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Course: MATH 171A, Spring 2012
School: UCSD
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Optimality Recap: conditions for LP Math 171A: Linear Programming Lecture 14 Farkas' Lemma and its Implications Philip E. Gill c 2011 LP minimize x c Tx Ax b subject to with A an m n matrix, b an m-vector and c an n-vector. The vector x is a solution of LP if and only if : (a) Ax b (b) c = AT for some vector 0, where Aa is the active a a a constraint matrix at x Monday, February 7th, 2011...

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Optimality Recap: conditions for LP Math 171A: Linear Programming Lecture 14 Farkas' Lemma and its Implications Philip E. Gill c 2011 LP minimize x c Tx Ax b subject to with A an m n matrix, b an m-vector and c an n-vector. The vector x is a solution of LP if and only if : (a) Ax b (b) c = AT for some vector 0, where Aa is the active a a a constraint matrix at x Monday, February 7th, 2011 http://ccom.ucsd.edu/~peg/math171a UCSD Center for Computational Mathematics Slide 2/41, Monday, February 7th, 2011 x2 Example: minimize 2x1 + x2 subject to the constraints: constraint #1: constraint #2: constraint #3: x1 + x2 1 x2 0 x1 0 a1 c a3 Written in the form min c Tx subject 1 2 c= , A=0 1 1 to Ax b, we get 1 1 b=0 1 , 0 0 x a2 a1 c x x1 UCSD Center for Computational Mathematics Slide 3/41, Monday, February 7th, 2011 At the point x = with Aa = 1 1 1 0 , ba = 1 0 0 1 , the active set is A = {1, 3} At the point x= with Aa = 1 0 1 1 , ba = 1 0 1 0 , the active set is A = {1, 2} Solving for the Lagrange multipliers gives AT a = c a x is optimal. 1 1 1 0 a = 2 1 a = 1 1 0 Solving for the Lagrange multipliers gives AT a = c a 1 1 0 1 a = 2 1 a = 2 -1 0 x is not optimal. UCSD Center for Computational Mathematics Slide 5/41, Monday, February 7th, 2011 UCSD Center for Computational Mathematics Slide 6/41, Monday, February 7th, 2011 Proof of Farkas' lemma Proof: We have two statements that we must show are equivalent: Result (Farkas' Lemma) (A) c Tp 0 for all p such that Aa p 0 if and only if (B) c = AT for some 0 a a a (A) (B) c Tp 0 for all p such that AT p 0 a c = AT for some 0 a a a We must show that (A) (B) and (B) (A) UCSD Center for Computational Mathematics Slide 7/41, Monday, February 7th, 2011 UCSD Center for Computational Mathematics Slide 8/41, Monday, February 7th, 2011 Next we must show that (A) (B). It is "EASY" to show that (B) (A) If (B) holds, then c = Then, c p= T This is the same as showing that: 0. If (B) does not hold, then (A) does not hold. We must show that AT a a for some a (AT )Tp a a = ( )T (Aa p) a 0 for all p with Aa p 0 (B) is not true (B) not true (A) not true (A) is not true c Tp 0 for all p with Aa p 0 (A) holds. there is no 0 such that c = AT a a a c Tp < 0 for some p such that Aa p 0 UCSD Center for Computational Mathematics Slide 9/41, Monday, February 7th, 2011 UCSD Center for Computational Mathematics Slide 10/41, Monday, February 7th, 2011 We must show that there is no 0 such that c = AT a a a c Tp < 0 for some p such that Aa p 0 Two possibilities for there being no 0 with c = AT . a a a Case I: Case II: c does not lie in range(AT ) a c range(AT ), but c = AT with 0 a a a a Case I: c does not lie in range(AT ) a Every c Rn can be written uniquely as c = cR + cN with cR range(AT ) and cN null(Aa ) a T The definition of cR and cN imply that cR cN = 0. If c range(AT ), then a c = cR + cN with cN = 0 In both cases we construct a p such that c Tp < 0 and Aa p 0. UCSD Center for Computational Mathematics Slide 11/41, Monday, February 7th, 2011 UCSD Center for Computational Mathematics Slide 12/41, Monday, February 7th, 2011 Define p = -cN . Then Aa p = -Aa cN = 0 since cN null(Aa ). c Tp = -c TcN = -(cR + cN )TcN T T = -cR cN - cN cN Case II: c range(AT ), but c = AT with 0 a a a a This is the "HARD" part of Farkas' lemma. However, it is "quite EASY" if we assume that the rows of Aa are linearly independent. = -cN cN T = - cN 2 <0 Aa = with linearly independent rows. p = -cN satisfies Aa p 0 and c Tp < 0. (A) does not hold. UCSD Center for Computational Mathematics Slide 13/41, Monday, February 7th, 2011 UCSD Center for Computational Mathematics Slide 14/41, Monday, February 7th, 2011 Case II: c range(AT ), but c = AT with 0 a a a a In this case, exists, but ( )s < 0 for some index s. a a Let p be a solution of Aa p = es , where es is the unit vector 0 . . . 