Lagrange对偶问题

The dual function provides a lower bound on the optimal value for convex optimization problems only.

对偶函数可以为任意优化问题的最优值提供下界。

The dual function of an optimization problem evaluated at some (λ,ν), with λ⪰0, is equal to 42. Enter a number that can represent the optimal value of the original problem.

因为对偶函数为优化问题的最优值提供下界，只有最优值大于等于42即可。

强弱对偶性

Consider a convex optimization problem

$\begin{array}{ll} \mbox{minimize} & f_0(x)\\ \mbox{subject to} & f_i(x) \leq b_i, \quad i=1, \ldots, m\\ & Ax = d \end{array}$

that satisfies Slater's constraint qualification.

Which of the following are true?

当原问题是凸问题，且满足Slater约束时，满足强对偶性，所以选项1正确。但是强对偶性成立，并不代表原问题和对有问题可行，选项2,3不正确。

几何解释

If the feasible set of an optimization problem is not convex, then the duality gap is always nonzero.

优化问题的可行集是凸集，大部分情况下对偶间隙是0，即强对偶性成立。但可行集不是凸集时，也可能对偶间隙为0.

优化条件

Consider a convex optimization problem

$\begin{array}{ll} \mbox{minimize} & f_0(x)\\ \mbox{subject to} & f_i(x) \leq 0, \quad i=1, \ldots, m\\ & Ax = d \end{array}$

that satisfies Slater's constraint qualification.

What is the value of $\lambda_1^{\star}$ if $f_1(x^\star) = -0.2$ ?

凸优化问题满足Slater约束条件，所以强对偶性成立，所以

$f_0(x^*)=g(\lambda ^*,v^*)=\underset{x }{inf}(f_0(x)+\sum _{i=1}^m \lambda_i^* f_i(x)+\sum _{i=1}^p v^*_ih_i(x))\\ \leq (f_0(x^*)+\sum _{i=1}^m \lambda_i^* f_i(x^*)+\sum _{i=1}^p v^*_ih_i(x^*))\\ \leq f_0(x^*)$

所以

$\sum _{i=1}^m \lambda_i^* f_i(x^*)+\sum _{i=1}^p v^*_ih_i(x^*)=0\Rightarrow \sum _{i=1}^m \lambda_i^* f_i(x^*)=0$

而 $f_i(x)\leq 0,\lambda^*_i\geq 0\Rightarrow f_i(x)\lambda_i^*=0,i=1,\cdots m$

所以当 $f_1(x^\star) = -0.2$ 时， $\lambda^*_1=0$

扰动及灵敏度分析

Consider the convex optimization problem

$\begin{array}{ll} \mbox{minimize} & f_0(x) \\ \mbox{subject to} & f_1(x) \leq s\\ & Ax = b, \end{array}$

with variable $x\in R^n$ , where s is some fixed real number. Let $\lambda ^*$ be an optimal dual variable (Lagrange multiplier) associated with the constraint $f_1(x)\leq s$ . Below we consider scenarios in which we change the value of s, and then solve the modified problem. We are interested in the optimal objective value of this modified problem, compared to the original one above.

If $\lambda ^*$ is large, then decreasing s

If $\lambda ^*$ is small, then increasing s

If $\lambda ^*$ =0, then increasing s

根据灵敏度分析：

如果 $\lambda_i^*$ 比较大，加强第i个约束，即 $u_i< 0$ ，减小s，则最优值 $P^*(u,l)$ 会大幅增加。
如果 $\lambda_i^*$ 比较小，放松第i个约束，即 $u_i> 0$ ，增大s，则最优值 $P^*(u,l)$ 不会减小太多。
如果 $\lambda_i^*$ 为0，第i个约束对最优值无影响。

例子

When you apply a transformation of an optimization problem (for example, introduce new variables and new constraints), the dual problem does not change.

广义不等式

Every element of a generalized Lagrange multiplier λ associated with a vector inequality constraint has to satisfy $\lambda _i \geq 0$ for dual feasibility.

