The Algebraic Technique for Solving LP s. Métodos Cuantitativos M. En C. Eduardo Bustos Farías 1

The Algebraic Technique for

Solving LP’s

Example Problem

Maximize Z = 5x ₁ + 2x ₂ + x ₃

subject to

x ₁ + 3x ₂ - x ₃ ≤ 6,

x ₂ + x ₃ ≤ 4,

3x ₁ + x 2 ≤ 7,

x ₁ , x ₂ , x ₃ ≥ 0.

Simplex and

Example Problem

Step 1. Convert to Standard Form a ₁₁ x ₁ + a ₁₂ x ₂ + ••• + a _1n x _n ≤ b ₁ ,

a ₂₁ x ₁ + a ₂₂ x ₂ + ••• + a _2n x _n ≥ b ₂ ,

a _m1 x ₁ + a _m2 x ₂ + ••• + a _mn x _n ≤ b _m ,

#

a ₁₁ x ₁ + a ₁₂ x ₂ + ••• + a _1n x _n + x x _{n+1 n+1} = b ₁ , a ₂₁ x ₁ + a ₂₂ x ₂ + ••• + a _2n x _n - x x _{n+2 n+2} = b ₂ ,

a _m1 x ₁ + a _m2 x ₂ + ••• + a _mn x _n + x x _{n+k n+k} = b _m , In our example problem:

x ₁ + 3x ₂ - x ₃ ≤ 6, x ₂ + x ₃ ≤ 4, 3x ₁ + x 2 ≤ 7, x ₁ , x ₂ , x ₃ ≥ 0.

x ₁ + 3x ₂ - x ₃ + x ₄ = 6,

x ₂ + x ₃ + x ₅ = 4,

3x ₁ + x 2 + x ₆ = 7,

x ₁ , x ₂ , x ₃ , x ₄ , x ₅ , x ₆ ≥ 0.

Simplex: Step 2

Step 2. Start with an initial basic feasible solution (b.f.s.) and set up the initial tableau.

In our example

Maximize Z = 5x ₁ + 2x ₂ + x ₃ x ₁ + 3x ₂ - x ₃ + x ₄ = 6,

x ₂ + x ₃ + x ₅ = 4, 3x ₁ + x 2 + x ₆ = 7,

x ₁ , x ₂ , x ₃ , x ₄ , x ₅ , x ₆ ≥ 0. ^c

5 2 1 0 0 0

c

Basis

x

x

x

x

x

x

Constants

0 x

1 3 -1 1 0 0 6

0 x

0 1 1 0 1 0 4

Step 2: Explanation

Adjacent Basic Feasible Solution

If we bring a nonbasic variable x _s into the basis, our system changes from the basis, x _b , to the following:

x ₁ + ā _1s x _s = x _r + ā _rs x _r =

x _m + ā _ms x _s =

b 1

b r

b s

#

#

x _i = b _i − a _is for i =1, …, m x _s = 1

x _j = 0 for j=m+1, ..., n and j≠s The new value of the objective function becomes:

∑ =

+

−

= ^m

1 i

s is

i

i ( b a ) c

c Z

Thus the change in the value of Z per unit increase in x _s is

c s = new value of Z - old value of Z

= ∑ ∑

=

=

− +

− ^m

1 i

i i m

1 i

s is

i

i ( b a ) c c b

c

∑ =

−

m

1 i

is i

s c a

c

This is the Inner Product rule

=

Simplex:

Step 3

Use the inner product rule to find the relative profit coefficients

c

5 2 1 0 0 0

c

Basis

x

x

x

x

x

x

Constants

0 x

1 3 -1 1 0 0 6

0 x

0 1 1 0 1 0 4

0 x

3 1 0 0 0 1 7

c row 5 2 1 0 0 0 Z=0

j B j

j c c P

c = −

c ₁ = 5 - 0(1) - 0(0) - 0(3) = 5 -> largest positive c ₂ = …. 2

c ₃ = …. 1

Simplex: Step 5

Apply the minimum ratio rule to determine the basic variable to leave the basis.

