The Simplex Method - Andrew David Joseph Hall · 2 Quantitative Methods for Economics and Business I The Simplex Method Formulate Constrained Maximization or Minimization Problem

1Quantitative Methods for Economics and Business I

The Simplex Method


The Simplex Method

Formulate Constrained Maximization or Minimization Problem

Convert to Standard Form

Convert to Canonical Form

Apply The Simplex Algorithm

Perform Sensitivity Analysis

Set Up the Tableau and the Initial Basic Feasible Solution


A Simple Example Revenue Maximization

The Nutty Wood Co. makes chairs and tables from Walnut and Oak.

The company wants to maximize revenue.

Management would be happy to make chairs alone, tables alone or both as it can sell as much as it can produce.


Nutty Wood Co. Revenue Maximization cont…

A table sells for $400, needs 5 cubic feet of Walnut and 2.5 cubic feet of Oak.

A chair which takes 2 cubic feet of Oak and 1 cubic foot of Walnut sells for $125.

Only 400 cubic feet of Walnut and only 250 cubic feet of Oak is available.


The Decision Variables

The decision to be made is how many tables and how many chairs to make, so the decision variables are:

Let c be the number of chairs built.

Let t be the number of tables made.

Sales prices are parameters as it is assumed that these are givens.


The Objective Function

What is the objective?

How do we express this in terms of Decision variables c and t?

What is the Objective Function?


The Objective Function

Maximize revenue is the objective.Revenue is Sales Price by Number Sold.Revenue = (125 * c) + (400 * t)So the Objective Function is:

Max Revenue =(125 * c) + (400 * t)


Constraints

A constraint is something that prevents us from doing something.

In this example we are limited by the amount of Oak and Walnut available.

The Oak constraint is that using 2.5 cubic feet for a table and 2 cubic feet for a chair we can use no more than 250 cubic feet.


Constraints cont…

The Walnut constraint is that using 5 cubic feet for a table and 1 cubic foot for a chair we can use no more than 400 cubic feet.

How do we express the constraints mathematically?


Constraints cont…

The Walnut constraint is that using 5 cubic feet for a table and 1 cubic foot for a chair we can use no more than 400 cubic feet.

So:

• (2 * c) + (2.5 * t) 250• (1 * c) + (5 * t) 400

are the constraints.


Bounds on the Problem

Nutty Wood Co. do not make fewer than zero chairs or tables.

How do we express this?


Bounds on the Problem

Nutty Wood Co. do not make fewer than zero chairs or tables.

We express this with lower bounds as follows:

• c 0• t 0

Note this allows for zero in both cases.


Write Down The Full Model


The Model

Maximize 125c + 400t (Revenue)

Subject to 2c + 2.5t 250 (Oak Constraint)

1c + 5t 400 (Walnut Constraint)

c 0 (Lower Bounds)

t 0


The Simplex MethodFormulate Constrained Maximization or Minimization Problem







Simple Mathematical Operations on Constraints

Any constraint may be multiplied or divided through by a positive number without changing the constraint.

• x + y => 2 is the same as 2x + 2y => 4


x + y => 2 is the same as 2x + 2y => 4

0.5 1 1.5 2

0.5

1

1.5

2

x

y

Equation 1: x + y => 2

0.5 1 1.5 2

0.5

1

1.5

2

x

y

Equation 2:2 x +2 y => 4


Simple Mathematical Operations on Constraints

When a “>”, “=>”, “<=“ or “<“ constraint is multiplied or divided through by a negative number, the direction of the constraint must be changed.


Multiply x + y => 2 by -1

(x + y => 2) * (-1) ⇒ -x – y <= -2 and they are equivalent.

0.5 1 1.5 2

0.5

1

1.5

2

x

y

Equation 1: x + y => 2

0.5 1 1.5 2

0.5

1

1.5

2

x

y

Equation 2:- x - y < = -2


Standard Form

A Linear Program is in Standard Form if:

• All the constraints are written as equalities.

• All variables are required to be non-negative.

This involves the addition of “Slack” variables and “Surplus” variables.


Adding Slack Variables

When a Constraint is in the form:

• x + y <= v : With bounds (x => 0, y => 0)

We add a “Slack Variable” to take up the slack between the value of (x + y) and the value of v.

Thus x + y + s = v.

Note that s => 0 since x + y <= v.


Adding Surplus Variables

When a Constraint is in the form:

• x + y => v: With bounds (x => 0, y => 0)

We add a “Surplus Variable” to account for the surplus left over when the value of v is deducted from the values of (x + y).

Thus x + y = v + s

Note that s => 0 since x + y => v.


