AMITY UNIVERSITY
Computer Oriented Statistical and
Optimization Methods
Submitted to:
Submitted by
:
Ms. upasana Sharma
Jasdeep Singh
BCA (III-A)
A1004814024
INDEX
S.No. Experiments Signature
1. Write a program to find mean, median and
mode.
2. Write a program to evaluate measure of
dispersion.
3. Write a program to find skewness.
4. Write a program to find Kurtosis.
5. Write a program to find regression line of y on x from any given set of points.
6. Write a program to find regression line of x on y from any given set of points.
7. Write a program to find the solution of linear equations using Simplex Method.
8. Write a program to find the solution of linear equations using Big M Method.
9. Write a program to find the solution of Transportation problem using North-West corner Method.
10. Write a program to find the solution of Transportation problem using Least Cost Method.
11. Write a program to find the solution of
Program 1. Write a program to find mean, median and mode.
Solution:-#include <iostream> using namespace std; void mode(int[ ], int); void mean(int[ ], int); void sort(int[ ], int); void median(int[ ], int); int main( )
{
int array[15]; float total, mode;
int n = 15;//number of elements in array //fill in the value of array
for(int i=0; i<n; i++){
cout << "fill in the "<< i+1 << " number. :"; cin >> array[i];
}
sort(array, n); return 0; }
void mean(int new_array[], int num){ //GET TOTAL & CALCULATE MEAN float total;
for(int i=0;i<num; i++){ total += new_array[i]; }
cout << "The mean is " << total/num << endl; mode(new_array, num);
}
void median(int new_array[], int num){
//CALCULATE THE MEDIAN (middle number) if(num % 2 != 0){// is the # of elements odd? int temp = ((num+1)/2)-1;
cout << "The median is " << new_array[temp] << endl; }
else{// then it's even! :)
cout << "The median is "<< new_array[(num/2)-1] << " and " << new_array[num/2] << endl;
}
mean(new_array, num); }
void mode(int new_array[], int num) { int* ipRepetition = new int[num];
// alocate a new array in memory of the same size (round about way of defining number of elements by a variable) for (int i = 0; i < num; i++) {
ipRepetition[i] = 0;//initialize each element to 0 int j = 0;//
while ((j < i) && (new_array[i] != new_array[j])) { if (new_array[i] != new_array[j]) { j++; } } (ipRepetition[j])++; } int iMaxRepeat = 0;
for (int i = 1; i < num; i++) {
if (ipRepetition[i] > ipRepetition[iMaxRepeat]) { iMaxRepeat = i;
} }
cout<< "The mode is " << new_array[iMaxRepeat] << endl; }
void sort(int new_array[], int num){ //ARRANGE VALUES
for(int x=0; x<num; x++){ for(int y=0; y<num-1; y++){
if(new_array[y]>new_array[y+1]){ int temp = new_array[y+1]; new_array[y+1] = new_array[y]; new_array[y] = temp; } } } cout << "List: ";
for(int i =0; i<num; i++){ cout << new_array[i] << " "; }
cout << "\n";
median(new_array, num); }
Program 2. Write a program to evaluate measure of dispersion. Solution:- #include <stdio.h> #include <math.h> #define MAXSIZE 10 void main() { float x[MAXSIZE]; int i, n;
float average, variance, std_deviation, sum = 0, sum1 = 0; printf("Enter the value of N \n");
scanf("%d", &n);
printf("Enter %d real numbers \n", n); for (i = 0; i < n; i++)
{
scanf("%f", &x[i]); }
/* Compute the sum of all elements */ for (i = 0; i < n; i++)
{