The new values of the basis variables:

x _i = b _i − a _is x _s for i = 1, ..., m

⎥ ⎦

⎢ ⎤

⎣

= ⎡

> is i 0 s a

a min b x

max

is

In our example:

c

5 2 1 0 0 0

c

Basis

x

x

x

x

x

x

Constants

0 x

1 3 -1 1 0 0 6

0 x

0 1 1 0 1 0 4

0 x

3 1 0 0 0 1 7

c row 5 2 1 0 0 0 Z=0

Row Basic Variable Ratio

1 x ₄ 6

2 x ₅ -

3 x ₆ 7/3

c

5 2 1 0 0 0

c

Basis

x

x

x

x

x

x

Constants

0 x

0 8/3 -1 1 0 0 11/3

0 x

0 1 1 0 1 0 4

5 x 1 1/3 0 0 0 1/3 7/3

Simplex:

Step 6

Perform the pivot operation to get the new tableau and the b.f.s.

c

5 2 1 0 0 0

c

Basis

x

x

x

x

x

x

Constants

0 x

1 3 -1 1 0 0 6

0 x

0 1 1 0 1 0 4

0 x

3 1 0 0 0 1 7

c row 5 2 1 0 0 0 Z=0

j B j

j c c P

c = −

c _B = (0 0 5) New iteration:

find entering

variable:

Final Tableau

c

5 2 1 0 0 0

c

Basis

x

x

x

x

x

x

Constants

0 x

0 8/3 -1 1 0 0 11/3

0 x

0 1 1 0 1 0 4

5 x

1 1/3 0 0 0 1/3 7/3

c row 0 1/3 1 0 0 -5/3 Z=35/3

c

5 2 1 0 0 0

c

Basis

x

x

x

x

x

x

Constants

0 x

0 4 0 1 1 0 23/3

1 x

0 1 1 0 1 0 4

5 x

1 1/3 0 0 0 1/3 7/3

c row 0 -2/3 0 0 -1 -5/3 Z=47/3 x ₃ enters basis, x ₅ leaves basis

Wrong value!

4 should be 11/3

Final Tableau

c

5 2 1 0 0 0

c

Basis

x

x

x

x

x

x

Constants

0 x

0 4 0 1 1 0 23/3

1 x

0 1 1 0 1 0 4

5 x

1 1/3 0 0 0 1/3 7/3

c row 0 -2/3 0 0 -1 -5/3 Z=47/3

X1= 7/3

X2 = 0

X3 = 4

X4 = 23/3

X5 = 0

Métodos Cuantitativos M. En C. Eduardo Bustos Farías 11

Exercise: Solving an LP-problem:

A California vintner has available 660 lbs of

Cabernet Sauvignon (CS) grapes, 1860 lbs of Pinot Noir (PN) grapes, and 2100 lbs of Barbera (B)

grapes. The vintner makes a Pinot Noir (PN) wine, which contains 20% CS, 60% PN, and 20% B grapes and sells $3 a bottle, and a Barbera (B) wine, which contains 10% CS, 20% PN, and 70% B grapes and sells for $2 a bottle. Assuming each bottle of wine requires 3 lbs of grapes, determine how many bottles of each type of wine should be produced to

maximize income.

Solving an LP-problem ^continued :

• Algebraic Technique

¾ The graphical method solving linear programming problems can be used for problems with two

variables (with some difficulty three)

¾ However, for problems were the number of

variables might run into hundreds or thousands, algebraic techniques must be used

¾ The simplex method, with the aid of the computer,

can solve these problems

As with the graphical procedure, the

simplex method finds the optimal corner- point solution of the set of feasible

solutions.

Regardless of the number of decision

variables and regardless of the number of constraints, the simplex method uses the key property of a linear programming

problem, which is:

• A linear programming problem always has an optimal solution occurring at a corner-point solution

Simplex method begins with a feasible

solution and tests whether or not it is

optimum.

Solution of a maximum-type linear programming problem by the

simplex algorithm involves the following steps:

1. Adding slack variables to convert the inequalities into equations

In the case of less-than-or-equal-to

constraints, slack variables are used to

increase the left-hand side to equal the right-

hand side limits of the constraint conditions.

Example: max Z = 3x ₁ + x ₂ st. 2x ₁ + x ₂ ≤ 8

2x ₁ + 3x ₂ ≤ 12

adding slacks and rewriting the object row:

2x ₁ + x ₂ + s ₁ = 8 2x ₁ + 3x ₂ + s ₂ = 12

-3x – x + Z = 0

2x ₁ + x ₂ + s ₁ = 8 2x ₁ + 3x ₂ + s ₂ = 12 -3x ₁ – x ₂ + Z = 0

2. Setting up the initial simplex tableau Form an augmented coefficient matrix:

⎥ ⎥

⎥

⎦

⎤

⎢ ⎢

⎢

⎣

⎡

−

− 3 1 0 0 1 0 12 0

1 0

3 2

8 0

0 1

1 2

2 1

2 1

2 1

Z s

s

Z s

s x

x

3. Finding an initial feasible solution

Simplex algorithm starts always from origin. In our case it means that the initial feasible solution is:

x ₁ = 0, x ₂ = 0, s ₁ = 8, s ₂ = 12.

