Solving Linear Program by Simplex Method The Concept Simplex method is an algebraic procedure However, its underlying concepts are geometric An iterative algorithm If possible, use the origin as the initial solution Optimality test looks at the scope of improvement of solution

The Simplex Method in a Nutshell An iterative procedure Initialization (Find initial solution) Is the current solution optimal? Yes Stop No Move to a better solution

Initial Assumptions All constraints are of the form ≤ All right-hand-side values are positive Initial basic feasible solution is taken as origin

The Augmented Form Set up the method first: Convert inequality constraints to equality constraints by adding slack variables Original Form Maximize Z = 10x1 + 15x2 + 20x3 subject to 2x1 + 4x2 + 6x3 ≤ 24 3x1 + 9x2 + 6x3 ≤ 30 x1,x2 ≥ 0 Augmented Form Maximize Z - 10x1 - 15x2 - 20x3 – 0s1 – 0s2 = 0 subject to 2x1 + 4x2 + 6x3 +s1 = 24 3x1 + 9x2 + 6x3 +s2 = 30 x1,x2, x3 , s1, s2 ≥ 0

Basic and Non-basic variables and BASIS In an LP Number of variables (n) > Number of equations (m) The difference (n – m) is the Non basic variables and the number of equations (m) is equal to basic variables The basic variables are on which appear once in the table with +1 as coefficient Non basic variables (set to 0) The basis is the set of basic variables If all basic variables are ≥ 0, we have a BFS Between two basic solutions, if their bases are the same except for one variable, then they are adjacent

Algebra of the Simplex Method Initialization Maximize (Z) Z - 10x1 - 15x2 - 20x3 – 0s1 – 0s2 = 0 subject to 2x1 + 4x2 + 6x3 +s1 = 24 3x1 + 9x2 + 6x3 +s2 = 30 x1,x2, x3 , s1, s2 ≥ 0 Find an initial basic feasible solution Remember from key concepts “If possible, use the origin as the initial Basic feasible solution” Choose the slack variables as basic variables (s1 and s2) and original variables be nonbasic (xi=0 where i =1,2,3…n)

The Entering Variable in a maximizing (minimizing) problem is the non basic variable with most negative (positive) coefficient in Z equation. A tie may be broken arbitrarily. Optimum is reached when all variables in coefficient in the Z equation are non-negative (non-positive) For both maximizing and minimizing problem the Leaving variable is the current basic variable having the smallest intercept (a minimum ratio with positive denominator) in the direction of the Entering variable (Entering variable column). A tie may be broken arbitrarily

Converting to tabular column Maximize Z - 10x1 - 15x2 - 20x3 – 0s1 – 0s2 = 0 subject to 2x1 + 4x2 + 6x3 +s1 = 24 3x1 + 9x2 + 6x3 +s2 = 30 x1,x2, x3 , s1, s2 ≥ 0 Initial table Basic variable x1 x2 x3 s1 s2 Solution Ratio Z -10 -15 -20 2 4 6 1 24 24/4 = 6 3 9 30 30/6 = 5

AFTER ITERATION 1: x1 x2 x3 s1 s2 Basic variable Solution Ratio Z -10/3 -5/3 10/3 80 1/3 2/3 1 1/6 4 12 5 -1 6

AFTER ITERATION 2: x1 x2 x3 s1 s2 Basic variable x1 x2 x3 s1 s2 Solution Ratio Z 15 10/3 100 -1 1 1/2 -1/3 2 5 6 Final solution as we have all NBV as Non negative Maximum Z = 100 with X1 = 6 and X3 = 2

Dual Simplex Method

Dual Simplex Method Suppose a “basic solution” satisfies the optimality conditions but not feasible, we apply dual simplex algorithm. In regular Simplex method, we start with a Basic Feasible solution (which is not optimal) and move towards optimality always retaining feasibility. In the dual simplex method, the exact opposite occurs. We start with an (more than) “optimal” solution (which is not feasible) and move towards feasibility always retaining optimality conditions. The algorithm ends once we obtain feasibility.

To start the dual Simplex method, the following three conditions are to be met: The objective function must satisfy the optimality conditions of the regular Simplex method. All the constraints must be of the type . 3. All variables should be  0. (Note: As in the simplex method, we have an identical conditions however, the RHS constants need NOT be  0). We use dual simplex when RHS constant < 0.

ITERATION PROCEDURE In any iteration, we first decide the “leaving” variable and then decide which variable should enter (the basis). Dual Feasibility Condition (Condition for a variable to leave the basis) The leaving variable, xi, is the basic variable having the most negative value in the Simplex table (ie the constant on RHS is most negative). Ties are broken arbitrarily. If all the basic variables are  0, the algorithm ends and we have obtained the optimal solution.

Dual Optimality Condition (Condition for a variable to Enter the basis) Entering variable is the basic variable with minimum ratio of modulus of (Z/a) i.e. having the minimum ratio of Z coefficient with corresponding coefficients in the Leaving variable row