Lecture slides by Kevin Wayne

Transcription

1 LINEAR PROGRAMMING I a refreshing example standard form fundamental questions geometry linear algebra simplex algorithm Lecture slides by Kevin Wayne Last updated on 7/25/17 11:09 AM

2 Linear programming Linear programming. Optimize a linear function subject to linear inequalities. (P) max c j x j n j=1 n s. t. a ij x j = b i 1 i m j=1 x j 0 1 j n (P) max c T x s. t. Ax = b x 0 2

3 Linear programming Linear programming. Optimize a linear function subject to linear inequalities. Generalizes: Ax = b, 2-person zero-sum games, shortest path, max flow, assignment problem, matching, multicommodity flow, MST, min weighted arborescence, Why significant? Design poly-time algorithms. Design approximation algorithms. Solve NP-hard problems using branch-and-cut. Ranked among most important scientific advances of 20th century. 3

4 LINEAR PROGRAMMING I a refreshing example standard form fundamental questions geometry linear algebra simplex algorithm Reference: The Allocation of Resources by Linear Programming, Scientific American, by Bob Bland

5 Brewery problem Small brewery produces ale and beer. Production limited by scarce resources: corn, hops, barley malt. Recipes for ale and beer require different proportions of resources. Beverage Corn (pounds) Hops (ounces) Malt (pounds) Profit () Ale (barrel) Beer (barrel) constraint How can brewer maximize profits? Devote all resources to ale: 34 barrels of ale 442 Devote all resources to beer: 32 barrels of beer barrels of ale, 29.5 barrels of beer barrels of ale, 28 barrels of beer 800 5

6 Brewery problem objective function Ale Beer max 13A + 23B s. t. 5A + 15B 480 4A + 4B A + 20B 1190 A, B 0 Profit Corn Hops Malt constraint decision variable 6

7 Reference 7

8 LINEAR PROGRAMMING I a refreshing example standard form fundamental questions geometry linear algebra simplex algorithm

9 Standard form of a linear program Standard form of an LP. Input: real numbers a ij, c j, b i. Output: real numbers x j. n = # decision variables, m = # constraints. Maximize linear objective function subject to linear inequalities. (P) max c j x j n j=1 n s. t. a ij x j = b i 1 i m j=1 x j 0 1 j n (P) max c T x s. t. Ax = b x 0 Linear. No x 2, x y, arccos(x), etc. Programming. Planning (term predates computer programming). 9

10 Brewery problem: converting to standard form Original input. max 13A + 23B s. t. 5A + 15B 480 4A + 4B A + 20B 1190 A, B 0 Standard form. Add slack variable for each inequality. Now a 5-dimensional problem. max 13A + 23B s. t. 5A + 15B + S C = 480 4A + 4B + S H = A + 20B + S M = 1190 A, B, S C, S H, S M 0 10

11 Equivalent forms Easy to convert variants to standard form. (P) max c T x s. t. Ax = b x 0 Less than to equality. x + 2y 3z 17 x + 2y 3z + s = 17, s 0 Greater than to equality. x + 2y 3z 17 x + 2y 3z s = 17, s 0 Min to max. min x + 2y 3z max x 2y + 3z Unrestricted to nonnegative. x unrestricted x = x + x, x + 0, x 0 11

12 LINEAR PROGRAMMING I a refreshing example standard form fundamental questions geometry linear algebra simplex algorithm

13 Fundamental questions LP. For A R m n, b R m, c R n, α R, does there exist x R n such that: Ax = b, x 0, c T x α? Q. Is LP in NP? Q. Is LP in co-np? Q. Is LP in P? Q. Is LP in P R? Blum Shub Smale model Input size. n = number of variables. m = number of constraints. L = number of bits to encode input. 13

14 LINEAR PROGRAMMING I a refreshing example standard form fundamental questions geometry linear algebra simplex algorithm

15 Brewery problem: feasible region Hops 4A + 4B 160 Malt 35A + 20B 1190 (0, 32) (12, 28) (26, 14) Corn 5A + 15B 480 Beer (0, 0) Ale (34, 0) 15

16 Brewery problem: objective function Profit (0, 32) (12, 28) 13A + 23B = 1600 (26, 14) Beer 13A + 23B = 800 (0, 0) Ale (34, 0) 13A + 23B =

