Chapter 2. Simplex method

1. Chapter 2. Simplex method • Geometric view : x2 (0,2) (1,2) (2,2/3) x1 (2,0)

2. Let a  Rn, b R. Geometric intuition for the solution sets of { x : a’x = 0 } { x : a’x  0 } { x : a’x  0 } { x : a’x = b } { x : a’x  b } { x : a’x  b }

3. { x : a’x  0 } { x : a’x = 0 } { x : a’x  0 } • Geometry in 2-D a 0

4. Let z be a (any) point satisfying a’x = b. Then { x : a’x = b } = { x : a’x = a’z } = { x : a’(x – z) = 0 } Hence x – z = y, where y is any solution to a’y = 0, or x = y + z. So x can be obtained by adding z to every point y satisfying Ay = 0. Similarly, for { x : a’x  b }, { x : a’x  b }. { x : a’x  b } a z 0 { x : a’x = b } { x : a’x  b } { x : a’x = 0 }

5. (2,2/3) • Points satisfying (halfspace) x2 (4,3) x1

6. Def: The set of points which can be described in the form is called a polyhedron. ( Intersection of finite number of halfspaces) Hence, linear programming is the problem of optimizing (maximize, minimize) a linear function over a polyhedron. • Thm: Polyhedron is a convex set. Pf) HW earlier.

7. Solving LP graphically x2 (0,2) (1,2) (2,2/3) x1 (2,0)

8. Properties of optimal solutions • Thm: If LP has a unique optimal solution, the optimal solution is an extreme point. Pf) Suppose x* is unique optimal solution and it is not extreme point of the feasible set. Then there exist feasible points y, z  x* such that x* = y +(1- )z for some 0 <  < 1. Then c’x* = c’y + (1- )c’z. If c’x*  c’y, then either c’y > c’x* or c’z > c’x*, hence contradiction to x* being optimal solution. If c’x* = c’y, y is also optimal solution. Contradiction to x* being unique optimal.  • Thm: Suppose polyhedron P has at least one extreme point. If LP over Phas an optimal solution, it has an extreme point optimal solution. Pf) not given here. 

9. Multiple optimal solutions x2 (0,2) (1,2) (2,2/3) x1 (2,0)

10. Obtaining extreme point algebraically x2 (0,2) (1,2) (2,2/3) x1 (2,0)

11. Suppose polyhedron is given (A: mxn). Extreme point of the polyhedron can be obtained by setting n of the inequalities as equations (coefficient vectors must be linearly independent) and obtaining the solution satisfying the equations. If the obtained point satisfies other inequalities, it is in P and it is an extreme point of the polyhedron • Enumeration : ( the number of ways to choose n inequalities (which hold at equalities) out of (m+n) inequalities.) • Algorithm strategy : from an extreme point, move to the neighboring extreme point which gives a better (precisely speaking, not worse) solution

12. Remark: There exists a polyhedron which is not full-dimensional. (extreme point is defined same as before.) x3 1 x1 1 This polyhedron is 2-dimensional. 1 x2

13. Geometric Idea of the Simplex Method • Any LP problem must be converted to a problem having only equations except the nonnegativity constraints if simplex method can be applied (details later) • Consider the LP problem max c’x, Ax = b, -x  0 A: m  n, full row rank(n m) P = { x : Ax=b, -x  0 } To define an extreme point of P, we need n equations. Since we already have mequations in Ax=b, (n - m) equations must come from -x  0, which means (n - m) variables are set to 0. Let A=[B:N], where N is the submatrix corresponding to the variables set to 0. Then we solve the system Bx = b for the remaining m variables. (Note that the coefficient matrix B must be nonsingluar so that the system of equations has a unique solution.)

14. Ex:extreme point ( 1, 0, 0 ) can be obtained from x1 + x2 + x3 = 1, x2 = 0, x3 = 0. Since ( 1, 0, 0 ) satisfies –x1 0, it is an extreme point. x3 1 x1 1 This polyhedron is 2-dimensional. 1 x2

15. (continued) Let A = [B : N] , B: m  m, nonsingular, N: m  (n - m), where N is the submatrix of A having columns associated with variables set to 0. Then an extreme point can be found by solving Ax = b, xN = 0.  [B : N] (xB: xN)’ = b  BxB + NxN = b, -xN = 0. (or BxB = b - NxN , -xN = 0. )  multiplying B-1 on both sides, we obtain B-1BxB + B-1NxN = B-1b or IxB + B-1NxN = B-1b, -xN = 0. Solution is xB = B-1b, xN = 0 This is the basic solution we mentioned earlier. By the choice of the variables we set at 0, we obtain different basic solutions (different extreme points). xB are called basic variables, and xN are called nonbasic variables. • If the obtained solution satisfies nonnegativity, xB = B-1b  0, we have a basic and feasible solution (satisfies nonnegativity of variables)

16. Coeff. Matrix for Basic solution m n-m xB Ax = b B N b m = xN -I 0 0 -xN = 0 n-m

17. Simplex method searches only basic feasible solutions, which is tantamount to searching the extreme points of the corresponding polyhedron until it finds an optimal solution.

18. Simplex method (algebraic interpretation) Add slack variables(여유변수) to each constraint to convert them to equations. (1) (2)

19. Hence we have a 1-1 mapping which maps each feasible point in (1) to a feasible point in (2) uniquely (and conversely) and the objective values are the same for the points. So solve (2) instead of (1). (Surplus variable (잉여변수) : a’x  b  a’x – xs =b, xs 0)

20. Remark: If LP includes equations, we need to convert each equation to two inequalities to express the problem in standard form as we have seen earlier. Then we may add slack or surplus variables to convert them to equations. However, this procedure will increase the number of constraints and variables. Equations in an LP can be handled directly without changing them to inequalities. Detailed method will be explained in Chap8. General LP Problems. For the time being, we assume that we follow the standard procedure to convert equations to inequalities.

21. Changes in the solution space when slack is added x2 x3 1 1 x1 1 1 x1 1 x2

22. Next let Then find solution to the following system which maximizes z (tableau form) In the text, dictionary form used, i.e. each dependent variable (including z) (called basic variable) is expressed as linear combinations of indep. var. (called nonbasic variable). (Note that, unlike the text, we place the objective function in the first row. Such presentation style is used more widely and we follow that convention)

23. From previous lectures, we know that if the polyhedron P has at least one extreme point and the LP over P has a finite optimal solution, the LP has an extreme point optimal solution. Also an extreme point of P for our problem is a basic feasible solution algebraically. We obtain a basic solution by setting x1 = x2 = x3 = 0 and finding the values of x4, x5, and x6 , which can be read directly from the dictionary. (also z values can be read.) If all values of x4, x5, and x6 are nonnegative, we obtain a basic feasible solution. • The equation for z may be regarded as part of the systems of equations, or we may think of it as a separate equation used to evaluate the objective value for the given solution.