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Theorem 2. Let \( f \in {C}^{1} \) be convex on the convex set \( \Omega \) . If there is a point \( {\mathbf{x}}^{ * } \in \Omega \) such that, for all \( \mathbf{y} \in \Omega ,\nabla f\left( {\mathbf{x}}^{ * }\right) \left( {\mathbf{y} - {\mathbf{x}}^{ * }}\right) \geq 0 \), then \( {\mathbf{x}}^{ * } \) is a global... | Proof. We note parenthetically that since \( \mathbf{y} - {\mathbf{x}}^{ * } \) is a feasible direction at \( {\mathbf{x}}^{ * } \), the given condition is equivalent to the first-order necessary condition stated in Sect. 7.1. The proof of the proposition is immediate, since by Proposition 4 of the last section \[ f\le... | Yes |
Theorem 3. Let \( f \) be a convex function defined on the bounded, closed convex set \( \Omega \) . If \( f \) has a maximum over \( \Omega \) it is achieved at an extreme point of \( \Omega \) . | Proof. Suppose \( f \) achieves a global maximum at \( {\mathbf{x}}^{ * } \in \Omega \) . We show first that this maximum is achieved at some boundary point of \( \Omega \) . If \( {\mathbf{x}}^{ * } \) is itself a boundary point, then there is nothing to prove, so assume \( {\mathbf{x}}^{ * } \) is not a boundary poin... | Yes |
Proposition 1 (Zero-Order Necessary Conditions). If \( {\mathbf{x}}^{ * } \) solves (7.14) under the stated convexity conditions, then there is a nonzero vector \( \lambda \in {E}^{n} \) such that \( {\mathbf{x}}^{ * } \) is a solution to the two problems:\n\n\[ \text{minimize}f\left( \mathbf{x}\right) + {\lambda }^{T}... | Proof. Problem (7.17) follows from (7.15) (with \( s = 1 \) ) and the fact that \( f\left( \mathbf{x}\right) \leq r \) for \( r \geq f\left( \mathbf{x}\right) \) . The value \( c \) is attained from above at \( \left( {{f}^{ * },{\mathbf{x}}^{ * }}\right) \) . Likewise (7.18) follows from (7.16) and the fact that \( {\... | No |
Consider a continuously differentiable function \( f \) of a single variable \( x \in {E}^{1} \) defined on the unit interval \( \left\lbrack {0,1}\right\rbrack \) which plays the role of \( \Omega \) here. The first problem (7.17) implies \( {f}^{\prime }\left( {x}^{ * }\right) = - \lambda \) . If the solution is at t... | These together are identical to the first-order conditions of Sect. 7.1. | No |
Proposition 2 (Zero-Order Sufficiency Conditions). If there is a \( \lambda \) such that \( {\mathbf{x}}^{ * } \in \Omega \) solves the problems (7.17) and (7.18), then \( {\mathbf{x}}^{ * } \) solves (7.14). | Proof. Suppose \( {\mathbf{x}}_{1} \) is any other point in \( \mathbf{\Omega } \) . Then from (7.17)\n\n\[ f\left( {\mathbf{x}}_{1}\right) + {\lambda }^{T}{\mathbf{x}}_{1} \geq f\left( {\mathbf{x}}^{ * }\right) + {\lambda }^{T}{\mathbf{x}}^{ * }.\]\n\nThis can be rewritten as\n\n\[ f\left( {\mathbf{x}}_{1}\right) - f\... | Yes |
As a special case, suppose that the mapping \( \mathbf{A} \) is a point-to-point mapping; that is, for each \( \mathbf{x} \in X \) the set \( \mathbf{A}\left( \mathbf{x}\right) \) consists of a single point in \( Y \) . Suppose also that \( \mathbf{A} \) is continuous at \( \mathbf{x} \in X \) . This means that if \( {... | The converse is, however, not true in general. | Yes |
Proposition. Let \( \mathbf{A} : X \rightarrow Y \) and \( \mathbf{B} : Y \rightarrow Z \) be point-to-set mappings. Suppose \( \mathbf{A} \) is closed at \( \mathbf{x} \) and \( \mathbf{B} \) is closed on \( \mathbf{A}\left( \mathbf{x}\right) \) . Suppose also that if \( {\mathbf{x}}_{k} \rightarrow \mathbf{x} \) and ... | Proof. Let \( {\mathbf{x}}_{k} \rightarrow \mathbf{x} \) and \( {\mathbf{z}}_{k} \rightarrow \mathbf{z} \) with \( {\mathbf{z}}_{k} \in \mathbf{C}\left( {\mathbf{x}}_{k}\right) \) . It must be shown that \( \mathbf{z} \in \mathbf{C}\left( \mathbf{x}\right) \) . Select \( {\mathbf{y}}_{k} \in \mathbf{A}\left( {\mathbf{x... | Yes |
In many respects condition (iii) of the theorem, the closedness of A outside the solution set, is the most important condition. The failure of many popular algorithms can be traced to nonsatisfaction of this condition. On the real line consider the point-to-point algorithm\n\n\[ A\\left( x\\right) = \\left\\{ \\begin{m... | However, starting from \( x > 1 \), the algorithm generates a sequence converging to \( x = 1 \) which is not a solution. The difficulty is that \( A \) is not closed at \( x = 1 \) . | Yes |
On the real line \( X \) consider the solution set to be empty, the descent function \( Z\left( x\right) = {e}^{-x} \), and the algorithm \( A\left( x\right) = x + 1 \). All conditions of the convergence theorem except (i) hold. The sequence generated from any starting condition diverges to infinity. | This is not strictly a violation of the conclusion of the theorem but simply an example illustrating that if no compactness assumption is introduced, the generated sequence may have no convergent subsequence. | Yes |
