EXAMPLE 3G used in Examples 3E and 3F.

In this example we shall use Newton's method to solve the same economic dispatch as
used in Examples 3E and 3F.
在這個例子中,我們應利用牛頓法來解決相同的經濟調度 3E 和 3F。
Given the generator cost functions found in Example 3A, solve for the economic
dispatch of generation with a total load of 800 MW.
鑑於發電機成本函數為例 3A 條中發現,解決在經濟調度的發電總負荷為 800MW。
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:
使用 α = 100 ,並開始 P1= 300 MW, P2 = 200 MW。和 P3 = 300 MW,
我們設置的初始值 λ 等於平均增量成本的發電機出發一代的價值觀。這就是:
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.
此值是 9.4484
所取得的梯度見表 3.2。該表顯示疊代無法解決此問題。
TABLE 3.2 Economic Dispatch by Gradient Method
表 3.2 經濟調度的梯度法
The gradient is the same as in Example 3E, the Hessian matrix is:
範例 3E 的 Hessian 矩陣是:
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:
在這個例子中,我們將簡單地設置初始 λ 等於 0 ,和最初的值 3E 將是一樣的。
梯度的 Lagrange 函數是:
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.
解決更正後的 X 向量,我們會取得和總發電成本 7738.8。任何進一步的步驟是