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1. Chapter 3 The Simplex Algorithm

2. Standard Form • Before the simplex algorithm can be used to solve an LP, the LP must be converted into a problem where all the constraints are equations and all variables are nonnegative. An LP in this form is said to be in standard form.

3. How to Convert an LP to Standard Form • A constraint i of an LP is a ≤ constraint: adding a slack variable (si ) to the i th constraint and adding the sign restriction si ≥ 0. A slack variable is the amount of the resource unused in the i th constraint • A constraint i of an LP is a ≥ constraint : subtracting the excess variable (ei ) from the i th constraint and adding the sign restriction ei ≥ 0. We define ei to be the amount by which i th constraint is over satisfied.

4. How to Convert an LP to Standard Form The decision variables are: x1 = number of deluxe belts produced weekly x2 = number of regular belts produced weekly the appropriate LP is: max z = 4x1 + 3x2 s.t. x1 + x2 ≤ 40 (leather constraint) 2x1 +x2 ≤ 60 (labor constraint) x1, x2 ≥ 0 Is it in standard form?

5. How to Convert an LP to Standard Form min z = 50 x1 + 20x2 + 30X2 + 80 x4 s.t. 400x1 + 200x2 + 150 x3 + 500x4 ≥ 500 3x1 + 2x2 ≥ 6 2x1 + 2x2 + 4x3 + 4x4 ≥ 10 2x1 + 4x2 + x3 + 5x4 ≥ 8 x1, x2, x3, x4 ≥ 0 Is it in standard form?

6. How to Convert an LP to Standard Form max z = 20x1 + 15x2 s.t. x1 ≤ 100 x2 ≤ 100 50x1 + 35x2 ≤ 6000 20x1 + 15x2 ≥ 2000 x1, x2 > 0 Is it in standard form?

7. Preview of the Simplex Algorithm Standard form: max ( or min) z = c1x1 + c2x2 + … +cnxn s.t. a11x1 + a12x2 + … + a1nxn =b1 a21x1 + a22x2 + … + a2nxn =b2 . . . . . . am1x1 + am2x2 + … + amnxn =bm xi ≥ 0 ( i = 1,2, …, n)

8. Consider a system Ax = b of m linear equations in n variables (where n ≥ m). A basic solution to Ax = b is obtained by setting n – m variables equal to 0 and solving for the remaining m variables. This assumes that setting the n – m variables equal to 0 yields a unique value for the remaining m variables, or equivalently, the columns for the remaining m variables are linearly independent.

9. Preview of the Simplex Algorithm To find a basic solution to Ax = b • choose a set of n – m variables (the nonbasic variables, or NBV) • set each of these variables equal to 0. • solve for the values of the n – (n – m) = m variables (the basic variables, or BV) that satisfy Ax = b.

10. Preview of the Simplex Algorithm Different choices of nonbasic variables will lead to different basic solutions. Example: x1 + x2 = 10 x1 + x3 = 8 Find 3 basic solutions

11. Preview of the Simplex Algorithm Any basic solution in which all variables are nonnegative is called a basic feasible solution ( or bfs). Theorem 1 The feasible region for any linear programming problem is a convex set. Also, if an LP has an optimal solution, there must be an extreme point of the feasible region that is optimal.

12. Preview of the Simplex Algorithm • Theorem 2 For any LP, there is a unique extreme point of the LP’s feasible region corresponding to each basic feasible solution. Also, there is at least one bfs corresponding to each extreme point in the feasible region.

13. Find bfs corresponding to extreme point max z = 4x1 + 3x2 s.t. x1 + x2 + s1 = 40 2x1 + x2 + s2 = 60 x1, x2, s1, s2 ≥ 0

14. Adjacent Basic Feasible Solutions For any LP with m constraints, two basic feasible solutions are said to be adjacent if their sets of basic variables have m – 1 basic variables in common.(only one basic variable is changed) From previous example, what is adjacent of point C ?

15. The Simplex Algorithm Procedure for maximization LPs Step 1 : Convert the LP to standard form Step 2 : Obtain a bfs (if possible) from the standard form Step 3 : Determine whether the current bfs is optimal Step 4 : If the current bfs is not optimal, determine which nonbasic variable should be come a basic variable and which basic variable should become a nonbasic variable to find a bfs with a better objective function value. Step 5 : Use ero’s to find a new bfs with a better objective function value. Go back to Step 3.

