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Example 3. Let \( {g}_{i}, i = 1,2,\ldots, m \) be continuous functions on \( {\mathcal{R}}^{n} \) . Suppose\n\n\[ S = \left\{ {\mathbf{x} \in {\mathcal{R}}^{n} \mid {g}_{i}\left( \mathbf{x}\right) \geq 0, i = 1,2,\ldots, m}\right\} \]\n\nis robust, and suppose the interior of \( S \) is\n\n\[ \operatorname{int}\left( ... | where \( c \) is a positive constant. This is a constrained problem, and indeed the constraint is somewhat more complicated than in the original problem (6). The advantage of this problem, however, is that it can be solved by using an unconstrained search technique. Since the value of the objective function approaches ... | Yes |
Consider the problem\n\n\\[ \n\\min \\;{x}_{1} + {x}_{2} \n\\]\n\n\\[ \n\\text{s.t.}\\; - {x}_{1}^{2} + {x}_{2} \geq 0 \n\\]\n\n\\[ \n{x}_{1} \geq 0\\text{.}\n\\] | Define the problem\n\n\\[ \n\\min r\\left( {\\mathbf{x}, c}\\right) = {x}_{1} + {x}_{2} - \\frac{1}{c}\\left\\lbrack {\\ln \\left( {-{x}_{1}^{2} + {x}_{2}}\\right) + \\ln {x}_{1}}\\right\\rbrack .\n\\]\n\nLet\n\n\\[ \n\\nabla r\\left( {\\mathbf{x}, c}\\right) = \\left( \\begin{matrix} 1 - \\frac{1}{c}\\left( {\\frac{-2... | Yes |
Theorem 1 The feasible region \( \mathcal{F} \) of \( \left( {QP}\right) \) is nonempty if and only if for any nonzero vector \( \mu \) with | 利用Farkas引理考虑约束\n\n\[\n\text{若}\mathop{\sum }\limits_{{i \in \mathcal{E} \cup \mathcal{I}}}{\mu }_{i}{a}_{i} = 0,\;{\mu }_{i} \geq 0, i \in \mathcal{I}\text{,}\n\]\n的择一系统\n\nthe following inequality holds:\n\n\[\n\text{则}\mathop{\sum }\limits_{{i \in \mathcal{E} \cup \mathcal{I}}}{\mu }_{i}{b}_{i} \leq 0\text{}.\n\] | No |
Theorem 2 Suppose that the matrix \( Q \) in (1) is symmetric and positive definite. If \( \left( {{\lambda }^{ * },{\mu }^{ * }}\right) \) is a KKT-pair of (2), then \( {x}^{ * } = {Q}^{-1}\left( {{A}^{T}{\lambda }^{ * } - c}\right) \) is the (unique) optimal point of (1). | Proof: Since \( \left( {{\lambda }^{ * },{\mu }^{ * }}\right) \) is a KKT-pair of (2), \n\n\[ \nA{Q}^{-1}{A}^{T}{\lambda }^{ * } - \left( {b + A{Q}^{-1}c}\right) = {\mu }^{ * } \n\] \n\n\[ \n{\lambda }_{i}^{ * } \geq 0,{\mu }_{i}^{ * } \geq 0,{\lambda }_{i}^{ * }{\mu }_{i}^{ * } = 0, i \in \mathcal{I} \n\] \n\n(3) \n\n... | Yes |
Theorem 3 Suppose that \( A \in {\mathcal{R}}^{m \times n} \) has row full rank, and \( {x}^{T}{Qx} > 0 \) holds for any nonzero vector \( x \) with \( {Ax} = 0 \) . Then the system (5) has one solution, denoted by \( \left( {{x}^{ * },{\lambda }^{ * }}\right) \) . Moreover, \( {x}^{ * } \) is a global optimal solution... | Proof\n\nThe matrix\n\n\[ \left( \begin{matrix} Q & - {A}^{T} \\ - A & 0 \end{matrix}\right) \]\n\nis nonsingular. Indeed, suppose \( \left( {x,\lambda }\right) \in {\mathcal{R}}^{n \times m} \) such that\n\n\[ {Qx} - {A}^{T}\lambda = 0,\;{Ax} = 0.\;\text{没有非 解故可逆} \]\n\nMultiplication of the first equation by \( {x}^{... | Yes |
Theorem 5 Let \( {d}^{k} \) solves (6). If \( {d}^{k} \neq 0 \) then \( f\left( {{x}^{k} + \alpha {d}^{k}}\right) < f\left( {x}^{k}\right) \) holds for any \( \alpha \in (0,1\rbrack \) . | proof: Obviously, \( {d}^{k} \) is an optimal solution for\n\n\[ \operatorname{\mathbf{m} \mathbf{i} \mathbf{n} }\;\frac{1}{2}{\left( {x}^{k} + d\right) }^{T}Q\left( {{x}^{k} + d}\right) + {c}^{T}\left( {{x}^{k} + d}\right) \]\n\n\[ \text{s.t.}\;{a}_{i}^{T}d = 0,\;i \in {W}_{k}\text{.} \]\n\nHence\n\n\[ f\left( {x}^{k}... | Yes |
Lemma 1 If \( {d}^{{k}_{1}} = {d}^{{k}_{2}} = 0 \) and \( {d}^{k} \neq 0,\forall {k}_{1} < k < {k}_{2} \), then \( {W}_{{k}_{1}} \neq {W}_{{k}_{2}} \) . | Proof: If \( {k}_{2} = {k}_{1} + 1 \) then \( {W}_{{k}_{2}} = {W}_{{k}_{1}} \smallsetminus \left\{ {i}_{{k}_{1}}\right\} \neq {W}_{{k}_{1}} \) . Now we assume that \( {k}_{2} > {k}_{1} + 1 \) . We only need to prove that when \( {k}_{2} = {k}_{1} + 2 \) , outgoing index \( {i}_{{k}_{1}} \) in Step \( {k}_{1}\left( {{d}... | Yes |
Theorem 6 Let \( \{ {x}^{k}\} \) be generated by the active set method. If the vectors \( \left\lbrack {{a}_{i},\text{ }i \in \mathcal{A}\left( {x}^{k}\right) }\right\rbrack \) are linearly independent, then the algorithm terminates at a KKT point of (1) after a finite number of iterations, or the objective function of... | Proof: Suppose that the objective function \( f \) is bounded below and the infinite sequence \( \{ {x}^{k}\} \) is generated by the active set method. There exists \( {k}_{0} > 0 \) such that \( {x}^{k} = {x}^{{k}_{0}},\forall k \geq {k}_{0} \) . That is, for any \( k \geq {k}_{0},{d}^{k} = 0 \) or \( {\alpha }_{k} = ... | Yes |
Theorem 1 Let \( \left( {{\mathbf{x}}^{ * },{v}^{ * }}\right) \) satisfy the second-order sufficiency condition for problem (2). Then there exists \( {c}^{ * } > 0 \) such that for any \( c \geq {c}^{ * } \) , \( {\mathbf{x}}^{ * } \) is a strict local minimum for the unconstrained problem\n\n\[ \min \phi \left( {\math... | If \( {\mathbf{x}}_{c} \) is a minimum point of \( \min \phi \left( {\mathbf{x}, c}\right) \) and \( {h}_{i}\left( {\mathbf{x}}_{c}\right) = 0, i \in \mathcal{E} \), then \( {\mathbf{x}}_{c} \) is a local optimal solution to (2). | No |
Lemma 1 Let \( B \) be an \( n \times n \) matrix and a vector \( \mathbf{b} \in {\mathcal{R}}^{n} \) . If \( {\mathbf{d}}^{T}B\mathbf{d} > 0 \) for any vector \( \mathbf{d} \in {\mathcal{R}}^{n} \) with \( \mathbf{d} \neq \mathbf{0} \) and \( {\mathbf{b}}^{T}\mathbf{d} = 0 \) , then there exists \( {c}^{ * } > 0 \) su... | Proof. Consider the two sets\n\n\[ \n{K}_{1} = \left\{ {\mathbf{d} \in {\mathcal{R}}^{n} \mid \parallel \mathbf{d}\parallel = 1}\right\} ,\;{K}_{2} = \left\{ {\mathbf{d} \in {K}_{1} \mid {\mathbf{d}}^{T}B\mathbf{d} \leq 0}\right\} .\n\]\n\nFor any vector \( z \neq 0 \) , the vector \( \mathbf{d} = z/\parallel z\paralle... | Yes |
Consider the problem\n\nminimize \( {c}_{1}{x}_{1} + {c}_{2}{x}_{2} + \cdots + {c}_{n}{x}_{n} \)\n\n\[ \n\text{subject to}{a}_{11}{x}_{1} + {a}_{12}{x}_{2} + \cdots + {a}_{1n}{x}_{n} \leq {b}_{1} \]\n\n\[ {a}_{21}{x}_{1} + {a}_{22}{x}_{2} + \cdots + {a}_{2n}{x}_{n}\; \leq {b}_{2} \]\n\n\[ \vdots \]\n\n\[ {a}_{m1}{x}_{1... | 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\n\[ {a... | Yes |
