Overall Idea of Simplex Method

Linear Programming

(LP)

The Simplex Method

Overall Idea of Simplex Method

 Simplex Method translates the geometric

definition of the extreme point into an algebraic definition

 Initial Step: all the constraints are put in a

standard form

 In standard form: all the constraints are

expressed as equations by augmenting

Overall Idea

 The conversion of inequality to equation

normally results in a set of simultaneous

equations in which the number of variables exceeds the number of equations

 The Equation might yield an infinite

number of solution points

Overall Idea

 Linear Algebra Theory:

A basic solution is obtained

 by setting to zero as many variable as difference

between the total number of variables and the total number of equation

 solving for the remaining variable

Standard LP Form

 To develop a general solution method, LP

Development of the Simplex

Method

 Two types of Simplex Method

Primal Simplex Method

_{Dual Simplex Method}

 Basically, the difference between the two

Standard LP Form

 The Properties of Standard LP Form

All constraints are equations (Primal Simplex Method requires a non-negative right-hand side)

_{All the variables are non-negative}

The Objective Function may be maximization

Standard LP Form:

Constraints

 A constraint of type () is converted to

equation by adding a slack variable to the

left side of the constraint

 For example:

 _{x1 + 2x2}  ₆

Standard LP Form:

Constraints

 A constraint of type () is converted to

equation by subtracting a surplus variable form the left side of the constraint

 For example:

 _{3x1 + 2x2 – 3x3}  ₅

Standard LP Form

Constraints

 The right side of an equation can always

be made non-negative by multiplying both sides by –1

 For example:

 2x₁ + 3x₂ – 7x₃ = –5

Standard LP Form

Constraints

 The direction of an inequality is reversed

when both sides are multiplied by –1

 For example:

Standard LP Form

Variables

 Unrestricted variable x_i can be expressed

in terms of two non-negative variables

 x_i = x_i’ – x_i”, where x_i’, x_i”  0  x_i’ > 0, x_i” = 0, and vice versa

Standard LP Form

Objective Function

 Maximization or Minimization

 Maximization = Minimization of the

negative of the function

 For example:

Standard LP Form

Example

 Write the following LP Model in standard

form:

 Minimize z = 2x₁ + 3X₂

 Subject to (with constraints):

 _{x1 + x2 = 10}

Basic Solution

 For m: number of equations

 _{And n: number of unknowns (variables)}  Basic solution is determined

 by setting n – m (number of) variables equal to zero

 The n – m variables are called the non-basic variables

 solving the m remaining variables

 The m remaining variables are called the basic variable

Basic Solution

Example

 2x₁ + x₂ + 4x₃ + x₄ = 2  x₁ + 2x₂ + 2x₃ + x₄ = 3

 _{m = 4}

 n = 2

 _{Basic solution is associated with m – n} (= 4 – 2 = 2) zero variables

Basic Solution

Example

 2x₁ + x₂ + 4x₃ + x₄ = 2  x₁ + 2x₂ + 2x₃ + x₄ = 3

 Set x₂ = 0 and x₄ = 0

 2x₁ + 4x₃ = 2

Basic Solution

Example

 2x₁ + x₂ + 4x₃ + x₄ = 2  x₁ + 2x₂ + 2x₃ + x₄ = 3

 _{Set x3 = 0 and x4 = 0 (non-basic variables)}  x₁ and x₂ are basic variables

 _{2x1 + x2 = 2}  x₁ + 2x₂ = 3  x₁ = 1/3

 x₂ = 4/3

 _{Basic feasible solution:}

Basic Solution

Example

 _{2x1 + x2 + 4x3 + x4 = 2}

 x₁ + 2x₂ + 2x₃ + x₄ = 3

 Set x₁ = 0 and x₂ = 0

 x₃ and x₄

 4x₃ + x₄ = 2

 _{2x3 + x4 = 3}

 x₃ = – 1/2

Basic Feasible Solution

 A basic solution is said to be feasible if all

its solution values are non-negative

Primal and Dual Simplex

 All iterations in Primal Simplex Method

are always associated to feasible basic

solutions only

 Primal Simplex Method deals with feasible

Primal and Dual Simplex

 Iterations in Dual Simplex Method end

only if the last iteration is infeasible

 Both methods yield feasible basic solution

Primal Simplex Method

The Reddy Mikks Company

 _{XE: tons produced daily of exterior paint}

 X_I: tons produced daily of interior paint

 Objective Function to be satisfy:

 Maximize z = 3X_E + 2X_I

 Subject to these constraints:

 X_E + 2X_I  6

 2X_E + X_I  8

 – X_E + X_I  1

 X_I  2

Primal Simplex Method

The Reddy Mikks Company

 Set the constraints to equations

 x_E + 2x_I  6  x_E + 2x_I + s_I = 6  2x_E + x_I  8  2x_E + x_I + s₂ = 8  – x_E + x_I  1  – x_E + x_I + s₃ = 1  x_I  2  x_I + s₄= 2

 x_I, x_E, s_I, s₂, s₃, s₄  0

 m = 4

Primal Simplex Method

The Reddy Mikks Company

 Basic Solution

 m = 4

 n = 2 (x_E and x_I)

 If x_E and x_I = 0 then

_{xE + 2xI + sI = 6}  _{sI = 6}

2x_E + x_I + s₂ = 8  s₂ = 8

_{– xE + xI + s3 = 1}  _{s3 = 1}

_{xI + s4= 2}  _{s4= 2}

Primal Simplex Method

The Reddy Mikks Company

 Convert the Objective Function

 Maximize z = 3x_E + 2x_I

 z – 3x_E – 2x_I = 0

 Include the slack variables

 Maximize

Primal Simplex Method

The Reddy Mikks Company

 Maximize

z – 3x_E – 2x_I + 0s_I + 0s₂ + 0s₃ + 0s₄ = 0

 _{Later (in the next sub-chapters), the slack and} the substitute variables are written as common variables

 _Maximize

Primal Simplex Method

The Reddy Mikks Company

 Incorporate the objective function with the

basic feasible solution

 x_E and x_I : entering variables

 s_I, s₂, s₃ ,and s₄ : leaving variables

 Start the iterations with the values of s_I, s₂,

s₃ ,and s₄ = zero

 Final Result of the iterations the values of

Primal Simplex Method

Easiness (a commentary)

 Each equation has a slack variable

 The right hand of all constraints are

Entering Variables in

Maximization and Minimization

Optimality Condition

 The entering variable in Maximization is

the non-basic variable with the most

negative coefficient in the z-equation

 The optimum of Maximization is reached

when all the non-basic coefficients are

non-negative (or zero)

Entering Variables in

Maximization and Minimization

Optimality Condition

 The entering variable in Minimization is the

non-basic variable with the most positive

coefficient in the z-equation

 The optimum of Minimization is reached

when all the non-basic coefficients are

non-positive (Zero)

Leaving Variables in

Maximization and Minimization

Feasibility Condition

 For both Maximization and Minimization,

the leaving variable is the current basic

variable having the smallest intercept

(minimum ratio with strictly positive

denominator) in the direction of the entering variable

Primal Simplex Method:

The Formal Iterative Steps

 Step 0: Using the standard form with all

non-negative right hand sides, determine a starting feasible solution

 Step 1: Select an entering variable from

Primal Simplex Method:

The Formal Iterative Steps

 Step 2: Select the leaving variable from

the current basic variables using the feasibility condition

 Step 3: Determine the value of the new

basic variables by making the entering and the leaving variable non basic

 Stop if the optimum solution is achieved;

Primal Simplex Method

The Reddy Mikks Company

 Incorporate the objective function with the

basic feasible solution

 x_E and x_I : entering variables

 s_I, s₂, s₃ ,and s₄ : leaving variables

 Start the iterations with the values of s_I, s₂,

s₃ ,and s₄ = zero

 Final Result of the iterations the values of

Primal Simplex Method

The Reddy Mikks Company

 Maximize

z – 3X_E – 2X_I + 0s_I + 0s₂ + 0s₃ + 0s₄ = 0

 x_E + 2x_I + s_I = 6  Raw Material A

 2x_E + x_I + s₂ = 8  Raw Material B

 – x_E + x_I + s₃ = 1  Demand x₁ #1

 x_I + s₄= 2 Demand x₁ #2

 Put all the equation in the Table (Tableau)

Primal Simplex Method:

The Tableau

2 1 0 0 0 1 0 0 s₄ 1 0 1 0 0 1 -1 0 s₃ 8 0 0 1 0 1 2 0 s₂ 6 0 0 0 1 2 1 0 s₁ 0 0 0 0 0 -2 -3 1 z Inter-cept solution s₄ s₃ s₂ s₁ x_I x_E z Basic

