Discrete Optimization 2010 Lecture 5 Min-Cost Flows & Total Unimodularity

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1 Discrete Optimization 2010 Lecture 5 Min-Cost Flows & Total Unimodularity Marc Uetz University of Twente m.uetz@utwente.nl Lecture 5: sheet 1 / 26 Marc Uetz Discrete Optimization

2 Outline 1 Min-Cost Flows 2 Polyhedra and Total Unimodularity Lecture 5: sheet 2 / 26 Marc Uetz Discrete Optimization

3 Minimum Cost Flow Problem In Matrix Notation... A = node arc incidence matrix, E = identity matrix c = vector of integer arc costs, w.l.o.g. c 0 [arc reversal] u = vector of integer arc capacities, u 0 b = vector of integer node balances, i V b i = 0 min c x s.t. Ax = b Ex u x 0 W.l.o.g. consider special case: b s = v, b t = v, b i = 0 otherwise Lecture 5: sheet 3 / 26 Marc Uetz Discrete Optimization

4 Optimality Condition I Theorem (Negative Cost Cycle Optimality Condition) A flow x (of value v) is a minimum cost flow if and only if the residual graph G(x) has no negative cost directed cycle. Lecture 5: sheet 4 / 26 Marc Uetz Discrete Optimization

5 Cycle Canceling Algorithm Algorithm 1: Cycle Canceling input : network (G, u) capacities u 0, costs c, flow value v, and nodes s, t V output: x = minimum-cost (s, t)-flow with value v compute (s, t)-flow x with value v [e.g., augmenting paths]; while ( negative cost dicycle C in G(x)) do x C = maximum flow along C; reduce cost by updating flow, x = x + x C ; update residual graph, i.e., re-compute G(x); Computation time is not polynomial in general (adapting the Ford-Fulkerson example for maximum flow) Lecture 5: sheet 5 / 26 Marc Uetz Discrete Optimization

6 Minimum Mean Cycle Algorithm Definition Given a network with arc costs c ij, a minimum mean cycle is a directed cycle minimizing the expression 1 C c ij (i,j) C 1 Find any (s, t)-flow flow x of value v [e.g., augmenting paths] 2 While ( negative cost dicycle(s) in G(x)) C = minimum mean cycle, x C = maximum flow along C x = x + x C update G(x) Implementable in O( n 2 m 2 log(nc) ) time, where C = max{c ij } note that this is polynomial time [min. mean cycle in O( nm )] Lecture 5: sheet 6 / 26 Marc Uetz Discrete Optimization

7 Optimality Condition II (Dual) Theorem (Reduced Cost Optimality Condition) A flow x (of value v) is a minimum cost flow if and only if for some set of node labels π(i), i V, the reduced cost optimality condition holds in the residual graph G(x): c π ij := c ij π(i) + π(j) 0 (i, j) G(x) Lecture 5: sheet 7 / 26 Marc Uetz Discrete Optimization

8 Intuition about Reduced Cost Condition π(i) π(j) c ij π(i) = min. cost of getting 1 unit flow into i in G(x) then the π s should better fulfill -inequality, π(j) π(i) + c ij c π ij = c ij π(i) + π(j) 0 Lecture 5: sheet 8 / 26 Marc Uetz Discrete Optimization

9 Successive Shortest Path Algorithm Algorithm 2: Successive Shortest Paths input : network (G, u) capacities u 0, costs c, flow value v, and nodes s, t V output: x = minimum-cost (s, t)-flow with value v Initialize x = 0, G(x) = G, π(i) = 0 i V ; while (value v(x) < v) do d(i) := shortest path distances in G(x) with arc lengths c π ij ; P = shortest (s, t)-path (w.r.t. distances d); x P = maximum/necessary flow on P; x = x + x P ; update residual graph, i.e., re-compute G(x); update node labels π, i.e., π(i) = π(i) d(i) i V ; Lecture 5: sheet 9 / 26 Marc Uetz Discrete Optimization

10 Analysis Invariant (requires a proof) Anytime c π ij 0 for all (i, j) G(x), hence anytime the flow is a min-cost flow (correctness thus follows from Reduced Cost Condition) Termination Each iteration augments flow by 1, at most v iterations Computation time Each iteration: O( n ) + 1 Dijkstra, so total: O( vn 2 ) (polynomial only if v polynomial) Lecture 5: sheet 10 / 26 Marc Uetz Discrete Optimization