不等式约束满足 $\lambda_i\succeq _{K_i^*}0$ ，即 $\lambda_i$ 满足对偶锥非负。

Numerical perturbation analysis example

Consider the quadratic program

$\begin{array}{ll}\mbox{minimize} & x_1^2 + 2x_2^2 -x_1 x_2 - x_1\\ \mbox{subject to} & x_1 + 2 x_2 \leq u_1 \\ & x_1 - 4x_2 \leq u_2, \\& x_1+ x_2 \geq -5, \end{array}$

with variables $x_1,x_2$ , and parameters $u_1,u_2$ .

(a) Solve this QP, for parameter values $u_1=-2,u_2=-3$ , to find optimal primal variable values $x_1^*$ and $x_2^*$ , and optimal dual variable values $\lambda_1^*,\lambda_2^*$ and $\lambda_3^*$ . Let $p^*$ denote the optimal objective value. Verify that the KKT conditions hold for the optimal primal and dual variables you found (within reasonable numerical accuracy).

Hint: See §§4.7 of the CVX users' guide to find out how to retrieve optimal dual variables. To specify the quadratic objective, use quad_form().

What is $x_2^*$ ? Enter your result rounded to two decimal places.

What is $\lambda_3^*$ ? Enter your result rounded to two decimal places.

KKT条件：

$x_1^*+2x_2^*+2\leq 0\\ x_1^*-4x_2^*+3 \leq 0\\ -x_1^*-x_2^*-5 \leq 0\\ \lambda_1^*,\lambda_2^*,\lambda^*_3 \geq 0\\ \lambda_1^*(x_1^*+2x_2^*+2)=0\\ \lambda_2^*(x_1^*-4x_2^*+3)=0\\ \lambda_3^*(-x_1^*-x_2^*-5)=0\\ 2x_1^*-x_2^*+\lambda_1^*+\lambda_2^*-\lambda_3^*-1=0\\ 4x_2^*-x_1^*+2\lambda_1^*-4\lambda_2^*-\lambda_3^*=0$

解方程组。

(b) We will now solve some perturbed versions of the QP, with

$u_1 = -2 + \delta_1, \qquad u_2 = -3 + \delta_2,$

where $\delta _1$ and $\delta _2$ each take values from {−0.1,0,0.1}. (There are a total of nine such combinations, including the original problem with δ1=δ2=0.) For each combination of $\delta _1$ and $\delta _2$ , make a prediction $p_{pred}^*$ of the optimal value of the perturbed QP, and compare it to $p_{exact}^*$ , the exact optimal value of the perturbed QP (obtained by solving the perturbed QP). Find the values that belong in the two righthand columns in a table with the form shown below. Check that the inequality $p_{pred}^*\leq p_{exact}^*$ holds.

不明白咋回事。

A simple example

Consider the optimization problem

$\begin{array}{ll} \mbox{minimize} & x^2 + 1 \\\mbox{subject to} & (x-2)(x-4) \leq 0,\end{array}$

with variable x∈R.

(a) Analysis of primal problem. What is the optimal value?

b) Lagrangian and dual function. Plot the objective $x^2+1$ versus x. On the same plot, show the feasible set, optimal point and value, and plot the Lagrangian L(x,λ)versus x for a few positive values of λ. Verify the lower bound property ( $p^\star \geq \inf_x L(x,\lambda)$ for λ≥0). Derive and sketch the Lagrange dual function g.

