Topic III The Simplex Method Setting up the Method Tabular Form Chapter(s): 4

Key Concepts The simplex method focuses solely on CPF solutions –For any problem with at least one optimal solution, finding one only requires finding a best CPF solution The simplex method is an iterative algorithm –Initialization –Optimality Test If no, perform an iteration to find a better solution If yes, stop Whenever possible, the initialization of the simplex method chooses the origin to be initial CPF solution

Key Concepts Given a CPF solution, it is much quicker computationally to gather information about its adjacent CPF solutions that about other solutions After the current CPF solution is identified, the method identifies the rate of improvement in Z that would be obtained by moving along an edge to an adjacent solution –Chooses to move along the one with the largest rate of improvement in Z If none of the edges give a positive rate of improvement, then the current CPF solution is optimal

Setting up the Simplex Method Convert the functional inequality constraints to equivalent equality constraints –Accomplished by introducing slack variables –An augmented solution is a solution for the original (decision) variables that has been augmented by the corresponding values of the slack variables Example x 1 ≤ 4 Adding slack variable gives x 1 + x 3 = 4 –Note that these are equivalent iff x 3 ≥ 0

Setting up the Simplex Method Original Model (from Topic II) –Maximizing Total profit, Z Maximize Z = 3x 1 + 5x 2 –Constraints x 1 ≤ 4 2x 2 ≤ 12 3x 1 + 2x 2 ≤ 18 –Other constraints x 1 ≥ 0 x 2 ≥ 0

Setting up the Simplex Method Augmented form of the model –Maximizing Total profit, Z Maximize Z = 3x 1 + 5x 2 –Constraints x 1 + x 3 = 4 2x 2 + x 4 = 12 3x 1 + 2x 2 + x 5 = 18 –Other constraints x j ≥ 0, for j = 1, 2, 3, 4, 5

Setting up the Simplex Method The system of functional constraints has 5 variables and 3 equations –Number of variables – number of equations = 5 – 3 = 2 2 Degrees of freedom in solving the system (as long as there aren’t any redundant equations) –Set any two variables to an arbitrary value to solve the three equation system The simplex method uses zero for this arbitrary value –The two variables set to zero are the nonbasic variables –The other three variables are the basic variables

Basic Solution A basic solution is an augmented corner-point solution –Properties of a basic solution Each variable is designated as either a nonbasic variable or a basic variable The number of basic variables equals the number of functional constraints The number of nonbasic variables equals the total number of variables minus the number of functional constraints

Basic Solution A basic solution is an augmented corner-point solution –Properties of a basic solution The nonbasic variables are set to zero The values of the basic variables are obtained as the simultaneous solution of the system of equations (functional constrains in augmented form) –The set is often referred to as the basis If the basic variables satisfy the nonnegativity constraints, the basic solution is a BF solution –A basic feasible (BF) solution is an augmented CPF solution

Basic Feasible (BF) Solutions Two BF solutions are adjacent if all but one of their nonbasic variables are the same –Note that all but one of their basic variables are also the same –Moving from the current BF solution to an adjacent one involves switching one variable from nonbasic to basic (and vice versa for one other variable) Adjust the values of the basic variables to satisfy the system of equations

The Simplex Method Step 1: Initialization –Choose x 1 and x 2 to be the nonbasic variables (the variables set to zero) Using system of equations, x 3, x 4, x 5 equal 4, 12, 18 Thus, the initial BF solution is (0, 0, 4, 12, 18)

The Simplex Method Step 2: Optimality Test –The objective function is Z = 3x 1 + 5x 2 Z = 0 for the initial BF solution –Rate of improvement for x 2 is more than x 1 (5 > 3) Increase x 2

