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Theorem 1. The Shapley value is given by\n\n\[ \n{\Phi }_{i}\left( v\right) \mathrel{\text{:=}} \mathop{\sum }\limits_{{S \subset N\smallsetminus \{ i\} }}\frac{\left| S\right| !\left( {n - \left| S\right| - 1}\right) !}{n!}\left( {v\left( {S\cup \{ i\} }\right) - v\left( S\right) }\right) \n\] | Proof. It's not difficult to check that it satisfies axioms 1 through 4. (DUM & ADD: easy; How about EFF & SYM?) | No |
Theorem 2. There exists a unique function \( \mathbf{\Phi } \) satisfying the Shapley axioms. | Proof.\n\nlet \( \varphi \) be an allocation rule satisfying EFF, NPP, SYM, and ADD.\n\nSince \( \varphi \) satisfies EFF, NPP, and SYM, we have that, for each \( i \in N \) , each \( \varnothing \neq S \subset N \), and each \( {\alpha }_{S} \in \mathbb{R} \) ,\n\n\[{\varphi }_{i}\left( {{\alpha }_{S}{w}^{S}}\right) =... | No |
Theorem 2 Let \( \Gamma \) be an extensive game with perfect information. For any behavior strategy profile \( b \in B \) of \( \Gamma, b \) is a subgame perfect equilibrium of \( \Gamma \) if and only if no one-shot deviation from \( b \) is profitable. One-shot deviation: 对某个子博弈, 这个子博弈开始节点的玩家只在该节点改變他的choice,造成跟 \( b ... | 证明概要: 反证法 - - 假设 \( b \) 不是SPE 则存在子博弈 \( {\Gamma }_{x} \) 和 \( i \) ,其中 \( {b}_{i} \) 不是 \( i \) 的最佳选择 \( \left( {\text{仍记}{b}_{x} = b}\right) \) O O O, 分） 清火掌 Game T&A 证明概要: 反证法例如 Among all such deviations, let \( {\widehat{b}}_{i} \) be the one that differs from \( {b}_{i} \) ![5f38e725-5816-491f-9911-7b64d12f9c09_69... | No |
Theorem 3 Every extenstive game \( \Gamma \) with behavior Nash perfect recall has at least one behavior strat- equilibrium egy profile \( b \in B \) that is a subgame perfect equilibrium of \( \Gamma \). | ## 定理三 證明概要\n\nSketch: Induction on the number of decomposable nodes in \( \Gamma \) .\n\nInductive basis: 如果都不能分割, 則任何每一個behavior Nash都是一個subgame perfect equilibrium。而根據perfect recall得知一定有behavior Nash, 所以OK。 Induction Step: 找一個可以分割的internal node \( x \) 。先替 \( {\Gamma }_{x} \) 找到一個subgame perfect equilibrium \( {b}_{... | No |
Theorem 3.7.5 (Finite horizon perfect folk theorem; adapted from Benoît and Krishna (1985)). Let \( G\left( {\delta, T}\right) \) be a finitely repeated game. Suppose that \( F \) is full dimensional and that, for each player \( i \in N \), there are two Nash equilibria \( {e}^{i} \) and \( {\widetilde{e}}^{t} \) of \(... | Proof. Omitted (see GSM-115) | No |
Theorem 4.5.1 (The revelation principle). Let \( \left( {\Theta ,\rho, M,\mathcal{S}, u}\right) \) be a mechanism and \( \widehat{m} \) a Bayesian equilibrium. Then, there is a direct mechanism in which there is a truthful equilibrium whose outcome coincides with the outcome of the original equilibrium, i.e., for each ... | Proof. Consider the direct mechanism \( \left( {\Theta ,\rho ,\bar{M},\overline{\mathcal{S}}, u}\right) \) defined as follows: \( \bar{M} \mathrel{\text{:=}} \mathop{\prod }\limits_{{i \in N}}{\Theta }_{i} \cup \{ r\} \) and \( \overline{\mathcal{S}} \) is defined, for each \( \theta \in \Theta \), as \( \overline{\mat... | Yes |
Lemma 5.3.2. Let \( \left( {F, d}\right) \in {B}^{N} \) and let \( z \mathrel{\text{:=}} \mathrm{{NA}}\left( {F, d}\right) \) . For each \( x \in {\mathbb{R}}^{N} \), let \( h\left( x\right) : = \mathop{\sum }\limits_{{i \in N}}\mathop{\prod }\limits_{{j \neq i}}\left( {{z}_{j} - {d}_{j}}\right) {x}_{i} \) . Then, for ... | Proof Suppose that there is \( x \in F \) with \( h\left( x\right) > h\left( z\right) \) for sufficiently small \( \varepsilon \in \left( {0,1}\right) ,{x}^{\varepsilon } \mathrel{\text{:=}} {\varepsilon x} + \left( {1 - \varepsilon }\right) z \in {F}_{d} \n\n\[ \n{g}^{d}\left( {x}^{\varepsilon }\right) = \mathop{\prod... | Yes |
Example 5.4.1. (Divide a million). A wealthy man dies and leaves one million euros to his three nephews, with the condition that at least two of them must agree on how to divide this amount among them; otherwise, the million euros will be burned. | \[ N = \{ 1,2,3\}, v\left( 1\right) = v\left( 2\right) = v\left( 3\right) = 0, \] \[ v\left( {12}\right) = v\left( {13}\right) = v\left( {23}\right) = v\left( N\right) = 1. \] | No |
Example 5.4.2. (The glove game). Three players are willing to divide the benefits of selling a pair of gloves. Player 1 has a left glove and players 2 and 3 have one right glove each. A left-right pair of gloves can be sold for one euro. | \[ N = \{ 1,2,3\} \] \[ v\left( 1\right) = v\left( 2\right) = v\left( 3\right) = v\left( {23}\right) = 0, \] \[ v\left( {12}\right) = v\left( {13}\right) = v\left( N\right) = 1\text{.} \] | Yes |
Proposition 5.5.2. Let \( v \in {S}^{N} \) be a simple game. Then, \( C\left( v\right) \neq \varnothing \) if and only if there is at least one veto player in \( v \) . Moreover, if \( C\left( v\right) \neq \varnothing \), then \( C\left( v\right) = \left\{ {x \in I\left( v\right) : }\right. \) for each nonveto player ... | Proof. Easy (GSM-115, Page 220) | No |
Theorem 1. ELS games are balanced. | Proof. Let \( B \) be a balanced collection of \( {2}^{N} \) and let \( {\lambda }_{S} \in {\mathbb{Q}}^{ + }, S \in B \), be the corresponding weights. It is sufficient to prove that \( \;\mathop{\sum }\limits_{{S \in B}}{\lambda }_{S}c\left( S\right) \geq c\left( N\right) \) . | No |
Theorem 1. The Shapley value is given by\n\n\[ \n{\Phi }_{i}\left( v\right) \mathrel{\text{:=}} \mathop{\sum }\limits_{{S \subset N\smallsetminus \{ i\} }}\frac{\left| S\right| !\left( {n - \left| S\right| - 1}\right) !}{n!}\left( {v\left( {S\cup \{ i\} }\right) - v\left( S\right) }\right) \n\] | Proof. It's not difficult to check that it satisfies axioms 1 through 4. (DUM & ADD: easy; How about EFF & SYM?) | No |
Theorem 2. There exists a unique function \( \mathbf{\Phi } \) satisfying the Shapley axioms. | Proof.\n\nlet \( \varphi \) be an allocation rule satisfying EFF, NPP, SYM, and ADD.\n\nSince \( \varphi \) satisfies EFF, NPP, and SYM, we have that, for each \( i \in N \) , each \( \varnothing \neq S \subset N \), and each \( {\alpha }_{S} \in \mathbb{R} \) ,\n\n\[{\varphi }_{i}\left( {{\alpha }_{S}{w}^{S}}\right) =... | No |
Problem 5. Portfolio Management Problem Recall portfolio management problem in Lecture Note #02. This time you need to consider an instance of the portfolio risk-minimization problem with two stocks. The expected mean and variance of the return on each share of stock 1 are 1 and 2 respectively, whereas the correspondin... | You need to assign your investment between these two assets reasonably to minimize the portfolio variance. | No |