sum = sum + x[i]; }
average = sum / (float)n;
/* Compute variance and standard deviation */ for (i = 0; i < n; i++)
{
sum1 = sum1 + pow((x[i] - average), 2); }
variance = sum1 / (float)n; std_deviation = sqrt(variance);
printf("Average of all elements = %.2f\n", average); printf("variance of all elements = %.2f\n", variance); printf("Standard deviation = %.2f\n", std_deviation); }
Program 3. Write a program to find skewness. Solution: #include <cstdlib> #include <iostream> #include <algorithm> #include <numeric> #include <vector> #include <math.h> using namespace std; int main() {
cout << "Enter number of elements." << endl;
int n, i, x; double avg, var = 0, skewness = 0, S, k = 0; cin >> n;
cout << "Enter elements." << endl; vector<int> v; int a[] = {2, 11, 8, 10, 1, 13, 9, 5, 9, 2, 10, 3, 8, 6, 17, 2, 10, 15, 14, 5}; for (i = 0; i < n; i++) { x = a[i]; v.push_back(x); } sort(v.begin(), v.end()); for (i = 0; i < n; i++) cout << v[i] << " ";
int sum = accumulate(v.begin(), v.end(), 0); avg = (double)sum/n;
cout << "\nMean = " << avg << endl; vector<int>::iterator it;
if (n % 2 != 0) {
x = (n+1)/2; it = v.begin() + x;
nth_element(v.begin(), it, v.end());
cout << "Median = " << *(v.begin() + x -1) << endl; } else { float m; x = (n)/2 - 3; it = v.begin() + x; m = (*(v.begin()+ x - 1) + (*(v.begin() + x - 2 )))/2; cout << "Median = " << m << endl;
for (i = 0; i < n; i++) {
var += (v[i] - avg)*(v[i] - avg); }
var = (double)(var)/(n - 1);
cout << "Variance = " << var << endl; S = (double)sqrt(var);
for (i = 0; i < n; i++)
skewness += (v[i] - avg)*(v[i] - avg)*(v[i] - avg); skewness = skewness/(n * S * S * S);
cout << "Skewness = " << skewness << endl; system("PAUSE");
return EXIT_SUCCESS; }
Program 4. Write a program to find Kurtosis. Solution:-#include <cstdlib> #include <iostream> #include <algorithm> #include <numeric> #include <vector> #include <math.h> using namespace std; int main() {
cout << "Enter number of elements." << endl;
int n, i, x; double avg, var = 0, skewness = 0, S, k = 0; cin >> n;
cout << "Enter elements." << endl; vector<int> v; int a[] = {2, 11, 8, 10, 1, 13, 9, 5, 9, 2, 10, 3, 8, 6, 17, 2, 10, 15, 14, 5}; for (i = 0; i < n; i++) { x = a[i]; v.push_back(x); } sort(v.begin(), v.end()); for (i = 0; i < n; i++) cout << v[i] << " ";
int sum = accumulate(v.begin(), v.end(), 0); avg = (double)sum/n;
cout << "\nMean = " << avg << endl; vector<int>::iterator it;
if (n % 2 != 0) {
x = (n+1)/2; it = v.begin() + x;
nth_element(v.begin(), it, v.end());
cout << "Median = " << *(v.begin() + x -1) << endl; }
else {
float m; x = (n)/2 - 3;
it = v.begin() + x;
m = (*(v.begin()+ x - 1) + (*(v.begin() + x - 2 )))/2; cout << "Median = " << m << endl;
}
for (i = 0; i < n; i++) {
var += (v[i] - avg)*(v[i] - avg); }
var = (double)(var)/(n - 1);
cout << "Variance = " << var << endl; S = (double)sqrt(var);
for (i = 0; i < n; i++)
skewness += (v[i] - avg)*(v[i] - avg)*(v[i] - avg); skewness = skewness/(n * S * S * S);
cout << "Skewness = " << skewness << endl; for (i = 0; i < n; i++)
k += (v[i] - avg)*(v[i] - avg)*(v[i] - avg)*(v[i] - avg); k = k/(n*S*S*S*S);
k -= 3;
cout << "Kurtosis = " << k << endl; system("PAUSE");
return EXIT_SUCCESS; }
Program 5. Write a program to find regression line of y on x from any given set of points.
