Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
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\[ M = \left( \begin{array}{ll} 0 & - 1 \\ 1 & - 1 \end{array}\right) \;\text{ and }\;q = \left( \begin{array}{l} 2 \\ 0 \end{array}\right) . \] | For this example the potential reduction algorithm constantly generates the solution \[ x = \left( {2;2}\right) \text{ and }s = \left( {0;0}\right) \] from virtually any interior starting point, avoiding the trivial solution \( x = 0 \) and \( s = q \) . | Yes |
\[ \mathcal{F} \) is nonempty since \( x = \left( {3;1}\right) \) is an interior feasible point; \( {\sum }^{ + } \) is empty since \( {x}_{1} - {x}_{2} > 1,{x}_{1} - {\pi }_{1} > 0,{x}_{2} - {\pi }_{2} > 0,{x}_{1} - {x}_{2} - 1 + {\pi }_{1} + 2{\pi }_{2} > 0 \) and \( 2{x}_{1} - 1 - {\pi }_{1} > 0 \) imply\n\n\[ \n{x}... | \[ \n= {x}^{T}\left( {{Mx} + q}\right) - {q}^{T}\pi \n\]\n\n\[ \n= {x}_{1}\left( {{x}_{1} - {x}_{2}}\right) + 2{x}_{1}{x}_{2} - {x}_{1} - {x}_{2} + {\pi }_{1} + {\pi }_{2} \n\]\n\n\[ \n= {x}_{1}^{2} + {x}_{1}{x}_{2} - 2{x}_{1} - {x}_{2} + 1 + \left( {{x}_{1} - {x}_{2} - 1 + {\pi }_{1} + 2{\pi }_{2}}\right) + \left( {{x... | Yes |
\[ {\left( a + b\right) }^{2} + {\left( a + c\right) }^{2} \geq {a}^{2} - {2bc}. \] | \[ {\left( a + b\right) }^{2} + {\left( a + c\right) }^{2} = 2{a}^{2} + {2a}\left( {b + c}\right) + {b}^{2} + {c}^{2} \] \[ = {a}^{2} - {2bc} + {a}^{2} + {2a}\left( {b + c}\right) + {\left( b + c\right) }^{2} \] \[ = {a}^{2} - {2bc} + {\left( a + b + c\right) }^{2} \geq {a}^{2} - {2bc}. \] | Yes |
Theorem 9.20 Let \( \psi \left( {{x}^{0},{s}^{0}}\right) \leq O\left( {n\log n}\right) \) and \( M \) be a P-matrix. Then, the potential reduction algorithm terminates at \( {\left( {x}^{k}\right) }^{T}{s}^{k} \leq \epsilon \) in iterations and each iteration uses at most \( O\left( {n}^{3}\right) \) arithmetic operati... | \[ O\left( {{n}^{2}\max \{ \left| \lambda \right| /\left( {\theta n}\right) ,1\} \log \left( {1/\epsilon }\right) }\right) \] | Yes |
Proposition 9.21 Let \( \rho > 0 \) and be fixed. Then, for \( M \) being a row-sufficient matrix and \( \left\{ {\left( {x, s}\right) \in \mathcal{F} : \psi \left( {x, s}\right) \leq {\psi }^{0}}\right\} \) being bounded, | Proof. It is easy to show that for any \( \left( {x, s}\right) \in \mathcal{F} \), \[ \parallel g\left( {x, s}\right) {\parallel }_{H}^{2} > 0. \] Moreover, for all \( \left( {x, s}\right) \in \mathcal{F},{x}^{T}s \geq \epsilon \) and \( \psi \left( {x, s}\right) \leq {\psi }^{0} \), \[ {\psi }^{0} \geq \psi \left( {x,... | Yes |
Lemma 9.22 The scaled gradient projection \( \begin{Vmatrix}{p}^{k}\end{Vmatrix} < 1 \) implies\n\n\[ \n{s}^{k} - {A}^{T}{\pi }^{k} > 0;\;{x}^{k} - {B}^{T}{\pi }^{k} > 0;\; - {C}^{T}{\pi }^{k} > 0, \n\]\n\nand\n\n\[ \n\frac{{2n} + d - \sqrt{{2n} + d}}{n + \rho }{\Delta }^{k} < \bar{\Delta } < \frac{{2n} + d + \sqrt{{2n... | Proof. The proof is by contradiction. It is obvious that if \( {s}^{k} - {A}^{T}{\pi }^{k} \ngtr 0 \) or \( {x}^{k} - {B}^{T}{\pi }^{k} \ngtr 0 \) or \( - {C}^{T}{\pi }^{k} \ngtr 0 \), then\n\n\[ \n\begin{Vmatrix}{p}^{k}\end{Vmatrix} \geq {\begin{Vmatrix}{p}^{k}\end{Vmatrix}}_{\infty } \geq 1 \n\]\n\nOn the other hand,... | Yes |
Theorem 9.23 For any given \( 0 < \epsilon \leq 1 \), let \( n + \rho = \left( {{2n} + d + \sqrt{{2n} + d}}\right) /\epsilon \) . Then, under Assumptions 9.5 and 9.6, the potential reduction algorithm terminates in at most \( O\left( {{\left( n + \rho \right) }^{2}\log \left( {1/\epsilon }\right) + \left( {n + \rho }\r... | Proof. The proof directly follows Lemma 9.22. If \( \begin{Vmatrix}{p}^{k}\end{Vmatrix} \geq 1 \) for all \( k \), then from (9.29)\n\n\[ \psi \left( {{x}^{k + 1},{s}^{k + 1},{z}^{k + 1}}\right) - \psi \left( {{x}^{k},{s}^{k},{z}^{k}}\right) \leq - O\left( {1/\left( {n + \rho }\right) }\right) . \]\n\nTherefore, in at ... | Yes |
Theorem 9.24 Let \( B{A}^{T} \) be negative semi-definite. Furthermore, let \( \rho = \) \( n + d + \sqrt{{2n} + d} \) . Then, the potential reduction algorithm generates a solution to the GLCP in \( O\left( {{\left( 2n + d\right) }^{2}\log \left( {1/\epsilon }\right) + \left( {{2n} + d}\right) d\log R}\right) \) itera... | Proof. Basically, we show that \( \begin{Vmatrix}{p}^{k}\end{Vmatrix} \geq 1 \) for all \( k \) if \( B{A}^{T} \) is negative semidefinite and \( \rho \geq n + d + \sqrt{{2n} + d} \) . We prove it by contradiction. Suppose \( \begin{Vmatrix}{p}^{k}\end{Vmatrix} < 1 \), then\n\n\[ \n{s}^{k} - {A}^{T}{\pi }^{k} > 0\;{x}^... | Yes |
\[ A = \left( \begin{array}{lll} 0 & 1 & {10} \\ 0 & 1 & 1 \\ 0 & 0 & 2 \end{array}\right) ,\;B = - I,\;C = \varnothing ,\;\text{ and }\;q = e. \] The starting point is \( {x}^{0} = \left( {2;2;2}\right) \). | The algorithm consistently generates the solution to the \( {LCP},{x}^{ * } = \left( {0;{0.5};{0.5}}\right) \) and \( {s}^{ * } = \left( {{4.5};0;0}\right) \). In this example, \( A \) is a so called \( {P}_{0} \) matrix, and it is indefinite. | Yes |