Problem 1. Let \( \Omega \subset {R}^{n} \) be a nonempty convex set. Show that \( \operatorname{int}\left( \Omega \right) = \operatorname{int}\left( {cl\Omega }\right) \) and \( \partial \Omega = \partial \left( {cl\Omega }\right) \) using the following conclusion in Page 34 of Lecture Note#03 淑芬讲义第七章10页 | \[ x \in \operatorname{int}\left( \Omega \right) ,\;y \in \operatorname{cl}\Omega ,\;\lambda \in \left( {0,1}\right) \; \Rightarrow \;{\lambda x} + \left( {1 - \lambda }\right) y \in \operatorname{int}\left( \Omega \right) . \] | No |
Problem 5. Consider the two-variable linear program with 6 inequality constraints:\n\n\[ \max \;3{x}_{1} + 5{x}_{2} \]\n\n\[ \text{s.t.}\;{x}_{1} \geq 0 \]\n\n\[ {x}_{2} \geq 0 \]\n\n\[ - {x}_{1} + {x}_{2} \leq {2.5} \]\n\n\[ {x}_{1} + 2{x}_{2} \leq 9 \]\n\n\[ {x}_{1} \leq 4 \]\n\n\[ {x}_{2} \leq 3 \] | a) Plot the constraints in a two-dimensional graph.\nb) Identify the extreme points of the feasible region.\nc) Identify the optimal solution point of the problem. | No |
Problem 4. Given the LP problem\n\n\\[ \n\\min \\; - 2{x}_{1} - {x}_{2} + {x}_{3} \n\\]\n\n\\[ \n\\text{s.t.}\\;{x}_{1} + {x}_{2} + 2{x}_{3} \\leq 6 \n\\]\n\n\\[ \n{x}_{1} + 4{x}_{2} - {x}_{3} \\leq 4 \n\\]\n\n\\[ \n{x}_{1},{x}_{2},{x}_{3} \\geq 0 \n\\] | and its optimal simplex tableau\n\n<table><thead><tr><th>Basic</th><th>Row</th><th>\\( {x}_{1} \\)</th><th>\\( {x}_{2} \\)</th><th>\\( {x}_{3} \\)</th><th>\\( {x}_{4} \\)</th><th>\\( {x}_{5} \\)</th><th>RHS</th></tr></thead><tr><td>-Z</td><td>(0)</td><td>0</td><td>6</td><td>0</td><td>\\( \\frac{1}{3} \\)</td><td>\\( \\... | Yes |
Problem 1. Let \( f \) be a convex function on \( {R}^{n} \), and let \( g \) be a convex nondecreasing function on \( R \) . [The nondecreasing property of \( g \) means that for all \( x \) and \( y \) in \( R, x \leq y \) implies \( g\left( x\right) \leq g\left( y\right) \) .]\n\n(a) Show that the composite function... | Solution: (a) Let \( x \) and \( y \) be any two points in \( {R}^{n} \) and let \( \alpha \in \left( {0,1}\right) \) be arbitrary. Then\n\n\[ f\left( {{\alpha x} + \left( {1 - \alpha }\right) y}\right) \leq {\alpha f}\left( x\right) + \left( {1 - \alpha }\right) f\left( y\right) . \]\n\nSince \( g \) is nondecreasing ... | Yes |
Given matrix \( A \in {\mathcal{R}}^{m \times n} \) and vectors \( b \in {\mathcal{R}}^{m} \) and \( c \in {\mathcal{R}}^{n} \), what’s the alternative system for\n\n\[ \n{Ax} = b,{A}^{T}y \leq c,{c}^{T}x - {b}^{T}y \leq 0, x \geq 0?\n\] | Solution: In light of Farkas' Lemma: the following two systems are alternative:\n\n\[ \n\text{(I)}{Ax} = b, x \geq 0\n\]\n\n(II) \( {A}^{T}y \leq 0,{b}^{T}y > 0 \) .\n\nThe desired alternative system is\n\n\[ \nA{x}^{\prime } - {b\tau } = 0,{A}^{T}{y}^{\prime } - {c\tau } \leq 0,{b}^{T}{y}^{\prime } - {c}^{T}{x}^{\prim... | Yes |
\[ \min \;\left| x\right| + \left| y\right| \] \[ \text{s.t.}\;x + y \geq 2,\;x \leq 3\text{.} \] | Solution: Let \[ {x}_{1} = \frac{\left| x\right| + x}{2},\;{x}_{2} = \frac{\left| x\right| - x}{2}, \] and \[ {x}_{3} = \frac{\left| y\right| + y}{2},\;{x}_{4} = \frac{\left| y\right| - y}{2}. \] Then, its standard form is \[ \min \;{x}_{1} + {x}_{2} + {x}_{3} + {x}_{4} \] \[ \text{s.t.}\;{x}_{1} - {x}_{2} + {x}_{3} - ... | Yes |
Problem 4. Consider the two-variable linear program with 6 inequality constraints:\n\n\[ \max \;3{x}_{1} + 5{x}_{2} \]\n\n\[ \text{s.t.}\;{x}_{1} \geq 0 \]\n\n\[ {x}_{2} \geq 0 \]\n\n\[ - {x}_{1} + {x}_{2} \leq {2.5} \]\n\n\[ {x}_{1} + 2{x}_{2} \leq 9 \]\n\n\[ {x}_{1} \leq 4 \]\n\n\[ {x}_{2} \leq 3 \] | Solution: (b) The extreme points are: \( \left( {0,0}\right) ,\left( {0,{2.5}}\right) ,\left( {{0.5},3}\right) ,\left( {3,3}\right) ,\left( {4,{2.5}}\right) ,\left( {4,0}\right) \) .\n\n(c) The optimal value is 24.5 at \( \left( {4,{2.5}}\right) \) . | Yes |
Problem 5. While solving a standard simplex form linear programming problem using the simplex method, we get the following tableau:\n\n<table><thead><tr><th></th><th>\\( {x}_{1} \\)</th><th>\\( {x}_{2} \\)</th><th>\\( {x}_{3} \\)</th><th>\\( {x}_{4} \\)</th><th>\\( {x}_{5} \\)</th><th></th></tr></thead><tr><td></td><td... | Solution: a) The necessary and sufficient condition is \\( {\\bar{c}}_{3} \\) and \\( {\\bar{c}}_{5} \\) are both nonnegative. | Yes |
Problem 1. 考虑投资组合风险管理问题。假设你有一笔可用资金(不防设一个单位)全部用来投资两种股票。由于投资收益的不确定性, 设投资第1种股票收益的期望和方差分别为1和2, 投资第2种股票收益的期望和方差分别为2和3 , 它们的协方差为 -2 。假设你的最小期望收益为1.4。试确定一个最优的投资方案(不允许卖空), 使得在一定的条件下你的投资风险最小。 | 回答下列问题:\n\n1) 建立该问题的优化模型。\n\n2) 利用其KKT条件求解该优化问题, 来确定你的最优投资方案。\n\n3) 利用MATLAB或LINGO编程求解。 | No |
设函数 \( f : {\mathcal{R}}^{n} \rightarrow \mathcal{R} \) 连续可微。考虑约束优化问题\n\n\[ \min \;f\left( x\right) \]\n\n\[ \text{s.t.}\;{Ax} = b\text{,} \]\n\n其中 \( A \in {\mathcal{R}}^{m \times n} \) 且 \( A \) 的秩是 \( m, b \in {\mathcal{R}}^{m} \) 。令 \( A = \left( {{A}_{B},{A}_{N}}\right), x = \left( \begin{matrix} {x}_{B} \\ {x}_{N... | 证明:\n\n1. \( x \) 是该优化问题的KKT点当且仅当 \( d = 0 \) 。\n\n2. 若 \( d \neq 0 \) ,则 \( d \) 是该优化问题在 \( x \) 处的可行下降方向。 | Yes |
设函数 \( f : {\mathcal{R}}^{n} \rightarrow \mathcal{R} \) 连续可微,矩阵 \( A, B \in {\mathcal{R}}^{m \times n} \) ,向量 \( a, b \in {\mathcal{R}}^{m} \) 。考虑约束优化问题\n\n(P) \( \min \;f\left( x\right) \)\n\n\[ \text{s.t.}\;{Ax} \geq a\text{,} \]\n\n\[ {Bx} = b\text{.} \]\n\n设 \( \bar{x} \) 是(P)的一个可行解,在 \( \bar{x} \) 处不等式约束 \( {Ax} \... | 1) \( \bar{x} \) 是优化问题(P)的KKT点当且仅当 \( \nabla f{\left( \bar{x}\right) }^{T}{d}^{ * } = 0 \) 。\n\n2) 若 \( \nabla f{\left( \bar{x}\right) }^{T}{d}^{ * } \neq 0 \) ,则 \( {d}^{ * } \) 是该优化问题(P)在 \( \bar{x} \) 处的可行下降方向。 | Yes |
Consider the equality-constrained optimization problem\n\n\[ \min \;f\left( x\right) \]\n\n(1)\n\n\[ \text{s.t.}\;{h}_{i}\left( x\right) = 0, i \in \mathcal{E}\text{.} \] | Let \( \left( {{x}^{ * },{v}^{ * }}\right) \) satisfy the second-order sufficiency condition for problem (1). Then there exists \( {c}^{ * } > 0 \) such that for any \( c \geq {c}^{ * },{x}^{ * } \) is a strict local minimum for the unconstrained problem\n\n\[ \min \phi \left( {x, c}\right) = f\left( x\right) - \mathop... | Yes |
Theorem 2 Let \( C \subset {\mathcal{R}}^{n} \) be a convex set. Then both \( {int}\left( C\right) \) and \( {cl}\left( C\right) \) are convex. | Proof. We first prove for any \( \lambda \in \left( {0,1}\right) \)\n\n\[ \mathbf{x} \in \operatorname{int}\left( C\right) ,\mathbf{y} \in \operatorname{cl}\left( C\right) \; \Rightarrow \;\mathbf{u} = \lambda \mathbf{x} + \left( {1 - \lambda }\right) \mathbf{y} \in \operatorname{int}\left( C\right) .\n\]\n\nSince \( \... | Yes |
Theorem 3 Let \( S \in {\mathcal{R}}^{n} \) be a nonempty convex set and \( \bar{\mathbf{x}} \in \partial S \) . Then there exists a vector \( \mathbf{a} \neq \mathbf{0} \) such that \( {\mathbf{a}}^{T}\mathbf{x} \leq {\mathbf{a}}^{T}\bar{\mathbf{x}} \) for each \( \mathbf{x} \in {cl}\left( S\right) \) . | Proof: Since \( \bar{\mathbf{x}} \in \partial S \) , there exists a sequence \( \{ {\mathbf{y}}^{k}\} \) not in \( {cl}\left( S\right) \) such that\n\n\( {\mathbf{y}}^{k} \rightarrow \bar{\mathbf{x}} \) . By the separating hyperplane theorem, corresponding to each \( {\mathbf{y}}^{k} \) there exists a vector \( {\mathb... | Yes |