11 Proof of invariant c π 0, by induction Start: c π = c 0, since we assumed arc costs 0, and π = 0 Step: Suppose c π ij = c ij π(i) + π(j) 0 before current iteration d(i) = shortest c π -path distances s i in G(x) send flow along P and update G(x) (possibly new arcs!) we update π(i) π(i) d(i) Claim: new c π ij = c ij π(i) + π(j) 0 for all (i, j) new G(x) Case 1, (i, j) not new: d( ) = shortest paths d(j) d(i) + c π ij d(j) d(i) + c ij π(i) + π(j) 0 c ij (π(i) d(i)) + (π(j) d(j)) = new c π ij Case 2, (i, j) new: flow was sent along (j, i) d(i) = d(j) + c π ji d(i) = d(j) + c ji π(j) + π(i) c ji = (π(i) d(i)) + (π(j) d(j)) Now, new c π ij = c ij (π(i) d(i)) + (π(j) d(j)) = c ij + c ji = 0 Lecture 5: sheet 11 / 26 Marc Uetz Discrete Optimization

12 Comparison of Algorithms Cycle Canceling starts with feasible flow (value v), possibly not optimal always maintains feasibility (anytime flow value v) works towards optimality Successive Shortest Paths starts with infeasible flow (value 0), but cost optimal always maintains optimality (for current flow value) works towards feasibility Lecture 5: sheet 12 / 26 Marc Uetz Discrete Optimization

13 Outline 1 Min-Cost Flows 2 Polyhedra and Total Unimodularity Lecture 5: sheet 13 / 26 Marc Uetz Discrete Optimization

14 Integrality of Min-Cost Flows Theorem Given integer input data, and given the problem is not unbounded (negative cost cycle with unbounded capacity), any Min-Cost Flow problem has an optimal solution with integer arc flows. Proof: Both algorithms solve the problem, and anytime augment flow in integer amounts. Lecture 5: sheet 14 / 26 Marc Uetz Discrete Optimization

15 Integrality of Primal & Dual Optimal Solutions Primal Min-Cost Flow min c t x s.t. Ax = b Ex u x 0 Dual of Min-Cost Flow ( ) max (b t, u t π ) α ( ) s.t. [A t π, E] c α α 0 If not unbounded, both problems have an integer optimal solution. Lecture 5: sheet 15 / 26 Marc Uetz Discrete Optimization

16 Integrality of the Dual of Min-Cost Flow ( ) π max (b, u) α ( ) s.t. [A t π, E] c α α 0 Observe that the constraints are π(i) π(j) α ij c ij α ij cij π Let x be flow by successive shortest path alg., define dual solution π(i) upon termination α ij = max{0, c π ij } [note: c π ij < 0 is possible, only if (i, j) G(x), so if x ij = u ij ] feasible: α 0 and α c π by definition optimal: it fulfills complementary slackness (Exercise) Lecture 5: sheet 16 / 26 Marc Uetz Discrete Optimization

17 LPs With No Integer Optimal Solutions!"#$%#$%&'(%()*+%,')%-&.%/#&+-)%0)'1)-2 Another LP with all data integer, even all 0/1 (which problem?):!"#$%&'(%)*'+,-."'#%./,-0'#%(.#1,-."2% %486% Only optimal solution is ( 1 9-,&%./,-0'#%.:;<+,-=<%='#1<%786>%?@*..)AB 2, 1 2, 1 2 ) with value 3 2, but any integer solution, e.g. (0, 0, 1), only has value 1 C"$%-",<D*'#%(.#1,-."%."#$%$-<#E(%='#1<%4> Question When has a linear programming problem integer optimal solutions? Lecture 5: sheet 17 / 26 Marc Uetz Discrete Optimization

18 Polyhedra & Polytopes Polyhedron P R n is polyhedron if it is the solution set of a finite number of linear inequalities, P = {x R n Ax b}, for A R m n, b R m. P Polytope A polyhedron that is bounded (does not contain any infinite ray ) P Lecture 5: sheet 18 / 26 Marc Uetz Discrete Optimization