\[ A = \left( \begin{array}{lll} 0 & 1 & {10} \\ 0 & 0 & 1 \\ 0 & 0 & 2 \end{array}\right) ,\;B = - I,\;C = \varnothing ,\;\text{ and }\;q = e. \] | The starting point is again \( {x}^{0} = \left( {2;2;2}\right) \) . The algorithm consistently generates a KKT point of the LCP, \( \bar{x} = \left( {0;\alpha ;1}\right) ,\bar{s} = \left( {9 + \alpha ;0;1}\right) \) and \( \bar{\pi } = \left( {0; - 3;1}\right) \) for some \( \alpha > 0 \) . Note that there is no soluti... | No |
Proposition 9.29 Let \( \underline{q} \mathrel{\text{:=}} \underline{q}\left( Q\right) ,\bar{q} \mathrel{\text{:=}} - \underline{q}\left( {-Q}\right) ,\underline{p} \mathrel{\text{:=}} \underline{p}\left( Q\right) ,\bar{p} \mathrel{\text{:=}} - \underline{p}\left( {-Q}\right) \) ,\n\n\( \left( {\bar{y},\bar{z}}\right) ... | Proof. The first and second statements are straightforward to verify. Let \( X = \underline{x}{\underline{x}}^{T} \in {\mathcal{M}}^{n} \) . Then \( X \succcurlyeq 0, d\left( X\right) \leq e \) ,\n\n\[ {A}_{i} \bullet X = {\underline{x}}^{T}{A}_{i}\underline{x} = \mathop{\sum }\limits_{{j = 1}}^{n}{a}_{ij}{\left( {\und... | Yes |
Lemma 9.30 Let \( u \) be uniformly distributed on the unit sphere in \( {\mathcal{R}}^{n} \) . Then,\n\n\[ \underline{q}\left( Q\right) = \mathop{\operatorname{minimize}}\limits_{m}{\mathrm{E}}_{u}\left( {\sigma {\left( {V}^{T}u\right) }^{T}{DQD\sigma }\left( {{V}^{T}u}\right) }\right) \]\n\n\[ \text{s.t.}\;{A}_{i} \b... | Proof. Since, for any feasible \( V,{D\sigma }\left( {{V}^{T}u}\right) \) is a feasible point for the QP problem, we have\n\n\[ \underline{q}\left( Q\right) \leq {\mathrm{E}}_{u}\left( {\sigma {\left( {V}^{T}u\right) }^{T}{DQD\sigma }\left( {{V}^{T}u}\right) }\right) .\n\nOn the other hand, for any fixed \( u \) with \... | Yes |
Theorem 9.32\n\n\\[ \n\\underline{q}\\left( Q\\right) = \\operatorname{infimum}\\frac{2}{\\pi }Q \\bullet \\left( {D\\arcsin \\left\\lbrack {{D}^{-1}X{D}^{-1}}\\right\\rbrack D}\\right) \n\\]\n\n\\[ \n\\text{s.t.}\\;{A}_{i} \\bullet X = {b}_{i}, i = 1,\\ldots, m\\text{,}\n\\]\n\n\\[ \nd\\left( X\\right) \\leq e, X \\su... | Proof. For any \\( X = {V}^{T}V \\succ 0, d\\left( X\\right) \\leq e \\), we have\n\n\\[ \n{\\mathrm{E}}_{u}\\left( {\\sigma \\left( {{v}_{i}^{T}u}\\right) \\sigma \\left( {{v}_{j}^{T}u}\\right) }\\right) = 1 - 2\\Pr \\left\\{ {\\sigma \\left( {{v}_{i}^{T}u}\\right) \\neq \\sigma \\left( {{v}_{j}^{T}u}\\right) }\\right... | Yes |
Corollary 9.34 Let \( X = {V}^{T}V \succ 0, d\left( X\right) \leq e,{A}_{i} \bullet X = {b}_{i}\left( {i = 1,\ldots, m}\right) \) , \( D = \operatorname{diag}\left( {\sqrt{{x}_{11}},\ldots ,\sqrt{{x}_{nn}}}\right) \), and \( \widehat{x} = {D\sigma }\left( {{V}^{T}u}\right) \) where \( u \) with \( \parallel u\parallel ... | \[ \mathop{\lim }\limits_{{X \rightarrow \underline{X}}}{\mathrm{E}}_{u}\left( {q\left( \widehat{x}\right) }\right) = \mathop{\lim }\limits_{{X \rightarrow \underline{X}}}\frac{2}{\pi }Q \bullet \left( {D\arcsin \left\lbrack {{D}^{-1}X{D}^{-1}}\right\rbrack D}\right) \leq \frac{2}{\pi }\underline{p} + \left( {1 - \frac... | Yes |
Theorem 9.35 Let \( \widehat{x} \) be randomly generated from \( \underline{X} \) . Then\n\n\[ \frac{{\mathrm{E}}_{u}q\left( \widehat{x}\right) - \underline{q}}{\bar{q} - \underline{q}} \leq \frac{\pi }{2} - 1 < 4/7 \] | Proof. Since\n\n\[ \bar{p} \geq \bar{q} \geq \frac{2}{\pi }\bar{p} + \left( {1 - \frac{2}{\pi }}\right) \underline{p} \geq \left( {1 - \frac{2}{\pi }}\right) \bar{p} + \frac{2}{\pi }\underline{p} \geq \underline{q} \geq \underline{p}, \]\n\nwe have, from Corollary 9.34,\n\n\[ \frac{{\mathrm{E}}_{u}q\left( \widehat{x}\r... | Yes |
Theorem 10.2 Given any initial point in \( {\mathcal{N}}_{\infty }^{ - }\left( \eta \right) \) for any constant \( \eta \in \) \( \left( {0,1}\right) \), Algorithm 10.1, with any constant \( 0 < \gamma < 1 \), will terminate in \( O\left( {{n}^{\frac{r + 1}{2r}}\log \left( {{\left( {x}^{0}\right) }^{T}{s}^{0}/\epsilon ... | As we can see that if \( r = n \) and \( n \) increases, the iteration complexity of the algorithm tends to \( O\left( {\sqrt{n}\log \left( {{\left( {x}^{0}\right) }^{T}{s}^{0}/\epsilon }\right) }\right) \) asymptotically. Furthermore, a popular choice for \( \gamma \) in practice is not a constant but \( \gamma = O\le... | No |
Theorem 10.3 Let \( B \) be an optimal basis for \( {LP}\left( {A,{b}^{k},{c}^{k}}\right) \) . Then, there is \( 0 < \bar{t} < \infty \) such that \( B \) must be also an optimal basis for the original \( {LP}\left( {A, b, c}\right) \) when \( {\left( {x}^{k}\right) }^{T}{s}^{k} \leq {2}^{-\bar{t}} \) . Furthermore, if... | This advocates for an application of the above basis identification procedure to the perturbed problem (10.18), since an optimal complementary solution to problem (10.18) is known, and it will be an optimal basis for (LP) when problem (10.18 is near (LP). | No |
Consider the problem\n\n\[ \n\\text{minimize}\;{c}_{1}{x}_{1} + {c}_{2}{x}_{2} + \\cdots + {c}_{n}{x}_{n} \n\]\n\n\[ \n\\text{subject to}\;{a}_{11}{x}_{1} + {a}_{12}{x}_{2} + \\cdots + {a}_{1n}{x}_{n} \\leq {b}_{1} \n\]\n\n\[ \n{a}_{21}{x}_{1} + {a}_{22}{x}_{2} + \\cdots + {a}_{2n}{x}_{n} \\leq {b}_{2} \n\]\n\n\[ \n{a}... | In this case the constraint set is determined entirely by linear inequalities. The problem may be alternatively expressed as\n\nminimize \( \\;{c}_{1}{x}_{1} + {c}_{2}{x}_{2} + \\cdots + {c}_{n}{x}_{n} \)\n\n\[ \n\\text{subject to}\;{a}_{11}{x}_{1} + {a}_{12}{x}_{2} + \\cdots + {a}_{1n}{x}_{n} + {y}_{1}\; = {b}_{1} \n\... | Yes |
