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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 \( {\mathbf{x}}_{\mathbf{1}} \in \Omega \) such that \( \mathbf{g}\left( {\mathbf{x}}_{\mathbf{1}}\right) < \mathbf{0} \). Then, if \( {\mathbf{x}}^{ * } \... | 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\[ A = \{ \left( {r,\mathbf{0}}\right) : r \geq f\left( \mathbf{x}\right) ,\mathbf{0} \geq \mathbf{g}\left( \mathbf{x}\right) ,\text{ for some }\mathbf{x} \in \Omega \} \]\... | 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... | Hence there is a Lagrange multiplier \\( \\mu \\geq 0 \\) (a scalar in this case) such that the solution \\( {\\mathbf{x}}^{ * } \\) to the quadratic program is also a solution to\n\n\\[ \n\\text{minimize}\\;{\\mathbf{x}}^{T}\\mathbf{Q}\\mathbf{x} + {\\mathbf{c}}^{T}\\mathbf{x} + \\mu \\left( {{\\mathbf{a}}^{T}\\mathbf... | 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, \( m \) -dimensional\n\nand p-dimensional functions on \( \Omega \) . Finally, suppose there are vectors \( {\mathbf{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 |
Consider the region shown in Fig. 12.2 together with the sequence of feasible points \( \left\{ {\mathbf{x}}_{k}\right\} \) and feasible directions \( \left\{ {\mathbf{d}}_{k}\right\} \) . We have \( {\mathbf{x}}_{k} \rightarrow {\mathbf{x}}^{ * } \) and \( {\mathbf{d}}_{k} \rightarrow {\mathbf{d}}^{ * } \) . Also from... | \[ \mathbf{M}\left( {{\mathbf{x}}_{k},{\mathbf{d}}_{k}}\right) = {\mathbf{x}}_{k + 1} \rightarrow {\mathbf{x}}^{ * },\;\mathbf{M}\left( {{\mathbf{x}}^{ * },{\mathbf{d}}^{ * }}\right) = \mathbf{y} \neq {\mathbf{x}}^{ * }. \] Thus \( \mathbf{M} \) is not closed at \( \left( {{\mathbf{x}}^{ * },{\mathbf{d}}^{ * }}\right) ... | Yes |
In the simplified method presented in Example 1, the feasible direction selection map \( \mathbf{D} \) is not closed. | This can be seen from Fig. 12.3 where the directions are shown for a convergent sequence of points, and the limiting direction is not equal to the direction at the limiting point. Basically, nonclosedness is caused in this case by the fact that the method used for generating the feasible direction changes suddenly when... | Yes |
Proposition 1. Let \( \mathbf{x}\left( t\right) ,0 \leq t \leq T \), be a geodesic on \( \Omega \) . Then\n\n\[ \frac{d}{dt}f\left( {\mathbf{x}\left( t\right) }\right) = {l}_{\mathbf{x}}\left( {\mathbf{x},\mathbf{\lambda }\left( \mathbf{x}\right) }\right) \dot{\mathbf{x}}\left( t\right) \]\n\n(39)\n\n\[ \frac{{d}^{2}}{... | 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. \[ q\left( {{c}_{k},{\mathbf{x}}_{k}}\right) \leq q\left( {{c}_{k + 1},{\mathbf{x}}_{k + 1}}\right) \] (5) \[ P\left( {\mathbf{x}}_{k}\right) \geq P\left( {\mathbf{x}}_{k + 1}\right) \] (6) \[ f\left( {\mathbf{x}}_{k}\right) \leq f\left( {\mathbf{x}}_{k + 1}\right) . \] (7) | Proof. \[ q\left( {{c}_{k + 1},{\mathbf{x}}_{k + 1}}\right) = f\left( {\mathbf{x}}_{k + 1}\right) + {c}_{k + 1}P\left( {\mathbf{x}}_{k + 1}\right) \geq f\left( {\mathbf{x}}_{k + 1}\right) + {c}_{k}P\left( {\mathbf{x}}_{k + 1}\right) \] \[ \geq f\left( {\mathbf{x}}_{k}\right) + {c}_{k}P\left( {\mathbf{x}}_{k}\right) = q... | Yes |
Lemma 2. Let \( {\mathbf{x}}^{ * } \) be a solution to problem (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 m... | When formulated with \( c \) we take \( c \) large (going to infinity); while when formulated with \( \\mu \) we take \( \\mu \) small (going to zero). Either way the result is a constrained problem, and indeed the constraint is somewhat more complicated than in the original problem (13). The advantage of this problem,... | Yes |
\[ P\left( \mathbf{x}\right) = \frac{1}{2}\mathop{\sum }\limits_{{i = 1}}^{p}{g}_{i}^{ + }{\left( \mathbf{x}\right) }^{2} = \frac{1}{2}{\left| {\mathbf{g}}^{ + }\left( \mathbf{x}\right) \right| }^{2}, \] | 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 |
In the penalty method we solve, for various \( {c}_{k} \), the unconstrained problem\n\n\[ \text{minimize} f\left( \mathbf{x}\right) + {c}_{k}P\left( \mathbf{x}\right) \text{.} \] | In view of this assumption, problem (19) will have its solution at a point \( {\mathbf{x}}_{k} \) satisfying\n\n\[ \nabla f\left( {\mathbf{x}}_{k}\right) + {c}_{k}\nabla \gamma \left( {{\mathbf{g}}^{ + }\left( {\mathbf{x}}_{k}\right) }\right) \nabla \mathbf{g}\left( {\mathbf{x}}_{k}\right) = \mathbf{0}, \]\n\nwhich can... | Yes |
For \( P\left( \mathbf{x}\right) = \frac{1}{2}{\left| {\mathbf{g}}^{ + }\left( \mathbf{x}\right) \right| }^{2} \), we have | \[ \mathbf{\Gamma }\left( {{\mathbf{g}}^{ + }\left( {\mathbf{x}}_{k}\right) }\right) = \left\lbrack \begin{matrix} {e}_{1} & 0 & \cdots & 0 \\ 0 & {e}_{2} & & 0 \\ 0 & \cdot & & \cdot \\ \cdot & \cdot & & \cdot \\ \cdot & & \cdot & \cdot \\ 0 & \cdots & 0 & {e}_{p} \end{matrix}\right\rbrack ,\] 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\[ {\left\lbrack \begin{array}{ll} {\mathbf{A}}_{1} & {\mathbf{A}}_{2} \\ {\mathbf{A}}_{2}^{T} & {\mathbf{A}}_{3} \end{array}\right\rbrack }^{-1} \]\n\n\[ = \left\lbrack \begin{matrix} {\left( {\mathbf{A}}_{1} - {\mathbf{A}}_{2}{\mathbf{A}}_{3}^{-1}{\mathbf{A}}_{2}^{T}\right) }^{-1} & - \... | 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 \( {\mathbf{\lambda }}_{k} \) to be the vector having \( i \) th component \( \frac{1}{{c}_{k}} \cdot \frac{1}{{g}_{i}{\left( {\mathbf{x}}_{k}\right) }^{2}} \) . Then\n\n(32)\ncan be written as\n\n\[ \nabla f\left( {\mathbf{x}}_{k}\right) + {\mathbf{\lambda }}_{k}^{T}\nabla \mathbf{g}\left( {\mathbf{x}}... | 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{... | Defining\n\n\\[ \n{\\lambda }_{\\mu, i} = \\mu \\frac{-1}{{g}_{i}\\left( {\\mathbf{x}}_{\\mu }\\right) }\n\\]\n\n(34) can be written as\n\n\\[ \n\\nabla f\\left( {\\mathbf{x}}_{\\mu }\\right) + {\\mathbf{\\lambda }}_{\\mu }^{T}\\nabla \\mathbf{g}\\left( {\\mathbf{x}}_{\\mu }\\right) = \\mathbf{0}.\n\\]\n\nFurther we ex... | 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 \( {\mathbf{\mu }}_{1},{\mathbf{\mu }}_{2} \) 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(... | Yes |