Theorem 6 If \( f : C \rightarrow \mathcal{R} \) is convex, then\n\n\[ f\left( {{\lambda }_{1}{\mathbf{x}}^{1} + \ldots + {\lambda }_{m}{\mathbf{x}}^{m}}\right) \leq {\lambda }_{1}f\left( {\mathbf{x}}^{1}\right) + \ldots + {\lambda }_{m}f\left( {\mathbf{x}}^{m}\right) \]\n\nfor any \( {\mathbf{x}}^{1},\ldots ,{\mathbf{... | ## Proof of Jensen's Inequality\n\nProof. For \( m = 2 \) the inequality is just the definition of convexity. Arguing\n\ninductively, we now assume \( m > 2 \) and that the inequality holds for \( m - 1 \)\n\npoints. For \( m \) points we have\n\n\[ {\lambda }_{1}{\mathbf{x}}^{1} + \ldots + {\lambda }_{m}{\mathbf{x}}^{... | Yes |
Theorem 7 Let \( f \in {C}^{1} \) be in a region containing the line segment \( \left\lbrack {\mathbf{x},\mathbf{y}}\right\rbrack \) . Then there is a \( \alpha \) , \( 0 \leq \alpha \leq 1 \) , such that\n\n\[ f\left( \mathbf{y}\right) = f\left( \mathbf{x}\right) + \nabla f\left( {\alpha \mathbf{x} + \left( {1 - \alph... | Furthermore, if \( f \in {C}^{2} \) then there is a \( \alpha \) , \( 0 \leq \alpha \leq 1 \) , such that\n\n\[ f\left( \mathbf{y}\right) = f\left( \mathbf{x}\right) + \nabla f\left( \mathbf{x}\right) \left( {\mathbf{y} - \mathbf{x}}\right) + \left( {1/2}\right) {\left( \mathbf{y} - \mathbf{x}\right) }^{T}{\nabla }^{2}... | Yes |
Theorem 8 Suppose \( X \subseteq {\mathcal{R}}^{n} \) is open, \( x \in X \) , and \( f : X \rightarrow \mathcal{R} \) is differentiable. Then\n\n\[ f\left( {x + h}\right) = f\left( x\right) + \nabla f\left( x\right) h + o\left( {\parallel h\parallel }\right) \text{as}h \rightarrow 0\text{.} \] | If \( f \in {C}^{2} \), then\n\n\[ f\left( {x + h}\right) = f\left( x\right) + \nabla f\left( x\right) h + \frac{1}{2}{h}^{T}{\nabla }^{2}f\left( x\right) h + o\left( {\parallel h{\parallel }^{2}}\right) \text{as}h \rightarrow 0\text{.} \] | No |
Theorem 9 Let \( f \in {C}^{1} \) . Then \( f \) is convex over a convex set \( \Omega \) if and only if the gradient inequality holds, i.e., | \[ f\left( \mathbf{y}\right) \geq f\left( \mathbf{x}\right) + \nabla f\left( \mathbf{x}\right) \left( {\mathbf{y} - \mathbf{x}}\right) \] for all \( \mathbf{x},\mathbf{y} \in \Omega \) . | No |
Theorem 11 Let \( A \in {\mathcal{R}}^{m \times n} \) and \( \mathbf{b} \in {\mathcal{R}}^{m} \) . The system \( \{ \mathbf{x} : A\mathbf{x} = \mathbf{b}\} \) has a solution if and only if that \( {A}^{T}\mathbf{y} = \mathbf{0} \) and \( {\mathbf{b}}^{T}\mathbf{y} \neq 0 \) has no solution. | Proof: We assume that there exists a vector \( \mathbf{y} \) such that \[ {A}^{T}\mathbf{y} = \mathbf{0},\;{\mathbf{b}}^{T}\mathbf{y} \neq 0. \] Let \( \overline{\mathbf{x}} \) be a solution of the system \( \{ \mathbf{x} : A\mathbf{x} = \mathbf{b}\} \) . Then \[ 0 \neq {\mathbf{b}}^{T}\mathbf{y} = {\overline{\mathbf{x... | Yes |
Theorem 12 (Farkas’ Lemma) Let \( A \in {\mathcal{R}}^{m \times n} \) and \( \mathbf{c} \in {\mathcal{R}}^{n} \) . The system \( \{ \mathbf{x} : A\mathbf{x} \leq \mathbf{0},{\mathbf{c}}^{T}\mathbf{x} > 0\} \) has a solution if and only if that \( {A}^{T}\mathbf{y} = \mathbf{c} \) and \( \mathbf{y} \geq \mathbf{0} \) ha... | Proof: Suppose, reasoning by contradiction, that the system\n\n\[ \left\{ {\mathbf{y} : {A}^{T}\mathbf{y} = \mathbf{c},\mathbf{y} \geq \mathbf{0}}\right\} \]\n\nhas a solution \( \overline{\mathbf{y}} \) . Let \( \overline{\mathbf{x}} \in \left\{ {\mathbf{x} : A\mathbf{x} \leq \mathbf{0},{\mathbf{c}}^{T}\mathbf{x} > 0}... | Yes |
Theorem 1 Consider the polyhedron in the LP standard form (LP). Then, a basic feasible solution and an extreme point are equivalent; the formal is algebraic and the latter is geometric. | ## Proof\n\nLet \( \mathbf{x} \) be a basic feasible solution of (LP) with respect to basis \( {A}_{B} \) . Then, the columns of \( {A}_{B} \) are linearly independent, and\n\n\[ \mathbf{x} = \left( \begin{array}{l} {\mathbf{x}}_{B} \\ {\mathbf{x}}_{N} \end{array}\right) ,{\mathbf{x}}_{B} \geq \mathbf{0},{\mathbf{x}}_{... | Yes |
Theorem 2 Given (LP) where \( A \) has full row rank \( m \) , (i) if there is a feasible solution, there is a basic feasible solution; (ii) if there is an optimal solution, there is an optimal basic solution. | ## Proof of (i)\n\nLet \( \mathbf{x} \in P \) and, without loss of generality, suppose that\n\n\( \mathbf{x} = \left( {{x}_{1};\ldots ;{x}_{k};0;\ldots ;0}\right) \), where \( {x}_{j} > 0 \) for \( j = 1,\ldots, k \) .\n\nIf \( {A}_{.1},\ldots ,{A}_{.k} \) are linearly independent, then \( k \leq m \) and \( \mathbf{x}... | Yes |
Theorem 3 If \( {\mathbf{r}}_{N} \geq \mathbf{0} \) at a feasible basis \( B \) , then the corresponding basic feasible solution \( \bar{\mathbf{x}} = \left( {{\bar{\mathbf{x}}}_{B};0}\right) \) is an optimal basic solution of (LP) and \( {A}_{B} \) is an optimal basis. | For any \( \mathbf{x} \in P \) , corresponding to the feasible basis \( B \) , we have\n\n\[ \n{\mathbf{x}}_{B} = \overline{\mathbf{b}} - {\bar{A}}_{N}{\mathbf{x}}_{N},\;{\mathbf{x}}_{N} \geq 0 \n\]\n\nand\n\n\[ \n{\mathbf{c}}^{T}\mathbf{x} = {\mathbf{c}}_{B}^{T}{\mathbf{x}}_{B} + {\mathbf{c}}_{N}^{T}{\mathbf{x}}_{N} \... | Yes |
Theorem 4 Let \( \mathbf{x} = \left( {{\mathbf{x}}_{B};{\mathbf{x}}_{N}}\right) \) be a basic feasible solution of (LP). If there exists a negative component of \( {\mathbf{r}}_{N} \) and the corresponding column vector of \( {\bar{A}}_{N} \) is nonpositive, then (LP) is unbounded. | ## Proof\n\nLet \( {\mathbf{r}}_{s} < 0 \) and \( {\bar{A}}_{.s} \leq \mathbf{0} \) where \( s \in N \) . For any sufficiently large positive number \( \alpha > 0 \), define \( \widetilde{\mathbf{x}} = \left( {{\widetilde{\mathbf{x}}}_{B};{\widetilde{\mathbf{x}}}_{N}}\right) \) with\n\n\[ \n{\widetilde{\mathbf{x}}}_{B}... | Yes |
Theorem 1 (Weak duality theorem) Let feasible regions \( {\mathcal{F}}_{p} \) and \( {\mathcal{F}}_{d} \) be non-empty. Then,\n\n\[ \n{\mathbf{c}}^{T}\mathbf{x} \geq {\mathbf{b}}^{T}\mathbf{y}\text{ where }\mathbf{x} \in {\mathcal{F}}_{p},\left( {\mathbf{y},\mathbf{s}}\right) \in {\mathcal{F}}_{d}. \n\] | \[ \n{\mathbf{c}}^{T}\mathbf{x} - {\mathbf{b}}^{T}\mathbf{y} = {\mathbf{c}}^{T}\mathbf{x} - {\left( A\mathbf{x}\right) }^{T}\mathbf{y} = {\mathbf{x}}^{T}\left( {\mathbf{c} - {A}^{T}\mathbf{y}}\right) = {\mathbf{x}}^{T}\mathbf{s} \geq 0. \n\] | Yes |
Theorem 4 For each optimal dual price vector for the linear program of the grand alliance, allocating each firm the value of its resource vector at those prices yields a profit allocation vector in the core. | Let \( {\mathbf{y}}^{ * } \) be any optimal dual solution and let\n\n\[ \mathbf{z} = \left( {{\left( {\mathbf{b}}^{1}\right) }^{T}{\mathbf{y}}^{ * },{\left( {\mathbf{b}}^{2}\right) }^{T}{\mathbf{y}}^{ * },\ldots ,{\left( {\mathbf{b}}^{\left| I\right| }\right) }^{T}{\mathbf{y}}^{ * }}\right) .\n\nThen, \( \mathbf{z} \) ... | Yes |