If the linear inequalities of Example 1 are reversed so that a typical inequality is\n\n\[ \n{a}_{i1}{x}_{1} + {a}_{i2}{x}_{2} + \cdots + {a}_{in}{x}_{n} \geq {b}_{i}, \n\]\n\nit is clear that this is equivalent to\n\n\[ \n{a}_{i1}{x}_{1} + {a}_{i2}{x}_{2} + \cdots + {a}_{in}{x}_{n} - {y}_{i} = {b}_{i} \n\]\nwith \( {y... | Variables, such as \( {y}_{i} \), adjoined in this fashion to convert a \ | No |
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 (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(3)\n\nwhere we require \( {u}_{1} \geq 0 \) and \( {v}_{1} \geq 0 \... | 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 (1) which has a nonzero coefficient for \( {x}_{1} \). Say, for example,\n\n\[ \n{a}_{i1}{x}_{1} + {a}_{i2}... | If this expression is substituted for \( {x}_{1} \) everywhere in (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 be... | 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 dietician 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}_{11}{x}_{1} + {a}_{12}{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 4 (A warehousing problem). Consider the problem of operating a warehouse, by buying and selling the stock of a certain commodity, in order to maximize profit over a certain length of time. The warehouse has a fixed capacity \( C \), and there is a cost \( r \) per unit for holding stock for one period. The pric... | If the constraints are written out explicitly for the case \( n = 3 \), they take the form\n\n<table><thead><tr><th></th><th></th><th></th><td rowspan=\ | No |
Example 6 (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 \( s \) to \[ \text{maximize}\;{\mathbf{\pi }}^{T}\mathbf{x} - s \] \[ \text{subject to}\;\mathbf{{Ax}} - \mathbf{1}s \leq \mathbf{0} \] \[ \mathbf{x} \leq \mathbf{q} \] \[ \mathbf{x} \geq \mathbf{0}, \] where 1 is the vector of all 1 's. Not... | Yes |
Corollary 1. If the convex set \( \mathbf{K} \) corresponding to (17) 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 \( \mathbf{K} \) corresponding to (17) 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 \( \mathbf{K} \) are a subset of these basic solutions. | Yes |
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 | This set is illustrated in Fig. 2.4. We see by inspection that this set has five extreme points. In order to compare this example with our general results we must introduce slack variables to yield the equivalent set in \( {E}^{5} \) :\n\n\[ \n{x}_{1} + \frac{8}{3}{x}_{2} + {x}_{3}\; = 4 \]\n\n\[ \n{x}_{1} + {x}_{2}\; ... | 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\[ \n\begin{array}{rrrrrrr} {x}_{1} & {x}_{2} & {x}_{3} & {x}_{4} & {x}_{5} & {x}_{6} & \\ 1 & 0 & 0 & \text{ (1) } & 1 & - 1 & 5 \\ 0 & 1 & 0 & 2 & - 3 & 1 & 3 \\ 0 & 0 & 1 & - 1 & 2 & - 1 & - 1 \end{array} \]\n\nThe circle indicated is our first pivot element and corresponds ... | 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{array}{rrrrrrr} {\\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 & ... | No |
Consider the system\n\n\[\n\\begin{array}{rrrrrrr} {\\mathbf{a}}_{1} & {\\mathbf{a}}_{2} & {\\mathbf{a}}_{3} & {\\mathbf{a}}_{4} & {\\mathbf{a}}_{5} & {\\mathbf{a}}_{6} & \\mathbf{b} \\\\\n1 & 0 & 0 & 2 & 4 & 6 & 4 \\\\\n0 & 1 & 0 & 1 & 2 & 3 & 3 \\\\\n0 & 0 & 1 & - 1 & 2 & 1 & 1 \\end{array}\n\]\n\nwhich has basis \( ... | The new tableau is\n\n\[\n\\begin{array}{rrrrrrr} {\\mathbf{a}}_{1} & {\\mathbf{a}}_{2} & {\\mathbf{a}}_{3} & {\\mathbf{a}}_{4} & {\\mathbf{a}}_{5} & {\\mathbf{a}}_{6} & \\mathbf{b} \\\\\n1/2 & 0 & 0 & 1 & 2 & 3 & 2 \\\\\n- 1/2 & 1 & 0 & 0 & 0 & 0 & 1 \\\\\n1/2 & 0 & 1 & 0 & 4 & 4 & 3 \\end{array}\n\]\n\nwith correspon... | 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\[ {x}_{1} \geq 0,\;{x}_{2} \geq 0,{x}_{3} \geq 0. \] | 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![10936b3a-73c6-4b1... | Yes |
Find a basic feasible solution to\n\n\[ 2{x}_{1} + {x}_{2} + 2{x}_{3} = 4 \]\n\n\[ 3{x}_{1} + 3{x}_{2} + {x}_{3} = 3 \]\n\n\[ {x}_{1} \geq 0,{x}_{2} \geq 0,{x}_{3} \geq 0. \] | 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## Initial tableau\n\nA basic feasible solution to the expanded system... | 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 |
We solve again Example 1 of Section 3.4. The vectors are listed once for reference\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} \\\\\n2 & 1 & 1 & 1 & 0 & 0 & 2 \\\\\n1 & 2 & 3 & 0 & 1 & 0 & 5 \\\\\n2 & 2 & 1 ... | We start with an initial basic feasible solution and corresponding \( {\\mathbf{B}}^{-1} \) as shown in the tableau below\n\n<table><thead><tr><th>Variable</th><th></th><th>\\( {\\mathbf{B}}^{-1} \\)</th><th>)</th><th>\\( {\\mathbf{x}}_{\\mathbf{B}} \\)</th></tr></thead><tr><td>4</td><td>1</td><td>0</td><td>0</td><td>2... | No |