Simplex method proceeds from this

corner point to an adjacent corner

point. Such corner points are called

4. Introducing basic and nonbasic

variables as the natural way to express basic feasible solutions

For any basic feasible solution (B.F.S.), the

variables held zero are called nonbasic variables and the other are called basic variables.

Moving form one B.F.S. to another means that one basic variable (we call it a departing or

exiting variable) becomes nonbasic and a

nonbasic variable (entering variable) becomes a

basic variable.

5. Choosing the proper pivot element to advance the solution and maintain the non-negativity of all variables

How to decide which variable to make

basic and which nonbasic (in other words in which direction to move from the current

B.F.S.)?

The natural direction is the one in which the value of the objective function

increases the most:

If in Z = 3x ₁ + x ₂ , x ₁ is allowed to become basic, x ₂ remains at 0 and Z = 3x ₁ ; thus for each one- unit increase in x ₁ , Z increases by 3 units. On the other hand, if x ₂ is allowed to become basic, Z will increase only by one unit if x ₂ is

increased by one unit. This is one way to

determine the pivot column.

Métodos Cuantitativos M. En C. Eduardo Bustos Farías 23

Choosing the proper pivot element continued

Another way to determine the pivot column is to examine the bottom row of the simplex

tableau:

⎥ ⎥

⎥

⎦

⎤

⎢ ⎢

⎢

⎣

⎡

−

− 3 1 0 0 1 0

12 0

1 0

3 2

8 0

0 1

1 2

2 1

2 1

2 1

Z s

s

Z s

s x

x

indicators entering

variable

The bottom row entries to the left of the vertical line are called indicators.

We choose the column with the most negative indicator as the pivot column.

Having chosen the pivot column, we must now determine the pivot row in order to know which element in our

simplex tableau is the pivoting element.

For that purpose we divide the right hand side entries of the simplex tableau by

corresponding entries of the pivot column.

Of the resulting quotients we choose the smallest (minimum quotient). So the pivot element is the intersection of the pivot

column and pivot row.

The pivot element is the intersection of the pivot column and pivot row:

⎥ ⎥

⎥

⎦

⎤

⎢ ⎢

⎢

⎣

⎡

−

− 3 1 0 0 1 0

12 0

1 0

3 2

8 0

0 1

1 2

2 1

2 1

2 1

Z s

s

Z s

s x

x _Quotients

8 ÷ 2 = 4 (smaller) 12 ÷ 2 = 6

departing variable

entering variable (most negative

indicator)

Since x ₁ and s ₂ will be the basic

variables in our new B.F.S, it would be convenient to change our previous

tableau by elementary row operations into a form where the values of x ₁ , s ₂ ,

and Z can be read off with ease just as in

the initial basic solution.

To do this we want to find a matrix

which is equivalent to the tableau above but which has the form:

⎥ ⎥

⎥

⎦

⎤

⎢ ⎢

⎢

⎣

⎡

? 1

0

?

? 0

? 0

1

?

? 0

? 0

0

?

? 1

2 1

2

1 x s s Z

x

where the question marks represent

6. Pivoting, which is done column wise.

We must transform the tableau to an equivalent matrix that has a 1 at the place of the pivot

element (pivot entry) and 0’s elsewhere in the column x ₁ .

⎥ ⎥

⎥

⎦

⎤

⎢ ⎢

⎢

⎣

⎡

−

− 3 1 0 0 1 0

12 0

1 0

3 2

8 0

0 1

1 2

Z s s

Z s

s x

x

2 1

2 1

2 1

departing *

variable

entering variable

By elementary row operations, we have:

⎤

⎡

⎥ ⎥

⎥

⎦

⎤

⎢ ⎢

⎢

⎣

⎡

−

−

⎥ ⎥

⎥

⎦

⎤

⎢ ⎢

⎢

⎣

⎡

−

−

4 0

0 1

Z s

s x

x

0 1

0 0

1 3

12 0

1 0

3 2

4 0

0 1

0 1

0 0

1 3

12 0

1 0

3 2

8 0

0 1

1 2

Z s

s x

x

1 1

2 1

2 1

2 1 2

1

2 1

2 1

½ R ₁

7. Continuing and recognizing when the algorithm terminates.

General termination rule: all values in the

indicator row are non-negative.

We have a new simplex tableau:

⎥ ⎥

⎥

⎦

⎤

⎢ ⎢

⎢

⎣

⎡

−

12 1

0 0

4 0

1 1

2 0

4 0

0 1

Z s x

Z s

s x

x

2 3 2

1

2 1 2

1

2 1

2 1

2 1

indicators

All values in the indicator row are non-

negative. Hence, we have found the optimal

solution for the problem.

Alternate Optimal solutions

For some LPs, more than one extreme point is optimal.

