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Theorem 1.1 A convex cone \( C \) is polyhedral if and only if it is finitely generated, that is, the cone is generated by a finite number of vectors \( {b}_{1},\ldots ,{b}_{m} \) : | \[ C = \operatorname{cone}\left( {{b}_{1},\ldots ,{b}_{m}}\right) \mathrel{\text{:=}} \left\{ {\mathop{\sum }\limits_{{i = 1}}^{m}{b}_{i}{y}_{i} : {y}_{i} \geq 0, i = 1,\ldots, m}\right\} . \] | Yes |
Example 1.5 The following polyhedral cone of \( {\mathcal{R}}^{2} \) in Figure 1.3 can be represented as either\n\n\[ C = \left\{ {\left( \begin{array}{l} 1 \\ 2 \end{array}\right) {y}_{1} + \left( \begin{array}{l} 2 \\ 1 \end{array}\right) {y}_{2} : {y}_{1},{y}_{2} \geq 0}\right\} \]\n\nor\n\[ C = \left\{ {x \in {\mat... | The two vectors in the former representation are two basic directional vectors. | No |
Example 1.6 Let \( C \) be a unit circle centered at the point \( \left( {1;1}\right) \) . That is, \( C = \left\{ {x \in {\mathcal{R}}^{2} : {\left( {x}_{1} - 1\right) }^{2} + {\left( {x}_{2} - 1\right) }^{2} \leq 1}\right\} \) . If \( y = \left( {2;0}\right), a = \left( {-1;1}\right) \) is a separating hyperplane vec... | It is worth noting that these separating hyperplanes are not unique. | No |
Theorem 1.11 (Weak duality theorem) Let \( {\mathcal{F}}_{p} \) and \( {\mathcal{F}}_{d} \) be non-empty. Then,\n\n\[ \n{c}^{T}x \geq {b}^{T}y\;\text{ where }\;x \in {\mathcal{F}}_{p},\left( {y, s}\right) \in {\mathcal{F}}_{d}.\n\] | This theorem shows that a feasible solution to either problem yields a bound on the value of the other problem. We call \( {c}^{T}x - {b}^{T}y \) the duality gap. From this we have important results. | No |
Theorem 1.15 (LP fundamental theorem) Given (LP) and (LD) where \( A \) has full row rank \( m \) , i) if there is a feasible solution, there is a basic feasible solution; ii) if there is an optimal solution, there is an optimal basic solution. | The above theorem reduces the task of solving a linear program to that searching over basic feasible solutions. By expanding upon this result, the simplex method, a finite search procedure, is derived. The simplex method is to proceed from one basic feasible solution (an extreme point of the feasible region) to an adja... | No |
Theorem 1.21 (Karush-Kuhn-Tucker theorem) Let \( {x}^{ * } \) be a relative (local) minimum solution for (NP) and suppose \( {x}^{ * } \) is a regular point for the constraints, meaning that the Jacobian matrix \( \nabla h\left( {x}^{ * }\right) \) and the gradient vectors \( \nabla {g}_{j}\left( {x}^{ * }\right) \) fo... | If \( f \) is convex, \( h \) is affine, and \( g \) is concave, then \( {x}^{ * } \) is optimal if and only if \( {x}^{ * } \) is a KKT point for (NP). Thus, the necessary condition becomes sufficient. | No |
Example 1.14 Consider the bisection method to locate a root of a continuous function \( f\left( x\right) : \mathcal{R} \rightarrow \mathcal{R} \) within interval \( \left\lbrack {0,1}\right\rbrack \), where \( f\left( 0\right) > 0 \) and \( f\left( 1\right) < 0 \) . The method calls the oracle to evaluate \( f\left( {1... | Thus, in \( \log \left( {1/\epsilon }\right) \) steps, we must have an approximate root whose distance to the root is less than \( \epsilon \) . Therefore, the bisection method is a polynomial algorithm. | Yes |
Consider the steepest descent algorithm for solving\n\n\\[ \n\\text{ minimize }\\;q\\left( x\\right) = \\left( {1/2}\\right) {x}^{T}{Qx} + {c}^{T}x, \n\\]\n\nwhere \\( Q \\in {\\mathcal{R}}^{n \\times n} \\) is positive definite. Denote by \\( {x}^{ * } = - {Q}^{-1}c \\in {\\mathcal{R}}^{n} \\) the minimizer of the pro... | It is well known that the algorithm generates a sequence of \\( \\left\\{ {x}^{k}\\right\\} \\) such that\n\n\\[ \n\\frac{{\\begin{Vmatrix}{x}^{k + 1} - {x}^{ * }\\end{Vmatrix}}_{Q}}{{\\begin{Vmatrix}{x}^{k} - {x}^{ * }\\end{Vmatrix}}_{Q}} \\leq \\frac{{\\lambda }_{n} - {\\lambda }_{1}}{{\\lambda }_{n} + {\\lambda }_{1... | Yes |
Let us randomly generate positive definite matrix \( Q \) in Example 1.15. Then, \( \xi \left( Q\right) \) is a random number, and the expected number of iterations to reach \( \frac{{\begin{Vmatrix}{x}^{k + 1} - {x}^{ * }\end{Vmatrix}}_{Q}}{{\begin{Vmatrix}{x}^{k} - {x}^{ * }\end{Vmatrix}}_{Q}} \leq \epsilon \) is bou... | \[ \log \left( {1/\epsilon }\right) E\left( \frac{1}{\log \left( {1/\xi \left( Q\right) }\right) }\right) \] | Yes |
Theorem 2.3 The Volumes of the new ellipsoids satisfy the inequality\n\n\[ V\left( {\underline{E}\left( {\Omega }^{ * }\right) }\right) \leq {0.843V}\left( {\underline{E}\left( \Omega \right) }\right) ,\;\text{ where }\; * = + \text{ or } - . \] | Thus, one can use the max-volume inscribing ellipsoid as an estimate of \( \Omega \) . These ellipsoids will shrink at a constant rate in the central-section method. The volume of \( \underline{E}\left( \Omega \right) \) is also bounded from below by \( {\pi }_{m}{r}^{m} \) and bounded from above by \( {\pi }_{m}{R}^{m... | No |
Theorem 2.5 The Volumes of the new ellipsoids satisfy the inequality\n\n\\[ \nV\\left( {\\bar{E}\\left( {E}^{ * }\\right) }\\right) \\leq \\exp \\left( {-{.5}/\\left( {m + 1}\\right) }\\right) V\\left( E\\right) ,\\;\\text{ where }\\; * = + \\text{ or } - .\n\\] | In algebra, let the ellipsoid\n\n\\[ \nE = \\left\\{ {y \\in {\\mathcal{R}}^{m} : {\\left( y - {y}^{e}\\right) }^{T}Q\\left( {y - {y}^{e}\\right) \\leq 1}\\right\\} .\n\\]\n\nThen, \\( \\bar{E} \\) containing the half ellipsoid \\( \\left\\{ {y \\in E : {a}^{T}y \\leq {a}^{T}{y}^{e}}\\right\\} \\) is given by\n\n\\[ \n... | Yes |
Let \( A = \\left( {1, - 1}\\right) \) and \( c = \\left( {1;1}\\right) \) . Then the set of \( \\Omega \) is the interval \( \\left\\lbrack {-1,1}\\right\\rbrack \) . Let \( {A}^{\\prime } = \\left( {1, - 1, - 1}\\right) \) and \( {c}^{\\prime } = \\left( {1;1;1}\\right) \) . Then the set of \( {\\Omega }^{\\prime } \... | Note that\n\n\[ d\\left( {-1/2,\\Omega }\\right) = \\left( {3/2}\\right) \\left( {1/2}\\right) = 3/4\\;\\text{ and }\\;\\mathcal{B}\\left( {-1/2,\\Omega }\\right) = \\log \\left( {3/4}\\right) ,\]\n\nand\n\n\[ d\\left( {-1/2,{\\Omega }^{\\prime }}\\right) = \\left( {3/}\\right) \\left( {1/2}\\right) \\left( {1/2}\\righ... | Yes |
Theorem 2.6 The analytic center \( {s}^{a} \in {\mathcal{S}}_{\Omega } \) is the center of the unique maximal-volume coordinate-aligned ellipsoid \( {E}_{{s}^{a}} \) inscribing the orthant \( {\mathcal{R}}_{ + }^{n} \) . Its intersection with \( {\mathcal{A}}_{\Omega } \) is contained by polytope \( {\mathcal{S}}_{\Ome... | Proof. The uniqueness of the analytic center is resulted from the fact that the potential function is strictly concave in the interior of the polytope and \( A \) has a full row-rank. Let \( {y}^{a} \) be the analytic center and \( {s}^{a} = c - {A}^{T}{y}^{a} \) , then there is \( {x}^{a} \) such that \( A{x}^{a} = 0 ... | Yes |
Theorem 2.7 Let \( \Omega \) and \( {\Omega }^{ + } \) be defined as the above. Then\n\n\[ \mathcal{B}\left( {\Omega }^{ + }\right) \leq \mathcal{B}\left( \Omega \right) - 1 \]\n\nor\n\n\[ \exp \left( {\mathcal{B}\left( {\Omega }^{ + }\right) }\right) \leq \exp \left( {-1}\right) \exp \left( {\mathcal{B}\left( \Omega \... | Proof. Since \( {y}^{a} \) is the analytic center of \( \Omega \), there exists \( {x}^{a} > 0 \) such that\n\n\[ {X}^{a}\left( {c - {A}^{T}{y}^{a}}\right) = e\;\text{ and }\;A{x}^{a} = 0. \]\n\n(2.3)\n\nRecall that \( e \) is the vector of all ones and \( {X}^{a} \) designates the diagonal matrix of \( {x}^{a} \) . Th... | Yes |