The sequence \( {r}_{k} = 1/k \) converges to zero arithmetically. | The convergence is of order one but it is not linear, since \( \mathop{\lim }\limits_{{k \rightarrow \infty }}\left( {{r}_{k + 1}/{r}_{k}}\right) = 1 \), that is, \( \beta \) is not strictly less than one. | Yes |
For the sequence \( {r}_{k} = {a}^{\left( {2}^{k}\right) },0 < a < 1 \), given in Example 2, we have | \[ {\left| {r}_{k}\right| }^{1/{2}^{k}} = a \] while \[ {\left| {r}_{k}\right| }^{1/{p}^{k}} = {a}^{{\left( 2/p\right) }^{k}} \rightarrow 1 \] for \( p > 2 \) . Thus the average order is two. | Yes |
Lemma 2. The iterative process (8.39) satisfies\n\n\[ E\left( {\mathbf{x}}_{k + 1}\right) = \left\{ {1 - \frac{{\left( {\mathbf{g}}_{k}^{T}{\mathbf{g}}_{k}\right) }^{2}}{\left( {{\mathbf{g}}_{k}^{T}\mathbf{Q}{\mathbf{g}}_{k}}\right) \left( {{\mathbf{g}}_{k}^{T}{\mathbf{Q}}^{-l}{\mathbf{g}}_{k}}\right) }}\right\} E\left... | Proof. The proof is by direct computation. We have, setting \( {\mathbf{y}}_{k} = {\mathbf{x}}_{k} - {\mathbf{x}}^{ * } \), \n\n\[ \frac{E\left( {\mathbf{x}}_{k}\right) - E\left( {\mathbf{x}}_{k + 1}\right) }{E\left( {\mathbf{x}}_{k}\right) } = \frac{2{\alpha }_{k}{\mathbf{g}}_{k}^{T}\mathbf{Q}{\mathbf{y}}_{k} - {\alph... | Yes |
For any \( {\mathbf{x}}_{0} \in {E}^{n} \) the method of steepest descent (8.39) converges to the unique minimum point \( {\mathbf{x}}^{ * } \) of \( f \) Furthermore, with \( E\left( \mathbf{x}\right) = \) \( \frac{1}{2}{\left( \mathbf{x} - {\mathbf{x}}^{ * }\right) }^{T}\mathbf{Q}\left( {\mathbf{x} - {\mathbf{x}}^{ *... | Proof. By Lemma 2 and the Kantorovich inequality\n\n\[ E\left( {\mathbf{x}}_{k + 1}\right) \leq \left\{ {1 - \frac{4aA}{{\left( A + a\right) }^{2}}}\right\} E\left( {\mathbf{x}}_{k}\right) = {\left( \frac{A - a}{A + a}\right) }^{2}E\left( {\mathbf{x}}_{k}\right) .\n\]\n\n## It follows immediately that \( E\left( {\math... | Yes |
Consider the earlier example of \( f\left( x\right) = {tx} - \ln x \) . | \[ \lambda \left( x\right) = {\left\lbrack {f}^{\prime }{\left( x\right) }^{2}/{f}^{\prime \prime }\left( x\right) \right\rbrack }^{\frac{1}{2}} = \left| {\left( {t - 1/x}\right) x}\right| = \left| {1 - {tx}}\right| . \]\n\nThen (8.70) gives\n\n\[ \left( {1 - t{x}^{ + }}\right) \leq 2{\left( 1 - tx\right) }^{2} \]\n\nA... | Yes |
Theorem 2. In the method of conjugate gradients we have\n\n\[ \nE\left( {\mathbf{x}}_{k + 1}\right) \leq \mathop{\max }\limits_{{\lambda }_{i}}{\left\lbrack 1 + {\lambda }_{i}{P}_{k}\left( {\lambda }_{i}\right) \right\rbrack }^{2}E\left( {\mathbf{x}}_{0}\right) \n\]\n\n(9.27)\n\nfor any polynomial \( {P}_{k} \) of degr... | This way of viewing the conjugate gradient method as an optimal process is exploited in the next section. We note here that it implies the far from obvious fact that every step of the conjugate gradient method is at least as good as a steepest descent step would be from the same point. To see this, suppose \( {\mathbf{... | Yes |
Consider the problem\n\n\\[ \n\\text{minimize}{x}_{1}{x}_{2} + {x}_{2}{x}_{3} + {x}_{1}{x}_{3} \n\\]\n\n\\[ \n\\text{subject to}{x}_{1} + {x}_{2} + {x}_{3} = 3\\text{.} \n\\] | The necessary conditions become\n\n\\[ \n{x}_{2} + {x}_{3} + \\lambda = 0 \n\\]\n\n\\[ \n{x}_{1}\; + {x}_{3} + \\lambda = 0 \n\\]\n\n\\[ \n{x}_{1} + {x}_{2}\; + \\lambda = 0. \n\\]\n\nThese three equations together with the one constraint equation give four equations that can be solved for the four unknowns \\( {x}_{1}... | Yes |
We seek to construct a cardboard box of maximum volume, given a fixed area of cardboard. Denoting the dimensions of the box by \( x, y, z \), the problem can be expressed as maximize \( xyz \) subject to \( (xy + yz + xz) = \frac{c}{2} \) where \( c > 0 \) is the given area of cardboard. | Introducing a Lagrange multiplier, the first-order necessary conditions are easily found to be \( yz + \lambda (y + z) = 0 \), \( xz + \lambda (x + z) = 0 \), and \( xy + \lambda (x + y) = 0 \) together with the constraint. Before solving these, let us note that the sum of these equations is \( (xy + yz + xz) + 2\lambd... | Yes |
If the value of mean is known to be \( m \) (by the physical situation), the maximum entropy argument suggests that the density should be taken as that which solves the following problem:\n\n\[ \text{ maximize }\; - \mathop{\sum }\limits_{{i = 1}}^{n}{p}_{i}\log \left( {p}_{i}\right) \]\n\n\[ \text{subject to}\mathop{\... | We begin by ignoring the nonnegativity constraints, believing that they may be inactive. Introducing two Lagrange multipliers, \( \lambda \) and \( \mu \), the Lagrangian is\n\n\[ l = \mathop{\sum }\limits_{{i = 1}}^{n}\left\{ {-{p}_{i}\log {p}_{i} + \lambda {p}_{i} + \mu {x}_{i}{p}_{i}}\right\} - \lambda - {\mu m}. \]... | Yes |
