Arrays, arrays in “for” loops

Question

I'm trying to calculate something properly. A user enters an amount of trees they want to plant, and there is a rule that a certain percent of trees die every single year. It says after 7 years the trees are sellable. To find the minimum number of seeds needed, you'd take however many die over the course of 7 years for each tree type, and add that to the initial number the user inputted, right?

Well, this is what I get when i enter 20,40,30 for each tree type respectively:

I get the same number of minimum seeds regardless. I assume this has to do with my index or something in my for loop. I'd just like to know how to properly calculate this.

The other issue: I am trying to format my output properly so that it tabs over for each variable so it doesn't look so ugly the way it does. I tried "\\t" but it did nothing. How do I fix this, and the calculations so that each iteration is actually kept track of in the end? should I be using decayRate[i] , desiredYield[i] etc instead of desiredYield[index] etc?

public class TreeCalc {

    public static void main(String[] args){

        //Initializes methods and arrays
        String[] treeTypes = new String[] {"Fir", "Pine", "Spruce"};
        int[] desiredYield = new int [treeTypes.length];
        double[] decayRate = new double[] {0.07, 0.12, 0.08};
        desiredYield = getYield(decayRate, desiredYield, treeTypes);
        int[] data = getCalculate(decayRate, desiredYield, treeTypes);

        printMessage(decayRate, desiredYield, treeTypes);
    }

    //Asks user to input # of trees for each tree type
    public static int[] getYield(double[]decayRate, int[] desiredYield,  String[]treeTypes) {

        int index= 0;
        for(int i=0;i < treeTypes.length;i++) {

            try {
                desiredYield[index] = Integer.parseInt(JOptionPane.showInputDialog("Please enter your desired yield     for: " +treeTypes[i]));
                //Catches any non-number input, displays error to user
            } catch(NumberFormatException e){
                desiredYield[index] = 0;
                JOptionPane.showMessageDialog(null, "Error: Please enter your desired yield for "+treeTypes[i]);
            }

            if (desiredYield[index] <= 0) {
                JOptionPane.showMessageDialog(null, "Error: Please enter your desired yield for "+treeTypes[i]);
            } else{
                index++;
            }
        }

        return desiredYield;
    }

    //Calculates totals and minimums
    public static int[] getCalculate(double[]decayRate, int[]desiredYield, String[]treeTypes){

        int totalSeeds =0;
        int totalTrees=0;
        int minSeeds=0;
        int index=0;

        //For each iteration of the loop, calculates totals/mins
        for(int i = 0; i < treeTypes.length; i++){
            minSeeds += (desiredYield[index] * (decayRate[index]*7)) +  desiredYield[index];
            totalSeeds += minSeeds;
            totalTrees += desiredYield[index];
        }

        return new int[] {totalSeeds, totalTrees, minSeeds};
    }

        //Prints the totals, minimums, and input from the user
    public static void printMessage(double[]decayRate, int[]desiredYield, String[]treeTypes){

        //Calls method and stores values within array
        int[]data = getCalculate(decayRate, desiredYield, treeTypes);
        int totalSeeds = data[0];
        int totalTrees = data[1];
        int minSeeds = data[2];

        //Report displays after user is done entering inputs
        String treeReport = "Tree Type | Desired Yield | Minimum Seeds | Total Seeds | Total Trees ";
        for(int i=0; i<treeTypes.length; i++){
            treeReport += "\n"+treeTypes[i] +  "  "+desiredYield[i] + "  "+minSeeds + "  "+totalSeeds + "   "+totalTrees;
        }

        JOptionPane.showMessageDialog(null, treeReport);
    }
}

Answer 1

No, your equation for the minimum number of seeds needed is not correct. You need something like M=Y/((1-r)^7) Where r is the proportion of trees that die each year, and Y is the desired yield. You solve for M, the minimum # of seeds.