Theorem 2.12 There is a unique minimal-volume coordinate-aligned ellipsoid \( {E}_{{x}^{a}} \), where \( {x}^{a} \in {\mathcal{X}}_{\Omega } \), whose intersection with \( {\mathcal{A}}_{\Omega } \) contains polytope \( {\mathcal{S}}_{\Omega } \) . Moreover, polytope \( {\mathcal{S}}_{\Omega } \) contains the intersect... | Proof. The uniqueness of \( {E}_{{x}^{a}} \) is already demonstrated earlier.\n\nTo prove \( {E}_{{x}^{a}} \) and \( {E}_{{s}^{a}} \) are concentric, we only need to prove that the center of \( {E}_{{x}^{a}} \) is the analytic center of \( {\mathcal{S}}_{\Omega } \) . Recall that \( \bar{s} \) is the center of \( {E}_{... | Yes |
i) \[ {\eta }_{d}\left( s\right) \leq \eta \left( {x, s}\right) \] and \[ {\eta }_{p}\left( x\right) \leq \eta \left( {x, s}\right) \] Conversely, if \( {\eta }_{d}\left( s\right) < 1 \) then there is an \( x\left( s\right) \in {\mathcal{X}}_{\Omega } \) such that \[ \eta \left( {x\left( s\right), s}\right) \leq {\eta ... | Proof. i) Given \( s > 0 \) we can verify that \( {\eta }_{d}\left( s\right) \) is the minimal value and \( x\left( s\right) \) is the minimizer of the least-squares problem \[ \text{minimize}\parallel {Sx} - e\parallel \] \[ \text{s.t.}\;{Ax} = 0\text{.} \] Since \( x \) in \( \eta \left( {x, s}\right) \) is a feasibl... | Yes |
Theorem 3.3 Let \( \\left( {y, s}\\right) \) be an interior point and \( {\\eta }_{d}\\left( s\\right) < 1 \) . Then,\n\n\[ \n{s}^{ + } > 0\\;\\text{ and }\\;{\\eta }_{d}\\left( {s}^{ + }\\right) \\leq {\\eta }_{d}{\\left( s\\right) }^{2}.\n\] | Proof. Note that from the proof of (i) of Theorem 3.2\n\n\[ \n{\\eta }_{d}\\left( {s}^{ + }\\right) = \\begin{Vmatrix}{{S}^{ + }x\\left( {s}^{ + }\\right) - e}\\end{Vmatrix} \\leq \\begin{Vmatrix}{{S}^{ + }x\\left( s\\right) - e}\\end{Vmatrix},\n\]\n\nwhere \( x\\left( s\\right) > 0 \) .\n\nBut\n\n\[ \n{\\begin{Vmatrix... | Yes |
Lemma 3.5 There exists a point \( {x}^{ + } > 0 \) such that\n\n\[ A{x}^{ + } = 0\;\text{ and }\;\begin{Vmatrix}{{X}^{ + }{s}^{ + } - e}\end{Vmatrix} \leq \eta + \left( {1 - \beta }\right) \left( {1 + \eta }\right) < 1, \]\n\nwhere \( {s}^{ + } = {c}^{k + 1} - {A}^{T}{y}^{k} \) . | Proof. Let \( {x}^{ + } = x\left( {s}^{k}\right) > 0 \) with \( {s}^{k} = {c}^{k} - {A}^{T}{y}^{k} \) . Then,\n\n\[ A{x}^{ + } = {Ax}\left( {s}^{k}\right) = 0 \]\n\nand\n\n\[ \begin{Vmatrix}{{x}^{ + }{s}^{k} - e}\end{Vmatrix} = {\eta }_{d}\left( {s}^{k}\right) \leq \eta . \]\n\nNote that \( {s}_{j}^{ + } = {s}_{j}^{k} ... | Yes |
Lemma 3.6 Let \( \left( {{y}^{k},{s}^{k}}\right) \) be an approximate center for \( {\Omega }^{k} \) with \( {\eta }_{d}\left( {s}^{k}\right) \leq \eta \) and let \( {\Omega }^{k + 1} \) be defined above. Then, if \( {c}_{i}^{k + 1} > {c}_{i} \) in the update | Proof. The proof of the first case is similar to Theorem 2.7 of Chapter 2. Let \( \left( {{y}^{a},{s}^{a}}\right) \) and \( \left( {{y}_{ + }^{a},{s}_{ + }^{a}}\right) \) be the centers of \( {\Omega }^{k} \) and \( {\Omega }^{k + 1} \), respectively. Note that \( A{\left( {S}^{a}\right) }^{-1}e = 0,{\left( {c}^{k}\rig... | No |
Theorem 3.9 The number of total iterations of the primal potential algorithm to generate an approximate analytic center of \( {x}^{a} \), starting from \( x \in {\mathcal{X}}_{\Omega }^{ \circ } \), is bounded by \( O\left( {\mathcal{P}\left( x\right) - \mathcal{P}\left( {x}^{a}\right) }\right) \) . | Since \( \mathcal{P}\left( x\right) \) represents the logarithmic volume of the coordinate-aligned ellipsoid centered at \( s\left( x\right) \) and containing \( {\mathcal{S}}_{\Omega } \), the primal potential algorithm generates a sequence of ellipsoids whose volumes decreases at a constant rate, see Figure 3.2. | No |
Theorem 3.11 If the starting point of the Newton procedure satisfies \( \eta \left( {x, s}\right) < 1 \), then\n\n\[ \n{x}^{ + } > 0,\;A{x}^{ + } = 0,\;{s}^{ + } > 0 \n\]\n\nand\n\n\[ \n\eta \left( {{x}^{ + },{s}^{ + }}\right) \leq \frac{\sqrt{2}\eta {\left( x, s\right) }^{2}}{4\left( {1 - \eta \left( {x, s}\right) }\r... | Proof. To prove the result we first see that\n\n\[ \n\begin{Vmatrix}{{X}^{ + }{s}^{ + } - e}\end{Vmatrix} = \begin{Vmatrix}{{D}_{x}{d}_{s}}\end{Vmatrix}. \n\]\n\nMultiplying the both sides of the first equation of (3.15) by \( {\left( XS\right) }^{-1/2} \), we see\n\n\[ \nD{d}_{x} + {D}^{-1}{d}_{s} = r \mathrel{\text{:... | No |
Theorem 3.13 Let \( {x}^{ + } \) and \( {s}^{ + } \) be defined in Lemma 3.12. Then, if \( \eta \left( {x, s}\right) \geq \) \( \eta \) for a positive constant \( \eta < 1 \), we can choose \( \alpha \) such that\n\n\[{\psi }_{n}\left( {x + \theta {d}_{x}, s + \theta {d}_{s}}\right) - {\psi }_{n}\left( {x, s}\right) \l... | Proof. Note that\n\n\[ \alpha \sqrt{\min \left( {Xs}\right) }\begin{Vmatrix}{{\left( XS\right) }^{-1/2}\left( {e - {Xs}}\right) }\end{Vmatrix} = \alpha \sqrt{\min \left( {Xs}\right) }\begin{Vmatrix}{{\left( XS\right) }^{-1/2}e - {\left( XS\right) }^{1/2}e}\end{Vmatrix}. \]\n\nLet \( z = {Xs} \) and \( {z}_{1} = \min \l... | Yes |
Corollary 4.3 Given the interior feasible point \( \left( {{x}^{0},{y}^{0},{s}^{0}}\right) \) with \( {\mu }^{0} = {\left( {x}^{0}\right) }^{T}{s}^{0}/n \), and consider the dual level set \( \Omega \left( z\right) \) where \( {z}^{0} \mathrel{\text{:=}} {b}^{T}{y}^{0} - \) \( {\mu }^{0} \leq z < {z}^{ * } \) . Let \( ... | \[ \mathcal{B}\left( {\Omega \left( z\right) }\right) \geq \mathcal{B}\left( {\Omega \left( {z}^{0}\right) }\right) + \left( {n + 1}\right) \log \frac{{c}^{T}\widehat{x} - z}{{c}^{T}{x}^{0} - {z}^{0}} - O\left( n\right) ,\] and for all \( x \in {\mathcal{F}}_{p} \) the primal potential \[ {\mathcal{P}}_{n + 1}\left( {x... | Yes |
Lemma 4.5 Let \( \beta = 1 - \eta /\sqrt{n} \) . Then, there exists a point \( {x}^{ + } > 0 \) such that\n\n\[ \left( {A, - b,\cdots , - b}\right) {x}^{ + } = 0\;\text{ and }\;\begin{Vmatrix}{{X}^{ + }{s}^{ + } - e}\end{Vmatrix} \leq \eta + \eta \left( {1 + \eta }\right) ,\]\n\nwhere\n\n\[ {s}^{ + } = \left( \begin{ma... | Proof. Let \( {x}^{ + } = x\left( {s}^{k}\right) > 0 \) of (3.3) for \( \Omega \left( {z}^{k}\right) \), which minimizes\n\n\[ \begin{matrix} \text{ minimize } & \begin{Vmatrix}{{S}^{k}x - e}\end{Vmatrix} \\ \text{ s.t. } & \left( {A, - b,\cdots , - b}\right) x = 0, x \in {\mathcal{R}}^{2n}. \end{matrix} \]\n\nSince \(... | Yes |
Lemma 4.6 Let \( \left( {{y}^{k},{s}^{k}}\right) \) be an approximate center for \( {\Omega }^{k} \) with \( {\eta }_{d}\left( {s}^{k}\right) \leq \eta \) and let \( \Omega \left( {z}^{k + 1}\right) \) be given as in Step 1 of Algorithm 4.2. Then, \[ \mathcal{B}\left( {\Omega \left( {z}^{k + 1}\right) }\right) \leq \ma... | Proof. The proof is very similar to Theorem 2.7 in Chapter 2. Let \( \left( {{y}^{a},{s}^{a}}\right) \) and \( \left( {{y}_{ + }^{a},{s}_{ + }^{a}}\right) \) be the centers of \( \Omega \left( {z}^{k}\right) \) and \( \Omega \left( {z}^{k + 1}\right) \), respectively. Note we have \[ {s}_{n + 1}^{a} = \cdots = {s}_{2n}... | Yes |
