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برمجة السمبلكس بلغة سي

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سـؤال

أريد شفرة مصدرية أو برنامج مكتوب بلغة سي لحل مسائل سمبلكس.

[عدل]

رد

ما يلي شفرة مصدرية (المصدر) بلغة سي:

#include <stdio.h>
#include <conio.h>
#define INFINITY 999
#define N 3
#define M 6
/************************************************************/
/***** Solves the LPP by "SIMPLEX" method i.e. by table *****/
/************************************************************/
void minimum(float *arr,int *arrminpos,int n);
/* Calculates the minimum valued position among the array arr having n elements. */
void display (float c[],float b[],float a[][M],int basic[]);
/* Display the table */
void displayframe(float c[M]);
/* Displays the frame of the table */
void calctemp(float *,float [][M],float [],int []);
/* Calculates Zj-Cj */
/*--------------------------------------------------------------------------*\
Cj        5        4        3        0        0        0        miniRatio
cB       xB        b        a1       a2       a3       a4       a5       a6    bi/aij
0        x4        5        2        3        1        1        0        0        2.5
0        x5        11       4        1        2        0        1        0        2.75
0        x6        8        3        4        2        0        0        1        2.66
----------------------------------------------------------------------------
Zj-Cj              -5       -4       -3       0        0        0
----------------------------------------------------------------------------
5        x1        2.5      1        1.5      0.5      0.5      0        0        5
0        x5        1        0        -5       0        -2       1        0        infinity
0        x6        105      0        -0.5     0.5      -1.5     0        1        1
----------------------------------------------------------------------------
Zj-Cj              0        3.5      -0.5     2.5      0        0
----------------------------------------------------------------------------
5        x1        2        1        2        0        2        0        -1
0        x5        1        0        -5       0        -2       1        0
3        x3        1        0        -1       1        -3       0        2
----------------------------------------------------------------------------
Zj-Cj              0        3        0        1        0        1
----------------------------------------------------------------------------
So the solution is :-
x1=2        x2=0        x3=1        x4=0        x5=1        x6=0
max(z) = 5*2 + 4*0 + 3*1 = 13.
\*--------------------------------------------------------------------------*/
 
void main()
{
  float c[M]={{5},{4},{3},{0},{0},{0}};
  /* Stores co-efficient of the objective function Max(z) */
  float a[N][M]={
    {2,3,1,1,0,0},
    {4,1,2,0,1,0},
    {3,4,2,0,0,1}
  };
  /* Stores the co-efficent of the constraints */
  float b[N]={{5},{11},{8}};
  /* Stores the values on RHS of constraints */
  float temp[M]={{0},{0},{0},{0},{0},{0}};
  /* Stores the values of Zj-Cj*/
  int tempminpos;      /* Stores the minimum valued position
              of {Zj-Cj} i.e. coming in variable */
  float miniratio[N];   /* Stores the value of the ratio b[i]/a[i][j] */
  int miniratiominpos;  /* Stores the minimum valued position of
              b[i]/a[i][j] i.e. going out variable */
  float key;          /* Stores the key element */
  int gooutcol;        /* Stores the column number which goes out */
  float z;          /* Stores the value of the objective function */
  float x[M];        /* Stores the value of the variables */
  int i,j;        /* Loop variables */
  int basic[N];        /* Stores the basic variable */
  int nonbasic[N];    /* Stores the non-basic variable */
  int flag=0;      /* Terminating variable */
  //clrscr();
  /*** Initializing basic variables to 3,4,5 i.e. x4,x5,x6 ***/
  for(i=0;i<N;i++)
  {
    basic[i]=(i+N);
    nonbasic[i]=i;
  }
  printf("\nMax z = c1x1 + c2x2 + c3x3\n");
  printf("\na11x1 + a12x2 + a13x3 <= b1\n");
  printf("\na21x1 + a22x2 + a23x3 <= b2\n");
  printf("\na31x1 + a31x2 + a32x3 <= b3\n");
  printf("\nEnter values of ci's\n");
  /*** Inputing requisite amount of data ***/
  for(i=0;i<N;i++)
  {
    printf("\nEnter c[%d]\t",i+1);
    scanf("%f",&c[i]);
  }
  printf("\nEnter values of ai's\n");
  for(i=0;i<N;i++)
  {
    for(j=0;j<N;j++)
    {
      printf("\nEnter a[%d][%d]\t",i+1,j+1);
      scanf("%f",&a[i][j]);
    }
  }
  printf("\nEnter values of bi's\n");
  for(i=0;i<N;i++)
  {
    printf("\nEnter b[%d]\t",i+1);
    scanf("%f",&b[i]);
  }
  /*** Calculation for actual table ***/
  while(flag==0)
  {
    z=0;
    calctemp(temp,a,c,basic);
    printf("\n");
 
