Chapter 4.11, Problem 4P

Explanation of Solution

Linear Programming problem

• ${\text{x}}_{\text{1}}$, ${\text{x}}_{\text{2}}$, ${\text{x}}_{\text{3}}$ , ${\text{x}}_{\text{4}}$_,${\text{s}}_{\text{1}}$, ${\text{s}}_{\text{2}}$ and ${\text{s}}_{\text{3}}$ are the basic variables.
• For solving a linear problem using simplex algorithm, first the linear problem is converted to its standard form.
• Then a basic feasible solution is found which is easy if all the constraints are less than or equal with non-negative right hand side.
• If all nonbasic variables have nonnegative coefficients in row 0 then that basic feasible solution is optimal.
• If any variable have negative coefficient in row 0 then the variable with the most negative coefficient is chosen to enter the basis.
• If one or more constraints have a negative right hand side then there is no longer a basic feasible solution.
• After the first iteration, the table is

z	x1	x2	x3	x4	s1	s2	s3	RHS
1	3	-1	6	0	0	0	0	0
0	9	1	-9	-2	1	0	0	0
0	1	1/3	-2	-1/3	0	1	0	0
0	-9	-1	9	2	0	0	1	1

• After the second iteration, ${\text{x}}_{\text{2}}$ enters in row 1. Hence the table is

z	x1	x2	x3	x4	s1	s2	s3	RHS
1	12	0	-3	-2	1	0	0	0
0	9	1	-9	-2	1	0	0	0
0	-2	0	1	1/3	-1/3	1	0	0
0	0	0	0	0	1	0	1	1

• After the third iteration, ${\text{x}}_{\text{3}}$ enters in row 2. Hence the table is

z	x1	x2	x3	x4	s1	s2	s3	RHS
1	6	0	0	-1	0	3	0	0
0	-9	1	0	1	-2	9	0	0
0	-2	0	1	1/3	-1/3	1	0	0
0	0	0	0	0	1	0	1	1

• After the fourth iteration, ${\text{x}}_{\text{4}}$ enters in row 1