A chain is suspended from two thin hooks that are 16 ft apart on a horizontal line as shown in Fig. 11.3. The chain itself consists of 20 links of stiff steel. Each link is one foot in length (measured inside). We wish to formulate the problem to determine the equilibrium shape of the chain. | The solution can be found by minimizing the potential energy of the chain. Let us number the links consecutively from 1 to 20 starting with the left end. We let link \( i \) span an \( x \) distance of \( {x}_{i} \) and a \( y \) distance of \( {y}_{i} \) . Then \( {x}_{i}^{2} + {y}_{i}^{2} = 1 \) . The potential energ... | Yes |
Suppose there are \( n \) securities indexed by \( i = 1,2 \) , \( \ldots, n \) . Each security \( i \) is characterized by its random rate of return \( {r}_{i} \) which has mean value \( {\bar{r}}_{i} \) . Its covariances with the rates of return of other securities are \( {\sigma }_{ij} \), for \( j = 1,2,\ldots, n \... | Markowitz introduced the concept of devising efficient portfolios which for a given expected rate of return \( \bar{r} \) have minimum possible variance. Such a portfolio is the solution to the problem\n\n\[ \mathop{\min }\limits_{{{w}_{i},{w}_{2},\ldots {w}_{n}}}\mathop{\sum }\limits_{{i, j = 1}}^{n}{w}_{i}{\sigma }_{... | Yes |
Consider the problem\n\n\\[ \n\\text{maximize}{x}_{1}{x}_{2} + {x}_{2}{x}_{3} + {x}_{1}{x}_{3} \n\\]\n\n\\[ \n\\text{subject to}{x}_{1} + {x}_{2} + {x}_{3} = 3\\text{.} \n\\] | In Example 1 of Sect. 11.4 it was found that \( {x}_{1} = {x}_{2} = {x}_{3} = 1,\\lambda = - 2 \) satisfy the first-order conditions. The matrix \( \\mathbf{F} + {\\mathbf{\\lambda }}^{T}\\mathbf{H} \) becomes in this case\n\n\\[ \nL = \\left\\lbrack \\begin{array}{lll} 0 & 1 & 1 \\\\ 1 & 0 & 1 \\\\ 1 & 1 & 0 \\end{arr... | Yes |
In the last section we considered\n\n\[ \nL = \left\lbrack \begin{array}{lll} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}\right\rbrack \n\]\n\nrestricted to \( M = \left\{ {\mathbf{y} : {y}_{1} + {y}_{2} + {y}_{3} = 0}\right\} \) . To obtain an explicit matrix representation on \( M \) let us introduce the orthonor... | This gives, upon expansion,\n\n\[ \n{\mathbf{E}}^{T}\mathbf{{LE}} = \left\lbrack \begin{array}{rr} - 1 & 0 \\ 0 & - 1 \end{array}\right\rbrack \n\]\n\nand hence \( \mathbf{L} \) restricted to \( M \) acts like the negative of the identity. | Yes |
\[ \text{extremize}{x}_{1} + {x}_{2}^{2} + {x}_{2}{x}_{3} + 2{x}_{3}^{2} \] \[ \text{subject to}\frac{1}{2}\left( {{x}_{1}^{2} + {x}_{2}^{2} + {x}_{3}^{2}}\right) = 1\text{.} \] | The first-order necessary conditions are \[ 1 + \;\lambda {x}_{1} = 0 \] \[ 2{x}_{2} + {x}_{3} + \lambda {x}_{2} = 0 \] \[ {x}_{2} + 4{x}_{3} + \lambda {x}_{3} = 0. \] One solution to this set is easily seen to be \( {x}_{1} = 1,{x}_{2} = 0,{x}_{3} = 0,\lambda = - 1 \) . Let us examine the second-order conditions at th... | Yes |
Example 3. Approaching Example 2 in this way and noting \( \mathbf{A} = \nabla \mathbf{h} = \left( {1,0,0}\right) \) we have | \[ {\mathbf{P}}_{A} = \mathbf{I} - \left\lbrack \begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right\rbrack {\left\lbrack \begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right\rbrack }^{T} = \left\lbrack \begin{array}{lll} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right\rbrack \]\n\nThen\n\n\[ {\mathbf{P}}_{A}\mathbf{L}{\math... | Yes |
Proposition 1. Suppose \( \Omega \) is convex, the function \( f \) is convex, and \( \mathbf{h} \) is affine. Then the primal function \( \omega \) is convex. | Proof. For simplicity of notation we assume that \( \Omega \) is the entire space \( X \) . Then we observe\n\n\[ \Omega \left( {\alpha {\mathbf{y}}_{1} + \left( {1 - \alpha }\right) {\mathbf{y}}_{2}}\right) = \inf \{ f\left( \mathbf{x}\right) : \mathbf{h}\left( \mathbf{x}\right) = \alpha {\mathbf{y}}_{1} + \left( {1 -... | Yes |
Consider the classic problem of finding the rectangle of maximum area while limiting the perimeter to a length of 4. | The regularity condition is met because it is possible to make the right hand side of the functional constraint slightly positive or slightly negative with nonnegative \( {x}_{1} \) and \( {x}_{2} \) . We know the answer to the problem is \( {x}_{1} = {x}_{2} = 1 \) . The Lagrange multiplier is \( \lambda = 1 \) . The ... | No |
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\\[ \... | 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. I | No |
Proposition 4. 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),... | Proof. 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\{ {\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... | 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\[ \n\\mathbf{x} \\geq \\mathbf{0}.\\text{.} \n\]\n\nLet \( \\Omega = \\{ \\mathbf{x} : \\mathbf{x} \\geq \\mathbf{0}\\} \... | \[ \n\\text{minimize} \\;\\mathbf{x}^{T}\\mathbf{Q}\\mathbf{x} + \\mathbf{c}^{T}\\mathbf{x} + \\mu \\left( {{\\mathbf{a}}^{T}\\mathbf{x} - b}\\right) \n\]\n\n\[ \n\\text{subject to} \\;\\mathbf{x} \\geq \\mathbf{0} \\;\\text{and} \\;\\mu \\left( {{\\mathbf{a}}^{T}\\mathbf{x}^{ * } - b}\\right) = 0\\text{.} \n\] | 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 |