Lemma 4.8 Let\n\n\\[ \n{\\mu }^{k} = \\frac{{\\left( {x}^{k}\\right) }^{T}{s}^{k}}{n} = \\frac{{c}^{T}{x}^{k} - {z}^{k}}{n}\\;\\text{ and }\\;\\mu = \\frac{{\\left( {x}^{k}\\right) }^{T}s\\left( {z}^{k}\\right) }{n}. \n\\]\n\nIf\n\n\\[ \n\\begin{Vmatrix}{p\\left( {z}^{k}\\right) }\\end{Vmatrix} < \\min \\left( {\\eta \... | Proof. The proof is by contradiction.\n\ni) If the first inequality of (4.15) is not true, then \\( \\exists j \\) such that \\( {s}_{j}\\left( {z}^{k}\\right) \\leq 0 \\)\n\nand\n\\[\n\\begin{Vmatrix}{p\\left( {z}^{k}\\right) }\\end{Vmatrix} \\geq 1 - \\frac{\\left( n + \\rho \\right) }{n{\\mu }^{k}}{x}_{j}{s}_{j}\\le... | Yes |
Theorem 4.9 Given \( \left( {{x}^{k},{y}^{k},{s}^{k}}\right) \in \mathcal{F} \) . Let \( \rho = \sqrt{n},{z}^{k} = {b}^{T}{y}^{k},{x}^{k + 1} \) be given by (4.10), and \( {y}^{k + 1} = y\left( {z}^{k}\right) \) in (4.13) and \( {s}^{k + 1} = s\left( {z}^{k}\right) \) in (4.12). Then, either\n\n\[ \n{\psi }_{n + \rho }... | Proof. If (4.14) does not hold, i.e.,\n\n\[ \n\begin{Vmatrix}{p\left( {z}^{k}\right) }\end{Vmatrix} \geq \min \left( {\eta \sqrt{\frac{n}{n + {\eta }^{2}}},1 - \eta }\right) ,\n\]\n\nthen\n\n\[ \n{\mathcal{P}}_{n + \rho }\left( {{x}^{k + 1},{z}^{k}}\right) - {\mathcal{P}}_{n + \rho }\left( {{x}^{k},{z}^{k}}\right) \leq... | Yes |
Corollary 4.10 Let \( \rho = \sqrt{n} \) . Then, Algorithm 4.3 terminates in at most \( O\left( {\sqrt{n}\log \left( {{c}^{T}{x}^{0} - {b}^{T}{y}^{0}}\right) /\epsilon }\right) \) iterations with\n\n\[ \n{c}^{T}{x}^{k} - {b}^{T}{y}^{k} \leq \epsilon .\n\] | Proof. In \( O\left( {\sqrt{n}\log \left( {{\left( {x}^{0}\right) }^{T}{s}^{0}/\epsilon }\right) }\right) \) iterations\n\n\[ \n- \sqrt{n}\log \left( {{\left( {x}^{0}\right) }^{T}{s}^{0}/\epsilon }\right) = {\psi }_{n + \rho }\left( {{x}^{k},{s}^{k}}\right) - {\psi }_{n + \rho }\left( {{x}^{0},{s}^{0}}\right) \n\]\n\n\... | Yes |
Lemma 4.11 Let the direction \( d = \left( {{d}_{x},{d}_{y},{d}_{s}}\right) \) be generated by equation (4.17) with \( \gamma = n/\left( {n + \rho }\right) \), and let\n\n\[ \theta = \frac{\alpha \sqrt{\min \left( {Xs}\right) }}{\begin{Vmatrix}{\left( XS\right) }^{-1/2}\left( \frac{{x}^{T}s}{\left( n + \rho \right) }e ... | Let \( v = {Xs} \) . Then, we can prove the following lemma (Exercise 4.8): | No |
Lemma 4.14 With the notations above,\n\ni)\n\n\[ \parallel {Pq}\parallel \leq \frac{\sqrt{2}}{4}\parallel r{\parallel }^{2} \]\n\nii)\n\n\[ - \frac{\parallel r{\parallel }^{2}}{4} \leq {p}_{j}{q}_{j} \leq \frac{{r}_{j}^{2}}{4}\text{ for each }j \]\n\niii)\n\n\[ \parallel {Pq}{\parallel }_{\infty }^{ - } \leq \frac{\par... | Lemma 4.14 has been partially proved in Theorem 3.11. We leave the rest proof of the lemma to the reader. | No |
Lemma 4.15 Let \( r \) be as above.\n\ni) If \( \gamma = 0 \), then \( \parallel r{\parallel }^{2} = {n\mu } \).\n\nii) If \( \eta \in \left( {0,1}\right) ,\gamma = 1 \) and \( \left( {x, s}\right) \in {\mathcal{N}}_{2}\left( \eta \right) \), then \( \parallel r{\parallel }^{2} \leq {\eta }^{2}\mu /\left( {1 - \eta }\r... | Proof. i) If \( \gamma = 0, r = - {\left( XS\right) }^{-{.5}}{Xs} \), so \( \parallel r{\parallel }^{2} = {x}^{T}s = {n\mu } \).\n\nii) Now \( r = {\left( XS\right) }^{-{.5}}\left( {{\mu e} - {Xs}}\right) \), so \( \parallel r\parallel \leq \frac{1}{\sqrt{\left( {1 - \eta }\right) \mu }}{\eta \mu } \), which yields the... | Yes |
Lemma 4.17 With the notation above, the step-size in the predictor step satisfies\n\n\[ \bar{\theta } \geq \frac{2}{1 + \sqrt{1 + 4\parallel {Pq}/\mu \parallel /\eta }}. \] | Proof. By (4.21) applied to the predictor step,\n\n\[ \parallel X\left( \theta \right) s\left( \theta \right) - \mu \left( \theta \right) e\parallel = \begin{Vmatrix}{\left( {1 - \theta }\right) \left( {{Xs} - {\mu e}}\right) + {\theta }^{2}{Pq}}\end{Vmatrix} \]\n\n\[ \leq \left( {1 - \theta }\right) \parallel {Xs} - {... | Yes |
Theorem 4.18 Let \( \eta = 1/4 \) . Then Algorithm 4.5 will terminate in at most \( O\left( {\sqrt{n}\log \left( {{\left( {x}^{0}\right) }^{T}{s}^{0}/\epsilon }\right) }\right) \) iterations with \[ {c}^{T}{x}^{k} - {b}^{T}{y}^{k} \leq \epsilon \] | Proof. Using Lemma 4.14(i) and Lemma 4.15(i), we have \[ \parallel {Pq}\parallel \leq \frac{\sqrt{2}}{4}\parallel r{\parallel }^{2} = \frac{\sqrt{2}}{4}{n\mu } \] so that \[ \bar{\theta } \geq \frac{2}{1 + \sqrt{1 + \sqrt{2}n/\eta }} = \frac{2}{1 + \sqrt{1 + 4\sqrt{2}n}} \] at each iteration. Then (4.26) and Lemma 4.17... | Yes |
Lemma 4.19 Let \( \eta \in \left( {0,1}\right) ,\gamma \in \left( {0,1}\right) \), and \( \mathcal{N} = {\mathcal{N}}_{\infty }\left( \eta \right) \) or \( {\mathcal{N}}_{\infty }^{ - }\left( \eta \right) \) . Let \( x, s, d \) and \( \bar{\theta } \) be as in the \( k \) th iteration of Algorithm 4.6, and define \( p,... | Proof. Suppose first \( \mathcal{N} = {\mathcal{N}}_{\infty }^{ - }\left( \eta \right) \) . Then, for each \( \theta \in \left\lbrack {0,{\theta }_{2}^{ - }}\right\rbrack ,\left( {4.20}\right) \) and (4.21) imply\n\n\[ X\left( \theta \right) s\left( \theta \right) - \mu \left( \theta \right) e \]\n\n\[ = \left( {1 - \t... | Yes |
Theorem 4.20 Let \( \eta \in \left( {0,1}\right) \) and \( \gamma \in \left( {0,1}\right) \) be constants with \( \gamma \leq 2(1 - \) \( \eta ) \) . Then Algorithm 4.6, with \( \mathcal{N} = {\mathcal{N}}_{\infty }\left( \eta \right) \) or \( {\mathcal{N}}_{\infty }^{ - }\left( \eta \right) \), will terminate in \( O\... | Proof. In either case, each iterate lies in \( {\mathcal{N}}_{\infty }^{ - }\left( \eta \right) \), where\n\n\[ \parallel {Pq}{\parallel }_{\infty }^{ - } \leq \parallel {Pq}{\parallel }_{\infty } \leq \parallel r{\parallel }^{2}/4 \leq n{\mu }^{k}/4, \]\n\nusing Lemma 4.14(iii) and Lemma 4.15(iii). Hence\n\n\[ {\theta... | Yes |
Proposition 4.21 The following statements hold for the affine scaling algorithm:\n\ni) \( {x}^{k} \in {\mathcal{F}}_{p} \) for all \( k \geq 0 \) ;\n\nii) The sequence of objective function values \( \left\{ {{c}^{T}{x}^{k}}\right\} \) strictly decreases and converges to a finite value;\n\niii) \( {X}^{k}{s}^{k} \right... | Proof. Assume that \( {x}^{k} \in {\mathcal{F}}_{p} \) . Using (4.30), we obtain\n\n\[ \n{\left( {X}^{k}\right) }^{-1}{x}^{k + 1} = e - \alpha \frac{{p}^{k}}{\max \left( {p}^{k}\right) } \geq e - {\alpha e} = \left( {1 - \alpha }\right) e > 0, \n\]\n\nfrom which i) follows. Using (4.30) and noting\n\n\[ \n{c}^{T}{X}^{k... | Yes |
For the affine scaling algorithm, the following statements hold.\ni) There exists a constant \( M\\left( {A, c}\\right) > 0 \) such that\n\n\[\\begin{Vmatrix}{{X}^{k}{p}^{k}}\\end{Vmatrix} \\leq M\\left( {A, c}\\right) {\\begin{Vmatrix}{p}^{k}\\end{Vmatrix}}^{2} = M\\left( {A, c}\\right) {c}^{T}{X}^{k}{p}^{k},\\;\\fora... | The proof of i) follows Exercise 4.13. Using (4.30), (4.31) and (4.32), we obtain\n\n\[\\begin{Vmatrix}{{x}^{k + 1} - {x}^{k}}\\end{Vmatrix} \\leq M\\left( {A, c}\\right) \\left( {{c}^{T}{x}^{k} - {c}^{T}{x}^{k + 1}}\\right) ,\\;\\forall k \\geq 0.\]\n\n\\( \\left( {4.33}\\right) \\)\n\nSince \\( \\left\\{ {{c}^{T}{x}^... | No |
