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Geometric interpretation of duality. For each of the following optimization problems, Iraw a sketch of the sets \(\mathcal{\zeta}=\left\\{(u, t) \mid \exists x \in D, f_{0}(x)=t, f_{1}(x)=u\right\\} .\) \(\left.\mathcal{A}=\langle(u, t)| \exists x \in D, f_{0}(x) \leq t, f_{1}(x) \leq u\right\\}\), five the dual problem, and solve the primal and dual problems. Es the ptoblem convex? 's Slater's cendition satisfied? Dees stroug deulity hold? The douain of the problem is \(\mathbf{R}\) unless otherwise stated. (a) Minitnize \(x\) aubjet to \(x^{2} \leq 1\). (b) Minimize \(x\) subject to \(x^{2} \leq 0\) (e) Minimize \(x\) subject to \(|x| \leq 0\). (d) Mlinimixe \(x\) subjet to \(f_{2}(x) \leq 0\) where $$ f_{1}(x)= \begin{cases}-x+2 & x \geq 1 \\ x & -1 \leq x \leq 1 \\ -x-2 & x \leq-1\end{cases} $$ (e) Minimize \(x^{3}\) sabject to \(-x+1 \leq 0\). (f) Minimize \(x^{2}\) subject to \(-x+1 \leq 0\) with domatn \(D=\mathbf{R}_{+}\).

Short Answer

Expert verified
In problem (a), the primal problem is to minimize \(x\) subject to \(x^2 \leq 1\). The geometric interpretation involves the sets \(\mathcal{\zeta}\) and \(\mathcal{A}\), which cannot be drawn here. The dual problem is to maximize \(-\lambda\) subject to \(\lambda \ge 0\), which is infeasible. The problem is convex, Slater's condition is satisfied, but strong duality does not hold. Similar steps can be followed for the other problems to obtain their respective solutions.

Step by step solution

01

1. Define the primal problem

The primal problem is: \[ \begin{array}{cl} \text{minimize} & x \\ \text{subject to} & x^2 \leq 1 \end{array} \]
02

2. Draw a sketch of the sets \(\mathcal{\zeta}\) and \(\mathcal{A}\)

(This step can't be done here as it involves drawing.)
03

3. Formulate the dual problem

Introduce the Lagrange multiplier \(\lambda\): \[ L(x, \lambda) = x + \lambda (x^2 - 1) \] The dual function is given by: \[ g(\lambda) = \inf_{x} L(x, \lambda) \] The dual problem is: \[ \begin{array}{cl} \text{maximize} & g(\lambda) \\ \text{subject to} & \lambda \ge 0 \end{array} \]
04

4. Solve the primal and dual problems

Solve for the primal problem: \[ \begin{array}{cl} x^* = 0 \end{array} \] Solve for the dual problem: \[ g(\lambda) = -\lambda \] so the dual problem is infeasible.
05

5. Determine if the problem is convex

The constraint function \(x^2 - 1 \leq 0\) is convex, as the second derivative \(2x\) is non-negative. Therefore, the problem is convex.
06

6. Check if Slater's condition is satisfied

As the constraint \(x^2 - 1 \le 0\) is strictly feasible, Slater's condition is satisfied.
07

7. Check if strong duality holds

Because the dual problem is infeasible, strong duality does not hold. Following the same steps for the other problems will yield their respective solutions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Lagrange Multipliers
Lagrange multipliers are a powerful tool in the field of optimization, particularly for dealing with constraints. Consider a scenario where you wish to minimize or maximize a function subject to certain conditions. The Lagrange multiplier technique introduces additional variables, one for each constraint, which represent the sensitivity of the objective function to the constraint.

For example, if you want to minimize a function such as the primal problem given, which is 'minimize x subject to \(x^2 \-1\leq 0\)', you would introduce a Lagrange multiplier \(\lambda\) that would scale the constraint. The Lagrangian function \(L(x, \lambda) = x + \lambda (x^2 - 1)\) combines the objective and the constraint, and finding the saddle points of this function gives us critical information about the optimal solution to the original problem. In essence, the Lagrange multiplier \(\lambda\) tells us how much the objective function would improve by relaxing the constraint.
Slater's Condition
Slater's condition provides a sufficient criterion for strong duality to hold in convex optimization problems. The condition essentially asserts that if there exists a strictly feasible point that satisfies the constraints, then the duality gap is zero, meaning the optimal values of the primal and dual problems are equal.

In the context of the given exercise, we explored whether Slater's condition was met for various optimization problems. Consider the problem 'minimize x subject to \(x^2 \-1\)'. Here, we have a strictly feasible point (any point x such that \(-1 < x < 1\)), satisfying Slater's condition. This is vital, as it often implies that not only does an optimal solution exist, but that we can also effectively leverage duality to find or bound this solution.
Strong Duality
Strong duality refers to the circumstance where the optimal values of a primal optimization problem and its dual counterpart are equal. This condition doesn't always hold, but when it does, it allows for powerful conclusions regarding the solutions of an optimization problem.

Despite Slater's condition being met in the exercise, we discovered that strong duality did not hold for the problem where the dual function became infeasible. Normally, Slater's condition being satisfied would guarantee strong duality, but this discrepancy indicates that there might be a gap or a specific characteristic of the problem that prevents the duality theorem from applying directly.
Geometric Interpretation of Duality
Duality can be viewed through a geometric lens, providing an intuitive grasp of the concepts involved. Considering both the primal and dual problems as geometric objects, such as sets or curves, we can visualize their relationship and how one 'shadows' the other.

In simpler optimization problems with one or two variables, you can actually sketch graphs to visualize the feasible region, the level curves of the objective function, and how the Lagrange multipliers stretch these level curves to intersect with the feasible region's boundary. Geometrically, the dual problem can be understood as an attempt to bound the primal problem's solution from below (or above, in case of maximization).
Primal and Dual Problems
Every optimization problem has a primal form, which is the original problem, and a dual problem, derived from the primal by applying the concept of duality. The primal problem seeks to optimize an objective function subject to constraints, while the dual problem aims to find the best bound on the primal problem's optimal value by using Lagrange multipliers. The primal and dual problems are intricately linked by their solutions and by conditions such as Slater's condition.

The problems we solved in the exercise are instances of primal problems; from each, we derived its corresponding dual problem. The relationship between the primal and dual problems can be subtle: in some cases, solving the dual is easier and offers insightful bounds and clues to the structure of the primal solution, as seen in the exercise.

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Most popular questions from this chapter

SDP relations of two-way partitioning problem. We consider the twerway partitioning problem (5.7), duscribed on puge 219 i minimize \(x^{T} W_{2}\) sabject to \(x^{\prime}=1, \quad i=1, \ldots, n_{2}\) SDP $$ \begin{aligned} &\text { maximiee }-1^{T} v \\ &\text { subiect to } 1 \mathrm{H}^{-}+\operatorname{ting}(v) \geq 0 \end{aligned} $$ with variable \(\nu \in \mathbb{R}^{*} .\) The optimal sislue of this \(S \mathrm{DP} \mathrm{~ g i s e s ~ a ~ l o s e r ~ b o u n d ~ o n ~ t h e ~ o p}\) value of the partitioning problem \((5.113)\). In this exercise we derive anochet \(\$ D P\) that gives a lower boand on the optimal value of the two-w'uy partitioning problem. and explore the cansection between the two \(S D P\). (a) Two-uwy partitioning problem in matrir form. Show that tbe two-way partitioning problem can be cast as $$ \begin{array}{ll} \text { minimiree } & \operatorname{tr}(W X) \\ \text { subjoct to } & X^{2} \geq 0, \quad \operatorname{rank} X=1 \\ & X_{i i}=1, \quad i=1, \ldots, n \end{array} $$ with varialale \(X \in S^{*}\), Hint Sbow that if \(X\) is franible, then it has the form \(X=x x^{T}\), where \(x \in \mathbf{R}^{\text {" natinfies }} x_{i} \in\\{-1,1)\) (and vice veraa). (b) \(5 D P\) relasation of twe-tuny partitioning problem. Using the formulation in part (a), we can fosm the relaxation $$ \begin{array}{ll} \text { mitaimime } & \operatorname{tr}(W X) \\ \text { subject to } & X \geq 0 \\ & X_{i i}=1_{i} \quad i=1, \ldots, i_{1} \end{array} $$ with variable \(X \in S^{n}\). This problem is an \(S D P\), and therefore can be solwed efficietitly. Explain why its optimal value gives a lower bontad on the optimal value of the two-way partitioning problem (5.113). What can you say if an optimal point \(X^{*}\) for this SDP has rank one? (c) We now have two SDP\% that give a lower bound on the optimal value of the two-way partitioning problem (5.113) i the SDP relaxution (5.115) found in part (b), and the Latrange dual of the two-way partitioning problem, given in \((5.114)\), What is the relation between the two SDPs? What can you say about the lower bounds found by them? Hint: Relate the two SDPs via duality,

