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Example 2 (Best Diagonal). As an alternative problem, consider minimizing the length of the diagonal of a rectangle subject to the perimeter being of length 4 . This problem can be formulated as\n\n\[ \n\\text{minimize}\\frac{1}{2}\\left( {{x}_{1}^{2} + {x}_{2}^{2}}\\right) \n\]\n\n\[ \n\\text{subject to}{x}_{1} + {x}_... | In this case the objective function is convex. The solution is \( {x}_{1} = {x}_{2} = 1 \) and the Lagrange multiplier is \( \\lambda = - 1 \) . The Lagrangian problem is\n\n\[ \n\\text{minimize}\\frac{1}{2}\\left( {{x}_{1}^{2} + {x}_{2}^{2}}\\right) - 1 \\cdot \\left( {{x}_{1} + {x}_{2} - 2}\\right) \n\]\n\n\[ \n\\tex... | Yes |
Proposition 3. Suppose \( \Omega \subset {\mathrm{E}}^{\mathrm{n}} \) is convex and \( f \) and \( \mathbf{g} \) are convex functions. Then the primal function \( \omega \) is also convex. | ## Proof. The proof parallels that of Proposition 1. One simply substitutes \( \mathbf{g}\left( \mathbf{x}\right) \leq \mathbf{0} \) for \( \mathbf{h}\left( \mathbf{x}\right) = \mathbf{y} \) throughout the series of inequalities. | No |
Assume \( \Omega \) is a convex subset of \( {E}^{n} \) and that \( f \) and \( \mathbf{g} \) are convex functions. Assume also that there is a point \( {\mathbf{x}}_{1} \in \Omega \) such that \( \mathbf{g}\left( {\mathbf{x}}_{1}\right) < \mathbf{0} \) . Then, if \( {\mathbf{x}}^{ * } \) solves (11.47), there is a vec... | Here is the proof outline. Let \( {f}^{ * } = f\left( {\mathbf{x}}^{ * }\right) \) . In this case define in \( {E}^{p + 1} \) the two sets\n\n\[ A = \{ \left( {r,\mathbf{0}}\right) : r \geq f\left( \mathbf{x}\right) ,\mathbf{0} \geq \mathbf{g}\left( \mathbf{x}\right) \text{, for some }\mathbf{x} \in \Omega \} \text{ an... | Yes |
Consider the quadratic program\n\n\\[ \n\\text{minimize}\\mathbf{x}^{T}\\mathbf{Q}\\mathbf{x} + \\mathbf{c}^{T}\\mathbf{x}\n\\]\n\n\\[ \n\\text{subject to}\\mathbf{a}^{T}\\mathbf{x} \\leq b\n\\]\n\n\\[ \n\\mathbf{x} \\geq \\mathbf{0}.\\text{.}\n\\] | Let \\( \\Omega = \\{ \\mathbf{x} : \\mathbf{x} \\geq \\mathbf{0}\\} \\) and \\( g\\left( \\mathbf{x}\\right) = \\mathbf{a}^{T}\\mathbf{x} - b \\) . Assume that the \\( n \\times n \\) matrix \\( \\mathbf{Q} \\) is positive definite, in which case the objective function is convex. Assuming that \\( b > 0 \\), the Slate... | Yes |
Proposition 5 (Sufficiency Conditions). Suppose \( f \) is a real-valued function on a set \( \Omega \subset \) \( {E}^{n} \) . Suppose also that \( \mathbf{h} \) and \( \mathbf{g} \) are, respectively, \( m \) -dimensionaland p-dimensional functions on \( \Omega \) . Finally, suppose there are vectors \( {\mathrm{x}}^... | Proof. Suppose there is \( {\mathbf{x}}_{1} \in \Omega \) with \( f\left( {\mathbf{x}}_{1}\right) < f\left( {\mathbf{x}}^{ * }\right) ,\mathbf{h}\left( {\mathbf{x}}_{1}\right) = \mathbf{h}\left( {\mathbf{x}}^{ * }\right) \), and \( \mathbf{g}\left( {\mathbf{x}}_{1}\right) \leq \) \( \mathbf{g}\left( {\mathbf{x}}^{ * }\... | Yes |
Example 2 (Simplified Zoutendijk Method). Another proposal solves a sequence of linear subprograms as follows. Given a feasible point, \( {\mathbf{x}}_{k} \), let \( I \) be the set of indices representing active constraints, that is, \( {\mathbf{a}}_{i}^{T}{\mathbf{x}}_{k} = {b}_{i} \) for \( i \in I \) . The directio... | There are two major shortcomings of feasible direction methods that require that they be modified in most cases. The first shortcoming is that for general problems there may not exist any feasible directions. If, for example, a problem had nonlinear equality constraints, we might find ourselves in the situation depicte... | No |
