In general, applications are I/O bound or processor bound, meaning that the process is waiting for either I/O or everything else is waiting for the processor to finish working.
With the speed of today's CPU's, most modern applications are NOT processor bound, meaning that we're typically not waiting for the processor to finish what it's doing. Usually we're waiting for some kind of I/O process to complete.
Many times we're I/O bound when we have insufficient memory and the operating system has to "swap" an application or parts of an application out to swap space (virtual memory) on the disk. Moving an application into and out of virtual memory becomes very time consuming and is often the reason people are advised to add more Random Access Memory (RAM) to make the computer run faster.
The verb for computer is compute.Other verbs are computes, computing and computed."I am computing"."We have computed the program""I like to compute".
{A program to compute the area of a triangle} {by Ogboin W. Meshach} Var; b,h:real; BEGIN Writeln('Triangle'); Write('Base: '); Readln(base); Write('Height: '); Readln(height); area:=0.5*base*height; Writeln('Area: ', area :0:2); End.
Write a c program to compute the surface area and volume of a cube
PRINT 2,3,5,7,11,13,17,19,23,29,31,37
This isn't a question!
Explain briefly the meaning of an input-output (1/0) bound program, and a processor (CPU) bound program.
there is no solution of this problem...........that's it..........
input output bound program is a program (or process in precise way), which spends most of time allocated to it for execution, on input/output devices and need very small CPU time for it.
What is the assembly program to generate a geometric series and compute its sum The inputs are the base root and the length of the series The outputs are the series elements and their sum?
Increasing CPU speed will maintain the current level of CPU utilization, in the case where the process mix is compute bound, or decrease CPU utilization, in the case where the process mix is I/O bound.
int x= 1; int y= 2;
$n = 10*(1+10)/2;