If a linear program is given in standard form except that one or more of the unknown variables is not required to be nonnegative, the problem can be transformed to standard form by either of two simple techniques. | To describe the first technique, suppose in (2.1), for example, that the restriction \( {x}_{1} \geq 0 \) is not present and hence \( {x}_{1} \) is free to take on either positive or negative values. We then write\n\n\[ {x}_{1} = {u}_{1} - {v}_{1} \]\n\n(2.3)\n\nwhere we require \( {u}_{1} \geq 0 \) and \( {v}_{1} \geq... | Yes |
A second approach for converting to standard form when \( {x}_{1} \) is unconstrained in sign is to eliminate \( {x}_{1} \) together with one of the constraint equations. Take any one of the \( m \) equations in (2.1) which has a nonzero coefficient for \( {x}_{1} \) . Say, for example,\n\n\[ \n{a}_{i1}{x}_{1} + {a}_{i... | If this expression is substituted for \( {x}_{1} \) everywhere in (2.1), we are led to a new problem of exactly the same form but expressed in terms of the variables \( {x}_{2},{x}_{3},\ldots ,{x}_{n} \) only. Furthermore, the \( i \) th equation, used to determine \( {x}_{1} \), is now identically zero and it too can ... | Yes |
\[ \text{minimize}{x}_{1} + 3{x}_{2} + 4{x}_{3} \] \[ \text{subject to}{x}_{1} + 2{x}_{2} + {x}_{3} = 5 \] \[ 2{x}_{1} + 3{x}_{2} + {x}_{3} = 6 \] \[ {x}_{2} \geq 0,\;{x}_{3} \geq 0. \] | Since \( {x}_{1} \) is free, we solve for it from the first constraint, obtaining \[ {x}_{1} = 5 - 2{x}_{2} - {x}_{3} \] (2.5) Substituting this into the objective and the second constraint, we obtain the equivalent problem (subtracting five from the objective) \[ \text{minimize}{x}_{2} + 3{x}_{3} \] \[ \text{subject t... | Yes |
How can we determine the most economical diet that satisfies the basic minimum nutritional requirements for good health? Such a problem might, for example, be faced by the dietitian of a large army. We assume that there are available at the market \( n \) different foods and that the \( j \) th food sells at a price \(... | If we denote by \( {x}_{j} \) the number of units of food \( j \) in the diet, the problem then is to select the \( {x}_{j} \) ’s to minimize the total cost\n\n\[ \n{c}_{1}{x}_{1} + {c}_{2}{x}_{2} + \cdots + {c}_{n}{x}_{n} \n\]\n\nsubject to the nutritional constraints\n\n\[ \n{a}_{i1}{x}_{1} + {a}_{i2}{x}_{2} + \cdots... | Yes |
Example 3 (The Transportation Problem). 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 un... | To formulate this problem as a linear programming problem, we set up the array shown below:\n\n\n\nThe \( i \) th row in this array defines the variables associated with the \( i \) th origin, while the \( j \) th colu... | Yes |
Example 7 (Combinatorial Auction). Suppose there are \( m \) mutually exclusive potential states and only one of them will be true at maturity. For example, the states may correspond to the winning horse in a race of \( m \) horses, or the value of a stock index, falling within \( m \) intervals. An auction organizer w... | Accompanying the order is a number \( {\pi }_{j} \) which is the price limit the participant is willing to pay for one unit of the order. Finally, the participant also declares the maximum number \( {q}_{j} \) of units he or she is willing to accept under these terms.\n\nThe auction organizer, after receiving these var... | Yes |
Corollary 1. If the convex set \( K \) corresponding to (2.18) is nonempty, it has at least one extreme point. | ## Proof. This follows from the first part of the Fundamental Theorem and the Equivalence Theorem above. I | No |
Corollary 3. The constraint set \( K \) corresponding to (2.18) possesses at most a finite number of extreme points. | Proof. There are obviously only a finite number of basic solutions obtained by selecting \( m \) basis vectors from the \( n \) columns of \( \mathbf{A} \) . The extreme points of \( K \) are a subset of these basic solutions. | Yes |
Consider the constraint set in \( {E}^{3} \) defined by\n\n\[ \n{x}_{1} + {x}_{2} + {x}_{3} = 1 \n\]\n\n\[ \n{x}_{1} \geq 0,{x}_{2} \geq 0,{x}_{3} \geq 0\text{.} \n\] | This set is illustrated in Fig. 2.3. It has three extreme points, corresponding to the three basic solutions to \( {x}_{1} + {x}_{2} + {x}_{3} = 1 \) . | No |
Consider the constraint set in \( {E}^{3} \) defined by\n\n\[ {x}_{1} + {x}_{2} + {x}_{3} = 1 \]\n\n\[ 2{x}_{1} + 3{x}_{2}\; = 1 \]\n\n\[ {x}_{1} \geq 0,{x}_{2} \geq 0,{x}_{3} \geq 0\text{.} \] | This set is illustrated in Fig. 2.4. It has two extreme points, corresponding to the two basic feasible solutions. Note that the system of equations itself has three basic solutions, \( \left( {2, - 1,0}\right) ,\left( {1/2,0,1/2}\right) ,\left( {0,1/3,2/3}\right) \), the first of which is not feasible. | No |
Example 3. Consider the constraint set in \( {E}^{2} \) defined in terms of the inequalities\n\n\[ \n{x}_{1} + \frac{8}{3}{x}_{2} \leq 4 \n\]\n\n\[ \n{x}_{1} + \;{x}_{2} \leq 2 \n\]\n\n\[ \n2{x}_{1}\; \leq 3 \n\]\n\n\[ \n{x}_{1} \geq 0,\;{x}_{2} \geq 0. \n\] | This set is illustrated in Fig. 2.5. 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\[ \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\[ \n\begin{matrix} {x}_{1} & {x}_{2} & {x}_{3} & {x}_{4} & {x}_{5} & {x}_{6} & \\ 1 & 0 & 0 & \ddots & 1 & - 1 & 5 \\ 0 & 1 & 0 & 2 & - 3 & 1 & 3 \\ 0 & 0 & 1 & - 1 & 2 & - 1 & - 1 \end{matrix} \]\n\nThe circle indicated is our first pivot element and corresponds to the replac... | Yes |
Suppose we wish to solve the simultaneous equations\n\n\\[ \n{x}_{1} + {x}_{2} - {x}_{3} = 5 \n\\]\n\n\\[ \n2{x}_{1} - 3{x}_{2} + {x}_{3} = 3 \n\\]\n\n\\[ \n- {x}_{1} + 2{x}_{2} - {x}_{3} = - 1\\text{.} \n\\] | To obtain an original basis, we form the augmented tableau\n\n\\[ \n\\begin{matrix} {\\mathbf{e}}_{1} & {\\mathbf{e}}_{2} & {\\mathbf{e}}_{3} & {\\mathbf{a}}_{1} & {\\mathbf{a}}_{2} & {\\mathbf{a}}_{3} & \\mathbf{b} \\ \\ 1 & 0 & 0 & 1 & 1 & - 1 & 5 \\ \\ 0 & 1 & 0 & 2 & - 3 & 1 & 3 \\ \\ 0 & 0 & 1 & - 1 & 2 & - 1 & - ... | No |
Consider the system\n\n\[\n\begin{matrix} {\mathbf{a}}_{1} & {\mathbf{a}}_{2} & {\mathbf{a}}_{3} & {\mathbf{a}}_{4} & {\mathbf{a}}_{5} & {\mathbf{a}}_{6} & \mathbf{b} \\ 1 & 0 & 0 & 2 & 4 & 6 & 4 \\ 0 & 1 & 0 & 1 & 2 & 3 & 3 \\ 0 & 0 & 1 & - 1 & 2 & 1 & 1 \end{matrix}\n\]\n\nwhich has basis \( {\mathbf{a}}_{1},{\mathbf... | \[4/2 = 2,3/1 = 3,1/ - 1 = - 1\]\n\nand select the smallest nonnegative one. This gives 2 as the pivot element. The new tableau is\n\n\[\n\begin{matrix} {\mathbf{a}}_{1} & {\mathbf{a}}_{2} & {\mathbf{a}}_{3} & {\mathbf{a}}_{4} & {\mathbf{a}}_{5} & {\mathbf{a}}_{6} & \mathbf{b} \\ 1/2 & 0 & 0 & 1 & 2 & 3 & 2 \\ - 1/2 & ... | Yes |