16. Example of simplex algorithm using previous problem max z = 4x1 + 3x2 s.t. x1 + x2 + s1 = 40 2x1 + x2 + s2 = 60 x1, x2, s1, s2 ≥ 0

17. Resource Desk Table Chair Lumber 8 board ft 6 board ft 1 board ft Finishing hours 4 hours 2 hours 1.5 hours Carpentry hours 2 hours 1.5 hours 0.5 hours Example of simplex algorithm

18. Example of simplex algorithm • At present, 48 board feet of lumber, 20 finishing hours, 8 carpentry hours are available. A desk sells for $60, a table for $30, and a chair for $ 20. • Dakota believes that demand for desks and chairs is unlimited, but at most 5 tables can be sold. • Since the available resources have already been purchased, Dakota wants to maximize total revenue.

19. Define: x1 = number of desks produced x2 = number of tables produced x3 = number of chairs produced. The LP is: max z = 60x1 + 30x2 + 20x3 s.t. 8x1 + 6x2 + x3 ≤ 48 (lumber constraint) 4x1 + 2x2 + 1.5x3 ≤ 20 (finishing constraint) 2x1 + 1.5x2 + 0.5x3 ≤ 8 (carpentry constraint) x2 ≤ 5 (table demand constraint) x1, x2, x3 ≥ 0

21. Alternate Optimal solutions For some LPs, more than one extreme point is optimal. If an LP has more than one optimal solution, it has multiple optimal solutions. If there is no nonbasic variable (NBV) with a zero coefficient in row 0 of the optimal tableau, the LP has a unique optimal solution. Even if there is a nonbasic variable with a zero coefficient in row 0 of the optimal tableau, it is possible that the LP may not have alternative optimal solutions.

22. Unbounded LPs For some LPs, there exist points in the feasible region for which z assumes arbitrarily large (in max problems) or arbitrarily small (in min problems) values. When this occurs, we say the LP is unbounded. An unbounded LP occurs in a max problem if there is a nonbasic variable with a negative coefficient in row 0 and there is no constraint that limits how large we can make this NBV. Specifically, an unbounded LP for a max problem occurs when a variable with a negative coefficient in row 0 has a non positive coefficient in each constraint

23. 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 problem below.

24. The Big M Method x1 = number of ounces of orange soda in a bottle of Oranj x2 = number of ounces of orange juice in a bottle of Oranj 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 of Oranj) x1, x2, > 0

25. The Big M Method Row 1: z - 2x1 - 3x2 = 0 Row 2: 0.5x1 + 0.25x2 + s1 = 4 Row 3: x1 + 3x2 - e2 = 20 Row 4: x1 + x2 = 10 In order to use the simplex method, a bfs is needed. To remedy the predicament, artificial variables are created. Row 1: z - 2x1 - 3x2 = 0 Row 2: 0.5x1 + 0.25x2 + s1 = 4 Row 3: x1 + 3x2 - e2 + a2 = 20 Row 4: x1 + x2 + a3 = 10

26. The Big M Method In the optimal solution, all artificial variables must be set equal to zero. In order to force artificial to be zero, we add –Mai to the objective function for each ai. M represents some very large number. The modified Bevco LP in standard form then becomes:

27. The Big M Method Row 1: z - 2x1 - 3x2 -Ma2 - Ma3 = 0 Row 2: 0.5x1 + 0.25x2 + s1 = 4 Row 3: x1 + 3x2 - e2 + a2 = 20 Row 4: x1 + x2 + a3 = 10 Modifying the objective function this way makes it extremely costly for an artificial variable to be positive. The optimal solution should force a2 = a3 =0.

28. Description of the Big M Method • Modify the constraints so that the rhs of each constraint is nonnegative. Identify each constraint that is now an = or ≥ const. • Convert each inequality constraint to standard form (add a slack variable for ≤ constraints, add an excess variable for ≥ consts). • For each ≥ or = constraint, add artificial variables. Add sign restriction ai≥ 0. • Let M denote a very large positive number. Add (for each artificial variable) Mai to min problem objective functions or -Mai to max problem objective functions. • Since each artificial variable will be in the starting basis, all artificial variables must be eliminated from row 0 before beginning the simplex. Remembering M represents a very large number, solve the transformed problem by the simplex.

29. The Big M Method If all artificial variables in the optimal solution equal zero, the solution is optimal. If any artificial variables are positive in the optimal solution, the problem is infeasible. The Bevco example continued: Initial Tableau

30. The Big M Method

31. The Big M Method