Consider the problem\n\n\\[ \n\\text{minimize}\\;\\frac{1}{2}{\\mathbf{x}}^{T}\\mathbf{Q}\\mathbf{x} \n\\]\n\n(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} + {... | Yes |
Example 4 (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\\text{.} \n\\]\n\nThe primal function \\( \\omega \\left( z... | Yes |
Lemma 1. The dual function \( \phi \) has gradient\n\n\[ \nabla \phi \left( \mathbf{\lambda }\right) = \mathbf{h}{\left( \mathbf{x}\left( \mathbf{\lambda }\right) \right) }^{T} \] | Proof. We have explicitly, from (13),\n\n\[ \phi \left( \mathbf{\lambda }\right) = f\left( {\mathbf{x}\left( \mathbf{\lambda }\right) }\right) + {\mathbf{\lambda }}^{T}\mathbf{h}\left( {\mathbf{x}\left( \mathbf{\lambda }\right) }\right) .\n\nThus\n\n\[ \nabla \phi \left( \mathbf{\lambda }\right) = \left\lbrack {\nabla ... | Yes |
Lemma 2. The Hessian of the dual function is\n\n\\[ \Phi \\left( \\mathbf{\\lambda }\\right) = - \\nabla \\mathbf{h}\\left( {\\mathbf{x}\\left( \\mathbf{\\lambda }\\right) }\\right) {\\mathbf{L}}^{-1}\\left( {\\mathbf{x}\\left( \\mathbf{\\lambda }\\right) ,\\mathbf{\\lambda }}\\right) \\nabla \\mathbf{h}{\\left( \\math... | Proof. The Hessian is the derivative of the gradient. Thus, by Lemma 1,\n\n\\[ \Phi \\left( \\mathbf{\\lambda }\\right) = \\nabla \\mathbf{h}\\left( {\\mathbf{x}\\left( \\mathbf{\\lambda }\\right) }\\right) \\nabla \\mathbf{x}\\left( \\mathbf{\\lambda }\\right) \\]\n\n(16)\n\nBy definition we have\n\n\\[ \\nabla f\\lef... | 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... | In this example we consider \( \mathbf{x} \) as the \( {2N} \) -dimensional vector of unknowns \( y\left( k\right), u\left( k\right), k = 1,2,\ldots, N \) . This vector is partitioned into the pairs \( {\mathbf{x}}_{k} = \) \( \left( {y\left( k\right), u\left( k\right) }\right) \) . The objective function is then clear... | Yes |
In Example 2 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 \( \mathbf{\lambda } \in {E}^{N} \) 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... | Yes |
Consider the simple quadratic problem studied in Section 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 \( \mathbf{d},\mathbf{\lambda } \) (with \( \mathbf{d} \neq \mathbf{0} \) ) be a solution to the quadratic program (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| \mat... | Proof. We have from the constraint equation\n\n\[ \mathbf{{Ad}} = \left( {1/c}\right) \widehat{\mathbf{\lambda }} - \mathbf{h}\left( \mathbf{x}\right) \]\n\nwhich yields\n\n\[ c{\mathbf{A}}^{T}\mathbf{A}\mathbf{d} = {\mathbf{A}}^{T}\widehat{\mathbf{\lambda }} - {\mathbf{{cA}}}^{T}\mathbf{h}\left( \mathbf{x}\right) \]\n... | Yes |
In solving (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{y} ... | 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\n\n\[ {\mathbf{d}}_{\mathbf{x}}{}^{T}{\mathbf{d}}_{\mathbf{s}} = {\left( {\mathbf{d}}_{\mathbf{x}},{\mathbf{d}}_{\mathbf{y}}\right) }^{T}\left( {{\mathbf{d}}_{\... | No |
To see that the problem (SDP) (that is, (56)) generalizes linear programing define \( \mathbf{C} = \operatorname{diag}\left\lbrack {{c}_{1},{c}_{2},\ldots ,{c}_{n}}\right\rbrack \), and let \( {\mathbf{A}}_{i} = \) \( \operatorname{diag}\left\lbrack {{a}_{i1},{a}_{i2},\ldots ,{a}_{in}}\right\rbrack \) for \( i = 1,2,\l... | \[ \text{minimize}{\mathbf{c}}^{T}\mathbf{x} \] \[ \text{subject to}\mathbf{{Ax}} = \mathbf{b} \] \[ \mathbf{x} \geq \mathbf{0}. \] | Yes |
Example 4 (The dual of binary quadratic optimization). Consider the semidefinite relaxation (57) for the binary quadratic problem. It's dual is\n\n\[ \n\\text{maximize}\\mathop{\\sum }\\limits_{{i = 1}}^{{n = 1}}{y}_{i} \n\]\n\n\[ \n\\text{subject to}\\mathop{\\sum }\\limits_{{j = i}}^{{n + 1}}{y}_{i}{\\mathbf{I}}_{i} ... | Note that\n\n\[ \n\\left\\lbrack \\begin{array}{ll} \\mathbf{Q} & \\mathbf{c} \\\\ {\\mathbf{c}}^{T} & 0 \\end{array}\\right\\rbrack - \\mathop{\\sum }\\limits_{{i = 1}}^{{n + 1}}{y}_{i}{\\mathbf{I}}_{i} \n\]\n\nis the Hessian matrix of the Lagrange function of the quadratic problem; see Chapter 11. | Yes |
The dual of the linear program (58) is\n\n\\[ \n\\text{maximum}\\;{\\mathbf{b}}^{T}\\mathbf{y} \n\\]\n\n\\[ \n\\text{subject to}\\;{\\mathbf{A}}^{T}\\mathbf{y} \\leq \\mathbf{c}\\text{.} \n\\] | It can be written as\n\n\\[ \n\\text{maximum}\\;{\\mathbf{b}}^{T}\\mathbf{y} \n\\]\n\n\\[ \n\\text{subject to}\\operatorname{diag}\\left( {\\mathbf{c} - {\\mathbf{A}}^{T}\\mathbf{y}}\\right) \\succcurlyeq \\mathbf{0} \n\\]\n\nwhere as usual \\( \\operatorname{diag}\\left( \\mathbf{c}\\right) \\) denotes the diagonal ma... | Yes |
Quadratic constraints can be transformed to linear semidefinite form by using the concept of Schur complements. To introduce this concept, consider the quadratic problem\n\n\\[ \n{\\text{ minimize }}_{\\mathbf{x}}\\;{\\mathbf{x}}^{T}\\mathbf{A}\\mathbf{x} + 2{\\mathbf{y}}^{T}{\\mathbf{B}}^{T}\\mathbf{x} + {\\mathbf{y}}... | Now consider a general quadratic constraint of the form\n\n\\[ \n{\\mathbf{x}}^{T}{\\mathbf{B}}^{T}\\mathbf{B}\\mathbf{x} - {\\mathbf{c}}^{T}\\mathbf{x} - d \\geq 0. \n\\]\n\n(61)\n\nThis is equivalent to\n\n\\[ \n\\left\\lbrack \\begin{matrix} \\mathbf{I} & \\mathbf{{Bx}} \\\\ {\\mathbf{x}}^{T}{\\mathbf{B}}^{T} & {\\m... | Yes |
