The matrix s sn is given show that the optimal xopt

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Unformatted text preview: ) chooses a vector of values x as a solution of the LP minimize cT x subject to Ax n b, n with variable x ∈ R . You can think of x ∈ R as a vector of activity levels, b ∈ Rm as a vector of requirements, and c ∈ Rn as a vector of costs or prices for the activities. With this interpretation, the LP above finds the cheapest set of activity levels that meet all requirements. (This interpretation is not needed to solve the problem.) We suppose that A is known, along with a set of data (b(1) , x(1) ), ..., (b(r) , x(r) ), where x(j ) is an optimal point for the LP, with b = b(j ) . (The solution of an LP need not be unique; all we say here is that x(j ) is an optimal solution.) Roughly speaking, we have samples of optimal decisions, for different values of requirements. You do not know the cost vector c. Your job is to compute the tightest possible bounds on the costs ci from the given data. More specifically, you are to find cmax and cmin , the maximum and i i minimum possible values for ci , consi...
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This note was uploaded on 09/10/2013 for the course C 231 taught by Professor F.borrelli during the Fall '13 term at University of California, Berkeley.

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