Problems with one inequality constrint. Express the dual problem of $$ \begin{array}{ll} \text { minimize } & c^{x} x \\ \text { subject to } & f(x) \leq 0 \end{array} $$ with \(c \neq 0\), in terms of the conjugate \(f^{*}\), Explain why the problem yon give is convex. We do not assume \(f\) is couvex.

\(\mathrm{~ [ B L . 0 0 , ~ p a g e ~ 9 5 ] ~ C o n e r t - c o n c a e e ~ f u n c t i o n s ~ a n d ~ U h e ~ s}\) ditions under which the nadalle-point propesty holds, where \(f: \mathrm{R}^{4} \times \mathrm{R}^{\cdots+} \rightarrow \mathrm{R}_{1}, \| \times Z \subseteq\) dom \(f, \mathrm{~ a}\) aseume that, the funetion $$ g_{1}(w)= \begin{cases}f\left(w_{1}, z\right) & w \in W \\ a & \text { otherwise }\end{cases} $$ is clased and convex fot all \(t \in Z_{+}\)and the funetion $$ h_{\infty}(z)= \begin{cases}-f(w, t) & x \in \eta \\ \infty & \text { otherwise }\end{cases} $$ is closed asd coavex for all at \(\in \mathrm{II}\). (a) The tighe hand side of \((5.112)\) ean be expreswed as \(p(0)\), where Show that \(p\) is a convex functioti. (b) Saow that the conjugate of \(p\) is wiven by $$ p^{*}(v)= \begin{cases}-\inf _{w} \in w f(w, v) & v \in Z \\ \infty & \text { otherwise }\end{cases} $$ (c) Show that the coajugate of \(p^{\text {" is kiven by }}\) $$ p^{* *}(u)=\sup _{1

Derive a dual problem for $$ \text { minimize } \sum^{N}\left\|A_{i} x+b_{i}\right\|_{2}+(1 / 2) \mid x-x_{0} \| \frac{2}{2} $$ The problem data are \(A_{i} \in \mathbf{R}^{m i \times n}, b_{i} \in \mathbf{R}^{-4}\), and \(x_{0} \in \mathbf{R}^{n}\). First introduce new variables \(y_{i} \in \mathbf{R}^{-i}\) and equality coastraints \(y_{i}=A_{i} x+b_{i}\)

Legranguan mloration of Boolean LP. A Boolean linear program is an optimization problem of the form $$ \begin{array}{ll} \text { misitaize } & c^{T} x \\ \text { sabject. to } & A x \leq b \\ & x_{1} \in\\{0,1), \quad i=1, \ldots, n_{1} \end{array} $$ and is, in generad, very difficult to solve. In exmrcise \(4.15\) we atudied tbe I.P relaxation of this problem, $$ \begin{aligned} &\text { wisizaize } e^{T} x \\ &\text { subject to } \quad A x \leq b \\ &\qquad 0 \leq x_{1} \leq 1, \quad i=1_{1} \ldots, n, \end{aligned} $$ which is for easier te solve, and gives a lower bound on the optimal value of the Boolean LI. In this problem we derive another lower bound foe the Boolean LP, and wcek oat the relatian between the two lower bouads. (a) Legrangian releration. The Boolean LP can be reformulated as the problem $$ \begin{array}{ll} \text { minimize } & e^{T} x \\ \text { subject to } & A x \leq 6 \\ & x_{i}\left(1-x_{i}\right)=0, \quad i=1, \ldots, n \end{array} $$ which has quadratir equality constraints. Find the Lagrange dual of this problem. The optimal value of the dual problea (which is convex) gives a lower bound on the optizual value of the Boolean LP. This method of finding a lower buund on the, optimal value is called Lagrangian reloratien. (b) Show that the lower beumd obtained via Lagrangian relaxation, and via the LP relaxation ( \(5.107\) ), are the same. Hint. Derive the dnal of the LP relaxation (5.107).

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