The diet problem, Example 1, Section 2.2, was the problem faced by a dietician trying to select a combination of foods to meet certain nutritional requirements at minimum cost. This problem has the form\n\nminimize \( \;{\mathbf{c}}^{T}\mathbf{x} \)\n\nsubject to \( \mathbf{{Ax}} \geq \mathbf{b} \)\n\n\[ \mathbf{x} \ge... | Imagine a pharmaceutical company that produces in pill form each of the nutrients considered important by the dietician. The pharmaceutical company tries to convince the dietician 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 2, Section 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 (6) and (7... | \[ \text{maximize}\mathop{\sum }\limits_{{i = 1}}^{m}{a}_{i}{u}_{i} + \mathop{\sum }\limits_{{j = 1}}^{n}{b}_{j}{v}_{j} \] \[ \text{subject to}\;{u}_{i} + {v}_{j} \leq {c}_{ij},\;i = 1,2,\ldots, m\text{,} \] \[ j = 1,2,\ldots, n\text{.} \] To interpret the dual problem, we imagine an entrepreneur who, feeling that he c... | Yes |
Lemma 1. (Weak Duality Lemma). If \( \mathbf{x} \) and \( \mathbf{\lambda } \) are feasible for (3) and (4), respectively, then \( {\mathbf{c}}^{T}\mathbf{x} \geq {\mathbf{\lambda }}^{T}\mathbf{b} \) . | Proof. We have\n\n\[ \n{\mathbf{\lambda }}^{T}\mathbf{b} = {\mathbf{\lambda }}^{T}\mathbf{A}\mathbf{x} \leq {\mathbf{c}}^{T}\mathbf{x} \n\] \n\nthe last inequality being valid since \( \mathbf{x} \geq \mathbf{0} \) and \( {\mathbf{\lambda }}^{T}\mathbf{A} \leq {\mathbf{c}}^{T} \) . | Yes |
Theorem 1 (Complementary slackness-asymmetric form). Let \( \mathbf{x} \) and \( \mathbf{\lambda } \) be feasible solutions for the primal and dual programs, respectively, in the pair (2). A necessary and sufficient condition that they both be optimal solutions is that \( {}^{ \dagger } \) for all \( i \n\n i) \( {x}_{... | Proof. If the stated conditions hold, then clearly \( \left( {{\mathbf{\lambda }}^{T}\mathbf{A} - {\mathbf{c}}^{T}}\right) \mathbf{x} = 0 \) . Thus \( {\mathbf{\lambda }}^{T}\mathbf{b} = \) \( {\mathbf{c}}^{T}\mathbf{x} \), and by the corollary to Lemma 1, Section 4.2, the two solutions are optimal. Conversely, if the ... | Yes |
Theorem 2 (Complementary slackness-symmetric form). Let \( \mathbf{x} \) and \( \mathbf{\lambda } \) be feasible solutions for the primal and dual programs, respectively, in the pair (1). A necessary and sufficient condition that they both be optimal solutions is that for all \( i \) and \( j \)\ni)\( {x}_{i} > 0 \Righ... | Proof. This follows by transforming the previous theorem. | No |
Consider the system of linear inequalities\n\n\[ \n{x}_{1} + 3{x}_{2} + 4{x}_{3} = 4 \]\n\n\[ \n2{x}_{1} + {x}_{2} + 3{x}_{3} = 6 \]\n\n(23)\n\n\[ \n{x}_{1} \geq 0,\;{x}_{2} \geq 0,\;{x}_{3} \geq 0. \]\n | By subtracting the second equation from the first and rearranging, we obtain\n\n\[ \n{x}_{1} = 2 + 2{x}_{2} + {x}_{3} \]\n\n(24)\n\nFrom this we observe that since \( {x}_{2} \) and \( {x}_{3} \) are nonnegative, the value of \( {x}_{1} \) is greater than or equal to 2 in any solution to the equalities. This means that... | Yes |
Theorem 1. The ellipsoid \( {E}_{k + 1} = \operatorname{ell}\left( {{\mathbf{y}}_{k + 1},{\mathbf{B}}_{k + 1}^{-1}}\right) \) defined as above is the ellipsoid of least volume containing \( \left( {1/2}\right) {E}_{k} \) . Moreover,\n\n\[ \n\frac{\operatorname{vol}\left( {E}_{k + 1}\right) }{\operatorname{vol}\left( {E... | 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\[ \n\frac{\operatorname{vol}\left( {E}_{k + 1}\right) }{\operatorname{vol}\left( {E}_{k}\right) } = \frac{\det \left( {\mat... | 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\( \min \)\n\[ \n- {x}_{1} \n\] \n\nsubject to \n\[ \n{x}_{1} + {x}_{3} = 1 \n\] \n\n\[ \n{x}_{2} + {x}_{4} = 1 \n\] \n\n\[ \n{x}_{1} \geq 0,{x}_{2} \geq 0,{x... | 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 2 linear constraint equations and the four equations \n\n\[ \n{y}_{1} + {s}_{1} = 1 \n\] \n\n\[ \n{y}_{2} + {s}_{2} = 0 \... | No |
Consider the dual of example 2. This is\n\n\\[ \n\\max {y}_{1} + {y}_{2} \n\\]\n\n\\[ \n\\text{subject to}{y}_{1} \\leq - 1 \n\\]\n\n\\[ \n{y}_{2} \\leq 0\\text{.} \n\\] | The solution to the dual barrier problem is easily found from the solution of the primal barrier problem to be\n\n\\[ \n{y}_{1}\\left( \\mu \\right) = - 1 - \\mu /{x}_{1}\\left( \\mu \\right) ,\\;{y}_{2} = - {2\\mu }. \n\\]\n\nAs \\( \\mu \\rightarrow 0 \\), we have \\( {y}_{1} \\rightarrow - 1,{y}_{2} \\rightarrow 0 \... | 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 (17)\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\log ... | Yes |
Theorem 1 Consider problems (HSDP).\n\n(i) (HSDP) has an optimal solution and its optimal solution set is bounded.\n\n(ii) The optimal value of (HSDP) is zero,\n\n\[ \left( {\mathbf{y},\mathbf{x},\tau ,\theta ,\mathbf{s},\kappa }\right) \in {\mathcal{F}}_{h}\;\text{ implies that }\;\left( {n + 1}\right) \theta = {\math... | Part (ii) of the theorem shows that as \( \theta \) goes to zero, the solution tends toward satisfying complementary slackness between \( \mathbf{x} \) and \( \mathbf{s} \) and between \( \tau \) and \( \kappa \) . Part (iii) shows that at a solution with \( \theta = 0 \), the complemenary slackness is strict in the se... | No |
