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# 数学代写| Duality 凸优化代考

## 凸优化代写

In Chapter 1, we considered the following optimization problem: With 80 feet of fencing forming the perimeter, what dimensions $x \times y$ form the rectangular garden of maximum area? We saw that the solution is when $x=y=20$ feet, so the solution is a square. This is not a linear programming problem, of course, since the area is given by the nonlinear function $A=x y$.

Some reflection might convince the reader that this problem can be phrased as a minimization problem instead: if a rectangular garden has 400 square feet area, what dimensions yield the least amount of fencing (i.e., smallest perimeter)? The solution is again $x=y=20$ feet, so the solution is a square. Even though the two problems are expressed differently, they both express the idea that a square is in some sense the most efficient rectangle. Among all rectangles of a fixed perimeter, the square has the maximum area; or, equivalently, among all rectangles of a fixed area, the square has the minimum perimeter.

This example suggests that there may be two ways to express a single optimization problem, as a maximization or as a minimization, depending on what part of the problem you view as the constraint(s) and what part you view as the objective function. When the optimization problem is a linear programming problem, we can make this idea very precise. Given a linear programming problem (which we refer to as the primal problem), there is another linear programming problem

maximization problem, the dual is a minimization, and when the primal is a minimization problem, the dual is a maximization. The problems are so closely related that, in a sense, they can be regarded as two ways to view the same problem, and have the “same” solution, just as in the previous example the solution to both problems about rectangles is a square.

The concept of a dual problem has important theoretical uses, and it also provides a tool to save work when doing certain computations. We are not concerned in this book with the theoretical aspects, leaving that for more advanced texts in programming and optimization. We do intend to exploit the computational aspects of duality. In Section 4.5.1, we present a technique for finding the dual problem, without worrying about what the decision variables of the dual problem mean. This is all one really needs to know about the dual in order to exploit the computational shortcuts mentioned before. The technique for finding the dual is mechanical and very easy, but it gives no hint as to why the two problems are connected conceptually.

The more interesting aspect of the theory is explaining how the two problems relate to one another. This is done in Section 4.5.2, where we show how to give an interpretation of the decision variables in the dual problem. To make an analogy, in Section 4.5.1, you learn how to drive a car, which you can easily do without knowing anything about combustion engines, while in Section 4.5.2, you learn how the car works. In subsequent sections of this book, only knowing how to drive is required.
4.5.1 The Construction of the Dual
We illustrate the technique of finding the dual problem using a maximization problem in standard form as the primal problem. Then the dual is always a minimization problem also in standard form. In Section 3.5, we discussed the following problem (a modified version of Example 3.12):

A farmer has 150 acres available to split between three crops: asparagus, beans, and corn. Each bag of asparagus seed is sufficient to plant 1 acre and requires 8 pounds of fertilizer. Each bag of bean seeds is sufficient to seed 2 acres and requires 10 pounds of fertilizer. Each bag of corn seed will plant 1 acre and requires 20 pounds of fertilizer. She has a total of $1,000 \mathrm{lbs}$. fertilizer to use on her two crops. The profit from a bag of asparagus seeds is $\$ 1,500$, and the profit from a bag of bean seeds is$\$2,000$. How many bags of each should she plant in order to maximize profit?
Let $x$ be the number of bags of asparagus seed.
Let $y$ be the number of bags of bean seed.
Let $z$ be the number of bags of corn seed.
Maximize: $P=1,500 x+2,000 y+3,700 z$
subject to:
\begin{aligned} &x+2 y+z \leq 150 \text { (acres land) } \ &8 x+10 y+20 z \leq 1,000 \text { (lbs. fertilizer) } \ &x \geq 0, y \geq 0, z \geq 0 \end{aligned}
Observe that this is a $2 \times 3$ problem in standard form. We begin by arranging all the information in this problem in a table or matrix. It is a modified version of the augmented coefficient matrix of the system of equations (if we replaced the inequalities by equations.) See the following table.

## 凸优化代考

4.5.1 对偶的构建

$$\开始{对齐} &x+2 y+z \leq 150 \text { (英亩土地) } \ &8 x+10 y+20 z \leq 1,000 \text { (磅肥料) } \ &x \geq 0, y \geq 0, z \geq 0 \end{对齐}$$

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