Overall Idea of Simplex Method

Linear Programming
(LP)
The Simplex Method
A. Primal 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
slack and surplus variables as necessary


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
 Extreme Points ↔ Basic Solutions


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
 resulting in a unique solution


Standard LP Form



To develop a general solution method, LP
problem should be put in a common
format

Development of the Simplex
Method


Two types of Simplex Method
 Primal

Simplex Method
 Dual Simplex Method


Basically, the difference between the two

method lies on the method of choosing the
initial setup

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
or minimization

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
x1 + 2x2 + s1 = 6,

s1  0

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
3x1 + 2x2 – 3x3 – s2 = 5,

s2  0

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

–2x1 – 3x2 +7x3 = +5

Standard LP Form
Constraints
The direction of an inequality is reversed
when both sides are multiplied by –1
 For example:


2x1 – x2  –5 = –2x1 + x2  5

Standard LP Form
Variables
Unrestricted variable xi can be expressed
in terms of two non-negative variables
 xi = xi’ – xi”, where xi’, xi”  0
 xi’ > 0, xi” = 0, and vice versa


Slack ↔Surplus

Standard LP Form
Objective Function
Maximization or Minimization
 Maximization = Minimization of the
negative of the function
 For example:


Maximize z = 5x1 + 2x2 + 3X3 equivalent to
Minimize (–z) = – 5x1 – 2x2 – 3X3

Standard LP Form
Example
Write the following LP Model in standard
form:
 Minimize z = 2x1 + 3X2
 Subject to (with constraints):








x1 + x2 = 10
–2 x1 +3 x2  –5
7 x1 – 2 x2  6
x1 unrestricted
x2  0

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
 resulting a unique solution

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
 Number of possible basic solution = mCn = 4C2 =
4!/2!2! = 6



Basic Solution
Example
2x1 + x2 + 4x3 + x4 = 2
 x1 + 2x2 + 2x3 + x4 = 3
 Set x2 = 0 and x4 = 0
 2x1 + 4x3 = 2
inconsistent
 x1 + 2x3 = 3


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:
 x1 =

1/3, x2 = 4/3, x3 = 0 and x4 = 0

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
infeasible
x2 = 4/3

Basic Feasible Solution
A basic solution is said to be feasible if all
its solution values are non-negative
 Else: infeasible basic solution


Primal and Dual Simplex
All iterations in Primal Simplex Method
are always associated to feasible basic
solutions only
 Primal Simplex Method deals with feasible
extreme points only


Primal and Dual Simplex
Iterations in Dual Simplex Method end
only if the last iteration is infeasible
 Both methods yield feasible basic solution
as stipulated by the non-negativity
condition of the LP model


Primal Simplex Method
The Reddy Mikks Company




XE: tons produced daily of exterior paint
XI: tons produced daily of interior paint
Objective Function to be satisfy:
 Maximize



z = 3XE + 2XI

Subject to these constraints:
+ 2XI  6
 2XE + XI  8
 – XE + XI  1
 XI  2
 XE, XI  0
 XE

Primal Simplex Method
The Reddy Mikks Company
Set the constraints to equations
 xE + 2xI  6 
xE + 2xI + sI = 6
 2xE + xI  8 
2xE + xI + s2 = 8
 – xE + xI  1 
– xE + xI + s3 = 1
 xI  2

xI + s4= 2
 xI, xE, sI, s2, s3, s4  0
m=4
n=2


Primal Simplex Method
The Reddy Mikks Company
Basic Solution
m=4
 n = 2 (xE and xI)
 If xE and xI = 0 then


 xE

+ 2xI + sI = 6
 2xE + xI + s2 = 8
 – xE + xI + s3 = 1
 x I + s 4= 2






sI = 6
s2 = 8
s3 = 1
s 4= 2

Basic
Feasible
Solution

Primal Simplex Method
The Reddy Mikks Company
Convert the Objective Function
 Maximize z = 3xE + 2xI
 z – 3xE – 2xI = 0
 Include the slack variables
 Maximize
z – 3xE – 2xI + 0sI + 0s2 + 0s3 + 0s4 = 0


Primal Simplex Method
The Reddy Mikks Company
Maximize
z – 3xE – 2xI + 0sI + 0s2 + 0s3 + 0s4 = 0
 Later (in the next sub-chapters), the slack and
the substitute variables are written as common
variables
 Maximize
z – 3xE – 2xI + 0x1 + 0x2 + 0x3 + 0x4 = 0