Example 8 The following semidefinite program has a duality gap: | \[ \mathbf{C} = \left\lbrack \begin{array}{lll} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right\rbrack ,{\mathbf{A}}_{1} = \left\lbrack \begin{array}{lll} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array}\right\rbrack ,{\mathbf{A}}_{2} = \left\lbrack \begin{array}{rrr} 0 & - 1 & 0 \\ - 1 & 0 & 0 \\ 0 & 0 & 2 \end{a... | Yes |
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\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}}{\... | Proof. Consider the set \( G = C - B \) . It is easily shown that \( G \) is convex and that 0 is not a relative interior point of \( G \) . Hence, Theorem 1 or Theorem 2 applies and gives the appropriate hyperplane. | No |
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 |
If a linear program is given in standard form except that one or more of the unknown variables is not required to be nonnegative, the problem can be transformed to standard form by either of two simple techniques. | To describe the first technique, suppose in (2.1), for example, that the restriction \( {x}_{1} \geq 0 \) is not present and hence \( {x}_{1} \) is free to take on either positive or negative values. We then write\n\n\[ {x}_{1} = {u}_{1} - {v}_{1} \]\n\n(2.3)\n\nwhere we require \( {u}_{1} \geq 0 \) and \( {v}_{1} \geq... | Yes |
A second approach for converting to standard form when \( {x}_{1} \) is unconstrained in sign is to eliminate \( {x}_{1} \) together with one of the constraint equations. Take any one of the \( m \) equations in (2.1) which has a nonzero coefficient for \( {x}_{1} \) . Say, for example,\n\n\[ \n{a}_{i1}{x}_{1} + {a}_{i... | If this expression is substituted for \( {x}_{1} \) everywhere in (2.1), we are led to a new problem of exactly the same form but expressed in terms of the variables \( {x}_{2},{x}_{3},\ldots ,{x}_{n} \) only. Furthermore, the \( i \) th equation, used to determine \( {x}_{1} \), is now identically zero and it too can ... | Yes |
As a specific instance of the above technique consider the problem\n\n\\[ \n\\text{minimize}{x}_{1} + 3{x}_{2} + 4{x}_{3} \n\\]\n\n\\[ \n\\text{subject to}{x}_{1} + 2{x}_{2} + {x}_{3} = 5 \n\\]\n\n\\[ \n2{x}_{1} + 3{x}_{2} + {x}_{3} = 6 \n\\]\n\n\\[ \n{x}_{2} \\geq 0,\\;{x}_{3} \\geq 0. \n\\] | Since \\( {x}_{1} \\) is free, we solve for it from the first constraint, obtaining\n\n\\[ \n{x}_{1} = 5 - 2{x}_{2} - {x}_{3} \n\\]\n\n(2.5)\n\nSubstituting this into the objective and the second constraint, we obtain the equivalent problem (subtracting five from the objective)\n\n\\[ \n\\text{minimize}{x}_{2} + 3{x}_{... | Yes |
How can we determine the most economical diet that satisfies the basic minimum nutritional requirements for good health? Such a problem might, for example, be faced by the dietitian of a large army. We assume that there are available at the market \( n \) different foods and that the \( j \) th food sells at a price \(... | If we denote by \( {x}_{j} \) the number of units of food \( j \) in the diet, the problem then is to select the \( {x}_{j} \) ’s to minimize the total cost\n\n\[ \n{c}_{1}{x}_{1} + {c}_{2}{x}_{2} + \cdots + {c}_{n}{x}_{n} \n\]\n\nsubject to the nutritional constraints\n\n\[ \n{a}_{i1}{x}_{1} + {a}_{i2}{x}_{2} + \cdots... | Yes |
Quantities \( {a}_{1},{a}_{2},\ldots ,{a}_{m} \), respectively, of a certain product are to be shipped from each of \( m \) locations and received in amounts \( {b}_{1},{b}_{2},\ldots ,{b}_{n} \), respectively, at each of \( n \) destinations. Associated with the shipping of a unit of product from origin \( i \) to des... | To formulate this problem as a linear programming problem, we set up the array shown below:\n\n\n\nThe \( i \) th row in this array defines the variables associated with the \( i \) th origin, while the \( j \) th colu... | Yes |
Example 7 (Combinatorial Auction). Suppose there are \( m \) mutually exclusive potential states and only one of them will be true at maturity. For example, the states may correspond to the winning horse in a race of \( m \) horses, or the value of a stock index, falling within \( m \) intervals. An auction organizer w... | This problem can be expressed alternatively as selecting \( \mathbf{x} \) and scalar \( s \) to \[ \text{maximize}{\pi }^{T}\mathbf{x} - s \] \[ \text{subject to}\mathbf{{Ax}} - \mathbf{1}s \leq \mathbf{0} \] \[ \mathbf{0} \leq \mathbf{x} \leq \mathbf{q} \] where \( \mathbf{1} \) is the vector of all 1’s. Notice that t... | Yes |
Corollary 1. If the convex set \( K \) corresponding to (2.18) is nonempty, it has at least one extreme point. | ## Proof. This follows from the first part of the Fundamental Theorem and the Equivalence Theorem above. | No |
Corollary 3. The constraint set \( K \) corresponding to (2.18) possesses at most a finite number of extreme points. | Proof. There are obviously only a finite number of basic solutions obtained by selecting \( m \) basis vectors from the \( n \) columns of \( \mathbf{A} \) . The extreme points of \( K \) are a subset of these basic solutions. | Yes |
Consider the constraint set in \( {E}^{3} \) defined by\n\n\[ \n{x}_{1} + {x}_{2} + {x}_{3} = 1 \n\]\n\n\[ \n{x}_{1} \geq 0,{x}_{2} \geq 0,{x}_{3} \geq 0\text{.} \n\] | This set is illustrated in Fig. 2.3. It has three extreme points, corresponding to the three basic solutions to \( {x}_{1} + {x}_{2} + {x}_{3} = 1 \) . | No |