17 Brewery problem: geometry Brewery problem observation. Regardless of objective function coefficients, an optimal solution occurs at a vertex. (0, 32) (12, 28) (26, 14) vertex Beer (0, 0) Ale (34, 0) 17

18 Convexity Convex set. If two points x and y are in the set, then so is λ x + (1 λ) y for 0 λ 1. convex combination not a vertex iff d 0 s.t. x ± d in set Vertex. A point x in the set that can t be written as a strict convex combination of two distinct points in the set. vertex x y convex not convex Observation. LP feasible region is a convex set. 18

19 Purificaiton Theorem. If there exists an optimal solution to (P), then there exists one that is a vertex. (P) max c T x s. t. Ax = b x 0 Intuition. If x is not a vertex, move in a non-decreasing direction until you reach a boundary. Repeat. xʹ = x + α * d x + d x x - d 19

20 Purificaiton Theorem. If there exists an optimal solution to (P), then there exists one that is a vertex. Pf. Suppose x is an optimal solution that is not a vertex. There exist direction d 0 such that x ± d P. A d = 0 because A(x ± d) = b. Assume c T d 0 (by taking either d or d). Consider x + λ d, λ > 0 : Case 1. [ there exists j such that d j < 0 ] Increase λ to λ * until first new component of x + λ d hits 0. x + λ * d is feasible since A(x + λ * d) = Ax = b and x + λ * y 0. x + λ * d has one more zero component than x. c T xʹ = c T (x + λ * d) = c T x + λ * c T d c T x. d k = 0 whenever x k = 0 because x ± d P 20

21 Purificaiton Theorem. If there exists an optimal solution to (P), then there exists one that is a vertex. Pf. Suppose x is an optimal solution that is not a vertex. There exist direction d 0 such that x ± d P. A d = 0 because A(x ± d) = b. Assume c T d 0 (by taking either d or d). Consider x + λ d, λ > 0 : Case 2. [d j 0 for all j ] x + λd is feasible for all λ 0 since A(x + λd) = b and x + λd x 0. As λ, c T (x + λd) because c T d < 0. if c T d = 0, choose d so that case 1 applies 21

22 LINEAR PROGRAMMING I a refreshing example standard form fundamental questions geometry linear algebra simplex algorithm

23 Intuition Intuition. A vertex in R m is uniquely specified by m linearly independent equations. 4A + 4B A + 20B 1190 (26, 14) 4A + 4B = A + 20B =

24 Basic feasible solution Theorem. Let P = { x : Ax = b, x 0 }. For x P, define B = { j : x j > 0 }. Then, x is a vertex iff A B has linearly independent columns. Notation. Let B = set of column indices. Define A B to be the subset of columns of A indexed by B. Ex. A = " % # &, b = " 7 % 16 # 0 & x = " 2 % 0 1 # 0 &, B = {1, 3}, A B = " 2 3% 7 2 # 0 0& 24

25 Basic feasible solution Theorem. Let P = { x : Ax = b, x 0 }. For x P, define B = { j : x j > 0 }. Then, x is a vertex iff A B has linearly independent columns. Pf. Assume x is not a vertex. There exist direction d 0 such that x ± d P. A d = 0 because A(x ± d) = b. Define Bʹ = { j : d j 0 }. A Bʹ has linearly dependent columns since d 0. Moreover, d j = 0 whenever x j = 0 because x ± d 0. Thus Bʹ B, so A Bʹ is a submatrix of A B. Therefore, A B has linearly dependent columns. 25

26 Basic feasible solution Theorem. Let P = { x : Ax = b, x 0 }. For x P, define B = { j : x j > 0 }. Then, x is a vertex iff A B has linearly independent columns. Pf. Assume A B has linearly dependent columns. There exist d 0 such that A B d = 0. Extend d to R n by adding 0 components. Now, A d = 0 and d j = 0 whenever x j = 0. For sufficiently small λ, x ± λ d P x is not a vertex. 26