Primal Simplex Method
The Reddy Mikks Company
Incorporate the objective function with the
basic feasible solution
 xE and xI : entering variables
 sI, s2, s3 ,and s4 : leaving variables
 Start the iterations with the values of sI, s2,
s3 ,and s4 = zero
 Final Result of the iterations the values of
xE and xI in z = zero


Primal Simplex Method
Easiness (a commentary)
Each equation has a slack variable
 The right hand of all constraints are nonnegative


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)
 A Tie may be broken arbitrarily

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)
 A Tie may be broken arbitrarily

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
 A Tie may be broken arbitrarily

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
among the current non-basic variables
using the optimality condition

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;
else Go to Step 1

Primal Simplex Method
The Reddy Mikks Company
Incorporate the objective function with the
basic feasible solution
 xE and xI : entering variables
 sI, s2, s3 ,and s4 : leaving variables
 Start the iterations with the values of sI, s2,
s3 ,and s4 = zero
 Final Result of the iterations the values of
xE and xI in z = zero


Primal Simplex Method
The Reddy Mikks Company
Maximize
z – 3XE – 2XI + 0sI + 0s2 + 0s3 + 0s4 = 0
 xE + 2xI + sI = 6
 Raw Material A
 2xE + xI + s2 = 8
 Raw Material B
 – xE + xI + s3 = 1  Demand x1 #1
 xI + s4= 2 Demand x1 #2
 Put all the equation in the Table (Tableau)
 Use Gauss-Jordan Method (Swapping)


Primal Simplex Method:
The Tableau
Basic

z

xE

xI

s1

s2

s3

s4

solution

z

1

-3

-2

0

0

0

0

0

s1

0

1

2

1

0

0

0

6

s2

0

2

1

0

1

0

0

8

s3

0

-1

1

0

0

1

0

1

s4

0

0

1

0

0

0

1

2

This column can be omitted

Intercept

Entering Column
Basic

xE

xI

s1

s2

s3

s4

z

-3

-2

0

0

0

0

Pivot

s1

1

2

1

0

0

0

6

6

Equa
-tion
(PE)

s2

2

1

0

1

0

0

8

4

s3

-1

1

0

0

1

0

1

-

s4

0

1

0

0

0

1

2

-

Leaving Variable

Pivot Element

solution Intercept
0

Gauss – Jordan
1.

2.

Pivot Equation
New Pivot Equation (NPE)= Old Pivot
Equation : Pivot Element
All other equations, including z
New Equation = (Old Equation) – (its
entering column coefficient  NPE)

Pivot Element = 2
Basic

xE

xI

s1

s2

s3

s4

solution

1

1/2

0

1/2

0

0

4

z
s1
NPE xE
s3
s4

z
Coefficient xE for z = -3
NPE

1

1/2

0

1/2

0

0

4

-3NPE -3

-3/2

0

-3/2

0

0

-12

Z
(OLD)

-3

-2

0

0

0

0

0

3NPE

-3

-3/2

0

-3/2

0

0

-12

Z
(NEW)

0

-1/2

0

3/2

0

0

12

Basic

xE

xI

s1

s2

s3

s4

solution

z

0

-1/2

0

3/2

0

0

12

1

1/2

0

1/2

0

0

4

s1
xE
s3
s4

s1
Coefficient xE for s1 = 1
NPE

1

1/2

0

1/2

0

0

4

1NPE

1

1/2

0

1/2

0

0

4

s1 (OLD)

1

2

1

0

0

0

6

1NPE

1

1/2

0

1/2

0

0

4

s1
(NEW)

0

3/2

1

-1/2

0

0

2

Basic

xE

xI

s1

s2

s3

s4

solution

z

0

-1/2

0

3/2

0

0

12

s1

0

3/2

1

-1/2

0

0

2

xE

1

1/2

0

1/2

0

0

4

s3
s4

s3
Coefficient xE for S3 = -1
NPE

1

1/2

0

1/2

0

0

4

-1NPE -1

-1/2

0

-1/2

0

0

-4

s3(OLD)

-1

1

0

0

1

0

1

-1NPE

-1

-1/2

0

-1/2

0

0

-4

s3
(NEW)

