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Alja [10]
3 years ago
13

We begin with a computer implemented in single-cycle implementation. When the stages are split by functionality, the stages do n

ot require exactly the same amount of time. The original machine had a clock cycle time of 7 ns. After the stages were split, the measured times were IF, 1 ns; ID, 1.5 ns; EX, 1 ns; MEM, 2 ns; and WB, 1.5 ns. The pipeline register delay is 0.1 ns.
a. What is the clock cycle time of the 5-stage pipelined machine?
b. If there is a stall every 4 instructions, what is the CPI of the new ?machine?
c. What is the speedup of the pipelined machine over the single- cycle machine?
d. If the pipelined machine had an infinite number of stages, what would its speedup be over the single-cycle machine?
Computers and Technology
1 answer:
vampirchik [111]3 years ago
8 0

<u>Answer:</u>

a) First, we need to determine the pipeline stage amounting to the maximum time. In the given case, the maximum time required is 2ns for MEM. In addition, the pipeline register delay=0.1 ns.

Clock cycled time of the pipelined machine= max time+delay

=2ns+0.1 ns

=2.1 ns

b) For any processor, ideal CPI=1. However, since there is a stall after every four instructions, the effective CPI of the new machine is specified by:

1+(1 / 4)=1.25

c) The speedup of pipelined machine over the single-cycle machine=avg time per instruction of single cycle/avg time per instruction of pipelined.

Single cycle processor:

CPI=1

Clock period=7 ns

Pipelined processor:

Clock period=2.1 ns

CPI=1.25

Therefore, speedup==7^{*} 1 /\left(2.1^{*} 1.25\right)

=7/2.625

= 2.67

d) As the number of stages approach infinity, the speedup=k where k is the number of stages in the machine.

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Explanation:

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3 years ago
Construct a class that will model a quadratic expression (ax^2 + bx + c). In addition to a constructor creating a quadratic expr
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Answer:

Following are the code to this question:

#include <iostream>//header file

#include<math.h>//header file

using namespace std;

class Quadratic//defining a class Quadratic  

{

private:

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public:

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return a;//return value a

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void setA(double a)//defining a set method to hold value in parameter

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this->a = a;//assigning value in a variable

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double getB() //defining a get method  

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return b;//return value b

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void setB(double b)//defining a set method to hold value in parameter  

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this->b = b;//assigning value in b variable

}

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{

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this->c = c;//assigning value in c variable

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double r2=(-b-sqrt(numberOfReal()))/(2*a);//holding root value r2

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else//else block

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}

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{

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{

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