Well, just how many basic solutions are there?!!!

Basic Solutions Let n = the number of variables m = the number of constraints After adding slack variables, there are n + m variables. Since a basic solution requires setting n variables to zero and solving uniquely for the remaining variables, there are ways to select the n variables to set equal to zero or equivalently the m basic variables. If n = 20 and m = 10 (a trivially small problem), then 30!/(20! 10!) = 30,045,015. (not all are feasible)

Basic Solutions Again! Augmented solution - solution to the problem after slack variables have been added. Basic solution - augmented corner point solution obtained by setting n variables to zero and solving for the remaining. Basic feasible solution (BFS) - feasible basic solution. Basic variables - variables that were solved for in the basic solution. Non-basic variables - variables set equal to zero in the basic solution.

Outline of the Simplex Algorithm Start at a basic feasible solution (BFS) (often the origin) Move to a better basic feasible solution the objective function improves Stop when the basic feasible solution is better than all adjacent basic feasible solutions. Solution is optimal I can do this!

Now - Presenting the Simplex Algorithm! Be the first in your neighborhood to master this exciting and wonderful numerical, recursive process for solving linear programs. Two students discussing the high points of the simplex algorithm.

The Simplex Tableau-1 Basic Eq Var z x1 x2 x3 x4 x5 RHS Max z - 6x1 - 4x2 = 0 Subj. to: x1 + x2 + x3 = 12 x1 - 2x2 + x4 = 6 x2 + x5 = 8 The Simplex Tableau-1 Basic Eq Var z x1 x2 x3 x4 x5 RHS

The Simplex Tableau-2 Basic Eq Var z x1 x2 x3 x4 x5 RHS Max z - 6x1 - 4x2 = 0 Subj. to: x1 + x2 + x3 = 12 x1 - 2x2 + x4 = 6 x2 + x5 = 8 The Simplex Tableau-2 Basic Eq Var z x1 x2 x3 x4 x5 RHS 0 z 1 -6 -4 0 0 0 0

The Simplex Tableau-3 Basic Eq Var z x1 x2 x3 x4 x5 RHS Max z - 6x1 - 4x2 = 0 Subj. to: x1 + x2 + x3 = 12 x1 - 2x2 + x4 = 6 x2 + x5 = 8 The Simplex Tableau-3 Basic Eq Var z x1 x2 x3 x4 x5 RHS 0 z 1 -6 -4 0 0 0 0 1 x3 0 1 1 1 0 0 12

The Simplex Tableau-4 Basic Eq Var z x1 x2 x3 x4 x5 RHS Max z - 6x1 - 4x2 = 0 Subj. to: x1 + x2 + x3 = 12 x1 - 2x2 + x4 = 6 x2 + x5 = 8 The Simplex Tableau-4 Basic Eq Var z x1 x2 x3 x4 x5 RHS 0 z 1 -6 -4 0 0 0 0 1 x3 0 1 1 1 0 0 12 2 x4 0 1 -2 0 1 0 6

The Simplex Tableau-5 Basic Eq Var z x1 x2 x3 x4 x5 RHS Max z - 6x1 - 4x2 = 0 Subj. to: x1 + x2 + x3 = 12 x1 - 2x2 + x4 = 6 x2 + x5 = 8 The Simplex Tableau-5 Basic Eq Var z x1 x2 x3 x4 x5 RHS 0 z 1 -6 -4 0 0 0 0 1 x3 0 1 1 1 0 0 12 2 x4 0 1 -2 0 1 0 6 3 x5 0 0 1 0 0 1 8

The Simplex Algorithm-1 Step 1: Select a new variable to enter the basis. Basic Eq Var z x1 x2 x3 x4 x5 RHS 0 z 1 -6 -4 0 0 0 0 1 x3 0 1 1 1 0 0 12 2 x4 0 1 -2 0 1 0 6 3 x5 0 0 1 0 0 1 8

The Simplex Algorithm-2 z = 6x1 + 4x2 Step 1: Select a new variable to enter the basis. Pick the non-basic variable having the greatest negative value. Basic Eq Var z x1 x2 x3 x4 x5 RHS 0 z 1 -6 -4 0 0 0 0 1 x3 0 1 1 1 0 0 12 2 x4 0 1 -2 0 1 0 6 3 x5 0 0 1 0 0 1 8