Minimum Ratio Test Step 2: Optimality Test –Minimum Ratio Test Objective is to determine which basic variable drops to zero first as the entering basic variable is increased The system of equations x 1 + x 3 = 4 No upper bound on increasing x 2 2x 2 + x 4 = 12 x 4 = 12 – 2x 2 Thus, x 2 ≤ 6 3x 1 + 2x 2 + x 5 = 18 x 5 = 18 – 2x 2 Thus, x 2 ≤ 9 Since the 2 nd equation restricts x 2 to 6, x 4 is the leaving basic variable for this iteration

Solve for New Solution Step 3: Solving for the new BF Solution –Original System Z – 3x 1 – 5x 2 = 0 x 1 + x 3 = 4 2x 2 + x 4 = 12 3x 1 + 2x 2 + x 5 = 18 –x 2 has replaced x 4 as the basic variable The pattern of coefficients of x 4 (0, 0, 1, 0) need to become the coefficients of x 2

Solve for New Solution Step 3: Solving for the new BF Solution –How Divide constraint equation 2 by 2 x 2 + ½x 4 = 6 Add 5 times this new equation to the objective function Z – 3x 1 + 5/2 x 4 = 30 Subtract 2 times new equation to constraint equation 3 3x 1 – x 4 + x 5 = 6

Solve for New Solution Step 3: Solving for the new BF Solution –New System Z – 3x 1 + 5/2 x 4 = 30 x 1 + x 3 = 4 x 2 + ½x 4 = 6 3x 1 – x 4 + x 5 = 6 –New BF Solution (0, 6, 4, 0, 6)

Next Iteration Next Iteration: Return to Step 2 Z = x 1 – 5/2 x 4 Z can be increased by increasing x 1, but not x 4 Thus, x 1 needs to be the next entering basic variable –Minimum Ratio Test x 1 + x 3 = 4 x 1 ≤ 4 x 2 + ½x 4 = 6 No upper bound on x 1 3x 1 – x 4 + x 5 = 6 x 1 ≤ 2 –x 5 is the leaving basic variable

Next Iteration The pattern of coefficients of x 5 (0, 0, 0, 1) needs to become the pattern for x 1 –Divide constraint equation 3 by 3 x 1 – 1/3 x 4 + 1/3 x 5 = 2 –Add 3 times this equation to objective function Z + 3/2 x 4 + x 5 = 36 –Subtract new equation from constraint equation 1 x 3 + 1/3 x 4 – 1/3 x 5 = 2

Next Iteration The pattern of coefficients of x 5 (0, 0, 0, 1) needs to become the pattern for x 1 –New System Z + 3/2 x 4 + x 5 = 36 x 3 + 1/3 x 4 – 1/3 x 5 = 2 x 2 + ½x 4 = 6 x 1 – 1/3 x 4 + 1/3 x 5 = 2 –New BF Solution (2, 6, 2, 0, 0)

Final Iteration Next iteration Z = 36 – 3/2 x 4 – x 5 –If either nonbasic variable x 4 or x 5 is increased, Z would decrease Thus, the current BF solution is optimal –Original variables: x 1 and x 2 x 1 = 2 x 2 = 6 –Maximum value of Z: 36

Tabular Form Basic Var EqZx1x1 x2x2 x3x3 x4x4 x5x5 Right Side Z x3x x4x x5x Start with initial equations

Tabular Form Iterations –Determine entering basic variable Select variable with negative coefficient with largest absolute value –If none, the algorithm is finished Draw box around column below this variable as the pivot column

Tabular Form Iterations –Minimum ratio test Select each coefficient in pivot column that is positive Divide each coefficient into corresponding right side entry Identify smallest ratio Basic variable for that row is leaving basic variable –Replace it by entering basic variable column of table –Box the row and call it the pivot row –The number in both pivot row and pivot column is pivot number

Tabular Form Iterations –New BF solution Divide pivot row by pivot number (use this total in next two steps) For each other row (including row 0) that has a negative coefficient in the pivot column –Add to this row the product of absolute value of this coefficient and new pivot row For each other row that has a positive coefficient in the pivot column –Subtract from this the product of its coefficient and the new pivot row