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(12.39)\n\n\[ \frac{{d}^{2}}{d{t}^{2}}f\... | 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 |
For the same situation as Example 1, we may use the logarithmic utility function\n\n\[ B\\left( \\mathbf{x}\\right) = - \\mathop{\\sum }\\limits_{{i = 1}}^{p}\\log \\left\\lbrack {-{g}_{i}\\left( \\mathbf{x}\\right) }\\right\\rbrack .\n\]\n\nThis is the barrier function commonly used in linear programming interior poin... | 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 |
Example 3. A general class of penalty functions is\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 \( \\varepsilon > 0 \) . | In view of this assumption, problem (13.19) will have its solution at a point \( {\\mathbf{x}}_{k} \) satisfying\n\n\[ \\nabla f\\left( {\\mathbf{x}}_{k}\\right) + {c}_{k}\\nabla \\gamma \\left( {{\\mathbf{g}}^{ + }\\left( {\\mathbf{x}}_{k}\\right) }\\right) \\nabla \\mathbf{g}\\left( {\\mathbf{x}}_{k}\\right) = \\math... | 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 |
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 ,\mathbf{x}}\right) = f\left( \m... | Defining\n\n\[ \n{\lambda }_{\mu, i} = \mu \frac{-1}{{g}_{i}\left( {\mathbf{x}}_{\mu }\right) }\n\]\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 that \( {\lambda }_{\mu } \r... | 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\\] | 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\[ \text{minimize}\frac{1}{2}{\mathbf{x}}^{T}\mathbf{Q}\mathbf{x} \]\n\n(14.5)\n\n\[ \text{subject to}\mathbf{{Bx}} - \mathbf{b} \leq \mathbf{0}\text{.} \] | The dual function is\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\nThis gives the necessary conditions\n\n\[ \mathbf{Q}\mathbf{x} + {\mathbf{B}}^{T}\mathbf{\mu } = \ma... | 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 \( \\omega \\left( z\\right) \)... | 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\\ma... | 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 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\[ c{\mathbf{A}}^{T}\mathbf{A}\mathbf{d} = {\mathbf{A}}^{T}\widehat{\lambda } - \mathbf{c}{\mathbf{A}}^{T}\mathbf{h}\left( \mathbf{x}\right) \]\n\nSolving t... | 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 and a constant \( c \) such that \( 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 be... | 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 |
A point \( \mathbf{x} \) of the polyhedral set \( P = \left\{ {\mathbf{x} \in {R}^{n} \mid \mathbf{{Ax}} = \mathbf{b},\mathbf{x} \geq \mathbf{0}}\right\} \) is an extreme point of \( P \) if and only if the columns of \( \mathbf{A} \) corresponding to the positive components of \( \mathbf{x} \) are linearly independent... | Proof. Without loss of generality, we may assume that the components of \( \mathbf{x} \) are zero except for the first \( p \) components, namely\n\n\[ \mathbf{x} = \left( \frac{\overline{\mathbf{x}}}{\mathbf{0}}\right) \;\text{ where }\;\overline{\mathbf{x}} = {\left( {x}_{1},\ldots ,{x}_{p}\right) }^{T} > \mathbf{0} ... | Yes |
Theorem 2.3 (Fundamental Theorem of Linear Programming). For a consistent linear program in its standard form with a feasible domain \( P \), the minimum objective value of \( \mathbf{z} = {\mathbf{c}}^{T}\mathbf{x} \) over \( P \) is either unbounded below or is achievable at least at one extreme point of \( P \) . | Proof. Let \( V = \left\{ {{\mathbf{v}}^{i} \in P \mid i \in I}\right\} \) be the set of all extreme points of \( P \) with a finite index set \( I \) . Since the problem is consistent, \( I \) is nonempty and there is at least one \( {\mathbf{v}}^{1} \in V \) . By the resolution theorem, \( P \) either has an external... | Yes |
Theorem 3.1. If\n\n\[ \n{\mathbf{x}}^{ * } = \left\lbrack \frac{{\mathbf{x}}_{B}^{ * }}{{\mathbf{x}}_{N}^{ * }}\right\rbrack = \left\lbrack \frac{{\mathbf{B}}^{-1}\mathbf{b}}{\mathbf{0}}\right\rbrack \geq \mathbf{0} \]\n\nis a basic feasible solution with nonnegative reduced costs vector \( \mathbf{r} \), given by Equa... | Moreover, we have developed a stopping rule based on the appearance of nonnegative reduced costs.\n\n## 3.3.1 Stopping the Simplex Method-Checking for Optimality\n\nLet \( {\mathbf{x}}^{ * } \) be a current basic feasible solution with \( \mathbf{B} \) being its corresponding basis, \( \mathbf{N} \) the nonbasis, \( \w... | No |
Theorem 3.2. Let\n\n\[ \n{\mathbf{x}}^{ * } = \left\lbrack \frac{{\mathbf{B}}^{-1}\mathbf{b}}{\mathbf{0}}\right\rbrack \n\]\n\nbe a basic feasible solution to the linear programming problem defined by (3.1) with basis B. If the reduced cost \( {r}_{q} < 0 \), for some nonbasic variable \( {x}_{q} \), then the edge dire... | Note that when \( {\mathbf{x}}^{ * } \) is nondegenerate, each edge direction is a feasible direction, therefore a positive step length \( \alpha \) can be chosen to translate current basic feasible solution along the direction \( {\mathbf{d}}^{q} \) to a distinct adjacent neighbor with improved objective value. Howeve... | Yes |
