Chapter 4.5, Problem 6P

Explanation of Solution

Simplex algorithm

• ${\text{x}}_{\text{1}}$, ${\text{x}}_{\text{2}}$, ${\text{s}}_{\text{1}}$, ${\text{s}}_{\text{2}}$ 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.
• The $RHS$ is the right hand side of the constraints in the linear problem.
• 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 iteration one, the initial table is

z	x1	x2	x3	s1	s2	s3	RHS
1	-1	-1	-1	0	0	0	0
0	1	2	2	1	0	0	20
0	2	1	2	0	1	0	20
0	2	2	1	0	0	1	20

• After the next iteration, now x₁ is entered in row 3. The resulting table is

z	x1	x2	x3	s1	s2	s3	RHS
1	0	0	-