Solution:
#include<iostream> Using namespace std; float mean(float *a, int n);
void deviation(float *a, float mean, int n, float *d, float *S); void main()
{
float a[20],b[20],dx[20],dy[20];
float sy=0,sx=0,mean_x=0,mean_y=0,sum_xy=0; float corr_coff=0,reg_coff_xy=0, reg_coff_yx=0; char type_coff[7];
int n=0,i=0; clrscr();
printf("Enter the value of n: "); scanf("%d",&n);
printf("Enter the values of x and y:\n"); for(i=0;i scanf("%f%f",&a[i],&b[i]); mean_x=mean(a,n); mean_y=mean(b,n); deviation(a,mean_x,n,dx,&sx); deviation(b,mean_y,n,dy,&sy); for(i=0;i sum_xy=sum_xy+dx[i]*dy[i]; corr_coff=sum_xy/(n*sx*sy);
printf("Enter the type of regression coefficient as 'x on y' or 'y on x': "); fflush(stdin); gets(type_coff); if(strcmp(type_coff,"x on y")==1) { reg_coff_xy=corr_coff*(sx/sy);
printf("\nThe value of linear regression coefficient is %f",reg_coff_xy);
}
else if(strcmp(type_coff,"y on x")==1) {
printf("\nThe value of linear regression coefficient is %f",reg_coff_yx);
} else
printf("\nEnter the correct type of regression coefficient."); getch();
}
float mean(float *a, int n) {
float sum=0, i=0;
for(i=0;i sum=sum+a[i]; sum=sum/n;
return (sum); }
void deviation(float *a, float mean, int n, float *d, float *s) { float sum=0,t=0; int i=0; for(i=0;i { d[i]=a[i]-mean; t=d[i]*d[i]; sum=sum+t; } sum=sum/n; *s=sqrt(sum); } OUTPUT:
Program 6. Write a program to find regression line of x on y from any given set of points.
Solution: #include <iostream> #include <iomanip> using namespace std; int main() { double x[7] = { 1.5, 2.4, 3.2, 4.8, 5.0, 7.0, 8.43 }; double y[7] = { 3.5, 5.3, 7.7, 6.2, 11.0, 9.5, 10.27 }; Maths::Regression::Linear A(7, x, y);
cout << " Slope = " << A.getSlope() << endl;
cout << "Intercept = " << A.getIntercept() << endl << endl; cout << "Regression coefficient = " << A.getCoefficient() << endl;
cout << endl << "Regression line values" << endl << endl; for (double i = 0.0; i <= 3; i += 0.6) { cout << "x = " << setw(3) << i << " y = " << A.getValue(i); cout << endl; } return 0; }
OUTPUT:
-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*- How many values You are Entering7 -*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*- Enter coressponding Elements X & Y
1 0.5 2 2.5 3 2 4 4 5 3.5 6 6 7 5.8 -*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*- Y=-0.371429+0.871429X
-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-Program 7. Write a program to find the solution of linear equations using Simplex Method.
Solution: #include <stdio.h> #include <math.h> #define CMAX 10 #define VMAX 10 int NC, NV, NOPTIMAL,P1,P2,XERR; double TS[CMAX][VMAX]; void Data() { double R1,R2; char R; int I,J;
printf("\n SIMPLEX METHOD\n\n");
printf(" MAXIMIZE (Y/N) ? "); scanf("%c", &R);
printf("\n NUMBER OF VARIABLES OF THE FUNCTION ? "); scanf("%d", &NV);
printf("\n NUMBER OF CONSTRAINTS ? "); scanf("%d", &NC); if (R == 'Y' || R=='y')
R1 = 1.0; else
R1 = -1.0;
printf("\n INPUT COEFFICIENTS OF THE FUNCTION:\n"); for (J = 1; J<=NV; J++) {