Consider the constraint set in \( {E}^{3} \) defined by\n\n\[ \n{x}_{1} + {x}_{2} + {x}_{3} = 1 \]\n\n\[ \n2{x}_{1} + 3{x}_{2}\; = 1 \]\n\n\[ \n{x}_{1} \geq 0,{x}_{2} \geq 0,{x}_{3} \geq 0\text{.} \]\n | This set is illustrated in Fig. 2.4. It has two extreme points, corresponding to the two basic feasible solutions. Note that the system of equations itself has three basic solutions, \( \left( {2, - 1,0}\right) ,\left( {1/2,0,1/2}\right) ,\left( {0,1/3,2/3}\right) \), the first of which is not feasible. | No |
Example 3. Consider the constraint set in \( {E}^{2} \) defined in terms of the inequalities\n\n\[ \n{x}_{1} + \frac{8}{3}{x}_{2} \leq 4 \]\n\n\[ \n{x}_{1} + \;{x}_{2} \leq 2 \]\n\n\[ \n2{x}_{1}\; \leq 3 \]\n\n\[ \n{x}_{1} \geq 0,\;{x}_{2} \geq 0. \]\n\nThis set is illustrated in Fig. 2.5. We see by inspection that thi... | The last example illustrates that even when not expressed in standard form the extreme points of the set defined by the constraints of a linear program correspond to the possible solution points. This can be illustrated more directly by including the objective function in the figure as well. Suppose, for example, that ... | Yes |
Consider the system in canonical form:\n\n\[ \n{x}_{1}\; + {x}_{4} + {x}_{5} - {x}_{6} = 5 \]\n\n\[ \n{x}_{2}\; + 2{x}_{4} - 3{x}_{5} + {x}_{6} = 3 \]\n\n\[ \n{x}_{3} - \;{x}_{4} + 2{x}_{5} - {x}_{6} = - 1. \]\n\nLet us find the basic solution having basic variables \( {x}_{4},{x}_{5},{x}_{6} \) . | We set up the coefficient array below:\n\n\[ \n\begin{matrix} {x}_{1} & {x}_{2} & {x}_{3} & {x}_{4} & {x}_{5} & {x}_{6} & \\ 1 & 0 & 0 & \ddots & 1 & - 1 & 5 \\ 0 & 1 & 0 & 2 & - 3 & 1 & 3 \\ 0 & 0 & 1 & - 1 & 2 & - 1 & - 1 \end{matrix} \]\n\nThe circle indicated is our first pivot element and corresponds to the replac... | Yes |
Suppose we wish to solve the simultaneous equations\n\n\\[ \n{x}_{1} + {x}_{2} - {x}_{3} = 5 \n\\]\n\n\\[ \n2{x}_{1} - 3{x}_{2} + {x}_{3} = 3 \n\\]\n\n\\[ \n- {x}_{1} + 2{x}_{2} - {x}_{3} = - 1\\text{.} \n\\] | To obtain an original basis, we form the augmented tableau\n\n\\[ \n\\begin{matrix} {\\mathbf{e}}_{1} & {\\mathbf{e}}_{2} & {\\mathbf{e}}_{3} & {\\mathbf{a}}_{1} & {\\mathbf{a}}_{2} & {\\mathbf{a}}_{3} & \\mathbf{b} \\ \\ 1 & 0 & 0 & 1 & 1 & - 1 & 5 \\ \\ 0 & 1 & 0 & 2 & - 3 & 1 & 3 \\ \\ 0 & 0 & 1 & - 1 & 2 & - 1 & - ... | No |
Consider the system\n\n\[ \n\begin{matrix} {\mathbf{a}}_{1} & {\mathbf{a}}_{2} & {\mathbf{a}}_{3} & {\mathbf{a}}_{4} & {\mathbf{a}}_{5} & {\mathbf{a}}_{6} & \mathbf{b} \\ 1 & 0 & 0 & 2 & 4 & 6 & 4 \\ 0 & 1 & 0 & 1 & 2 & 3 & 3 \\ 0 & 0 & 1 & - 1 & 2 & 1 & 1 \end{matrix} \n\]\n\nwhich has basis \( {\mathbf{a}}_{1},{\math... | \[ 4/2 = 2,3/1 = 3,1/ - 1 = - 1 \]\n\nand select the smallest nonnegative one. This gives 2 as the pivot element. The new tableau is\n\n\[ \n\begin{matrix} {\mathbf{a}}_{1} & {\mathbf{a}}_{2} & {\mathbf{a}}_{3} & {\mathbf{a}}_{4} & {\mathbf{a}}_{5} & {\mathbf{a}}_{6} & \mathbf{b} \\ 1/2 & 0 & 0 & 1 & 2 & 3 & 2 \\ - 1/2... | Yes |
Maximize \( 3{x}_{1} + {x}_{2} + 3{x}_{3} \) subject to\n\n\[ 2{x}_{1} + {x}_{2} + {x}_{3} \leq 2 \]\n\n\[ {x}_{1} + 2{x}_{2} + 3{x}_{3} \leq 5 \]\n\n\[ 2{x}_{1} + 2{x}_{2} + {x}_{3} \leq 6 \]\n\n\[ {x}_{1} \geq 0,{x}_{2} \geq 0,{x}_{3} \geq 0\text{.} \]\n | To transform the problem into standard form so that the simplex procedure can be applied, we change the maximization to minimization by multiplying the objective function by minus one, and introduce three nonnegative slack variables \( {x}_{4},{x}_{5},{x}_{6} \) . We then have the initial tableau\n\n![3fdd0720-d576-4dc... | Yes |
Find a basic feasible solution to\n\n\[ \n2{x}_{1} + {x}_{2} + 2{x}_{3} = 4 \n\]\n\n\[ \n3{x}_{1} + 3{x}_{2} + {x}_{3} = 3 \n\]\n\n\[ \n{x}_{1} \geq 0,{x}_{2} \geq 0,{x}_{3} \geq 0\text{.} \n\] | We introduce artificial variables \( {x}_{4} \geq 0,{x}_{5} \geq 0 \) and an objective function \( {x}_{4} + {x}_{5} \) . The initial tableau is\n\n\n\nA basic feasible solution to the expanded system is given by the a... | Yes |
Consider the problem\n\n\\[ \n\\text{minimize}4{x}_{1} + {x}_{2} + {x}_{3} \n\\]\n\n\\[ \n\\text{subject to}2{x}_{1} + {x}_{2} + 2{x}_{3} = 4 \n\\]\n\n\\[ \n3{x}_{1} + 3{x}_{2} + {x}_{3} = 3 \n\\]\n\n\\[ \n{x}_{1} \\geq 0,\\;{x}_{2} \\geq 0,\\;{x}_{3} \\geq 0. \n\\] | There is no basic feasible solution apparent, so we use the two-phase method. The first phase was done in Example 1 for these constraints, so we shall not repeat it here. We give only the final tableau with the columns corresponding to the artificial variables deleted, since they are not used in phase II. We use the ne... | Yes |
Example 3 (A Free Variable Problem).\n\n\\[ \n\\text{minimize} - 2{x}_{1} + 4{x}_{2} + 7{x}_{3} + {x}_{4} + 5{x}_{5} \n\\]\n\n\\[ \n\\text{subject to} - {x}_{1} + {x}_{2} + 2{x}_{3} + {x}_{4} + 2{x}_{5} = 7 \n\\]\n\n\\[ \n- {x}_{1} + 2{x}_{2} + 3{x}_{3} + {x}_{4} + {x}_{5} = 6 \n\\]\n\n\\[ \n- {x}_{1} + {x}_{2} + {x}_{... | Since \\( {x}_{1} \\) is free, it can be eliminated, as described in Chap. 2, by solving for \\( {x}_{1} \\) in terms of the other variables from the first equation and substituting everywhere else. This can all be done with the simplex tableau as follows:  be equal to the total supply (which is also equal to the total demand). Then let \( {x}_{ij} = {a}_{i}{b}_{j}/S \) for \( i = 1,2,\ldots, m;j = 1,2,\ldots, n \... | Yes |
