A furniture Company produces tables and chairs. Each table takes four hours of labor from the carpentry department and two hours of labor from the finishing department. Each chair requires three hours of carpentry and one hour of finishing. During the current week, 260 hours of carpentry time are available and 120 hours of finishing time. Each table produced gives a profit of $60 and each chair a profit of $40. How many chairs and tables should be made?

1) Propose an LP formulation for this model

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2) Solve this model using the graphical method

3) Solve this model using the simplex method

clearly explain all steps

The Simplex Method

The geometric method of solving linear programming problems

presented before. The graphical method is useful only for

problems involving two decision variables and relatively few

problem constraints.

What happens when we need more decision variables and more problem constraints?

We use an algebraic method called the simplex method, which

was developed by George B. DANTZIG (1914-2005) in 1947 while

on assignment with the U.S. Department of the air force.

Standard Maximization Problems in

Standard Form

A linear programming problem is said to be a standard maximization problem in

standard form if its mathematical model is of the following form:

Maximize the objective function

Z max = P = c1x1 + c2 x2 + … + cn xn

Subject to problem constraints of the form

a1x1 + a2 x2 + … + an xn b

,b 0

With non-negative constraints

x1, x2 ,…, xn 0

Slack Variables

“A mathematical representation of surplus

resources.” In real life problems, it’s unlikely

that all resources will be used completely, so

there usually are unused resources.

Slack variables represent the unused resources

between the left-hand side and right-hand

side of each inequality.

Basic and Nonbasic Variables

Basic variables are selected arbitrarily with the restriction that

there be as many basic variables as there are equations. The

remaining variables are non-basic variables.

x1 + 2 x2 + s1

= 32

3×1 + 4 x2 + s2 = 84

This system has two equations, we can select any two of the four

variables as basic variables. The remaining two variables are

then non-basic variables. A solution found by setting the two

non-basic variables equal to 0 and solving for the two basic

variables is a basic solution. If a basic solution has no negative

values, it is a basic feasible solution.

SIMPLEX METHOD

Step-1

Write the

standard

maximization

problem in

standard form,

introduce slack

variables to form

the initial system,

and write the

initial tableau.

Step 2

Are there

any

negative

indicators

in the

bottom

row?

STOP

The optimal solution has been found.

Step-3

Select

the

pivot

column

Step 4

Are there

any positive

elements in

the pivot

column

above the

dashed

line?

STOP

The linear programming problem has

no optimal solution

Simplex algorithm for standard maximization problems

Step-5

Select

the pivot

element

and

perform

the pivot

operatio

n

To solve a linear programming problem in standard form, use the following steps.

1- Convert each inequality in the set of constraints to an equation by adding slack

variables.

2- Create the initial simplex tableau.

3- Select the pivot column. ( The column with the “most negative value” element

in the last row.)

4- Select the pivot row. (The row with the smallest non-negative result when the

last element in the row is divided by the corresponding in the pivot column.)

5-Use elementary row operations calculate new values for the pivot row so that

the pivot is 1 (Divide every number in the row by the pivot number.)

6- Use elementary row operations to make all numbers in the pivot column equal

to 0 except for the pivot number. If all entries in the bottom row are zero or

positive, this the final tableau. If not, go back to step 3.

7- If you obtain a final tableau, then the linear programming problem has a

maximum solution, which is given by the entry in the lower-right corner of

the tableau.

Pivot

Pivot Column: The column of the tableau

representing the variable to be entered into

the solution mix.

Pivot Row: The row of the tableau representing

the variable to be replaced in the solution mix.

Pivot Number: The element in both the pivot

column and the pivot row.

Simplex Tableau

Most real-world problems are too complex to

solve graphically. They have too many corners

to evaluate, and the algebraic solutions are

lengthy. A simplex tableau is a way to

systematically evaluate variable mixes in order

to find the best one.

Initial Simplex Tableau

All variables

Basic variables

Solution

coefficients

0

EXAMPLE

The Cannon Hill furniture Company produces tables

and chairs. Each table takes four hours of labor

from the carpentry department and two hours of

labor from the finishing department. Each chair

requires three hours of carpentry and one hour

of finishing. During the current week, 240 hours

of carpentry time are available and 100 hours of

finishing time. Each table produced gives a profit

of $70 and each chair a profit of $50. How many

chairs and tables should be made?

STEP 1

All information about example

Resource

Table s ( x1 )

Chairs (x2 )

Constraints

Carpentry (hr)

4

3

240

Finishing (hr)

2

1

100

$70

$50

Unit Profit

Objective Function

Carpentry Constraint

Finishing Constraint

Non-negativity conditions

P = 70 x1 + 50 x2

4 x1 + 3×2 240

2 x1 + 1×2 100

x1, x2 0

The first step of the simplex method requires that each inequality

be converted into an equation. ”less than or equal to”

inequalities are converted to equations by including slack

variables.