Theorem 1 The optmal basis of \( {LP}\left( 0\right) \) remains optimal for \( {LP}\left( \lambda \right) \) if and only if 最优基不变的充要条件 | \[ \begin{array}{l} {A}_{B}^{-1}\left( {\mathbf{b} + \lambda \mathbf{d}}\right) \geq \mathbf{0}\;\mathbf{{and}}\;\left( {\mathbf{c} + \lambda \mathbf{g}}\right) - {\mathbf{A}}^{\mathbf{T}}{\left( {\mathbf{A}}_{\mathbf{B}}^{\mathbf{T}}\right) }^{-\mathbf{1}}{\left( \mathbf{c} + \lambda \mathbf{g}\right) }_{\mathbf{B}} \... | Yes |
Theorem 1 Given matrix \( A \in {\mathcal{R}}^{m \times n} \) where \( n > m \) , take a convex polyhedral cone \( C = \{ A\mathbf{x} : \mathbf{x} \geq \mathbf{0}\} \) . Then for any \( \mathbf{b} \in C \) , | \[ \mathbf{b} = \mathop{\sum }\limits_{{i = 1}}^{d}{\mathbf{a}}_{{j}_{i}}{x}_{{j}_{i}},\;{x}_{{j}_{i}} \geq 0,\forall i \] for some linearly independent vectors \( {\mathbf{a}}_{{j}_{1}},\ldots ,{\mathbf{a}}_{{j}_{d}} \) chosen from \( {\mathbf{a}}_{1},\ldots ,{\mathbf{a}}_{n} \) . 极大无关组 | Yes |
Theorem 2 (Carathéodory’s Theorem) Let \( \Omega \subseteq {\mathcal{R}}^{n} \) and \( x \in \mathrm{{co}}\left( \Omega \right) \). Then there exist at most \( n + 1 \) points in \( \Omega \) such that \( x \) can be expressed as their convex combination, that is, there exist \( {x}^{1},\ldots ,{x}^{p} \in \Omega \) su... | ## Proof of Carathéodory's Theorem - Let \( {\mathbf{x}}^{1},\ldots ,{\mathbf{x}}^{m} \in {\mathcal{R}}^{n}\left( {m \geq n + 2}\right) \) and \[ \mathbf{x} = \mathop{\sum }\limits_{{i = 1}}^{m}{\alpha }_{i}{\mathbf{x}}^{i},\;\mathop{\sum }\limits_{{i = 1}}^{m}{\alpha }_{i} = 1,{\alpha }_{i} \geq 0\left( {i = 1,\ldots,... | Yes |
Theorem 3 (Separating hyperplane theorem) Let \( C \subset {\mathcal{R}}^{n} \) be a closed convex set, and let \( \mathbf{b} \notin C \) . Then there is a vector \( \mathbf{a} \neq \mathbf{0} \) such that\n\n\[ \mathbf{a} \bullet \mathbf{b} > \mathop{\sup }\limits_{{\mathbf{x} \in C}}\mathbf{a} \bullet \mathbf{x} \] | 有一个充要条件 \( \exists \bar{x} \in C \)\n\nSt. \( {\left( b - \bar{x}\right) }^{\top }\left( {x - \bar{x}}\right) \leq 0 \)\n\n\[ \forall x \in C \] | No |
Theorem 6 Let \( f \in {C}^{1} \) . Then \( f \) is concave over a convex set \( \Omega \) if and only if the gradient inequality holds, i.e., | \[ f\left( \mathbf{y}\right) \leq f\left( \mathbf{x}\right) + \nabla f\left( \mathbf{x}\right) \left( {\mathbf{y} - \mathbf{x}}\right) \] for all \( \mathbf{x},\mathbf{y} \in \Omega \) . | No |
Theorem 8 A function \( f \) is quasi-concave if and only if for any \( \mathbf{x},\mathbf{y} \in C \) and any \( \alpha \in \left( {0,1}\right) \)\n\n\[ f\left( {\alpha \mathbf{x} + \left( {1 - \alpha }\right) \mathbf{y}}\right) \geq \min \{ f\left( \mathbf{x}\right), f\left( \mathbf{y}\right) \} . \]\n\n(1)\n\nIn oth... | Proof: First, suppose \( f \) is quasi-concave. For any \( \mathbf{x},\mathbf{y} \in C \) and any\n\n\( \alpha \in \left( {0,1}\right) \) , we may assume, WLOG, \( f\left( \mathbf{y}\right) \geq f\left( \mathbf{x}\right) \) . Then, by the definition,\n\n\( \mathbf{x},\mathbf{y} \in U\left( {f;\mathbf{x}}\right) \) . Si... | Yes |
Theorem 10 (Weak duality theorem) Let feasible regions \( {\mathcal{F}}_{p} \) and \( {\mathcal{F}}_{d} \) be non-empty. Then,\n\n\[{\mathbf{c}}^{T}\mathbf{x} \geq {\mathbf{b}}^{T}\mathbf{y}\text{ where }\mathbf{x} \in {\mathcal{F}}_{p},\left( {\mathbf{y},\mathbf{s}}\right) \in {\mathcal{F}}_{d}.\] | This theorem shows that a feasible solution to either problem yields a bound on the value of the other problem. We call \( {\mathbf{c}}^{T}\mathbf{x} - {\mathbf{b}}^{T}\mathbf{y} \) the duality gap. | No |
Theorem 1 (Carathéodory’s Theorem) Let \( \Omega \subseteq {\mathcal{R}}^{n} \) and \( x \in \operatorname{co}\left( \Omega \right) \) . Then there exist at most \( n + 1 \) points in \( \Omega \) such that \( x \) can be expressed as their convex combination, that is, there exist \( {x}^{1},\ldots ,{x}^{p} \in \Omega ... | ## Proof of Carathéodory's Theorem\n\n- Let \( {\mathbf{x}}^{1},\ldots ,{\mathbf{x}}^{m} \in {\mathcal{R}}^{n}\left( {m \geq n + 2}\right) \) and\n\n\[ \mathbf{x} = \mathop{\sum }\limits_{{i = 1}}^{m}{\alpha }_{i}{\mathbf{x}}^{i},\;\mathop{\sum }\limits_{{i = 1}}^{m}{\alpha }_{i} = 1,{\alpha }_{i} \geq 0\left( {i = 1,\... | Yes |
Theorem 2 Given matrix \( A \in {\mathcal{R}}^{m \times n} \) where \( n > m \) , take a convex polyhedral cone \( C = \{ A\mathbf{x} : \mathbf{x} \geq \mathbf{0}\} \) . Then for any \( \mathbf{b} \in C \) , \n\n\[ \n\mathbf{b} = \mathop{\sum }\limits_{{i = 1}}^{d}{\mathbf{a}}_{{j}_{i}}{x}_{{j}_{i}},\;{x}_{{j}_{i}} \ge... | ## Proof of Theorem 2 \n\nLet \( \mathbf{b} \in C \) and, without loss of generality, suppose that \( \mathbf{b} = A\mathbf{x} \) and \n\n\( \mathbf{x} = \left( {{x}_{1};\ldots ;{x}_{k};0;\ldots ;0}\right) \), where \( {x}_{j} > 0 \) for \( j = 1,\ldots, k \) . \n\nIf \( {A}_{.1},\ldots ,{A}_{.k} \) are linearly indepe... | Yes |
Theorem 3 (Separating hyperplane theorem) Let \( C \subset {\mathcal{R}}^{n} \) be a closed convex set, and let \( \mathbf{b} \notin C \) . Then there is a vector \( \mathbf{a} \neq \mathbf{0} \) such that | \[ \mathbf{a} \bullet \mathbf{b} > \mathop{\sup }\limits_{{\mathbf{x} \in C}}\mathbf{a} \bullet \mathbf{x} \] | Yes |
Theorem 6 Let \( f \in {C}^{1} \) . Then \( f \) is convex over a convex set \( \Omega \) if and only if the gradient inequality holds, i.e., | \[ f\left( \mathbf{y}\right) \geq f\left( \mathbf{x}\right) + \nabla f\left( \mathbf{x}\right) \left( {\mathbf{y} - \mathbf{x}}\right) \] for all \( \mathbf{x},\mathbf{y} \in \Omega \) . | No |
Theorem 8 A function \( f \) is quasi-concave (quasi-convex) if and only if for any\n\n\( \mathbf{x},\mathbf{y} \in C \) and any \( \alpha \in \left( {0,1}\right) \) ,\n\n\[ f\left( {\alpha \mathbf{x} + \left( {1 - \alpha }\right) \mathbf{y}}\right) \geq \min \{ f\left( \mathbf{x}\right), f\left( \mathbf{y}\right) \} .... | Proof: First, suppose \( f \) is quasi-concave. For any \( \mathbf{x},\mathbf{y} \in C \) and any\n\n\( \alpha \in \left( {0,1}\right) \) , we may assume, WLOG, \( f\left( \mathbf{y}\right) \geq f\left( \mathbf{x}\right) \) . Then, by the definition,\n\n\( \mathbf{x},\mathbf{y} \in U\left( {f;\mathbf{x}}\right) \) . Si... | Yes |
Theorem 10 (Weak duality theorem) Let feasible regions \( {\mathcal{F}}_{p} \) and \( {\mathcal{F}}_{d} \) be non-empty. Then,\n\n\[ \n{\mathbf{c}}^{T}\mathbf{x} \geq {\mathbf{b}}^{T}\mathbf{y}\text{ where }\mathbf{x} \in {\mathcal{F}}_{p},\left( {\mathbf{y},\mathbf{s}}\right) \in {\mathcal{F}}_{d}.\n\] | This theorem shows that a feasible solution to either problem yields a bound on the value of the other problem. We call \( {\mathbf{c}}^{T}\mathbf{x} - {\mathbf{b}}^{T}\mathbf{y} \) the duality gap. | No |