Let \( \mathbf{x}\left( t\right) ,0 \leq t \leq T \), be a geodesic on \( \mathbf{\Omega } \) . Then\n\n\[ \frac{d}{dt}f\left( {\mathbf{x}\left( t\right) }\right) = {l}_{\mathbf{x}}\left( {\mathbf{x},\lambda \left( \mathbf{x}\right) }\right) \dot{\mathbf{x}}\left( t\right) \]\n\n\[ \frac{{d}^{2}}{d{t}^{2}}f\left( {\mat... | Proof. We have\n\n\[ \frac{d}{dt}f\left( {\mathbf{x}\left( t\right) }\right) = \nabla f\left( {\mathbf{x}\left( t\right) }\right) \dot{\mathbf{x}}\left( t\right) = {l}_{\mathbf{x}}\left( {\mathbf{x},\mathbf{\lambda }\left( \mathbf{x}\right) }\right) \dot{\mathbf{x}}\left( t\right) ,\]\n\nthe second equality following f... | Yes |
Lemma 1.\n\n\[ q\left( {{c}_{k},{\mathbf{x}}_{k}}\right) \leq q\left( {{c}_{k + 1},{\mathbf{x}}_{k + 1}}\right) \]\n\n(13.5)\n\n\[ P\left( {\mathbf{x}}_{k}\right) \geq P\left( {\mathbf{x}}_{k + 1}\right) \]\n\n(13.6)\n\n\[ f\left( {\mathbf{x}}_{k}\right) \leq f\left( {\mathbf{x}}_{k + 1}\right) . \]\n\n(13.7) | Proof.\n\n\[ q\left( {{c}_{k + 1},{\mathbf{x}}_{k + 1}}\right) = f\left( {\mathbf{x}}_{k + 1}\right) + {c}_{k + 1}P\left( {\mathbf{x}}_{k + 1}\right) \geq f\left( {\mathbf{x}}_{k + 1}\right) + {c}_{k}P\left( {\mathbf{x}}_{k + 1}\right) \]\n\n\[ \geq f\left( {\mathbf{x}}_{k}\right) + {c}_{k}P\left( {\mathbf{x}}_{k}\righ... | Yes |
Lemma 2. Let \( {\mathbf{x}}^{ * } \) be a solution to problem (13.1). Then for each \( k \)\n\n\[ f\left( {\mathbf{x}}^{ * }\right) \geq q\left( {{c}_{k},{\mathbf{x}}_{k}}\right) \geq f\left( {\mathbf{x}}_{k}\right) . \]\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 |
Let \( {g}_{i}, i = 1,2,\ldots, p \) be continuous functions on \( {E}^{n} \) . Suppose\n\n\[ S = \\left\\{ {\\mathbf{x} : {g}_{i}\\left( \\mathbf{x}\\right) \\leq 0, i = 1,2,\\ldots, p}\\right\\} .\n\]\n\nis robust, and suppose the interior of \( S \) is the set of \( \\mathbf{x} \) ’s where \( {g}_{i}\\left( \\mathbf... | It is illustrated in one dimension for \( {g}_{1} = x - a,{g}_{2} = x - b \) in Fig. 13.3. | No |
For the same situation as Example 1, we may use the logarithmic utility function\n\n\[ B\\left( \\mathbf{x}\\right) = - \\mathop{\\sum }\\limits_{{i = 1}}^{p}\\log \\left\\lbrack {-{g}_{i}\\left( \\mathbf{x}\\right) }\\right\\rbrack . \]\n\nThis is the barrier function commonly used in linear programming interior point... | The barrier method is quite analogous to the penalty method. Let \( \\left\{ {c}_{k}\\right\} \) be a sequence tending to infinity such that for each \( k, k = 1,2,\\ldots ,{c}_{k} \\geq 0,{c}_{k + 1} > {c}_{k} \). Define the function\n\n\[ r\\left( {c,\\mathbf{x}}\\right) = f\\left( \\mathbf{x}\\right) + \\frac{1}{c}B... | No |
\[ P\left( \mathbf{x}\right) = \frac{1}{2}\mathop{\sum }\limits_{{i = 1}}^{p}{g}_{i}^{ + }{\left( \mathbf{x}\right) }^{2} = \frac{1}{2}{\left| {\mathbf{g}}^{ + }\left( \mathbf{x}\right) \right| }^{2}, \] | which is without doubt the most popular penalty function. In this case \( \gamma \) is one-half times the identity quadratic form on \( {E}^{p} \), that is, \( \gamma \left( \mathbf{y}\right) = \frac{1}{2}{\left| \mathbf{y}\right| }^{2} \) . | No |