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 assumme 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 \( \... | Yes |
A transportation problem always has a solution, but there is exactly one redundant equality constraint. When any one of the equality constraints is dropped, the remaining system of \( n + m - 1 \) equality constraints is linearly independent. | The existence of a solution and a redundancy were established above. The sum of all origin constraints minus the sum of all destination constraints is identically zero. It follows that any constraint can be expressed as a linear combination of the others, and hence any one constraint can be dropped.\n\nSuppose that one... | 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\nClearly a nonsingular lower triangular matrix is triangular according to the above definition. A nonsingular upper triangular matrix is... | Yes |
Basis Triangularity Theorem. Every basis of the transportation problem is triangular. | Proof. Refer to the system of constraints (3). 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 Section 6.1, delete any one of the equations to eliminate the redundancy. From the res... | Yes |
As an illustration of the Basis Triangularity Theorem, consider the basis selected by the Northwest Corner Rule in Example 1 of Section 6.2. This basis is represented below, except that only the basic variables are indicated, not their values. | A row in a basis matrix corresponds to an equation in the original system and is associated with a constraint either on a row or column sum in the solution array. In this example the equation corresponding to the first column sum contains only one basis variable, \( {x}_{11} \) . The value of this variable can be found... | Yes |
Corollary. If the row and column sums of a transportation problem are integers, then the basic variables in any basic solution are integers. | 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), it follows that the process of back substitution will simply involve repeated additions and subtractions of the given row and column sums. No multiplication is required. It therefore... | 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 \( \mathbf{y} = {\mathbf{B}}^{-1}\mathbf{d} \) are either \( 0, + 1 \), or -1 . | Proof. Let \( \mathbf{y} \) be the solution to the equation \( \mathbf{{By}} = \mathbf{d} \) . Then \( \mathbf{y} \) is the representation of \( \mathbf{d} \) in terms of the basis. This equation can be solved by Cramer’s rule as\n\n\[ \n{y}_{k} = \frac{\det {\mathbf{B}}_{k}}{\det \mathbf{B}} \n\]\n\nwhere \( {\mathbf{... | 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 \( {u}_{i} + {v}_{j} \) from \( {c}_{ij} \) . I... | Yes |
Any basic feasible solution of the assignment problem has every \( {x}_{ij} \) equal to either zero or one. | Proof. According to the corollary of the Basis Triangularity Theorem, all basic variables in any basic solution are integers. Clearly, no variable can exceed 1 because the right-hand sides of the constraint equations are all 1 . Therefore, all variables must be either zero or one. I | 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 th... | No |
A common problem in economic theory is the determination of the best way to combine various inputs in order to produce a certain commodity. There is a known production function \( f\left( {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right) \) that gives the amount of the commodity produced as a function of the amounts \( {x}_{i} ... | 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\[ q\frac{\partial f}{\partial {x}_{i}}\left( {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right) = {p}_{i},\;i = 1,2,\ldots, n. \]\n\nThese equations can be inter... | 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 3 (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 |
Example 4 (Control). Dynamic problems, where the variables correspond to actions taken at a sequence of time instants, can often be formulated as unconstrained optimization problems. As an example suppose that the position of a large object is controlled by a series of corrective control forces. The error in position (... | The problem can be converted to an unconstrained problem by eliminating the \( {x}_{k} \) variables, \( k = 1,2,\ldots, n \), from the objective. It is readily seen that\n\n\[ \n{x}_{k} = {x}_{0} + {u}_{0} + {u}_{1} + \cdots + {u}_{k - 1}. \n\]\n\nThe objective can therefore be rewritten as\n\n\[ \nJ = \mathop{\sum }\l... | 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 Section 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 |
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 b... | Yes |
Proposition 3 (Second-order sufficient conditions-unconstrained case).\n\nLet \( f \in {C}^{2} \) be a function defined on a region in which the point \( {\mathbf{x}}^{ * } \) is an interior point. Suppose in addition that\n\ni) \( \nabla f\left( {\mathbf{x}}^{ * }\right) = \mathbf{0} \)\n\n(7)\n\nii) \( \mathbf{F}\lef... | 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\[ f\left( {{\math... | 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 \l... | 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. I | 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(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 off 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 |