Problem 1. Show that the following problem is unbounded.\n\n\[ \n\\max \\;{x}_{1} + {x}_{2} \n\]\n\n\[ \n\\text{s.t.}\\;{x}_{1} - {x}_{2} - {x}_{3} = 1 \n\]\n\n\[ \n- {x}_{1} + {x}_{2} + 2{x}_{3} \\geq 1 \n\]\n\n\[ \n{x}_{1},{x}_{2},{x}_{3} \\geq 0 \n\] | Solution: It is clear that the primal problem is feasible since \( x = \\left( {3;0;2}\\right) \) is a feasible solution. Its dual problem is\n\n\[ \n\\min \\;{y}_{1} + {y}_{2} \n\]\n\n\[ \n\\text{s.t.}\\;{y}_{1} - {y}_{2} \\geq 1 \n\]\n\n\[ \n- {y}_{1} + {y}_{2} \\geq 1 \n\]\n\n\[ \n- {y}_{1} + 2{y}_{2} \\geq 0 \n\]\n... | Yes |
Problem 2. Let \( \Omega \subseteq {R}^{n} \) be a nonempty convex set and \( \bar{\Omega } \) be its closure. Show that\n\n\[ \text{int}\Omega = \text{int}\bar{\Omega }\text{and}\partial \Omega = \partial \bar{\Omega }\text{.} \] | Solution: Clearly, \( \mathsf{{int}}\Omega \subseteq \mathsf{{int}}\bar{\Omega } \) . We will prove that \( \mathsf{{int}}\bar{\Omega } \subseteq \mathsf{{int}}\Omega . \)\n\nLet \( x \in \mathbf{{int}}\bar{\Omega } \) . If \( \mathbf{{int}}\Omega \neq \varnothing \) , take \( y \in \mathbf{{int}}\Omega \) , then there... | Yes |
Problem 3. Consider the following linear program\n\n\\[ \max \\;{10}{x}_{1} + 7{x}_{2} + {30}{x}_{3} + 2{x}_{4} \\]\n\n\\[ \text{s.t.}\\;{x}_{1} - 6{x}_{3} + {x}_{4} \\leq - 2 \\]\n\n\\[ {x}_{1} + {x}_{2} + 5{x}_{3} - {x}_{4} \\leq - 7 \\]\n\n\\[ {x}_{2},{x}_{3},{x}_{4} \\leq 0\\text{.} \\] | Solution: The dual problem is\n\n\\[ \min \\; - 2{y}_{1} - 7{y}_{2} \\]\n\n\\[ \text{s.t.}\\;{y}_{1} + {y}_{2} = {10}\\text{,} \\]\n\n\\[ {y}_{2} \\leq 7 \\]\n\n\\[ - 6{y}_{1} + 5{y}_{2} \\leq {30} \\]\n\n\\[ {y}_{1} - {y}_{2} \\leq 2 \\]\n\n\\[ {y}_{1} \\geq 0,{y}_{2} \\geq 0 \\]\n\nIt is easy to see that \\( {y}_{1}^... | Yes |
Consider the following problem.\n\n\\[ \n\\text{maximize}\\;Z = 2{x}_{1} - 4{x}_{2} \n\\]\n\n\\[ \n\\text{ subject to }\\;{x}_{1} - {x}_{2} \\leq 1 \n\\]\n\n\\[ \n{x}_{1} \\geq 0 \n\\]\n\n\\[ \n{x}_{2} \\geq 0 \n\\] | Solution: (a) The Dual problem is\n\n\\[ \n\\text{minimize}\\;y \n\\]\n\n\\[ \n\\text{subject to}\\;y \\geq 2 \n\\]\n\n\\[ \ny \\leq 4 \n\\]\n\n\\[ \ny \\geq 0\\text{.} \n\\]\n\nBy inspection, we can find the optimal solution \\( {y}^{ * } = 2 \\) .\n\n(b) Let \\( \\left( {{x}_{1}^{ * },{x}_{2}^{ * }}\\right) \\) be th... | Yes |
Theorem 1 (Carathéodory’s Theorem) Let \( \Omega \subseteq {\mathcal{R}}^{n} \) and \( x \in \operatorname{co}\left( \Omega \right) \) . Then there exist at most \( n + 1 \) points in \( \Omega \) such that \( x \) can be expressed as their convex combination, that is, there exist \( {x}^{1},\ldots ,{x}^{p} \in \Omega ... | ## Proof of Carathéodory's Theorem\n\n- Let \( {\mathbf{x}}^{1},\ldots ,{\mathbf{x}}^{m} \in {\mathcal{R}}^{n}\left( {m \geq n + 2}\right) \) and\n\n\[ \mathbf{x} = \mathop{\sum }\limits_{{i = 1}}^{m}{\alpha }_{i}{\mathbf{x}}^{i},\;\mathop{\sum }\limits_{{i = 1}}^{m}{\alpha }_{i} = 1,{\alpha }_{i} \geq 0\left( {i = 1,\... | Yes |
Theorem 2 Given matrix \( A \in {\mathcal{R}}^{m \times n} \) where \( n > m \) , take a convex polyhedral cone \( C = \{ A\mathbf{x} : \mathbf{x} \geq \mathbf{0}\} \) . Then for any \( \mathbf{b} \in C \) , \[ \mathbf{b} = \mathop{\sum }\limits_{{i = 1}}^{d}{\mathbf{a}}_{{j}_{i}}{x}_{{j}_{i}},\;{x}_{{j}_{i}} \geq 0,\f... | ## Proof of Theorem 2\n\nLet \( \mathbf{b} \in C \) and, without loss of generality, suppose that \( \mathbf{b} = A\mathbf{x} \) and \( \mathbf{x} = \left( {{x}_{1};\ldots ;{x}_{k};0;\ldots ;0}\right) \), where \( {x}_{j} > 0 \) for \( j = 1,\ldots, k \) .\n\nIf \( {A}_{.1},\ldots ,{A}_{.k} \) are linearly independent,... | Yes |
Theorem 3 (Separating hyperplane theorem) Let \( C \subset {\mathcal{R}}^{n} \) be a closed convex set, and let \( \mathbf{b} \notin C \) . Then there is a vector \( \mathbf{a} \neq \mathbf{0} \) such that | \[ \mathbf{a} \bullet \mathbf{b} > \mathop{\sup }\limits_{{\mathbf{x} \in C}}\mathbf{a} \bullet \mathbf{x} \] | Yes |
Theorem 6 Let \( f \in {C}^{1} \) . Then \( f \) is convex over a convex set \( \Omega \) if and only if the gradient inequality holds, i.e., | \[ f\left( \mathbf{y}\right) \geq f\left( \mathbf{x}\right) + \nabla f\left( \mathbf{x}\right) \left( {\mathbf{y} - \mathbf{x}}\right) \] for all \( \mathbf{x},\mathbf{y} \in \Omega \) . | No |
Theorem 8 A function \( f \) is quasi-concave (quasi-convex) if and only if for any\n\n\( \mathbf{x},\mathbf{y} \in C \) and any \( \alpha \in \left( {0,1}\right) \), \n\n\[ \nf\left( {\alpha \mathbf{x} + \left( {1 - \alpha }\right) \mathbf{y}}\right) \geq \min \{ f\left( \mathbf{x}\right), f\left( \mathbf{y}\right) \}... | Proof: First, suppose \( f \) is quasi-concave. For any \( \mathbf{x},\mathbf{y} \in C \) and any\n\n\( \alpha \in \left( {0,1}\right) \), we may assume, WLOG, \( f\left( \mathbf{y}\right) \geq f\left( \mathbf{x}\right) \). Then, by the definition,\n\n\( \mathbf{x},\mathbf{y} \in U\left( {f;\mathbf{x}}\right) \). Since... | Yes |
Theorem 10 (Weak duality theorem) Let feasible regions \( {\mathcal{F}}_{p} \) and \( {\mathcal{F}}_{d} \) be non-empty. Then,\n\n\[{\mathbf{c}}^{T}\mathbf{x} \geq {\mathbf{b}}^{T}\mathbf{y}\text{ where }\mathbf{x} \in {\mathcal{F}}_{p},\left( {\mathbf{y},\mathbf{s}}\right) \in {\mathcal{F}}_{d}.\] | This theorem shows that a feasible solution to either problem yields a bound on the value of the other problem. We call \( {\mathbf{c}}^{T}\mathbf{x} - {\mathbf{b}}^{T}\mathbf{y} \) the duality gap. | No |
Theorem 14 (Strict complementarity theorem) If (LP) and (LD) both have feasible solutions then both problems have a pair of strictly complementary solutions \( {\mathbf{x}}^{ * } \geq \mathbf{0} \) and \( {\mathbf{s}}^{ * } \geq \mathbf{0} \) meaning\n\n\[ \n{X}^{ * }{\mathbf{s}}^{ * } = 0\text{ and }{\mathbf{x}}^{ * }... | ## Proof of strict complementarity theorem\n\nWe only need to show that exactly one of the following holds:\n\neither (i) (LD) has an optimal solution with \( {s}_{i}^{ * } > 0 \)\n\nor (ii) (LP) has an optimal solution with \( {x}_{i}^{ * } > 0 \) .\n\nSuppose now (i) is not satisfied. That is, there is no optimal sol... | Yes |
Problem 1. Show that the following problem is unbounded.\n\n\\[ \n\\max \\;{x}_{1} + {x}_{2} \n\\]\n\n\\[ \n\\text{s.t.}\\;{x}_{1} - {x}_{2} - {x}_{3} = 1 \n\\]\n\n\\[ \n- {x}_{1} + {x}_{2} + 2{x}_{3} \\geq 1 \n\\]\n\n\\[ \n{x}_{1},{x}_{2},{x}_{3} \\geq 0 \n\\] | Solution: It is clear that the primal problem is feasible since \\( x = \\left( {3;0;2}\\right) \\) is a feasible solution. Its dual problem is\n\n\\[ \n\\min \\;{y}_{1} + {y}_{2} \n\\]\n\n\\[ \n\\text{s.t.}\\;{y}_{1} - {y}_{2} \\geq 1 \n\\]\n\n\\[ \n- {y}_{1} + {y}_{2} \\geq 1 \n\\]\n\n\\[ \n- {y}_{1} + 2{y}_{2} \\geq... | Yes |
