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CaHeK987 [17]
3 years ago
12

Server farms such as Google and Yahoo! provide enough compute capacity for the highest request rate of the day. Imagine that mos

t of the time these servers operate at only 60% capacity. Assume further that the power does not scale linearly with the load; that is, when the servers are operating at 60% capacity, they consume 90% of maximum power. The servers could be turned off, but they would take too long to restart in response to more load. A new system has been proposed that allows for a quick restart but requires 20% of the maximum power while in this "barely alive" state. a. How much power savings would be achieved by turning off 60% of the servers? b. How much power savings would be achieved by placing 60% of the servers in the "barely alive" state? c. How much power savings would be achieved by reducing the voltage by 20% and frequency by 40%? d. How much power savings would be achieved by placing 30% of the servers in the "barely alive" state and 30% off?

Computers and Technology
1 answer:
EleoNora [17]3 years ago
3 0

Answer:

a) Power saving = 26.7%

b) power saving = 48%

c) Power saving = 61%

d) Power saving = 25.3%

Explanation:

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const2013 [10]

Answer:

b. connect two or more network segments that use different data link protocols

Explanation:

A switch is an analog device that allows to interconnect networks, it is possibly one of the devices with a higher level of scalability. It is important to clarify that a Switch does not only provide connectivity with other networks, nor does it provide connectivity with the Internet, for this purpose a router is necessary.  

6 0
3 years ago
in python, using the simplest form the print_seconds function so that it prints the total amount of seconds given the hours, min
PolarNik [594]

Answer:

I'm guessing you want a function so...

def print_seconds(hours, minutes, seconds):

seconds += 3600 * hours + 60 * minutes

print(seconds)

return 0

Hope this helps. :)

4 0
3 years ago
Que implicaciones sociales puede tener la robótica
wolverine [178]
Según estos estudios, el impacto de los robots en la productividad ya se compara con la contribución de las máquinas de vapor en el pasado. ... Algunos de los aumentos en la productividad que se derivan de la densificación de los robots se comparten con los trabajadores mediante el pago de salarios más altos.
8 0
3 years ago
Different methods of developing useful information from large data bases are dealt with under
oee [108]

Answer:

The answer to this question is option "d".

Explanation:

In this question, the answer is data mining. Because Data mining is a technique that finds a piece of new information in a lot of the data. In the data mining collecting information from data is hopefully both new and useful. Data mining help us to discover new patterns and relationship in data to help make better decisions. It is used everywhere like Television and radio, Banking, Retail, and business. That's why data mining is useful.

4 0
3 years ago
Create a Binary Expressions Tree Class and create a menu driven programyour program should be able to read multiple expressions
solong [7]

Answer:

Explanation:

Program:

#include<iostream>

#include <bits/stdc++.h>

using namespace std;

//check for operator

bool isOperator(char c)

{

switch(c)

{

case '+': case '-': case '/': case '*': case '^':

return true;

}

return false;

}

//Converter class

class Converter

{

private:

string str;

public:

//constructor

Converter(string s):str(s){}

//convert from infix to postfix expression

string toPostFix(string str)

{

stack <char> as;

int i, pre1, pre2;

string result="";

as.push('(');

str = str + ")";

for (i = 0; i < str.size(); i++)

{

char ch = str[i];

if(ch==' ') continue;

if (ch == '(')

as.push(ch);

else if (ch == ')')

{

while (as.size() != 0 && as.top() != '('){

result = result + as.top() + " ";

as.pop();

}

as.pop();

}

else if(isOperator(ch))

{

while (as.size() != 0 && as.top() != '(')

{

pre1 = precedence(ch);

pre2 = precedence(as.top());

if (pre2 >= pre1){

result = result + as.top() + " ";

as.pop();

}

else break;

}

as.push(ch);

}

else

{

result = result + ch;

}

}

while(as.size() != 0 && as.top() != '(') {

result += as.top() + " ";

as.pop();

}

return result;

}

//return the precedence of an operator

int precedence(char ch)

{

int choice = 0;

switch (ch) {

case '+':

choice = 0;

break;

case '-':

choice = 0;

break;

case '*':

choice = 1;

break;

case '/':

choice = 1;

break;

case '^':

choice = 2;

default:

choice = -999;

}

return choice;

}

};

//Node class

class Node

{

public:

string element;

Node *leftChild;

Node *rightChild;

//constructors

Node (string s):element(s),leftChild(nullptr),rightChild(nullptr) {}

Node (string s, Node* l, Node* r):element(s),leftChild(l),rightChild(r) {}

};

//ExpressionTree class

class ExpressionTree

{

public:

//expression tree construction

Node* covert(string postfix)

{

stack <Node*> stk;

Node *t = nullptr;

for(int i=0; i<postfix.size(); i++)

{

if(postfix[i]==' ') continue;

string s(1, postfix[i]);

t = new Node(s);

if(!isOperator(postfix[i]))

{

stk.push(t);

}

else

{

Node *r = nullptr, *l = nullptr;

if(!stk.empty()){

r = stk.top();

stk.pop();

}

if(!stk.empty()){

l = stk.top();

stk.pop();

}

t->leftChild = l;

t->rightChild = r;

stk.push(t);

}

}

return stk.top();

}

//inorder traversal

void infix(Node *root)

{

if(root!=nullptr)

{

cout<< "(";

infix(root->leftChild);

cout<<root->element;

infix(root->rightChild);

cout<<")";

}

}

//postorder traversal

void postfix(Node *root)

{

if(root!=nullptr)

{

postfix(root->leftChild);

postfix(root->rightChild);

cout << root->element << " ";

}

}

//preorder traversal

void prefix(Node *root)

{

if(root!=nullptr)

{

cout<< root->element << " ";

prefix(root->leftChild);

prefix(root->rightChild);

}

}

};

//main method

int main()

{

string infix;

cout<<"Enter the expression: ";

cin >> infix;

Converter conv(infix);

string postfix = conv.toPostFix(infix);

cout<<"Postfix Expression: " << postfix<<endl;

if(postfix == "")

{

cout<<"Invalid expression";

return 1;

}

ExpressionTree etree;

Node *root = etree.covert(postfix);

cout<<"Infix: ";

etree.infix(root);

cout<<endl;

cout<<"Prefix: ";

etree.prefix(root);

cout<<endl;

cout<< "Postfix: ";

etree.postfix(root);

cout<<endl;

return 0;

}

3 0
4 years ago
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