Lemma 4.23 The set \( \left\{ {x \in {\mathcal{F}}_{p} : {x}_{Z} = 0}\right\} \) is a face of \( {\mathcal{F}}_{p} \) where the objective function is constant, and for an \( x \in {\mathcal{F}}_{p} \), we have\n\n\[ \n{c}^{T}\left( {x - \bar{x}}\right) = {\bar{s}}^{T}\left( {x - \bar{x}}\right) = {\bar{s}}_{Z}^{T}{x}_{... | This lemma shows that \( \left| Z\right| > 0 \) . We now show that \( \bar{x} \) is an optimal solution of problem (LP) if the step-size \( \alpha \in (0,2/3\rbrack \) . An important ingredient used in the proof of these results is the following (partial) potential function. | No |
Lemma 4.24 For the affine scaling algorithm,\n\n\\[ \n\\frac{{c}^{T}{x}^{k} - \\bar{z}}{\\mathop{\\sum }\\limits_{{j \\in Z}}{x}_{j}^{k}} \\geq \\frac{1}{\\sqrt{\\left| Z\\right| }M\\left( {A, c}\\right) },\\;\\forall k \\geq 0.\n\\] | Proof. By Theorem 4.22(iii), we know that for some \\( M\\left( {A, c}\\right) \\geq 0 \\) ,\n\n\\[ \n\\begin{Vmatrix}{{x}^{k} - \\bar{x}}\\end{Vmatrix} \\leq M\\left( {A, c}\\right) \\left( {{c}^{T}{x}^{k} - \\bar{z}}\\right) ,\\;\\forall k \\geq 0.\n\\]\n\nSince \\( {\\bar{x}}_{Z} = 0 \\), this implies\n\n\\[ \n\\beg... | Yes |
Theorem 4.26 For the affine scaling algorithm, assume that \( \alpha \leq 2/3 \) . Then the limit \( \bar{x} \) of the primal sequence \( \left\{ {x}^{k}\right\} \) is an optimal solution lying in the interior of the optimal face of (LP). | Proof. Assume by contradiction that \( \bar{x} \) is not an optimal solution lying in the relative interior of the optimal face of (LP). By Theorem 4.25, it follows that\n\n\[ \mathop{\lim }\limits_{{k \rightarrow \infty }}{\mathcal{P}}_{Z}\left( {{x}^{k},\bar{z}}\right) = - \infty \]\n\n\( \left( {4.37}\right) \)\n\nN... | Yes |
Lemma 5.1\n\n\[ \n{\begin{Vmatrix}{\left( {D}^{t}\right) }^{-1}\left( {x}^{t} - {d}^{t}\right) \end{Vmatrix}}_{\infty } \leq {0.1}\;\text{ for any }\;t = 0,1,\ldots \n\] | Proof. The proof of the first inequality is straightforward. Let\n\n\[ \n{\sigma }^{t} = {\begin{Vmatrix}{\left( {D}^{t}\right) }^{-1}\left( {x}^{t} - {d}^{t}\right) \end{Vmatrix}}_{1}\;\text{ for }\;t = 0,1,\ldots \n\]\n\nThen, for \( t = 1,2,\ldots \)\n\n\[ \n{\sigma }^{t} = \mathop{\sum }\limits_{{i \in {I}^{t}}}\fr... | Yes |
Theorem 5.3 Given an interior solution \( {x}^{k} \) and \( {s}^{k} \) in the solution sequence generated by any of interior-point algorithms possessing property (5.7) below, define\n\n\[ \n{\sigma }^{k} = \left\{ {j : {x}_{j}^{k} \geq {s}_{j}^{k}}\right\} \n\]\n\nThen, for \( K = O\left( {\sqrt{n}\left( {\log \left( {... | Proof. For simplicity, we use \( \xi = \xi \left( {A, b, c}\right) \) . For a given \( j \in {P}^{ * } \), let \( \left( {{x}^{ * },{s}^{ * }}\right) \) be a complementarity solution such that \( {x}_{j}^{ * } \) is maximized on the primal optimal face \( {\Omega }_{p} \), i.e, \( {x}_{j}^{ * } \geq {\xi }_{p}\left( {A... | Yes |
Theorem 5.4 All \( O\left( {\sqrt{n}\log \left( {R/\epsilon }\right) }\right) \) -iteration polynomial-time interior-point algorithms discussed earlier, coupled with the termination procedure, will generate an optimal solution in \( O\left( {\sqrt{n}\left( {\log \left( {{R\zeta }\left( {A, b, c}\right) /{\xi }^{2}\left... | \[ R \leq {2}^{L},\;\zeta \left( {A, b, c}\right) \leq {2}^{L},\;\text{ and }\;\xi \left( {A, b, c}\right) \geq {2}^{-L}, \] where \( L \) is the size of the \( {LP} \) data. Thus, \( {\sigma }^{k} = {P}^{ * } \) and an exact solution will be generated in \( O\left( {\sqrt{n}L}\right) \) iterations and \( O\left( {{n}^... | Yes |
Proposition 5.5 Let \( \widetilde{A} \in {R}^{p \times p} \) be skew-symmetric, and let \( \widetilde{b} = - \widetilde{c} \in {R}^{p} \) . Then the problem\n\n\[\n\begin{matrix} \text{ (SDP) } & \text{ minimize } & {\widetilde{c}}^{T}\widetilde{u} \\ & \text{ s.t. } & \widetilde{A}\widetilde{u} \geq \widetilde{b},\wid... | The advantage of self-duality is that we can apply a primal-dual interior-point algorithm to solve the self-dual problem without doubling the dimension of the linear system solved at each iteration. | No |
Theorem 5.6 Consider problems (HSDP) and (HSDD).\n\ni) (HSDD) has the same form as (HSDP), i.e., (HSDD) is simply (HSDP) with \( \\left( {y, x,\\tau ,\\theta }\\right) \) being replaced by \( \\left( {{y}^{\\prime },{x}^{\\prime },{\\tau }^{\\prime },{\\theta }^{\\prime }}\\right) \) .\n\nii) (HSDP) has a strictly feas... | Proof. In what follows, denote the slack vector and scalar in (HSDD) by \( {s}^{\\prime } \) and \( {\\kappa }^{\\prime } \), respectively. The proof of (i) is based on the skew-symmetry of the linear constraint system of (HSDP). We omit the details. Result (ii) can be easily verified. Then (iii) is due to the self-dua... | Yes |
Theorem 5.7 Let \( \left( {{y}^{ * },{x}^{ * },{\tau }^{ * },{\theta }^{ * } = 0,{s}^{ * },{\kappa }^{ * }}\right) \) be a strictly self complementary solution for (HSDP).\n\ni) (LP) has a solution (feasible and bounded) if and only if \( {\tau }^{ * } > 0 \) . In this case, \( {x}^{ * }/{\tau }^{ * } \) is an optimal ... | Proof. If (LP) and (LD) are both feasible, then they have a complementary solution pair \( \bar{x} \) and \( \left( {\bar{y},\bar{s}}\right) \) for (LP) and (LD), such that\n\n\[{\left( \bar{x}\right) }^{T}\bar{s} = 0.\]\n\nLet\n\n\[ \alpha = \frac{n + 1}{{e}^{T}\bar{x} + {e}^{T}\bar{s} + 1} > 0.\]\n\nThen one can veri... | Yes |
Theorem 5.9 Consider problem (HSDP).\n\ni) For any \( \mu > 0 \), there is a unique \( \left( {y, x,\tau ,\theta, s,\kappa }\right) \) in \( {\mathcal{F}}_{h}^{0} \), such that\n\n\[ \left( \begin{matrix} {Xs} \\ {\tau \kappa } \end{matrix}\right) = {\mu e} \]\n\nii) Let \( \left( {{d}_{y},{d}_{x},{d}_{\tau },{d}_{\the... | Proof. For any \( \mu > 0 \), there is a unique feasible point \( \left( {y, x,\tau ,\theta, s,\kappa }\right) \) for (HSDP) and a unique feasible point \( \left( {{y}^{\prime },{x}^{\prime },{\tau }^{\prime },{\theta }^{\prime },{s}^{\prime },{\kappa }^{\prime }}\right) \) for (HSDD) such that\n\n\[ X{s}^{\prime } = {... | Yes |
Consider the example where\n\n\[ A = \\left( \\begin{array}{lll} - 1 & 0 & 0 \\end{array}\\right) ,\\;b = 1,\\;\\text{ and }\\;c = \\left( \\begin{array}{lll} 0 & 1 & - 1 \\end{array}\\right) .\n\]\n\nThen, could a strictly self-complementary solution generated for (HSDP) with\n\n\[ {c}^{T}{x}^{ * } = 1 > 0,\\;{b}^{T}{... | \[ {y}^{ * } = 2,\\;{x}^{ * } = {\\left( 0,2,1\\right) }^{T},\\;{\\tau }^{ * } = 0,\\;{\\theta }^{ * } = 0,\\;{s}^{ * } = {\\left( 2,0,0\\right) }^{T},\\;{\\kappa }^{ * } = 1 \]\n\nOf course, if the algorithm generates instead \( {x}^{ * } = \) \( {\\left( 0,1,2\\right) }^{T} \), then we get demonstrated infeasibility ... | Yes |
Theorem 5.12. If (LP) possesses an optimal solution, then\n\n\[ \n{\tau }^{k} \geq \frac{1 - {2\eta }}{\left( {e}^{T}\bar{x} + {e}^{T}\bar{s} + 1\right) }\;\text{ for all }\;k, \]\n\nwhere \( \bar{x} \) and \( \left( {\bar{y},\bar{s}}\right) \) are any optimal solution pair for (LP) and (LD); otherwise,\n\n\[ \n{\kappa... | Proof. Note that the sequence generated by the predictor-corrector algorithm is in \( \mathcal{N}\left( {2\eta }\right) \). Note that\n\n\[ \n{y}^{ * } = \alpha \bar{y},\;{x}^{ * } = \alpha \bar{x},\;{\tau }^{ * } = \alpha ,\;{\theta }^{ * } = 0,\;{s}^{ * } = \alpha \bar{s},\;{\kappa }^{ * } = 0, \]\n\nwhere\n\n\[ \n\a... | Yes |