Adding Surplus Variables cont…

We then shuffle “s” to the left hand side of the equality by subtracting s from both sides, giving:

x + y – s = v

x=> 0, y =>0, and s => 0


Conversion to Standard Form, Nutty Wood Co.

maximize 125c + 400t (Revenue)

Subject to 2c + 2.5t 250 (Oak Constraint) 1c + 5t 400 (Walnut Constraint)

≤≤

c 0 (Lower Bounds) t 0

≥≥


Oak Constraint (1)

This is a “<=” constraint, so we need to add a surplus variable (S1).

1

1

maximize + + (Revenue)125c 400t 0S

Subject to + + 250 (Oak Constraint)2c 2.5t S 1c + 5t 400 (Wal

=≤

1

nut Constraint) c 0 (Lower Bounds) t 0 0S

≥≥≥


Walnut Constraint (2)

This is “<=” constraint, so we need to add a surplus variable (S2).

1 2

1

2

maximize + + + (Revenue)125c 400t 0S 0S

Subject to + + 250 (Oak Constraint)2c 2.5t S 1c + 5t + = 400 (Walnut CoS

=

1 2

nstraint) c 0 (Lower Bounds) t 0 0, 0S S

≥≥≥ ≥


Conventional Presentation

Use x1, x2, x3 etc…in place of c, t, x, y or z.

Slack or Surplus Variables are denoted by:• “S” concatenated with the constraint number

Hence S1 and S3 in the example.


Negative Values on RHS

If any RHS values are negative we multiply through the equality by (-1) to make the RHS value positive.

e.g.

• 2x1 – 3x2 – S = - 12

becomes• -2x1 + 3x2 + S = 12.









Canonical Form

A Linear Program is in Canonical Form if:

• It is in Standard Form, and

• For each constraint, there exists a variable that appears only in the constraint, and its coefficient in that constraint is “+1”.


Example of Canonical Form

1 2 1 2

1 2 1

1 2 2

maximize + + + (Revenue)125x 400x 0S 0S

Subject to + + 250 (Oak Constraint)2.5x 1S2x + + = 400 (Walnut5x 1S1x

=

1

2

1 2

Constraint) 0 (Lower Bounds)x 0x 0, 0S S

≥≥≥ ≥


Nutty Wood Co. in Full Canonical Form

1 2 1 2

1 2 1

1 2 2



=

1

2

1 2


≥≥≥ ≥









Setting up The Simplex Tableau

To set up the Simplex Tableau we will:

• Identify the Basic and Non-basic variables.

• Set up a Basic Feasible Solution.

• Look at the form of the Simplex Tableau, and

• Enter the LP and values of the Initial Basic Solution into the Simplex Tableau.


Basic and Non-basic Variables

Recall from the definition of the Canonical Form.

• “For each constraint, there exists a variable that appears only in the constraint, and its coefficient in that constraint is “+1”. “ These variables are the initial Basic Variables.

• The other variables are the Non-basic Variables.


Basic and Non-basic Variables cont…In our example we added S1 and S2 to satisfy the conditions for the Canonical Form. These are the Basic Variables.

The Non-basic Variables are x1 and x2.

1 2 1 2

1 2 1

1 2 2



=

1

2

1 2


≥≥≥ ≥


Setting up the Basic (Feasible) Solution

A Basic Feasible Solution for an LP in the Canonical form is one where:

• The Non-basic variables are set to zero, and

• The Basic variables take on the values of the RHS of their constraint.


Setting up the Basic (Feasible) Solution cont…

A Basic Solution is a Basic Feasible Solution if:

• All of the variables (Basic and Non-basic) are greater than or equal to zero.


The Initial Basic Feasible Solution to Nutty Wood Co.

S1 = 250, S2 = 400.x1 = 0, x2 = 0.The Objective Function Value:• z = 0

1 2 1 2

1 2 1

1 2 2



=

1

2

1 2


≥≥≥ ≥


The Form of The Simplex Tableau

Basis Cj Bi

ZjSum of

(RHS * Cj ) for all j

Cj - Zj

Objective Function

Coefficients of Basic

Variables

Names of the Basic Variables

Sum of (Each Column * Cj Column).

Cj row value less Zj row value.

Objective Function Coefficients

Names of the Variables

Coefficients of the Left Hand Sides of the Constraints.

Right Hand Side Values of

Constraints


Names of Variables

X1 X2 S1 S2

Basis Cj Bi

ZjSum of


Cj - Zj




Constraints

Objective Function


Variables

Names of the Basic Variables




Basic Variables

X1 X2 S1 S2

Basis Cj Bi

S1

S2

ZjSum of


Cj - Zj






Constraints

Objective Function


Variables



1 2 1 2

1 2 1 2

maximize + + + = Revenue125x 400x 0S 0S- - - - + Revenue = 0 125x 400x 0S 0S

X1 X2 S1 S2

Basis Cj -125 -400 0 0 Bi

S1 0

S2 0

ZjSum of


Cj - Zj


Constraints





Coefficients and RHS Values of the Constraints1 2 1

1 2 2

+ + 250 (Oak Constraint)2.5x 1S2x + + = 400 (Walnut Constraint)5x 1S1x

=

X1 X2 S1 S2

Basis Cj -125 -400 0 0 Bi

S1 0 2 2.5 1 0 250

S2 0 1 5 0 1 400

ZjSum of


Cj - Zj




The Zj Row

X1 X2 S1 S2

Basis Cj -125 -400 0 0 Bi

S1 0 2 2.5 1 0 250

S2 0 1 5 0 1 400

Zj 0 0 0 0Sum of


Cj - Zj Cj row value less Zj row value.