    /*** Determining the incoming column ***/
 
    minimum(temp,&tempminpos,M);
    display(c,b,a,basic);
    printf("\nZj-Cj\t\t\t");
    for(i=0;i<M;i++)
      printf("%.4g\t",temp[i]);
    printf("\n\n");
    for(i=0;i<N;i++)
    {
      x[basic[i]]=b[i];
      x[nonbasic[i]]=0;
      printf("x[%d]=%g\n",basic[i]+1,b[i]);
    }
    for(i=0;i<N;i++)
      z=z+c[i]*x[i];
    printf("Max(z) = %g",z);
 
    /*** Determining the outgoing column ***/
 
    for(i=0;i<N;i++)
    {
      if(a[i][tempminpos]==0)
      {
        miniratio[i]=INFINITY;
        continue;
      }
      if(a[i][tempminpos]<0)
      {
        miniratio[i]=INFINITY;
        continue;
      }
      miniratio[i]=b[i]/a[i][tempminpos];
    }
    minimum(miniratio,&miniratiominpos,N);
    for(i=0;i<N;i++)
      if(miniratiominpos==i)
        gooutcol=basic[i];
    printf("\nComing in variable = X%d\t",tempminpos+1);
    printf("Going out variable = X%d\n",gooutcol+1);
 
    /*** Changing the basic and non-basic variable ***/
 
    basic[miniratiominpos]=tempminpos;
    nonbasic[tempminpos]=gooutcol;
 
    /*** Performing the operations to bring similar expressions in
    in-coming variable as out-going variable by row operations ***/
 
    key=a[miniratiominpos][tempminpos];
    b[miniratiominpos]=b[miniratiominpos]/key;
    for(i=0;i<M;i++)
      a[miniratiominpos][i]=a[miniratiominpos][i]/key;
    for(i=0;i<N;i++)
    {
      if(miniratiominpos==i)
        continue;
      key=a[i][tempminpos];
      for(j=0;j<M;j++)
      {
        a[i][j]=a[i][j]-a[miniratiominpos][j]*key;
      }
      b[i]=b[i]-b[miniratiominpos]*key;
    }
    getch();
 
    /*** Terminating condition ***/
 
    for(i=0;i<M;i++)
    {
      flag=1;
      if(temp[i]<0)
      {
        flag=0;
        break;
      }
    }
  }
  printf("\nPress any key to exit...\n");
  getch();
}
void calctemp(float *temp,float a[N][M],float c[M],int basic[N])
{
  int i,j;
  for(i=0;i<M;i++)
  {
    temp[i]=0;
    for(j=0;j<N;j++)
      temp[i]=temp[i]+c[basic[j]]*a[j][i];
    temp[i]=temp[i]-c[i];
  }
}
void minimum(float *arr,int *arrminpos, int n)
{
  int i;
  float arrmin;
  arrmin=arr[0];
  *arrminpos=0;
  for(i=0;i<n;i++)
    if(arr[i]<arrmin)
    {
      arrmin=arr[i];
      *arrminpos=i;
    }
    printf("\n%d\n",*arrminpos);
}
void display (float c[N],float b[N],float a[N][M],int basic[N])
{
  int i,j;
  displayframe(c);
  for(i=0;i<N;i++)
  {
    printf("\n%.4g\tX%d\t%.4g\t",c[basic[i]],basic[i]+1,b[i]);
    for(j=0;j<M;j++)
      printf("%.4g\t",a[i][j]);
    printf("\n");
  }
}
void displayframe(float c[M])
{
  printf("\t\tc[j]\t");
  printf("%g\t%g\t%g\t%g\t%g\t%g\n",c[0],c[1],c[2],c[3],c[4],c[5]);
  printf("\nc[b]\tB\tb\ta1\ta2\ta3\ta4\ta5\ta6\n");
}

