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In this section we shall discuss an alternative approach to optimization introduced by J. B. Rosen [R-4, 5, 6] which does not involve the solution of nonlinear two-point boundary-value problems. Rosen’s method, called gradient projection, is an iterative numerical procedure for finding an extremum of a function of several variables that are required to satisfy various constraining relations. If the function to be extremized (called the objective function) and the constraints are linear functions of the variables, the optimization problem is referred to as a linear programming problem; when nonlinear terms are present in the constraining relations or in the objective function, the problem is referred to as a nonlinear programming problem.
We shall first discuss gradient projection as it applies to nonlinear programming problems that have linear constraints, but nonlinear objective functions. Then we shall show how the gradient projection algorithm can be used to solve optimal control problems.
Minimization of Functions by the Gradient Projection Method
Example 6.6-1. To begin, let us consider a simple example. Let $f$ be a function of two variables $y_1$ and $y_2$ and $f\left(y_1, y_2\right)$ denote the value of $f$ at the point $\left(y_1, y_2\right)$. The problem is to find the point $\left(y_1^, y_2^\right)$ where $f$ has its minimum value. The variables $y_1$ and $y_2$ are required to satisfy the linear inequality constraints
\begin{aligned} y_1 & \geq 0 \ y_2 & \geq 0 \ 2 y_1 & -5 y_2+10 \geq 0 \ -4 y_1 & -7 y_2+22.5 \geq 0 \ -9 y_1 & -2 y_2+26.5 \geq 0 \end{aligned}

## 数学代写|优化理论代写Optimization Theory代考|Calculation Requirements

Let us now discuss the calculations that are required by the gradient projection algorithm.
The Gradient. It is assumed that the expression for the function to be minimized is known. The components of the gradient vector are found by taking the partial derivatives of $f$ with respect to $y_1, y_2, \ldots, y_K$. For example, if
$$f(\mathbf{y})=y_1^2-80 y_1+1600+y_2^2-100 y_2,$$
then
$$\frac{\partial f^{(i)}}{\partial \mathbf{y}} \triangleq \frac{\partial f}{\partial \mathbf{y}}\left(\mathbf{y}^{(i)}\right)=\left[2 y_1^{(i)}-80,2 y_2^{(i)}-100\right]^T$$
is the gradient at the point $\mathbf{y}^{(i)} .-\partial f^{(n)} / \partial \mathbf{y}$ is obtained by changing the sign of each component of $\partial f^{(i)} / \partial \mathbf{y}$.
The Projection Matrix. From Eq. (6.6-9) it is seen that to determine the projection matrix $\mathbf{P}_q$ at some point $\mathbf{y}^{(i)}$, it is first necessary to find the matrix $\mathbf{N}_q$. This is done by forming the $L$ vector $\lambda\left(\mathbf{y}^{(i)}\right)=\mathbf{N}_L^T \mathbf{y}^{(i)}-\mathbf{v}_L$ of (6.6-3b) and checking the sign of each component of $\boldsymbol{\lambda}$. Since $y^{(i)}$ is assumed to be an admissible point, each component of $\lambda$ must be nonnegative. If $\lambda_j$ (the $j$ th component of $\lambda$ ) is zero, the unit vector $\mathbf{n}_j$ is to be included in $\mathbf{N}_q$; if $\lambda_j>0$, $\mathbf{n}_j$ is not included in $\mathbf{N}_q$. Once $\mathbf{N}_q$ is known, the matrices $\mathbf{N}_q^T \mathbf{N}_q$ and $\left[\mathbf{N}_q^T \mathbf{N}_q\right]^{-1}$ can be found and the projection matrix formed by using Eq. (6.6-9); that is,
$$\mathbf{P}_q=\mathbf{I}-\mathbf{N}_q\left[\mathbf{N}_q^T \mathbf{N}_q\right]^{-1} \mathbf{N}_q^T .$$
Subsequently, we shall see that only one vector $\mathbf{n}_q$ is added to, or dropped from, $\mathbf{N}_q$ at each stage of the iterative procéss; in addition to simplifying the determination of $\mathbf{N}_q$, this allows the matrix $\left[\mathbf{N}_q^r \mathbf{N}_q\right]^{-1}$ to be found by using recurrence relations $\dagger$ that do not require matrix inversion.

## 优化理论代写

\begin{aligned} y_1 & \geq 0 \ y_2 & \geq 0 \ 2 y_1 & -5 y_2+10 \geq 0 \ -4 y_1 & -7 y_2+22.5 \geq 0 \ -9 y_1 & -2 y_2+26.5 \geq 0 \end{aligned}

## 数学代写|优化理论代写Optimization Theory代考|Calculation Requirements

$$f(\mathbf{y})=y_1^2-80 y_1+1600+y_2^2-100 y_2,$$

$$\frac{\partial f^{(i)}}{\partial \mathbf{y}} \triangleq \frac{\partial f}{\partial \mathbf{y}}\left(\mathbf{y}^{(i)}\right)=\left[2 y_1^{(i)}-80,2 y_2^{(i)}-100\right]^T$$

$$\mathbf{P}_q=\mathbf{I}-\mathbf{N}_q\left[\mathbf{N}_q^T \mathbf{N}_q\right]^{-1} \mathbf{N}_q^T .$$

## Matlab代写

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