TS[1][J+1] = R2 * R1; }
printf(" Right hand side ? "); scanf("%lf", &R2); TS[1][1] = R2 * R1;
for (I = 1; I<=NC; I++) {
printf("\n CONSTRAINT #%d:\n", I); for (J = 1; J<=NV; J++) {
printf(" #%d ? ", J); scanf("%lf", &R2); TS[I + 1][J + 1] = -R2;
}
printf(" Right hand side ? "); scanf("%lf", &TS[I+1][1]); }
printf("\n\n RESULTS:\n\n");
for(J=1; J<=NV; J++) TS[0][J+1] = J;
for(I=NV+1; I<=NV+NC; I++) TS[I-NV+1][0] = I; } void Pivot(); void Formula(); void Optimize(); void Simplex() { e10: Pivot(); Formula(); Optimize();
if (NOPTIMAL == 1) goto e10; } void Pivot() { double RAP,V,XMAX; int I,J; XMAX = 0.0; for(J=2; J<=NV+1; J++) { if (TS[1][J] > 0.0 && TS[1][J] > XMAX) { XMAX = TS[1][J]; P2 = J; } } RAP = 999999.0;
for (I=2; I<=NC+1; I++) { if (TS[I][P2] >= 0.0) goto e10; V = fabs(TS[I][1] / TS[I][P2]); if (V < RAP) { RAP = V; P1 = I; } e10:;} V = TS[0][P2]; TS[0][P2] = TS[P1][0]; TS[P1][0] = V; }
void Formula() {; int I,J;
for (I=1; I<=NC+1; I++) { if (I == P1) goto e70; for (J=1; J<=NV+1; J++) { if (J == P2) goto e60; TS[I][J] -= TS[P1][J] * TS[I][P2] / TS[P1][P2]; e60:;} e70:;} TS[P1][P2] = 1.0 / TS[P1][P2]; for (J=1; J<=NV+1; J++) { if (J == P2) goto e100; TS[P1][J] *= fabs(TS[P1][P2]); e100:;}
for (I=1; I<=NC+1; I++) { if (I == P1) goto e110; TS[I][P2] *= TS[P1][P2]; e110:;} } void Optimize() { int I,J;
for (I=2; I<=NC+1; I++) if (TS[I][1] < 0.0) XERR = 1; NOPTIMAL = 0; if (XERR == 1) return; for (J=2; J<=NV+1; J++) if (TS[1][J] > 0.0) NOPTIMAL = 1; } void Results() { int I,J;
if (XERR == 0) goto e30;
printf(" NO SOLUTION.\n"); goto e100; e30:for (I=1; I<=NV; I++)
for (J=2; J<=NC+1; J++) { if (TS[J][0] != 1.0*I) goto e70;
printf(" VARIABLE #%d: %f\n", I, TS[J][1]); e70: ;}
printf("\n ECONOMIC FUNCTION: %f\n", TS[1][1]); e100:printf("\n"); } void main() { Data(); Simplex(); Results(); }
OUTPUT:
MAXIMIZE (Y/N) ? y
NUMBER OF VARIABLES OF THE FUNCTION ? 2 NUMBER OF CONSTRAINTS ? 3
INPUT COEFFICIENTS OF THE FUNCTION: #1 ? 4
#2 ? 10
Right hand side ? 0
CONSTRAINT #1: #1 ? 2
#2 ? 1
Right hand side ? 50 CONSTRAINT #2:
#1 ? 2 #2 ? 5
Right hand side ? 100 CONSTRAINT #3:
#1 ? 2 #2 ? 3
Right hand side ? 90
RESULTS:
VARIABLE #2: 20.000000
ECONOMIC FUNCTION: 200.000000 8
Program 8. Write a program to find the solution of linear equations using Big M Method.
Solution: #include <stdio.h> int bk=1,g,l; void max(); void min(); int main() { int i;
printf("\nNOTE: SLACK VARIABLES HELP IN FORMING THE UNIT MAQTRIX. BUT
OBJECTIVE\n");
printf("FUNCTIONS WITH <= WILL PROVIDE SLACK VARIABLES BUT THOSE
WITH >= WILL NOT.\n");
printf("HENCE TO MAKE A UNIT MATRIX, WE MUST TAKE HELP OF ARTIFICIAL
VARIABLES. THIS\n");
printf("INCLUDE ANOTHER COEFFICIENT CALLED BIG-M. “);
printf("\n OBJECTIVE? Maximization(1) or Minimization(2) or any other to Exit...:\n"); scanf("%d",&i); if (i==1) max(); else if (i==2) min();
else return 0; } void min() { int i,j,row,col,kc,kr,sub,var,flag=-1,k; double sum,max,ba,kn; double cj[20],basis[20],c[20],colmat[20],tab[20][20]; bk=0;