Basic Eq Var z x1 x2 x3 x4 x5 RHS 0 z 1 -6 -4 0 0 0 0 Step 2a: Select a basic variable to leave the basis. Basic Eq Var z x1 x2 x3 x4 x5 RHS 0 z 1 -6 -4 0 0 0 0 1 x3 0 1 1 1 0 0 12 2 x4 0 1 -2 0 1 0 6 3 x5 0 0 1 0 0 1 8

Basic Eq Var z x1 x2 x3 x4 x5 RHS Ratio 0 z 1 -6 -4 0 0 0 0 Step 2b: Select a basic variable to leave the basis. Pick the basic variable having the smallest ratio of the RHS divided by the corresponding positive coefficient from the incoming variable. Ratio 12/1 6/1 Basic Eq Var z x1 x2 x3 x4 x5 RHS 0 z 1 -6 -4 0 0 0 0 1 x3 0 1 1 1 0 0 12 2 x4 0 1 -2 0 1 0 6 3 x5 0 0 1 0 0 1 8

Basic Eq Var z x1 x2 x3 x4 x5 RHS Ratio 0 z 1 -6 -4 0 0 0 0 Step 2c: Select a basic variable to leave the basis. Pick the basic variable having the smallest ratio of the RHS divided by the corresponding positive coefficient from the incoming variable. Ratio 12/1 6/1 1x1 - 2x2 + x4 = 6 Basic Eq Var z x1 x2 x3 x4 x5 RHS 0 z 1 -6 -4 0 0 0 0 1 x3 0 1 1 1 0 0 12 2 x4 0 1 -2 0 1 0 6 3 x5 0 0 1 0 0 1 8 pivot point

Step 3a: Use row operations to find the new basic solution. Eq Var z x1 x2 x3 x4 x5 RHS 0 z 1 -6 -4 0 0 0 0 1 x3 0 1 1 1 0 0 12 2 x4 0 1 -2 0 1 0 6 3 x5 0 0 1 0 0 1 8 Basic Eq Var z x1 x2 x3 x4 x5 RHS 0 z 1 x3 2 x1 3 x5

Step 3b: Use row operations to find the new basic solution. Eq Var z x1 x2 x3 x4 x5 RHS 0 z 1 -6 -4 0 0 0 0 1 x3 0 1 1 1 0 0 12 2 x4 0 1 -2 0 1 0 6 3 x5 0 0 1 0 0 1 8 Basic Eq Var z x1 x2 x3 x4 x5 RHS 0 z 1 x3 2 x1 3 x5 0 1 -2 0 1 0 6

Step 3c: Use row operations to find the new basic solution. Eq Var z x1 x2 x3 x4 x5 RHS 0 z 1 -6 -4 0 0 0 0 1 x3 0 1 1 1 0 0 12 2 x4 0 1 -2 0 1 0 6 3 x5 0 0 1 0 0 1 8 Basic Eq Var z x1 x2 x3 x4 x5 RHS 0 z 1 x3 2 x1 3 x5 0 1 -2 0 1 0 6 0 0 1 0 0 1 8

Step 3d: Use row operations to find the new basic solution. Eq Var z x1 x2 x3 x4 x5 RHS 0 z 1 -6 -4 0 0 0 0 1 x3 0 1 1 1 0 0 12 2 x4 0 1 -2 0 1 0 6 3 x5 0 0 1 0 0 1 8 Basic Eq Var z x1 x2 x3 x4 x5 RHS 0 z 1 x3 2 x1 3 x5 0 0 3 1 -1 0 6 0 1 -2 0 1 0 6 0 0 1 0 0 1 8

Step 3e: Use row operations to find the new basic solution. Eq Var z x1 x2 x3 x4 x5 RHS 0 z 1 -6 -4 0 0 0 0 1 x3 0 1 1 1 0 0 12 2 x4 0 1 -2 0 1 0 6 3 x5 0 0 1 0 0 1 8 Basic Eq Var z x1 x2 x3 x4 x5 RHS 0 z 1 x3 2 x1 3 x5 1 0 -16 0 6 0 36 0 0 3 1 -1 0 6 0 1 -2 0 1 0 6 0 0 1 0 0 1 8