27 Basic feasible solution Theorem. Given P = { x : Ax = b, x 0 }, x is a vertex iff there exists B { 1,, n } such B = m and: A B is nonsingular. x B = A B -1 b 0. x N = 0. basic feasible solution Pf. Augment A B with linearly independent columns (if needed). x = " # " % " 7 % A = , b = 16 # & # 0 & % " 2 3 0, B = { 1, 3, 4 }, A B = # & % & Assumption. A R m n has full row rank. 27

28 Basic feasible solution: example Basic feasible solutions. {B, S H, S M } (0, 32) Basis {A, B, S M } (12, 28) Infeasible {A, B, S H } (19.41, 25.53) max 13A + 23B s. t. 5A + 15B + S C = 480 4A + 4B + S H = A + 20B + S M = 1190 A, B, S C, S H, S M 0 Beer {A, B, S C } (26, 14) {S H, S M, S C } (0, 0) Ale {A, S H, S C } (34, 0) 28

29 Fundamental questions LP. For A R m n, b R m, c R n, α R, does there exist x R n such that: Ax = b, x 0, c T x α? Q. Is LP in NP? A. Yes. Number of vertices ( ) nm. C(n, m) = n m Cramer s rule can check a vertex in poly-time. Cramer s rule. For B R n n invertible, b R n, the solution to Bx = b is given by: x i = det(b i ) det(b) replace ith column of B with b 29

30 LINEAR PROGRAMMING I a refreshing example standard form fundamental questions geometry linear algebra simplex algorithm

31 Simplex algorithm: intuition Simplex algorithm. [George Dantzig 1947] Move from BFS to adjacent BFS, without decreasing objective function. replace one basic variable with another edge Greedy property. BFS optimal iff no adjacent BFS is better. Challenge. Number of BFS can be exponential! 31

32 Simplex algorithm: initialization max Z subject to 13A + 23B Z = 0 5A + 15B + S C = 480 4A + 4B + S H = A + 20B + S M = 1190 A, B, S C, S H, S M 0 Basis = {S C, S H, S M } A = B = 0 Z = 0 S C = 480 S H = 160 S M =

33 Simplex algorithm: pivot 1 max Z subject to 13A + 23B Z = 0 5A + 15B + S C = 480 4A + 4B + S H = A + 20B + S M = 1190 A, B, S C, S H, S M 0 Basis = {S C, S H, S M } A = B = 0 Z = 0 S C = 480 S H = 160 S M = 1190 Substitute: B = 1/15 (480 5A S C ) max Z subject to 16 3 A S C Z = A + B + 1 S 3 15 C = A 4 15 S C + S H = A 4 3 S C + S M = 550 A, B, S C, S H, S M 0 Basis = {B, S H, S M } A = S C = 0 Z = 736 B = 32 S H = 32 S M =

34 Simplex algorithm: pivot 1 max Z subject to 13A + 23B Z = 0 5A + 15B + S C = 480 4A + 4B + S H = A + 20B + S M = 1190 A, B, S C, S H, S M 0 Basis = {S C, S H, S M } A = B = 0 Z = 0 S C = 480 S H = 160 S M = 1190 Q. Why pivot on column 2 (or 1)? A. Each unit increase in B increases objective value by 23. Q. Why pivot on row 2? A. Preserves feasibility by ensuring RHS 0. min ratio rule: min { 480/15, 160/4, 1190/20 } 34

35 Simplex algorithm: pivot 2 max Z subject to 16 3 A S C Z = A + B + 1 S 3 15 C = A 4 15 S C + S H = A 4 3 S C + S M = 550 A, B, S C, S H, S M 0 Basis = {B, S H, S M } A = S C = 0 Z = 736 B = 32 S H = 32 S M = 550 Substitute: A = 3/8 (32 + 4/15 S C S H ) max Z subject to S C 2 S H Z = 800 B S C S H = 28 A 1 10 C S H = C 85 8 S H + S M = 110 A, B, S C, S H, S M 0 Basis = {A, B, S M } S C = S H = 0 Z = 800 B = 28 A = 12 S M =