Maximize \( 3{x}_{1} + {x}_{2} + 3{x}_{3} \) subject to\n\n\[ 2{x}_{1} + {x}_{2} + {x}_{3} \leq 2 \]\n\n\[ {x}_{1} + 2{x}_{2} + 3{x}_{3} \leq 5 \]\n\n\[ 2{x}_{1} + 2{x}_{2} + {x}_{3} \leq 6 \]\n\n\[ {x}_{1} \geq 0,{x}_{2} \geq 0,{x}_{3} \geq 0\text{.} \]\n | To transform the problem into standard form so that the simplex procedure can be applied, we change the maximization to minimization by multiplying the objective function by minus one, and introduce three nonnegative slack variables \( {x}_{4},{x}_{5},{x}_{6} \) . We then have the initial tableau\n\n![839b7907-f889-4cb... | 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\[ {x}_{1} \geq 0,{x}_{2} \geq 0,{x}_{3} \geq 0\text{.} \] | We introduce artificial variables \( {x}_{4} \geq 0,{x}_{5} \geq 0 \) and an objective function \( {x}_{4} + {x}_{5} \). The initial tableau is\n\n\n\nA basic feasible solution to the expanded system is given by the ar... | Yes |
Consider the problem\n\n\\[ \n\\text{minimize}4{x}_{1} + {x}_{2} + {x}_{3} \n\\]\n\n\\[ \n\\text{subject to}2{x}_{1} + {x}_{2} + 2{x}_{3} = 4 \n\\]\n\n\\[ \n3{x}_{1} + 3{x}_{2} + {x}_{3} = 3 \n\\]\n\n\\[ \n{x}_{1} \\geq 0,\\;{x}_{2} \\geq 0,\\;{x}_{3} \\geq 0. \n\\] | There is no basic feasible solution apparent, so we use the two-phase method. The first phase was done in Example 1 for these constraints, so we shall not repeat it here. We give only the final tableau with the columns corresponding to the artificial variables deleted, since they are not used in phase II. We use the ne... | Yes |
Example 3 (A Free Variable Problem).\n\n\\[ \n\\text{minimize} - 2{x}_{1} + 4{x}_{2} + 7{x}_{3} + {x}_{4} + 5{x}_{5} \n\\]\n\n\\[ \n\\text{subject to} - {x}_{1} + {x}_{2} + 2{x}_{3} + {x}_{4} + 2{x}_{5} = 7 \n\\]\n\n\\[ \n- {x}_{1} + 2{x}_{2} + 3{x}_{3} + {x}_{4} + {x}_{5} = 6 \n\\]\n\n\\[ \n- {x}_{1} + {x}_{2} + {x}_{... | Since \\( {x}_{1} \\) is free, it can be eliminated, as described in Chap. 2, by solving for \\( {x}_{1} \\) in terms of the other variables from the first equation and substituting everywhere else. This can all be done with the simplex tableau as follows:  be equal to the total supply (which is also equal to the total demand). Then let \( {x}_{ij} = {a}_{i}{b}_{j}/S \) for \( i = 1,2,\ldots, m;j = 1,2,\ldots, n \... | Yes |
A basic feasible solution constructed by the Northwest corner Rule is shown below for Example 1 of the last section. | In the first step, at the upper left-hand corner, a maximum of 10 units could be allocated, since that is all that was required by column 1 . This left \( {30} - {10} = {20} \) units required in the first row. Next, moving to the second cell in the top row, the remaining 20 units were allocated. At this point the row 1... | No |
Basis Triangularity Theorem. Every basis of the transportation problem is triangular. | Proof. Refer to the system of constraints (3.36). Let us change the sign of the top half of the system; then the coefficient matrix of the system consists of entries that are either +1, -1 , or 0 . Following the result of the theorem in Sect. 3.7, delete any one of the equations to eliminate the redundancy. From the re... | Yes |
Corollary. If the row and column sums of a transportation problem are integers, then the basic variables in any basic solution are integers. | The importance of triangularity is, of course, the associated method of back substitution for the solution of a triangular system of equations, as discussed in Appendix C. Moreover, since any basis matrix is triangular and all nonzero elements are equal to one (or minus one if the signs of some equations are changed), ... | Yes |
Theorem. Let \( \mathbf{B} \) be a basis from \( \mathbf{A} \) (ignoring one row), and let \( \mathbf{d} \) be another column. Then the components of the vector \( \mathbf{w} = {\mathbf{B}}^{-1}\mathbf{d} \) are either \( 0, + 1 \), or -1 . | Proof. Let \( \mathbf{w} \) be the solution to the equation \( \mathbf{{Bw}} = \mathbf{d} \) . Then \( \mathbf{w} \) is the representation of \( \mathbf{d} \) in terms of the basis. This equation can be solved by Cramer’s rule as\n\n\[ \n{w}_{k} = \frac{\det {\mathbf{B}}_{k}}{\det \mathbf{B}} \]\n\nwhere \( {\mathbf{B}... | No |
We can now completely solve the problem that was introduced in Example 1 of the first section. The requirements and a first basic feasible solution obtained by the Northwest Corner Rule are shown below. The plus and minus signs indicated on the array should be ignored at this point, since they cannot be computed until ... | The cost coefficients of the problem are shown in the array below, with the circled cells corresponding to the current basic variables. The simplex multipliers, computed by row and column scanning, are shown as well. The relative cost coefficients are found by subtracting \( {u}_{j} + {v}_{j} \) from \( {c}_{ij} \). In... | Yes |
To illustrate the method of dealing with degeneracy, consider a modification of Example 3, with the fourth row sum changed from 60 to 20 and the fourth column sum changed from 80 to 40 . Then the initial basic feasible solution found by the Northwest Corner Rule is degenerate. | An \( \varepsilon \) is placed in the array for the zero-valued basic variable as shown below:\n\n\n\nThe relative cost coefficients will be the same as in Example 3, and hence again \( {x}_{43} \) should be chosen to ... | Yes |
The diet problem, Example 1, Sect. 2.2, was the problem faced by a dietitian trying to select a combination of foods to meet certain nutritional requirements at minimum cost. This problem has the form\n\n\[ \n\\text{minimize}{\\mathbf{c}}^{T}\\mathbf{x} \n\]\n\n\[ \n\\text{subject to}\\mathbf{{Ax}} \\geq \\mathbf{b},\\... | Imagine a pharmaceutical company that produces in pill form each of the nutrients considered important by the dietitian. The pharmaceutical company tries to convince the dietitian to buy pills, and thereby supply the nutrients directly rather than through purchase of various foods. The problem faced by the drug company... | Yes |