البرنامج التالي أيضاً لحل مسائل البرمجة الخطية بطريقة التبسيط المزدوجة:

#include <stdio.h>
#include <conio.h>
#define INFINITY 999
#define N 3
#define M 6
/************************************************************/
/***** Solves the LPP by "SIMPLEX" method i.e. by table *****/
/************************************************************/
void minimum(float *arr,int *arrminpos,int n);
/* Calculates the minimum valued position among the array arr having n elements. */
void display (float c[],float b[],float a[][M],int basic[]);
/* Display the table */
void displayframe(float c[M]);
/* Displays the frame of the table */
void calctemp(float *,float [][M],float [],int []);
/* Calculates Zj-Cj */
/*--------------------------------------------------------------------------*\
Cj        5        4        3        0        0        0        miniRatio
cB       xB        b        a1       a2       a3       a4       a5       a6    bi/aij
0        x4        5        2        3        1        1        0        0        2.5
0        x5        11       4        1        2        0        1        0        2.75
0        x6        8        3        4        2        0        0        1        2.66
----------------------------------------------------------------------------
Zj-Cj              -5       -4       -3       0        0        0
----------------------------------------------------------------------------
5        x1        2.5      1        1.5      0.5      0.5      0        0        5
0        x5        1        0        -5       0        -2       1        0        infinity
0        x6        105      0        -0.5     0.5      -1.5     0        1        1
----------------------------------------------------------------------------
Zj-Cj              0        3.5      -0.5     2.5      0        0
----------------------------------------------------------------------------
5        x1        2        1        2        0        2        0        -1
0        x5        1        0        -5       0        -2       1        0
3        x3        1        0        -1       1        -3       0        2
----------------------------------------------------------------------------
Zj-Cj              0        3        0        1        0        1
----------------------------------------------------------------------------
So the solution is :-
x1=2        x2=0        x3=1        x4=0        x5=1        x6=0
max(z) = 5*2 + 4*0 + 3*1 = 13.
\*--------------------------------------------------------------------------*/
 
void main()
{
  float c[M]={{5},{4},{3},{0},{0},{0}};
  /* Stores co-efficient of the objective function Max(z) */
  float a[N][M]={
    {2,3,1,1,0,0},
    {4,1,2,0,1,0},
    {3,4,2,0,0,1}
  };
  /* Stores the co-efficent of the constraints */
  float b[N]={{5},{11},{8}};
  /* Stores the values on RHS of constraints */
  float temp[M]={{0},{0},{0},{0},{0},{0}};
  /* Stores the values of Zj-Cj*/
  int tempminpos;      /* Stores the minimum valued position
              of {Zj-Cj} i.e. coming in variable */
  float miniratio[N];   /* Stores the value of the ratio b[i]/a[i][j] */
  int miniratiominpos;  /* Stores the minimum valued position of
              b[i]/a[i][j] i.e. going out variable */
  float key;          /* Stores the key element */
  int gooutcol;        /* Stores the column number which goes out */
  float z;          /* Stores the value of the objective function */
  float x[M];        /* Stores the value of the variables */
  int i,j;        /* Loop variables */
  int basic[N];        /* Stores the basic variable */
  int nonbasic[N];    /* Stores the non-basic variable */
  int flag=0;      /* Terminating variable */
  //clrscr();
  /*** Initializing basic variables to 3,4,5 i.e. x4,x5,x6 ***/
  for(i=0;i<N;i++)
  {
    basic[i]=(i+N);
    nonbasic[i]=i;
  }
  printf("\nMax z = c1x1 + c2x2 + c3x3\n");
  printf("\na11x1 + a12x2 + a13x3 <= b1\n");
  printf("\na21x1 + a22x2 + a23x3 <= b2\n");
  printf("\na31x1 + a31x2 + a32x3 <= b3\n");
  printf("\nEnter values of ci's\n");
  /*** Inputing requisite amount of data ***/
  for(i=0;i<N;i++)
  {
    printf("\nEnter c[%d]\t",i+1);
    scanf("%f",&c[i]);
  }
  printf("\nEnter values of ai's\n");
  for(i=0;i<N;i++)
  {
    for(j=0;j<N;j++)
    {
      printf("\nEnter a[%d][%d]\t",i+1,j+1);
      scanf("%f",&a[i][j]);
    }
  }
  printf("\nEnter values of bi's\n");
  for(i=0;i<N;i++)
  {
    printf("\nEnter b[%d]\t",i+1);
    scanf("%f",&b[i]);
  }
  /*** Calculation for actual table ***/
  while(flag==0)
  {
    z=0;
    calctemp(temp,a,c,basic);
    printf("\n");
 