Suppose s1 carpentry hours and s2 finishing hours remain unused

in a week. The constraints become;

4 x1 + 3×2 + s1 = 240

2 x1 + x2 + s2 = 100

or

4 x1 + 3×2 + s1 + 0s2 = 240

2 x1 + x2 + 0s1 + s2 = 100

As unused hours result in no profit, the slack variables can be

included in the objective function with zero coefficients:

P = 70 x1 + 50 x2 + 0s1 + 0s2

P − 70 x1 − 50 x2 − 0s1 − 0s2 = 0

The problem can now be considered as solving a system of 3 linear

equations involving the 5 variables x1, x2 , s1, s2 , P in such a way

that P has the maximum value;

4 x1 + 3×2 + s1 + 0s2 = 240

2 x1 + x2 + 0s1 + s2 = 100

P − 70 x1 − 50 x2 − 0s1 − 0s2 = 0

Now, the system of linear equations can be written in matrix form

or as a 3×6 augmented matrix. The initial tableau is;

STEP 2

Basic

Variables

x1

x2

S1

S2

P

Right

Hand

Side

S1

4

3

1

0

0

240

S2

P

2

-70

1

-50

0

0

1

0

0

1

100

0

The tableau represents the initial solution;

x1 = 0, x2 = 0, s1 = 240, s2 = 100, P = 0

The slack variables S1 and S2 form the initial solution mix. The initial

solution assumes that all avaliable hours are unused. i.e. The slack variables

take the largest possible values.

Variables in the solution mix are called basic variables. Each basic

variables has a column consisting of all 0’s except for a single 1.

all variables not in the solution mix take the value 0.

The simplex process, a basic variable in the solution mix is

replaced by another variable previously not in the solution

mix. The value of the replaced variable is set to 0.

STEP 3

Select the pivot column (determine which variable to enter into the

solution mix). Choose the column with the “most negative”

element in the objective function row.

Basic

Variables

x1

x2

S1

S2

P

Right

hand

side

S1

S2

P

4

2

-70

3

1

-50

1

0

0

0

1

0

0

0

1

240

100

0

Pivot column

x1 should enter into the solution mix because each unit of x1 (a table)

contributes a profit of $70 compared with only $50 for each unit of x1 (a

chair)

Step 4

No, There aren’t any positive elements in the

pivot column above the dashed line.

We can go on step 5

STEP 5

Select the pivot row (determine which variable to replace in the solution mix).

Divide the last element in each row by the corresponding element in the

pivot column. The pivot row is the row with the smallest non-negative

result.

Enter

Exit

Basic

Variables

x1

x2

S1

S2

P

S1

S2

P

4

2

-70

3

1

-50

1

0

0

0

1

0

0

0

1

Right

hand

side

240 240/ 4 = 60

100 100/ 2 = 50

0

Pivot row

Pivot column

Pivot number

Should be replaced by x1 in the solution mix. 60 tables can be made with 240

unused carpentry hours but only 50 tables can be made with 100 finishing

hours. Therefore we decide to make 50 tables.

Now calculate new values for the pivot row. Divide every number in the row

by the pivot number.

Basic

Variables

x1

x2

S1

S2

P

S1

x1

P

4

1

-70

3

1/2

-50

1

0

0

0

1/2

0

0

0

1

Right

hand

side

240

50

0

R2

2

Use row operations to make all numbers in the pivot column equal to 0 except

for the pivot number which remains as 1.

Basic

Variables

x1

x2

S1

S2

P

S1

x1

P

0

1

0

1

1/2

-15

1

0

0

-2

1/2

35

0

0

1

Right

hand

side

40

−4.R2 + R1

50

3500

70.R2 + R3

If 50 tables are made, then the unused carpentry hours are reduced by 200

hours (4 h/table multiplied by 50 tables); the value changes from 240 hours to 40

hours. Making 50 tables results in the profit being increased by $3500; the value

changes from $0 to $3500.

x1 = 50, x2 = 0, s1 = 40, s2 = 0, P = 3500

In this case,

Now repeat the steps until there are no negative numbers in the last row.

Select the new pivot column. x2 should enter into the solution mix.

Select the new pivot row. S1 should be replaced by x2 in the solution mix.

Enter

Exit

Basic

Variables

x1

x2

S1

S2

P

S1

0

1

1

-2

0

Right

hand

side

40

x1

P

1

0

1/2

-15

0

0

1/2

35

0

1

50

3500

40/1 = 40

50/ 0,5 = 100

New pivot row

New pivot

column

Calculate new values for the pivot row. As the pivot number is already 1,

there is no need to calculate new values for the pivot row.

Use row operations to make all numbers in the pivot column equal to

except for the pivot number.