19 Faces Let polyhedron P = {x R n Ax b} be given. Supporting Hyperplane Let 0 w R n, r R, then {x w t x = r} is a hyperplane in R n. Inequality w t x r is called valid for P if P {x w t x r} and a supporting hyperplane if it is valid for P and P {x w t x = r} Face Let {x w t x = r} be a supporting hyperplane for P, then the intersection F = P {x w t x = r} is called a (proper) face of P. Theorem (see Literature Proposition 6.7) If F is a minimal face of P = {x Ax b} (by inclusion), then F = {x A o x = b o } for some subsystem A o, b o of Ax b. Lecture 5: sheet 19 / 26 Marc Uetz Discrete Optimization

20 Vertices (I) Definition A minimal face of polyhedron P that consist of only one point {x} is called a vertex. Theorem (see Literature Proposition 6.9) Point x P is a vertex of P if and only if it cannot be written as convex combination of two other points of P ( cornerpoint of P). A polyhedron need not have vertices, for example P = {(x 1, x 2 ) R 2 x 1 0} Lecture 5: sheet 20 / 26 Marc Uetz Discrete Optimization

21 Vertices (II) Theorem (see Literature Proposition 6.7) Any vertex x of a polyhedron P = {x R n Ax b} is defined by n linearly independent rows of A (which are fulfilled with equality, so A o x = b o for a non-singular subsystem A o, b o of Ax b). Consequence Polyhedron {x R n Ax b} has a vertex if and only if rank(a) = n. Exercise: Show that polyhedron P {x x 0} has a vertex Lecture 5: sheet 21 / 26 Marc Uetz Discrete Optimization

22 Polytopes Theorem (Minkowski) A polytope is equal to the convex hull of its vertices, x 1,..., x k, so P = {x R n Ax b} = conv{x 1,..., x k }. P If Q is convex, and Q contains all vertices of polytope P, P Q. Lecture 5: sheet 22 / 26 Marc Uetz Discrete Optimization

23 Polytopes Associated to LPs Polytope described by!"#$%"&' ()*+'),-(.'+*.$*%/'*0*1."2'*(3'451#(%(')6 6 inequalities given above "-*7*'8%-'9'*&"(3%)*:;<6<6<=6;>6<6<=6;<6>6<=6;<6<6>=6;?6?6?=@ 5 vertices: {(0, 0, 0), (1, 0, 0), (0, 1, 0), (0, 0, 1), ( 1 2, 1 2, 1 2 )} A"%(,'B*C3'*"D*%/'*'8%-'9'*&"(3%)*()3E%*(3%'F-1#6*1#%/"5F/* Note: 1##*,"'DD(,('3%)*(3*%/'*#(3'1-*(3'451#(%(')*1-' not all vertices are integer, although all coefficients are Lecture 5: sheet 23 / 26 Marc Uetz Discrete Optimization

24 Question Under what conditions on A and b is it true that all vertices of P = {x R n Ax b, x 0} happen to be integer? Because then, for any c the linear program max c t x s.t. Ax b, x 0 (if not infeasible or unbounded) has an integer optimal solution as the set of optimal solutions of P is a face of {x Ax b, x 0} Can even solve integer linear programming problems by solving an LP only (say with Simplex), because an optimum solution (if it exists) also occurs at a vertex which happens to be integer. Answer: If A is totally unimodular, and b integer Lecture 5: sheet 24 / 26 Marc Uetz Discrete Optimization

25 (Totally) Unimodular Matrices Definition 1 An integer square matrix B Z n n is unimodular if det(b) = ±1 2 An integer matrix A Z m n is totally unimodular (TU) if each square submatrix B of A has det(b) {0, ±1}. Some easy facts: entries of a TU matrix are {0, ±1} by definition (1 1 submatrices) if A is TU, adding or deleting a row or column vector (0,..., 1,..., 0), the result is again TU if A is TU, multiplying any row/column by 1, the result is again TU Lecture 5: sheet 25 / 26 Marc Uetz Discrete Optimization

26 Square Submatrix Back Lecture 5: sheet 26 / 26 Marc Uetz Discrete Optimization