Topic IIIThe Simplex Method

Setting up the MethodTabular 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

• Examplex1 ≤ 4

• Adding slack variable gives x1 + x3 = 4–Note that these are equivalent iff x3 ≥ 0

Setting up the Simplex Method

• Original Model (from Topic II)– Maximizing Total profit, Z

Maximize Z = 3x1 + 5x2

– Constraintsx1 ≤ 42x2 ≤ 123x1 + 2x2 ≤ 18

– Other constraintsx1 ≥ 0x2 ≥ 0

Setting up the Simplex Method

• Augmented form of the model– Maximizing Total profit, Z

Maximize Z = 3x1 + 5x2

– Constraintsx1 + x3 = 42x2 + x4 = 123x1 + 2x2 + x5 = 18

– Other constraintsxj ≥ 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 x1 and x2 to be the nonbasic variables (the

variables set to zero)• Using system of equations, x3, x4, x5 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 = 3x1 + 5x2

• Z = 0 for the initial BF solution

– Rate of improvement for x2 is more than x1 (5 > 3)• Increase x2

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 equationsx1 + x3 = 4

No upper bound on increasing x2

2x2 + x4 = 12x4 = 12 – 2x2

Thus, x2 ≤ 6

3x1 + 2x2 + x5 = 18x5 = 18 – 2x2

Thus, x2 ≤ 9

• Since the 2nd equation restricts x2 to 6, x4 is the leaving basic variable for this iteration

Solve for New Solution

• Step 3: Solving for the new BF Solution– Original System

Z – 3x1 – 5x2 = 0x1 + x3 = 42x2 + x4 = 123x1 + 2x2 + x5 = 18

– x2 has replaced x4 as the basic variable•The pattern of coefficients of x4 (0, 0, 1, 0) need to become the coefficients of x2

Solve for New Solution

• Step 3: Solving for the new BF Solution– How

• Divide constraint equation 2 by 2x2 + ½x4 = 6

• Add 5 times this new equation to the objective functionZ – 3x1 + 5/2 x4 = 30

• Subtract 2 times new equation to constraint equation 33x1 – x4 + x5 = 6

Solve for New Solution

• Step 3: Solving for the new BF Solution– New System

Z – 3x1 + 5/2 x4 = 30x1 + x3 = 4x2 + ½x4 = 63x1 – x4 + x5 = 6

– New BF Solution(0, 6, 4, 0, 6)

Next Iteration

• Next Iteration: Return to Step 2Z = 30 + 3x1 – 5/2 x4

• Z can be increased by increasing x1, but not x4

• Thus, x1 needs to be the next entering basic variable

–Minimum Ratio Testx1 + x3 = 4

x1 ≤ 4

x2 + ½x4 = 6No upper bound on x1

3x1 – x4 + x5 = 6x1 ≤ 2

–x5 is the leaving basic variable

Next Iteration

• The pattern of coefficients of x5 (0, 0, 0, 1) needs to become the pattern for x1

– Divide constraint equation 3 by 3x1 – 1/3 x4 + 1/3 x5 = 2

– Add 3 times this equation to objective functionZ + 3/2 x4 + x5 = 36

– Subtract new equation from constraint equation 1x3 + 1/3 x4 – 1/3 x5 = 2

Next Iteration

• The pattern of coefficients of x5 (0, 0, 0, 1) needs to become the pattern for x1

– New SystemZ + 3/2 x4 + x5 = 36x3 + 1/3 x4 – 1/3 x5 = 2x2 + ½x4 = 6x1 – 1/3 x4 + 1/3 x5 = 2

– New BF Solution(2, 6, 2, 0, 0)

Final Iteration

• Next iterationZ = 36 – 3/2 x4 – x5

– If either nonbasic variable x4 or x5 is increased, Z would decrease•Thus, the current BF solution is optimal

– Original variables: x1 and x2

x1 = 2x2 = 6

– Maximum value of Z: 36

Tabular Form

Basic Var

Eq Z x1 x2 x3 x4 x5 Right Side

Z 0 1 -3 -5 0 0 0 0

x3 1 0 1 0 1 0 0 4

x4 2 0 0 2 0 1 0 12

x5 3 0 3 2 0 0 1 18

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