Gauss – Jordan

1. Pivot Equation

New Pivot Equation (NPE)= Old Pivot Equation : Pivot Element

2. All other equations, including z

New Equation = (Old Equation) – (its

s₄ s₃

4 0

0 1/2

0 1/2

1 x_E

s₁ z

solution s₄

s₃ s₂

s₁ x_I

x_E Basic

Pivot Element = 2

4 0 0 1/2 0 1/2 1 NPE 0 0 0 0 0 -2 -3 Z (OLD)

Coefficient x_E for z = -3

-12 0 0 -3/2 0 -3/2 -3 -3NPE

-12 0 0 -3/2 0 -3/2 -3 3NPE

s₄ s₃

4 0

0 1/2

0 1/2

1 x_E

s₁

12 0

0 3/2

0 -1/2

0 z

solution s₄

s₃ s₂

s₁ x_I

4 0 0 1/2 0 1/2 1 NPE

Coefficient x_E for s₁ = 1

4 0 0 1/2 0 1/2 1 1NPE

s₁ 6 0 0 0 1 2 1 s₁ (OLD)

4 0 0 1/2 0 1/2 1 1NPE

4 0 0 1/2 0 1/2 1 NPE

Coefficient x_E for S₃ = -1

-4 0 0 -1/2 0 -1/2 -1 -1NPE

s₃ 5 0 1 1/2 0 3/2 0 s₃ (NEW) 1 0 1 0 0 1 -1 s₃(OLD)

4 0 0 1/2 0 1/2 1 NPE

Coefficient x_E for S₄ = 0

0 0 0 0 0 0 0 0NPE

2 5 4 2 12 solution 0 1 0 0 1 0 s₄ 0 1 1/2 0 3/2 0 s₃ 0 0 1/2 0 1/2 1 x_E 0 0 -1/2 1 3/2 0 s₁ 0 0 3/2 0 -1/2 0 z s₄ s₃ s₂ s₁ x_I x_E Basic

s₄ s₃ x_E

4/3 0

0 -1/3

2/3 1

0 x_i

z

solution s₄

s₃ s₂

s₁ x_I

x_E Basic

Pivot Element = 3/2

4/3 0 0 -1/3 2/3 1 0 NPE 12 0 0 3/2 0 -1/2 0 Z (OLD)

Coefficient x_I for z = -1/2

-2/3 0 0 1/6 -1/3 -1/2 0 -1/2NPE

s₄ s₃ x_E

4/3 0

0 -1/3

2/3 1

0 x_i

12 2_/ 3 0

0 4/3

1/3 0

0 z

solution s₄

s₃ s₂

s₁ x_I

4/3 0 0 -1/3 2/3 1 0 NPE 4 0 0 1/2 0 1/2 1 x_E (OLD)

Coefficient x_i for x₃ = 1/2

-2/3 0 0 -1/6 1/3 1/2 0 1/2NPE

4/3 0 0 -1/3 2/3 1 0 NPE 5 0 1 1/2 0 3/2 0 s₃ (OLD)

Coefficient x_i for s₃ = 3/2

2 0 0 -1/2 1 3/2 0 3/2NPE

s₄ 3 0 1 1 -1 0 0 s₃ 10/3 0 0 2/3 -1/3 0 1 x_E 4/3 0 0 -1/3 2/3 1 0 x_i

4/3 0 0 -1/3 2/3 1 0 NPE 2 0 1 0 0 1 0 s₄ (OLD)

Coefficient x_i for s₄ = 1

4/3 0 0 -1/3 2/3 1 0 1NPE

2/3 0 1 1/3 -2/3 0 0 s₄ 3 0 1 1 -1 0 0 s₃ 10/3 0 0 2/3 -1/3 0 1 x_E 4/3 0 0 -1/3 2/3 1 0 x_i

12 2_/ 3 0 0 4/3 1/3 0 0 z solution s₄ s₃ s₂ s₁ x_I x_E Basic

Interpretation of the Simplex

Tableau

 Information from the tableau

 Optimum Solution

 Status of Resources

 Dual Prices (Unit worth of resources) and Reduced Cost

 Sensitivity of the optimum solution to the changes of in

 availability of resources

Interpretation of the Simplex

Tableau: Optimum Solution

2/3 0 1 1/3 -2/3 0 0 s₄ 3 0 1 1 -1 0 0 s₃ 10/3 0 0 2/3 -1/3 0 1 x_E 4/3 0 0 -1/3 2/3 1 0 x_i

Interpretation of the Simplex

Tableau: Optimum Solution

 The Reddy Mikks Company

 Optimum Solution

 x₁ = 4/3

 x₂ = 10/3

 Objective Function:

_{Maximize z = 3XE + 2XI}

Interpretation of the Simplex

Tableau: Status of Resources

abundant s > 0

scarce s = 0

Interpretation of the Simplex

Tableau: Status of Resources

2/3 0 1 1/3 -2/3 0 0 s₄ 3 0 1 1 -1 0 0 s₃ 10/3 0 0 2/3 -1/3 0 1 x_E 4/3 0 0 -1/3 2/3 1 0 x_i

Interpretation of the Simplex

Tableau: Status of Resources

abundant Limit of demand for

interior paint s₄ = 2/3

scarce Raw material B

s₂ = 0

scarce Raw material A

s₁ = 0

s₃ = 3 Slack Variable

abundant Limit of excess of

interior over exterior paint

Interpretation of the Simplex

Tableau: Status of Resources

 Which of the scarce resources should be

given priority in the allocation of additional funds to improve profit most

advantageously?

 Compare the dual price of the scarce

Interpretation of the Simplex

Tableau: Dual Price

2/3 0 1 1/3 -2/3 0 0 s₄ 3 0 1 1 -1 0 0 s₃ 10/3 0 0 2/3 -1/3 0 1 x_E 4/3 0 0 -1/3 2/3 1 0 x_i

y₄ = 0 Limit of demand

for interior paint s₄ = 0

y₂ =4/3 thousand dollars per ton of material B

Raw material B s₂ = 4/3

y₁ = 1/3 thousand dollars per ton of material A

Raw material A s₁ = 1/3

s₃ = 0 Slack Variable

y₃ = 0 Limit of excess of

interior over exterior paint

Dual Price function (y)

Interpretation of the Simplex

Tableau: Maximum Change in

Resource Availability

 Maximum Change in Resource Availability

of Resource i = D_i

 Two Cases:

_{Di > 0}

4/3 2

6 x_i

12 2_/ 3 12

0 z

2

(Optimum) 1

0

10/3 4

8 x_E

Right-Side Element (Solution) in Iteration

Equation

2/3 2

2 s₄

3 5

4/3 2

6 x_i (raw material A)

12 2_/ 3 12

0 z

2

(Optimum) 1

0

10/3 4

8 x_E (raw material B)

Right-Side Element (Solution) in Iteration Equation

2/3 2

2 s₄ (Demand x_I #2)

3 5

4/3 + ? 2 + D₁

6 + D₁ 1 (raw material A)

12 2_/

3 + ? 12

0 z

2

(Optimum) 1

0

10/3 + ? 4

8 2 (raw material B)

Right-Side Element (Solution) in Iteration Equation

2/3 + ? 2

2 4 (Demand x_I #2)

3 + ? 5

2/3 0 1 1/3 -2/3 0 0 s₄ 3 0 1 1 -1 0 0 s₃ 10/3 0 0 2/3 -1/3 0 1 x_E 4/3 0 0 -1/3 2/3 1 0 x_i

12 2_/ 3 0 0 4/3 1/3 0 0 z solution s₄ s₃ s₂ s₁ x_I x_E Basic

Right-Side Element (Solution) in Iteration

Equation

4/3 + 2/3 D₁ 2/3

1 (raw material A)

12 2_/

3 + 1/3 D1 1/3

z

2

(Optimum) s₁

10/3 – 1/3 D₁ -1/3

2 (raw material B)

2/3 – 2/3D₁ -2/3

4 (Demand x_I #2)

3 - 1D₁ -1

D₁  -2 D₁  1

Overall

Satisfied D₁  1

2/3 – 2/3D₁  0

Case

D₁ < 0 D₁ > 0

D₁  3 D₁  10 Satisfied

-2  D₁  1 3 - 1D₁  0

10/3 – 1/3 D₁  0 4/3 + 2/3 D₁  0

Satisfied Satisfied

-2  D₁  1

Any change outside this range (i.e.

decreasing raw material A by more than 2 tons or increasing raw material A by more

Maximum Change in Resource

Availability

 -2  D₁  1  feasible solution

 Any change outside this range (i.e. decreasing raw material A by more than 2 tons or increasing raw material A by more than 1 ton) will lead to

infeasibility and a new set of basic variables (See Chapter of Sensitivity Analysis)

Interpretation of the Simplex

Tableau: Maximum Change in

Marginal Profit/Cost

 Changing the coefficients in z-row

 Case 1: in accordance with basic variables  _{Case 2: in accordance with non-basic}

variables

 Maximum change in marginal profit/cost =

d_i

 For example

Interpretation of the Simplex

Tableau: Maximum Change in

Marginal Profit/Cost

2/3 0 1 1/3 -2/3 0 0 s₄ 3 0 1 1 -1 0 0 s₃ 10/3 0 0 2/3 -1/3 0 1 x_E 4/3 0 0 -1/3 2/3 1 0 x_i