Example 3. A general class of penalty functions is\n\n\[ \n P\left( \mathbf{x}\right) = \mathop{\sum }\limits_{{i = 1}}^{p}{\left( {g}_{i}^{ + }\left( \mathbf{x}\right) \right) }^{\varepsilon } \]\n\nfor some \( \varepsilon > 0 \) . | We define\n\n\[ \n \nabla {g}_{i}^{ + }\left( \mathbf{x}\right) = \left\{ \begin{array}{ll} \nabla {g}_{i}\left( \mathbf{x}\right) & \text{ if }{g}_{i}\left( \mathbf{x}\right) \geq 0 \\ \mathbf{0} & \text{ if }{g}_{i}\left( \mathbf{x}\right) < 0 \end{array}\right. \]\n\n(13.20)\n\nand, of course, \( \nabla {\mathbf{... | Yes |
For \( P\left( \mathbf{x}\right) = \frac{1}{2}{\left| {\mathbf{g}}^{ + }\left( \mathbf{x}\right) \right| }^{2} \), what is the form of \( \mathbf{\Gamma }\left( {{\mathbf{g}}^{ + }\left( {\mathbf{x}}_{k}\right) }\right) \)? | \[ \mathbf{\Gamma }\left( {{\mathbf{g}}^{ + }\left( {\mathbf{x}}_{k}\right) }\right) = \left\lbrack \begin{matrix} {e}_{1} & 0 & \cdots & 0 \\ 0 & {e}_{2} & 0 & \\ 0 & \cdot & & \cdot \\ \cdot & \cdot & \cdot & \\ \cdot & & \cdot & \cdot \\ 0 & \cdots & 0 & {e}_{p} \end{matrix}\right\rbrack ,\] where \[ {e}_{i} = \left... | Yes |
Lemma 1. Let \( \mathbf{A}\left( c\right) \) be a symmetric matrix written in partitioned form\n\n\[ \mathbf{A}\left( c\right) = \left\lbrack \begin{array}{ll} {\mathbf{A}}_{1}\left( c\right) & {\mathbf{A}}_{2}\left( c\right) \\ {\mathbf{A}}_{2}^{T}\left( c\right) & {\mathbf{A}}_{3}\left( c\right) \end{array}\right\rbr... | Proof. We have the identity\n\n\[ {\left\lbrack \begin{matrix} {\mathbf{A}}_{1}{\mathbf{A}}_{2} \\ {\mathbf{A}}_{2}^{T}{\mathbf{A}}_{3} \end{matrix}\right\rbrack }^{-1} = \left\lbrack \begin{matrix} {\left( {\mathbf{A}}_{1} - {\mathbf{A}}_{2}{\mathbf{A}}_{3}^{-1}{\mathbf{A}}_{2}^{T}\right) }^{-1} & - \left( {{\mathbf{A... | Yes |
The barrier objective\n\n\[ r\left( {{c}_{k},\mathbf{x}}\right) = f\left( \mathbf{x}\right) - \frac{1}{{c}_{k}}\mathop{\sum }\limits_{{i = 1}}^{p}\frac{1}{{g}_{i}\left( \mathbf{x}\right) }\n\]\nhas its minimum at a point \( {\mathbf{x}}_{k} \) satisfying\n\n\[ \nabla f\left( {\mathbf{x}}_{k}\right) + \frac{1}{{c}_{k}}\... | Thus, we define \( {\lambda }_{k} \) to be the vector having \( i \) th component \( \frac{1}{{c}_{k}} \cdot \frac{1}{{g}_{i}{\left( {\mathbf{x}}_{k}\right) }^{2}} \) . Then (13.32) can be written as\n\n\[ \nabla f\left( {\mathbf{x}}_{k}\right) + {\lambda }_{k}^{T}\nabla \mathbf{g}\left( {\mathbf{x}}_{k}\right) = \math... | Yes |
Example 2. Let us use the logarithmic barrier function\n\n\\[ \nB\\left( \\mathbf{x}\\right) = - \\mathop{\\sum }\\limits_{{i = 1}}^{p}\\log \\left\\lbrack {-{g}_{i}\\left( \\mathbf{x}\\right) }\\right\\rbrack \n\\]\n\nIn this case we will define the barrier objective in terms of \\( \\mu \\) as\n\n\\[ \nr\\left( {\\mu... | Defining\n\n\\[ \n{\\lambda }_{\\mu, i} = \\mu \\frac{-1}{{g}_{i}\\left( {\\mathbf{x}}_{\\mu }\\right) }\n\\]\n\n\n(13.34) can be written as\n\n\\[ \n\\nabla f\\left( {\\mathbf{x}}_{\\mu }\\right) + {\\lambda }_{\\mu }^{T}\\nabla \\mathbf{g}\\left( {\\mathbf{x}}_{\\mu }\\right) = \\mathbf{0}.\n\\]\n\nFurther we expect ... | Yes |
Consider the simple quadratic problem\n\n\\[ \n\\text{minimize}2{x}^{2} + {2xy} + {y}^{2} - {2y} \n\\]\n\n\\[ \n\\text{subject to}x = 0\\text{.} \n\\]\n\n(13.64) | It is easy to solve this problem directly by substituting \\( x = 0 \\) into the objective. This leads immediately to \\( x = 0, y = 1 \\) . | Yes |