36 Simplex algorithm: optimality Q. When to stop pivoting? A. When all coefficients in top row are nonpositive. Q. Why is resulting solution optimal? A. Any feasible solution satisfies system of equations in tableaux. In particular: Z = 800 S C 2 S H, S C 0, S H 0. Thus, optimal objective value Z* 800. Current BFS has value 800 optimal. max Z subject to S C 2 S H Z = 800 B S C S H = 28 A 1 10 C S H = C 85 8 S H + S M = 110 A, B, S C, S H, S M 0 Basis = {A, B, S M } S C = S H = 0 Z = 800 B = 28 A = 12 S M =

37 Simplex tableaux: matrix form Initial simplex tableaux. c B T x B + c N T x N = Z A B x B + A N x N = b x B, x N 0 Simplex tableaux corresponding to basis B. (c N T c B T A B 1 A N ) x N = Z c B T A B 1 b I x B + A 1 B A N x N = A 1 B b x B, x N 0 subtract c B T A B -1 times constraints multiply by A B -1 x B = A B -1 b 0 x N = 0 basic feasible solution c N T c B T A B -1 A N 0 optimal basis 37

38 Simplex algorithm: corner cases Simplex algorithm. Missing details for corner cases. Q. What if min ratio test fails? Q. How to find initial basis? Q. How to guarantee termination? 38

39 Unboundedness Q. What happens if min ratio test fails? all coefficients in entering column are nonpositive max Z subject to + 2x x 5 Z = 2 x 1 4x 4 8x 5 = 3 x 2 + 5x 4 12x 5 = 4 x 3 = 5 x 1, x 2, x 3, x 4, x 5 0 A. Unbounded objective function. Z = x 5 " # x 1 x 2 x 3 x 4 x 5 % & = " # 3 + 8x x % & 39

40 Phase I simplex method Q. How to find initial basis? (P) max c T x s. t. Ax = b x 0 A. Solve (P ), starting from basis consisting of all the z i variables. m ( P ") max z i i=1 s. t. A x + I z = b x, z 0 Case 1: min > 0 (P) is infeasible. Case 2: min = 0, basis has no z i variables okay to start Phase II. Case 3a: min = 0, basis has z i variables. Pivot z i variables out of basis. If successful, start Phase II; else remove linear dependent rows. 40

41 Simplex algorithm: degeneracy Degeneracy. New basis, same vertex. Degenerate pivot. Min ratio = 0. max Z subject to 3 x x x 2 6 6x 7 Z = 0 x x 4 4 8x 5 x 6 + 9x 7 = 0 x x 4 12x x 6 + 3x 7 = 0 x 3 + x 6 = 1 x 1, x 2, x 3, x 4, x 5, x 6, x

42 Simplex algorithm: degeneracy Degeneracy. New basis, same vertex. Cycling. Infinite loop by cycling through different bases that all correspond to same vertex. Anti-cycling rules. Bland s rule: choose eligible variable with smallest index. Random rule: choose eligible variable uniformly at random. Lexicographic rule: perturb constraints so nondegenerate. 42

43 Lexicographic rule Intuition. No degeneracy no cycling. Perturbed problem. much much greater, say ε i = δ i for small δ ( P ") max c T x s. t. Ax = b + ε x 0 where ε = # ε 1 % % ε 2 %! % ε n & ( ( ( (, such that ε 1 ε 2 # ε n Lexicographic rule. Apply perturbation virtually by manipulating ε symbolically: ε ε 2 + 8ε ε ε 2 + 3ε 3 43

44 Lexicographic rule Intuition. No degeneracy no cycling. Perturbed problem. much much greater, say ε i = δ i for small δ ( P ") max c T x s. t. Ax = b + ε x 0 where ε = # ε 1 % % ε 2 %! % ε n & ( ( ( (, such that ε 1 ε 2 # ε n Claim. In perturbed problem, x B = A B -1 (b + ε) is always nonzero. Pf. The j th component of x B is a (nonzero) linear combination of the components of b + ε contains at least one of the ε i terms. Corollary. No cycling. which can t cancel 44

45 Simplex algorithm: practice Remarkable property. In practice, simplex algorithm typically terminates after at most 2(m + n) pivots. but no polynomial pivot rule known Issues. Avoid stalling. Choose the pivot. Maintain sparsity. Ensure numerical stability. Preprocess to eliminate variables and constraints. Commercial solvers can solve LPs with millions of variables and tens of thousands of constraints. 45