Example 2 (Dual of the Transportation Problem). The transportation problem, Example 3, Sect. 2.2, is the problem, faced by a manufacturer, of selecting the pattern of product shipments between several fixed origins and destinations so as to minimize transportation cost while satisfying demand. Referring to (4.6) and (4... | To interpret the dual problem, we imagine an entrepreneur who, feeling that he can ship more efficiently, comes to the manufacturer with the offer to buy his product at the plant sites (origins) and sell it at the warehouses (destinations). The product price that is to be used in these transactions varies from point to... | Yes |
Lemma 1 (Weak Duality Lemma). If \( \mathbf{x} \) and \( \mathbf{y} \) are feasible for (4.3) and (4.4), respectively, then \( {\mathbf{c}}^{T}\mathbf{x} \geq {\mathbf{y}}^{T}\mathbf{b} \) . | Proof. We have\n\n\[ \n{\mathbf{y}}^{T}\mathbf{b} = {\mathbf{y}}^{T}\mathbf{A}\mathbf{x} \leq {\mathbf{c}}^{T}\mathbf{x} \n\] \n\nthe last inequality being valid since \( \mathbf{x} \geq \mathbf{0} \) and \( {\mathbf{y}}^{T}\mathbf{A} \leq {\mathbf{c}}^{T} \) . I | Yes |
The ellipsoid \( {E}_{k + 1} = E\left( {{\mathbf{y}}_{k + 1},{\mathbf{B}}_{k + 1}^{-1}}\right) \) defined as above is the ellipsoid of least volume containing \( \left( {1/2}\right) {E}_{k} \) . Moreover, | \[ \frac{\operatorname{vol}\left( {E}_{k + 1}\right) }{\operatorname{vol}\left( {E}_{k}\right) } = \frac{\det \left( {\mathbf{B}}_{k + 1}^{1/2}\right) }{\det \left( {\mathbf{B}}_{k}^{1/2}\right) } \] For simplicity, by a change of coordinates, we may take \( {\mathbf{B}}_{k} = \mathbf{I} \) . Then \( {\mathbf{B}}_{k + ... | Yes |
Consider the set \( \mathcal{S} \) defined by \( {x}_{i} \geq 0,\left( {1 - {x}_{i}}\right) \geq 0 \), for \( i = 1,2,\ldots, n \). This is \( \mathcal{S} = {\left\lbrack 0,1\right\rbrack }^{n} \), the unit cube in \( {E}^{n} \). | The analytic center can be found by differentiation to be \( {x}_{i} = 1/2 \), for all \( i \). Hence, the analytic center is identical to what one would normally call the center of the unit cube. | Yes |
Consider the problem of maximizing \( {x}_{1} \) within the unit square \( \mathcal{S} = {\left\lbrack 0,1\right\rbrack }^{2} \). The problem is formulated as\n\n\[ \min \; - {x}_{1} \]\n\n\[ \text{s.t.}\;{x}_{1} + {x}_{3} = 1 \]\n\n\[ {x}_{2} + {x}_{4} = 1 \]\n\n\[ {x}_{1} \geq 0,{x}_{2} \geq 0,{x}_{3} \geq 0,{x}_{4} ... | Here \( {x}_{3} \) and \( {x}_{4} \) are slack variables for the original problem to put it in standard form. The optimality conditions for \( \mathbf{x}\left( \mu \right) \) consist of the original two linear constraint equations and the four equations\n\n\[ {y}_{1} + {s}_{1} = - 1,{y}_{2} + {s}_{2} = 0,{y}_{1} + {s}_... | No |
Consider the dual of Example 2. This is\n\n\\[ \max \\;{y}_{1} + {y}_{2} \\]\n\n\\[ \text{subject to}{y}_{1} \leq - 1 \\]\n\n\\[ {y}_{2} \leq 0\\text{.} \\] | The solution to the dual barrier problem is easily found from the solution of the primal barrier\nproblem to be\n\n\\[ {y}_{1}\left( \mu \right) = - 1 - \mu /{x}_{1}\left( \mu \right) ,{y}_{2} = - {2\mu }. \\]\n\nAs \\( \mu \\rightarrow 0 \\), we have \\( {y}_{1} \\rightarrow - 1,{y}_{2} \\rightarrow 0 \\), which is th... | Yes |
Proposition 1. Suppose the feasible sets of the primal and dual programs contain interior points. Then the primal-dual central path \( \left( {\mathbf{x}\left( \mu \right) ,\mathbf{y}\left( \mu \right) ,\mathbf{s}\left( \mu \right) }\right) \) exists for all \( \mu ,0 \leq \mu < \infty \) . Furthermore, \( \mathbf{x}\l... | Let \( \left( {\mathbf{x}\left( \mu \right) ,\mathbf{y}\left( \mu \right) ,\mathbf{s}\left( \mu \right) }\right) \) be on the primal-dual central path. Then from (5.9) it follows that\n\n\[ \n{\mathbf{c}}^{T}\mathbf{x} - {\mathbf{y}}^{T}\mathbf{b} = {\mathbf{y}}^{T}\mathbf{A}\mathbf{x} + {\mathbf{s}}^{T}\mathbf{x} - {\... | Yes |
Theorem 2. The algorithm above terminates in at most \( O\left( {\rho \log \left( {n/\varepsilon }\right) }\right) \) iterations with\n\n\[ \frac{{\left( {\mathbf{s}}_{k}\right) }^{T}{\mathbf{x}}_{k}}{{\left( {\mathbf{s}}_{0}\right) }^{T}{\mathbf{x}}_{0}} \leq \varepsilon \] | Proof. Note that after \( k \) iterations, we have from (5.16)\n\n\[ {\psi }_{n + \rho }\left( {{\mathbf{x}}_{k},{\mathbf{s}}_{k}}\right) \leq {\psi }_{n + \rho }\left( {{\mathbf{x}}_{0},{\mathbf{s}}_{0}}\right) - k \cdot \delta \leq \rho \log \left( {{\left( {\mathbf{s}}_{0}\right) }^{T}{\mathbf{x}}_{0}}\right) + n\lo... | Yes |
Theorem 2. Let \( \left( {{\mathbf{y}}^{ * },{\mathbf{x}}^{ * },{\tau }^{ * },{\theta }^{ * } = 0,{\mathbf{s}}^{ * },{\kappa }^{ * }}\right) \) be a strictly-self complementary solution for (HSDP).\n\n(i) (LP) has a solution (feasible and bounded) if and only if \( {\tau }^{ * } > 0 \) . In this case, \( {\mathbf{x}}^{... | Proof. We prove the second statement. We first assume that one of (LP) and (LD) is infeasible, say (LD) is infeasible. Then there is some certificate \( \overline{\mathbf{x}} \geq \mathbf{0} \) such that \( \mathbf{A}\overline{\mathbf{x}} = \mathbf{0} \) and \( {\mathbf{c}}^{T}\overline{\mathbf{x}} = - 1 \) . Let \( \l... | Yes |
Example 2 (Binary Quadratic Optimization). Consider a binary quadratic maximization problem\n\n\\[ \n\\text{maximize}\\mathbf{x}^{T}\\mathbf{Q}\\mathbf{x} + 2\\mathbf{c}^{T}\\mathbf{x}\n\\]\n\n\\[ \n\\text{subject to}x_{j} = \\{ 1, - 1\\} \\text{, for all}j = 1,\\ldots, n\\text{,}\n\\]\n\nwhich is a difficult nonconvex... | Since \\( \\left\\lbrack \\begin{matrix} \\mathbf{x} \\ x_{n + 1} \\end{matrix}\\right\\rbrack \\left\\lbrack \\begin{matrix} \\mathbf{x} \\ x_{n + 1} \\end{matrix}\\right\\rbrack ^{T} \\) forms a positive-semidefinite matrix (with rank equal to 1),\n\na semidefinite relaxation of the problem is defined as\n\n\\[ \nz^{... | Yes |
Theorem 1. The interior of the followings convex cones are given as:\n\n- The interior of the non-negative orthant cone is the set of all vectors where every entry is positive.\n\n- The interior of the positive semidefinite cone is the set of all positive definite matrices.\n\n- The interior of p-order cone is the set ... | We give a sketch of the proof for the second order cone, i.e., \( p = 2 \) . Let \( \\left( {\\bar{u};\\overline{\\mathbf{x}}}\\right) \\neq \\mathbf{0} \) be any second-order cone point but \( \\bar{u} = \\left| \\overline{\\mathbf{x}}\\right| \) . Then, we can choose a dual cone (also the second-order cone) point \( ... | Yes |