Lemma 5.15 Suppose that at the kth iteration, for all \( j = 1,\ldots, n \) ,\n\n\[ \left| {{\left( {d}_{x}\right) }_{j}{\left( {d}_{s}\right) }_{j} - \eta {d}_{x}^{T}{d}_{s}/n}\right| \leq \eta \;\text{ and }\;\left| {{d}_{x}^{T}{d}_{s}}\right| \leq \eta .\n\]\n\nThen,\n\n\[ \bar{\alpha } \geq \min \left\{ {1,\frac{\g... | Note that if \( \left( {{y}^{k},{x}^{k},{s}^{k}}\right) \) is a feasible pair, then \( {d}_{x}^{T}{d}_{s} = 0 \) and the conditions of the above lemma hold for \( \eta = O\left( {n{\mu }^{k}}\right) \). Nevertheless, we can prove that the conditions hold in general for \( \eta = O\left( {{n}^{2}{\mu }^{k}}\right) \), i... | No |
Lemma 5.16 At the kth iteration, we have\n\n\[ \n{D}^{-1}{d}_{x} = - {\theta }^{k}P{D}^{-1}\left( {{x}^{0} - {u}^{0}}\right) + {\theta }^{k}\left( {I - P}\right) D\left( {{s}^{0} - {w}^{0}}\right) \n\]\n\n\[ \n- \left( {I - P}\right) {\left( {X}^{k}{S}^{k}\right) }^{-{.5}}\left( {{X}^{k}{s}^{k} - \gamma {\mu }^{k}e}\ri... | From the definition of \( \rho \) and \( {\rho }^{0} \), we have\n\n\[ \n{\begin{Vmatrix}\left( {u}^{0},{w}^{0}\right) \end{Vmatrix}}_{\infty } \leq {\rho }^{0} \leq \rho . \n\]\n\nThus,\n\n\[ \n- {\rho e} \leq {x}^{0} - {u}^{0} \leq {2\rho e}\;\text{ and }\; - {\rho e} \leq {s}^{0} - {w}^{0} \leq {2\rho e}. \n\]\n\nTh... | Yes |
Lemma 6.1 Let \( {\bar{E}}_{j} \) be the complement of \( {E}_{j}, j = 1,2,\ldots \) Then, if the probability\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {\mathop{\sum }\limits_{{j = 1}}^{n}\Pr \left( {\bar{E}}_{j}\right) }\right) = 0 \]\n\nthen \( {E}_{1} \cap {E}_{2} \cap \ldots \cap {E}_{m} \) is tru... | Proof. The proof simply follows\n\n\[ \Pr \left( {{E}_{1} \cap {E}_{2} \cap \ldots \cap {E}_{m}}\right) \geq \Pr \left( {{E}_{1} \cap {E}_{2} \cap \ldots \cap {E}_{n}}\right) \geq 1 - \mathop{\sum }\limits_{{j = 1}}^{n}\Pr \left( {\bar{E}}_{j}\right) \]\n\nfor any \( 1 \leq m \leq n \) . Since \( \mathop{\lim }\limits_... | Yes |
Theorem 6.3 Let \( \xi = \parallel r{\parallel }_{\infty }/\parallel r\parallel \) . Then, with the assumption above,\ni)\n\[ \Pr \left( {\parallel {Pq}\parallel \leq \frac{\parallel r{\parallel }^{2}}{4}{\left( 2{\xi }^{2} + \frac{6.5}{n}\right) }^{1/2}}\right) \rightarrow 1\text{ as }n \rightarrow \infty ; \]\nii)\n\... | ## 6.1.2 Proof of the theorem\nNow we sketch the proof of Theorem 6.3. The proof of (i) is long and technical and hence we omit it here. However, we will prove a slightly weaker version of (i) at the end of this section.\n\nBecause \( p \) and \( q \) are homogeneous of degree 1 in \( \parallel r\parallel \), we assume... | No |
Lemma 6.5 Let \( F = \left\lbrack {g, H}\right\rbrack \) be an orthogonal matrix. If \( v \) is uniformly distributed on the unit sphere in \( {\mathcal{R}}^{n - 1} \), \[ \Pr \left( {\parallel {Hv}{\parallel }_{\infty } \leq \sqrt{3\frac{\log n}{n}}}\right) \rightarrow 1\text{ as }n \rightarrow \infty . \] | Proof. Since \( v \) is uniformly distributed on the unit sphere in \( {\mathcal{R}}^{n - 1} \), it can be generated as follows: \( v = \lambda /\parallel \lambda \parallel \), where \( \lambda \sim N\left( {0, I}\right) \) (the standard normal distribution in \( {\mathcal{R}}^{n - 1} \) ). Hence we wish to obtain an u... | Yes |
Corollary 6.12 The probability that system (6.5) has an interior feasible point is\n\n\[ \n{p}_{nd} = \frac{\left( \begin{matrix} n - 1 \\ 0 \end{matrix}\right) + \ldots + \left( \begin{matrix} n - 1 \\ d - 1 \end{matrix}\right) }{{2}^{n - 1}}. \n\] | Proof. From the strictly complementary property of linear systems, this probability is the probability that the dual\n\n\[ \ns = {A}^{T}y \geq 0,\;s \neq 0 \n\]\n\n(6.11)\n\nis infeasible. However, the latter probability is exact\n\n\[ \n1 - {p}_{nm} = {p}_{nd} \n\] | No |
Theorem 6.13 With probability one, exactly one of systems (6.5) and (6.11) is empty and the other has an interior feasible point. | Proof. The probability that (6.11) has an interior feasible solution is \( {p}_{nm} \) . Note that these two events are exclusive, and \( {p}_{nd} + {p}_{nm} = 1 \) . | Yes |
Theorem 6.16 With high probability the homogeneous random linear feasibility problem (6.5) can be solved in \( O\left( {\sqrt{n}\log n}\right) \) iterations. | Proof. From Theorem 6.13, with probability one either \( A = {A}_{P}^{ * } \) or \( {z}^{ * } > 0 \) for problem (6.6) associated with (6.5). Then from Theorem 6.15, with high probability, there exists positive primal variables or positive dual slacks (not both) satisfying condition (6.9) or (6.10). Thus, the theorem f... | No |
Lemma 6.18 Let \( \zeta = \left( {1 - \eta }\right) {\xi }_{h}^{ * }/\left( {n + 1}\right) \) . Then in order to obtain \( {v}_{j}^{k} < \zeta \leq \) \( {u}_{j}^{k}, j \in {\sigma }_{h}^{ * } \), and \( {u}_{j}^{k} < \zeta \leq {v}_{j}^{k}, j \notin {\sigma }_{h}^{ * } \), it suffices to have | \[ {\theta }^{k} < \frac{1 - \eta }{{\left( n + 1\right) }^{2}}{\left( {\xi }_{h}^{ * }\right) }^{2} \] | Yes |
Lemma 6.21 Consider an instance of Todd’s degenerate model, and let \( \widehat{\xi } \) and \( \widehat{\rho } \) be as in (6.27). Suppose that \( k \) is large enough so that the following inequality is satisfied:\n\n\[ \n{\theta }^{k} \leq \frac{\left( {1 - \eta }\right) {\widehat{\xi }}^{2}}{{\left( n + 1\right) }^... | Using the criterion in the previous lemma, we can terminate the algorithm once (6.28) holds. From \( {\theta }^{0} = 1,\left( {6.15}\right) \), and (6.28), this definitely happens if\n\n\[ \n{\theta }^{k} \leq {\left( 1 - \frac{2}{1 + \sqrt{1 + 4\sqrt{2}\left( {n + 1}\right) }}\right) }^{k} \leq \frac{\left( {1 - \eta ... | Yes |
The sequence with \( {r}^{k} = 1/k \) converges to zero. | The convergence is of order one but it is not linear, since \( \lim \left( {{r}^{k + 1}/{r}^{k}}\right) = 1 \), that is, \( \beta \) is not strictly less than one. | No |
Consider the predictor-corrector algorithm described in Section 4.5. We will show that this \( O\left( {\sqrt{n}\log \left( {R/\epsilon }\right) }\right) \) -iteration algorithm actually forces quadratic convergence of the duality gap \( {r}^{k} \mathrel{\text{:=}} {\left( {x}^{k}\right) }^{T}{s}^{k} > 0 \) to zero. | We follow the same notations in Section 4.5. At the \( k \) th predictor step, let \( {\mu }^{k} = {\left( {x}^{k}\right) }^{T}{s}^{k}/n,\left( {{d}_{x},{d}_{s}}\right) \mathrel{\text{:=}} d\left( {{x}^{k},{s}^{k},0}\right) \), and\n\n\[{\delta }^{k} = \frac{{D}_{x}{d}_{s}}{{\mu }^{k}}.\]\n\n(7.1)\n\nIf \( {\theta }^{k... | Yes |
Lemma 7.2 If \( {d}_{x} \) and \( {d}_{s} \) satisfy the equation\n\n\[ \nX{d}_{s} + S{d}_{x} = - {Xs} \]\n\nand the inequality\n\n\[ \n{\left( {d}_{x}\right) }^{T}{d}_{s} \geq 0 \]\n\nthen,\n\n\[ \n{\begin{Vmatrix}{D}^{-1}{d}_{x}\end{Vmatrix}}^{2} + {\begin{Vmatrix}D{d}_{s}\end{Vmatrix}}^{2} \leq {x}^{T}s \]\n\nwhere ... | Proof. Multiplying the diagonal matrix \( {\left( XS\right) }^{-1/2} \) on both sides of the equation, we have\n\n\[ \n{D}^{-1}{d}_{x} + D{d}_{s} = - {\left( XS\right) }^{1/2}e. \]\n\nTaking the norms of both sides,\n\n\[ \n{\begin{Vmatrix}{D}^{-1}{d}_{x} + D{d}_{s}\end{Vmatrix}}^{2} = {x}^{T}s \]\n\nor\n\n\[ \n{\begin... | Yes |