0 es = 1 row s 0 . . . 0 The equations Aa p = es are compatible (why?). Moreover, p is a feasible descent direction, as we now show: c Tp = (AT a )Tp = (T Aa )p = T (Aa p) = T es = (a )s < 0 a a a a Hence we have constructed a direction p such that c Tp < 0 and we are done! and Aa p ( = es ) 0 UCSD Center for Computational Mathematics Slide 15/41, Monday, February 7th, 2011 UCSD Center for Computational Mathematics Slide 16/41, Monday, February 7th, 2011 Implications of Farkas' Lemma In the proof we defined a feasible descent direction p such that Aa p = es where s is an index such that ( )s < 0 a The direction p satisfies: Result (Farkas' Lemma) (A) c Tp 0 for all p such that Aa p 0 if and only if (B) c = AT for some 0 a a a ajTp = 0 for j = s, with ajT the jth row of Aa T 1 for j = s, with as the sth row of Aa A x step + p keeps all active constraint residuals fixed at zero except the sth, which increases. i.e., x + p "stays on" all the active constraints and "moves off" of a constraint with a negative component of a . UCSD Center for Computational Mathematics Slide 17/41, Monday, February 7th, 2011 UCSD Center for Computational Mathematics Slide 18/41, Monday, February 7th, 2011 x2 a1 c a3 At the point x= with Aa = a2 a1 c 1 0 , the active set is A = {1, 2} x 1 0 1 1 , ba = 1 0 Solving for the Lagrange multipliers gives AT a = c a x1 x 1 1 0 1 a = 2 1 a = 2 -1 UCSD Center for Computational Mathematics Slide 20/41, Monday, February 7th, 2011 x2 a = 2 -1 s=2 We can construct a feasible descent direction p such that Aa p = e2 1 0 1 1 p= 0 1 p= -1 1 a2 p a1 c Observe that T a1 p = 0, T a2 p = 1, c Tp = 2 1 -1 1 = -1 x1 p is a feasible descent direction. we "move off" constraint #2, but "stay on" constraint #1. UCSD Center for Computational Mathematics Slide 21/41, Monday, February 7th, 2011 Complications of degeneracy Unfortunately, defining a feasible descent direction from Farkas' Lemma is known as a theorem of the alternative EITHER OR c = AT for some 0 a a a there is a p such that c Tp < 0 and Aa p 0 Aa p = es will not work if the active constraints are dependent, i.e., Aa = i.e., if Aa defines a degenerate vertex. In this case, Aa p = es is incompatible and no p exists. UCSD Center for Computational Mathematics Slide 23/41, Monday, February 7th, 2011 UCSD Center for Computational Mathematics Slide 24/41, Monday, February 7th, 2011 Example: minimize 2x1 + x2 subject to the constraints: constraint #1: constraint #2: constraint #3: constraint #4: x2 a4 x1 + x2 1 x2 0 x1 0 x1 + 2x2 1 to Ax b, we get 1 1 0 1 , b= 0 0 2 1 x a2 Written in the form min c Tx c= 2 1 , subject 1 0 A= 1 1 a1 c x x1 UCSD Center for Computational Mathematics Slide 25/41, Monday, February 7th, 2011 At the point x= with 1 0 The equations Aa p = e3 are , the active set is 1 1 , 2 A = {1, 2, 4} 1 Aa = 0 1 p1 + p2 = 0 p2 = 0 p1 + 2p2 = 1 Yet p = -1 1 1 1 2 incompatible equations! p1 p2 0 =0 1 1 Aa = 0 1 1 ba = 0 1 Solving AT a = c for the Lagrange multipliers gives a 1 1 0 1 1 2 2 1 3 a = 0 -1 s=3 a = is still a feasible descent direction. UCSD Center for Computational Mathematics Slide 27/41, Monday, February 7th, 2011 UCSD Center for Computational Mathematics Slide 28/41, Monday, February 7th, 2011 x2 a4 x -1 p= 1 a2 a1 c There is no direction that moves off one active constraint while staying on all the others. x x1 UCSD Center for Computational Mathematics Slide 30/41, Monday, February 7th, 2011 Geometric ideas behind Farkas' proof Consider the active-set matrix at x = 1 Aa = 0 1 i.e., a1 = 1 1 , a2 = 0 1 and a4 = 1 2 1 1 2 1 0 2 1 y2 a4 with c= a2 a1 c In this case c CN (AT ), so the corresponding x is not optimal. a y1 UCSD Center for Computational Mathematics Slide 31/41, Monday, February 7th, 2011 y2 a4 As CN (AT ) is convex, we can draw a line through the origin that a separates every point y CN (AT ) from c. a This line has the equation p Ty = 0 with normal vector p. i.e., p1 y1 + p2 y2 = 0 p p1 y1 + p2 y2 = 0 a2 a1 c y1 UCSD Center for Computational Mathematics