Fcfs small program

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Your machine would be able to perform 5 X 8 embroideries. The hoop size on this machine is recognized by the machine itself, depending of the embroidery file size. So it the embroidery is small, it will indicate that the small hoop is needed.

But U can use the large hoop, position the embroidery where U want then perform another embroidery on the sams hoop.

TRICK : the embroidery size exceeds the hoop? reduce it in X and or Y using this function on the machine

Firstly this may be a bit over my head, but, i will try to help u.
CPU scheduling, is all about what the the CPU's priority's are .( Or tasks that come first.) Have you tried a block diagram. ie. a bit like a pyramid Ok got as far as MLQF then done a search on internet.
Well done good one..

Question 4: Consider the following set of processes that arrive in the ready queue at the same time:

Process CPU time

P1 2

P2 1

P3 4

P4 3

P5 1

P6 2



Consider the following scheduling algorithms: FCFS, SJF and Round Robin (quantum = 1)

(i) What is turnaround time of each process for each of the above

scheduling algorithms?

(ii) What is the waiting time of each process for each of the above

algorithms?

well threading does this for you, look at the thread class and related classess in .net

Nice try! Find an upper-class student to give you one-on-one tutoring with your homework.

// crt_clock.c // This example prompts for how long // the program is to run and then continuously // displays the elapsed time for that period. // #include <stdio.h> #include <stdlib.h> #include <time.h> void sleep( clock_t wait ); int main( void ) { long i = 6000000L; clock_t start, finish; double duration; // Delay for a specified time. printf( "Delay for three seconds\n" ); sleep( (clock_t)3 * CLOCKS_PER_SEC ); printf( "Done!\n" ); // Measure the duration of an event. printf( "Time to do %ld empty loops is ", i ); start = clock(); while( i-- ) ; finish = clock(); duration = (double)(finish - start) / CLOCKS_PER_SEC; printf( "%2.1f seconds\n", duration ); } // Pauses for a specified number of milliseconds. void sleep( clock_t wait ) { clock_t goal; goal = wait + clock(); while( goal > clock() ) ; }

Reprogram one at a time with others unplugged...no power no programming will be received...pretty simple.