Artificial Variables, 2-Phase and Big M Methods For the example, go to p. 16.

Artificial Variables, 2-Phase and Big M Methods Facts: To start, we need a canonical form If we have a  constraint with a nonnegative right-hand side, it will contain an obvious basic variable (which?) after introducing a slack var. If we have an equality constraint, it contains no obvious basic variable If we have a  constraint with a nonnegative right-hand side, it contains no obvious basic variable even after introducing a surplus var. © 2008 MGS

Compare! 2x + 3y  5  2x + 3y + s = 5, s  0 (s basic) 2x + 3y = 5  ??????? Infeasible if x=y=0! 2x + 3y  5  2x + 3y - s = 5, s  0 (??????) Infeasible if x=y=0! ?????????????????? © 2008 MGS

One Equality??? 2x + 3y = 5  2x + 3y + a = 5, a = 0 (E) (a is basic, but it should be 0!) How do we force a = 0? This is of course not feasible if x=y=0, as 0+0+0 5! © 2008 MGS

One Equality??? 2x + 3y = 5  2x + 3y + a = 5, a = 0 (E) (a basic, but it should be 0!) How do we force a = 0? This is of course not feasible if x=y=0, as 0+0+0 5 Idea: ! solve a first problem with Min {a | constraint (E) + (a  0) + other constraints }! © 2008 MGS

Artificial Variables Notice: In an equality constraint, the extra variable is called an artificial variable. For instance, in 2x + 3y + a = 5, a = 0 (E) a is an artificial variable. © 2008 MGS

One Inequality  ??? 2x + 3y  5  2x + 3y - s = 5, s  0 (I) s could be the basic variable, but it should be  0 and for x=y=0, it is -5 ! How do we force s  0? ? © 2008 MGS

how? One Inequality  ??? 2x + 3y  5  2x + 3y - s = 5, s  0 (I) s could be the basic variable, but it should be  0 and it is -5 for x=y=0! How do we force s  0? By making it 0! how? © 2008 MGS

One Inequality  ??? 2x + 3y  5  2x + 3y - s = 5, s  0 (E) s could be basic, but it should be  0 and it is -5 for x=y=0! How do we force s  0? By making it 0! But we have to start with a canonical form… so treat is as an equality constraint! 2x + 3y - s + a = 5, s  0, a  0 and Min a © 2008 MGS

Artificial Variables Notice: In a  inequality constraint, the extra variable a is called an artificial variable. For instance, in 2x + 3y – s + a = 5, s  0, a  0 (E) a is an artificial variable. In a sense, we allow temporarily a small amount of cheating, but in the end we cannot allow it! © 2008 MGS

What if we have many such = and  constraints? 7x - 3y – s1 + a1 = 6, s1,a1  0 (I) 2x + 3y + a2 = 5, a2  0 (II) a1 and a2 are artificial variables, s1 is a surplus variable. One minimizes their sum: Min {a1+a2 | a1, a2  0, (I), (II), other constraints} i.e., one minimizes the total amount of cheating! © 2008 MGS

Then What? We have two objectives: Get a “feasible” canonical form Maximize our original problem Two methods: 2-phase method (phase 1, then phase 2) big M method © 2008 MGS

2-Phase Method Phase I: find a BFS Minimize the sum of the artificial variables If min = 0, we have found a BFS If min > 0, then we cannot find a solution with no “cheating”… the original problem is infeasible Phase 2: solve original LP Start from the phase 1 BFS, and maximize the original objective function. © 2008 MGS

Big-M Method Combine both objectives : (1) Min i ai (2) Max j cj xj into a single one: (3) Max – M i ai + j cj xj where M is a large number, larger than anything subtracted from it. If one minimizes j cj xj then the combined objective function is Min M i ai + j cj xj © 2008 MGS

The simplex method algorithm requires a starting bfs. The Big M Method The simplex method algorithm requires a starting bfs. Previous problems have found starting bfs by using the slack variables as our basic variables. If an LP have ≥ or = constraints, however, a starting bfs may not be readily apparent. In such a case, the Big M method may be used to solve the problem. Consider the following problem. © 2003 Brooks/Cole

Example Bevco manufactures an orange-flavored soft drink called Oranj by combining orange soda and orange juice. Each orange soda contains 0.5 oz of sugar and 1 mg of vitamin C. Each ounce of orange juice contains 0.25 oz of sugar and 3 mg of vitamin C. It costs Bevco 2¢ to produce an ounce of orange soda and 3¢ to produce an ounce of orange juice. Bevco’s marketing department has decided that each 10-oz bottle of Oranj must contain at least 30 mg of vitamin C and at most 4 oz of sugar. Use linear programming to determine how Bevco can meet the marketing department’s requirements at minimum cost. © 2003 Brooks/Cole

The Big M Method Let x1 = number of ounces of orange soda in a bottle of Oranj x2 = number of ounces of orange juice in a bottle of Oranj x1 x2

The Big M Method Letting x1 = number of ounces of orange soda in a bottle of Oranj x2 = number of ounces of orange juice in a bottle of Oranj The LP is: min z = 2x1 + 3x2 st 0.5x1 + 0.25x2 ≤ 4 (sugar constraint) x1 + 3x2 ≥ 20 (Vitamin C constraint) x1 + x2 = 10 (10 oz in 1 bottle) x1, x2  0 The LP in standard form is shown on the next slide. © 2003 Brooks/Cole

The Big M Method Row 1:-z + 2x1 + 3x2 = 0 Row 2: 0.5x1 + 0.25x2 + s1 = 4 Row 3: x1 + 3x2 - s2 = 20 Row 4: x1 + x2 = 10 The LP in standard form has z and s1 which could be used for BVs but row 2 would violate sign restrictions and row 3 no readily apparent basic variable. In order to use the simplex method, a bfs is needed. To remedy the predicament, artificial variables are created. The variables will be labeled according to the row in which they are used as seen below. The basic variables are in red: Row 1:-z + 2x1 + 3x2 = 0 Row 2: 0.5x1 + 0.25x2 + s1 = 4 Row 3: x1 + 3x2 - s2 + a2 = 20 Row 4: x1 + x2 + a3 = 10 © 2003 Brooks/Cole

The Big M Method If all artificial variables in the optimal solution equal zero, the solution is optimal. If some artificial variables are positive in the optimal solution, the problem is infeasible. The Bevco example continued: Initial Tableau © 2003 Brooks/Cole

4.10 – The Big M Method

The Big M Method