printf("\n\nTAKE \"M\" AS A BIT LARGER THAN ANY LARGEST NUMBER IN THE
printf("\nTOTAL NUMBER OF >= EQUALITIES IN SUBJECTIVE FUNCTIONS:\n");
scanf("%d", &g);
printf("\nTOTAL NUMBER OF <= EQUALITIES IN SUBJECTIVE FUNCTIONS:\n");
scanf("%d", &l);
printf("\nTOTAL NUMBER OF \"ARTIFICIAL VARIABLES\": %d\n",g); printf("ENTER THE TOTAL NUMBER OF VARIABLES IN OBJECTIVE FUNCTION:\n");
scanf("%d", &var);
printf("\nSTART ENTERING THE COEFFICIENTS OF THE OBJECTIVE FUNCTION:\n");
for(i=0;i<var;i++) {
scanf("%lf", &cj[i]); }
printf("\nENTER TOTAL NUMBER OF SUBJECTIVE FUNCTIONS:\n"); scanf("%d", &sub);
//DECIDING TABLE DIMENSION col=var+4;
row=sub+2;
for(i=1;i<row-1;i++) {
printf("\nENTER ELEMENTS IN %d-th ROW:\n",i); for(j=4;j<col;j++) {scanf("%lf",&tab[i][j]); } }
printf("\nENTER THE BASIS:\n"); for(i=0;i<sub;i++)
scanf("%lf", &basis[i]);
printf("\nENTER THE VALUES OF \"C\" FROM OBJECTIVE FUNCTION\n");
for(i=0;i<sub;i++) scanf("%lf", &c[i]);
printf("\nENTER THE COLUMN MATRIX:\n"); for(i=0;i<sub;i++)
scanf("%lf", &colmat[i]); //INITIALIZING THE TABLE for(i=1;i<row-1;i++) tab[i][0]=i; for(i=1;i<row-1;i++) { tab[i][1]=basis[i-1]; tab[i][2]=c[i-1]; tab[i][3]=colmat[i-1]; }
printf("\nPRINTING THE MATRIX YOU HAVE INSERTED:\n"); printf("---\n"); for(i=1;i<row-1;i++) { for(j=0;j<col;j++) { printf("%.3lf\t",tab[i][j]); } printf("\n"); } printf("---\n"); for(i=4;i<col;i++) tab[row-1][i]=0; tab[row-1][3]=0;
//STARTING THE ITERATION for(k=0;k<10;k++) { //INITIALIAZING FLAG flag=-1.00; //Z0 for(i=1;i<row-1;i++) tab[row-1][3]=tab[row-1][3]+tab[i][3]*tab[i][2]; //Zj-Cj for(i=4;i<col;i++) { sum=0; for(j=1;j<row-1;j++) sum=tab[j][i]*tab[j][2]+sum; tab[row-1][i]=sum-cj[i-4]; } //FINDING MAXIMUM IN Zj-Cj max=tab[row-1][4]; kc=4; for(i=4;i<col;i++) { if(max<tab[row-1][i])
{
max=tab[row-1][i]; kc=i;
} }
//FINDING b/a RATIO for(j=1;j<row-1;j++) { if(tab[j][kc]>0) { ba=(colmat[j-1]/tab[j][kc]); goto pop; } } pop: kr=j; for(;j<row-1;j++) {
if((tab[j][kc]>0) && ((colmat[j-1]/tab[j][kc])<ba)) kr=j;
}
//SWAPPING KEY COLUMN WITH BASIS tab[kr][1]=kc-3;
kn=tab[kr][kc]; tab[kr][2]=cj[kc-4];
//DIVIDING OTHER ROWS BY THE FORMULA for(i=1;i<row-1;i++) { if(i==kr) continue; else { for(j=3;j<kc;j++) tab[i][j]=tab[i][j]-((tab[i][kc]*tab[kr][j])/kn); } } printf("\n\n"); for(i=1;i<row-1;i++) { if(i==kr) continue; else { for(j=kc+1;j<col;j++) tab[i][j]=tab[i][j]-((tab[i][kc]*tab[kr][j])/kn); }
} for(i=1;i<row-1;i++) { if(i==kr) continue; else tab[i][kc]=tab[i][kc]-((tab[i][kc]*tab[kr][kc])/kn); }
//DIVIDING KEY ROW BY KEY NUMBER for(i=3;i<col;i++)
tab[kr][i]=tab[kr][i]/kn;
//CHECKING IF Zj-Cj ARE ALL NEGATIVE for(i=4;i<col;i++) { if(tab[row-1][i]>0) { flag=1; } }
//BREAKING THE LOOP if(flag==-1)
goto sos; }
sos:
printf("\nTHE SOLUTION IS...\n"); for(i=1;i<row-1;i++) printf("X%d=%lf\n",(int)tab[i][1],tab[i][3]); } void max() { int i,j,row,col,kc,kr,sub,var,flag=-1,k; double sum,max,ba,kn; double cj[20],basis[20],c[20],colmat[20],tab[20][20]; bk=0;