Problem 2. Let \( \Omega \subseteq {R}^{n} \) be a nonempty convex set and \( \bar{\Omega } \) be its closure. Show that\n\n\[ \n\text{int}\Omega = \text{int}\bar{\Omega }\text{and}\partial \Omega = \partial \bar{\Omega }\text{.} \n\] | Solution: Clearly, \( \mathsf{{int}}\Omega \subseteq \mathsf{{int}}\bar{\Omega } \) . We will prove that \( \mathsf{{int}}\bar{\Omega } \subseteq \mathsf{{int}}\Omega . \)\n\nLet \( x \in \mathbf{{int}}\bar{\Omega } \) . If \( \mathbf{{int}}\Omega \neq \varnothing \) , take \( y \in \mathbf{{int}}\Omega \) , then there... | Yes |
Theorem 2 Assume \( f \) is \( {\mathcal{C}}^{\mathcal{1}} \) on \( S \) . If \( {x}^{ * } \in S \) is a local min of \( f \) , then\n\n\[\n\nabla f{\left( {x}^{ * }\right) }^{T}d \geq 0\n\]\n可行方向与下降方向交集为空\n\nfor all feasible directions \( d \) at \( {x}^{ * } \) . 必要非充分条件 | Proof. Let \( d \in {\mathcal{R}}^{n} \) be a feasible direction at \( {x}^{ * } \) and define \( g\left( t\right) = f\left( {{x}^{ * } + {td}}\right) \non \( \left\lbrack {0,\bar{t}}\right\rbrack \) . We have \( g\left( t\right) \geq g\left( 0\right) \) for \( t \) sufficiently small, and\n\n\[{g}^{\prime }\left( t\ri... | Yes |
Theorem 3 Assume \( f \) is \( {\mathcal{C}}^{2} \) on \( S \) . If \( {x}^{ * } \in S \) is a local min of \( f \) , then for any feasible direction \( d \in {\mathcal{R}}^{n} \) at \( {x}^{ * } \)\n\n\[\n\text{1)}\nabla f{\left( {x}^{ * }\right) }^{T}d \geq 0\text{.\}\n\]\n\n## 必要非充分条件\n\n2) \( \nabla f{\left( {x}^{ ... | Proof. For 2), using Taylor's second-order approximation, we have\n\n\[ g\left( t\right) = g\left( 0\right) + {g}^{\prime }\left( 0\right) t + \frac{1}{2}{g}^{\prime \prime }\left( 0\right) {t}^{2} + o\left( {t}^{2}\right) .\n\]\n\nSince \( {g}^{\prime }\left( 0\right) = \nabla f{\left( {x}^{ * }\right) }^{T}d = 0 \) a... | No |
Theorem 6 Suppose that \( f \) is twice differentiable at \( \bar{\mathbf{x}} \) . If \( \nabla f\left( \bar{\mathbf{x}}\right) = \mathbf{0} \) and the Hessian matrix \( H\left( \bar{\mathbf{x}}\right) \) is positive definite, then \( \bar{\mathbf{x}} \) is a strictly local minimizer of (UP). 二阶 充分条件 | Proof. Since \( f \) is twice differentiable at \( \bar{\mathbf{x}} \) and \( \nabla f\left( \bar{\mathbf{x}}\right) = \mathbf{0} \) , we must have, for each \( \mathbf{x} \in {\mathcal{R}}^{n} \) ,\n\n\[ f\left( \mathbf{x}\right) = f\left( \overline{\mathbf{x}}\right) + \frac{1}{2}{\left( \mathbf{x} - \overline{\mathb... | Yes |
Theorem 7 If the functions \( f \) is convex and continuously differentiable, then \( \bar{\mathbf{x}} \) is a global minimizer of (UP) if and only if\n\n\[ \nabla f\left( \overline{\mathbf{x}}\right) = \mathbf{0}. \] | 由此结合凸函数梯度不等式易见为\n\n局部最优解 | No |
Theorem 10 At a regular point \( {x}^{ * } \) of the surface \( S \) defined by \( \mathbf{h}\left( \mathbf{x}\right) = \mathbf{0} \) the tangent plane is equal to\n\n\[ H = \left\{ {d \mid \nabla \mathbf{h}\left( {\mathbf{x}}^{ * }\right) \mathbf{d} = \mathbf{0}}\right\} \] | Proof. It is clear that \( T\left( {x}^{ * }\right) \subset H \) whether \( {x}^{ * } \) is regular or not, for any curve\n\n\( x\left( t\right) \) passing through \( {x}^{ * } \) at \( t = {t}^{ * } \) having derivative \( \dot{x}\left( {t}^{ * }\right) \) such that\n\n\( \nabla \mathbf{h}\left( {\mathbf{x}}^{ * }\rig... | Yes |
Lemma 1 Suppose that hypotheses of The Lagrange Theorem hold, the system\n\n\[ \nabla \mathbf{h}\left( \overline{\mathbf{x}}\right) \mathbf{d} = \mathbf{0},\;\nabla \mathbf{f}{\left( \overline{\mathbf{x}}\right) }^{\mathbf{T}}\mathbf{d} < \mathbf{0} \]\n\n(3)\n\nhas no solution. | Proof. Assume by contradiction that \( \mathbf{d} \) solves the above system. Then there exists a curve \( \mathbf{x}\left( t\right) \) on \( S \) such that \( \bar{\mathbf{x}} = \mathbf{x}\left( \bar{t}\right) \) and \( \dot{\mathbf{x}}\left( \bar{t}\right) = \mathbf{d} \) . Then\n\n\[ f\left( {\mathbf{x}\left( t\righ... | Yes |
Theorem 11 Suppose that \( \overline{\mathbf{x}} \) is a local minimum of (EP) and that it is a regular point of these constraints. Then there is a \( \bar{y} \in {E}^{m} \) 二阶必要条件 \[ \nabla f\left( \overline{\mathbf{x}}\right) - \mathop{\sum }\limits_{{i = 1}}^{m}{\bar{y}}_{i}\nabla {h}_{i}\left( \overline{\mathbf{x}}... | proof. From elementary calculus it is clear that for every twice differentiable curve \( \mathbf{x}\left( t\right) \) on the constraint surface \( S \) through \( \bar{\mathbf{x}} \) (with \( \mathbf{x}\left( 0\right) = \bar{\mathbf{x}} \) ) we have \[ 0 \leq \frac{{d}^{2}}{d{t}^{2}}f\left( {\mathbf{x}\left( t\right) }... | Yes |
Theorem 12 Let \( \overline{\mathrm{x}} \) be a feasible point of (EP) such that the Jacobian matrix \( \nabla \mathbf{h}\left( \bar{\mathbf{x}}\right) \) has rank \( m \) . Suppose there exist scalars \( {\bar{y}}_{1},\ldots ,{\bar{y}}_{m} \) such that\n\n\[ \nabla f\left( \overline{\mathbf{x}}\right) - \mathop{\sum }... | 某种意义上的严格正定 | No |
Theorem 3 Assume \( f \) is \( {\mathcal{C}}^{2} \) on \( S \) . If \( {x}^{ * } \in S \) is a local min of \( f \) , then for any feasible direction \( d \in {\mathcal{R}}^{n} \) at \( {x}^{ * } \)\n\n\[\n\text{1)}\nabla f{\left( {x}^{ * }\right) }^{T}d \geq 0\text{.\n\]\n\n## 必要非充分条件\n\n2) \( \nabla f{\left( {x}^{ * ... | Proof. For 2), using Taylor's second-order approximation, we have\n\n\[\ng\left( t\right) = g\left( 0\right) + {g}^{\prime }\left( 0\right) t + \frac{1}{2}{g}^{\prime \prime }\left( 0\right) {t}^{2} + o\left( {t}^{2}\right) .\n\]\n\nSince \( {g}^{\prime }\left( 0\right) = \nabla f{\left( {x}^{ * }\right) }^{T}d = 0 \) ... | No |
Theorem 6 Suppose that \( f \) is twice differentiable at \( \bar{\mathbf{x}} \) . If \( \nabla f\left( \bar{\mathbf{x}}\right) = \mathbf{0} \) and the Hessian matrix \( H\left( \bar{\mathbf{x}}\right) \) is positive definite, then \( \bar{\mathbf{x}} \) is a strictly local minimizer of (UP). 二阶 充分条件 | Proof. Since \( f \) is twice differentiable at \( \bar{\mathbf{x}} \) and \( \nabla f\left( \bar{\mathbf{x}}\right) = \mathbf{0} \) , we must have, for each \( \mathbf{x} \in {\mathcal{R}}^{n} \) ,\n\n\[ f\left( \mathbf{x}\right) = f\left( \overline{\mathbf{x}}\right) + \frac{1}{2}{\left( \mathbf{x} - \overline{\mathb... | Yes |
Theorem 14 (F.John [1948]) If \( \overline{\mathbf{x}} \) is a local minimizer for (IP) in which the functions \( f \) and \( {c}_{i}, i = 1,\ldots, m \) are differentiable, then there exists a set of nonnegative scalars \( {\bar{y}}_{0},{\bar{y}}_{1},\ldots ,{\bar{y}}_{m} \) not all of which are zero such that\n\n\[ \... | ## Proof\n\nWe may assume that \( \mathcal{A}\left( \bar{\mathbf{x}}\right) \neq \varnothing \) . Since \( \bar{\mathbf{x}} \) is a local minimizer of (IP), the system\n\n\[ \n{v}^{T}\nabla f\left( \overline{\mathbf{x}}\right) < 0 \n\]\n\n\[ \n{v}^{T}\nabla {c}_{i}\left( \overline{\mathbf{x}}\right) > 0\text{ for all }... | Yes |
