- Introduction to Operations Research Peng Zhang April, 5 School of Computer Science and Technology, Shandong University, Ji nan 5, China. Email: algzhang@sdu.edu.cn.
Introduction
Overview of the Operations Research Modeling Approach
Introduction to Linear Programming
4 Solving Linear Programming Problems: The Simplex Method 4
5 The Theory of the Simplex Method 5
5-5. FOUNDATIONS OF THE SIMPLEX METHOD Properties of CPF Solutions Property : (a) If there is exactly one optimal solution, then it must be a CPF solution. (b) If there are multiple optimal solutions (and a bounded feasible region), then at least two must be adjacent CPF solutions. Property : There are only a finite number of CPF solutions. Suppose in the standard form of an LP there are m functional constraints and n variable constraints. Each CPF solution is the simultaneous solution of a system of n out
5- of the m + n constraint boundary equations. So, the number of different combinations of m + n equations taken n at a time is ( ) m + n (m + n)! =. n m!n! Property : Suppose the problem possesses at least one optimal solution. If a CPF solution has no adjacent CPF solutions that are better (as measured by Z), then there are no better CPF solutions anywhere. Therefore, such a CPF solution is guaranteed to be an optimal solution. set. This property holds since the feasible region of any LP is a convex Definition. Suppose S R n is a set of vectors in n dimensional Euclidean space. If for any x S, y S, and any λ,, it always
5- holds λx + ( λ)y S, then S is a convex set. Theorem. The feasible region D = {x R n Ax = b, x } of an LP is a convex set. Proof. Let x, y be any two vectors in D, and λ be any number in,. Define w = λx + ( λ)y. We need only to prove w D. Since x, y, and λ, we know w. Since Ax = b, Ay = b, for w we have Aw = λax + ( λ)ay = λb + ( λ)b = b, concluding the theorem.
5-4 By the above theorem, the following feasible region would never occur for any LP.
5-5 Extensions to the Augmented Form of the Problem Suppose the augmented form of an LP is as follows. max c x + c x + + c n x n s.t. a x + a x + + a n x n + x n+ = b a x + a x + + a n x n + x n+ = b a m x + a m x + + a mn x n + x n+m = b m x j, j We have the following facts for this form. Each basic solution has m basic variables, and the rest of the variables are nonbasic variables set equal to zero.
5-6 After the nonbasic variables are set to zero, the values of the basic variables are given by the simultaneous solution of the system of m equations for the problem, This basic solution is the augmented CP solution whose n defining equations are those indicated by the nonbasic variables.
5-7 5. THE SIMPLEX METHOD IN MATRIX FORM Suppose the problem in the augmented form (as well as the artificial form) is max s.t. c T x Ax = b x In the Simplex method, the system is expressed as ct Z =. A x b Fix a bfs. Denote by B the submatrix of A consisting of the columns corresponding the basic variables, called the basic matrix.
5-8 We interchangeably use the terms basic matrix and basis, since they are one-to-one corresponding. When the basic matrix is determined, the system can be expressed as ct N ct B N B Z x N x B = b. In the Simplex method, the system is in the proper form from Gaussian elimination. This is obtained by a series of elementary row operations on the simplex tableau, which is equivalent to premultiply the both sides of the above equation by ct B B : B
5-9 ct B B N c T N B N I Let A = B A = ζ T = = = Z x N x B = ct B B b B b B N I, b = B b, and c T B B N c T N c T B B N c T N ct B B B c T B c T B B N c T B B B c T. = c T BB A c T. So, the system in simplex tableau can be expressed as ζt Z = ct B b. A x b
5- The simplex tableau is thus b.v. x rhs Z ζ T c T B b x B A b The Simplex Algorithm Input: An augmented form (as well as artificial form) of the problem. Choose the initial basic matrix, getting the initial tableau and the initial bfs. while the current bfs is not optimal do Determine the entering basic variable x j by selecting the column with ζ j < having the largest ζ j. 4 Determine the leaving basic variable x i by applying the minimum
5- { } ratio test: i = arg min bi ā ij ā ij >. If A j, then the problem is unbounded and the algorithm exits. 5 Solve for the new bfs by using elementary row operations. 6 endwhile 7 return the current bfs as an optimal solution. When starting the Simplex method, the prepared simplex tableau is b.v. x rhs Z c T A b Once we determine the basic matrix, we can convert the prepared tableau into the initial tableau by using elementary row operations.
5- Example. The solving procedure of Simplex on the Wyndor Glass problem interpreted by the matrix form. max x + 5x s.t. x 4 x x + x 8 x j, j
5- Solution. The augmented form of the problem. max x + 5x s.t. x + x = 4 x + x 4 = x + x + x 5 = 8 x j, j Obtain the augmented form, getting the initial simplex tableau.
5-4 b.v. Z x x x x 4 x 5 rhs Z 5 x 4 x 4 B = x B = x 5 8 A A 4 A 5 = x x 4 x 5 = b, ct B = A = B A = A, b = B b = b., B =..
5-5 ζ T = c T B B A c T = c T = c T, Z = c T B B b =. b.v. Z x x x x 4 x 5 rhs B = x B = Z 5 x 4 x 6 x 5 6 A A A 5 = x x x 5 = b, ct B = 5, B =..
5-6 A = B A = b = B b = 4 4 = 6. 8 6 =,
5-7 ζ T = c T BB A c T = = = Z = c T B b = 5 5 5 5 5 5 5, 4 6 6 =.
5-8 b.v. Z x x x x 4 x 5 rhs B = x B = Z 6 x x 6 x A A A = x x x = b, ct B = 5, B =..
5-9 A = B A = b = B b = 4 = 6. 8 =,
5- ζ T = c T BB A c T = = = Z = c T B b = 5 5 5, 5 6 = 6. 5