printf("\n\nTAKE \"M\" AS A BIT SMALLER THAN ANY SMALLEST NUMBER IN
THE EQUATIONS...\n\n\n");
printf("\nTOTAL NUMBER OF >= EQUALITIES IN SUBJECTIVE FUNCTIONS:\n");
scanf("%d", &g);
printf("\nTOTAL NUMBER OF <= EQUALITIES IN SUBJECTIVE FUNCTIONS:\n");
scanf("%d", &l);
printf("\nTOTAL NUMBER OF \"ARTIFICIAL VARIABLES\": %d\n",g); printf("ENTER THE TOTAL NUMBER OF VARIABLES IN OBJECTIVE FUNCTION:\n");
scanf("%d", &var);
printf("\nSTART ENTERING THE COEFFICIENTS OF THE OBJECTIVE FUNCTION:\ n"); for(i=0;i<var;i++) { scanf("%lf", &cj[i]); }
printf("\nENTER TOTAL NUMBER OF SUBJECTIVE FUNCTIONS:\n"); scanf("%d", &sub);
//DECIDING TABLE DIMENSION col=var+4;
row=sub+2;
for(i=1;i<row-1;i++) {
printf("\nENTER ELEMENTS IN %d-th ROW:\n",i); for(j=4;j<col;j++)
{
scanf("%lf",&tab[i][j]); }
}
printf("\nENTER THE BASIS:\n"); for(i=0;i<sub;i++)
scanf("%lf", &basis[i]);
printf("\nENTER THE VALUES OF \"C\" FROM OBJECTIVE FUNCTION\n");
for(i=0;i<sub;i++) scanf("%lf", &c[i]);
printf("\nENTER THE COLUMN MATRIX:\n"); for(i=0;i<sub;i++)
scanf("%lf", &colmat[i]); //INITIALIZING THE TABLE for(i=1;i<row-1;i++) tab[i][0]=i; for(i=1;i<row-1;i++) { tab[i][1]=basis[i-1]; tab[i][2]=c[i-1]; tab[i][3]=colmat[i-1]; }
printf("\nPRINTING THE MATRIX YOU HAVE INSERTED:\n"); printf("---\n");
for(i=1;i<row-1;i++) {
for(j=0;j<col;j++) {
printf("%.3lf\t",tab[i][j]); } printf("\n"); } printf("---\n"); for(i=4;i<col;i++) tab[row-1][i]=0; tab[row-1][3]=0;
//STARTING THE ITERATION for(k=0;k<10;k++) { //INITIALIAZING FLAG flag=-1.00; //Z0 for(i=1;i<row-1;i++) tab[row-1][3]=tab[row-1][3]+tab[i][3]*tab[i][2]; //Zj-Cj for(i=4;i<col;i++) { sum=0; for(j=1;j<row-1;j++) sum=tab[j][i]*tab[j][2]+sum; tab[row-1][i]=sum-cj[i-4]; } //FINDING MAXIMUM IN Zj-Cj max=tab[row-1][4]; kc=4; for(i=4;i<col;i++) { if(max<tab[row-1][i]) { max=tab[row-1][i]; kc=i; } }
//FINDING b/a RATIO for(j=1;j<row-1;j++) { if(tab[j][kc]>0) { ba=(colmat[j-1]/tab[j][kc]); goto pop; } } pop: kr=j;
for(;j<row-1;j++) {
if((tab[j][kc]>0) && ((colmat[j-1]/tab[j][kc])<ba)) kr=j;
}
//SWAPPING KEY COLUMN WITH BASIS tab[kr][1]=kc-3;
kn=tab[kr][kc]; tab[kr][2]=cj[kc-4];
//DIVIDING OTHER ROWS BY THE FORMULA for(i=1;i<row-1;i++) { if(i==kr) continue; else { for(j=3;j<kc;j++) tab[i][j]=tab[i][j]-((tab[i][kc]*tab[kr][j])/kn); } } printf("\n\n"); for(i=1;i<row-1;i++) { if(i==kr) continue; else { for(j=kc+1;j<col;j++) tab[i][j]=tab[i][j]-((tab[i][kc]*tab[kr][j])/kn); } } for(i=1;i<row-1;i++) { if(i==kr) continue; else tab[i][kc]=tab[i][kc]-((tab[i][kc]*tab[kr][kc])/kn); }
//DIVIDING KEY ROW BY KEY NUMBER for(i=3;i<col;i++)
tab[kr][i]=tab[kr][i]/kn;
//CHECKING IF Zj-Cj ARE ALL POSITIVE for(i=4;i<col;i++)
{
if(tab[row-1][i]<0) flag=1;
}
//BREAKING THE LOOP if(flag==-1)
goto sos; }
sos:
printf("\nTHE SOLUTION IS...\n"); for(i=1;i<row-1;i++)
printf("X%d=%lf\n",(int)tab[i][1],tab[i][3]); }
Program 9. Write a program to find the solution of
Transportation problem using North-West corner Method.