Theorem 15 Let \( \overline{\mathbf{x}} \) be a local minimizer for (P). Assume the functions \( {c}_{i} \) are differentiable at \( \bar{\mathbf{x}} \) for all \( i \in \mathcal{I} \) , and the functions \( {h}_{i} \) are continuously differentiable at \( \bar{\mathbf{x}} \) for all \( i \in \mathcal{E} \) . If all th... | ## Proof of The KKT Theorem for (P) Lemma 2 Let \( \overline{\mathbf{x}} \) | No |
Lemma 2 Let \( \overline{\mathbf{x}} \) be a local minimizer for \( \left( P\right) \) and a regular point of the surface \( \{ \mathbf{x} \in {\mathcal{R}}^{n}|\mathbf{h}\left( \mathbf{x}\right) = \mathbf{0}\} \) . If the functions \( {c}_{i} \) are continuous at \( \bar{\mathbf{x}} \) for all \( i \notin \mathcal{A}\... | Proof. Assume by contradiction that there exists \( d \in \mathcal{D} \cap \mathcal{F}. \) By Theorem \( {10} \) ,\nthere is a curve \( x\left( t\right) \) on the surface passing through \( \bar{\mathbf{x}} \) such that \( x\left( 0\right) = \bar{\mathbf{x}} \) and\n\n\( \dot{x}\left( 0\right) = d \) .\n\nFor \( i \in ... | Yes |
Theorem 16 (Second-Order Necessary Condition) Suppose the functions \( f,\mathbf{c},\mathbf{h} \in {\mathcal{C}}^{\mathbf{2}} \) and the hypotheses in Theorem 15 hold, then there exist (unique) multipliers \( {\lambda }_{1},\ldots ,{\lambda }_{m},{\mu }_{1},\ldots ,{\mu }_{\ell } \) such that \( \bar{\bar{\mathbf{x}}} ... | \[ \mathcal{T}\left( \bar{\mathbf{x}}\right) = \left\{ {\mathbf{d} : {\operatorname{\nabla c}}_{i}\left( \bar{\mathbf{x}}\right) \mathbf{d} = 0\text{ }\mathbf{{forall}}\text{ }i \in \mathcal{A}\left( \bar{x}\right) ,\text{ }\nabla \mathbf{h}\left( \bar{\mathbf{x}}\right) \mathbf{d} = \mathbf{0}}\right\} . \] | No |
Theorem 17 (Second-Order Sufficient Condition) Let the functions \( f,{c}_{i}, i \in \mathcal{I},{h}_{i}, i \in \mathcal{E} \) be twice continuously differentiable and \( {\mathbf{x}}^{ * } \) be a feasible solution to (P). If there exist vectors \( {w}^{ * } \) and \( {v}^{ * } \) such that \( \left( {{\mathbf{x}}^{ *... | ## Proof If \( {\mathbf{x}}^{ * } \) is not a strict local minimum point, then there exists a sequence \( \{ {\mathbf{x}}^{k}\} \subset S \) such that \( {\mathbf{x}}^{k} \neq {\mathbf{x}}^{ * },{\mathbf{x}}^{k} \rightarrow {\mathbf{x}}^{ * } \) and \( f\left( {\mathbf{x}}^{k}\right) \leq f\left( {\mathbf{x}}^{ * }\rig... | Yes |
Theorem 18 Consider the convex program \( \min \{ f\left( \mathbf{x}\right) ,\mathbf{x} \in \Omega \} \) , where \( \Omega \) is a closed convex set. Then \( {\mathbf{x}}^{ * } \in \Omega \) is an optimal solution if and only if\n\n\[ \nabla f\left( {\mathbf{x}}^{ * }\right) \left( {\mathbf{x} - {\mathbf{x}}^{ * }}\rig... | Proof. For any \( \mathbf{x} \in \Omega \) , if \( \nabla f\left( {\mathbf{x}}^{ * }\right) \left( {\mathbf{x} - {\mathbf{x}}^{ * }}\right) \geq 0 \) , then we have by the convexity of \( f \) ,\n\n\[ f\left( \mathbf{x}\right) \geq f\left( {\mathbf{x}}^{ * }\right) + \nabla f\left( {\mathbf{x}}^{ * }\right) \left( {\ma... | Yes |
Corollary 1 If \( f \) is differentiable convex functions, then the (first-order) KKT optimality conditions are necessary and sufficient for the global optimality of a feasible solution for linearly constrained optimization. | Let \( \left( {\overline{\mathbf{x}},\overline{\mathbf{y}}}\right) \) be a KKT pair for (CP) in which \( \overline{\mathbf{x}} \) is a feasible solution. Consider the Lagrangian function \( L\left( {\mathbf{x},\mathbf{y}}\right) = f\left( \mathbf{x}\right) - {\mathbf{y}}^{T}\mathbf{c}\left( \mathbf{x}\right) \) associa... | Yes |
Theorem 20 If Slater \( {CQ} \) is satisfied for \( \left( {CP}\right) \), then for any \( \mathbf{x} \in \mathcal{F} \) there exists a feasible direction. | Proof. By hypothesis, there exists \( \widehat{\mathbf{x}} \in \mathcal{F} \) such that \( {c}_{i}\left( \widehat{\mathbf{x}}\right) > 0 \) for all \( i \in \mathcal{I} \) . For any \( \mathbf{x} \in \mathcal{F} \) , let \( \mathbf{d} = \widehat{\mathbf{x}} - \mathbf{x} \) . The convexity of \( - {c}_{i}, i \in \mathca... | Yes |
Theorem 21 If Slater CQ holds for (CP), then the optimal solution is the KKT point. | Proof. Let \( {\mathbf{x}}^{ * } \) be the optimal solution of (CP). By Theorem 13 and the geometric optimality condition,\n\n\[ \left\{ {\mathbf{d} \in {\mathbb{R}}^{n} \mid \nabla {c}_{i}{\left( {\mathbf{x}}^{ * }\right) }^{T}\mathbf{d} > 0,\;i \in \mathcal{A}\left( {\mathbf{x}}^{ * }\right) }\right\} \neq \varnothin... | Yes |
Theorem 22 Let \( \\mathbf{x} \) be a feasible solution of Problem \( \\mathbf{P} \) and let \( \\left( {\\mathbf{u},\\mathbf{v}}\\right) \) be a feasible solution of Problem D. Then \( f\\left( \\mathbf{x}\\right) \\geq \\theta \\left( {\\mathbf{u},\\mathbf{v}}\\right) \) . | Proof. By the definition of \( \\theta \) and since \( \\mathbf{x} \\in {\\mathcal{F}}_{p} \), we have\n\n\[ \n\\theta \\left( {\\mathbf{u},\\mathbf{v}}\\right) \\leq f\\left( \\mathbf{x}\\right) - {\\mathbf{u}}^{T}\\mathbf{c}\\left( \\mathbf{x}\\right) - {\\mathbf{v}}^{T}\\mathbf{h}\\left( \\mathbf{x}\\right) \\leq \\... | Yes |
Theorem 23 Let \( \mathcal{X} \subset {\mathcal{R}}^{n} \) be a nonempty convex set, and let \( f \) and \( - {c}_{1},\ldots , - {c}_{m} \) be convex functions. Suppose that the following CQ holds true: \[ \{ \mathbf{x} \in \mathcal{X} : \mathbf{c}\left( \mathbf{x}\right) > \mathbf{0}\} \neq \varnothing \text{,有内点} \] ... | Furthermore, if the \( \inf \) is finite, then \( \sup \{ \theta \left( \mathbf{u}\right) : \mathbf{u} \geq \mathbf{0}\} \) is achieved at \( \bar{\mathbf{u}} \) with \( \bar{\mathbf{u}} \geq \mathbf{0} \) . If the \( \inf \) is achieved at \( \bar{\mathbf{x}} \), then \( {\bar{\mathbf{u}}}^{T}\mathbf{c}\left( \bar{\ma... | Yes |
Lemma 4 Let \( \mathcal{X} \subset {\mathcal{R}}^{n} \) be a nonempty convex set, and let \( f \) and \( - {c}_{1},\ldots , - {c}_{m} \) be convex functions. Consider the following two systems:\n\nSystem I: \( f\left( \mathbf{x}\right) < 0,\mathbf{c}\left( \mathbf{x}\right) \geq \mathbf{0} \) for some \( \mathbf{x} \in... | Proof: Suppose that System I has no solution, then\n\n\[ \left( {0,\mathbf{0}}\right) \notin \Omega \mathrel{\text{:=}} \{ \left( {\alpha ,\mathbf{y}}\right) : \alpha > f\left( \mathbf{x}\right) ,\mathbf{y} \leq \mathbf{c}\left( \mathbf{x}\right) \text{ for some }\mathbf{x} \in \mathcal{X}\} . \]\n\nNoting that \( \mat... | Yes |