Proposition 1. The dual function is concave on the region where it is finite. | Proof. Suppose \( {\mu }_{1},{\mu }_{2} \) are in the finite region, and let \( 0 \leq \alpha \leq 1 \) . Then\n\n\[ \phi \left( {\alpha {\mathbf{\mu }}_{1} + \left( {1 - \alpha {\mathbf{\mu }}_{2}}\right) }\right) = \inf \left\{ {f\left( \mathbf{x}\right) + {\left( \alpha {\mathbf{\mu }}_{1} + \left( 1 - \alpha \right... | Yes |
Consider the problem\n\n\[ \n\\text{minimize}\\frac{1}{2}{\\mathbf{x}}^{T}\\mathbf{Q}\\mathbf{x} \n\]\n\n(14.5)\n\n\[ \n\\text{subject to}\\mathbf{{Bx}} - \\mathbf{b} \\leq \\mathbf{0}\\text{.} \n\] | The dual function is\n\n\[ \n\\phi \\left( \\mathbf{\\mu }\\right) = \\mathop{\\min }\\limits_{\\mathbf{x}}\\frac{1}{2}{\\mathbf{x}}^{T}\\mathbf{Q}\\mathbf{x} + {\\mathbf{\\mu }}^{T}\\left( {\\mathbf{B}\\mathbf{x} - \\mathbf{b}}\\right) . \n\]\n\nThis gives the necessary conditions\n\n\[ \n\\mathbf{Q}\\mathbf{x} + {\\m... | Yes |
Example 2 (Integer Solutions). Duality gaps may arise if the object function or the constraint functions are not convex. A gap may also arise if the underlying set is not convex. This is characteristic, for example, of problems in which the components of the solution vector are constrained to be integers. For instance,... | It is clear that the solution is \\( {x}_{1} = 1,{x}_{2} = 0 \\), with objective value \\( {f}^{ * } = 1 \\) . To put this problem in the standard form we have discussed, we write the constraint as\n\n\\[ \n- {x}_{1} - {x}_{2} + 1/2 \\leq z,\\;\\text{ where }z = 0. \n\\]\n\nThe primal function \\( \\omega \\left( z\\ri... | Yes |
Lemma 1. The dual function \( \phi \) has gradient\n\n\[ \nabla \phi \left( \lambda \right) = \mathbf{h}{\left( \mathbf{x}\left( \lambda \right) \right) }^{T} \] | Proof. We have explicitly, from (14.13),\n\n\[ \phi \left( \lambda \right) = f\left( {\mathbf{x}\left( \lambda \right) }\right) + {\lambda }^{T}\mathbf{h}\left( {\mathbf{x}\left( \lambda \right) }\right) .\n\nThus\n\n\[ \nabla \phi \left( \lambda \right) = \left\lbrack {\nabla f\left( {\mathbf{x}\left( \lambda \right) ... | Yes |
Lemma 2. The Hessian of the dual function is\n\n\[ \mathbf{\Phi }\left( \lambda \right) = - \nabla \mathbf{h}\left( {\mathbf{x}\left( \lambda \right) }\right) {\mathbf{L}}^{-1}\left( {\mathbf{x}\left( \lambda \right) ,\lambda }\right) \nabla \mathbf{h}{\left( \mathbf{x}\left( \lambda \right) \right) }^{T}. \] | Proof. The Hessian is the derivative of the gradient. Thus, by Lemma 1,\n\n\[ \Phi \left( \lambda \right) = \nabla \mathbf{h}\left( {\mathbf{x}\left( \lambda \right) }\right) \nabla \mathbf{x}\left( \lambda \right) \]\n\nBy definition we have\n\n\[ \nabla f\left( {\mathbf{x}\left( \lambda \right) }\right) + {\lambda }^... | Yes |
Consider the problem in two variables\n\n\\[ \n\\text{minimize}\; - {xy} \n\\]\n\n\\[ \n\\text{subject to}\\;{\\left( x - 3\\right) }^{2} + {y}^{2} = 5\\text{.}\n\\] | The first-order necessary conditions are\n\n\\[ \n- y + \\left( {{2x} - 6}\\right) \\lambda = 0 \n\\]\n\n\\[ \n- x + {2y\\lambda } = 0 \n\\]\n\ntogether with the constraint. These equations have a solution at\n\n\\[ \nx = 4,\\;y = 2,\\;\\lambda = 1. \n\\]\n\n\nThe Hessian of the corresponding Lagrangian is\n\n\\[ \n\\m... | Yes |