$\begin{array}{ll}\mbox{maximize} &\underline{\hspace{1.5cm}} \\ \mbox{subject to} & \lambda \geq 0.\end{array}$

The correct equation that fills the blank spot is

解：

$L(x,\lambda)=x^2+1+\lambda((x-2)(x-4))=x^2+1+\lambda x^2-6\lambda x+8\lambda$

对x进行极小化， $\bigtriangledown _xL(x,\lambda)=2(1+\lambda)x-6\lambda$

令其为0，解得 $x=\frac{3\lambda}{1+\lambda}$

所以

$g(\lambda)=\underset{x}{inf}L(x,\lambda)\\ =\underset{x}{inf}L(\frac{3\lambda}{1+\lambda},\lambda)=\frac{9\lambda^2}{(1+\lambda)^2}+1+\lambda \frac{9\lambda^2}{(1+\lambda)^2}-6\lambda\frac{3\lambda}{1+\lambda}+8\lambda\\ =\frac{-9\lambda^2}{(1+\lambda)}+1+8\lambda$

Note that the dual problem is convex. What is the dual optimal solution $\lambda^*$ ?

解：带入 $x=2=3\lambda/(1+\lambda)$ ，求出 $\lambda$ 。

Note that strong duality holds.

(d) Sensitivity analysis. Let $p^*(u)$ denote the optimal value of the problem

$\begin{array}{ll} \mbox{minimize} & x^2 + 1 \\\mbox{subject to} & (x-2)(x-4) \leq u, \end{array}$

as a function of the parameter u.

We can show that

$p^\star(u) = \left\{ \begin{array}{ll}\infty & u < -1 \\\underline{\hspace{1.5cm}}& -1 \leq u \leq 8 \\1 & u \geq 8. \end{array} \right.$

The correct equation that fills the blank spot is

Note that $dp^\star(0)/du = -\lambda^\star$ .

$L(x,\lambda)=x^2+1+\lambda((x-2)(x-4)-u)=x^2+1+\lambda x^2-6\lambda x+(8+u)\lambda$

KKT条件：

$x^2-6x+8-u\leq 0\\ \lambda \geq 0 \\ \lambda (x^2-6x+8-u)=0 \\ 2x+\lambda(2x-6)=0$

从上图可看出，当 $-1\leq u \leq 8$ ，最优值解时两函数图像在左侧的交点，求得 $x=\frac{6-\sqrt{6^2-4*1*(8-u)}}{2}=\frac{3-\sqrt{36-32+4u}}{2}=3-\sqrt{1+u}$

带入目标函数，得到最优值为 $x^2+1=(3-\sqrt{1+u})^2+1=1+u-6\sqrt{1+u}+9+1=11-6\sqrt{1+u}$

Lagrangian relaxation of Boolean LP

A Boolean linear program is an optimization problem of the form

$\begin{array}{ll}\mbox{minimize} & c^T x \\\mbox{subject to} & Ax \preceq b \\& x_i \in \{0,1 \}, \quad i=1,\ldots,n,\end{array}$

and is, in general, very difficult to solve. In exercise 4.15 we studied the LP relaxation of this problem,

$\begin{array}{ll}\mbox{minimize} & c^T x \\ \mbox{subject to} & Ax \preceq b\\ & 0 \leq x_i \leq 1, \quad i=1,\ldots,n, \end{array}$

which is far easier to solve, and gives a lower bound on the optimal value of the Boolean LP. In this problem we derive another lower bound for the Boolean LP, and work out the relation between the two lower bounds.

(a) Lagrangian relaxation. The Boolean LP can be reformulated as the problem

$\begin{array}{ll}\mbox{minimize} & c^T x \\\mbox{subject to} & Ax \preceq b \\ & x_i(1-x_i)=0, \quad i=1,\ldots,n,\end{array}$

which has quadratic equality constraints. The Lagrange dual of this problem is

$\begin{array}{ll}\mbox{maximize} & \underline{\hspace{1.5cm}} \\\mbox{subject to} & \mu \succeq 0.\end{array}$

The correct equation that fills the blank spot is

解：

$g(\mu_i,v_i)=\underset{x}{inf}L(x_i,\mu_i,v_i)=\underset{x}{inf}c_ix_i+\mu_i^T(a_ix_i-b_i)+v_ix_i(x_i-1)\\ =\underset{x}{inf}(c_i+\mu_i^Ta_i)x_i-\mu_ib_i\\$

$g(\mu,v)=-\mu^Tb+\sum_{i=1}^n \underset{x}{inf} (c_i+\mu_i^Ta_i)x_i =-\mu^Tb+\sum_{i=1}^n min \left \{ 0,c_i+a_i^T\mu\right \}$

The optimal value of the dual problem (which is convex) gives a lower bound on the optimal value of the Boolean LP. This method of finding a lower bound on the optimal value is called Lagrangian relaxation.