If an LP has more than one optimal solution, it has multiple optimal solutions.

If there is no nonbasic variable (NBV) with a zero

coefficient in row 0 of the optimal tableau, the LP has a unique optimal solution. Even if there is a nonbasic variable with a zero coefficient in row 0 of the optimal tableau, it is possible that the LP may not have

alternative optimal solutions.

Unbounded LPs

For some LPs, there exist points in the feasible region for which z assumes arbitrarily large (in max

problems) or arbitrarily small (in min problems) values. When this occurs, we say the LP is

unbounded.

An unbounded LP occurs in a max problem if there is a nonbasic variable with a negative coefficient in row 0 and there is no constraint that limits how large we can make this NBV.

Specifically, an unbounded LP for a max problem occurs

when a variable with a negative coefficient in row 0 has a

The LINDO Computer Package

LINDO (Linear Interactive and Discrete Optimizer) is a user friendly computer package that can be used to solve linear, integer, and quadratic programming problems.

The Dakota Furniture LP will solved using LINDO as an example.

LINDO permits the user to name the variables so they are defined:

DESKS = number of desk produced TABLES = number of tables produced CHAIRS = number of chairs produced

LINDO assumes all variables are nonnegative, so nonnegative constraints are not necessary. To be consistent with LINDO, the objective function row is labeled row 1 and constraints rows 2-5.

View the LINDO Help file for syntax questions

The LINDO Computer Package

To enter this problem into LINDO make sure the screen contains a blank window (work area) with the word “untitled” at the top of the work area. If necessary, a new window can be opened by selecting New from the file menu or by clicking on the File button.

The first statement in a LINDO model is always the objective function. MAX or MIN directs LINDO to solve a maximization or minimization problem.

Enter the constraints by typing SUBJECT TO (or st) on the next

line. Then enter the constraints. LINDO assumes the < symbol

means ≤ and the > symbol means ≥. There is no need to insert

the asterisk symbol between coefficients and variables to indicate

multiplication. The Dakota input is shown on the next slide

The LINDO Computer Package

The LINDO Computer Package

To save the file (LP formulation) for future use, select SAVE from the File menu and when asked, replace the * symbol with the name of your choice. The file will be saved with the name you select with the suffix .LTX. DO NOT type over the .LTX suffix. Save using any path available.

To solve the model, from the Solve menu select the SOLVE command or click the red bulls eye button.

When asked if you want to do a range (sensitivity) analysis, choose no. Range or sensitivity analysis will be discussed AFTER. when the solution is complete, a display showing the status of the Solve command will be present. View the displayed information and select CLOSE.

The data input window will now be overlaying the Reports Window.

Click anywhere in the Reports Window to bring it to the foreground.

4.6 – The LINDO Computer Package

The LINDO Computer Package

The LINDO output shows:

• LINDO found the optimum solution after 2 iterations (pivots)

• The optimal z-value = 280

• VALUE gives the value of the variable in the optimal LP solution.

Thus the optimal solution calls for production of 2 desks, 0 tables, and 8 chairs.

• SLACK OR SURPLUS gives (by constraint row) the value of slack or excess in the optimal solution.

• REDUCED COST gives the coefficient in row 0 of the optimal tableau (in a max problem). The reduced cost of each basic variables must be 0. For NBVs, the reduced cost is the amount by which the optimal z-value will be reduced if the NBV is

increased by 1 unit (and all other NBVs remain 0).

Ejemplo 1. Tecnología Agrícola, S.A.

Maximización

Tecnología Agrícola, S.A. es una compañía fabricante de fertilizantes. El

gerente desea planear la combinación de sus dos mezclas a fin de obtener

las mayores utilidades. Las mezclas son

El modelo de programación lineal para este problema es:

Maximizar Z=18.5 X ₁ + 20 X ₂ Sujeto a

0.05 X ₁ + 0.05 X ₂ <= 1100

0.05 X ₁ + 0.10 X ₂ <= 1800

0.10 X ₁ + 0.05 X ₂ <= 2000

X ₁ , X ₂ >= 0

Conversión a la forma estándar (inecuación a ecuación):