Theorem 3.4. Let\n\n\\[ \n{\\mathbf{x}}^{ * } = \\left\\lbrack \\frac{{\\mathbf{B}}^{-1}\\mathbf{b}}{\\mathbf{0}}\\right\\rbrack \n\\]\n\nbe a basic feasible solution to the linear programming problem defined by (3.1) with basis B. If a reduced cost \\( {r}_{q} < 0 \\) is found for some nonbasic variable \\( {x}_{q} \\... | Note that if \\( {\\mathbf{d}}^{q} \\) is a feasible direction and \\( \\alpha > 0 \\), then the adjacent basic feasible solution obtained in Theorem 3.4 represents a distinct extreme point with improved objective value. However, when \\( \\alpha = 0 \\), we stay at the same extreme point with the same objective value. | No |
Minimize $x_1$ subject to $\\epsilon x_1 - x_2 - x_3 = \\epsilon \\; \\left( \\epsilon > 0 \\right)$ and $x_1, x_2, x_3 \\geq 0$ | The associated big- $M$ problem can be stated as follows:\n\n$\\text{Mimimize} x_1 + M x_4$\n\n$\\text{subject to} \\epsilon x_1 - x_2 - x_3 + x_4 = \\epsilon$\n\n$x_1, x_2, x_3, x_4 \\geq 0$\n\nIt is clear that $ \\overline{\\mathbf{x}}^T = \\left( 1, 0, 0, 0 \\right) $ and $ \\widehat{\\mathbf{x}}^T = \\left( 0, 0, 0... | Yes |
\[ \text{Minimize}\; - \frac{3}{4}{x}_{4} + {20}{x}_{5} - \frac{1}{2}{x}_{6} + 6{x}_{7} \] \[ \text{subject to}{x}_{1} + \frac{1}{4}{x}_{4} - 8{x}_{5} - {x}_{6} + 9{x}_{7}\; = 0 \] \[ {x}_{2} + \frac{1}{2}{x}_{4} - {12}{x}_{5} - \frac{1}{2}{x}_{6} + 3{x}_{7} = 0 \] \[ {x}_{3} + {x}_{6}\; = 1 \] \[ {x}_{1},{x}_{2},{x}_{... | In the exercise, it can be verified that the optimal solution is given by \( {x}_{1} = \) \( 3/4,{x}_{2} = {x}_{3} = 0,{x}_{4} = 1,{x}_{5} = 0,{x}_{6} = 1,{x}_{7} = 0 \) with an optimal objective value \( - 5/4 \) . However, if we start with a basis \( \left\{ {{x}_{1},{x}_{2},{x}_{3}}\right\} \) and follow the previou... | No |
\[ \text{Minimize }\; - 3{x}_{1} - 2{x}_{2} \] \[ \text{subject to}{x}_{1} + {x}_{2} + {x}_{3} = {40} \] \[ 2{x}_{1} + {x}_{2} + {x}_{4} = {60} \] \[ {x}_{1} + {x}_{5}\; = {30} \] \[ {x}_{1},{x}_{2},{x}_{3},{x}_{4},{x}_{5} \geq 0 \] | For the revised simplex method, we have \[ \mathbf{A} = \left\lbrack \begin{array}{lllll} 1 & 1 & 1 & 0 & 0 \\ 2 & 1 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 & 1 \end{array}\right\rbrack ,\;\mathbf{b} = \left\lbrack \begin{array}{l} {40} \\ {60} \\ {30} \end{array}\right\rbrack ,\;\text{ and }\;{\mathbf{c}}^{T} = \left\lbrack \beg... | Yes |
Lemma 4.1. Given a primal linear program, the dual problem of the dual linear program becomes the original primal problem. | Proof. Let us start with problem (4.1). Its dual problem (4.3) may be expressed as\n\n\[ \text{-Minimize}z = - {\mathbf{b}}^{T}\mathbf{w} \]\n\n(4.4a)\n\n\[ \text{subject to}{\mathbf{A}}^{T}\mathbf{w} \leq \mathbf{c};\;\mathbf{w}\text{unrestricted} \]\n\n(4.4b)\n\nSince \( \mathbf{w} \) is unrestricted, we may represen... | Yes |
Theorem 4.1 (Weak Duality Theorem of LP). If \( {\mathbf{x}}^{0} \) is a primal feasible solution and \( {\mathbf{w}}^{0} \) is dual feasible, then \( {\mathbf{c}}^{T}{\mathbf{x}}^{0} \geq {\mathbf{b}}^{T}{\mathbf{w}}^{0} \) . | Proof. The dual feasibility of \( {\mathbf{w}}^{0} \) implies that \( \mathbf{c} \geq {\mathbf{A}}^{T}{\mathbf{w}}^{0} \) . For \( {\mathbf{x}}^{0} \) is primal feasible, we know \( {\mathbf{x}}^{0} \geq \mathbf{0} \) and, hence, \( {\mathbf{x}}^{0T}\mathbf{c} \geq {\mathbf{x}}^{0T}{\mathbf{A}}^{T}{\mathbf{w}}^{0} \) .... | Yes |
Corollary 4.1.1. If \( {\mathbf{x}}^{0} \) is primal feasible, \( {\mathbf{w}}^{0} \) is dual feasible, and \( {\mathbf{c}}^{T}{\mathbf{x}}^{0} = {\mathbf{b}}^{T}{\mathbf{w}}^{0} \) , then \( {\mathbf{x}}^{0} \) and \( {\mathbf{w}}^{0} \) are optimal solutions to the respective problems. | Proof. Theorem 4.1 indicates that \( {\mathbf{c}}^{T}\mathbf{x} \geq {\mathbf{b}}^{T}{\mathbf{w}}^{0} = {\mathbf{c}}^{T}{\mathbf{x}}^{0} \), for each primal feasible solution \( \mathbf{x} \) . Thus \( {\mathbf{x}}^{0} \) is an optimal solution to the primal problem. A similar argument holds for the dual problem. | Yes |
Corollary 4.1.2. If the primal problem is unbounded below, then the dual problem is infeasible. | Proof. Whenever the dual problem has a feasible solution \( {\mathbf{w}}^{0} \), the weak duality theorem prevents the primal objective from falling below \( {\mathbf{b}}^{T}{\mathbf{w}}^{0} \) . | Yes |
Theorem 4.3 (Farka's lemma) The system\n\n\\[ \n\\mathrm{{Ax}} = \\mathrm{b},\\;\\mathrm{x} \\geq 0 \n\\]\n\n(4.10)\n\nhas no solution if and only if the system\n\n\\[ \n{\\mathbf{A}}^{T}\\mathbf{w} \\leq \\mathbf{0},\\;{\\mathbf{b}}^{T}\\mathbf{w} > 0 \n\\]\n\n(4.11)\n\nhas a solution. | Proof. Consider the (primal) linear program\n\n\\[ \n\\text{Minimize}{\\mathbf{0}}^{T}\\mathbf{x} \n\\]\n\n\\[ \n\\text{subject to}\\mathbf{{Ax}} = \\mathbf{b},\\;\\mathbf{x} \\geq \\mathbf{0} \n\\]\n\nand its dual\n\nMaximize \\( {\\mathbf{b}}^{T}\\mathbf{w} \\)\n\nsubject to \\( {\\mathbf{A}}^{T}\\mathbf{w} \\leq \\m... | Yes |