Example 2. Problems involving a series of decisions made at distinct times are often separable. For illustration, consider the problem of scheduling water release through a dam to produce as much electric power as possible over a given time interval while satisfying constraints on acceptable water levels. A discrete-ti... | ## Decomposition\n\nSeparable problems are ideally suited to dual methods, because the required unconstrained minimization decomposes into small subproblems. To see this we recall that the generally most difficult aspect of a dual method is evaluation of the dual function. For a separable problem, if we associate \( \\... | Yes |
In Example 1 using duality with respect to the equality constraints we denote the dual variables by \( \lambda \left( k\right), k = 1,2,\ldots, N \) . The \( k \) th subproblem becomes\n\n\[ \mathop{\max }\limits_{\substack{{c \leq y\left( k\right) \leq d} \\ {0 \leq u\left( k\right) } }}\{ f\left( {y\left( k\right), u... | which is a two-dimensional optimization problem. Selection of \( \lambda \in {E}^{N} \) decomposes the problem into separate problems for each time period. The variable \( \lambda \left( k\right) \) can be regarded as a value, measured in units of power, for water at the beginning of period \( k \) . The \( k \) th sub... | Yes |
Consider again the problem of finding the equilibrium position of the hanging chain considered in Example 4, Sect. 11.3, and Example 1, Sect. 12.7. The problem is\n\n\\[ \n\\text{ minimize }\\mathop{\\sum }\\limits_{{i = 1}}^{n}{c}_{i}{y}_{i}\n\\]\n\n\\[ \n\\text{subject to}\\mathop{\\sum }\\limits_{{i = 1}}^{n}{y}_{i}... | This problem is locally convex, since as shown in Sect. 12.7 the Hessian of the Lagrangian is positive definite. The dual function is accordingly\n\n\\[ \n\\phi \\left( {\\lambda ,\\mu }\\right) = \\min \\mathop{\\sum }\\limits_{{i = 1}}^{n}\\left\\{ {{c}_{i}{y}_{i} + \\lambda {y}_{i} + \\mu \\sqrt{1 - {y}_{i}^{2}}}\\r... | Yes |
Consider the simple quadratic problem studied in Sect. 13.8\n\n\\[ \n\\text{minimize}\\;2{x}^{2} + {2xy} + {y}^{2} - {2y} \n\\]\n\n\\[ \n\\text{subject to}\\;x = 0\\text{.} \n\\] | The augmented Lagrangian for this problem is\n\n\\[ \n{l}_{c}\\left( {x, y,\\lambda }\\right) = 2{x}^{2} + {2xy} + {y}^{2} - {2y} + {\\lambda x} + \\frac{1}{2}c{x}^{2}. \n\\]\n\nThe minimum of this can be found analytically to be \\( x = - \\left( {2 + \\lambda }\\right) /\\left( {2 + c}\\right), y = \\) \\( \\left( {4... | Yes |
Proposition 2. For any \( c > 0 \), let \( \mathbf{d},\lambda \) (with \( \mathbf{d} \neq \mathbf{0} \) ) be a solution to the quadratic program (15.45). Then \( \mathbf{d} \) is a descent direction of the function \( P\left( \mathbf{x}\right) = f\left( \mathbf{x}\right) + \left( {1/2}\right) c{\left| \mathbf{h}\left( ... | Proof. We have from the constraint equation\n\n\[ \mathbf{{Ad}} = \left( {1/c}\right) \widehat{\lambda } - \mathbf{h}\left( \mathbf{x}\right) \]\n\nwhich yields\n\n\[ c{\mathbf{A}}^{T}\mathbf{A}\mathbf{d} = {\mathbf{A}}^{T}\widehat{\mathbf{\lambda }} - \mathbf{c}{\mathbf{A}}^{T}\mathbf{h}\left( \mathbf{x}\right) \]\n\n... | Yes |
In solving (15.55) for \( \left( {{\mathbf{d}}_{\mathbf{x}},{\mathbf{d}}_{\mathbf{y}},{\mathbf{d}}_{\mathbf{s}}}\right) \), let \( \gamma = n/\left( {n + \rho }\right) < 1 \) for fixed \( \rho \geq \sqrt{n} \) and assign \( {\mathbf{x}}^{ + } = \mathbf{x} + \alpha {\mathbf{d}}_{\mathbf{x}},{\mathbf{y}}^{ + } = \mathbf{... | The proof of the theorem is also similar to that for linear programming; see Exercise 12. Notice that, since \( \mathbf{Q} \) is positive semidefinite, we have \[ {\left. {\mathbf{d}}_{\mathbf{x}}\right. }^{T}{\mathbf{d}}_{\mathbf{s}} = {\left( {\mathbf{d}}_{\mathbf{x}},{\mathbf{d}}_{\mathbf{y}}\right) }^{T}\left( {{\m... | No |
