# EXAMPLE 3G used in Examples 3E and 3F. ```EXAMPLE 3G
In this example we shall use Newton's method to solve the same economic dispatch as
used in Examples 3E and 3F.

EXAMPLE 3E
Given the generator cost functions found in Example 3A, solve for the economic
dispatch of generation with a total load of 800 MW.

Using α=100 and starting from P1 = 300MW, P2 = 200 MW. and P3 =300MW, we
set the initial value of λ equal to the average of the incremental costs of the
generators at their starting generation values. That is:

This value is 9.4484.
The progress of the gradient search is shown in Table 3.2. The table shows that the
iterations have led to no solution at all.

TABLE 3.2 Economic Dispatch by Gradient Method

The gradient is the same as in Example 3E, the Hessian matrix is:

In this example, we shall simply set the initial λ equal to 0, and the initial generation
values will be the same as in Example 3E as well. The gradient of the Lagrange
function is:

The Hessian matrix is
Solving for the correction to the x vector and making the correction, we obtain
and a total generation cost of 7738.8. Note that no further steps are necessary as the
Newton's method has solved in one step. When the system of equations making up the
generation cost functions are quadratic, and no generation limits are reached, the
Newton's method will solve in one step.

```