Theorem 2 (Farkas' Lemma for CLP). We have\n\n- Consider set\n\n\[ \n{\mathcal{F}}_{p} \mathrel{\text{:=}} \{ \mathbf{X} : \mathcal{A}\mathbf{X} = \mathbf{b},\mathbf{X} \in K\} .\n\]\n\nSuppose that there exists a vector \( \overset{ \circ }{\mathbf{y}} \) such that \( - {\overset{ \circ }{\mathbf{y}}}^{T}\mathcal{A} \... | Proof. We prove the first statement of the theorem. We prove the first part. It is clear that \( C \) is a convex set. To prove that \( C \) is a closed set, we need to show that if \( {\mathbf{y}}^{k} \mathrel{\text{:=}} \mathcal{A}{\mathbf{X}}^{k} \in {E}^{m} \) for \( {\mathbf{X}}^{k} \in K, k = 1,\ldots \), converg... | Yes |
Consider the semidefinite relaxation (6.3) for the binary quadratic maximization problem. It's dual is\n\n\\[ \n\\text{ minimize }\\mathop{\\sum }\\limits_{{j = 1}}^{{n + 1}}{y}_{j}\n\\]\n\n\\[ \n\\text{subject to}\\mathop{\\sum }\\limits_{{j = 1}}^{{n + 1}}{y}_{j}{\\mathbf{I}}_{j} - \\mathbf{S} = \\left\\lbrack \\begi... | Note that\n\n\\[ \n\\mathop{\\sum }\\limits_{{j = 1}}^{{n + 1}}{y}_{j}{\\mathbf{I}}_{j} - \\left\\lbrack \\begin{matrix} \\mathbf{Q} & \\mathbf{c} \\\\ {\\mathbf{c}}^{T} & 0 \\end{matrix}\\right\\rbrack\n\\]\n\nis exactly the Hessian matrix of the Lagrange function of the quadratic maximization problem; see Chap. 11. T... | No |
Example 4 (Euclidean Facility Location). This problem is to determine the location of a facility serving \( n \) clients placed in a Euclidean space, whose known locations are denoted by \( {\mathbf{a}}_{j} \in {E}^{d}, j = 1,\ldots, n \) . The location of the facility would minimize\nthe sum of the Euclidean distances... | The problem can be reformulated as\n\n\[ \n\text{minimize}\;\mathop{\sum }\limits_{{j = 1}}^{n}{\delta }_{j} \n\]\n\n\[ \n\text{subject to}{\mathbf{s}}_{j} + \mathbf{f} = {\mathbf{a}}_{j},\forall j = 1,\ldots, n\text{,} \n\]\n\n\[ \n\left| {\mathbf{s}}_{j}\right| \leq {\delta }_{j},\;\forall j = 1,\ldots, n. \n\]\n\nTh... | Yes |
Example 6. The following semidefinite program has a duality gap: | The primal minimal objective value is 0 achieved by\n\n\[ \mathbf{X} = \left\lbrack \begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{array}\right\rbrack \]\n\nand the dual maximal objective value is -2 achieved by \( \mathbf{y} = \left\lbrack {0, - 1}\right\rbrack \) ; so the duality gap is 2 . | Yes |
Proposition 1. Let \( {\mathbf{X}}^{ * } \) and \( \left( {{\mathbf{y}}^{ * },{\mathbf{S}}^{ * }}\right) \) be any optimal SDP solution pair with zero duality gap. Then complementarity of \( {\mathbf{X}}^{ * } \) and \( {\mathbf{S}}^{ * } \) implies\n\n\[ \operatorname{rank}\left( {\mathbf{X}}^{ * }\right) + \operatorn... | Furthermore, is there an optimal (dual) \( {\mathbf{S}}^{ * } \) such that \( \operatorname{rank}{\mathbf{S}}^{ * } \geq d \), then the rank of any optimal (primal) \( {\mathbf{X}}^{ * } \) is bounded above by \( n - d \), where integer \( 0 \leq d \leq n \) ; and the converse is also true. | No |
The following are barrier function for each of the convex cones. | - The \( n \) -dimensional non-negative orthant \( {E}_{ + }^{n} \) :\n\n\[ B\left( \mathbf{x}\right) = - \mathop{\sum }\limits_{{j = 1}}^{n}\log \left( {x}_{j}\right) \]\n\n- The \( n \) -dimensional semidefinite cone \( {\mathcal{S}}_{ + }^{n} \) :\n\n\[ B\left( \mathbf{X}\right) = - \log \left( {\det \mathbf{X}}\rig... | 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\\[ \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\\]\n\nThere are no constraints, so \\( \\Omega = {E}^{2} \\) . | Setting the partial derivatives of \\( f \\) equal to zero yields the two equations\n\n\\[ \n2{x}_{1} - {x}_{2} = 0\n\\]\n\n\\[ \n- {x}_{1} + 2{x}_{2} = 3\\text{. }\n\\]\n\nThese have the unique solution \\( {x}_{1} = 1,{x}_{2} = 2 \\), which is a global minimum point of \\( f \\) . | Yes |
Consider the problem\n\n\\[ \n\\text{minimize}f\\left( {{x}_{1},{x}_{2}}\\right) = {x}_{1}^{2} - {x}_{1} + {x}_{2} + {x}_{1}{x}_{2} \n\\]\n\n\\[ \n\\text{subject to}\\{x}_{1} \\geq 0,\\{x}_{2} \\geq 0\\text{.} \n\\] | This problem has a global minimum at \\( {x}_{1} = \\frac{1}{2},{x}_{2} = 0 \\) . At this point\n\n\\[ \n\\frac{\\partial f}{\\partial {x}_{1}} = 2{x}_{1} - 1 + {x}_{2} = 0 \n\\]\n\n\\[ \n\\frac{\\partial f}{\\partial {x}_{2}} = 1 + {x}_{1} = \\frac{3}{2} \n\\]\n\nThus, the partial derivatives do not both vanish at the... | Yes |
Recall the classification problem where we have vectors \( {\mathbf{a}}_{i} \in {E}^{d} \) for \( i = 1,2,\ldots ,{n}_{1} \) in a class, and vectors \( {\mathbf{b}}_{j} \in {E}^{d} \) for \( j = 1,2,\ldots ,{n}_{2} \) not. Then we wish to find \( \mathbf{y} \in {E}^{d} \) and a number \( \beta \) such that\n\n\[ \n\fra... | The problem can be cast as a unconstrained optimization problem, called the max-likelihood,\n\n\[ \n{\operatorname{maximize}}_{\mathbf{y},\beta }\left( {\mathop{\prod }\limits_{i}\frac{\exp \left( {{\mathbf{a}}_{i}^{T}\mathbf{y} + \beta }\right) }{1 + \exp \left( {{\mathbf{a}}_{i}^{T}\mathbf{y} + \beta }\right) }}\righ... | Yes |
A common problem in economic theory is the determination of the best way to combine various inputs in order to maximize a utility function \( f\left( {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right) \) (in the monetary unit) of the amounts \( {x}_{j} \) of the inputs, \( i = 1,2,\ldots, n \) . The unit prices of the inputs are... | The first-order necessary conditions are that the partial derivatives with respect to the \( {x}_{i} \) ’s each vanish. This leads directly to the \( n \) equations\n\n\[ \frac{\partial f}{\partial {x}_{i}}\left( {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right) = {p}_{i}, i = 1,2,\ldots, n. \]\n\nThese equations can be interpr... | Yes |
Example 3 (Parametric Estimation). 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}\ri... | This is a quadratic expression in the coefficients \( \mathbf{a} \) . To find a compact representation for this objective we define \( {q}_{ij} = \mathop{\sum }\limits_{{k = 1}}^{m}{\left( {x}_{k}\right) }^{i + j},{b}_{j} = \mathop{\sum }\limits_{{k = 1}}^{m}g\left( {x}_{k}\right) {\left( {x}_{k}\right) }^{j} \) and \(... | Yes |