Theorem 4.4 (Complementary slackness theorem). Let \( \mathbf{x} \) be a primal feasible solution and \( \mathbf{w} \) be a dual feasible solution to a symmetric pair of linear programs. Then \( \mathbf{x} \) and \( \mathbf{w} \) become an optimal solution pair if and only if the complementary slackness conditions\n\n\... | As to the primal-dual pair of linear programs in the standard form, i.e.,\n\n\[ \n\\text{Minimize}{\\mathbf{c}}^{T}\\mathbf{x} \n\]\n\n\( \\left( P\\right) \)\n\n\[ \n\\text{subject to}\\mathbf{{Ax}} = \\mathbf{b},\\;\\mathbf{x} \\geq \\mathbf{0} \n\]\n\n\[ \n\\text{Maximize}{\\mathbf{b}}^{T}\\mathbf{w} \n\]\n\n\( \\le... | Yes |
Theorem 4.5 (K-K-T optimality conditions for LP). Given a linear programming problem in its standard form, vector \( \mathbf{x} \) is an optimal solution to the problem if, and only if, there exist vectors \( \mathbf{w} \) and \( \mathbf{r} \) such that\n\n(1) \( \mathrm{{Ax}} = \mathrm{b},\;\mathrm{x} \geq 0 \) (prima... | ## Example 4.3\n\nLet us consider Example 3.4. When the revised simplex method terminates, it can be found\n\nthat \( \mathbf{x} = {\left\lbrack \begin{array}{llll} {20} & {20} & 0 & 0 \end{array}\right\rbrack }^{T},\mathbf{w} = {\left\lbrack \begin{array}{ll} - 1 & - 1 \end{array}\right\rbrack }^{T} \), and \( \mathbf... | Yes |
Lemma 4.2. Let \( \\mathbf{M} \) be an \( n \\times n \) nonsingular matrix and \( \\mathbf{u},\\mathbf{v} \) be two \( n \) -dimensional column vectors. If \( \\omega = 1 + {\\mathbf{v}}^{T}{\\mathbf{M}}^{-1}\\mathbf{u} \\neq 0 \), then the matrix \( \\left( {\\mathbf{M} + {\\mathbf{{uv}}}^{T}}\\right) \) is nonsingul... | Proof.\n\n\[ \n\\left\\lbrack {{\\mathbf{M}}^{-1} - \\left( \\frac{1}{\\omega }\\right) {\\mathbf{M}}^{-1}{\\mathbf{{uv}}}^{T}{\\mathbf{M}}^{-1}}\\right\\rbrack \\left\\lbrack {\\mathbf{M} + {\\mathbf{{uv}}}^{T}}\\right\\rbrack \n\]\n\n\[ \n= \\mathbf{I} - \\left( \\frac{1}{\\omega }\\right) {\\mathbf{M}}^{-1}{\\mathbf... | Yes |
Lemma 4.3. If the restricted primal problem has an optimal solution with zero objective value, then the solution must be an optimal solution to the original problem. | Proof. Assume that\n\n\[\n\left\lbrack \frac{{\mathbf{x}}_{T}^{ * }}{{\mathbf{x}}_{a}^{ * }}\right\rbrack\n\]\n\nis an optimal solution to the restricted problem with zero objective value. Since the optimal objective value of the restricted primal problem is zero, we have \( {\mathbf{x}}_{a}^{ * } = \mathbf{0} \) in it... | Yes |
Lemma 4.4. Let \( \mathbf{B} \) be an optimal basis to the original linear programming problem. If \( \overline{\mathbf{x}} \), essentially defined by (4.62), is nonnegative, then it is an optimal solution to the new linear programming problem with the additional constraint. | Proof. Since the basic solution \( \overline{\mathbf{x}} \) is nonnegative, it is a basic feasible solution. In order to declare it is an optimal solution, we need to show the reduced cost for each nonbasic variable is nonnegative, i.e., \[ {c}_{q} - {\left\lbrack \frac{{\mathbf{c}}_{\mathbf{B}}}{\mathbf{0}}\right\rbra... | Yes |
Example 5.1 (Klee-Minty's example)\n\nFor \( 0 < \delta < 1/2 \) ,\n\nMaximize \( {x}_{n} \)\n\n\[ \n\text{subject to}\;0 \leq {x}_{1} \leq 1 \n\]\n\n(5.5a)\n\n\[ \n\delta {x}_{i - 1} \leq {x}_{i} \leq 1 - \delta {x}_{i - 1},\;i = 2,3,\ldots, n \n\]\n\n(5.5b)\n\n\[ \n{x}_{i} \geq 0,\;i = 1,2,\ldots, n. \n\]\n\n(5.5c) | Obviously the origin point is a basic feasible solution. If we start with the origin and apply the largest reduction rule to the entering nonbasic variables, the simplex method takes \( {2}^{n} - 1 \) iterations to visit every extreme point of the feasible domain. For \( n = 2 \) and \( n = 3 \), Figures 5.1 and 5.2 il... | No |
In Figure 5.3, \( \mathbf{E} = \mathbf{S}\left( {\mathbf{0},1}\right) \) is the 2-dimensional unit sphere, the shaded area is \( \frac{1}{2}\mathbf{E} \) given by the intersection of \( \mathbf{E} \) with the halfspace \( \left\{ {\left( {{x}_{1},{x}_{2}}\right) \in {R}^{2} \mid {x}_{1} \geq 0}\right\} \) . Passing the... | The center of \( \widetilde{\mathbf{E}} \) is at \( \left( {1/3,0}\right) \) and the defining matrix \[ \mathbf{A} = \left\lbrack \begin{matrix} 3/2 & 0 \\ 0 & \sqrt{3}/2 \end{matrix}\right\rbrack \;\text{ with }\det \left( \mathbf{A}\right) = 3\sqrt{3}/4 \] | Yes |
Lemma 5.2. The smallest ellipsoid \( \mathbf{E} \) containing a convex polyhedral set \( \mathbf{P} \) has its center in \( \mathbf{P} \) . | Lemma 5.2 actually suggests an iterative scheme to solve the system of inequalities (5.6). Here is the basic idea: if part of the solution set of (5.6) forms a convex polyhedron \( \mathbf{P} \) and it is contained in an ellipsoid \( {\mathbf{E}}_{k} \) at the \( k \) th iteration, then we could check the center of \( ... | No |