Z = 18.5X1+20X2+0S1+0S2+0S3 Sujeto a

0.05 X1 +0.05 X2+S1+0S2+0S3 =1100 0.05 X1 +0.10 X2+0S1+S2+0S3 = 1800 0.10 X1 +0.05 X2+0S1+0S2+S3 = 2000

X1, X2, S1, S2, S3 =0

Métodos Cuantitativos M. En C. Eduardo Bustos Farías 45

La tabla del simplex algebraico

Variables no básicas

Variables

básicas X1 X2 S1 S2 S3 _solución ^Valor

S1 aij aij 1 0 0 b1

S2 aij aij 0 1 0 b2

S3 aij aij 0 0 1 b3

Zj 0 0 0 0 0 0

Cj-Zj C1 C2 0 0 0

Variables básicas

Zj= contribución que se pierde por unidad fabricada Cj-Zj= costo de oportunidad

Si= variables de holgura

Xi= variables de decisión

Elaborar la tabla simplex que permita obtener la primera solución básica

factible

Variables

básicas X1 X2 S1 S2 S3 Valor

solución

S1 0.05 0.05 1 0 0 1100

S2

S3

Zj

Cj-Zj

Z = 18.5X1+20X2+0S1+0S2+0S3 Sujeto a

0.05 X1 +0.05 X2+S1+0S2+0S3 =1100 0.05 X1 +0.10 X2+0S1+S2+0S3 = 1800 0.10 X1 +0.05 X2+0S1+0S2+S3 = 2000

X1, X2, S1, S2, S3 >=0

Llenado de la tabla:

Renglón S1

Variables

básicas X1 X2 S1 S2 S3 Valor

solución

S1 0.05 0.05 1 0 0 1100

S2 0.05 0.1 0 1 0 1800

S3

Zj

Z = 18.5X1+20X2+0S1+0S2+0S3 Sujeto a

0.05 X1 +0.05 X2+S1+0S2+0S3 =1100 0.05 X1 +0.10 X2+0S1+S2+0S3 = 1800 0.10 X1 +0.05 X2+0S1+0S2+S3 = 2000

X1, X2, S1, S2, S3 >=0

Llenado de la tabla:

Renglón S2

Variables

básicas X1 X2 S1 S2 S3 Valor

solución

S1 0.05 0.05 1 0 0 1100

S2 0.05 0.1 0 1 0 1800

S3 0.1 0.05 0 0 1 2000

Zj

Cj-Zj

Z = 18.5X1+20X2+0S1+0S2+0S3 Sujeto a

0.05 X1 +0.05 X2+S1+0S2+0S3 =1100 0.05 X1 +0.10 X2+0S1+S2+0S3 = 1800 0.10 X1 +0.05 X2+0S1+0S2+S3 = 2000

X1, X2, S1, S2, S3 >=0

Llenado de la tabla:

Renglón S3

Variables

básicas X1 X2 S1 S2 S3 Valor

solución

S1 0.05 0.05 1 0 0 1100

S2 0.05 0.1 0 1 0 1800

S3 0.1 0.05 0 0 1 2000

Zj 0 0 0 0 0 0

Z = 18.5X1+20X2+0S1+0S2+0S3 Sujeto a

0.05 X1 +0.05 X2+S1+0S2+0S3 =1100 0.05 X1 +0.10 X2+0S1+S2+0S3 = 1800 0.10 X1 +0.05 X2+0S1+0S2+S3 = 2000

X1, X2, S1, S2, S3 >=0

Llenado del

Renglón Zj

Variables

básicas X1 X2 S1 S2 S3 Valor

solución

S1 0.05 0.05 1 0 0 1100

S2 0.05 0.1 0 1 0 1800

S3 0.1 0.05 0 0 1 2000

Zj 0 0 0 0 0 0

Cj-Zj 18.5 20 0 0 0

Z = 18.5X1+20X2+0S1+0S2+0S3 Sujeto a

0.05 X1 +0.05 X2+S1+0S2+0S3 =1100 0.05 X1 +0.10 X2+0S1+S2+0S3 = 1800 0.10 X1 +0.05 X2+0S1+0S2+S3 = 2000

X1, X2, S1, S2, S3 >=0

Llenado de la tabla:

Renglón

Cj-Zj

Determinar la variable que

ingresa y la que la sale.

Regla de entrada (criterio de optimalidad):

Entra aquella variable no básica con la mayor ganancia unitaria (en el caso de MAX) o el menor costo unitario (en el caso de MIN).

En nuestro ejemplo: comparo los valores

de la fila Cj-Zj, determino la columna

pivote, entra X2 por que 20 es mayor

que 18.5

PASO 1. Columna pivote Entra X2 = 20 > ^{x1 = 18.5}

Entra X2 a la base

Variables

básicas

X1 X2 ^{S1 S2 S3 Valor}

solución

S1 0.05 0.05 1 0 0 1100

S2 0.05 0.1 0 1 0 1800

S3 0.1 0.05 0 0 1 2000

Zj 0 0 0 0 0 0

Cj-Zj 18.5 20 ^{0 0 0}

Métodos Cuantitativos M. En C. Eduardo Bustos Farías 55

Regla de salida (criterio de factibilidad):

Sale aquella variable cuyo resultado de dividir el valor solución entre el coeficiente de la columna pivote sea menor.

