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Representación gráfica y punto óptimo en Optimización Lineal - Prof. Mar Molinero, Apuntes de Administración de Empresas

La resolución de un problema de optimización lineal mediante el método gráfico y la tablaau simplex. El problema consiste en maximizar la función objetivo de un lp y determinar el punto óptimo en la región admisible. Se explica el proceso de representación gráfica de la región admisible, la localización del punto óptimo y la evaluación de la función objetivo en todos los puntos óptimos. Además, se agrega la tablaau simplex y se aplican las reglas de optimidad y feasibilidad para obtener la solución final. El documento incluye la resolución gráfica y la tablaau simplex, explicando cada paso detalladamente.

Tipo: Apuntes

2013/2014

Subido el 18/01/2014

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OPERATIONAL RESEARCH. FIRST IN CLASS TEST
Consider the following Linear Programming problem
0 0
33
0
Subject to
Maximise
21
21
21
210
0
+
+=
xx
xx
xx
xxx
x
Graphically represent the feasible region for the solution of the problem. 10 marks.
Indicate where in the feasible region could the optimal solution be found. 10 marks.
Evaluate the objective function in all the possible optimal points and find the optimum. 20
marks.
Add slack variables and write the SIMPLEX tableau LP associated with the problem. 10 marks.
Choose and initial basic feasible solution, and apply the optimality and the feasibility rules to
move one step towards the optimal solution. Optimality rule 10 marks. Feasibility rule 10
marks. Explain what you have done. 10 marks.
Solve the problem and check that your final solution coincides with the solution you obtained
using the graphical method. One Simplex iteration 10 marks. Correct solution found 10
marks.
Resolution:
ܺ
= ܺ
+ ܺ
−ܺ
+ ܺ
= 0
ܺ
= 3
The solution can be found in the corners of
the feasible region (A, B, C)
A (0,0): ܺ
= ܺ
+ ܺ
= 0 + 0 = 0
B (0,1): ܺ
= ܺ
+ ܺ
= 0 + 1 = 1
C (
,
): ܺ
= ܺ
+ ܺ
=
+
=
= 3
In this case, the optimal option is the corner
C because it gives us a value of 3 (the higher
value of the three corners)
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OPERATIONAL RESEARCH. FIRST IN CLASS TEST

Consider the following Linear Programming problem

Subject to

Maximise

1 2

1 2

1 2

0 1 2

0

x x

x x

x x

x x x

x

Graphically represent the feasible region for the solution of the problem. 10 marks.

Indicate where in the feasible region could the optimal solution be found. 10 marks.

Evaluate the objective function in all the possible optimal points and find the optimum. 20 marks.

Add slack variables and write the SIMPLEX tableau LP associated with the problem. 10 marks.

Choose and initial basic feasible solution, and apply the optimality and the feasibility rules to move one step towards the optimal solution. Optimality rule 10 marks. Feasibility rule 10 marks. Explain what you have done. 10 marks.

Solve the problem and check that your final solution coincides with the solution you obtained using the graphical method. One Simplex iteration 10 marks. Correct solution found 10 marks.

Resolution:

The solution can be found in the corners of the feasible region (A, B, C)

A (0,0): ᡐ⡨ = ᡐ⡩ + ᡐ⡰ = 0 + 0 = 0

B (0,1): ᡐ⡨ = ᡐ⡩ + ᡐ⡰ = 0 + 1 = 1

C (⡱⡰ , ⡱⡰): ᡐ⡨ = ᡐ⡩ + ᡐ⡰ = ⡱⡰ + ⡱⡰ = ⡴⡰ = 3

In this case, the optimal option is the corner C because it gives us a value of 3 (the higher value of the three corners)

Now we add the slack variables and perform the Simplex tableau LP associated with the problem:

ᡐ⡨ − ᡐ⡩ − ᡐ⡰ = 0 −ᡐ⡩ + ᡐ⡰ + ᡅ⡩ = 0 3ᡐ⡩ − ᡐ⡰ + ᡅ⡰ = 3

Basic X 0 X 1 X 2 S 1 S 2 Sol Ratio Sum Eq.

X 0 1 - 1 - 1 0 0 0 - 1 E 1

S 1 0 - 1 1 1 0 0 0/- 1 1 E 2

S 2 0 3 - 1 0 1 3 3/3 6 E 3

X 0 1 0 - 4/3 0 1/3 1 1 E 4 =E 1 +E 6 (1=-1+2)

S 1 0 0 2/3 1 1/3 1 3/2 3 E 5 =E 2 +E 6 (3=1+2)

X 1 0 1 - 1/3 0 1/3 1 3/- 1 2 E 6 =1/3 E 3 (2=1/3 *6)

X 0 1 0 0 2 1 3 7 E 7 =E 4 +4/3E 8 (7=1+[4/3]*[9/2])

X 2 0 0 1 3/2 1/2 3/2 9/2 E 8 =3/2 E 5 (9/2=9/2*3)

X 1 0 1 0 1/2 1/2 3/2 7/2 E 9 =E 6 +1/3 E 8 (7/2=2+1/3*9/2)

Since at the beginning we have two negative numbers, we chose one of them (in this case X 2 give us directly to the solution, so we made the difficult way).

Since there are some ratios that are not positive/bigger than zero, we eliminate the unique one that gives us a positive non-zero value.

And we end up with the same solution as in the graph.