We now establish the most important structural property of the transportation problem: the triangularity of all bases. This property simplifies the process of solution of a system of equations whose coefficient matrix corresponds to a basis, and thus leads to efficient implementation of the simplex method. | Definition. A nonsingular square matrix \( \mathbf{M} \) is said to be triangular if by a permutation of its rows and columns it can be put in the form of a lower triangular matrix.\n\nThere is a simple and useful procedure for determining whether a given matrix \( \mathbf{M} \) is triangular:\n\nStep 1. Find a row wit... | Yes |
Basis Triangularity Theorem. Every basis of the transportation problem is triangular. | Proof. Refer to the system of constraints (3.36). Let us change the sign of the top half of the system; then the coefficient matrix of the system consists of entries that are either +1, -1 , or 0 . Following the result of the theorem in Sect. 3.7, delete any one of the equations to eliminate the redundancy. From the re... | Yes |
Corollary. If the row and column sums of a transportation problem are integers, then the basic variables in any basic solution are integers. | The importance of triangularity is, of course, the associated method of back substitution for the solution of a triangular system of equations, as discussed in Appendix C. Moreover, since any basis matrix is triangular and all nonzero elements are equal to one (or minus one if the signs of some equations are changed), ... | Yes |
Theorem. Let \( \mathbf{B} \) be a basis from \( \mathbf{A} \) (ignoring one row), and let \( \mathbf{d} \) be another column. Then the components of the vector \( \mathbf{w} = {\mathbf{B}}^{-1}\mathbf{d} \) are either \( 0, + 1 \), or -1 . | Proof. Let \( \mathbf{w} \) be the solution to the equation \( \mathbf{{Bw}} = \mathbf{d} \) . Then \( \mathbf{w} \) is the representation of \( \mathbf{d} \) in terms of the basis. This equation can be solved by Cramer’s rule as\n\n\[ \n{w}_{k} = \frac{\det {\mathbf{B}}_{k}}{\det \mathbf{B}} \n\]\nwhere \( {\mathbf{B}... | No |
We can now completely solve the problem that was introduced in Example 1 of the first section. The requirements and a first basic feasible solution obtained by the Northwest Corner Rule are shown below. The plus and minus signs indicated on the array should be ignored at this point, since they cannot be computed until ... | The cost coefficients of the problem are shown in the array below, with the circled cells corresponding to the current basic variables. The simplex multipliers, computed by row and column scanning, are shown as well. The relative cost coefficients are found by subtracting \( {u}_{j} + {v}_{j} \) from \( {c}_{ij} \). In... | Yes |
To illustrate the method of dealing with degeneracy, consider a modification of Example 3, with the fourth row sum changed from 60 to 20 and the fourth column sum changed from 80 to 40 . Then the initial basic feasible solution found by the Northwest Corner Rule is degenerate. | An \( \varepsilon \) is placed in the array for the zero-valued basic variable as shown below:\n\n\n\nThe relative cost coefficients will be the same as in Example 3, and hence again \( {x}_{43} \) should be chosen to ... | Yes |
The diet problem, Example 1, Sect. 2.2, was the problem faced by a dietitian trying to select a combination of foods to meet certain nutritional requirements at minimum cost. This problem has the form\n\n\[ \n\\text{minimize}{\\mathbf{c}}^{T}\\mathbf{x} \n\]\n\n\[ \n\\text{subject to}\\mathbf{{Ax}} \\geq \\mathbf{b},\\... | Imagine a pharmaceutical company that produces in pill form each of the nutrients considered important by the dietitian. The pharmaceutical company tries to convince the dietitian to buy pills, and thereby supply the nutrients directly rather than through purchase of various foods. The problem faced by the drug company... | Yes |
Example 2 (Dual of the Transportation Problem). The transportation problem, Example 3, Sect. 2.2, is the problem, faced by a manufacturer, of selecting the pattern of product shipments between several fixed origins and destinations so as to minimize transportation cost while satisfying demand. Referring to (4.6) and (4... | To interpret the dual problem, we imagine an entrepreneur who, feeling that he can ship more efficiently, comes to the manufacturer with the offer to buy his product at the plant sites (origins) and sell it at the warehouses (destinations). The product price that is to be used in these transactions varies from point to... | Yes |
Lemma 1 (Weak Duality Lemma). If \( \mathbf{x} \) and \( \mathbf{y} \) are feasible for (4.3) and (4.4), respectively, then \( {\mathbf{c}}^{T}\mathbf{x} \geq {\mathbf{y}}^{T}\mathbf{b} \) . | Proof. We have\n\n\[ \n{\mathbf{y}}^{T}\mathbf{b} = {\mathbf{y}}^{T}\mathbf{A}\mathbf{x} \leq {\mathbf{c}}^{T}\mathbf{x} \n\] \n\nthe last inequality being valid since \( \mathbf{x} \geq \mathbf{0} \) and \( {\mathbf{y}}^{T}\mathbf{A} \leq {\mathbf{c}}^{T} \) . | Yes |
Theorem 1. The ellipsoid \( {E}_{k + 1} = E\left( {{\mathbf{y}}_{k + 1},{\mathbf{B}}_{k + 1}^{-1}}\right) \) defined as above is the ellipsoid of least volume containing \( \left( {1/2}\right) {E}_{k} \) . Moreover, | Proof. We shall not prove the statement about the new ellipsoid being of least volume, since that is not necessary for the results that follow. To prove the remainder of the statement, we have\n\n\[ \frac{\operatorname{vol}\left( {E}_{k + 1}\right) }{\operatorname{vol}\left( {E}_{k}\right) } = \frac{\det \left( {\mathb... | No |
