{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

anay_cen_web_art

anay_cen_web_art - Notes for Class Analytic Center Newtons...

This preview shows pages 1–5. Sign up to view the full content.

Notes for Class: Analytic Center, Newton’s Method, and Web-Based ACA Robert M. Freund April 8, 2004 c 2004 Massachusetts Institute of Technology. 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
1 The Analytic Center of a Polyhedral System Given a polyhedral system of the form: Ax b , Mx = g , the analytic center is the solution of the following optimization problem: m (ACP:) maximize x,s s i i =1 s.t. Ax + s = b s 0 Mx = g . This is easily seen to be the same as: m (ACP:) minimize x,s ln( s i ) i =1 s.t. Ax + s = b s 0 Mx = g . The analytic center possesses a very nice “centrality” property. Suppose that s ) is the analytic center. Define the following matrix: x, ˆ s 1 ) 2 0 . . . 0 0 s 2 ) 2 . . . 0 ˆ S 2 := . . . . . . . . . . . . . 0 0 . . . s m ) 2 2
x ^ P E out E in Figure 1: Illustration of the Ellipsoid construction at the analytic center. Next define the following sets: P := { x | Mx = g, Ax b } x ) T A T S ˆ 2 A ( x ˆ E IN := x | Mx = g, ( x ˆ x ) 1 x ) T A T S ˆ 2 A ( x ˆ E OUT := x | Mx = g, ( x ˆ x ) m x, ˆ Theorem 1.1 If s ) is the analytic center, then: E IN P E OUT . This theorem is illustrated in Figure 1. The theorem is actually pretty easy to prove. 3

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Proof: Suppose that x E IN , and let s = b Ax . Since Mx = g , we need only prove that s 0 to show that x P . By construction of E IN , s satisfies ( s s ˆ) T S ˆ 2 ( s s ˆ) 1, where ˆ = b A ˆ s x . This in turn can be written as: m ( s i s ˆ i ) 2 1 , 2 i =1 s ˆ i whereby we see that each s i must satisfy s i 0. Therefore Ax b and so x P . We can write the optimality conditions (KKT conditions) for problem ACP as: S ˆ 1 e + λ = 0 0 + A T λ + M T u = 0 A ˆ s x + ˆ = b M ˆ x = g , where e = (1 , . . . , 1) T , i.e., the e is the vector of ones. From this we can derive the following fact: if ( x, s ) is feasible for problem ACP, then T S ˆ 1 T ˆ T e s = e S 1 ( b Ax ) = λ T b + u T Mx = λ T b + u g .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern