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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
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.
- Fall '13
- The Aeneid