Lemma 7.3 If \( {d}_{x} \) and \( {d}_{s} \) are obtained from the linear system (7.4) and \( \mu = {x}^{T}s/n \), then\n\n\[ \begin{Vmatrix}{\left( {d}_{x}\right) }_{Z}\end{Vmatrix} = O\left( \mu \right) \;\text{ and }\;\begin{Vmatrix}{\left( {d}_{s}\right) }_{P}\end{Vmatrix} = O\left( \mu \right) . \] | Proof. From Lemma 7.2 and relation (7.6), we obtain\n\n\[ \begin{Vmatrix}{\left( {d}_{x}\right) }_{Z}\end{Vmatrix} = \begin{Vmatrix}{{D}_{Z}{D}_{Z}^{-1}{\left( {d}_{x}\right) }_{Z}}\end{Vmatrix} \]\n\n\[ \leq \begin{Vmatrix}{D}_{Z}\end{Vmatrix}\begin{Vmatrix}{{D}_{Z}^{-1}{\left( {d}_{x}\right) }_{Z}}\end{Vmatrix} \]\n\... | Yes |
Lemma 7.4 If \( {d}_{x} \) and \( {d}_{s} \) are obtained from the linear system (7.4), then \( {\left( {d}_{x}\right) }_{P} \) is the solution to the (weighted) least-squares problem\n\n\[ \mathop{\min }\limits_{u}\;\left( {1/2}\right) {\begin{Vmatrix}{D}_{P}^{-1}u\end{Vmatrix}}^{2} \]\n\n\[ \text{s.t.}\;{A}_{P}u = - ... | Proof. From (7.7), we see that\n\n\[ {x}_{P} + {\left( {d}_{x}\right) }_{P} \in R\left( {{D}_{P}^{2}{A}_{P}^{T}}\right) \]\n\n(7.8)\n\nSince \( {s}_{P}^{ * } = 0 \) for all optimal \( {s}^{ * } \), we must have \( {c}_{P} \in R\left( {A}_{P}^{T}\right) \) . Thus,\n\n\[ {s}_{P} = {c}_{P} - {A}_{P}^{T}y \in R\left( {A}_{... | Yes |
Theorem 7.6 Let \( \\left\\{ \\left( {{x}^{k},{s}^{k}}\\right) \\right\\} \) be the sequence generated by Algorithm 4.5. Then,\n\ni) the Algorithm has iteration complexity \( O\\left( {\\sqrt{n}\\log \\left( {R/\\epsilon }\\right) }\\right) \) ;\n\nii) \( 1 - {\\theta }^{k} = O\\left( {{\\left( {x}^{k}\\right) }^{T}{s}... | Proof. The proof of (i) is in Theorem 4.18, which also establishes\n\n\\[ \n\\mathop{\\lim }\\limits_{{k \\rightarrow \\infty }}{\\mu }^{k} = 0 \n\\]\n\nFrom relation (7.1), Lemma 7.3 and Theorem 7.5 we have\n\n\\[ \n\\begin{Vmatrix}{\\delta }^{k}\\end{Vmatrix} = \\begin{Vmatrix}{{D}_{x}{d}_{s}/{\\mu }^{k}}\\end{Vmatri... | Yes |
Lemma 7.7 If \( {\theta }^{k} \mathrel{\text{:=}} \bar{\theta } \) is the largest \( \theta \) such that \( \left( {x\left( \theta \right), s\left( \theta \right) }\right) \in {\mathcal{N}}_{2}\left( {\eta + \tau }\right) \) with \( 0 < \eta \leq 1/4 \) and \( 0 < \tau \leq \eta \), then | \[ {\theta }^{k} \geq \frac{2}{1 + \sqrt{1 + 4\begin{Vmatrix}{\delta }^{k}\end{Vmatrix}/\tau }}. \] | Yes |
Lemma 7.8 If \( {d}_{x} \) and \( {d}_{s} \) are obtained from the linear system (7.19), and \( \mu = {x}^{T}s/n \), then\n\n\[ \begin{Vmatrix}{\left( {d}_{x}\right) }_{Z}\end{Vmatrix} = O\left( \mu \right) \;\text{ and }\;\begin{Vmatrix}{\left( {d}_{s}\right) }_{P}\end{Vmatrix} = O\left( \mu \right) . | The proofs of \( \begin{Vmatrix}{\left( {d}_{x}\right) }_{P}\end{Vmatrix} = O\left( \mu \right) \) and \( \begin{Vmatrix}{\left( {d}_{s}\right) }_{Z}\end{Vmatrix} = O\left( \mu \right) \) are, again, more involved. We first note\n\n\[ S\left( {x + {d}_{x}}\right) = - X{d}_{s}, \]\n\n\[ X\left( {s + {d}_{s}}\right) = - ... | No |
Lemma 7.9 Let \( M \) be a PSD matrix, partitioned as in (7.21). Then \( {M}_{PP}{x}_{P} = 0 \) if and only if \( {M}_{PP}^{T}{x}_{P} = 0 \) . Furthermore, \( {M}_{PP}{x}_{P} = 0 \) implies that \( \left( {{M}_{ZP} + {M}_{PZ}^{T}}\right) {x}_{P} = 0 \) . | Proof. Let \( x = {\left( {x}_{P}^{T},{0}^{T}\right) }^{T} \) . If either \( {M}_{PP}{x}_{P} = 0 \) or \( {M}_{PP}^{T}{x}_{P} = 0 \), then \( {x}^{T}{Mx} = 0 \), so \( x \) is a global minimizer of the quadratic form \( {y}^{T}{My} \) . Consequently \( \left( {M + {M}^{T}}\right) x = 0 \), which is exactly\n\n\[ \left(... | Yes |
Lemma 7.11 If \( {d}_{x} \) and \( {d}_{s} \) are obtained from the linear system (7.19), and \( \mu = {x}^{T}s/n \), then \( u = {\left( {d}_{x}\right) }_{P} \) and \( v = {\left( {d}_{s}\right) }_{Z} \) are the solutions to the (weighted) least-squares problem\n\n\[ \mathop{\min }\limits_{{u, v}}\;\left( {1/2}\right)... | Proof. Note that from (7.19), \( u = {\left( {d}_{x}\right) }_{P}, v = {\left( {d}_{s}\right) }_{Z} \) is certainly feasible in the problem (7.24). Next, from (7.19) and (7.20), we see that\n\n\[ {x}_{P} + {\left( {d}_{x}\right) }_{P} = - {D}_{P}^{2}{M}_{B} \cdot {d}_{x} \]\n\n\[ {s}_{Z} + {\left( {d}_{s}\right) }_{Z} ... | Yes |
Proposition 7.13 There is a monotone LCP problem, where a strict complementarity solution does not exist, for which the predictor-corrector algorithm or affine scaling algorithm possesses no superlinear convergence. | Proof. Consider the simple monotone LCP with \( n = 1, M = 1 \) and \( q = 0 \) . The unique complementarity solution is \( s = x = 0 \), which is not strictly complementary. Note that the feasible solution \( s = x = \epsilon \) is a perfectly centered pair for any \( \epsilon > 0 \) . The direction in the predictor s... | Yes |
Theorem 7.15 Let \( {\left( {x}^{K}\right) }^{T}{s}^{K} \) be small enough. Then, Variant 2 generates a sequence \( \left\{ {{x}^{k},{s}^{k}}\right\} \) with \( k \geq K \) such that\ni) the order of the convergence of the complementary gap to zero equals at least 2,\nii) \( \left\{ {{x}^{k},{s}^{k}}\right\} \) is a Ca... | Proof. At the \( k \) th iteration \( \left( {k \geq K \mathrel{\text{:=}} 1}\right) \) we have from (7.29)\n\n\[{\left( {x}^{k + 1}\right) }^{T}{s}^{k + 1} \leq \frac{R{\left( {\left( {x}^{k}\right) }^{T}{s}^{k}\right) }^{2}}{{\tau }_{k}} = R{\left( {\left( {x}^{k}\right) }^{T}{s}^{k}\right) }^{2}{2}^{k}/\eta ,\]\n\no... | Yes |
Lemma 8.1 For all \( k = m,\ldots, n \) ,\n\n\[ \mathcal{P}\left( {x,{\Omega }^{k}}\right) \geq \log {2k}\log \left( \epsilon \right) + {2k}\log \left( {2k}\right) . \] | Proof. From (8.1), there is a \( y \) such that\n\n\[ {c}^{k} - {\left( {A}^{k}\right) }^{T}y \geq {\epsilon e}. \]\n\nThus, from the inequality in Section 2.3.2,\n\n\[ \mathcal{P}\left( {x,{\Omega }^{k}}\right) \geq \mathcal{B}\left( {y,{\Omega }^{k}}\right) + {2k}\log \left( {2k}\right) \]\n\n\[ = \mathop{\sum }\limi... | Yes |
Theorem 8.2 For any \( k = m,\ldots, n \), the number of total potential reduction iterations to compute the sequence of \( \left( {3/4}\right) \) -approximate analytic centers, \( \left\{ {\widehat{y}}^{t}\right\} \), of all \( {\Omega }^{t}, t = m,\ldots, k \), is bounded by \( O\left( {k\log \left( {{\bar{w}}^{k}/\e... | Proof. From inequalities (8.2),(8.3) and Lemma 8.1, we have for \( k = \) \( m,\ldots, n, \n\n\[ \n\left( {1/6}\right) {T}^{k} \leq \mathcal{P}\left( {{x}^{k}\left( 0\right) ,{\Omega }^{k}}\right) - \mathcal{P}\left( {{x}^{k}\left( {T}^{k}\right) ,{\Omega }^{k}}\right) \n\] \n\n\[ \n\leq {2k}\log \left( {2k}\right) + {... | Yes |
Lemma 8.3 Denote by \( \left( {\bar{y},\bar{s}}\right) \) the analytic center of \( \Omega \) and let\n\n\[ \n\bar{r} = \sqrt{{a}^{T}{\left( A{\bar{S}}^{-2}{A}^{T}\right) }^{-1}a}.\n\]\n\nThen the max-potential of \( {\Omega }^{ + } \)\n\n\[ \n\mathcal{B}\left( {\Omega }^{ + }\right) \leq \mathcal{B}\left( \Omega \righ... | Proof. Denote by \( {\bar{y}}^{ + } \) the analytic center for \( {\Omega }^{ + } \) . Let \( {\bar{s}}^{ + } = c - {A}^{T}{\bar{y}}^{ + } > 0 \) and \( {\bar{s}}_{n + 1}^{ + } = {a}^{T}{y}^{k} - {a}^{T}{\bar{y}}^{ + } \) . Then we have\n\n\[ \n{\bar{s}}_{n + 1}^{ + } = {a}^{T}\left( {{y}^{k} - {\bar{y}}^{ + }}\right)\... | Yes |
Lemma 8.4 For all \( k \geq 0 \) ,\n\n\[ \mathcal{B}\left( {\Omega }^{k}\right) \geq \left( {{2m} + k}\right) \log \epsilon \] | Proof. From (8.9), \( \Gamma \subset {\Omega }^{k} \) . Thus \( {\Omega }^{k} \) contains a full dimensional ball with \( \epsilon \) radius. Let the center of this ball be \( \bar{y} \) . Then \( {c}^{k} - {\left( {A}^{k}\right) }^{T}\bar{y} \geq {\epsilon e} \) ; thus\n\n\[ \mathcal{B}\left( {\Omega }^{k}\right) = \m... | Yes |