Theorem 24 If \( \left( {\overline{\mathbf{x}},\overline{\mathbf{y}}}\right) \) is a saddle point of \( L \), then \( \overline{\mathbf{x}} \) solves (IP) and \( \overline{\mathbf{y}} \) solves its Lagrangian dual. | Proof. The vector \( \overline{\mathbf{x}} \) is feasible for (IP). Indeed, if \( c\left( \overline{\mathbf{x}}\right) \) has a negative\n\ncomponent, then the inequality \( L\left( {\bar{\mathbf{x}},\mathbf{y}}\right) \leq L\left( {\bar{\mathbf{x}},\bar{\mathbf{y}}}\right) \) for all \( \mathbf{y} \geq \mathbf{0} \) c... | Yes |
Theorem 26 (Eisenberg and Gale 1959) Optimal dual vector of equality constraints is an equilibrium price vector. | ## Optimality Conditions of the aggregated problem\n\n\[ \n{w}_{i}\frac{{u}_{ij}}{{\mathbf{u}}_{i}^{T}{\mathbf{x}}_{i}} \leq {p}_{j},\;\forall i, j \]\n\n\[ \n\mathop{\sum }\limits_{{i \in B}}{x}_{ij} = 1,\;\forall j \]\n\n\[ \n\begin{matrix} {w}_{i}\frac{{u}_{ij}{x}_{ij}}{{\mathbf{u}}_{i}^{T}{\mathbf{x}}_{i}} & = & {p... | Yes |
Theorem 2 If \( g \) is twice continuously differentiable and \( {x}^{ * } \) is a zero of \( g \) at which\n\n\( {g}^{\prime }\left( {x}^{ * }\right) \neq 0 \) , then provided that \( \left| {{x}^{0} - {x}^{ * }}\right| \) is sufficiently small, the sequence\n\ngenerated by the iteration\n\n\[ \n{x}^{k + 1} = {x}^{k} ... | ## Proof\n\nSince \( {g}^{\prime }\left( {x}^{ * }\right) \neq 0 \), when \( {x}^{k} \) is sufficiently close to \( {x}^{ * },{g}^{\prime }\left( {x}^{k}\right) \neq 0 \) . By the Taylor series expansion of \( g \) ,\n\n\[ \n0 = g\left( {x}^{ * }\right) = g\left( {x}^{k}\right) + {g}^{\prime }\left( {x}^{k}\right) \lef... | Yes |
Theorem 3 Consider the purely quadratic problem \( \min f\left( \mathbf{x}\right) \) . Let \( \{ {\mathbf{d}}_{0},{\mathbf{d}}_{1},\ldots ,{\mathbf{d}}_{n - 1}\} \) be a set of nonzero \( Q \) -orthogonal vectors. For any \( {\mathbf{x}}^{0} \in {\mathcal{R}}^{n} \) the sequence \( \{ {\mathbf{x}}^{k}\} \) generated ac... | Proof Since \( {\mathbf{d}}_{k}^{\prime }s \) are linearly independent, we can write \( {\mathbf{x}}^{ * } - {\mathbf{x}}^{0} = {\beta }_{0}{\mathbf{d}}_{0} + {\beta }_{1}{\mathbf{d}}_{1} + \ldots + {\beta }_{n - 1}{\mathbf{d}}_{n - 1} \) for some set of \( {\beta }_{k}^{\prime }s \) . We multiply by \( {\mathbf{d}}_{k... | Yes |
Theorem 4 Consider the purely quadratic problem \( \min f\left( \mathbf{x}\right) \) . The conjugate gradient algorithm (1) terminates at most \( n \) steps. Moreover, for any \( 1 \leq i \leq m \) , where \( m \leq n \) , we have\n\n(a) \( {\mathbf{d}}_{i}^{T}Q{\mathbf{d}}_{j} = 0,\;j = 1,2,\ldots, i - 1 \) .\n\n(b) \... | ## Proof of Conjugate Gradient Theorem\n\nBy induction. When \( i = 1 \) , (c) holds. When \( i = 2 \) , \( {\mathbf{d}}_{2}^{T}Q{\mathbf{d}}_{1} = 0 \) implies that (a) holds. By the exact line search rule, \( {\mathbf{g}}_{\mathbf{2}}^{\mathbf{T}}{\mathbf{d}}_{\mathbf{1}} = \mathbf{0} \), and then (b) holds. Since\n\... | Yes |
Theorem 5 Let \( G \) be a fixed symmetric matrix and suppose that \( {s}_{0},{s}_{1},\ldots ,{s}_{k} \) are given vectors. Define the vectors \( {y}_{i} = G{s}_{i}, i = 0,1,\ldots, k \) . Starting with any initial symmetric matrix \( {H}_{0} \) let\n\n\[ \n{H}_{i + 1} = {H}_{i} + \frac{\left( {{s}_{i} - {H}_{i}{y}_{i}... | Proof: The proof is by induction. Suppose it is true for \( {H}_{k} \) and \( i \leq k - 1 \) . The relation was shown above to be true for \( {H}_{k + 1} \) and \( i = k \) . For \( i < k \)\n\n\[ \n{H}_{k + 1}{y}_{i} = {H}_{k}{y}_{i} + {p}_{k}\left( {{s}_{k}^{T}{y}_{i} - {y}_{k}^{T}{H}_{k}{y}_{i}}\right) \n\]\n\nwher... | Yes |
Theorem 6 Let \( {H}_{k} \) be a positive definite symmetric matrix, then \( {H}_{k + 1} \succ \mathbf{0} \Leftrightarrow {s}_{k}^{T}{y}_{k} > 0 \) . | Proof: For any \( x \in {\mathcal{R}}^{n} \) we have \( {x}^{T}{H}_{k + 1}x = {x}^{T}{H}_{k}x + \frac{{\left( {x}^{T}{s}_{k}\right) }^{2}}{{s}_{k}^{T}{y}_{k}} - \frac{{\left( {x}^{T}{H}_{k}{y}_{k}\right) }^{2}}{{y}_{k}^{T}{H}_{k}{y}_{k}}. \) Defining \( a = {H}_{k}^{1/2}x \) and \( b = {H}_{k}^{1/2}{y}_{k} \), we may r... | Yes |
Theorem 7 If \( f \) is quadratic with positive definite Hessian \( F \) , then for the DFP method with exact line search\n\n\[ \n{s}_{i}^{T}F{s}_{j} = 0,\;0 \leq i < j \leq k \n\]\n\n(8)\n\n\[ \n{H}_{k + 1}F{s}_{i} = {s}_{i},\;0 \leq i \leq k. \n\]\n\n(9) | Proof: We note that for the quadratic case 存疑二\n\n\[ \n{y}_{k} = {g}_{k + 1} - {g}_{k} = F{x}^{k + 1} - F{x}^{k} = F{s}_{k}. \n\]\n\n(10)\n\nAlso\n\n\[ \n{H}_{k + 1}F{s}_{k} = {H}_{k + 1}{y}_{k} = {s}_{k} \n\]\n\n(11)\n\nfrom the DFP formulation.\n\nWe now prove by induction. From (11) we see that (8) and (9) are true ... | Yes |
Example 1. Suppose \( S \) is defined by\n\n\[ S = \\left\\{ {\\mathbf{x} \\in {\\mathcal{R}}^{n} \\mid {g}_{i}\\left( \\mathbf{x}\\right) \\geq 0, i \\in \\mathcal{I},\\;{h}_{i}\\left( \\mathbf{x}\\right) = 0, i \\in \\mathcal{E}}\\right\\} .\n\]\n\nA very useful penalty function in this case is\n\n\[ P\\left( \\mathb... | The procedure of the penalty function method is: Let \( \\left\\{ {c}_{k}\\right\\} \) be a sequence tending to infinity such that for each \( k,{c}_{k} > 0 \) and \( {c}_{k + 1} > {c}_{k} \) . For each \( k \) solve the problem\n\n\[ \\min \;q\\left( {\\mathbf{x},{c}_{k}}\\right) \]\n\nobtaining a solution point \( {\... | No |
Lemma 1 Let \( \{ {\mathbf{x}}^{k}\} \) be the sequence generated by the penalty method. Then,\n\n(i) \( q\left( {{\mathbf{x}}^{k},{c}_{k}}\right) \leq q\left( {{\mathbf{x}}^{k + 1},{c}_{k + 1}}\right) \),\n\n(ii) \( P\left( {\mathbf{x}}^{k}\right) \geq P\left( {\mathbf{x}}^{k + 1}\right) \),\n\n(iii) \( f\left( {\math... | Proof. Since \( {c}_{k + 1} > {c}_{k} \) and \( P \) is a penalty function,\n\n\[ \begin{matrix} q\left( {{\mathbf{x}}^{k + 1},{c}_{k + 1}}\right) & = & f\left( {\mathbf{x}}^{k + 1}\right) + {c}_{k + 1}P\left( {\mathbf{x}}^{k + 1}\right) \geq f\left( {\mathbf{x}}^{k + 1}\right) + {c}_{k}P\left( {\mathbf{x}}^{k + 1}\rig... | Yes |
Lemma 2 Let \( {\mathbf{x}}^{ * } \) be a solution of problem (1). Then for each \( k \)\n\n\[ f\left( {\mathbf{x}}^{ * }\right) \geq q\left( {{\mathbf{x}}^{k},{c}_{k}}\right) \geq f\left( {\mathbf{x}}^{k}\right) . \]\n | Proof.\n\n\[ f\left( {\mathbf{x}}^{ * }\right) = f\left( {\mathbf{x}}^{ * }\right) + {c}_{k}P\left( {\mathbf{x}}^{ * }\right) \geq f\left( {\mathbf{x}}^{k}\right) + {c}_{k}P\left( {\mathbf{x}}^{k}\right) \geq f\left( {\mathbf{x}}^{k}\right) . \]\n | Yes |
Theorem 1 Let \( \{ {\mathbf{x}}^{k}\} \) be the sequence generated by the penalty method. Then, any limit point of the sequence is a solution of problem (1). | Proof. Suppose the subsequence \( \{ {\mathbf{x}}^{k}{\} }_{k \in \mathcal{K}} \) is a convergent subsequence of\n\n\( \left\{ {\mathbf{x}}^{k}\right\} \) having limit \( \overline{\mathbf{x}} \) . Then by the continuity of \( f \), we have\n\n\[ \mathop{\lim }\limits_{{k \rightarrow \infty, k \in \mathcal{K}}}f\left( ... | Yes |
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