+

=

*

*


The Objective Function Value

X1 X2 S1 S2

Basis Cj -125 -400 0 0 Bi

S1 0 2 2.5 1 0 250

S2 0 1 5 0 1 400

Zj 0 0 0 0 0

Cj - Zj Cj row value less Zj row value.

0 * 250+

0 * 400=0


The Cj – Zj Row

X1 X2 S1 S2

Basis Cj -125 -400 0 0 Bi

S1 0 2 2.5 1 0 250

S2 0 1 5 0 1 400

Zj 0 0 0 0 0

Cj - Zj -125 - 0= -125

-400 - 0 = -400

0 - 0 = 0

0 - 0 = 0


The Initial Basic Feasible Solution in the Simplex Tableau

X1 X2 S1 S2

Basis Cj -125 -400 0 0 Bi

S1 0 2 2.5 1 0 250

S2 0 1 5 0 1 400

Zj 0 0 0 0 0

Cj - Zj -125 -400 0 0









The Simplex Algorithm

1. Find the Entering Variable

2. Calculate: By how much can the Entering Variable be increased.

3. Pivot

Variable to Enter?STOP

No

Yes


The Meaning of the Cj, Zj and (Cj – Zj) Rows

Cj value – The gross increase in the Objective Function Value, given a one unit increase in the amount of that variable.

Zj value – The gross decrease in the Objective Function Value, given a one unit increase in the amount of that variable.

Cj-Zj – The net effect on the Objective Function Value of a one unit increase in the amount of that variable.


Find the Entering VariableIn a maximization we want to increase the Objective Function Value as much as possible, so:

Select the variable for which the net effect is the greatest

In our example, (-400) is largest (Cj-Zj), so x2 is the Entering Variable.

Zj 0 0 0 0 0

Cj - Zj -125 -400 0 0


Which row has the smallest quotient X2/Bi?

250/2.5 = 100

400/5 = 80

So S2 will be the variable leaving the basis since the smallest quotient is 80.

X2

Basis Cj -400 Bi

S1 0 2.5 250

S2 0 5 400

Zj 0 0

Cj - Zj -400


Pivoting

Calculate New Row to Replace Pivot Row

Calculate Other Rows in New Tableau

Replace Leaving Variable with Entering Variable in the Basis Column

Update the Objective Function Coefficient for the Entering Variable

Calculate new Zj Values for Each Column

Calculate New (Cj - Zj) Values for Each Column


X1 X2 S1 S2

Basis Cj -125 -400 0 0 Bi

S1 0 2 2.5 1 0 250

S2 0 1 5 0 1 400

Zj 0 0 0 0 0

Cj - Zj -125 -400 0 0

Pivot Row, Column and Element


Calculate “New Row” to Replace Pivot Row

Divide the pivot row by the pivot element:-X1 X2 S1 S2

Basis Cj -125 -400 0 0 Bi

S1 0 2 2.5 1 0 250

S2 0 1 / 5=1/5

5 / 5 = 1

0 / 5=0

1 / 5=1/5

400 / 5=80

Zj 0

Cj - Zj


1/5 * 10/4

= 10/20

1 * 10/4= 10/4

0 * 10/4= 0

1/5 * 10/4

= 10/20

80 * 10/40= 200

2 - 10/20= 3/2

2.5 - 2.5= 0

1 - 0=1

0 -10/20= -1/2

250 - 200= 50

a. S2 Row multiplied by Pivot Column Element of S1 row.

b. S1 Row minus row a. above.

Calculate New Row 2Multiply the New Row for X2 by the value of the coefficient in the Pivot Column and S2 Row and subtract this from S2 Row to calculate the New S2Row.

X1 X2 S1 S2

Basis Cj -125 -400 0 0 Bi

S1 0 2 2.5 1 0 250

S2 0 1/5 1 0 1/5 80


X1 X2 S1 S2

Basis Cj -125 -400 0 0 Bi

S1 0 3/2 0 1 -1/2 50

S2 0 1/5 1 0 1/5 80

Zj 0

Cj - Zj

Replace Row2 Basic Variable with Entering Variable

Also check that the result is in Canonical form. If not review Leaving Variable.