Lemma 8.5 Let \( s = {c}^{k} - {\left( {A}^{k}\right) }^{T}y \) for any \( y \in {\Omega }^{k} \). Then\ni) \( 0 \leq {s}_{j} \leq 1,\;j = 1,\ldots ,{2m} \)\nii) \( 0 \leq {s}_{j} \leq \sqrt{m},\;j = {2m} + 1,\ldots ,{2m} + k \) . | Proof. For \( j = 1,\ldots, m,{s}_{j} = 1 - {y}_{j} \) ; since \( 0 \leq {y}_{j} \leq 1,0 \leq {s}_{j} \leq 1 \).\n\nFor \( j = m + 1,\ldots ,{2m},{s}_{j} = {y}_{j - m} \) ; since \( 0 \leq {y}_{j - m} \leq 1,0 \leq {s}_{j} \leq 1 \).\n\nFor \( j = {2m} + 1,\ldots ,{2m} + k \),\n\n\[{s}_{j} = {a}_{j - {2m}}^{T}{y}^{j -... | Yes |
Lemma 8.6 Let \( s = {c}^{k} - {\left( {A}^{k}\right) }^{T}y \) for any \( y \in {\Omega }^{k} \) and \( {B}^{0} = {8I},{B}^{k + 1} = \) \( {B}^{k} + \frac{1}{m}{a}_{k + 1}{a}_{k + 1}^{T} \) . Then \[ {A}^{k}{S}^{-2}{\left( {A}^{k}\right) }^{T} \succcurlyeq {B}^{k} \] that is, \[ {A}^{k}{S}^{-2}{\left( {A}^{k}\right) }... | Proof. Let \( Y = \operatorname{diag}\left( y\right) \) . Then \[ {A}^{k}{S}^{-2}{\left( {A}^{k}\right) }^{T} = {Y}^{-2} + {\left( I - Y\right) }^{-2} + \mathop{\sum }\limits_{{j = 1}}^{k}\frac{{a}_{j}{a}_{j}^{T}}{{\left( {s}_{{2m} + j}\right) }^{2}} \] \[ \succcurlyeq {Y}^{-2} + {\left( I - Y\right) }^{-2} + \frac{1}{... | Yes |
Lemma 8.8\n\n\[ \mathop{\sum }\limits_{{j = 0}}^{k}{\left( {\omega }^{j}\right) }^{2} \leq 2{m}^{2}\log \left( {1 + \frac{k + 1}{8{m}^{2}}}\right) . \] | Proof. Note that\n\n\[ \det {B}^{k + 1} = \det \left( {{B}^{k} + \frac{1}{m}{a}_{k + 1}{a}_{k + 1}^{T}}\right) = \left( {1 + \frac{{\left( {\omega }^{k}\right) }^{2}}{m}}\right) \det {B}^{k}. \]\n\nThus\n\n\[ \log \det {B}^{k + 1} = \log \det {B}^{k} + \log \left( {1 + \frac{{\left( {\omega }^{k}\right) }^{2}}{m}}\righ... | Yes |
Theorem 8.9 The cutting plane algorithm stops with a feasible solution as soon as \( k \) satisfies:\n\n\[ \frac{{\epsilon }^{2}}{m} \geq \frac{\frac{1}{2} + {2m}\log \left( {1 + \frac{k + 1}{8{m}^{2}}}\right) }{{2m} + k + 1}\exp \left( {-{2\delta }\frac{k + 1}{k + 1 + {2m}}}\right) . \] | Proof. From relation (8.10) and Lemma 8.4,\n\n\[ \left( {{2m} + k + 1}\right) \log \epsilon \leq \mathcal{B}\left( {\Omega }^{k + 1}\right) \]\n\n\[ \leq \mathcal{B}\left( {\Omega }^{0}\right) + \frac{1}{2}\mathop{\sum }\limits_{{j = 0}}^{k}\log {\left( {\bar{r}}^{j}\right) }^{2} - \left( {k + 1}\right) \delta \]\n\n\[... | Yes |
Lemma 8.13 Let\n\n\\[ \n{\\mu }^{k} = \\frac{{S}^{k} \\bullet {X}^{k}}{n} = \\frac{C \\bullet {X}^{k} - {z}^{k}}{n}\\;\\text{ and }\\;\\mu = \\frac{S\\left( {z}^{k}\\right) \\bullet {X}^{k}}{n}. \n\\]\n\nIf\n\n\\[ \n\\begin{Vmatrix}{P\\left( {z}^{k}\\right) }\\end{Vmatrix} < \\min \\left( {\\beta \\sqrt{\\frac{n}{n + {... | Proof. The proof is by contradiction. For example, if the first inequality of (8.17) is not true, then \\( {\\left( {X}^{k}\\right) }^{.5}S\\left( {z}^{k}\\right) {\\left( {X}^{k}\\right) }^{.5} \\) has at least one eigenvalue less than or equal to zero, and\n\n\\[ \n\\begin{Vmatrix}{P\\left( {z}^{k}\\right) }\\end{Vma... | No |
Theorem 8.14 Given \( {X}^{k} \in {\mathcal{F}}_{p} \) and \( \left( {{y}^{k},{S}^{k}}\right) \in {\mathcal{F}}_{d} \), let \( \rho = \sqrt{n},{z}^{k} = {b}^{T}{y}^{k} \) , \( {X}^{k + 1} \) be given by (8.11), and \( {y}^{k + 1} = y\left( {z}^{k}\right) \) in (8.14) and \( {S}^{k + 1} = S\left( {z}^{k}\right) \) in (8... | Proof. If (8.16) does not hold, i.e.,\n\n\[ \begin{Vmatrix}{P\left( {z}^{k}\right) }\end{Vmatrix} \geq \min \left( {\beta \sqrt{\frac{n}{n + {\beta }^{2}}},1 - \beta }\right) ,\]\n\nthen, since \( \psi \left( {{X}^{k + 1},{S}^{k}}\right) - \psi \left( {{X}^{k},{S}^{k}}\right) = \mathcal{P}\left( {{X}^{k + 1},{z}^{k}}\r... | Yes |
Corollary 8.15 Let \( \rho = \sqrt{n} \) . Then, Algorithm 8.2 terminates in at most \( O\left( {\sqrt{n}\log \left( {C \bullet {X}^{0} - {b}^{T}{y}^{0}}\right) /\epsilon }\right) \) iterations with\n\n\[ C \bullet {X}^{k} - {b}^{T}{y}^{k} \leq \epsilon . \] | Proof. In \( O\left( {\sqrt{n}\log \left( {{S}^{0} \bullet {X}^{0}/\epsilon }\right) }\right) \) iterations\n\n\[ - \sqrt{n}\log \left( {{S}^{0} \bullet {X}^{0}/\epsilon }\right) = \psi \left( {{X}^{k},{S}^{k}}\right) - \psi \left( {{X}^{0},{S}^{0}}\right) \]\n\n\[ \geq \sqrt{n}\log {S}^{k} \bullet {X}^{k} + n\log n - ... | Yes |
Lemma 8.16 Let the direction \( {d}_{X},{d}_{y} \) and \( {d}_{S} \) be generated by equation (8.18) with \( \gamma = n/\left( {n + \rho }\right) \), and let\n\n\[ \theta = \frac{\alpha }{{\begin{Vmatrix}{V}^{-1/2}\end{Vmatrix}}_{\infty }\begin{Vmatrix}{\frac{I \bullet V}{n + \rho }{V}^{-1/2} - {V}^{1/2}}\end{Vmatrix}}... | Applying Lemma 4.12 to \( v \in {\mathcal{R}}^{n} \) as the vector of the \( n \) eigenvalues of \( V \), we can prove the following lemma: | No |
Lemma 8.20 Let \( r \in {\mathcal{R}}^{n} \) . Assume that \( \left( {{x}^{1},{s}^{1}}\right) \in {S}_{+ + }\left( {{\theta }^{1}r}\right) \) and \( \left( {{x}^{2},{s}^{2}}\right) \in \) \( {S}_{+ + }\left( {{\theta }^{2}r}\right) \) where \( {\theta }^{1} \) and \( {\theta }^{2} \) are two real numbers. Then | \[ \left( {{\theta }^{2} - {\theta }^{1}}\right) {r}^{T}\left( {{x}^{2} - {x}^{1}}\right) \leq {\left( {x}^{2} - {x}^{1}\right) }^{T}\left( {{s}^{2} - {s}^{1}}\right) . \] | Yes |
Lemma 8.21 Let \( r \in {\mathcal{R}}^{n} \) and \( {\theta }^{0} \leq {\theta }^{1} \) . Assume \( {S}_{+ + }\left( {{\theta }^{0}r}\right) \neq \varnothing \) and \( {S}_{+ + }\left( {{\theta }^{1}r}\right) \neq \varnothing \) . Then, for every \( \delta > 0 \), the union of\n\n\[ \n{C}_{+ + }\left( {{\theta r},\delt... | Proof. Let \( \left( {{x}^{0},{s}^{0}}\right) \in {S}_{+ + }\left( {{\theta }^{0}r}\right) \) and \( \left( {{x}^{1},{s}^{1}}\right) \in {S}_{+ + }\left( {{\theta }^{1}r}\right) \), and\n\n\[ \n\max \left( {{\left( {s}^{0}\right) }^{T}{x}^{0},{\left( {s}^{1}\right) }^{T}{x}^{1}}\right) \leq \delta .\n\]\n\nLet \( \thet... | Yes |
Theorem 8.23 \( {R}_{+ + } \) is an open convex subset of \( {\mathcal{R}}^{n} \) . | Proof. The openness has been discussed earlier. Let\n\n\[ \n{\theta }^{ * } = \inf \{ \theta : \left\lbrack {\theta ,1}\right\rbrack \subset \Theta \} .\n\]\n\nSince \( 1 \in \Theta \), we know by Lemma 8.22 that \( {\theta }^{ * } < 1 \) and \( {\theta }^{ * } \notin \Theta \) . If \( {\theta }^{ * } < 0 \) , \( {\the... | Yes |
Lemma 8.24 Let \( \nabla f \) be positive semi-definite in \( {\mathcal{R}}_{ + }^{n} \) . Then \( \nabla \psi \) is positive semi-definite in \( {\mathcal{R}}_{ + }^{n + 1} \), i.e. given \( \left( {x;\tau }\right) > 0 \) , \[ {\left( {d}_{x};{d}_{\tau }\right) }^{T}\nabla \psi \left( {x,\tau }\right) \left( {{d}_{x};... | Proof. \[ {\left( {d}_{x};{d}_{\tau }\right) }^{T}\nabla \psi \left( {x,\tau }\right) \left( {{d}_{x};{d}_{\tau }}\right) \] \[ = {d}_{x}^{T}\nabla f\left( {x/\tau }\right) {d}_{x} - {d}_{x}^{T}\nabla f\left( {x/\tau }\right) x\left( {{d}_{\tau }/\tau }\right) \] \[ - \left( {{d}_{\tau }/\text{ tau }}\right) {x}^{T}\na... | Yes |
Theorem 8.25 Let \( \psi \) be given by (8.23). Then,\ni) \( \psi \) is a continuous homogeneous function in \( {\mathcal{R}}_{ + }^{n + 1} \) with degree 1 and for any \( \left( {x;\tau }\right) \in {\mathcal{R}}_{ + }^{n + 1} \)\n\n\[{\left( x;\tau \right) }^{T}\psi \left( {x,\tau }\right) = 0\]\n\nand\n\n\[{\left( x... | Proof. The proof of (i) is straightforward. | No |
