Basic solutions in standard form polyhedra


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1 Recap, and outline of Lecture Previously min s.t. c x Ax = b x Converting any LP into standard form () Interpretation and visualization () Full row rank assumption on A () Basic solutions and bases in standard form polyhedra started Today Basic solutions and bases in standard form polyhedra Degeneracy in standard form polyhedra Adjacent solutions and adjacent bases Optimality conditions (for standard form LPs) IOE 5: Linear Programming I, Fall 2 Outline of Lecture Page Basic solutions in standard form polyhedra Assumptions A m n, b m Ax = b, x Ax = b has at least one solution A has full row rank (and hence m n) Recall: To construct a basic solutions, need to choose n linearly independent constraints to be active. At a basic solution for a problem in standard form: Ax = b give us m linearly independent active constraints (At least) n m of the constraints xj need to be active The resulting system of linear equations needs to have a unique solution! IOE 5: Linear Programming I, Fall 2 Basic solutions and bases Page 2
2 Example Arbitrarily picking n m sign constraints to be active might not result in a basic solution! 2 8 A = 2 6 4, b = n = 7, m = 4; need (at least) 3 sign constraints active for a BS. Try x = x 2 = x 3 =(a BFS) 2. Try x = x 2 = x 4 =(a BS, but not a BFS) 3. Try x = x 3 = x 6 =(not a BS: no solutions) 4. Try x 4 = x 5 = x 6 =(not a BS: multiple solutions, e.g., x =2, x 2 =5, x 3 =.5, x 7 =, or x =, x 2 =5.5, x 3 =.75, x 7 =.5) IOE 5: Linear Programming I, Fall 2 Basic solutions and bases Page 3 Basic solutions in standard form polyhedra Theorem 2.4 Consider the polyhedron represented by constraints Ax = b and x and assume that the m n matrix A has linearly independent rows. A vector x n is a basic solution if and only if we have Ax = b and there exist indices B(),...,B(m) suchthat: (a) The columns A B(),...,A B(m) are linearly independent; (b) if j = B(),...,B(m), then x j =. Proof of the if part: Suppose x satisfies (a) and (b). Then x satisfies m A B(i) x B(i) = b, i= x j =, j = B(),...,B(m) Above system has a unique solution (since A B(),...,A B(m) are linearly independent) Therefore, x is a BS IOE 5: Linear Programming I, Fall 2 Basic solutions and bases Page 4
3 Basic solutions in standard form polyhedra Theorem 2.4 Consider the polyhedron represented by constraints Ax = b and x and assume that the m n matrix A has linearly independent rows. A vector x n is a basic solution if and only if we have Ax = b and there exist indices B(),...,B(m) suchthat: (a) The columns A B(),...,A B(m) are linearly independent; (b) if j = B(),...,B(m), then x j =. Proof of the only if part: Suppose x is a BS. Let x B(),...,x B(k) be the nonzero components of x (k m) The following system has a unique solution (since x is a BS): k A B(i) x B(i) = b, i= x j =, j = B(),...,B(k) Hence, A B(),...,A B(k) are linearly independent If k < m, can find additional columns A B(k+),...,A B(m) so that columns A B(),...,A B(m) are linearly independent With this selection of B(),...,B(m), x satisfies (a) and (b) IOE 5: Linear Programming I, Fall 2 Basic solutions and bases Page 5 Procedure for constructing basic solutions of problems in standard form. Choose m linearly independent columns A B(),...,A B(m) 2. Let x i =foralli = B(),...,B(m) 3. Solve the system of m equations Ax = b for the unknowns x B(),...,x B(m) Example: A = , b = Let B() = 4, B(2) =, B(3) = 6, B(4) = x 3 = x 5 = x 7 = 3. Solve x B() + 2 x B(2) + x B(3) + x B(4) = x 4 = x B() =, x = x B(2) =3, x 6 = x B(3) =, x 2 = x B(4) =6 IOE 5: Linear Programming I, Fall 2 Basic solutions and bases Page
4 Terminology of BSs for standard form systems If x is a basic solution, and x B(),...,x B(m) are as above, Columns A B(),...,A B(m) basic columns; theyforma basis of m. The matrix B = A B() A B(2) A B(m) is the basis matrix Variables x B =(x B(),...,x B(m) ) basic variables; the remaining variables are nonbasic Unique solution of Bx B = b is x B = B b A BFS synthesizes the target vector b as a (nonnegative) linear combination of basic columns of A IOE 5: Linear Programming I, Fall 2 Basic solutions and bases Page 7 Degeneracy in standard form polyhedra Definition 2. A basic solution x n is said to be degenerate if more than n of the constraints are active at x. Definition 2. Consider the standard from polyhedron P = {x n Ax = b, x } and let x be a basic solution. Let A m n have full row rank. The vector x is a degenerate basic solution if more than n m of the components of x are zero. Degeneracy is not a purely geometric property! It depends on problem representation. P = {(x, x 2, x 3 ):x x 2 =, x + x 2 +2x 3 =2, x, x 2, x 3 } vs. P = {(x, x 2, x 3 ):x x 2 =, x + x 2 +2x 3 =2, x, x 3 } (,, ) is degenerate in the st representation, but not the 2nd. IOE 5: Linear Programming I, Fall 2 Basic solutions and bases Page 8
5 Correspondence between bases and basic solutions Two bases are distinct or different if they involve different set of indices {B(),...,B(m)}. Two different bases may correspond to the same basic solution. In the second part of Theorem 2.4, if the number of nonzero components in a BS is < m (i.e., the basic solution is degenerate), we might have a choice of which columns of A to use to complete the basis However, different basic solutions correspond to different bases because a basis uniquely determines the corresponding basic solution IOE 5: Linear Programming I, Fall 2 Basic solutions and bases Page 9 Adjacent basic solutions and adjacent bases Definition: Adjacent basic solutions Two distinct basic solution to a set of linear constraints in n are adjacent if we can find n linearly independent constraints that are active at both of them. Definition: Adjacent bases In a standard form problem, two bases are adjacent if they share all but one basic column. Adjacent basic solutions can always be obtained from adjacent bases If two adjacent bases lead to two different basic solutions, then these solutions are adjacent. IOE 5: Linear Programming I, Fall 2 Basic solutions and bases Page
6 Towards an algorithm for solving LPs Typical algorithm for solving an optimization problem: Find a feasible solution Check if there is a nearby feasible solution with a lower cost. If so, move to that solution and repeat. If no terminate; the current solution is a locally optimal solution i.e., feasible solution with the lowest cost among nearby feasible solutions Observations In linear programming, a locally optimal solution is also globally optimal Need an approach to determine a direction of cost decrease...or establish that none exists Definition 3. Let x be an element of a polyhedron P. Avectord n is said to be a feasible direction at x, if there exists a positive scalar θ for which x + θd P. IOE 5: Linear Programming I, Fall 2 Optimality conditions Page Basic directions Let x be a BFS, with B(),...,B(m) being the indices of the basic variables x i = for every nonbasic variable B =[A B(),...,A B(m) ] corresponding basis matrix x B =(x B(),...,x B(m) ) = B b. The jth basic direction: Let j = B(),...,B(m) Idea: move away from x by setting x j = θ >, and keeping all other nonbasic variables at. Formally: move from x to x + θd, where dj = di =foralli = j, B(),...,B(m) Need A(x + θd) =b, i.e.,ad = =Ad = n i= A id i = m i= A B(i)d B(i) + A j = Bd B + A j d B = B A j IOE 5: Linear Programming I, Fall 2 Optimality conditions Page 2
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