Consider the equation \( {x}_{1}^{2} + {x}_{2} = 0 \) . | A solution is \( {x}_{1} = 0,{x}_{2} = 0 \) . However, in a neighborhood of this solution there is no function \( \phi \) such that \( {x}_{1} = \) \( \phi \left( {x}_{2}\right) \) . At this solution condition (iii) of the implicit function theorem is violated. At any other solution, however, such a \( \phi \) exists. | Yes |
Proposition 1. Convex sets in \( {E}^{n} \) satisfy the following relations:\n\ni) If \( C \) is a convex set and \( \beta \) is a real number; the set\n\n\[ \n{\beta C} = \{ \mathbf{x} : \mathbf{x} = \beta \mathbf{c},\mathbf{c} \in C\}\n\]\n\nis convex.\n\nii) If \( C \) and \( D \) are convex sets, then the set\n\n\[... | The proofs of these three properties follow directly from the definition of a convex set and are left to the reader. | No |
Proposition 2. Let \( \mathbf{a} \) be a nonzero n-dimensional column vector, and let \( c \) be a real number. The set\n\n\[ H = \left\{ {\mathbf{x} \in {E}^{n} : {\mathbf{a}}^{T}\mathbf{x} = c}\right\} \]\n\nis a hyperplane in \( {E}^{n} \) . | Proof. It follows directly from the linearity of the equation \( {\mathbf{a}}^{T}\mathbf{x} = c \) that \( H \) is a linear variety. Let \( {\mathbf{x}}_{1} \) be any vector in \( H \) . Translating by \( - {\mathbf{x}}_{1} \) we obtain the set \( M = H - {\mathbf{x}}_{1} \) which is a linear subspace of \( {E}^{n} \) ... | Yes |
Proposition 3. Let \( H \) be a hyperplane in \( {E}^{n} \) . Then there is a nonzero \( n \) - dimensional vector\n\nand a constant \( c \) such that\n\n\[ H = \left\{ {\mathbf{x} \in {E}^{n} : {\mathbf{a}}^{T}\mathbf{x} = c}\right\} . \] | Proof. Let \( {\mathbf{x}}_{1} \in H \) and translate by \( - {\mathbf{x}}_{1} \) obtaining the set \( M = H - {\mathbf{x}}_{1} \) . Since \( H \) is a hyperplane, \( M \) is an \( \left( {n - 1}\right) \) -dimensional subspace. Let \( \mathbf{a} \) be any nonzero vector that is orthogonal to this subspace, that is, a ... | Yes |
Theorem 1. Let \( C \) be a convex set and let \( \mathbf{y} \) be a point exterior to the closure of \( C \) . Then there is a vector \( \mathbf{a} \) such that \( {\mathbf{a}}^{T}\mathbf{y} < \mathop{\inf }\limits_{{\mathbf{x} \in C}}{\mathbf{a}}^{T}\mathbf{x} \) . | Proof. Let\n\n\[ \delta = \mathop{\inf }\limits_{{\mathbf{x} \in C}}\left| {\mathbf{x} - \mathbf{y}}\right| > 0 \]\n\nThere is an \( {\mathbf{x}}_{0} \) on the boundary of \( C \) such that \( \left| {{\mathbf{x}}_{0} - \mathbf{y}}\right| = \delta \) . This follows because the continuous function \( f\left( \mathbf{x}\... | Yes |
Theorem 2. Let \( C \) be a convex set and let \( \mathbf{y} \) be a boundary point of \( C \) . Then there is a hyperplane containing \( \mathbf{y} \) and containing \( C \) in one of its closed half spaces. | Proof. Let \( \left\{ {\mathbf{y}}_{k}\right\} \) be a sequence of vectors, exterior to the closure of \( C \), converging to \( \mathbf{y} \) . Let \( \left\{ {\mathbf{a}}_{k}\right\} \) be the sequence of corresponding vectors constructed according to Theorem 1, normalized so that \( \left| {\mathbf{a}}_{k}\right| = ... | Yes |