Lemma 5.3. If the system of inequalities (5.6) has any solution, then it has a solution \( \mathbf{x} \in {R}^{n} \) such that\n\n\[ - {2}^{L} \leq {x}_{j} \leq {2}^{L},\;j = 1,2,\ldots, n \]\n\nwhere \( L \) is the input size given by (5.3) with \( {c}_{j} = 0 \) for all \( j \) . | Hence we can define \( {\mathbf{E}}_{0} = \mathbf{S}\left( {\mathbf{0},{2}^{2L}}\right) \), which is an \( n \) -dimensional sphere with radius equal to \( {2}^{2L} \) . In this case, the convex polyhedron \( \mathbf{P} \) defined by (5.6) and (5.13) is contained in \( {\mathbf{E}}_{0} \) to let us proceed with the ite... | No |
Lemma 5.4. If the system of inequalities (5.6) has a solution, then the volume of its solutions inside the cube \( \\left\\{ {\\mathbf{x} \\in {R}^{n}\\left| \\right| {x}_{i} \\mid \\leq {2}^{L}, i = 1,\\ldots, n}\\right\\} \) is at least \( {2}^{-\\left( {n + 1}\\right) L} \) . | Hence we can terminate the iterative scheme when \( \\operatorname{vol}\\left( {\\mathbf{E}}_{k}\\right) < {2}^{-\\left( {n + 1}\\right) L} \). In this case, (5.6) has no solution. | No |
Lemma 6.1. In Karmarkar's algorithm, let\n\n\\[ \n\\mathbf{y} = \\frac{\\mathbf{e}}{n} + \\frac{\\alpha }{n}\\left( \\frac{\\mathbf{d}}{\\parallel \\mathbf{d}\\parallel }\\right) \\;\\text{ for some }0 \\leq \\alpha \\leq 1 \n\\]\n\nwhere\n\n\\[ \n\\mathbf{d} = - \\left\\lbrack {\\mathbf{I} - {\\mathbf{B}}_{k}^{T}{\\le... | Proof. Note that the direction vector \\( \\mathbf{d} \\) is obtained as the projection of the negative cost vector \\( - {\\mathbf{c}}^{T}{\\mathbf{X}}_{k} \\), hence \\( {\\mathbf{c}}^{T}{\\mathbf{X}}_{k}\\mathbf{d} = - \\parallel \\mathbf{d}{\\parallel }^{2} \\) . Then we have\n\n\\[ \n{\\mathbf{c}}^{T}{\\mathbf{X}}... | Yes |
Lemma 6.2. If \( \mathbf{y} \in {\mathbf{S}}^{\prime }\left( {\frac{\mathbf{e}}{n},\frac{\alpha }{n}}\right) \) then\n\n\[ - \mathop{\sum }\limits_{{j = 1}}^{n}{\log }_{e}{y}_{j} \leq - \mathop{\sum }\limits_{{j = 1}}^{n}{\log }_{e}\left( \frac{1}{n}\right) + \frac{{\alpha }^{2}}{2{\left( 1 - \alpha \right) }^{2}} \] | Proof. Since\n\[ \mathbf{y} \in {\mathbf{S}}^{\prime }\left( {\frac{\mathbf{e}}{n},\frac{\alpha }{n}}\right) \]\n\nwe know\n\[ {y}_{j} \geq \frac{1}{n} - \frac{\alpha }{n} \]\n\nand hence \( n{y}_{j} \geq 1 - \alpha \), for \( j = 1,2,\ldots, n \) . Taking the Taylor series expansion of \( {\log }_{e}\left( {1 + \left(... | Yes |
Theorem 6.1. Under the assumptions (A1) and (A2), if a step-length is chosen to be \( \alpha = 1/3 \), then Karmarkar’s algorithm stops in \( O\left( {nL}\right) \) iterations. | The computational work at each iteration of Karmarkar's algorithm is dominated by inverting the matrix \( {\mathbf{B}}_{k}{\mathbf{B}}_{k}^{T} \) . A simpleminded direct implementation with exact arithmetic requires \( O\left( {n}^{3}\right) \) elementary operations to find the inverse matrix. Hence the total complexit... | Yes |
Theorem 6.2. Under the assumptions (A1) and (A2), if the iterates \( \left\{ {\mathbf{x}}^{k}\right\} \) defined in Karmarkar’s algorithm converge to a nondegenerate basic feasible solution \( {\mathbf{x}}^{ * } \) of problem (6.1), then \( \left\{ \left( {{\mathbf{w}}^{k},{z}^{k}}\right) \right\} \) defined by (6.34) ... | Proof. Let \( {\overline{\mathbf{X}}}^{ * } \) be the principal submatrix of \( {\mathbf{X}}^{ * } \) corresponding to the basic variables in \( {\mathbf{x}}^{ * } \) and\n\n\[ \left\lbrack \begin{array}{l} \overline{\mathbf{A}} \\ {\mathbf{e}}^{T} \end{array}\right\rbrack \]\n\nbe the basis matrix of the given linear ... | Yes |
Lemma 6.3. In applying the modified Karmarkar's algorithm with a given cost vector \( \widehat{\mathbf{c}} \in {R}^{n} \) and explicit constraint matrix \( \widehat{\mathbf{A}} \) such that \( \widehat{\mathbf{A}}\mathbf{e} = \mathbf{0} \), let \( {\mathbf{d}}^{k} = - {P}_{\widehat{\mathbf{B}}}\widehat{\mathbf{c}} \) ,... | Proof. Since \( {\mathbf{d}}^{k} \) is the projection of \( - \widehat{\mathbf{c}} \), we have \( {\begin{Vmatrix}{\mathbf{d}}^{k}\end{Vmatrix}}^{2} = {\widehat{\mathbf{c}}}^{T}{P}_{\widehat{\mathbf{B}}}\widehat{\mathbf{c}} = - {\widehat{\mathbf{c}}}^{T}{\mathbf{d}}^{k} \), and\n\n\[{\widehat{\mathbf{c}}}^{T}\left( {\f... | Yes |
Theorem 6.3. Let \( {\mathbf{w}}^{ * } \) be the unique maximum of the concave function \( h\left( {\mathbf{w};\mu }\right) \) with \( \mu > 0 \) . If \( {\mathbf{x}}^{ * } \) is defined by Equation (6.57), then\n\n\[ h\left( {{\mathbf{w}}^{ * };\mu }\right) = - \mu {\log }_{e}\left\{ {\mathop{\sum }\limits_{{j = 1}}^{... | Notice that, for \( \mathbf{x} \geq \mathbf{0} \) and \( {\mathbf{e}}^{T}\mathbf{x} = 1 \) ,\n\n\[ - \frac{1}{e} \leq {x}_{j}{\log }_{e}{x}_{j} \leq 0 \]\n\nConsequently, we know \( h\left( {{\mathbf{w}}^{ * };\mu }\right) \) approaches \( {\mathbf{c}}^{T}{\mathbf{x}}^{ * } \) as \( \mu \) goes to 0 . Hence, when \( \m... | Yes |
Lemma 7.3. If the linear programming problem (7.1) is bounded below and its objective function is not constant, then the sequence \( \left\{ {{\mathbf{c}}^{T}{\mathbf{x}}^{k} \mid k = 1,2,\ldots }\right\} \) is well-defined and strictly decreasing. | Proof. This is a direct consequence of Lemmas 7.1, 7.2, and Equation (7.12). | No |