    /*** Determining the incoming column ***/
 
    minimum(temp,&tempminpos,M);
    display(c,b,a,basic);
    printf("\nZj-Cj\t\t\t");
    for(i=0;i<M;i++)
      printf("%.4g\t",temp[i]);
    printf("\n\n");
    for(i=0;i<N;i++)
    {
      x[basic[i]]=b[i];
      x[nonbasic[i]]=0;
      printf("x[%d]=%g\n",basic[i]+1,b[i]);
    }
    for(i=0;i<N;i++)
      z=z+c[i]*x[i];
    printf("Max(z) = %g",z);
 
    /*** Determining the outgoing column ***/
 
    for(i=0;i<N;i++)
    {
      if(a[i][tempminpos]==0)
      {
        miniratio[i]=INFINITY;
        continue;
      }
      if(a[i][tempminpos]<0)
      {
        miniratio[i]=INFINITY;
        continue;
      }
      miniratio[i]=b[i]/a[i][tempminpos];
    }
    minimum(miniratio,&miniratiominpos,N);
    for(i=0;i<N;i++)
      if(miniratiominpos==i)
        gooutcol=basic[i];
    printf("\nComing in variable = X%d\t",tempminpos+1);
    printf("Going out variable = X%d\n",gooutcol+1);
 
    /*** Changing the basic and non-basic variable ***/
 
    basic[miniratiominpos]=tempminpos;
    nonbasic[tempminpos]=gooutcol;
 
    /*** Performing the operations to bring similar expressions in
    in-coming variable as out-going variable by row operations ***/
 
    key=a[miniratiominpos][tempminpos];
    b[miniratiominpos]=b[miniratiominpos]/key;
    for(i=0;i<M;i++)
      a[miniratiominpos][i]=a[miniratiominpos][i]/key;
    for(i=0;i<N;i++)
    {
      if(miniratiominpos==i)
        continue;
      key=a[i][tempminpos];
      for(j=0;j<M;j++)
      {
        a[i][j]=a[i][j]-a[miniratiominpos][j]*key;
      }
      b[i]=b[i]-b[miniratiominpos]*key;
    }
    getch();
 
    /*** Terminating condition ***/
 
    for(i=0;i<M;i++)
    {
      flag=1;
      if(temp[i]<0)
      {
        flag=0;
        break;
      }
    }
  }
  printf("\nPress any key to exit...\n");
  getch();
}
void calctemp(float *temp,float a[N][M],float c[M],int basic[N])
{
  int i,j;
  for(i=0;i<M;i++)
  {
    temp[i]=0;
    for(j=0;j<N;j++)
      temp[i]=temp[i]+c[basic[j]]*a[j][i];
    temp[i]=temp[i]-c[i];
  }
}
void minimum(float *arr,int *arrminpos, int n)
{
  int i;
  float arrmin;
  arrmin=arr[0];
  *arrminpos=0;
  for(i=0;i<n;i++)
    if(arr[i]<arrmin)
    {
      arrmin=arr[i];
      *arrminpos=i;
    }
    printf("\n%d\n",*arrminpos);
}
void display (float c[N],float b[N],float a[N][M],int basic[N])
{
  int i,j;
  displayframe(c);
  for(i=0;i<N;i++)
  {
    printf("\n%.4g\tX%d\t%.4g\t",c[basic[i]],basic[i]+1,b[i]);
    for(j=0;j<M;j++)
      printf("%.4g\t",a[i][j]);
    printf("\n");
  }
}
void displayframe(float c[M])
{
  printf("\t\tc[j]\t");
  printf("%g\t%g\t%g\t%g\t%g\t%g\n",c[0],c[1],c[2],c[3],c[4],c[5]);
  printf("\nc[b]\tB\tb\ta1\ta2\ta3\ta4\ta5\ta6\n");
}

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