Give the lagrange dual problem, and find the optimal solution and optimal value of the dual problem. What is the optimal duality gap?
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Answer:
Step-by-step explanation:
In mathematical optimization theory, duality or the duality principle is the principle that optimization problems may be viewed from either of two perspectives, the primal problem or the dual problem. The solution to the dual problem provides a lower bound to the solution of the primal (minimization) problem.[1] However in general the optimal values of the primal and dual problems need not be equal. Their difference is called the duality gap. For convex optimization problems, the duality gap is zero under a constraint qualification condition.
Dual problem
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Usually the term "dual problem" refers to the Lagrangian dual problem but other dual problems are used – for example, the Wolfe dual problem and the Fenchel dual problem. The Lagrangian dual problem is obtained by forming the Lagrangian of a minimization problem by using nonnegative Lagrange multipliers to add the constraints to the objective function, and then solving for the primal variable values that minimize the original objective function. This solution gives the primal variables as functions of the Lagrange multipliers, which are called dual variables, so that the new problem is to maximize the objective function with respect to the dual variables under the derived constraints on the dual variables (including at least the nonnegativity constraints).
In general given two dual pairs of separated locally convex spaces {\displaystyle \left(X,X^{*}\right)} \left(X,X^{*}\right) and {\displaystyle \left(Y,Y^{*}\right)} \left(Y,Y^{*}\right) and the function {\displaystyle f:X\to \mathbb {R} \cup \{+\infty \}} f:X\to \mathbb {R} \cup \{+\infty \}, we can define the primal problem as finding {\displaystyle {\hat {x}}} {\hat {x}} such that {\displaystyle f({\hat {x}})=\inf _{x\in X}f(x).\,} f({\hat {x}})=\inf _{x\in X}f(x).\, In other words, if {\displaystyle {\hat {x}}} {\hat {x}} exists, {\displaystyle f({\hat {x}})} f({\hat {x}}) is the minimum of the function {\displaystyle f} f and the infimum (greatest lower bound) of the function is attained.
If there are constraint conditions, these can be built into the function {\displaystyle f} f by letting {\displaystyle {\tilde {f}}=f+I_{\mathrm {constraints} }} {\tilde {f}}=f+I_{\mathrm {constraints} } where {\displaystyle I_{\mathrm {constraints} }} {\displaystyle I_{\mathrm {constraints} }} is a suitable function on {\displaystyle X} X that has a minimum 0 on the constraints, and for which one can prove that {\displaystyle \inf _{x\in X}{\tilde {f}}(x)=\inf _{x\ \mathrm {constrained} }f(x)} {\displaystyle \inf _{x\in X}{\tilde {f}}(x)=\inf _{x\ \mathrm {constrained} }f(x)}. The latter condition is trivially, but not always conveniently, satisfied for the characteristic function (i.e. {\displaystyle I_{\mathrm {constraints} }(x)=0} {\displaystyle I_{\mathrm {constraints} }(x)=0} for {\displaystyle x} x satisfying the constraints and {\displaystyle I_{\mathrm {constraints} }(x)=\infty } {\displaystyle I_{\mathrm {constraints} }(x)=\infty } otherwise). Then extend {\displaystyle {\tilde {f}}} {\tilde {f}} to a perturbation function {\displaystyle F:X\times Y\to \mathbb {R} \cup \{+\infty \}} F:X\times Y\to \mathbb {R} \cup \{+\infty \} such that {\displaystyle F(x,0)={\tilde {f}}(x)} F(x,0)={\tilde {f}}(x).[2]
The duality gap is the difference of the right and left hand sides of the inequality
{\displaystyle \sup _{y^{*}\in Y^{*}}-F^{*}(0,y^{*})\leq \inf _{x\in X}F(x,0),\,} \sup _{y^{*}\in Y^{*}}-F^{*}(0,y^{*})\leq \inf _{x\in X}F(x,0),\,
where {\displaystyle F^{*}} F^{*} is the convex conjugate in both variables and {\displaystyle \sup } \sup denotes the supremum (least upper bound
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