Theorem 2. Let \( f \in {C}^{1} \) be convex on the convex set \( \Omega \) . If there is a point \( {\mathbf{x}}^{ * } \in \Omega \) such that, for all \( \mathbf{y} \in \Omega ,\nabla f\left( {\mathbf{x}}^{ * }\right) \left( {\mathbf{y} - {\mathbf{x}}^{ * }}\right) \geq 0 \), then \( {\mathbf{x}}^{ * } \) is a global... | Proof. We note parenthetically that since \( \mathbf{y} - {\mathbf{x}}^{ * } \) is a feasible direction at \( {\mathbf{x}}^{ * } \) , the given condition is equivalent to the first-order necessary condition stated in Section 7.1. The proof of the proposition is immediate, since by Proposition 4 of the last section\n\n\... | Yes |
Theorem 3. Let \( f \) be a convex function defined on the bounded, closed convex set \( \Omega \) . If \( f \) has a maximum over \( \Omega \) it is achieved at an extreme point of \( \Omega \) . | Proof. Suppose \( f \) achieves a global maximum at \( {\mathbf{x}}^{ * } \in \Omega \) . We show first that this maximum is achieved at some boundary point of \( \Omega \) . If \( {\mathbf{x}}^{ * } \) is itself a boundary point, then there is nothing to prove, so assume \( {\mathbf{x}}^{ * } \) is not a boundary poin... | Yes |
Proposition 1 (Zero-order necessary conditions). If \( {\mathbf{x}}^{ * } \) solves (14) under the stated convexity conditions, then there is a nonzero vector \( \mathbf{\lambda } \in {E}^{n} \) such that \( {\mathbf{x}}^{ * } \) is a solution to the two problems:\n\n\[ \text{minimize}f\left( \mathbf{x}\right) + {\math... | Proof. Problem (17) follows from (15) (with \( s = 1 \) ) and the fact that \( f\left( \mathbf{x}\right) \leq r \) for \( r \geq f\left( \mathbf{x}\right) \) . The value \( c \) is attained from above at \( \left( {{f}^{ * },{\mathbf{x}}^{ * }}\right) \) . Likewise (18) follows from (16) and the fact that \( {\mathbf{x... | Yes |
Consider a continuously differentiable function \( f \) of a single variable \( x \in {E}^{1} \) defined on the unit interval \( \left\lbrack {0,1}\right\rbrack \) which plays the role of \( \Omega \) here. The first problem (17) implies \( {f}^{\prime }\left( {x}^{ * }\right) = - \lambda \). | If the solution is at the left end of the interval (at \( x = 0 \) ) then the second problem (18) implies that \( \lambda \leq 0 \) which means that \( {f}^{\prime }\left( {x}^{ * }\right) \geq 0 \) . The reverse holds if \( {x}^{ * } \) is at the right end. These together are identical to the first-order conditions of... | No |
Proposition 2 (Zero-order sufficiency conditions). If there is a \( \mathbf{\lambda } \) such that \( {\mathbf{x}}^{ * } \in \Omega \) solves the problems (17) and (18), then \( {\mathbf{x}}^{ * } \) solves (14). | Proof. Suppose \( {\mathbf{x}}_{1} \) is any other point in \( \Omega \) . Then from (17)\n\n\[ f\left( {\mathbf{x}}_{1}\right) + {\mathbf{\lambda }}^{T}{\mathbf{x}}_{1} \geq f\left( {\mathbf{x}}^{ * }\right) + {\mathbf{\lambda }}^{T}{\mathbf{x}}^{ * }.\]\n\nThis can be rewritten as\n\n\[ f\left( {\mathbf{x}}_{1}\right... | Yes |
As a special case, suppose that the mapping A is a point-to-point mapping; that is, for each \( \mathbf{x} \in X \) the set \( \mathbf{A}\left( \mathbf{x}\right) \) consists of a single point in \( Y \) . Suppose also that \( \mathbf{A} \) is continuous at \( \mathbf{x} \in X \) . This means that if \( {\mathbf{x}}_{k}... | The converse is, however, not true in general. | No |
Proposition. Let \( \mathbf{A} : X \rightarrow Y \) and \( \mathbf{B} : Y \rightarrow Z \) be point-to-set mappings. Suppose \( \mathbf{A} \) is closed at \( \mathbf{x} \) and \( \mathbf{B} \) is closed on \( \mathbf{A}\left( \mathbf{x}\right) \) . Suppose also that if \( {\mathbf{x}}_{k} \rightarrow \mathbf{x} \) and ... | Proof. Let \( {\mathbf{x}}_{k} \rightarrow \mathbf{x} \) and \( {\mathbf{z}}_{k} \rightarrow \mathbf{z} \) with \( {\mathbf{z}}_{k} \in \mathbf{C}\left( {\mathbf{x}}_{k}\right) \) . It must be shown that \( \mathbf{z} \in \mathbf{C}\left( \mathbf{x}\right) \) . Select \( {\mathbf{y}}_{k} \in \mathbf{A}\left( {\mathbf{x... | Yes |
In many respects condition (iii) of the theorem, the closedness of A outside the solution set, is the most important condition. The failure of many popular algorithms can be traced to nonsatisfaction of this condition. On the real line consider the point-to-point algorithm\n\n\[ A\\left( x\\right) = \\left\\{ \\begin{a... | However, starting from \( x > 1 \), the algorithm generates a sequence converging to \( x = 1 \) which is not a solution. The difficulty is that \( A \) is not closed at \( x = 1 \) . | Yes |
On the real line \( X \) consider the solution set to be empty, the descent function \( Z\left( x\right) = {e}^{-x} \), and the algorithm \( A\left( x\right) = x + 1 \) . All conditions of the convergence theorem except (i) hold. The sequence generated from any starting condition diverges to infinity. | This is not strictly a violation of the conclusion of the theorem but simply an example illustrating that if no compactness assumption is introduced, the generated sequence may have no convergent subsequence. | Yes |
