Chapter 4, Problem 3RP

Explanation of Solution

Optimal solution:

Consider the following linear programing problem:

$minz=5x_1-x_2$

Subject to the constraints:

  $\begin{array}{l}2x_1+x_2=6\\ x_1+x_2\le 4\\ x_1,x_{2 }\ge 0\end{array}$

Use $\text{Max z = -Min}\left(\text{-z}\right)$ to reduce the given problem to a minimization problem as: ${\text{-(min z'= -5x}}_{\text{1}}{\text{+x}}_{\text{2}}\text{)}$

Subject to the constraints:

$\begin{array}{l}2x_1+x_2=6\\ x_1+x_2\le 4\\ x_1+2x_2\le 4\\ x_1,x_{2 }\ge 0\end{array}$

Add slack variables s₁,s₂ and artificial variable a₁ to get:

-(min w′ = -5x₁+x₂+a₁)

Subject to the constraints:

$\begin{array}{l}{\text{2x}}_{\text{1}}{\text{+x}}_{\text{2}}{\text{+a}}_{\text{1}}\text{= 6}\\ {\text{x}}_{\text{1}}{\text{+x}}_{\text{2}}{\text{+s}}_{\text{1}}\text{= 4}\\ {\text{x}}_{\text{1}}{\text{+2nx}}_{\text{2}}{\text{+s}}_{\text{2}}\text{=5}\\ {\text{x}}_{\text{1}}{\text{,x}}_{\text{2}}{\text{,s}}_{\text{1}}{\text{,s}}_{\text{2}}{\text{,a}}_{\text{1}}\ge \text{0}\end{array}$

Two Phase Method:

Phase I linear programming problem is,

  ${\text{min w'= a}}_{\text{1}}$

Subject to the constraints:

$\begin{array}{l}2x_1+x_2+a_1=6\\ x_1+x_2+s_1=4\\ x_1+2x_2+s_2=5\\ x_1,x_2,s_1,s_2,a_1\ge 0\end{array}$

The initial simplex table is given below:

	w′	x1	x2	a1	s1	s2	rhs	basic variable
R0	1	0	0	-1	0	0	0	w′=0
R1	0	2	1	1	0	0	6	a1=6
R2	0	1	1	0	1	0	4	s1=4
R3	0	1	2	0	0	1	5	s2=5

• Since, the basic variable a₁ value in R₀ is non-zero, therefore, do the transformations

$R_0\to R_0+R_1$  to get:

	w′	x1	x2	a1	s1	s2	rhs	basic variable
R0	1	0	0	-1	0	0	0	w′=0
R1	0	2	1	1	0	0	6	a1=6
R2	0	1	1	0	1	0	4	s1=4
R3	0	1	2	0	0	1	5	s2=5

Since the highest positive entry 2 in R₀ corresponds to x₁, x₁ enters the basis.

	w′	x1	x2	a1	s1	s2	rhs	ratio
R0	1	2	1	0	0	0	6	-
R1	0	2	1	1	0	0	6	3*
R2	0	1	1	0	1	0	4	4
R3	1	2	0	0	0	1	5	5

Apply the simplex method further:

	w′	x1	x2	a1	s1	s2	rhs	basic variable
R0	1	0	0	-1	0	0	0	w′=0
R1	0	1	$\frac{1}{2}$	$\frac{1}{2}$	0	0	6	x1 = 3
R2	0	0	$\frac{1}{2}$	$-\frac{1}{2}$	1	0	1	s1=1
R3	0	0	$\frac{3}{2}$	$-\frac{1}{2}$	0	1	2	s2=2

• Optimally reached for phase 1. Proceed to phase 2 with the actual objective function