Consider the problem of maximizing \( {x}_{1} \) within the unit square \( \mathcal{S} = {\left\lbrack 0,1\right\rbrack }^{2} \). The problem is formulated as\n\n\[ \min \; - {x}_{1} \]\n\n\[ \text{s.t.}\;{x}_{1} + {x}_{3} = 1 \]\n\n\[ {x}_{2} + {x}_{4} = 1 \]\n\n\[ {x}_{1} \geq 0,{x}_{2} \geq 0,{x}_{3} \geq 0,{x}_{4} ... | Here \( {x}_{3} \) and \( {x}_{4} \) are slack variables for the original problem to put it in standard form. The optimality conditions for \( \mathbf{x}\left( \mu \right) \) consist of the original two linear constraint equations and the four equations\n\n\[ {y}_{1} + {s}_{1} = - 1,{y}_{2} + {s}_{2} = 0,{y}_{1} + {s}_... | No |
Consider the dual of Example 2. This is\n\n\\[ \max \\;{y}_{1} + {y}_{2} \\]\n\n\\[ \text{subject to}{y}_{1} \leq - 1 \\]\n\n\\[ {y}_{2} \leq 0\\text{.} \\] | The solution to the dual barrier problem is easily found from the solution of the primal barrier\nproblem to be\n\n\\[ {y}_{1}\left( \mu \right) = - 1 - \mu /{x}_{1}\left( \mu \right) ,{y}_{2} = - {2\mu }. \\]\n\nAs \\( \mu \\rightarrow 0 \\), we have \\( {y}_{1} \\rightarrow - 1,{y}_{2} \\rightarrow 0 \\), which is th... | Yes |
Theorem 2. The algorithm above terminates in at most \( O\left( {\rho \log \left( {n/\varepsilon }\right) }\right) \) iterations with\n\n\[ \frac{{\left( {\mathbf{s}}_{k}\right) }^{T}{\mathbf{x}}_{k}}{{\left( {\mathbf{s}}_{0}\right) }^{T}{\mathbf{x}}_{0}} \leq \varepsilon \] | Proof. Note that after \( k \) iterations, we have from (5.16)\n\n\[ {\psi }_{n + \rho }\left( {{\mathbf{x}}_{k},{\mathbf{s}}_{k}}\right) \leq {\psi }_{n + \rho }\left( {{\mathbf{x}}_{0},{\mathbf{s}}_{0}}\right) - k \cdot \delta \leq \rho \log \left( {{\left( {\mathbf{s}}_{0}\right) }^{T}{\mathbf{x}}_{0}}\right) + n\lo... | Yes |
Theorem 2. Let \( \left( {{\mathbf{y}}^{ * },{\mathbf{x}}^{ * },{\tau }^{ * },{\theta }^{ * } = 0,{\mathbf{s}}^{ * },{\kappa }^{ * }}\right) \) be a strictly-self complementary solution for (HSDP).\n\n(i) (LP) has a solution (feasible and bounded) if and only if \( {\tau }^{ * } > 0 \) . In this case, \( {\mathbf{x}}^{... | Proof. We prove the second statement. We first assume that one of (LP) and (LD) is infeasible, say (LD) is infeasible. Then there is some certificate \( \overline{\mathbf{x}} \geq \mathbf{0} \) such that \( \mathbf{A}\overline{\mathbf{x}} = \mathbf{0} \) and \( {\mathbf{c}}^{T}\overline{\mathbf{x}} = - 1 \) . Let \( \l... | Yes |
The followings are all (closed) convex cones. | - The \( n \) -dimensional non-negative orthant, \( {E}_{ + }^{n} = \left\{ {\mathbf{x} \in {E}^{n} : \mathbf{x} \geq \mathbf{0}}\right\} \), is a convex cone.\n- The set of all \( n \) -dimensional symmetric positive semidefinite matrices, denoted by \( {\mathcal{S}}_{ + }^{n} \), is a convex cone, called the positive... | Yes |
Example 2 (Binary Quadratic Optimization). Consider a binary quadratic maximization problem\n\n\[ \n\\text{maximize}{\\mathbf{x}}^{T}\\mathbf{Q}\\mathbf{x} + 2{\\mathbf{c}}^{T}\\mathbf{x} \n\]\n\n\[ \n\\text{subject to}{x}_{j} = \\{ 1, - 1\\} \\text{, for all}j = 1,\\ldots, n\\text{,}\n\]\n\nwhich is a difficult noncon... | Since \( \\left\\lbrack \\begin{matrix} \\mathbf{x} \\ {x}_{n + 1} \\end{matrix}\\right\\rbrack {\\left\\lbrack \\begin{matrix} \\mathbf{x} \\ {x}_{n + 1} \\end{matrix}\\right\\rbrack }^{T} \) forms a positive-semidefinite matrix (with rank equal to 1),\n\na semidefinite relaxation of the problem is defined as\n\n\[ \n... | Yes |
The interior of the second order cone is the set of \( \\left\\{ {\\left( {u;\\mathbf{x}}\\right) \\in {E}^{n + 1} : u > {\\left| \\mathbf{x}\\right| }_{2}}\\right\\} \) . | We give a sketch of the proof for the second order cone, i.e., \( p = 2 \) . Let \( \\left( {\\bar{u};\\overline{\\mathbf{x}}}\\right) \\neq \\mathbf{0} \) be any second-order cone point but \( \\bar{u} = \\left| \\overline{\\mathbf{x}}\\right| \) . Then, we can choose a dual cone (also the second-order cone) point \( ... | Yes |
Theorem 2 (Farkas' Lemma for CLP). We have\n\n- Consider set\n\n\[ \n{\mathcal{F}}_{p} \mathrel{\text{:=}} \{ \mathbf{X} : \mathcal{A}\mathbf{X} = \mathbf{b},\mathbf{X} \in K\} .\n\]\n\nSuppose that there exists a vector \( \overset{ \circ }{\mathbf{y}} \) such that \( - {\overset{ \circ }{\mathbf{y}}}^{T}\mathcal{A} \... | Proof. We prove the first statement of the theorem. We prove the first part. It is clear that \( C \) is a convex set. To prove that \( C \) is a closed set, we need to show that if \( {\mathbf{y}}^{k} \mathrel{\text{:=}} \mathcal{A}{\mathbf{X}}^{k} \in {E}^{m} \) for \( {\mathbf{X}}^{k} \in K, k = 1,\ldots \), converg... | Yes |
Consider the semidefinite relaxation (6.3) for the binary quadratic maximization problem. It's dual is\n\n\\[ \n\\text{ minimize }\\mathop{\\sum }\\limits_{{j = 1}}^{{n + 1}}{y}_{j}\n\\]\n\n\\[ \n\\text{subject to}\\mathop{\\sum }\\limits_{{j = 1}}^{{n + 1}}{y}_{j}{\\mathbf{I}}_{j} - \\mathbf{S} = \\left\\lbrack \\begi... | Note that\n\n\\[ \n\\mathop{\\sum }\\limits_{{j = 1}}^{{n + 1}}{y}_{j}{\\mathbf{I}}_{j} - \\left\\lbrack \\begin{matrix} \\mathbf{Q} & \\mathbf{c} \\\\ {\\mathbf{c}}^{T} & 0 \\end{matrix}\\right\\rbrack\n\\]\n\nis exactly the Hessian matrix of the Lagrange function of the quadratic maximization problem; see Chap. 11. T... | No |
