

% Create a program to plot the motion of the ping pong ball with drag.

% The program will take inputs for velocity in m/s and angle of launch.

% note that with a high speed camera I estmated a launch speed for a ping

% pong ball could be 30 m/s.

clear all

close all

clc

velStart=input('input the initial velocity in m/sec: ');

angle=20:5:80;

% convert the degrees into radians so MATLAB likes it

values_angle=angle*pi./180;

rangeAngle=length(values_angle);

hold on

%Lines 1 - 14 are fine

for i=1:rangeAngle

% set initial position and time

x=[];

y=[];

time=[];

vel=velStart;

VelocityX=[];

VelocityY=[];

x(1)=0; % meters

y(1)=.001; % meters

time(1)=0; % seconds

mass=.00247; %kg ping pong ball .00247 Kg

g=-9.8; % m/sec^2

c=0.0005; % coefficient of drag where Re is between 10^3 and 10^5

% so I can load an array for plotting

% start to increment the motion

index=1;

thisAngle=values_angle(i);

velx=vel*cos(thisAngle);

VelocityX(1)=velx;

vely=vel*sin(thisAngle);

VelocityY(1)=vely;

% set a time step

deltaTime=.001; % seconds

height=y(1);

flag=0;

velFinalX = 0;

velFinalY = 0;

%Don't see any issues here

while height>0 % check that the ball has not hit ground yet (1e-3 to speed things up, It hangs because it never leaves this line, check how height is calculated.)

index = index+1;

% *******************************************************

% break velocity into its components

if index == 2

velx = vel*cos(thisAngle);

vely = vel*sin(thisAngle);

else

velx = velFinalX;

vely = velFinalY;

end

% *******************************************************

% use an if/else statement to check to see if the ball is moving

% down (negative). If it is, then drag has an opposite sign

% as gravity in the acceleration formula. Otherwise gravity

% and drag have the same sign. Calculate the new acceleration in the y.

if vely < 0

adragy = g + (c*vely.^2)/mass;

velFinalY = vely + adragy.*deltaTime;

else

adragy = g - (c*vely.^2)/mass;

velFinalY = vely + adragy.*deltaTime;

end

% *******************************************************

% Now calculate the acceleration in the x .

adragx = (-c*velx.^2)/mass;

% *******************************************************

% *****************************************************

% calculate the new velocity at the end of the time step

% this will have X and Y components, so you need a variable

% for each. One is velFinalX and the other is velFinalY.

velFinalX = velx + adragx.*deltaTime;

% velFinalY = vely + adragy.*deltaTime;

% *******************************************************

% ******************************************************

% Get a new velocity vector and angle given the X and Y

% The velocity is the variable "vel" and angle is "angle"

vel = sqrt(velFinalX.^2 + velFinalY.^2);

angle = atan(velFinalY/velFinalX);

%*******************************************************

% now save my values at this time step

VelocityX(index)=velFinalX;

VelocityY(index)=velFinalY;

% and distance numbers

distX=VelocityX(index).*deltaTime;

distY=VelocityY(index).*deltaTime;

% save distance values

x(index)=x(index-1)+distX;

y(index)=y(index-1)+distY;

height=y(index);

time(index)=time(index-1)+deltaTime;

%lines 105-107 are good

if distY<0&&flag<=1

if height<.25

disp(distY)

fprintf('goal height detected at distance %.2f meters.',x(index));

flag =2;

LandingDistance(i)=x(index);

end

plot(x,y)

title('distance traveled by ping pong ball in meters')

xlabel('horizontal distance traveled (meters)')

ylabel('vertical distance traveled (meters)')

end

hold off

Explanation:

7 0

4 years ago