Max z = 6x1 + 4x2 x1 + x2 <= 12 x1 -2x2 <= 6 x2 <= 8 x2 x1 Subj. to: x1 + x2 <= 12 x1 -2x2 <= 6 x2 <= 8 x2 12 8 (4,8) z z (10,2) x1 6 12 -3

Now Sally, you want to pick the new variable to enter the basis. 2nd Iteration-1 z = 6x1 + 4x2 Now Sally, you want to pick the new variable to enter the basis. Basic Eq Var z x1 x2 x3 x4 x5 RHS 0 z 1 0 -16 0 6 0 36 1 x3 0 0 3 1 -1 0 6 2 x1 0 1 -2 0 1 0 6 3 x5 0 0 1 0 0 1 8

2nd Iteration-2 Basic Eq Var z x1 x2 x3 x4 x5 RHS 0 z 1 0 -16 0 6 0 36 Pick the non-basic variable having the greatest negative value. Basic Eq Var z x1 x2 x3 x4 x5 RHS 0 z 1 0 -16 0 6 0 36 1 x3 0 0 3 1 -1 0 6 2 x1 0 1 -2 0 1 0 6 3 x5 0 0 1 0 0 1 8

2nd Iteration-3 Basic Eq Var z x1 x2 x3 x4 x5 RHS Ratio 6/3 8/1 Find minimum ratio Basic Eq Var z x1 x2 x3 x4 x5 RHS 0 z 1 0 -16 0 6 0 36 1 x3 0 0 3 1 -1 0 6 2 x1 0 1 -2 0 1 0 6 3 x5 0 0 1 0 0 1 8

2nd Iteration-4 Basic Eq Var z x1 x2 x3 x4 x5 RHS Ratio 6/3 8/1 Find minimum ratio Basic Eq Var z x1 x2 x3 x4 x5 RHS 0 z 1 0 -16 0 6 0 36 1 x3 0 0 3 1 -1 0 6 2 x1 0 1 -2 0 1 0 6 3 x5 0 0 1 0 0 1 8 Pivot point

Basic Basic 2nd Iteration-5 Eq Var z x1 x2 x3 x4 x5 RHS

Basic Basic 2nd Iteration-6 Eq Var z x1 x2 x3 x4 x5 RHS 0 0 1 1/3 -1/3 0 2

Basic Basic 2nd Iteration-7 Eq Var z x1 x2 x3 x4 x5 RHS 0 0 1 1/3 -1/3 0 2 0 1 0 2/3 1/3 0 10

Basic Basic 2nd Iteration-8 Eq Var z x1 x2 x3 x4 x5 RHS 0 0 1 1/3 -1/3 0 2 0 1 0 2/3 1/3 0 10 0 0 0 -1/3 1/3 1 6

Basic Basic 2nd Iteration-9 Eq Var z x1 x2 x3 x4 x5 RHS 1 0 0 16/3 2/3 0 68 0 0 1 1/3 -1/3 0 2 0 1 0 2/3 1/3 0 10 0 0 0 -1/3 1/3 1 6

Basic 2nd Iteration-10 Clearly, this solution must be optimal. What does it mean to be optimal? Basic Eq Var z x1 x2 x3 x4 x5 RHS 0 z 1 x2 2 x1 3 x5 1 0 0 16/3 2/3 0 68 0 0 1 1/3 -1/3 0 2 0 1 0 2/3 1/3 0 10 0 0 0 -1/3 1/3 1 6

Max z = 6x1 + 4x2 x1 + x2 <= 12 x1 -2x2 <= 6 x2 <= 8 x2 x1 Subj. to: x1 + x2 <= 12 x1 -2x2 <= 6 x2 <= 8 x2 12 8 (4,8) z z (10,2) x1 6 12 -3

After the Simplex What can go wrong? Minimization rather than maximization Unbounded solutions Alternate solutions No feasible solution

What’s Next? It just doesn’t get any better than this! Wanna bet! Wait until you see sensitivity analysis.

Can we work another of these simplex problems next class Can we work another of these simplex problems next class? I think I understand how to find the most negative value and the smallest ratio, but I get dizzy when I pivot.