Basic

Variables

x1

x2

S1

S2

P

x2

x1

P

0

1

0

1

0

0

1

-1/2

15

-2

3/2

5

0

0

1

Right

hand

side

40

30

4100

1

− .R1 + R2

2

15.R1 + R3

If 40 chairs are made, then the number of tables are reduced by

20 tables (1/2 table/chair multiplied by 40 chairs); the value

changes from 50 tables to 30 tables. The replacement of 20

tables by 40 chairs results in the profit being increased by

$600; the value changes from $3500 to $4100.

As the last row contains no negative numbers, this solution gives

the maximum value of P.

Result

This simplex tableau represents the optimal

solution to the LP problem and is interpreted

as:

x1 = 30,

x2 = 40,

s1 = 0,

s2 = 0

and profit or P=$4100

The optimal solution (maximum profit to be

made) is to company 30 tables and 40 chairs

for a profit of $4100.

Example-2

A farmer owns a 100 acre farm and plans to plant at

most three crops. The seed for crops A,B, and C costs

$40, $20, and $30 per acre, respectively. A maximum

of $3200 can be spent on seed. Crops A,B, and C

require 1,2, and 1 workdays per acre, respectively,

and there are maximum of 160 workdays available. If

the farmer can make a profit of $100 per acre on

crop A, $300 per acre on crop B, and $200 per acre

on crop C, how many acres of each crop should be

planted to maximize profit?

The Dual Problem: Minimization with

problem constraints of the form ≥

• Linear programming problems exist in pairs. That is in

linear programming problem, every maximization

problem is associated with a minimization problem.

Conversely, associated with every minimization problem

is a maximization problem. Once we have a problem

with its objective function as maximization, we can write

by using duality relationship of linear programming

problems, its minimization version. The original linear

programming problem is known as primal problem, and

the derived problem is known as dual problem.

Thus, the dual problem uses exactly the same parameters

as the primal problem, but in different locations. To

highlight the comparison, now look at these same two

problems in matrix notation.

Primal Problem

Dual Problem

Minimize

Z=cx

Maximize

W=yb

Subject to

Ax≥b

Subject to

yAc

and

x≥0

And

Primal problem

A=

𝑎11

𝑏11

𝑎12

𝑏12

𝑎13

𝑏13

𝑐11

𝑐12

𝑐13

y≥0

Dual problem

T

A=

𝑎11

𝑎12

𝑏11

𝑏12

𝑐11

𝑐12

𝑎13

𝑏13

𝑐13

As an example,

Primal Problem in

algebraic form

Minimize

C=3×1+5×2

Subject to

and

x1 ≥ 4

2×2 ≥ 12

3×1+2×2 ≥18

Primal problem

1

A=

0

Dual problem

4

0

2

12

3

2

18

3

5

1

x1≥0, x2≥0

Consequently, (1) the parameters for a constraint

in either problem are the coefficients of a variable

in the other problem and (2) the coefficients for

the objective function of either problem are the

right sides for the other problem.

T

A=

1

0

0

2

3

2

3

5

4

12

18

1

Dual Problem in algebraic

form

Maximize

Z=4y1+12y2+18y3

Subject to

and

y1+3y3 3

2y2+2y3 5

y1≥0 , y2≥0 ,y3≥0

Summary

Primal

Dual

(a) Maximize.

Minimize

(b) Objective Function.

Right hand side.

(c) Right hand side.

Objective function.

(d) i th row of input-output

coefficients.

i th column of input output

coefficients.

(e) j th column of input-output

coefficients.

j the row of input-output

coefficients.

WORKED – OUT PROBLEM 1

The procedure for forming the dual problem is

summarized in the box below:

Formation of the Dual Problem

Given a minimization problem with problem

constraints,

Step 1. Use the coefficients and constants in the

problem constraints and the objective function to

form a matrix A with the coefficients of the objective function in the last row.

Step 2. Interchange the rows and columns of

T

matrix A to form the matrix A , the transpose of A.

T

Step 3. Use the rows of A to form a maximization

problem with problem constraints.

Forming the Dual Problem

Minimize C =

subject to

40×1 + 12×2 + 40×3

2×1 + x2 + 5×3 ≥ 20

4×1 + x2 + x3 ≥ 30

x1, x2, X3 ≥ 0

WORKED –OUT PROBLEM 2

Form the dual problem:

Minimize C = 16 x1 + 9×2 + 21×3

subject to

x1 + x2 + 3×3 ≥ 12

2×1 + x2 +x3 ≥ 16

x1, x2, x3 ≥ 0

Solution of Minimization Problems

ORIGINAL PROBLEM (1)

DUAL PROBLEM (2)

Minimize C = 16×1 + 45×2

Maximize P = 50y1 + 27y2

subject to 2×1 + 5×2 ≥ 50

x1 + 3×2 ≥ 27

x1, x2 ≥ 0

subject to 2y1 + y2 16

5y1 + 3y2 45

y1,y2 ≥ 0

MINIMIZATION PROBLEMS

– The Dual Form

– Graphical Approach

– Solution of Minimization Problems with

Simplex Method

– A Transportation Problem

– The Big M method

– Minimization by The Big M Method

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