En nuestro ejemplo sale S2, ya que al

dividir es el que tiene menor valor

positivo (ceros y negativos e infintos

no cuentan)

Sale S2 de la base: Divido la

columna de valor solución entre X2 (la variable que entra a la base)

Variables

básicas

X1 X2 S1 S2 S3 Valor

solución

S1 0.05 0.05 1 0 0 1100

S2 0.05 0.1 ^{0 1 0 1800}

S3 0.1 0.05 0 0 1 2000

Zj 0 0 0 0 0 0

Cj-Zj 18.5 20 0 0 0

1100/0.05=22000

1800/0.1=18000

2000/0.05=40000

Métodos Cuantitativos M. En C. Eduardo Bustos Farías 57

Se identifica el elemento pivote:

intersección entre x2 y s2 (la variable que sale y la que entra a la base)

Variables

básicas

X1 X2 S1 S2 S3 Valor

solución

S1 0.05 0.05 1 0 0 1100

X2 0.05 0.1 0 1 0 1800

S3 0.1 0.05 0 0 1 2000

Zj 0 0 0 0 0 0

Cj-Zj 18.5 20 0 0 0

Sustituyo S2 por X2

Se reestructuran los valores de la

tabla 1

Transformo el renglón X2, que contiene el elemento pivote en uno: lo multiplico por su

inverso multiplicativo (10)

Variables

básicas

X1 X2 S1 S2 S3 Valor

solución

S1 0.05 0.05 1 0 0 1100

X2 10(0.05) 10(0.1) 10(0) 10(1) 10(0) 10(1800)

S3 0.1 0.05 0 0 1 2000

Zj 0 0 0 0 0 0

Cj-Zj 18.5 20 0 0 0

Quedando la tabla como sigue:

Variables

básicas

X1 X2 S1 S2 S3 Valor

solución

S1 0.05 0.05 1 0 0 1100

X2 0.5 1 0 10 0 18000

S3 0.1 0.05 0 0 1 2000

Zj 0 0 0 0 0 0

Cj-Zj 18.5 20 0 0 0

Los valores de las celdas de las variables básicas de la columna pivote (X2) deben valer cero y

también la Cj-Zj:

Variables

básicas

X1 X2 S1 S2 S3 Valor

solución

S1 0.05 0.05 1 0 0 1100

X2 0.5 1 0 10 0 18000

S3 0.1 0.05 0 0 1 2000

Zj 0 0 0 0 0 0

Cj-Zj 18.5 20 0 0 0

Uso el renglón pivote

Variables

básicas

X1 X2 S1 S2 S3 Valor

solución

S1 0.05 0.05 1 0 0 1100

X2 0.5 1 0 10 0 18000

S3 0.1 0.05 0 0 1 2000

Zj 0 0 0 0 0 0

Cj-Zj 18.5 20 0 0 0

Empezamos con el renglón S1, quiero convertir el 0.05 en cero. Busco que número multiplicado por uno y sumado con 0.05 da cero.

Variables

básicas

X1 X2 S1 S2 S3 Valor

solución

S1 0.05

0.05 ^{1 0 0 1100}

X2 0.5 1 0 10 0 18000

S3 0.1 0.05 0 0 1 2000

Zj 0 0 0 0 0 0

Cj-Zj 18.5 20 0 0 0

Multiplico por -0.05 el renglón pivote y lo sumo al renglón S1

Variables

básicas X1 X2 S1 S2 S3 Valor

solución

S1 (0.5)(-0.05) + 0.05

(1)(-0.05) + 0.05

(0)(-0.05) + 1

(10)(-0.05) + 0

(0)(-0.05) +

0

(18000)(-0.05) +

1100

X2 0.5 1 0 10 0 18000

S3 0.1 0.05 0 0 1 2000

Zj 0 0 0 0 0 0

Cj-Zj 18.5 20 0 0 0

Resultado los nuevos valores del renglón S1 en la Tabla 2 del simplex

Variables

básicas X1 X2 S1 S2 S3 Valor

solución

S1 0.025 0 1 -0.5 0 200

X2 0.5 1 0 10 0 18000

S3 0.1 0.05 0 0 1 2000

Zj 0 0 0 0 0 0

Cj-Zj 18.5 20 0 0 0

Ahora el renglón S3: quiero convertir en cero 0.05, busco qué número multiplicado por uno y

sumado a 0.05 da cero.

Variables

básicas X1 X2 S1 S2 S3 Valor

solución

S1 0.025 0 1 -0.5 0 200

X2 0.5 1 0 10 0 18000

S3 0.1

0.05 ^{0 0 1 2000}

Zj 0 0 0 0 0 0

Cj-Zj 18.5 20 0 0 0

Multiplico por el inverso aditivo de 0.05.