Solution: #include<stdio.h> #include<iostream.h> void main() { int a[10][10],m,n,sum=0,f=0,e=0,d,c; cout<<"enter the no. of companies"; cin>>n;
cout<<"enter the no. of warehouses"; cin>>m;
for(f=1;f<=m;f++) {
for(e=1;e<=n;e++) {
cout<<"enter values of warehouse"<<m; cout<<"and company"<<n;
}
for(f=1;f<=m;f++) {
cout<<"enter value of supply"; cin>>c[f];
}
for(e=1;e<=n;e++) {
cout<<"enter value of demand"; cin>>d[e]; } while(c[f]<d[e]) { if(d[f]<c[e]) { c[e]=c[e]-d[f]; sum=sum+(c[e][f]*[f]) f=f+1; } else { if(d[f]>c[e]) { d[f]=d[f]-c[e]; sum=sum+(c[e][f]*c[e]) e=e+1; } else { if(d[f]==c[e]) { d[f]=d[f]-c[e]; sum=sum+(c[e][f]*c[e]) e=e+1; f=f+1; } } } }
cout<<"the optimal cost"; cin>>sum;
} }
Program 10. Write a program to find the solution of Transportation problem using Least Cost Method.
Solution: #include<conio.h> #include<stdio.h> void main() { int c[20] [20],i,j,min,m,n,b,d,c2,c1,p,q,dem[20],sup[20],rf[20],cf[20],sum= 0; clrscr();
printf("\nEnter the row & column:"); scanf("%d%d",&m,&n);
printf("\nEnter the cost:"); for(i=0;i<m;i++)
{ for(j=0;j<n;j++) scanf("%d",&c[i][j]); }
printf("\nEnter the demand:"); for(i=0;i<n;i++)
scanf("%d",&dem[i]);
printf("\nEnter the supply:"); for(i=0;i<m;i++)
scanf("%d",&sup[i]); printf("\nMatrix:\n"); for(i=0;i<m;i++) { for(j=0;j<n;j++)
printf(" %d ",c[i][j]); printf("%d",sup[i]); printf("\n"); } for(j=0;j<n;j++) printf("%d ",dem[j]); for(i=0;i<m;i++) rf[i]=0; for(i=0;i<n;i++) cf[i]=0; b=m;d=n; while(b>0&&d>0) { min=1000; for(i=0;i<m;i++) { if(rf[i]!=1) { for(j=0;j<n;j++) { if(cf[j]!=1) { if(min>c[i][j]) { min=c[i][j]; p=i; q=j; } } } } } if(sup[p]<dem[q]) c1=sup[p]; else c1=dem[q]; for(i=0;i<m;i++) { if(rf[i]!=1) { for(j=0;j<n;j++) { if(cf[j]!=1) { if(min==c[i][j]) { if(sup[i]<dem[j]) c2=sup[i]; else c2=dem[j]; if(c2>c1) { c1=c2; p=i; q=j; } } }
} } } printf("\n %d %d %d ",min,p,q); if(sup[p]<dem[q]) { sum=sum+c[p][q]*sup[p]; dem[q]-=sup[p]; rf[p]=1; b--; } else if(sup[p]>dem[q]) { sum=sum+c[p][q]*dem[q]; sup[p]-=dem[q]; cf[q]=1; d--; } else if(sup[p]==dem[q]) { sum=sum+c[p][q]*sup[p]; rf[p]=1; cf[q]=1; b--; d--; } printf("\n %d",sum); } printf("\n\n %d",sum); getch(); } OUTPUT:
Program 11. Write a program to find the solution of Transportation problem using Vogel’s Method.