Example 4 (Selection Problem). It is often necessary to select an assortment of factors to meet a given set of requirements. An example is the problem faced by an electric utility when selecting its power-generating facilities. The level of power that the company must supply varies by time of the day, by day of the wee... | Assuming that the solution is interior to the constraints, by setting the partial derivatives equal to zero, we obtain the two equations\n\n\[ {b}_{1} + \left( {{c}_{1} - {c}_{2}}\right) h\left( {x}_{1}\right) + \left( {{c}_{2} - {c}_{3}}\right) h\left( {{x}_{1} + {x}_{2}}\right) = 0 \]\n\n\[ {b}_{2} + \left( {{c}_{2} ... | Yes |
Proposition 1 (Second-Order Necessary Conditions). Let \( \Omega \) be a subset of \( {E}^{n} \) and let \( f \in {C}^{2} \) be a function on \( \Omega \) . If \( {\mathbf{x}}^{ * } \) is a relative minimum point of \( f \) over \( \Omega \), then for any \( \mathbf{d} \in {E}^{n} \) that is a feasible direction at \( ... | Proof. The first condition is just Proposition 1, and the second applies only if \( \nabla f\left( {\mathbf{x}}^{ * }\right) \mathbf{d} = 0 \) . In this case, introducing \( \mathbf{x}\left( \alpha \right) = {\mathbf{x}}^{ * } + \alpha \mathbf{d} \) and \( g\left( \alpha \right) = f\left( {\mathbf{x}\left( \alpha \righ... | Yes |
For the same problem as Example 2 of Sect. 7.1, we have for \( \mathbf{d} = \) \( \left( {{d}_{1},{d}_{2}}\right) \)\n\n\[ \nabla f\left( {\mathbf{x}}^{ * }\right) \mathbf{d} = \frac{3}{2}{d}_{2} \]\n\nThus condition (ii) of Proposition 1 applies only if \( {d}_{2} = 0 \) . | In that case we have \( {\mathbf{d}}^{T}{\nabla }^{2}f\left( {\mathbf{x}}^{ * }\right) \mathbf{d} = 2{d}_{1}^{2} \geq 0 \), so condition (ii) is satisfied. | No |
Proposition 2 (Second-Order Necessary Conditions-Unconstrained Case). Let \( {\mathbf{x}}^{ * }{be} \) an interior point of the set \( \Omega \), and suppose \( {\mathbf{x}}^{ * } \) is a relative minimum point over \( \Omega \) of the function \( f \in {C}^{2} \) . Then\n\n\[ \text{i)}\nabla f\left( {\mathbf{x}}^{ * }... | For notational simplicity we often denote \( {\mathbf{\nabla }}^{2}f\left( \mathbf{x}\right) \), the \( n \times n \) matrix of the second partial derivatives of \( f \), the Hessian of \( f \), by the alternative notation \( \mathbf{F}\left( \mathbf{x}\right) \) . Condition (ii) is equivalent to stating that the matri... | Yes |
Consider the problem\n\n\\[ \n\\text{minimize}f\\left( {{x}_{1},{x}_{2}}\\right) = {x}_{1}^{3} - {x}_{1}^{2}{x}_{2} + 2{x}_{2}^{2} \n\\]\n\n\\[ \n\\text{subject to}{x}_{1} \\geq 0,\\;{x}_{2} \\geq 0\\text{.} \n\\] | If we assume that the solution is in the interior of the feasible set, that is, if \\( {x}_{1} > 0,{x}_{2} > 0 \\), then the first-order necessary conditions are\n\n\\[ \n3{x}_{1}^{2} - 2{x}_{1}{x}_{2} = 0,\\; - {x}_{1}^{2} + 4{x}_{2} = 0.\n\\]\n\nThere is a solution to these at \\( {x}_{1} = {x}_{2} = 0 \\) which is a... | Yes |
Proposition 3 (Second-Order Sufficient Conditions—Unconstrained Case). Let \( f \in {C}^{2} \) be function defined on a region in which the point \( {\mathbf{x}}^{ * } \) is an interior point. Suppose in addition that\n\n\[ \n\text{i)}\nabla f\left( {\mathrm{x}}^{ * }\right) = \mathbf{0} \n\]\n\n(7.7)\n\n\[ \n\text{ii)... | Proof. Since \( \mathbf{F}\left( {\mathbf{x}}^{ * }\right) \) is positive definite, there is an \( a > 0 \) such that for all \( \mathbf{d},{\mathbf{d}}^{T}\mathbf{F}\left( {\mathbf{x}}^{ * }\right) \) \( \mathbf{d} \geq a{\left| \mathbf{d}\right| }^{2} \) . Thus by the Taylor’s Theorem (with remainder)\n\n\[ \nf\left(... | Yes |
Proposition 1. Let \( {f}_{1} \) and \( {f}_{2} \) be convex functions on the convex set \( \Omega \) . Then the function \( {f}_{1} + {f}_{2} \) is convex on \( \Omega \) . | Proof. Let \( {\mathbf{x}}_{1},{\mathbf{x}}_{2} \in \Omega \), and \( 0 < \alpha < 1 \) . Then\n\n\[ \n{f}_{1}\left( {\alpha {\mathbf{x}}_{1} + \left( {1 - \alpha }\right) {\mathbf{x}}_{2}}\right) + {f}_{2}\left( {\alpha {\mathbf{x}}_{1}}\right) + \left( {1 - \alpha }\right) {\mathbf{x}}_{2})\n\]\n\n\[ \n\leq \alpha \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. | No |
Proposition 3. Let \( f \) be a convex function on a convex set \( \Omega \) . The set \( {\Gamma }_{c} = \{ \mathbf{x} : \mathbf{x} \in \) \( \Omega, f\left( \mathbf{x}\right) \leq c\} \) is convex for every real number \( c \) . | Proof. Let \( {\mathbf{x}}_{1},{\mathbf{x}}_{2} \in {\Gamma }_{c} \) . Then \( f\left( {\mathbf{x}}_{1}\right) \leq c, f\left( {\mathbf{x}}_{2}\right) \leq c \) and for \( 0 < \alpha < 1 \) ,\n\n\[ f\left( {\alpha {\mathbf{x}}_{1} + \left( {1 - \alpha }\right) {\mathbf{x}}_{2}}\right) \leq {\alpha f}\left( {\mathbf{x}}... | Yes |
Proposition 4. Let \( f \in {C}^{1} \) . Then \( f \) is convex over a convex set \( \Omega \) if and only if\n\n\[ f\left( \mathbf{y}\right) \geq f\left( \mathbf{x}\right) + \nabla f\left( \mathbf{x}\right) \left( {\mathbf{y} - \mathbf{x}}\right) \]\n\n(7.9)\n\nfor all \( \mathbf{x},\mathbf{y} \in \mathbf{\Omega } \) ... | Proof. First suppose \( f \) is convex. Then for all \( \alpha ,0 \leq \alpha \leq 1 \) ,\n\n\[ f\left( {\alpha \mathbf{y} + \left( {1 - \alpha }\right) \mathbf{x}}\right) \leq {\alpha f}\left( \mathbf{y}\right) + \left( {1 - \alpha }\right) f\left( \mathbf{x}\right) .\n\nThus for \( 0 < \alpha \leq 1 \)\n\n\[ \frac{f\... | Yes |
Proposition 5. Let \( f \in {C}^{2} \) . Then \( f \) is convex over a convex set \( \Omega \) containing an interior point if and only if the Hessian matrix \( \mathbf{F} \) of \( f \) is positive semidefinite throughout \( \Omega \) . | Proof. By Taylor's theorem we have\n\n\[ f\left( \mathbf{y}\right) = f\left( \mathbf{x}\right) = \mathbf{\nabla }f\left( \mathbf{x}\right) \left( {\mathbf{y} - \mathbf{x}}\right) + \frac{1}{2}{\left( \mathbf{y} - \mathbf{x}\right) }^{T}\mathbf{F}\left( {\mathbf{x} + \alpha \left( {\mathbf{y} - \mathbf{x}}\right) }\righ... | Yes |
Theorem 1. Let \( f \) be a convex function defined on the convex set \( \Omega \) . Then the set \( \Gamma \) where \( f \) achieves its minimum is convex, and any relative minimum of \( f \) is a global minimum. | Proof. If \( f \) has no relative minima the theorem is valid by default. Assume now that \( {c}_{0} \) is the minimum of \( f \) . Then clearly \( \Gamma = \left\{ {\mathbf{x} : f\left( \mathbf{x}\right) \leq {c}_{0},\mathbf{x} \in \Omega }\right\} \) and this is convex by Proposition 3 of the last section.\n\nSuppose... | Yes |