Minimize - 2{x}_{1} + {x}_{2} | subject to{x}_{1} - {x}_{2} + {x}_{3} = {15}\n\n{x}_{2} + {x}_{4} = {15}\n\n{x}_{1},{x}_{2},{x}_{3},{x}_{4} \geq 0\n\nIn this case,\n\n\[ \mathbf{A} = \left\lbrack \begin{array}{rrrr} 1 & - 1 & 1 & 0 \\ 0 & 1 & 0 & 1 \end{array}\right\rbrack ,\;\mathbf{b} = {\left\lbrack \begin{array}{ll} {15} & {15} \end{array}\right\... | Yes |
Lemma 7.4. When the primal affine scaling algorithm applies, \( \mathop{\lim }\limits_{{k \rightarrow \infty }}{\mathbf{X}}_{k}{\mathbf{r}}^{k} = \mathbf{0} \) . | Proof. Since \( \left\{ {{\mathbf{c}}^{T}{\mathbf{x}}^{k}}\right\} \) is monotonically decreasing and bounded below (by the first assumption), the sequence converges. Hence Equations (7.12) and (7.9) imply that\n\n\[ 0 = \mathop{\lim }\limits_{{k \rightarrow \infty }}\left( {{\mathbf{c}}^{T}{\mathbf{x}}^{k} - {\mathbf{... | Yes |
Theorem 7.1. If \( \left\{ {\mathbf{x}}^{k}\right\} \) converges, then \( {\mathbf{x}}^{ * } = \mathop{\lim }\limits_{{k \rightarrow \infty }}{\mathbf{x}}^{k} \) is an optimal solution to problem (7.1). | Proof. We prove this result by contradiction. First notice that when \( \left\{ {\mathbf{x}}^{k}\right\} \) converges to \( {\mathbf{x}}^{ * },{\mathbf{x}}^{ * } \) must be primal feasible. However, let us assume that \( {\mathbf{x}}^{ * } \) is not primal optimal.\n\nSince \( {\mathbf{r}}^{k}\left( \cdot \right) \) is... | Yes |
Theorem 7.2. The sequence \( \left\{ {\mathbf{x}}^{k}\right\} \) generated by the primal affine scaling algorithm is convergent. | Proof. Since the feasible domain is nonempty, closed, and bounded, owing to compactness the sequence \( \left\{ {\mathbf{x}}^{k}\right\} \) has at least one accumulation point in \( P \), say \( {\mathbf{x}}^{ * } \) . Our objective is to show that \( {\mathbf{x}}^{ * } \) is also the only accumulation point of \( \lef... | Yes |
Lemma 7.5. Under the assumptions (A1) and (A2), both problem \( \left( {\mathrm{P}}_{\mu }\right) \) and system (7.76) have a unique solution. | Observe that system (7.76) also provides the necessary and sufficient conditions (the K-K-T conditions) for \( \left( {\mathbf{w}\left( \mu \right) ;\mathbf{s}\left( \mu \right) }\right) \) being a maximum solution of the following program \( \left( {\mathrm{D}}_{\mu }\right) \) : \[ \text{ Maximize }{\mathbf{b}}^{T}\m... | Yes |
Lemma 7.6. Under the assumptions (A1)-(A3), as \( \mu \rightarrow 0,\mathbf{x}\left( \mu \right) \) converges to the optimal solution of program \( \left( \mathrm{P}\right) \) and \( \left( {\mathbf{w}\left( \mu \right) ;\mathbf{s}\left( \mu \right) }\right) \) converges to the optimal solution of program (D). | For \( \mu > 0 \), we let \( \Gamma \) denote the curve, or path, consisting of the solutions of system (7.76), i.e., \[ \Gamma = \{ \left( {\mathbf{x}\left( \mu \right) ;\mathbf{w}\left( \mu \right) ;\mathbf{s}\left( \mu \right) }\right) \mid \left( {\mathbf{x}\left( \mu \right) ;\mathbf{w}\left( \mu \right) ;\mathbf{... | Yes |
Theorem 7.3. If the step-length \( {\beta }^{k} < 1 \) at the \( k \) th iteration, then\n\n\[ \n{\beta }^{k} \geq \frac{4\left( {\sigma - \tau }\right) }{n\left( {1 - {2\sigma } + {\theta }^{k}{\sigma }^{2}}\right) } \geq \frac{4\left( {\sigma - \tau }\right) }{n\left( {1 + {\sigma }^{2}}\right) {\theta }^{k}}\n\]\n\n... | Proof. Let us define\n\n\[ \n{\eta }^{k}\left( \beta \right) = {\phi }_{\min }^{k} + \beta \left( {{\mu }^{k} - {\phi }_{\min }^{k}}\right) - {\begin{Vmatrix}{\mathbf{r}}^{k}\left( {\mu }^{k}\right) \end{Vmatrix}}^{2}\left( {{\beta }^{2}/4}\right)\n\]\n\n(7.99)\n\nwhere \( {\mathbf{r}}^{k}\left( {\mu }^{k}\right) \) is... | Yes |
Theorem 7.4. Let \( {\mathbf{x}}^{ * } \) and \( \left( {{\mathbf{w}}^{ * };{\mathbf{s}}^{ * }}\right) \) be optimal solutions of the original problems (P) and (D). In addition to (7.112a) and (7.112b), suppose that\n\n\[ \lambda > {\left( {\mathbf{A}}^{T}{\mathbf{w}}^{0} + {\mathbf{s}}^{0} - \mathbf{c}\right) }^{T}{\m... | Proof. Since \( {\mathbf{x}}^{ * } \) is feasible to (P), if we further define that \( {x}_{n + 1}^{ * } = 0 \) and \( {x}_{n + 2}^{ * } = \lambda - \) \( {\left( {\mathbf{A}}^{T}{\mathbf{w}}^{0} + {\mathbf{s}}^{0} - \mathbf{c}\right) }^{T}{\mathbf{x}}^{ * } \), then \( \left( {{\mathbf{x}}^{ * },{x}_{n + 1}^{ * },{x}_... | Yes |
Lemma 8.1. Assume that \( m \leq n \) and \( \mathbf{A} \) is an \( \left( {m \times n}\right) \) -dimensional matrix with full row rank. If \( \mathbf{U} \) is an \( \left\lbrack {n \times \left( {n - m}\right) }\right\rbrack \) -dimensional matrix of full rank such that \( \mathbf{{AU}} = \mathbf{0} \) , then\n\n\[ \... | Proof. For each \( \mathbf{x} \in {R}^{n} \), since matrix \( \mathbf{A} \) has full row rank, \( \mathbf{x} \) can be decomposed\n\nas\n\n\[ \mathbf{x} = {\mathbf{A}}^{T}{\mathbf{u}}^{1} + \mathbf{U}{\mathbf{u}}^{2} \]\n\nwhere \( {\mathbf{u}}^{1} \in {R}^{m} \) and \( {\mathbf{u}}^{2} \in {R}^{n - m} \) . Hence \( \m... | Yes |
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