Theorem 3. Let \( B \) and \( C \) be convex sets with no common relative interior points. (That is the only common points are boundary points.) Then there is a hyperplane separating \( B \) and D. In particular, there is a nonzero vector \( \mathbf{a} \) such that \( \mathop{\sup }\limits_{{\mathbf{b} \in B}}{\mathbf{... | Proof. Consider the set \( G = C - B \) . It is easily shown that \( G \) is convex and that \( \mathbf{0} \) is not a relative interior point of \( G \) . Hence, Theorem 1 or Theorem 2 applies and gives the appropriate hyperplane. | Yes |
Lemma 1. Let \( C \) be a convex set, \( H \) a supporting hyperplane of \( C \), and \( T \) the intersection of \( H \) and \( C \) . Every extreme point of \( T \) is an extreme point of \( C \) . | Proof. Suppose \( {\mathbf{x}}_{0} \in T \) is not an extreme point of \( C \) . Then \( {\mathbf{x}}_{0} = \alpha {\mathbf{x}}_{1} + \left( {1 - \alpha }\right) {\mathbf{x}}_{2} \) for some \( {\mathbf{x}}_{1},{\mathbf{x}}_{2} \in C,{\mathbf{x}}_{1} \neq {\mathbf{x}}_{2},0 < \alpha < 1 \) . Let \( H \) be described as... | Yes |
Theorem 4. A closed bounded convex set in \( {E}^{n} \) is equal to the closed convex hull of its extreme points. | Proof. The proof is by induction on the dimension of the space \( {E}^{n} \) . The statement is easily seen to be true for \( n = 1 \) . Suppose that it is true for \( n - 1 \) . Let \( C \) be a closed bounded convex set in \( {E}^{n} \), and let \( K \) be the closed convex hull of the extreme points of \( C \) . We ... | 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\nSince \( \m... | 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 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\[ f\left( {x + h}\right) = f\left( x\right) + \nabla f\left( x\right) h + \frac{1}{2}{h}^{T}{\nabla }^{2}f\left( x\right) h + o\left( {\parallel h{\parallel }^{2}}\right) \text{ as }h \rightarrow 0. \] | 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 \( m \) ,\n\n(i) if there is a feasible solution, there is a basic feasible solution;\n\n(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 (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 \( \mathbf{x},\mathbf{y} \in C \) and any \( \alpha \in \left( {0,1}\right) \), \( 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 \( \alpha \in \left( {0,1}\right) \), we may assume, WLOG, \( f\left( \mathbf{y}\right) \geq f\left( \mathbf{x}\right) \). Then, by the definition, \( \mathbf{x},\mathbf{y} \in U\left( {f;\mathbf{x}}\right) \). Since \( U\... | 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 |
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 |
In \( {\mathcal{R}}^{2} \) let \( h\left( {{x}_{1},{x}_{2}}\right) = {x}_{1} \) . Then \( h\left( x\right) = 0 \) yields the \( {x}_{2} \) axis, and every point on that axis is regular. If instead we put \( h\left( {{x}_{1},{x}_{2}}\right) = {x}_{1}^{2} \) , again \( S \) is the \( {x}_{2} \) axis but now no point on t... | Indeed in this case \( H = {\mathcal{R}}^{2} \), while the tangent plane is the \( {x}_{2} \) axis. | No |
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 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 |
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}}} ... | \[ \mathcal{T}\left( \bar{\mathbf{x}}\right) = \left\{ {\mathbf{d} : \mathcal{T}{c}_{i}\left( \bar{\mathbf{x}}\right) \mathbf{d} = 0\text{ }\mathbf{{forall}}\text{ }i \in \mathcal{A}\left( \bar{x}\right) ,\mathcal{T}\mathbf{h}\left( \bar{\mathbf{x}}\right) \mathbf{d} = \mathbf{0}}\right\} . \] | 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 (CP), 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{\m... | No |