Theorem 2. Let \( f \in {C}^{1} \) be convex on the convex set \( \Omega \) . If there is a point \( {\mathbf{x}}^{ * } \in \Omega \) such that, for all \( \mathbf{y} \in \Omega ,\nabla f\left( {\mathbf{x}}^{ * }\right) \left( {\mathbf{y} - {\mathbf{x}}^{ * }}\right) \geq 0 \), then \( {\mathbf{x}}^{ * } \) is a global... | Proof. We note parenthetically that since \( \mathbf{y} - {\mathbf{x}}^{ * } \) is a feasible direction at \( {\mathbf{x}}^{ * } \), the given condition is equivalent to the first-order necessary condition stated in Sect. 7.1. The proof of the proposition is immediate, since by Proposition 4 of the last section \[ f\le... | Yes |
Theorem 3. Let \( f \) be a convex function defined on the bounded, closed convex set \( \Omega \) . If \( f \) has a maximum over \( \Omega \) it is achieved at an extreme point of \( \Omega \) . | Proof. Suppose \( f \) achieves a global maximum at \( {\mathbf{x}}^{ * } \in \Omega \) . We show first that this maximum is achieved at some boundary point of \( \Omega \) . If \( {\mathbf{x}}^{ * } \) is itself a boundary point, then there is nothing to prove, so assume \( {\mathbf{x}}^{ * } \) is not a boundary poin... | Yes |
Proposition 1 (Zero-Order Necessary Conditions). If \( {\mathbf{x}}^{ * } \) solves (7.14) under the stated convexity conditions, then there is a nonzero vector \( \lambda \in {E}^{n} \) such that \( {\mathbf{x}}^{ * } \) is a solution to the two problems:\n\n\[ \text{minimize}f\left( \mathbf{x}\right) + {\lambda }^{T}... | Proof. Problem (7.17) follows from (7.15) (with \( s = 1 \) ) and the fact that \( f\left( \mathbf{x}\right) \leq r \) for \( r \geq f\left( \mathbf{x}\right) \) . The value \( c \) is attained from above at \( \left( {{f}^{ * },{\mathbf{x}}^{ * }}\right) \) . Likewise (7.18) follows from (7.16) and the fact that \( {\... | No |
Consider a continuously differentiable function \( f \) of a single variable \( x \in {E}^{1} \) defined on the unit interval \( \left\lbrack {0,1}\right\rbrack \) which plays the role of \( \Omega \) here. The first problem (7.17) implies \( {f}^{\prime }\left( {x}^{ * }\right) = - \lambda \) . | If the solution is at the left end of the interval (at \( x = 0 \) ) then the second problem (7.18) 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 ... | No |
Proposition 2 (Zero-Order Sufficiency Conditions). If there is a \( \lambda \) such that \( {\mathbf{x}}^{ * } \in \Omega \) solves the problems (7.17) and (7.18), then \( {\mathbf{x}}^{ * } \) solves (7.14). | Proof. Suppose \( {\mathbf{x}}_{1} \) is any other point in \( \mathbf{\Omega } \) . Then from (7.17)\n\n\[ f\left( {\mathbf{x}}_{1}\right) + {\lambda }^{T}{\mathbf{x}}_{1} \geq f\left( {\mathbf{x}}^{ * }\right) + {\lambda }^{T}{\mathbf{x}}^{ * }.\]\n\nThis can be rewritten as\n\n\[ f\left( {\mathbf{x}}_{1}\right) - f\... | Yes |
As a special case, suppose that the mapping \( \mathbf{A} \) is a point-to-point mapping; that is, for each \( \mathbf{x} \in X \) the set \( \mathbf{A}\left( \mathbf{x}\right) \) consists of a single point in \( Y \) . Suppose also that \( \mathbf{A} \) is continuous at \( \mathbf{x} \in X \) . This means that if \( {... | The converse is, however, not true in general. | 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{m... | However, starting from \( x > 1 \), the algorithm generates a sequence converging to \( x = 1 \) which is not a solution. The difficulty is that \( A \) is not closed at \( x = 1 \) . | Yes |
On the real line \( X \) consider the solution set to be empty, the descent function \( Z\left( x\right) = {e}^{-x} \), and the algorithm \( A\left( x\right) = x + 1 \) . All conditions of the convergence theorem except (i) hold. The sequence generated from any starting condition diverges to infinity. | This is not strictly a violation of the conclusion of the theorem but simply an example illustrating that if no compactness assumption is introduced, the generated sequence may have no convergent subsequence. | Yes |
The sequence with \( {r}_{k} = {a}^{\left( {2}^{k}\right) } \) for \( 0 < a < 1 \) converges to zero with order two, since \( {r}_{k + 1}/{r}_{k}^{2} = 1 \) . | since \( {r}_{k + 1}/{r}_{k}^{2} = 1 \) | Yes |
The sequence \( {r}_{k} = 1/k \) converges to zero arithmetically. | The convergence is of order one but it is not linear, since \( \mathop{\lim }\limits_{{k \rightarrow \infty }}\left( {{r}_{k + 1}/{r}_{k}}\right) = 1 \), that is, \( \beta \) is not strictly less than one. | Yes |
For the sequence \( {r}_{k} = {a}^{\left( {2}^{k}\right) },0 < a < 1 \), given in Example 2, we have | \[ {\left| {r}_{k}\right| }^{1/{2}^{k}} = a \] while \[ {\left| {r}_{k}\right| }^{1/{p}^{k}} = {a}^{{\left( 2/p\right) }^{k}} \rightarrow 1 \] for \( p > 2 \) . Thus the average order is two. | Yes |
Lemma 1. Let \( f\left( \mathbf{x}\right) \) be differentiable everywhere and satisfy the (first-order) \( \beta \) -Lipschitz condition. Then, for any two points \( \mathbf{x} \) and \( \mathbf{y} \)\n\n\[ f\left( \mathbf{x}\right) - f\left( \mathbf{y}\right) - \nabla f\left( \mathbf{y}\right) \left( {\mathbf{x} - \ma... | ## Then we prove | No |
Lemma 2. The iterative process (8.39) satisfies\n\n\[ E\left( {\mathbf{x}}_{k + 1}\right) = \left\{ {1 - \frac{{\left( {\mathbf{g}}_{k}^{T}{\mathbf{g}}_{k}\right) }^{2}}{\left( {{\mathbf{g}}_{k}^{T}\mathbf{Q}{\mathbf{g}}_{k}}\right) \left( {{\mathbf{g}}_{k}^{T}{\mathbf{Q}}^{-l}{\mathbf{g}}_{k}}\right) }}\right\} E\left... | Proof. The proof is by direct computation. We have, setting \( {\mathbf{y}}_{k} = {\mathbf{x}}_{k} - {\mathbf{x}}^{ * } \), \n\n\[ \frac{E\left( {\mathbf{x}}_{k}\right) - E\left( {\mathbf{x}}_{k + 1}\right) }{E\left( {\mathbf{x}}_{k}\right) } = \frac{2{\alpha }_{k}{\mathbf{g}}_{k}^{T}\mathbf{Q}{\mathbf{y}}_{k} - {\alph... | Yes |
For any \( {\mathbf{x}}_{0} \in {E}^{n} \) the method of steepest descent (8.39) converges to the unique minimum point \( {\mathbf{x}}^{ * } \) of \( f \) Furthermore, with \( E\left( \mathbf{x}\right) = \) \( \frac{1}{2}{\left( \mathbf{x} - {\mathbf{x}}^{ * }\right) }^{T}\mathbf{Q}\left( {\mathbf{x} - {\mathbf{x}}^{ *... | Proof. By Lemma 2 and the Kantorovich inequality\n\n\[ E\left( {\mathbf{x}}_{k + 1}\right) \leq \left\{ {1 - \frac{4aA}{{\left( A + a\right) }^{2}}}\right\} E\left( {\mathbf{x}}_{k}\right) = {\left( \frac{A - a}{A + a}\right) }^{2}E\left( {\mathbf{x}}_{k}\right) . \]\n\n![839b7907-f889-4cb0-8fa0-7192729aac93_245_0.jpg]... | Yes |