Example 4 (Euclidean Facility Location). This problem is to determine the location of a facility serving \( n \) clients placed in a Euclidean space, whose known locations are denoted by \( {\mathbf{a}}_{j} \in {E}^{d}, j = 1,\ldots, n \) . The location of the facility would minimize\nthe sum of the Euclidean distances... | The problem can be reformulated as\n\n\[ \n\text{minimize}\;\mathop{\sum }\limits_{{j = 1}}^{n}{\delta }_{j} \n\]\n\n\[ \n\text{subject to}{\mathbf{s}}_{j} + \mathbf{f} = {\mathbf{a}}_{j},\forall j = 1,\ldots, n\text{,} \n\]\n\n\[ \n\left| {\mathbf{s}}_{j}\right| \leq {\delta }_{j},\;\forall j = 1,\ldots, n. \n\]\n\nTh... | Yes |
The following semidefinite program has a duality gap: | The primal minimal objective value is 0 achieved by\n\n\[ \mathbf{X} = \left\lbrack \begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{array}\right\rbrack \]\n\nand the dual maximal objective value is -2 achieved by \( \mathbf{y} = \left\lbrack {0, - 1}\right\rbrack \) ; so the duality gap is 2 . | Yes |
Proposition 1 (First-Order Necessary Conditions). Let \( \Omega \) be a subset of \( {E}^{n} \) and let \( f \in {C}^{1} \) be a function on \( \Omega \) . If \( {\mathbf{x}}^{ * } \) is a relative minimum point of \( f \) over \( \Omega \), then for any \( \mathbf{d} \in {E}^{n} \) that is a feasible direction at \( {... | Proof. For any \( \alpha ,0 \leq \alpha \leq \bar{\alpha } \), the point \( \mathbf{x}\left( \alpha \right) = {\mathbf{x}}^{ * } + \alpha \mathbf{d} \in \Omega \) . For \( 0 \leq \alpha \leq \bar{\alpha } \) define the function \( g\left( \alpha \right) = f\left( {\mathbf{x}\left( \alpha \right) }\right) \) . Then \( g... | Yes |
Consider the problem\n\n\[ \text{minimize}f\left( {{x}_{1},{x}_{2}}\right) = {x}_{1}^{2} - {x}_{1}{x}_{2} + {x}_{2}^{2} - 3{x}_{2}\text{.} \]\n\nThere are no constraints, so \( \Omega = {E}^{2} \) . | Setting the partial derivatives of \( f \) equal to zero yields the two equations\n\n\[ 2{x}_{1} - {x}_{2} = 0 \]\n\n\[ - {x}_{1} + 2{x}_{2} = 3\text{. } \]\n\nThese have the unique solution \( {x}_{1} = 1,{x}_{2} = 2 \), which is a global minimum point of \( f \) . | Yes |
Consider the problem\n\n\\[ \n\\text{minimize}f\\left( {{x}_{1},{x}_{2}}\\right) = {x}_{1}^{2} - {x}_{1} + {x}_{2} + {x}_{1}{x}_{2} \n\\]\n\n\\[ \n\\text{subject to}\;{x}_{1} \geq 0,\;{x}_{2} \geq 0\\text{.} \n\\] | This problem has a global minimum at \( {x}_{1} = \\frac{1}{2},{x}_{2} = 0 \) . At this point\n\n\\[ \n\\frac{\\partial f}{\\partial {x}_{1}} = 2{x}_{1} - 1 + {x}_{2} = 0 \n\\]\n\n\\[ \n\\frac{\\partial f}{\\partial {x}_{2}} = 1 + {x}_{1} = \\frac{3}{2} \n\\]\n\nThus, the partial derivatives do not both vanish at the s... | No |
Recall the classification problem where we have vectors \( {\mathbf{a}}_{i} \in {E}^{d} \) for \( i = 1,2,\ldots ,{n}_{1} \) in a class, and vectors \( {\mathbf{b}}_{j} \in {E}^{d} \) for \( j = \) \( 1,2,\ldots ,{n}_{2} \) not. Then we wish to find \( \mathbf{y} \in {E}^{d} \) and a number \( \beta \) such that\n\n\[ ... | The problem can be cast as a unconstrained optimization problem, called the max-likelihood,\n\n\[ \n{\operatorname{maximize}}_{\mathbf{y},\beta }\left( {\mathop{\prod }\limits_{i}\frac{\exp \left( {{\mathbf{a}}_{i}^{T}\mathbf{y} + \beta }\right) }{1 + \exp \left( {{\mathbf{a}}_{i}^{T}\mathbf{y} + \beta }\right) }}\righ... | Yes |
A common problem in economic theory is the determination of the best way to combine various inputs in order to maximize a utility function \( f\left( {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right) \) (in the monetary unit) of the amounts \( {x}_{j} \) of the inputs, \( i = 1,2,\ldots, n \) . The unit prices of the inputs are... | The first-order necessary conditions are that the partial derivatives with respect to the \( {x}_{i} \) ’s each vanish. This leads directly to the \( n \) equations\n\n\[ \frac{\partial f}{\partial {x}_{i}}\left( {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right) = {p}_{i}, i = 1,2,\ldots, n. \]\n\nThese equations can be interpr... | Yes |
A common use of optimization is for the purpose of function approximation. Suppose, for example, that through an experiment the value of a function \( g \) is observed at \( m \) points, \( {x}_{1},{x}_{2},\ldots ,{x}_{m} \) . Thus, values \( g\left( {x}_{1}\right), g\left( {x}_{2}\right) ,\ldots, g\left( {x}_{m}\right... | This is a quadratic expression in the coefficients \( \mathbf{a} \) . To find a compact representation for this objective we define \( {q}_{ij} = \mathop{\sum }\limits_{{k = 1}}^{m}{\left( {x}_{k}\right) }^{i + j},{b}_{j} = \mathop{\sum }\limits_{{k = 1}}^{m}g\left( {x}_{k}\right) {\left( {x}_{k}\right) }^{j} \) and \(... | Yes |
Example 4 (Selection Problem). It is often necessary to select an assortment of factors to meet a given set of requirements. An example is the problem faced by an electric utility when selecting its power-generating facilities. The level of power that the company must supply varies by time of the day, by day of the wee... | Assuming that the solution is interior to the constraints, by setting the partial derivatives equal to zero, we obtain the two equations\n\n\[ {b}_{1} + \left( {{c}_{1} - {c}_{2}}\right) h\left( {x}_{1}\right) + \left( {{c}_{2} - {c}_{3}}\right) h\left( {{x}_{1} + {x}_{2}}\right) = 0 \]\n\n\[ {b}_{2} + \left( {{c}_{2} ... | Yes |
Proposition 1 (Second-Order Necessary Conditions). Let \( \Omega \) be a subset of \( {E}^{n} \) and let \( f \in {C}^{2} \) be a function on \( \Omega \). If \( {\mathbf{x}}^{ * } \) is a relative minimum point of \( f \) over \( \Omega \), then for any \( \mathbf{d} \in {E}^{n} \) that is a feasible direction at \( {... | Proof. The first condition is just Proposition 1, and the second applies only if \( \nabla f\left( {\mathbf{x}}^{ * }\right) \mathbf{d} = 0 \) . In this case, introducing \( \mathbf{x}\left( \alpha \right) = {\mathbf{x}}^{ * } + \alpha \mathbf{d} \) and \( g\left( \alpha \right) = f\left( {\mathbf{x}\left( \alpha \righ... | Yes |