0

3/2

0

1/2

1

0

5

Basic

xE

xI

s1

s2

s3

s4

solution

z

0

-1/2

0

3/2

0

0

12

s1

0

3/2

1

-1/2

0

0

2

xE

1

1/2

0

1/2

0

0

4

s3

0

3/2

0

1/2

1

0

5

s4

s4
Coefficient xE for S4 = 0
NPE

1

1/2

0

1/2

0

0

4

0NPE

0

0

0

0

0

0

0

s4

0

1

0

0

0

1

2

0NPE

0

0

0

0

0

0

0

s4
(NEW)

0

1

0

1

0

2

0

Result of Iteration 1
Basic

xE

xI

s1

s2

s3

s4

solution

z

0

-1/2

0

3/2

0

0

12

s1

0

3/2

1

-1/2

0

0

2

xE

1

1/2

0

1/2

0

0

4

s3

0

3/2

0

1/2

1

0

5

s4

0

1

0

0

1

0

2

Entering Column
Basic

xE

xI

s1

s2

s3

s4

Pivot

z

0

-1/2

0

3/2

0

0

Equa
-tion
(PE)

s1

0

3/2

1

-1/2

0

0

2

4/3

XE

1

1/2

0

1/2

0

0

4

8

s3

0

3/2

0

1/2

1

0

5

10/3

s4

0

1

0

0

1

0

2

2

Leaving Variable

Pivot Element

solution Intercept
12

Pivot Element = 3/2
Basic

xE

xI

s1

s2

s3

s4

solution

0

1

2/3

-1/3

0

0

4/3

z
NPE

xi
xE
s3
s4

z
Coefficient xI for z = -1/2
NPE

0

1

2/3 -1/3

-1/2NPE

0

-1/2 -1/3

Z
(OLD)

0

-1/2

-1/2NPE

0

-1/2 -1/3

Z
(NEW)

0

0

0

1/3

0

0

4/3

0

0

-2/3

0

0

12

1/6

0

0

-2/3

4/3

0

0

12 2/3

1/6

3/2

Basic

xE

xI

s1

s2

s3

s4

solution

z

0

0

1/3

4/3

0

0

12 2/3

xi

0

1

2/3

-1/3

0

0

4/3

xE
s3
s4

xE
Coefficient xi for x3 = 1/2
NPE

0

1

2/3 -1/3

0

0

4/3

1/2NPE

0

1/2

1/3

-1/6

0

0

-2/3

xE
(OLD)

1

1/2

0

1/2

0

0

4

1/2NPE

0

1/2 1/3

-1/6

0

0

-2/3

xE
(NEW)

1

0

2/3

0

0

10/3

-1/3

Basic

xE

xI

s1

s2

s3

s4

solution

z

0

0

1/3

4/3

0

0

12 +2/3

xi

0

1

2/3

-1/3

0

0

4/3

xE

1

0

-1/3

2/3

0

0

10/3

s3
s4

s3
Coefficient xi for s3 = 3/2
NPE

0

1

3/2NPE

0

3/2

s3
(OLD)

0

3/2

3/2NPE

0

s3
(NEW)

0

2/3 -1/3

0

0

4/3

-1/2

0

0

2

0

1/2

1

0

3/2

1

-1/2

0

0

2

0

-1

1

1

0

3

1

5

Basic

xE

xI

s1

s2

s3

s4

solution

z

0

0

1/3

4/3

0

0

12 2/3

xi

0

1

2/3

-1/3

0

0

4/3

xE

1

0

-1/3

2/3

0

0

10/3

s3

0

0

-1

1

1

0

3

s4

s4
Coefficient xi for s4 = 1
NPE

0

1

2/3 -1/3

0

0

4/3

1NPE

0

1

2/3

-1/3

0

0

4/3

0

1

0

0

1

0

2

0

1

2/3

-1/3

0

0

4/3

0

0

-2/3

1/3

1

0

2/3

s4
(OLD)
1NPE
s4
(NEW)

Result of Iteration 2 = END

Basic

xE

xI

s1

s2

s3

s4

solution

z

0

0

1/3

4/3

0

0

12 2/3

xi

0

1

2/3

-1/3

0

0

4/3

xE

1

0

-1/3

2/3

0

0

10/3

s3

0

0

-1

1

1

0

3

s4

0

0

-2/3

1/3

1

0

2/3

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
marginal profit/cost (objective function coefficients)
The usage of the resources by the model activities