Consider the earlier example of \( f\left( x\right) = {tx} - \ln x \) . | \[ \lambda \left( x\right) = {\left\lbrack {f}^{\prime }{\left( x\right) }^{2}/{f}^{\prime \prime }\left( x\right) \right\rbrack }^{\frac{1}{2}} = \left| {\left( {t - 1/x}\right) x}\right| = \left| {1 - {tx}}\right| . \]\n\nThen (8.70) gives\n\n\[ \left( {1 - t{x}^{ + }}\right) \leq 2{\left( 1 - tx\right) }^{2} \]\n\nA... | Yes |
Theorem 2. In the method of conjugate gradients we have\n\n\[ E\left( {\mathbf{x}}_{k + 1}\right) \leq \mathop{\max }\limits_{{\lambda }_{i}}{\left\lbrack 1 + {\lambda }_{i}{P}_{k}\left( {\lambda }_{i}\right) \right\rbrack }^{2}E\left( {\mathbf{x}}_{0}\right) \]\n\nfor any polynomial \( {P}_{k} \) of degree \( k \), wh... | 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\[ \n{x}_{1} + {x}_{2}\; + \\lambda = 0. \n\]\n\nThese three equations together with the one constraint equation give four equations that can be solved for the four unknowns \( {x}_{1},{x}_{2... | Yes |
We seek to construct a cardboard box of maximum volume, given a fixed area of cardboard.\n\nDenoting the dimensions of the box by \( x, y, z \), the problem can be expressed as\n\n\[ \n\\text{maximize}{xyz} \n\]\n\n\[ \n\\text{subject to}\\;\\left( {{xy} + {yz} + {xz}}\\right) = \\frac{c}{2}\\text{,} \n\]\n\nwhere \( c... | Introducing a Lagrange multiplier, the first-order necessary conditions are easily found to be\n\n\[ \n{yz} + \\lambda \\left( {y + z}\\right) = 0 \n\]\n\n\[ \n{xz} + \\lambda \\left( {x + z}\\right) = 0 \n\]\n\n\[ \n{xy} + \\lambda \\left( {x + y}\\right) = 0 \n\]\n\ntogether with the constraint. Before solving these,... | Yes |
If the value of mean is known to be \( m \) (by the physical situation), the maximum entropy argument suggests that the density should be taken as that which solves the following problem:\n\n\[ \text{ maximize }\; - \mathop{\sum }\limits_{{i = 1}}^{n}{p}_{i}\log \left( {p}_{i}\right) \]\n\n\[ \text{subject to}\mathop{\... | We begin by ignoring the nonnegativity constraints, believing that they may be inactive. Introducing two Lagrange multipliers, \( \lambda \) and \( \mu \), the Lagrangian is\n\n\[ l = \mathop{\sum }\limits_{{i = 1}}^{n}\left\{ {-{p}_{i}\log {p}_{i} + \lambda {p}_{i} + \mu {x}_{i}{p}_{i}}\right\} - \lambda - {\mu m}. \]... | Yes |
A chain is suspended from two thin hooks that are 16 ft apart on a horizontal line as shown in Fig. 11.3. The chain itself consists of 20 links of stiff steel. Each link is one foot in length (measured inside). We wish to formulate the problem to determine the equilibrium shape of the chain. | The solution can be found by minimizing the potential energy of the chain. Let us number the links consecutively from 1 to 20 starting with the left end. We let link \( i \) span an \( x \) distance of \( {x}_{i} \) and a \( y \) distance of \( {y}_{i} \) . Then \( {x}_{i}^{2} + {y}_{i}^{2} = 1 \) . The potential energ... | Yes |
Suppose there are \( n \) securities indexed by \( i = 1,2 \) , \( \ldots, n \) . Each security \( i \) is characterized by its random rate of return \( {r}_{i} \) which has mean value \( {\bar{r}}_{i} \) . Its covariances with the rates of return of other securities are \( {\sigma }_{ij} \), for \( j = 1,2,\ldots, n \... | The overall rate of return of a portfolio is \( r = \mathop{\sum }\limits_{{i = 1}}^{n}{w}_{i}{\bar{r}}_{i} \) and variance \( {\sigma }^{2} = \mathop{\sum }\limits_{{i, j = 1}}^{n} \) \( {w}_{i}{\sigma }_{ij}{w}_{j} \)\n\nMarkowitz introduced the concept of devising efficient portfolios which for a given expected rate... | Yes |
Consider the problem\n\n\\[ \n\\text{maximize}{x}_{1}{x}_{2} + {x}_{2}{x}_{3} + {x}_{1}{x}_{3} \n\\]\n\n\\[ \n\\text{subject to}{x}_{1} + {x}_{2} + {x}_{3} = 3\\text{.} \n\\] | In Example 1 of Sect. 11.4 it was found that \\( {x}_{1} = {x}_{2} = {x}_{3} = 1,\\lambda = - 2 \\) satisfy the first-order conditions. The matrix \\( \\mathbf{F} + {\\mathbf{\\lambda }}^{T}\\mathbf{H} \\) becomes in this case\n\n\\[ \nL = \\left\\lbrack \\begin{array}{lll} 0 & 1 & 1 \\\\ 1 & 0 & 1 \\\\ 1 & 1 & 0 \\end... | Yes |
In the last section we considered\n\n\[ \nL = \left\lbrack \begin{array}{lll} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}\right\rbrack \n\]\n\nrestricted to \( M = \left\{ {\mathbf{y} : {y}_{1} + {y}_{2} + {y}_{3} = 0}\right\} \) . To obtain an explicit matrix representation on \( M \) let us introduce the orthonor... | \[ \n{\mathbf{E}}^{T}\mathbf{{LE}} = \left\lbrack \begin{array}{rr} - 1 & 0 \\ 0 & - 1 \end{array}\right\rbrack \n\]\n\nand hence \( \mathbf{L} \) restricted to \( M \) acts like the negative of the identity. | Yes |
\[ \text{extremize}{x}_{1} + {x}_{2}^{2} + {x}_{2}{x}_{3} + 2{x}_{3}^{2} \] \[ \text{subject to}\frac{1}{2}\left( {{x}_{1}^{2} + {x}_{2}^{2} + {x}_{3}^{2}}\right) = 1\text{.} \] | The first-order necessary conditions are \[ 1 + \;\lambda {x}_{1} = 0 \] \[ 2{x}_{2} + {x}_{3} + \lambda {x}_{2} = 0 \] \[ {x}_{2} + 4{x}_{3} + \lambda {x}_{3} = 0. \] One solution to this set is easily seen to be \( {x}_{1} = 1,{x}_{2} = 0,{x}_{3} = 0,\lambda = - 1 \) . Let us examine the second-order conditions at th... | Yes |
Example 3. Approaching Example 2 in this way and noting \( \mathbf{A} = \nabla \mathbf{h} = \left( {1,0,0}\right) \) we have | \[ {\mathbf{P}}_{A} = \mathbf{I} - \left\lbrack \begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right\rbrack {\left\lbrack \begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right\rbrack }^{T} = \left\lbrack \begin{array}{lll} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right\rbrack \]\n\nThen\n\n\[ {\mathbf{P}}_{A}\mathbf{L}{\math... | Yes |
Proposition 1. Suppose \( \Omega \) is convex, the function \( f \) is convex, and \( \mathbf{h} \) is affine. Then the primal function \( \omega \) is convex. | Proof. For simplicity of notation we assume that \( \Omega \) is the entire space \( X \) . Then we observe\n\n\[ \Omega \left( {\alpha {\mathbf{y}}_{1} + \left( {1 - \alpha }\right) {\mathbf{y}}_{2}}\right) = \inf \{ f\left( \mathbf{x}\right) : \mathbf{h}\left( \mathbf{x}\right) = \alpha {\mathbf{y}}_{1} + \left( {1 -... | Yes |
Consider the classic problem of finding the rectangle of maximum area while limiting the perimeter to a length of 4. | The 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 ne... | Yes |
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