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Unformatted text preview: −∆xT ∇f (x). You don’t really need
λ for anything; you can work with λ2 instead. (This is important for reasons described
• There can be small numerical errors in the Newton step ∆xnt that you compute. When
x is nearly optimal, the computed value of λ2 , i.e., λ2 = −∆xT ∇f (x), can actually be
(slightly) negative. If you take the squareroot to get λ, you’ll get a complex number,
and you’ll never recover. Moreover, your line search will never exit. However, this only
happens when x is nearly optimal. So if you exit on the condition λ2 /2 ≤ 10−6 , everything
will be ﬁne, even when the computed value of λ2 is negative.
• For the line search, you must ﬁrst multiply the step size t by β until x + t∆xnt is feasible
(i.e., strictly positive). If you don’t, when you evaluate f you’ll be taking the logarithm
of negative numbers, and you’ll never recover.
(b) LP solver with strictly feasible starting point. Using the centering code from part (a), implement a barrier method to solve the standard form LP
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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