For the same problem as Example 2 of Sect. 7.1, we have for \( \mathbf{d} = \) \( \left( {{d}_{1},{d}_{2}}\right) \)\n\n\[ \nabla f\left( {\mathbf{x}}^{ * }\right) \mathbf{d} = \frac{3}{2}{d}_{2} \] | Thus condition (ii) of Proposition 1 applies only if \( {d}_{2} = 0 \) . In that case we have \( {\mathbf{d}}^{T}{\nabla }^{2}f\left( {\mathbf{x}}^{ * }\right) \mathbf{d} = 2{d}_{1}^{2} \geq 0 \), so condition (ii) is satisfied. | No |
Proposition 2 (Second-Order Necessary Conditions-Unconstrained Case). Let \( {\mathbf{x}}^{ * }{be} \) an interior point of the set \( \Omega \), and suppose \( {\mathbf{x}}^{ * } \) is a relative minimum point over \( \Omega \) of the function \( f \in {C}^{2} \) . Then\n\n\[ \text{i)}\nabla f\left( {\mathbf{x}}^{ * }... | For notational simplicity we often denote \( {\mathbf{\nabla }}^{2}f\left( \mathbf{x}\right) \), the \( n \times n \) matrix of the second partial derivatives of \( f \), the Hessian of \( f \), by the alternative notation \( \mathbf{F}\left( \mathbf{x}\right) \) . Condition (ii) is equivalent to stating that the matri... | Yes |
Consider the problem\n\n\\[ \n\\text{minimize}f\\left( {{x}_{1},{x}_{2}}\\right) = {x}_{1}^{3} - {x}_{1}^{2}{x}_{2} + 2{x}_{2}^{2} \n\\]\n\n\\[ \n\\text{subject to}{x}_{1} \\geq 0,\\;{x}_{2} \\geq 0\\text{.} \n\\] | If we assume that the solution is in the interior of the feasible set, that is, if \\( {x}_{1} > 0,{x}_{2} > 0 \\), then the first-order necessary conditions are\n\n\\[ \n3{x}_{1}^{2} - 2{x}_{1}{x}_{2} = 0,\\; - {x}_{1}^{2} + 4{x}_{2} = 0.\n\\]\n\nThere is a solution to these at \\( {x}_{1} = {x}_{2} = 0 \\) which is a... | Yes |
Proposition 3 (Second-Order Sufficient Conditions—Unconstrained Case). Let \( f \in {C}^{2} \) be function defined on a region in which the point \( {\mathbf{x}}^{ * } \) is an interior point. Suppose in addition that\n\n\[ \n\text{i)}\nabla f\left( {\mathrm{x}}^{ * }\right) = \mathbf{0} \n\]\n\n(7.7)\n\n\[ \n\text{ii)... | Proof. Since \( \mathbf{F}\left( {\mathbf{x}}^{ * }\right) \) is positive definite, there is an \( a > 0 \) such that for all \( \mathbf{d},{\mathbf{d}}^{T}\mathbf{F}\left( {\mathbf{x}}^{ * }\right) \) \( \mathbf{d} \geq a{\left| \mathbf{d}\right| }^{2} \) . Thus by the Taylor’s Theorem (with remainder)\n\n\[ \nf\left(... | Yes |
Proposition 1. Let \( {f}_{1} \) and \( {f}_{2} \) be convex functions on the convex set \( \Omega \) . Then the function \( {f}_{1} + {f}_{2} \) is convex on \( \Omega \) . | Proof. Let \( {\mathbf{x}}_{1},{\mathbf{x}}_{2} \in \Omega \), and \( 0 < \alpha < 1 \) . Then\n\n\[ \n{f}_{1}\left( {\alpha {\mathbf{x}}_{1} + \left( {1 - \alpha }\right) {\mathbf{x}}_{2}}\right) + {f}_{2}\left( {\alpha {\mathbf{x}}_{1}}\right) + \left( {1 - \alpha }\right) {\mathbf{x}}_{2}) \n\]\n\n\[ \n\leq \alpha \... | No |
Proposition 2. Let \( f \) be a convex function over the convex set \( \Omega \) . Then the function af is convex for any \( a \geq 0 \) . | Proof. Immediate. | No |
Proposition 3. Let \( f \) be a convex function on a convex set \( \Omega \) . The set \( {\Gamma }_{c} = \{ \mathbf{x} : \mathbf{x} \in \) \( \Omega, f\left( \mathbf{x}\right) \leq c\} \) is convex for every real number \( c \) . | Proof. Let \( {\mathbf{x}}_{1},{\mathbf{x}}_{2} \in {\Gamma }_{c} \) . Then \( f\left( {\mathbf{x}}_{1}\right) \leq c, f\left( {\mathbf{x}}_{2}\right) \leq c \) and for \( 0 < \alpha < 1 \) ,\n\n\[ f\left( {\alpha {\mathbf{x}}_{1} + \left( {1 - \alpha }\right) {\mathbf{x}}_{2}}\right) \leq {\alpha f}\left( {\mathbf{x}}... | Yes |
Proposition 4. Let \( f \in {C}^{1} \) . Then \( f \) is convex over a convex set \( \Omega \) if and only if\n\n\[ f\left( \mathbf{y}\right) \geq f\left( \mathbf{x}\right) + \nabla f\left( \mathbf{x}\right) \left( {\mathbf{y} - \mathbf{x}}\right) \]\n\n(7.9)\n\nfor all \( \mathbf{x},\mathbf{y} \in \mathbf{\Omega } \) ... | Proof. First suppose \( f \) is convex. Then for all \( \alpha ,0 \leq \alpha \leq 1 \) ,\n\n\[ f\left( {\alpha \mathbf{y} + \left( {1 - \alpha }\right) \mathbf{x}}\right) \leq {\alpha f}\left( \mathbf{y}\right) + \left( {1 - \alpha }\right) f\left( \mathbf{x}\right) .\n\nThus for \( 0 < \alpha \leq 1 \)\n\n\[ \frac{f\... | Yes |
Proposition 5. Let \( f \in {C}^{2} \) . Then \( f \) is convex over a convex set \( \Omega \) containing an interior point if and only if the Hessian matrix \( \mathbf{F} \) of \( f \) is positive semidefinite throughout \( \Omega \) . | Proof. By Taylor's theorem we have\n\n\[ f\left( \mathbf{y}\right) = f\left( \mathbf{x}\right) = \mathbf{\nabla }f\left( \mathbf{x}\right) \left( {\mathbf{y} - \mathbf{x}}\right) + \frac{1}{2}{\left( \mathbf{y} - \mathbf{x}\right) }^{T}\mathbf{F}\left( {\mathbf{x} + \alpha \left( {\mathbf{y} - \mathbf{x}}\right) }\righ... | Yes |
Theorem 1. Let \( f \) be a convex function defined on the convex set \( \Omega \) . Then the set \( \Gamma \) where \( f \) achieves its minimum is convex, and any relative minimum of \( f \) is a global minimum. | Proof. If \( f \) has no relative minima the theorem is valid by default. Assume now that \( {c}_{0} \) is the minimum of \( f \) . Then clearly \( \Gamma = \left\{ {\mathbf{x} : f\left( \mathbf{x}\right) \leq {c}_{0},\mathbf{x} \in \Omega }\right\} \) and this is convex by Proposition 3 of the last section.\n\nSuppose... | Yes |
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