Interpretation of the Simplex
Tableau: Optimum Solution
Basic

xE

xI

s1

s2

s3

s4

solution

z

0

0

1/3

4/3

0

0

12 2/3

xi

0

1

2/3

-1/3

0

0

4/3

xE

1

0

-1/3

2/3

0

0

10/3

s3

0

0

-1

1

1

0

3

s4

0

0

-2/3

1/3

1

0

2/3

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
z = 3(4/3) + 2(10/3) = 12 2/3

Interpretation of the Simplex
Tableau: Status of Resources
Slack Variable

Status of Resource

s=0

scarce

s>0

abundant

Interpretation of the Simplex
Tableau: Status of Resources
Basic

xE

xI

s1

s2

s3

s4

solution

z

0

0

1/3

4/3

0

0

12 2/3

xi

0

1

2/3

-1/3

0

0

4/3

xE

1

0

-1/3

2/3

0

0

10/3

s3

0

0

-1

1

1

0

3

s4

0

0

-2/3

1/3

1

0

2/3

Interpretation of the Simplex
Tableau: Status of Resources
Slack
Variable

Resource

s1 = 0

Raw material A

Status of
Resource
scarce

s2 = 0

Raw material B

scarce

Limit of excess of
interior over exterior
paint
Limit of demand for
interior paint

abundant

s3 = 3
s4 = 2/3

abundant

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
resources


Interpretation of the Simplex
Tableau: Dual Price
Dual Price
Basic

xE

xI

s1

s2

s3

s4

solution

z

0

0

1/3

4/3

0

0

12 2/3

xi

0

1

2/3

-1/3

0

0

4/3

xE

1

0

-1/3

2/3

0

0

10/3

s3

0

0

-1

1

1

0

3

s4

0

0

-2/3

1/3

1

0

2/3

Slack
Variable

Resource

Dual Price function
(y)

Raw material A

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

s2 = 4/3

Raw material B

y2 =4/3 thousand
dollars per ton of
material B
y3 = 0

s3 = 0

Limit of excess of
interior over
exterior paint

s4 = 0

Limit of demand
for interior paint

y4 = 0

s1 = 1/3

Interpretation of the Simplex
Tableau: Maximum Change in
Resource Availability
Maximum Change in Resource Availability
of Resource i = Di
 Two Cases:


 Di

>0
 Di < 0

Right-Side Element (Solution) in
Iteration
Equation
2
0
1
(Optimum)
z

0

12

12 2/3

xi

6

2

4/3

xE

8

4

10/3

s3

1

5

3

s4

2

2

2/3

Equation

Right-Side Element
(Solution) in Iteration
0

1

2
(Optimum)

z

0

12

12 2/3

xi (raw material A)

6

2

4/3

xE (raw material B)

8

4

10/3

s3 (Demand xI #1)

1

5

3

s4 (Demand xI #2)

2

2

2/3

Equation

Right-Side Element
(Solution) in Iteration
0

1

2
(Optimum)

z

0

12

12 2/3 + ?

1 (raw material A)

6 + D1

2 + D1

4/3 + ?

2 (raw material B)

8

4

10/3 + ?

3 (Demand xI #1)

1

5

3+?

4 (Demand xI #2)

2

2

2/3 + ?

Coefficients for D1

Basic

xE

xI

s1

s2

s3

s4

solution

z

0

0

1/3

4/3

0

0

12 2/3

xi

0

1

2/3

-1/3

0

0

4/3

xE

1

0

-1/3

2/3

0

0

10/3

s3

0

0

-1

1

1

0

3

s4

0

0

-2/3

1/3

1

0

2/3

Right-Side Element (Solution)
in Iteration
Equation
s1

2
(Optimum)

z

1/3

12 2/3 + 1/3 D1

1 (raw material A)

2/3

4/3 + 2/3 D1

2 (raw material B)

-1/3

10/3 – 1/3 D1

3 (Demand xI #1)

-1

3 - 1D1

4 (Demand xI #2)

-2/3

2/3 – 2/3D1

Case
D1 > 0

D1 < 0

4/3 + 2/3 D1  0

Satisfied

D1  -2

10/3 – 1/3 D1  0

D1  10

Satisfied

3 - 1D1  0

D1  3

Satisfied

2/3 – 2/3D1  0

D1  1

Satisfied

Overall

D1  1

D1  -2

-2  D1  1

-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
than 1 ton) will lead to infeasibility and a new
set of basic variables (Chapter of Sensitivity
Analysis)

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)
Max and Min of raw material A with dual price y1
= 1/3 when Max = 6 + 1 = 7 and Min = 6 – 2 = 4