Enter X2’s Coefficient From the Objective Function in Cj Column, Row 2

X1 X2 S1 S2

Basis Cj -125 -400 0 0 Bi

S1 0 3/2 0 1 -1/2 50

X2 -400 1/5 1 0 1/5 80

Zj 0

Cj - Zj


Recalculate the Zj and (Cj – Zj) Rows as Before

X1 X2 S1 S2

Basis Cj -125 -400 0 0 Bi

S1 0 3/2 0 1 -1/2 50

X2 -400 1/5 1 0 1/5 80

Zj0 * 3/2 + -400 *1/5

=- 80

0 * 0 + -400 * 1 = -400

1 * 0 + -400 * 0

= 0

0 * -1/2 + -400 *1/5

=-800

Cj - Zj-125 - -

80=-45

-400 - -400 = 0

0 - 0 = 0 0 - - 80=80


Is There Another Entering Variable??

Yes, X1 has a (Cj – Zj) value of -45 and is the most negative value.

X1 X2 S1 S2

Basis Cj -125 -400 0 0 Bi

S1 0 3/2 0 1 -1/2 50

X2 -400 1/5 1 0 1/5 80

Zj -80 -400 0 -80 0

Cj - Zj -45 0 0 80


Which row has the smallest quotient X2/Bi?

50/(3/2) = 33.33

80/(1/5) = 400

So S1 will be the variable leaving the basis since the smallest quotient is 33.33.

X1

Basis Cj -125 Bi

S1 0 3/2/

50

X2 -400 1/5 / 80

Zj -80 0

Cj - Zj -45


Calculate “New Row” (*) to Replace Pivot Row

Divide the pivot row by the pivot element:-

X1 X2 S1 S2

Basis Cj -125 -400 0 0 Bi

S1 0 3/2 / 3/2

0 / 3/2

1 / 3/2

-1/2 / 3/2

50 / 3/2

X2 -400 1/5 1 0 1/5 80

Zj -80 -400 0 -80 0

Cj - Zj -45 0 0 80


Calculate “New Row” (*) to Replace Pivot Row

X1 X2 S1 S2

Basis Cj -125 -400 0 0 Bi

S1 0 1 0 2/3 -1/3 33.33

X2 -400 1/5 1 0 1/5 80

Zj -80 -400 0 -80 0

Cj - Zj -45 0 0 80


Calculate New Row 2Multiply the New Row by the value of the coefficient in the Pivot Column and Row 2 and subtract this from Row 2 to calculate the New Row 2.

X1 X2 S1 S2

Basis Cj -125 -400 0 0 Bi

S1 0 1 0 2/3 -1/3 33.33

X2 -400 1/5 1 0 1/5 80

1 * 1/5= 1/5

0 * 1/5= 0

2/3 * 1/5

= 2/15

-1/3 * 1/5

= -1/15

100/3 * 1/5= 100/15 = 6.67

1/5 - 1/5= 0

1 - 0= 1

0 - 2/15=-2/15

3/15 - -1/15

= 4/15

80 - 100/15= 73.33

a. S1 Row multiplied by Pivot Column Element of X2 row.

b. X2 Row minus row a. above.



X1 X2 S1 S2

Basis Cj -125 -400 0 0 Bi

X1 -125 1 0 2/3 -1/3 33.33

X2 -400 0 1 -2/15 4/15 73.33

Zj-125 * 1 + -400 * 0 =

-125

-125 * 0 + -400 * 1

= -400

(-125*2/3) + (-400 * -

2/15) = -30

(-125 * -1/3) + (-

400 * 4/15)= - 65

(-125*33.33) + (-400*73.33)=

-33,500

Cj - Zj -125 - - 125 = 0

-400 - - 400 = 0

0 - - 30 = 30

0 - -65= 65



All (Cj-Zj) values are positive so no entering variable…All Bi are positive...

X1 X2 S1 S2

Basis Cj -125 -400 0 0 Bi

X1 -125 1 0 2/3 -1/3 33.33

X2 -400 0 1 -2/15 4/15 73.33

Zj -125 -400 -30 -65 -33,500

Cj - Zj 0 0 30 65



We have a solution we can read from the TableauX1 number of chairs = 33.33X2 number of tables = 73.33Revenue = $33,500S1 = 0 , S2 = 0 : They are Non-Basic and Non-Basic Variableshave value zero!

X1 X2 S1 S2

Basis Cj -125 -400 0 0 Bi

X1 -125 1 0 2/3 -1/3 33.33

X2 -400 0 1 -2/15 4/15 73.33

Zj -125 -400 -30 -65 -33,500

Cj - Zj 0 0 30 65

Documents

The Simplex Method - Andrew David Joseph Hall · 2 Quantitative Methods for Economics and Business I The Simplex Method Formulate Constrained Maximization or Minimization Problem