Computer Speed Suppose a program has a certain number of statements, which on av

Question

Computer Speed Suppose a program has a certain number of statements, which on average execute many times in repetitions, that is, statementExecutions = (number of statements) Times (average repetition count). A compiler translates this into object-code instructions. Since source-code statements often include more than one explicit primitive operation, and there are often many invisible primitive operations in the background, the number of object-code instructions is typically much greater than the number of source code statements. We can represent these effects with the multiplier, instructionsPerStatement.

Explanation / Answer

Program:

// import the required files

import java.io.*;

import java.util.*;

import java.lang.management.*;

public class computerspeed

{

     public static void main(String[] args)

     {

          // idle process start time

          long idealstartTime = System.nanoTime();

          // named constands

          double SOURCE_CODE = 10000;

          double AVERAGE_INSTRUCTIONS= 20.0;

          double CLOCK_RATE= 2000.0;

          double AVERAGE_INSTRUCTIONS_CYCLE= 1.0;

          double FIRST_LEVEL_CACHE_HIT= 0.99;

          double FIRST_LEVEL_CACHE_SWAP =0.0010;

          double SECOND_LEVEL_CACHE_HIT=0.999;

          double SECOND_LEVEL_CACHE_SWAP=0.0050;

          double MAIN_MEMORY= 0.9999;

          double MAIN_MEMORY_PAGE=4096;

          double DISK_SPEED=500.0;

          double DISK_TRACK=400000;

          // execution process start time

          long processstartTime = System.nanoTime();

System.out.println("Assumed executions of source-code statements:"+SOURCE_CODE);

System.out.println("Assumed average instructions/statement:"+AVERAGE_INSTRUCTIONS);

System.out.println("Assumed clock rate in megahertz:"+CLOCK_RATE);

System.out.println("Assumed average instructions/cycle:"+AVERAGE_INSTRUCTIONS_CYCLE);

System.out.println("Assumed first-level cache hit fraction:"+FIRST_LEVEL_CACHE_HIT);

System.out.println("Assumed first-level cache swap time in microsec: "+FIRST_LEVEL_CACHE_SWAP);

System.out.println("Assumed second-level cache hit fraction:"+SECOND_LEVEL_CACHE_HIT);

System.out.println("Assumed second-level cache swap time in microsec:"+SECOND_LEVEL_CACHE_SWAP);

System.out.println("Assumed main memory hit fraction: "+MAIN_MEMORY);

System.out.println("Assumed main memory page size in bytes/page:"+MAIN_MEMORY_PAGE);

System.out.println("Assumed disk speed in revolutions/sec:"+DISK_SPEED);

System.out.println("Assumed disk track length in bytes/revolution:"+DISK_TRACK);

          // process end time

          long stopTime = System.nanoTime();

          // total execution time

          long tottime=stopTime-idealstartTime;

          // process execution time

          long processtime=stopTime-processstartTime;

          // idle time

          long idletime=tottime-processtime;

          // calculated time

System.out.println("Ideal execution time = " +idletime+" nano seconds.");

System.out.println("Expected execution time = " + processtime + " nano seconds.");

     }

}

Result:

Assumed executions of source-code statements: 10000.0

Assumed average instructions/statement: 20.0

Assumed clock rate in megahertz: 2000.0

Assumed average instructions/cycle: 1.0

Assumed first-level cache hit fraction: 0.99

Assumed first-level cache swap time in microsec: 0.0010

Assumed second-level cache hit fraction: 0.999

Assumed second-level cache swap time in microsec: 0.0050

Assumed main memory hit fraction: 0.9999

Assumed main memory page size in bytes/page: 4096.0

Assumed disk speed in revolutions/sec: 500.0

Assumed disk track length in bytes/revolution: 400000.0

Ideal execution time = 1400 nano seconds.

Expected execution time = 23831454 nano seconds.

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