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 =
di
 For example


d1 : Maximum change profit in
accordance with xE

Interpretation of the Simplex
Tableau: Maximum Change in
Marginal Profit/Cost
Basic

xE

xI

s1

s2

s3

s4

solution

z

0

0

1/3

4/3

0

0

12 2/3

xi

0

1

2/3

-1/3

0

0

4/3

xE

1

0

-1/3

2/3

0

0

10/3

s3

0

0

-1

1

1

0

3

s4

0

0

-2/3

1/3

1

0

2/3

Case 1 : Basic Variables
Objective Function: z = 3xE – 2xI
Basic

xE

xI

s1

z

0

0

xi

0

1

2/3

xE

1

0

s3

0

s4

0

s2

s3 s4

solution

0

0

12 2/3+10/3 d1

-1/3

0

0

4/3

-1/3

2/3

0

0

10/3

0

-1

1

1

0

3

0

-2/3

1/3

1

0

2/3

1/3 - 1/3d1 4/3+2/3 d1

Objective Function: z = 3xE – 2xI
Coefficient xE = cE =3
1/3 – 1/3 d1  0  d1  1

4/3 + 2/3 d1  0  d1  –2
– 2  d1  1  3 – 2  cE  1 + 1
 1  cE  4

1  cE  4
Current optimum remains unchanged for the range
of cE [1. 4]
The value of z will change according to the
expression: 12 2/3+10/3 d1, – 2  d1  1

Case 2: Non - Basic Variables
 Changes in their original objective
coefficients can only affect only their zequation coefficient and nothing else
 Corresponding column is not pivoted
 In general, the change di of the original
objective of a non-basic variable always
results in decreasing the objective
coefficient in the optimum tableau by the
same amount


Maximize z = 5xE + 2xI
 Objective Function


 Maximize




z = 5xE + 2xI
cI = 2  cI = 2 + d2

Coefficient xI




1/2 – d2  0 
cI  2 + 1/2
cI  5/3

d2  1/2

Reduced
Cost

Maximize z = 5xE + 2xI

Optimum
Solution

Basic

xE

xI

s1

s2

s3

s4

solution

z

0

1/2

0

5/2

0

0

20

s1

0

3/2

1

-1/2

0

0

2

xE

1

1/2

0

1/2

0

0

4

s3

0

3/2

0

1/2

1

0

5

s4

0

1

0

0

0

1

2

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
variable


Interpretation of the Simplex
Tableau: Reduced Cost
Reduced Cost = cost of resources to
produce per unit (xi) input – its revenue
per unit output
 Reduced Cost > 0  cost > revenue
No economic advantage in producing the
output
 Reduced cost < 0  cost < revenue
candidate for becoming positive in the
optimum solution


Interpretation of the Simplex
Tableau: Reduced Cost
Reduced
Cost

Maximize z = 5xE + 2xI

Optimum
Solution

Basic

xE

xI

s1

s2

s3

s4

solution

z

0

1/2

0

5/2

0

0

20

s1

0

3/2

1

-1/2

0

0

2

xE

1

1/2

0

1/2

0

0

4

s3

0

3/2

0

1/2

1

0

5

s4

0

1

0

0

0

1

2

Interpretation of the Simplex
Tableau: Reduced Cost
Optimum objective function
 z + 1/2 xI + 5/2 s1 = 20
 z = 20 – 1/2 xI – 5/2 s1


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
variable


Interpretation of the Simplex
Tableau: Reduced Cost
Reduced Cost = cost of resources to
produce per unit (xi) input – its revenue
per unit output
 Reduced Cost > 0  cost > revenue
No economic advantage in producing the
output
 Reduced cost < 0  cost < revenue
candidate for becoming positive in the
optimum solution


Interpretation of the Simplex
Tableau: Reduced Cost
Reduced
Cost

Maximize z = 5xE + 2xI

Optimum
Solution

Basic

xE

xI

s1

s2

s3

s4

solution

z

0

1/2

0

5/2

0

0

20

s1

0

3/2

1

-1/2

0

0

2

xE

1

1/2

0

1/2

0

0

4

s3

0

3/2

0

1/2

1

0

5

s4

0

1

0

0

0

1

2

Interpretation of the Simplex
Tableau: Reduced Cost
Optimum objective function
 z + 1/2 xI + 5/2 s1 = 20
 z = 20 – 1/2 xI – 5/2 s1


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
while #2 is related to market condition in which
other factors influence

The End


This is the end of Chapter 3A