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"Optimization Solutions"
Optimization Bisection Method
Minimum =
      Value =

f(x) = 1.6 x3 + 3x2 - 2x
[ Interval Low: ] [ Interval High: ] [ Tolerance: 1.0e-5]




IMPLEMENTATION
Bisection Method Optimization

The bisection method for finding the minimum starts with an interval that contains the minimum and then divides that interval into two parts to zoom in on the minimum location.

Algorithm Creation

The steps to apply the bisection method to find the minimum of the function f(x) are listed below,

  • Choose xa and xb as two guesses for the minimum such that f'(xa)f'(xb) < 0

  • Set xm = x(a + xb) / 2 as the mid point between xa and xb.

  • Repeat the above steps until the specified accuracy is reached.



Testing the Bisection Method

We use the bisection method to find the minimum of a nonlinear function. To test it out as defined above, a new TestBisection() static method has been added and executed. Supporting code and methods are not shown.

           static void TestBisection();
              {
                 ListBox1.Items.Clear();
                 ListBox2.Items.Clear();
                 double result = Optimization.Bisection(f, t1, t2, 1.0e-5);
                 ListBox1.Items.Add("x = " + result.ToString());
                 ListBox2.Items.Add("f(x) = " + f(result).ToString());
              }

In order to test the bisection method a nonlinear function f(x)=1.6x3+3x2-2x was created. We can now test the minimum and function values based on different intervals. The user can manipulate interval values as desired



Other Implementations...


Object-Oriented Implementation
Graphics and Animation
Sample Applications
Ore Extraction Optimization
Vectors and Matrices
Complex Numbers and Functions
Ordinary Differential Equations - Euler Method
Ordinary Differential Equations 2nd-Order Runge-Kutta
Ordinary Differential Equations 4th-Order Runge-Kutta
Higher Order Differential Equations
Nonlinear Systems
Numerical Integration
Numerical Differentiation
Function Evaluation


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