Question

Resource Allocation The representative of Health Department of Texas wants to investigate the optimal allocation of organ procurement organizations (OPOs) for organ transplant Location data Demand x oint # ???? 2 2 3 4 10 10.5 11 18 18 21 24 28.5 3 25 26 25.5 4 30 32.5 -5 2.5 6 -15 4 10 12 13 OPO2 6 OPO114 OPO1 30.5-5.5 OPO2 20.5 5.5 OP03 | 25.5 | 3.5

***The program language is Java***

Answer 1

import java.io.*;
import java.util.*;

public class Demo241{

     public static void main(String[] args){

         double[] x = {0,9,10,10.5,11,18,18,21,24,28.5,25,26,25.5,30,32.5};
         double[] y = {0,1,8,11,-5,2.5,-7,-4,-15,3,-9,-3,4,-6,-5};

         double[] opox = {30.5,20.5,25.5};
         double[] opoy = {-5.5,-5.5,3.5};
         int[] assign = new int[15];
         Scanner sc = new Scanner(System.in);
         int n = -1;
         int invalidcount = 0;
         while(true){
             System.out.print("Choose an allocation rule: 1- nearest rule, or 2 - farthest rule:");
             n = Integer.parseInt(sc.nextLine());
             if (n != 1 && n != 2){
                invalidcount++;
                System.out.println("Invalid choice\n");
                if (invalidcount == 3){
                   System.out.println("Terminating.....");
                   return;
                }
             }
             else {
                 break;
             }
          
         }
         System.out.print("Allocation Results: [");
         if (n == 1){
            for (int i = 0; i<15; i++){
                double min = 1000;
                int mini = 0;
                for (int j = 0; j<3; j++){
                    double d = Math.sqrt((x[i] - opox[j]) * (x[i] - opox[j]) + (y[i] - opoy[j]) * (y[i] - opoy[j]));
                    if (d < min){
                       min = d;
                       mini = j;
                    }
                }
                assign[i] = mini + 1;
           
                if (i < 14)
                   System.out.print(assign[i] + ",");
                else
                   System.out.print(assign[i] );
            }
         }
         if (n == 2){
            for (int i = 0; i<15; i++){
                double max = 0;
                int maxi = 0;
                for (int j = 0; j<3; j++){
                    double d = Math.sqrt((x[i] - opox[j]) * (x[i] - opox[j]) + (y[i] - opoy[j]) * (y[i] - opoy[j]));
                    if (d > max){
                       max = d;
                       maxi = j;
                    }
                }
                assign[i] = maxi + 1;
           
                if (i < 14)
                   System.out.print(assign[i] + ",");
                else
                   System.out.print(assign[i] );
             }
         }
         System.out.println("]");
        
     }
}

/*

Q2 :

OPO1 can cover only 50 miles

The modification will be as follows:

1. The for loop will be modified. Here we are calculating the distance between the demand point and the OPO center. We need to place a check there
   that if the distance is more than 50 miles from OPO1 , then OPO1 will not be considered for neares and farthest consideration for this demand point.

   The code modification is as follows: (Shown for nearest calculation but it will be same for the relevant for code for the farthest calculation

            for (int i = 0; i<15; i++){
                double min = 1000;
                int mini = 0;
                for (int j = 0; j<3; j++){
                    double d = Math.sqrt((x[i] - opox[j]) * (x[i] - opox[j]) + (y[i] - opoy[j]) * (y[i] - opoy[j]));
                    if ( j == 0 && d > 50){
                        continue;
                    }
                    if (d < min){
                       min = d;
                       mini = j;
                    }
                }
                assign[i] = mini + 1;
           
                if (i < 14)
                   System.out.print(assign[i] + ",");
                else
                   System.out.print(assign[i] );
           }

Q3.

OPO2 can cover demand points in the north of it.

1. The for loop will be modified. Here we are calculating the distance between the demand point and the OPO center. We need to place a check there
   that if the y coordinate of the demand point is less than the y-coordinate of OPO2 , then OPO2 will not be considered for neares and farthest consideration for this    demand point.

   The code modification is as follows: (Shown for nearest calculation but it will be same for the relevant for code for the farthest calculation)
            for (int i = 0; i<15; i++){
                double min = 1000;
                int mini = 0;
                for (int j = 0; j<3; j++){
                    if ( j == 1 && y[i] <= opoy[j]){
                        continue;
                    }
                    double d = Math.sqrt((x[i] - opox[j]) * (x[i] - opox[j]) + (y[i] - opoy[j]) * (y[i] - opoy[j]));
                    if (d < min){
                       min = d;
                       mini = j;
                    }
                }
                assign[i] = mini + 1;
           
                if (i < 14)
                   System.out.print(assign[i] + ",");
                else
                   System.out.print(assign[i] );
           }

Q4.

Total demand points covered should not exceed 10.

1. The for loop will be modified. So in this case we will run the loop for only upto 10 as only 10 demand points can be considered with all OPOs
   The modifed code is as follows: (Shown for nearest calculation but it will be same for the relevant for code for the farthest calculation)

           
            for (int i = 0; i<10; i++){
                double min = 1000;
                int mini = 0;
                for (int j = 0; j<3; j++){
                    double d = Math.sqrt((x[i] - opox[j]) * (x[i] - opox[j]) + (y[i] - opoy[j]) * (y[i] - opoy[j]));
                    if ( j == 1 && y[i] <= opoy[j]){
                        continue;
                    }
                    if (d < min){
                       min = d;
                       mini = j;
                    }
                }
                assign[i] = mini + 1;
           
                if (i < 14)
                   System.out.print(assign[i] + ",");
                else
                   System.out.print(assign[i] );
           }

  

*/