Chapter 4.7, Problem 1P

Explanation of Solution

Linear Programming problem

• ${\text{x}}_{\text{1}}$, ${\text{x}}_{\text{2}}$, ${\text{s}}_{\text{1}}$, ${\text{s}}_{\text{2}}$ 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.
• 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.
• Hence after iteration 1 the first table obtained is

z	x1	x2	s1	s2	s3	RHS
1	-4	-2	0	0	0	0
0	2	1	1	0	0	100
0	1	1	0	1	0	80
0	1	0	0	0	1	40

• After iteration 2 the next table is

z	x1	x2	s1	s2	s3	RHS
1	0	-2	0	0	4	160
0	0	1	1	0	-2	20
0	0<