2/3 0 1 1/3 -2/3 0 0 s₄ 3 0 1 1 -1 0 0 s₃ 10/3 0 0 2/3 -1/3 0 1 x_E 4/3 0 0 -1/3 2/3 1 0 x_i

12 2_/

3+10/3 d1 0

0 4/3+2/3 d₁

1/3 - 1/3d₁ 0 0 z solution s₄ s₃ s₂ s₁ x_I x_E Basic

Case 1 : Basic Variables

Objective Function: z = 3x_E – 2x_I Coefficient x_E = c_E =3

1/3 – 1/3 d₁  0  d₁  1

4/3 + 2/3 d₁  0  d₁  –2

1  c_E  4

Current optimum remains unchanged for the range of c_E [1. 4]

The value of z will change according to the expression: 12 2_/

 Case 2: Non - Basic Variables

 Changes in their original objective

coefficients can only affect only their z-equation coefficient and nothing else

 Corresponding column is not pivoted

 In general, the change d_i of the original

objective of a non-basic variable always results in decreasing the objective

 Maximize z = 5x_E + 2x_I

 Objective Function

Maximize z = 5x_E + 2x_I

 c_I = 2  c_I = 2 + d₂

 Coefficient x_I

 _{1/2 – d2}  ₀  _d2  _1/2

 _cI  _{2 + 1/2}

2 1 0 0 0 1 0 s₄ 5 0 1 1/2 0 3/2 0 s₃ 4 0 0 1/2 0 1/2 1 x_E 2 0 0 -1/2 1 3/2 0 s₁ 20 0 0 5/2 0 1/2 0 z solution s₄ s₃ s₂ s₁ x_I x_E Basic

Maximize z = 5x_E + 2x_{I Optimum}

Solution Reduced

Interpretation of the Simplex

Tableau: Reduced Cost

 Reduced Cost = the optimum objective

coefficients of the non-basic variables

 Reduced Cost = net rate of decrease in

the optimum objective value resulting from increasing of the associated non-basic

Interpretation of the Simplex

Tableau: Reduced Cost

 Reduced Cost = cost of resources to

produce per unit (x_i) input – its revenue

per unit output

 Reduced Cost > 0  cost > revenue

No economic advantage in producing the output

 Reduced cost < 0  cost < revenue

Interpretation of the Simplex

Tableau: Reduced Cost

2 1 0 0 0 1 0 s₄ 5 0 1 1/2 0 3/2 0 s₃ 4 0 0 1/2 0 1/2 1 x_E 2 0 0 -1/2 1 3/2 0 s₁ 20 0 0 5/2 0 1/2 0 z solution s₄ s₃ s₂ s₁ x_I x_E Basic

Maximize z = 5x_E + 2x_I

Optimum Solution Reduced

Interpretation of the Simplex

Tableau: Reduced Cost

 Optimum objective function

Interpretation of the Simplex

Tableau: Reduced Cost

 Reduced Cost = the optimum objective

coefficients of the non-basic variables

 Reduced Cost = net rate of decrease in

the optimum objective value resulting from increasing of the associated non-basic

Interpretation of the Simplex

Tableau: Reduced Cost

 Reduced Cost = cost of resources to

produce per unit (x_i) input – its revenue

per unit output

 Reduced Cost > 0  cost > revenue

No economic advantage in producing the output

 Reduced cost < 0  cost < revenue

Interpretation of the Simplex

Tableau: Reduced Cost

2 1 0 0 0 1 0 s₄ 5 0 1 1/2 0 3/2 0 s₃ 4 0 0 1/2 0 1/2 1 x_E 2 0 0 -1/2 1 3/2 0 s₁ 20 0 0 5/2 0 1/2 0 z solution s₄ s₃ s₂ s₁ x_I x_E Basic

Maximize z = 5x_E + 2x_I

Optimum Solution Reduced

Interpretation of the Simplex

Tableau: Reduced Cost

 Optimum objective function

Interpretation of the Simplex

Tableau: Reduced Cost

 Unused economic activity = non-basic variable can become economically viable in two ways: (Logically)

 #1 by decreasing its per unit use of the resource  #2 by increasing its per unit revenue (through

price increase)

 Combination of the two ways

 _{#1 is more viable than #2 since #1 is in} accordance with efficiency

The End