Consider the point-to-set algorithm \( A \) defined by the graph in Fig. 7.8 and given explicitly on \( X = \left\lbrack {0,1}\right\rbrack \) by\n\n\[ A\left( x\right) = \left\{ \begin{array}{ll} \lbrack 0, x) & 1 \geq x > 0 \\ 0 & x = 0, \end{array}\right. \]\n\nwhere \( \lbrack 0, x) \) denotes a half-open interval ... | The sequence defined by\n\n\[ {x}_{0} = 1 \]\n\n\[ {x}_{k + 1} = {x}_{k} - \frac{1}{{2}^{k + 2}} \]\n\nsatisfies \( {x}_{k + 1} \in A\left( {x}_{k}\right) \) but it can easily be seen that \( {x}_{k} \rightarrow \frac{1}{2} \notin \Gamma \) . The difficulty here, of course, is that the algorithm \( A \) is not closed o... | Yes |
For the sequence \( {r}_{k} = {a}^{\left( {2}^{k}\right) },0 < a < 1 \), given in Example 2, we have | \[ {\left| {r}_{k}\right| }^{1/{2}^{k}} = a \] while \[ {\left| {r}_{k}\right| }^{1/{p}^{k}} = {a}^{{\left( 2/p\right) }^{k}} \rightarrow 1 \] for \( p > 2 \) . Thus the average order is two. | Yes |
Proposition. Let \( f \) and \( g \) be two error functions satisfying \( f\left( {\mathbf{x}}^{ * }\right) = g\left( {\mathbf{x}}^{ * }\right) = 0 \) and, for all \( \mathbf{x} \), a relation of the form\n\n\[ 0 \leq {a}_{1}g\left( \mathbf{x}\right) \leq f\left( \mathbf{x}\right) \leq {a}_{2}g\left( \mathbf{x}\right) ... | Proof. The statement is easily seen to be symmetric in \( f \) and \( g \) . Thus we assume \( \left\{ {\mathbf{x}}_{k}\right\} \) is linearly convergent with average convergence ratio \( \beta \) with respect to \( f \), and will prove that the same is true with respect to \( g \) . We have\n\n\[ \beta = \overline{\ma... | Yes |
Lemma 1. The iterative process (32) satisfies\n\n\[ E\left( {\mathbf{x}}_{k + 1}\right) = \left\{ {1 - \frac{{\left( {\mathbf{g}}_{k}^{T}{\mathbf{g}}_{k}\right) }^{2}}{\left( {{\mathbf{g}}_{k}^{T}\mathbf{Q}{\mathbf{g}}_{k}}\right) \left( {{\mathbf{g}}_{k}^{T}{\mathbf{Q}}^{-1}{\mathbf{g}}_{k}}\right) }}\right\} E\left( ... | Proof. The proof is by direct computation. We have, setting \( {\mathbf{y}}_{k} = {\mathbf{x}}_{k} - {\mathbf{x}}^{ * } \), \n\n\[ \frac{E\left( {\mathbf{x}}_{k}\right) - E\left( {\mathbf{x}}_{k + 1}\right) }{E\left( {\mathbf{x}}_{k}\right) } = \frac{2{\alpha }_{k}{\mathbf{g}}_{k}^{T}\mathbf{Q}{\mathbf{y}}_{k} - {\alph... | Yes |
Consider the earlier example of \( f\left( x\right) = \) \( {tx} - \ln x \) . | \[ \lambda \left( x\right) = {\left\lbrack {f}^{\prime }{\left( x\right) }^{2}/{f}^{\prime \prime }\left( x\right) \right\rbrack }^{\frac{1}{2}} = \left| {\left( {t - 1/x}\right) x}\right| = \left| {1 - {tx}}\right| . \]\n\nThen (56) gives\n\n\[ \left( {1 - t{x}^{ + }}\right) \leq 2{\left( 1 - tx\right) }^{2} \]\n\nAct... | No |
Theorem 2. In the method of conjugate gradients we have\n\n\[ \nE\left( {\mathbf{x}}_{k + 1}\right) \leq \mathop{\max }\limits_{{\lambda }_{i}}{\left\lbrack 1 + {\lambda }_{i}{P}_{k}\left( {\lambda }_{i}\right) \right\rbrack }^{2}E\left( {\mathbf{x}}_{0}\right) \n\]\n\n(27)\n\nfor any polynomial \( {P}_{k} \) of degree... | This way of viewing the conjugate gradient method as an optimal process is exploited in the next section. We note here that it implies the far from obvious fact that every step of the conjugate gradient method is at least as good as a steepest descent step would be from the same point. To see this, suppose \( {\mathbf{... | No |
Consider the problem\n\n\\[ \n\\text{minimize}\\,{x}_{1}{x}_{2} + {x}_{2}{x}_{3} + {x}_{1}{x}_{3} \n\\]\n\n\\[ \n\\text{subject to}{x}_{1} + {x}_{2} + {x}_{3} = 3\\text{.}\n\\] | The necessary conditions become\n\n\\[ \n{x}_{2} + {x}_{3} + \\lambda = 0 \n\\]\n\n\\[ \n{x}_{1}\\, + {x}_{3} + \\lambda = 0 \n\\]\n\n\\[ \n{x}_{1} + {x}_{2}\\, + \\lambda = 0.\\]\n\nThese three equations together with the one constraint equation give four equations that can be solved for the four unknowns \\( {x}_{1},... | Yes |
We seek to construct a cardboard box of maximum volume, given a fixed area of cardboard. | Denoting the dimensions of the box by \( x, y, z \), the problem can be expressed as\n\n\[ \text{maximize}{xyz} \]\n\n\[ \text{ subject to }\;\left( {{xy} + {yz} + {xz}}\right) = \frac{c}{2}, \]\n\nwhere \( c > 0 \) is the given area of cardboard. Introducing a Lagrange multiplier, the first-order necessary conditions ... | Yes |
If the value of mean is known to be \( m \) (by the physical situation), the maximum entropy argument suggests that the density should be taken as that which solves the following problem:\n\n\[ \text{ maximize } - \mathop{\sum }\limits_{{i = 1}}^{n}{p}_{i}\log \left( {p}_{i}\right) \]\n\n\[ \text{subject to}\mathop{\su... | We begin by ignoring the nonnegativity constraints, believing that they may be inactive. Introducing two Lagrange multipliers, \( \lambda \) and \( \mu \), the Lagrangian is\n\n\[ l = \mathop{\sum }\limits_{{i = 1}}^{n}\left\{ {-{p}_{i}\log {p}_{i} + \lambda {p}_{i} + \mu {x}_{i}{p}_{i}}\right\} - \lambda - {\mu m}. \]... | Yes |
