The SIMPLEX Method: Identifying Optimal Solutions in Linear Programming, Study notes of Operational Research

Download The SIMPLEX Method: Identifying Optimal Solutions in Linear Programming and more Study notes Operational Research in PDF only on Docsity! The SIMPLEX Method - Development Every LP is in exactly one of the following states: 1. Feasible with a unique optimum solution. 2. Feasible with infinitely many optima. 3. Feasible, with no optimum solution (because the objective is unbounded). 4. Infeasible, and hence with no optimum solution. Assume we are in States 1 or 2. Then the following are true: A. The LP has at least one optimal corner point (or extreme point). 1) If in State 1, exactly one extreme point is optimal. 2) If in State 2, at least two adjacent (neighboring) extreme points are optimal. B. The number of extreme points is finite. C. If the objective function at some extreme point is as good as or better than at all of its adjacent extreme point, then this extreme point is optimal for the LP. It follows then that we can use the following iterative algorithmic procedure: STEP 0 (Initialization): Find an initial extreme point and make it the current candidate (if one cannot be found the LP is in state 4, i.e. it is infeasible – so STOP). STEP 1 (Stopping Criterion Check): Is the objective at the current extreme point at least as good or better than it is at all of its adjacent (neighboring) extreme points? If so this must be the optimal extreme point (via C) – so STOP. If not, go to Step 2. STEP 2 (Iterative Step): At least one of the adjacent extreme points is better – so make it the current candidate and go to Step 1. QUESTIONS: 1) How to find an initial extreme point? 2) What is the algebraic characterization of an “extreme point?” 3) What is the algebraic characterization of adjacent extreme points, i.e., how to move from an extreme point to one of its neighbors (in Step 2)? Note that n=6 and m=4. 1) An (infeasible) solution is [x1=70, x2=50, S1=-50, S2=-30, S3=0, S4=0] 2) A feasible solution is [x1=20, x2=20, S1=60, S2=50, S3=50, S4=30] Basic Solution: Suppose we fix (n-m) out of the n variables at zero, and try to solve the system of m equations in the remaining m variables. If a solution to this exists, then it is called a basic solution. Solution (1) on the previous page is an example of a basic solution. Basic Feasible Solution: A basic solution that also satisfies nonnegativity is called a basic feasible solution (BFS). An example of a BFS is [x1=20, x2=50, S1=0, S2=20, S3=50, S4=0]. Note that (1) is not a BFS even though it is a basic solution! Original LP has exactly Original LP is infeasible one feasible solution – it Feasible Infeasible must also be the optimal one! Augmented System One Solution No Solution Infinitely many solutions Basic Nonbasic Basic Nonbasic In a basic solution, the (n-m) variables that are chosen to be fixed at zero are called the nonbasic variables and the remaining m variables are called basic variables. Note that we can choose (n-m) out of the n variables in !)!( ! mmn n m n mn n − =      =      − different ways. Thus we can get a maximum of this many different basic solutions (some of which will also be basic feasible solutions) – in practice all of these may not exist since in some cases, the resulting m x m system may not have a solution! In our example, n=6 and m=4 so that 15 !4)!46( !6 = − =      − mn n The 15 combinations of nonbasic variables are: 1) x1, x2 2) x1, S1 6) x2, S1 3) x1, S2 7) x2, S2 10) S1, S2 4) x1, S3 8) x2, S3 11) S1, S3 13) S2, S2 5) x1, S4 9) x2, S4 12) S1, S4 14) S2, S4 15) S3, S4 Try to locate each of these on the graphical representation! FACT: Each BFS corresponds to an extreme point of the feasible region. X2 Max 20X1+10X2 S3=0, X1=70 st X1+2X2 ≤ 120 (Constr. 1, S1) X1+X2 ≤ 90 (Constr. 2, S2) 90 X1 ≤ 70 (Constr. 3, S3) S2=0, X1+X2=90 X2 ≤ 50 (Constr. 4, S4) X1, X2 ≥ 0 S1=S4=0 S2=S4=0 S3=S4=0 S4=0, X2=50 50 II III S1=S2=0 X1 *=70 IV S1=S3=0 X2 *=20 Feasible Region Opt. Val. =1600 V S2=S3=0 S1=0, X1+2X2=120 10 I VI X1 10 20 30 40 50 60 70 80 90 100 110 120 1. , : BASIC SOLUTIONS (13 of them). Note that there should be ( 4 6 )=15 of these, but only 13 exist because X1=70 is parallel to the X2-axis and X2=50 is parallel to the X1-axis. 2. : BASIC FEASIBLE SOLUTION: (6 of the above 13, numbered I, II, III, IV, V, VI) 3. : Contour of the objective function corresponding to a value of 1600 4. : Set of all Feasible solutions to the LP