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\nwhere... | 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 f... | Yes |
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( \mathbf{x}\right) = \mathop{\sum }\limi... | For large \( c \) it is clear that the minimum point of problem (2) will be in a region where \( P \) is small. It is expected that the region is \( S. \) Ideally then, as \( c \rightarrow \infty \) the solution point of (2) will converge to a solution of (1).\n\nThe procedure of the penalty function method is: Let \( ... | Yes |
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 |
Example 3. Let \( {g}_{i}, i = 1,2,\ldots, m \) be continuous functions on \( {\mathcal{R}}^{n} \) . Suppose\n\n\[ S = \left\{ {\mathbf{x} \in {\mathcal{R}}^{n} \mid {g}_{i}\left( \mathbf{x}\right) \geq 0, i = 1,2,\ldots, m}\right\} \]\n\nis robust, and suppose the interior of \( S \) is\n\n\[ \operatorname{int}\left( ... | The barrier method is quite analogous to the penalty method. 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\[ \min r\left( {\mathbf{x},{c}_{k}}\right) \]\n\nobtaining a solution point \( ... | Yes |
Consider the problem\n\n\\[ \n\\min \\;{x}_{1} + {x}_{2} \n\\]\n\n\\[ \n\\text{s.t.}\\; - {x}_{1}^{2} + {x}_{2} \geq 0 \n\\]\n\n\\[ \n{x}_{1} \geq 0\\text{.}\n\\] | Define the problem\n\n\\[ \n\\min r\\left( {\\mathbf{x}, c}\\right) = {x}_{1} + {x}_{2} - \\frac{1}{c}\\left\\lbrack {\\ln \\left( {-{x}_{1}^{2} + {x}_{2}}\\right) + \\ln {x}_{1}}\\right\\rbrack . \n\\]\n\nLet\n\n\\[ \n\\nabla r\\left( {\\mathbf{x}, c}\\right) = \\left( \\begin{matrix} 1 - \\frac{1}{c}\\left( {\\frac{-... | Yes |
Theorem 1 The feasible region \( \mathcal{F} \) of \( \left( {QP}\right) \) is nonempty if and only if for any nonzero vector \( \mu \) with | 利用Farkas引理考虑约束\n\n\[\n\text{若}\mathop{\sum }\limits_{{i \in \mathcal{E} \cup \mathcal{I}}}{\mu }_{i}{a}_{i} = 0,\;{\mu }_{i} \geq 0, i \in \mathcal{I}\text{,}\n\]\n的择一系统\n\nthe following inequality holds:\n\n\[\n\text{则}\mathop{\sum }\limits_{{i \in \mathcal{E} \cup \mathcal{I}}}{\mu }_{i}{b}_{i} \leq 0\text{.}\n\] | No |
Theorem 2 Suppose that the matrix \( Q \) in (1) is symmetric and positive definite. If \( \left( {{\lambda }^{ * },{\mu }^{ * }}\right) \) is a KKT-pair of (2), then \( {x}^{ * } = {Q}^{-1}\left( {{A}^{T}{\lambda }^{ * } - c}\right) \) is the (unique) optimal point of (1). | Proof: Since \( \left( {{\lambda }^{ * },{\mu }^{ * }}\right) \) is a KKT-pair of (2), \n\n\[ \nA{Q}^{-1}{A}^{T}{\lambda }^{ * } - \left( {b + A{Q}^{-1}c}\right) = {\mu }^{ * } \n\] \n\n\[ \n{\lambda }_{i}^{ * } \geq 0,{\mu }_{i}^{ * } \geq 0,{\lambda }_{i}^{ * }{\mu }_{i}^{ * } = 0, i \in \mathcal{I} \n\] \n\n(3) \n\n... | Yes |
Theorem 3 Suppose that \( A \in {\mathcal{R}}^{m \times n} \) has row full rank, and \( {x}^{T}{Qx} > 0 \) holds for any nonzero vector \( x \) with \( {Ax} = 0 \) . Then the system (5) has one solution, denoted by \( \left( {{x}^{ * },{\lambda }^{ * }}\right) \) . Moreover, \( {x}^{ * } \) is a global optimal solution... | Proof\n\nThe matrix\n\n\[ \left( \begin{matrix} Q & - {A}^{T} \\ - A & 0 \end{matrix}\right) \]\n\nis nonsingular. Indeed, suppose \( \left( {x,\lambda }\right) \in {\mathcal{R}}^{n \times m} \) such that\n\n\[ {Qx} - {A}^{T}\lambda = 0,\;{Ax} = 0.\;\text{没有非 解故可逆} \]\n\nMultiplication of the first equation by \( {x}^{... | Yes |
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