Multiplico el renglón pivote por -0.05 y lo sumo al renglón S3

Variables

básicas X1 X2 S1 S2 S3 Valor

solución

S1 0.025 0 1 -0.5 0 200

X2 0.5 1 0 10 0 18000

S3 (0.5)(-0.05) +

0.1

(1)(-0.05) + 0.05

(0)(-0.05) + 0

(10)(-0.05) + 0

(0)(-0.05) + 1

(18000)(-0.05)

+ 2000

Zj 0 0 0 0 0 0

Cj-Zj 18.5 20 0 0 0

Resultado

Variables

básicas X1 X2 S1 S2 S3 Valor

solución

S1 0.025 0 1 -0.5 0 200

X2 0.5 1 0 10 0 18000

S3 0.075 0 0 -0.5 1 1100

Zj 0 0 0 0 0 0

Cj-Zj 18.5 20 0 0 0

Ahora vamos a transformar el renglón Zj.

Multiplico el renglón pivote por 20 y lo sumo a Zj.

Variables

básicas X1 X2 S1 S2 S3 Valor

solución

S1 0.025 0 1 -0.5 0 200

X2 0.5 1 0 10 0 18000

S3 0.075 0 0 -0.5 1 1100

Zj 0 0 0 0 0 0

Cj-Zj 18.5

20 ^{0 0 0}

Las operaciones son:

Variables

básicas X1 X2 S1 S2 S3 Valor

solución

S1 0.025 0 1 -0.5 0 200

X2 0.5 1 0 10 0 18000

S3 0.075 0 0 -0.5 1 1100

Zj (0.5)(20) + 0

(1)(20) + 0

(0)(20) + 0

(10)(20) + 0

(0)(20) + 0

(18000)(20) +

0

El resultado es:

Variables

básicas X1 X2 S1 S2 S3 Valor

solución

S1 0.025 0 1 -0.5 0 200

X2 0.5 1 0 10 0 18000

S3 0.075 0 0 -0.5 1 1100

Zj 10 20 0 200 0 360000

Cj-Zj 18.5 20 0 0 0

Ahora seguimos con Cj-Zj, quiero convertir el 20 en cero.

Variables

básicas X1 X2 S1 S2 S3 Valor

solución

S1 0.025 0 1 -0.5 0 200

X2 0.5 1 0 10 0 18000

S3 0.075 0 0 -0.5 1 1100

Zj 10 20 0 200 0 360000

Cj-Zj 18.5 0 0 0

Multiplico el renglón pivote por -20 y lo sumo a Cj-Zj

Variables

básicas X1 X2 S1 S2 S3 Valor

solución

S1 0.025 0 1 -0.5 0 200

X2 0.5 1 0 10 0 18000

S3 0.075 0 0 -0.5 1 1100

Zj 10 20 0 200 0 360000

Cj-Zj (0.5)(-20) + 18.5

(1)(-20) + 20

(0)(-20) + 0

(10)(-20) + 0

(0)(-20)

+

0

Resultado

Variables

básicas X1 X2 S1 S2 S3 Valor

solución

S1 0.025 0 1 -0.5 0 200

X2 0.5 1 0 10 0 18000

S3 0.075 0 0 -0.5 1 1100

Zj 10 20 0 200 0 360000

Cj-Zj

Tabla 2 (Resumen)

Variables

básicas

X1 X2 S1 S2 S3 Valor

solución

S1 0.025 0 1 -0.5 0 200

X2 0.5 1 0 10 0 18000

S3 0.075 0 0 -0.5 1 1100

Zj 10 20 0 200 0 360000

Cj-Zj 8.5 0 0 -200 0

La solución se puede mejorar ya que aún hay valores positivos en el renglón Cj-Zj de las variables no básicas.

Variables

básicas

X1 X2 S1 S2 S3 Valor

solución

S1 0.025 0 1 -0.5 0 200

X2 0.5 1 0 10 0 18000

S3 0.075 0 0 -0.5 1 1100

Zj 10 20 0 200 0 360000

Cj-Zj 8.5 0 0 -200 0

Determinar la variable que

ingresa y la que la sale.

Regla de entrada (criterio de optimalidad):

Entra aquella variable no básica con la mayor ganancia unitaria (en el caso de MAX) o el menor costo unitario (en el caso de MIN).