Theorem 8.26 Consider (HMCP).\n\ni) For any \( 0 < \theta \leq 1 \), there exists a strictly positive point \( (x > 0,\tau > 0, s > \) \( 0,\kappa > 0) \) such that\n\n\[ \left( \begin{array}{l} s \\ \kappa \end{array}\right) - \psi \left( {x,\tau }\right) = \left( \begin{matrix} s - {\tau f}\left( {x/\tau }\right) \\ ... | Proof. We prove (i). Again, the set\n\n\[ {H}_{+ + } \mathrel{\text{:=}} \left\{ {\left( \begin{array}{l} s \\ \kappa \end{array}\right) - \psi \left( {x,\tau }\right) : \;\left( {x,\tau, s,\kappa }\right) > 0}\right\} \]\n\nis open and convex. We have \( \left( {{r}^{0};{z}^{0}}\right) \in {H}_{+ + } \) by constructio... | Yes |
Lemma 8.27 . The direction \( \left( {{d}_{x},{d}_{s}}\right) \) satisfies\n\n\[ \n{d}_{x}^{T}{d}_{s} = {d}_{x}^{T}\nabla \psi \left( {x}^{k}\right) {d}_{x} + \eta \left( {1 - \eta - \gamma }\right) \left( {n + 1}\right) {\mu }^{k}.\n\] | Proof. Premultiplying each side of (8.42) by \( {d}_{x}^{T} \) gives\n\n\[ \n{d}_{x}^{T}{d}_{s} - {d}_{x}^{T}\nabla \psi \left( {x}^{k}\right) {d}_{x} = - \eta {d}_{x}^{T}\left( {{s}^{k} - \psi \left( {x}^{k}\right) }\right) .\n\]\n\n(8.44)\n\n\n\nMultiplying each side of (8.42) by \( {x}^{k} \) and using (8.41) give\n... | Yes |
Lemma 8.28 . Consider the new iterate \( \left( {{x}^{ + },{s}^{ + }}\right) \) given by (8.46) and (8.47).\n\ni) \( {r}^{ + } = \left( {1 - {\alpha \eta }}\right) {r}^{k} \)\n\nii) \( \left( {x}^{ + }\right) {s}^{ + } = {\left( {x}^{k}\right) }^{T}{s}^{k}\left( {1 - \alpha \left( {1 - \gamma }\right) }\right) + {\alph... | Proof. From (8.47)\n\n\[ \n{r}^{ + } = {s}^{ + } - \psi \left( {x}^{ + }\right) \]\n\n\[ \n= \left( {1 - {\alpha \eta }}\right) {r}^{k}\text{.} \]\n\n\nNext we prove (ii) Using (8.41), (8.43), and Lemma 8.27, we have\n\n\[ \n{\left( {x}^{ + }\right) }^{T}{s}^{ + } \]\n\n\[ \n= {\left( {x}^{ + }\right) }^{T}\left( {{s}^... | Yes |
Theorem 8.29 Assume that \( \psi \) is scaled Lipschitz with \( v = {v}_{\psi } \) and at iteration \( k \)\n\n\[ \begin{Vmatrix}{{X}^{k}{s}^{k} - {\mu }^{k}e}\end{Vmatrix} \leq \beta {\mu }^{k},\;{\mu }^{k} = \frac{{\left( {x}^{k}\right) }^{T}{s}^{k}}{n + 1}, \]\n\nwhere\n\n\[ \beta = \frac{1}{3 + 4{v}_{\psi }\left( {... | Proof. It follows from Lemma 8.28 that \( {\mu }^{ + } = \gamma {\mu }^{k} \) . From (8.43) we further\n\nhave\n\n\[ {S}^{k}{d}_{x} + {X}^{k}{d}_{s} = - {X}^{k}{s}^{k} + {\mu }^{ + }e. \]\n\nHence,\n\n\[ {D}^{-1}{d}_{x} + D{d}_{s} = - {\left( {X}^{k}{S}^{k}\right) }^{-1/2}\left( {{X}^{k}{s}^{k} - {\mu }^{ + }e}\right) ... | Yes |
Lemma 9.2 Let \( {\gamma }^{0} < {\gamma }^{1} < {\gamma }^{ * } \) . Then, \[ \mathcal{B}\left( {\gamma }^{0}\right) > \mathcal{B}\left( {\gamma }^{1}\right) \] | Proof. Let \( {y}^{1} \) be the analytic center of \( \Gamma \left( {\gamma }^{1}\right) \) . Then, since \[ \left( {B - {\gamma }^{0}A}\right) {y}^{1} = \left( {B - {\gamma }^{1}A}\right) {y}^{1} + \left( {{\gamma }^{1} - {\gamma }^{0}}\right) A{y}^{1} \] and \( \left( {{\gamma }^{1} - {\gamma }^{0}}\right) A{y}^{1} \... | Yes |
Lemma 9.4 Let\n\n\\[ \n{\\Delta \\gamma } = \\beta \\min \\left\\{ {\\frac{1}{\\begin{Vmatrix}{X}^{0}A{Y}^{0}e\\end{Vmatrix}},{\\frac{1}{\\begin{Vmatrix}{Y}^{0}{A}^{T}{X}^{0}e\\end{Vmatrix}}}}\\right\\} ,\n\\]\n\nfor some constant \\( 0 < \\beta < 1 \\) and let \\( {\\gamma }^{1} = {\\gamma }^{0} + {\\Delta \\gamma } \... | Proof. The two equations in systems (9.5) and (9.6) for \\( \\gamma = {\\gamma }^{1} \\) can easily be verified. The inequality for the norm can be proved from relation (9.4) and the choice of \\( {\\Delta \\gamma } \\) .\n\n\\[ \n\\parallel \\bar{X}\\bar{s} - e{\\parallel }^{2} = {\\begin{Vmatrix}{X}^{0}\\left( {s}^{0... | Yes |
Lemma 9.6 Let positive constants \( \delta \) and \( \beta \) satisfy \( \delta + \sqrt{2}\beta < 1 \) . Let \( \left( {{x}^{0},{z}^{0}}\right) \) and \( \left( {{s}^{0},{y}^{0}}\right) \) be a \( \delta \) -approximate center pair for \( \Gamma \left( {\gamma }^{0}\right) \) . Let \( {\Delta \gamma },{\gamma }^{1} \) ... | \[ \begin{Vmatrix}\left( \begin{matrix} \bar{X}\bar{s} - e \\ \bar{Z}\bar{y} - e \end{matrix}\right) \end{Vmatrix} \leq \delta + \sqrt{2}\beta < 1 \] and \[ \left( {\bar{s},\bar{y}}\right) > 0\;\text{ and }\;\left( {\bar{x},\bar{z}}\right) > 0. \] | Yes |
Proposition 9.9 Any limit point of\n\n\[ \mathop{\lim }\limits_{{k \rightarrow \infty }}\frac{{x}^{k}}{{e}^{T}{x}^{k}} \]\n\nwhere \( {x}^{k} \) is generated by Algorithm 9.1, is a solution for the dual NEG problem with \( \eta = {\eta }^{ * } \) under Assumptions 9.1,9.2, and 9.4. | Proof. For simplicity, we assume that \( \left( {{x}^{k},{y}^{k},{s}^{k},{z}^{k}}\right) \) is exactly centered, i.e., it satisfies relations (9.4), (9.5), and (9.6). Since at least one component of \( {s}^{k} \) converges to zero, \( {e}^{T}{x}^{k} = {e}^{T}{\left( {S}^{k}\right) }^{-1}e \) tends to \( + \infty \) . M... | Yes |
Theorem 9.10 The potential reduction algorithm with \( \rho = \theta \left( n\right) > 0 \) solves the LCP for which \( \gamma \left( {M, q,\epsilon }\right) > 0 \) in\n\n\[ O\left( \frac{{\psi }^{0} + \rho \log \left( {1/\epsilon }\right) - n\log n}{\xi \left( {\gamma \left( {M, q,\epsilon }\right) }\right) }\right) \... | Proof. Since \( \mathcal{F} \) is nonempty, by solving a linear program in polynomial time, we can find an approximate analytic center \( \left( {{x}^{0},{s}^{0}}\right) \) of \( \mathcal{F} \) . Due to (9.15),(9.16), and (9.17) the potential function is reduced by \( O\left( {\xi \left( {\gamma \left( {M, q,\epsilon }... | Yes |
Proposition 9.12 Let \( \rho \geq n \) . Then, for \( M \) being a diagonal and PSD matrix, and any \( q \in {\mathcal{R}}^{n} \) ,\n\n\[ \gamma \left( {M, q,\epsilon }\right) \geq n \] | Proof. If \( M \) is diagonal and PSD, then the matrix\n\n\[ I - \left( {X{M}^{T} - S}\right) {\left( {S}^{2} + M{X}^{2}{M}^{T}\right) }^{-1}\left( {{MX} - S}\right) \]\n\nis diagonal. It is also PSD since the \( j \) th diagonal component is\n\n\[ 1 - \frac{{\left( {M}_{jj}{x}_{j} - {s}_{j}\right) }^{2}}{{s}_{j}^{2} +... | Yes |
Lemma 9.14 \( \begin{Vmatrix}{p}^{k}\end{Vmatrix} < 1 \) implies\n\n\[ \n{s}^{k} + {M}^{T}{\pi }^{k} > 0;\;{x}^{k} - {\pi }^{k} > 0 \]\n\nand\n\n\[ \n\frac{{2n} - \sqrt{2n}}{n + \rho }{\Delta }^{k} < \bar{\Delta } < \frac{{2n} + \sqrt{2n}}{n + \rho }{\Delta }^{k} \]\n\nwhere \( \bar{\Delta } = {\left( {x}^{k}\right) }^... | Proof. The proof is by contradiction. Let \( \bar{s} = {s}^{k} + {M}^{T}{\pi }^{k} \) and \( \bar{x} = {x}^{k} - {\pi }^{k} \) . It is obvious that if \( \bar{s} \ngtr 0 \) or \( \bar{x} \ngtr 0 \), then\n\n\[ \n{\begin{Vmatrix}{p}^{k}\end{Vmatrix}}^{2} \geq 1 \]\n\nOn the other hand, we have\n\n\[ \n{\begin{Vmatrix}{p... | Yes |
Proposition 9.16 Let\n\n\\[ \n\\left\\{ {\\pi : {x}^{T}s - {q}^{T}\\pi > 0, x - \\pi > 0}\\right. \\text{and}\\left. {s + {M}^{T}\\pi > 0\\text{for some}\\left( {x, s}\\right) \\in \\mathcal{F}}\\right\\} \n\\]\n\nbe empty for an LCP \\( \\left( {M, q}\\right) \\) . Then, for \\( 0 < \\rho \\leq n - \\sqrt{2n} \\)\n\n\... | Proof. The proof again results from Lemma 9.14. | No |
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