A chain is suspended from two thin hooks that are 16 feet apart on a horizontal line as shown in Fig. 11.3. The chain itself consists of 20 links of stiff steel. Each link is one foot in length (measured inside). We wish to formulate the problem to determine the equilibrium shape of the chain. | The solution can be found by minimizing the potential energy of the chain. Let us number the links consecutively from 1 to 20 starting with the left end. We let link \( i \) span an \( x \) distance of \( {x}_{i} \) and a \( y \) distance of \( {y}_{i} \) . Then \( {x}_{i}^{2} + {y}_{i}^{2} = 1 \) . The potential energ... | Yes |
The portfolio problem is to allocate total available wealth among these \( n \) securities, allocating a fraction \( {w}_{i} \) of wealth to the security \( i \) . | Markowitz introduced the concept of devising efficient portfolios which for a given expected rate of return \( \bar{r} \) have minimum possible variance. Such a portfolio is the solution to the problem\n\n\[ \mathop{\min }\limits_{{{w}_{i},{w}_{2},\ldots ,{w}_{n}}}\mathop{\sum }\limits_{{i, j = 1}}^{n}{w}_{i}{\sigma }_... | Yes |
Consider the problem\n\n\\[ \n\\text{maximize} \\; {x}_{1}{x}_{2} + {x}_{2}{x}_{3} + {x}_{1}{x}_{3} \n\\]\n\n\\[ \n\\text{subject to} {x}_{1} + {x}_{2} + {x}_{3} = 3. \n\\] | In Example 1 of Section 11.4 it was found that \\( {x}_{1} = {x}_{2} = {x}_{3} = 1, \\lambda = - 2 \\) satisfy the first-order conditions. The matrix \\( \\mathbf{F} + {\\mathbf{\\lambda }}^{T}\\mathbf{H} \\) becomes in this case\n\n\\[ \nL = \\left\\lbrack \\begin{array}{lll} 0 & 1 & 1 \\\\ 1 & 0 & 1 \\\\ 1 & 1 & 0 \\... | Yes |
In the last section we considered\n\n\[ \nL = \left\\lbrack \begin{array}{lll} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}\\right\\rbrack \n\]\n\nrestricted to \( M = \\left\\{ \\mathbf{y} : {y}_{1} + {y}_{2} + {y}_{3} = 0\\right\\} \) . To obtain an explicit matrix representation on \( M \) let us introduce the or... | This gives, upon expansion,\n\n\[ \n\\mathbf{E}^{T}\\mathbf{{LE}} = \\left\\lbrack \\begin{array}{rr} - 1 & 0 \\ 0 & - 1 \\end{array}\\right\\rbrack \n\]\n\nand hence \( \\mathbf{L} \) restricted to \( M \) acts like the negative of the identity. | Yes |
\[ \text{extremize}\;{x}_{1} + {x}_{2}^{2} + {x}_{2}{x}_{3} + 2{x}_{3}^{2} \] \[ \text{subject to}\frac{1}{2}\left( {{x}_{1}^{2} + {x}_{2}^{2} + {x}_{3}^{2}}\right) = 1\text{.} \] | \[ 1 + \;\lambda {x}_{1} = 0 \] \[ 2{x}_{2} + {x}_{3} + \lambda {x}_{2} = 0 \] \[ {x}_{2} + 4{x}_{3} + \lambda {x}_{3} = 0. \] One solution to this set is easily seen to be \( {x}_{1} = 1,{x}_{2} = 0,{x}_{3} = 0,\lambda = - 1 \) . Let us examine the second-order conditions at this solution point. The Lagrangian matrix ... | Yes |
Example 3. Approaching Example 2 in this way we have\n\n\\[ \np\left( \lambda \right) \equiv \det \left\lbrack \begin{matrix} 0 & 1 & 0 & 0 \\ - 1 & - \left( {1 + \lambda }\right) & 0 & 0 \\ 0 & 0 & \left( {1 - \lambda }\right) & 1 \\ 0 & 0 & 1 & \left( {3 - \lambda }\right) \end{matrix}\right\rbrack \\]\n\nThis determ... | The result is\n\n\\[ \np\left( \lambda \right) = \left( {1 - \lambda }\right) \left( {3 - \lambda }\right) - 1, \\]\n\nwhich is identical to that found earlier. | No |
Proposition 1. Suppose \( \Omega \) is convex, the function \( f \) is convex, and \( \mathbf{h} \) is affine. Then the primal function \( \omega \) is convex. | Proof. For simplicity of notation we assume that \( \Omega \) is the entire space \( X \) . Then we observe \[ \omega \left( {\alpha {\mathbf{y}}_{1} + \left( {1 - \alpha }\right) {\mathbf{y}}_{2}}\right) = \inf \left\{ {f\left( \mathbf{x}\right) : \mathbf{h}\left( \mathbf{x}\right) = \alpha {\mathbf{y}}_{1} + \left( {... | Yes |
Consider the classic problem of finding the rectangle of maximum area while limiting the perimeter to a length of 4. | The problem can be formulated as\n\n\[ \text{minimize}\; - {x}_{1}{x}_{2} \]\n\n\[ \text{subject to}{x}_{1} + {x}_{2} - 2 = 0 \]\n\n\[ {x}_{1} \geq 0,\;{x}_{2} \geq 0. \]\n\nThe regularity condition is met because it is possible to make the right hand side of the functional constraint slightly positive or slightly nega... | Yes |
Example 2 (Best diagonal). As an alternative problem, consider minimizing the length of the diagonal of a rectangle subject to the perimeter being of length 4 . This problem can be formulated as\n\n\\[ \n\\text{minimize}\\;\\frac{1}{2}\\left( {{x}_{1}^{2} + {x}_{2}^{2}}\\right)\n\\]\n\n\\[ \n\\text{subject to}{x}_{1} +... | In this case the objective function is convex. The solution is \\( {x}_{1} = {x}_{2} = 1 \\) and the Lagrange multiplier is \\( \\lambda = - 1 \\) . The Lagrangian problem is\n\n\\[ \n\\text{minimize}\\;\\frac{1}{2}\\left( {{x}_{1}^{2} + {x}_{2}^{2}}\\right) - 1 \\cdot \\left( {{x}_{1} + {x}_{2} - 2}\\right)\n\\]\n\n\\... | Yes |
Proposition 3. Suppose \( \Omega \subset {E}^{n} \) is convex and \( f \) and \( \mathbf{g} \) are convex functions. Then the primal function \( \omega \) is also convex. | Proof. The proof parallels that of Proposition 1. One simply substitutes \( \mathbf{g}\left( \mathbf{x}\right) \leq \mathbf{0} \) for \( \mathbf{h}\left( \mathbf{x}\right) = \mathbf{y} \) throughout the series of inequalities. | No |
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