En nuestro ejemplo: comparo los valores

de la fila Cj-Zj, determino la columna

Variable que entra a la base: X1

Variables

básicas

X1 X2 S1 S2 S3 Valor

solución

S1 0.025 0 1 -0.5 0 200

X2 0.5 1 0 10 0 18000

S3 0.075 0 0 -0.5 1 1100

Zj 10 20 0 200 0 360000

Cj-Zj 8.5 0 0 -200 0

Regla de salida (criterio de factibilidad):

Sale aquella variable cuyo resultado de dividir el valor solución entre el coeficiente de la columna pivote sea menor.

En nuestro ejemplo sale S1, ya que al

dividir es el que tiene menor valor

Variable que sale de la base S1:

divido la columna de valor solución entre X1

Variables

básicas X1 X2 S1 S2 S3 Valor

solución

S1 0.025 0 1 -0.5 0 200

X2 0.5 1 0 10 0 18000

S3 0.075 0 0 -0.5 1 1100

Zj 10 20 0 200 0 360000

Cj-Zj 8.5 0 0 -200 0

200/0.025=8000 18000/0.5=36000

1100/0.075=14666.6

Identifico el elemento pivote: la intersección de variable que entra (X1) y la variable que

sale de la base (S1)

Variables

básicas

X1 X2 S1 S2 S3 Valor

solución

S1 0.025 0 1 -0.5 0 200

X2 0.5 1 0 10 0 18000

S3 0.075 0 0 -0.5 1 1100

Zj 10 20 0 200 0 360000

Cj-Zj 8.5 0 0 -200 0

Convierto el elemento pivote en uno:

multiplico por su inverso multiplicativo (1/.025=40) el renglón S1

Variables

básicas

X1 X2 S1 S2 S3 Valor

solución

S1 (40)(0.025) (40)(0) (40)(1) (40)(-0.5) (40)(0) (40)(200)

X2 0.5 1 0 10 0 18000

S3 0.075 0 0 -0.5 1 1100

Zj 10 20 0 200 0 360000

Cj-Zj 8.5 0 0 -200 0

Resultando

Variables

básicas

X1 X2 S1 S2 S3 Valor

solución

X1 ^{1 0 40 -20 0 8000}

X2 0.5 1 0 10 0 18000

S3 0.075 0 0 -0.5 1 1100

Zj 10 20 0 200 0 360000

Cj-Zj 8.5 0 0 -200 0

Transformo los valores de la tabla para los renglones X2, S3, Zj y Cj-Zj

Variables

básicas

X1 X2 S1 S2 S3 Valor

solución

X1 ^{1 0 40 -20 0 8000}

X2 0.5 1 0 10 0 18000

S3 0.075 0 0 -0.5 1 1100

Zj 10 20 0 200 0 360000

Cj-Zj 8.5 0 0 -200 0

Operaciones

Variables

básicas

X1 X2 S1 S2 S3 Valor

solución

X1 ^{1 0 40 -20 0 8000}

X2 (1)(-0.5)+0.5 (0)(-0.5)+1 (40)(-0.5)+0 (-20)(-0.5)+10 (0)(-0.5)+0

S3

(0) (-0.075)+0 (40) (-0.075)+0

(0) (-0.075)+1

Zj (1)(8.5)+10 (0)(8.5)+20 (40)(8.5)+0 (-20)(8.5)+200 (0)(8.5)+0

Cj-Zj (1)(-8.5)+ 8.5 (0)(-8.5)+ 0 (40)(-8.5)+ 0 (-20)(-8.5) (0)(-8.5)+ 0

Tabla 3 (final)

Variables

básicas

X1 X2 S1 S2 S3 Valor

solución

X1 1 0 40 -20 0 8000

X2 0 1 -20 20 0 14000

S3 0 0 -3 1 1 500

Zj 18.5 20 340 30 0 428000

Cj-Zj 0 0 -340 -30 0

La solución no puede mejorar, ya que no hay valores positivos en el renglón Cj-Zj de las variables no

básicas, por lo que se ha alcanzado una solución factible óptima.

Variables

básicas

X1 X2 S1 S2 S3 Valor

solución

X1 1 0 40 -20 0 8000

X2 0 1 -20 20 0 14000

S3 0 0 -3 1 1 500

Zj 18.5 20 340 30 0 428000

Cj-Zj 0 0 -340 -30 0

Resultados finales

Variables

básicas

X1 X2 S1 S2 S3 Valor

solución

X1 1 0 40 -20 0 8000

X2 0 1 -20 20 0 14000

S3 0 0 -3 1 1 500

Zj 18.5 20 340 30 0 428000

Cj-Zj 0 0 -340 -30 0

Solución

X1 = 8000 X2 = 14000 S1 = 0

S2 = 0

S3 = 500 (hay una holgura del tercer recurso)

Z = $ 428,000

Las 3 tablas del problema