Solution:
#include<stdio.h> #include<conio.h> void sort(int a[],int n) { int temp,j,k;
for(j=0;j<n;j++)
{ for(k=j+1;k<n;k++) { if(a[j]>a[k])
a[j]=a[k]; a[k]=temp; } }}} void main() { int i,j,b,p,d,k,m,n,c[20][20],cf[20],rf[20],a[20],cp[20],rp[20]; int sup[20],dem[20],max,min,s,t,sum=0; clrscr();
printf("\nEnter the row & column:"); scanf("%d%d",&m,&n);
printf("\nEnter the cost:"); for(i=0;i<m;i++)
{ for(j=0;j<n;j++) scanf("%d",&c[i][j]); }
printf("\nEnter the demand:"); for(i=0;i<n;i++)
scanf("%d",&dem[i]);
printf("\nEnter the supply:"); for(i=0;i<m;i++) scanf("%d",&sup[i]); printf("\nMatrix:\n"); for(i=0;i<m;i++) { for(j=0;j<n;j++) printf(" %d ",c[i][j]); printf("%d",sup[i]); printf("\n"); } for(j=0;j<n;j++) printf("%d ",dem[j]); for(i=0;i<m;i++) rf[i]=0; for(i=0;i<n;i++) cf[i]=0; b=m,d=n; while(b>0&&d>0) { for(i=0;i<m;i++) rp[i]=-1; for(i=0;i<n;i++) cp[i]=-1; for(i=0;i<m;i++) { k=0; if(rf[i]!=1) { for(j=0;j<n;j++) { if(cf[j]!=1) a[k++]=c[i][j]; }
if(k==1) rp[i]=a[0]; else { sort(a,k); rp[i]=a[1]-a[0]; }} } for(i=0;i<n;i++) { k=0; if(cf[i]!=1) { for(j=0;j<m;j++) { if(rf[j]!=1) a[k++]=c[j][i]; } if(k==1) cp[i]=a[0]; else { sort(a,k); cp[i]=a[1]-a[0]; }} for(i=0;i<m;i++) a[i]=rp[i]; for(j=0;j<n;j++) a[i+j]=cp[j]; max=a[0]; p=0; for(i=1;i<m+n;i++) { if(max<a[i]) { max=a[i]; p=i; }} printf("\n\n %d %d",max,p); min=1000; if(p>m-1) { p=p-m; if(cf[p]!=1) { for(i=0;i<m;i++) { if(rf[i]!=1) { if(min>c[i][p]) { min=c[i][p]; s=i; t=p; }}}} } else { if(rf[p]!=1) { for(i=0;i<n;i++) { if(cf[i]!=1)
{ if(min>c[p][i]) { min=c[p][i]; s=p; t=i; }}}} } printf("\n\n %d %d %d",min,s,t); if(sup[s]<dem[t]) { sum+=c[s][t]*sup[s]; dem[t]-=sup[s]; rf[s]=1; b--; } else if(sup[s]>dem[t]) { sum+=c[s][t]*dem[t]; sup[s]-=dem[t]; cf[t]=1; d--; } else if(sup[s]==dem[t]) { sum+=c[s][t]*dem[t]; cf[t]=1; rf[s]=1; b--; d--; } } printf("\n\n %d ",sum); getch(); } OUTPUT:
Output: Output of Vogel Approximation Method
5 3 6 2 19 (1) 4 7 9 1 37 (3) 3 4 7 5 34 (1) 16 18 31 25
(1) (1) (1) (1) 5 3 6 1111 19 (2) 4 7 9 1111 12 (3) 3 4 7 1111 34 (1) 16 18 31 0 (1) (1) (1) (0) 5 3 6 1111 19 (2) 1111 1111 1111 1111 0 (0) 3 4 7 1111 34 (1) 4 18 31 0 (2) (1) (1) (0) ---5 1111 6 1111 1 (1) 1111 1111 1111 1111 0 (0) 3 1111 7 1111 34 (4) 4 0 31 0 (2) (0) (1) (0) ---1111 ---1111 6 ---1111 1 (1105) 1111 1111 1111 1111 0 (0) 1111 1111 7 1111 30 (1104) 0 0 31 0 (0) (0) (1) (0) ---1111 ---1111 ---1111 ---1111 0 (0) 1111 1111 1111 1111 0 (0) 1111 1111 7 1111 30 (1104) 0 0 30 0 (0) (0) (1104) (0) ---1111 ---1111 ---1111 ---1111 0 (0) 1111 1111 1111 1111 0 (0) 1111 1111 1111 1111 0 (0) 0 0 0 0
(0) (0) (0) (0) ---5 3|18 6|1 2 4|12 7 9 1|25 3|4 4 7|30 5 1111 1111 7 1111 30 (1104) X[1][2] = 18 X[1][3] = 1 X[2][1] = 12 X[2][4] = 25 X[3][1] = 4 X[3][3] = 30