(b) The lower bound obtained via Lagrangian relaxation, and via the LP relaxation are the same. The dual of the LP relaxation is

$\begin{array}{ll}\mbox{maximize} & \underline{\hspace{1.5cm}} \\\mbox{subject to} & A^T u - v+ w + c = 0 \\ & u\succeq 0, v\succeq 0, w\succeq 0.\end{array}$

The correct equation that fills the blank spot is

$g(u,v,w)=\underset{x}{inf}L(x,u,v,w)=\underset{x}{inf}c^Tx+u^T(Ax-b)+v^T(-x)+w^T(x-\boldsymbol{1})\\ =\underset{x}{inf}(c^T+u^TA-v^T+w^T)x-w^T\boldsymbol{1}-u^Tb\\ =-w^T\boldsymbol{1}-u^Tb\\$

Option Price Bounds

In this problem we use the methods and results of Example 5.10 to give bounds on the arbitrage-free price of an option. (See Exercise 5.38 for a simple version of option pricing.) We will use all the notation and definitions from Example 5.10.

We consider here options on an underlying asset (such as a stock); these have a payoff or value that depends on S, the value of the underlying asset at the end of the investment period. We will assume that the underying asset can only take on m different values, $S^{(1)}\cdots,S^{(m)}$ . These correspond to the m possible scenarios or outcomes described in Example 5.10.

A risk-free asset has value r>1 in every scenario.

A put option at strike price K gives the owner the right to sell one unit of the underlying stock at price K. At the end of the investment period, if the stock is trading at a price S, then the put option has payoff $(K-S)_+=max \left\{0,K-S \right \}$ (K−S)+=max{0,K−S} (since the option is exercised only if K>S). Similarly a call option at strike price K gives the buyer the right to buy a unit of stock at price K. A call option has payoff $(S-K)_+=max \left\{0,S-K \right \}$ .

A collar is an option with payoff

$\begin{cases} C-S_0 & S>C\\ S-S_0 & F \leq S \leq C\\ F-S_0 & S \le F \end{cases}$

where F is the floor, C is the cap and $S_0$ is the price of the underlying at the start of the investment period. This option limits both the upside and downside of payoff.

Now we consider a specific problem. The price of the risk-free asset, with r=1.05, is 1. The price of the underlying asset is $S_0=1$ . We will use m=200 scenarios, with $S^{(i)}$ uniformly spaced from $S^{(1)}=0.5$ to $S^{(200)}=2$ . The following options are traded on an exchange, with prices listed below.

Type	Strike	Price
Call	1.1	0.06
Call	1.2	0.03
Put	0.8	0.02
Put	0.7	0.01

A collar with floor F=0.9 and cap C=1.15 is not traded on an exchange. Find the range of prices for this collar, consistent with the absence of arbitrage and the prices given above. Enter your results rounded to three decimal places.

Lower bound =

Upper bound =

来源：https://blog.csdn.net/wangchy29/article/details/87070999

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凸优化第五章对偶作业题

Lagrange对偶问题

强弱对偶性

几何解释

优化条件

扰动及灵敏度分析

例子

广义不等式

Numerical perturbation analysis example

A simple example

Lagrangian relaxation of Boolean LP

Option Price Bounds

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最新文章

热门文章

凸优化第五章对偶 作业题

Lagrange对偶问题

强弱对偶性

几何解释

优化条件

扰动及灵敏度分析

例子

广义不等式

Numerical perturbation analysis example

A simple example

Lagrangian relaxation of Boolean LP

Option Price Bounds

凸优化第五章对偶 作业题相关推荐

最新文章

热门文章

凸优化第五章对偶作业题

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