Slide 33/41, Monday, February 7th, 2011 The hyperplane splits Rn into two half-spaces: In the general case, p Ty = 0 separates p Ty H1 = {y : p Ty 0} = 0 is called a separating hyperplane H2 = {y : p Ty < 0} from c. With the vector p pointing "into" the half-space H1 . By definition, we have CN (AT ) H1 a c H2 CN (AT ) a The existence of this hyperplane is the "beef" of Farkas' lemma (see the class notes). UCSD Center for Computational Mathematics Slide 35/41, Monday, February 7th, 2011 UCSD Center for Computational Mathematics Slide 36/41, Monday, February 7th, 2011 y2 a4 p H1 = {y : p Ty 0} H2 = {y : p Ty < 0} H1 a2 a1 H2 p1 y1 + p2 y2 = 0 Since CN (AT ) H1 , any y CN (AT ) satisfies p Ty 0. a a Now a1 , a2 and a4 must be in CN (AT ), so that a T a1 CN (AT ) a1 p 0 a T a2 CN (AT ) a2 p 0 a T a4 CN (AT ) a4 p 0 a c y1 Aa p 0 UCSD Center for Computational Mathematics Slide 38/41, Monday, February 7th, 2011 H1 = {y : p Ty 0} H2 = {y : p Ty < 0} Since c H2 , it must hold that p Tc < 0 (i.e., c Tp < 0 ) Putting all this together implies that the normal vector p satisfies c Tp < 0 and Aa p 0 One particular separating hyperplane may be found by finding the point in CN (AT ) that is closest to c. a Then, define p = cp - c UCSD Center for Computational Mathematics Slide 39/41, Monday, February 7th, 2011 UCSD Center for Computational Mathematics Slide 40/41, Monday, February 7th, 2011 y2 a4 p H1 a2 a1 H2 cp c y1
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UCSD - MATH - 171A
Recap: Farkas' lemmaMath 171A: Linear ProgrammingLecture 15 Testing for OptimalityPhilip E. Gillc 2011c Tp 0 for all p such that Aa p 0 if and only if c = AT for some 0 a a ahttp:/ccom.ucsd.edu/~peg/math171aWednesday, February 9th, 2011UCSD Center
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UCSD - MATH - 171A
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Math 171A: Linear ProgrammingInstructor: Philip E. GillWinter Quarter 2011Homework Assignment #2 Due Friday January 21, 2011 The starred exercises require the use of Matlab. Remember that it is necessary to do all the Matlab assignments to obtain credi
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Math 171A: Linear ProgrammingInstructor: Philip E. GillWinter Quarter 2011Homework Assignment #3 Due Friday January 28, 2011 Remember that the first midterm exam will be held in class on Wednesday, January 26. Starred exercises require the use of Matla
UCSD - MATH - 171A
Math 171A: Linear ProgrammingInstructor: Philip E. GillWinter Quarter 2011Homework Assignment #4 Due Friday February 4, 2011 Starred exercises require the use of Matlab. Exercise 4.1. Suppose that the constant vector c is such that cTp 0 for all p such
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Math 171A: Linear ProgrammingInstructor: Philip E. GillWinter Quarter 2011Homework Assignment #5 Due Friday February 11, 2011 Starred exercises require the use of Matlab. Exercise 5.1. Consider the set of inequality constraints Ax b, where 1 1 4 0 3 1
UCSD - MATH - 171A
Math 171A: Linear ProgrammingInstructor: Philip E. GillWinter Quarter 2011Homework Assignment #6 Due Friday February 18, 2011 Starred exercises require the use of Matlab. Exercise 6.1. (a) Consider the matrix of active constraints -1 -8 4 -4 2 0 Aa = -
UCSD - MATH - 171A
Math 171A: Linear ProgrammingInstructor: Philip E. GillWinter Quarter 2011Homework Assignment #7 Due Friday February 25, 2011 Starred exercises require the use of Matlab. Exercise 7.1. Consider the linear program of minimizing cTx subject to the genera
UCSD - MATH - 171A
Math 171A: Linear ProgrammingInstructor: Philip E. GillWinter Quarter 2011Homework Assignment #8 Due Friday March 4, 2011 The second midterm exam will be held in class on Wednesday, March 2. Starred exercises require the use of Matlab. Exercise 8.1. Co
UCSD - MATH - 171A
Math 171A: Linear ProgrammingInstructor: Philip E. GillWinter Quarter 2011Homework Assignment #9 Due Friday March 11, 2011 The final will be held Friday, March 18, 11:30am-2:30pm. Starred exercises require the use of Matlab. Exercise 9.1. Let a denote
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Cal Poly - MCB - 32
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