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Program to overcome the forward dependency using block scheduling using the most equitable distribution of work

Posted By: Joseph Evans     Category: C Programming     Views: 1172

Overcome the forward dependency using block scheduling using the most equitable distribution of work (i.e. considering excess points)

Code for Program to overcome the forward dependency using block scheduling using the most equitable distribution of work in C Programming

# include <stdio.h>
# include "/usr/include/sys/types.h"
# include "/usr/include/sys/shm.h"
# include "/usr/include/sys/ipc.h"
# include "/usr/include/sys/sem.h"
# include "forkjoin.h"
# include "sharedlib.h"
# include "spinlock.h"
# include "barrier.h"int main()
{
    int arrSize=11;
    int *arr;
    int arrold[11];            // For Holding boundary Cellsint minIndex[11];        // For Holding the Start of Each Blockint maxIndex[11];        // For Holding the End of Each Blockint id;                // Process Identificationint nProc=3;            // Number of Processesint *bararr;            // Barrier Arrayint shmidbararr,shmidarr;    // shmid for Shared Variablesint blocksize = arrSize / nProc;    // Calculating Block Sizeint index,iCount;    
    
    // Calculating excesspoints// When Number of Array Elements to Process not divisible by // Number of Processes, excesspoints occursint excesspoints=arrSize - (blocksize * nProc);
    
    arr=(int*)sshared(sizeof(int)*10,&shmidarr);
    
    printf("Number of Processes : %d\n",nProc);
    printf("Number of Excesspoints      : %d\n",excesspoints);
    
    printf("Original Array ...\n");    
    for(iCount=1;iCount<=arrSize+1;iCount++)
    {
        arr[iCount]=iCount;    // Assigning Array Values
        printf("arr[%d] : %d\n",iCount,arr[iCount]);
    }
    bararr=(int*)sshared(sizeof(int)* 4,&shmidbararr);    
    
    // Initialize Barrier Array
    barrier_init(bararr,nProc);
            
    id=process_fork(nProc);
    
    if(excesspoints ==0)
    {
        // if no excesspoints
        minIndex[id]=(id * blocksize) + 1;
        maxIndex[id]=(id * blocksize) + blocksize;
    }
    else
    {        
        if(id < excesspoints)        
        {
        // Process From ID 0 to (excesspoints-1) Will handle 1 more element,// So that there is most equitable distribution of  work
            minIndex[id]=(id * (blocksize + 1)) + 1;
            maxIndex[id]=minIndex[id] + blocksize;
        }
        else
        {
            minIndex[id]=(id * blocksize) + 1 + excesspoints;
            maxIndex[id]=minIndex[id] + blocksize - 1;
        }
        
    }
    
    
    // Copying Boundary Values
    arrold[id] = arr[maxIndex[id]+1];
    
    // Calling Barrier        
    barrier(bararr);

                
    for(iCount=minIndex[id];iCount<maxIndex[id];iCount++)
    {
        arr[iCount]=arr[iCount+1];
    }

    // Assigning Boundary Values
    arr[maxIndex[id]] = arrold[id];
        
    process_join(nProc,id);
    
    printf("After Copy ...\n");
    for(iCount=1;iCount<=arrSize;iCount++)
    {
        printf("arr[%d] = %d\n",iCount,arr[iCount]);
    }
    cleanup_memory(&shmidbararr);
    cleanup_memory(&shmidarr);
    return 0;
}
/* Output
[divyen@localhost pp-tw3]$ cc -o Prog05 Prog05.c
[divyen@localhost pp-tw3]$ ./Prog05
Number of Processes : 3
Number of Excesspoints : 2
Original Array ...
arr[1] : 1
arr[2] : 2
arr[3] : 3
arr[4] : 4
arr[5] : 5
arr[6] : 6
arr[7] : 7
arr[8] : 8
arr[9] : 9
arr[10] : 10
arr[11] : 11
arr[12] : 12
After Copy ...
arr[1] = 2
arr[2] = 3
arr[3] = 4
arr[4] = 5
arr[5] = 6
arr[6] = 7
arr[7] = 8
arr[8] = 9
arr[9] = 10
arr[10] = 11
arr[11] = 12

*/
  
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Joseph Evans
Joseph Evans author of Program to overcome the forward dependency using block scheduling using the most equitable distribution of work is from London, United Kingdom.
 
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