I want to solve a following problem: $$x_1+2x_2+3x_3\to \text{min}$$ $$x_1+3x_3\geq3$$ $$-x_1+x_2+3x_3\leq 4$$ $$x_!+2x_2-3x_3\leq -6$$ $$x_1\geq0$$ $$x_2\geq0$$ $$x_3\geq0$$ I tried using simplex method but it lead me to a solution that is not in the feasible region. Any ideas on how to solve this problem?
2026-03-27 07:47:53.1774597673
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Linear programming excercise
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You could try to solve it using the python scipy stack: scipy.optimization.linprog
You will discover that there is no solution to your problem - it is infeasible. This can be seen by adding together your second and third inequality constraints, as pointed out by @caverac
Here is an example python script to show this:
#!/usr/bin/env python
from scipy.optimize import linprog
def main():
print('\n')
print('*********************************')
print('Infeasible Example:')
print(' http://math.stackexchange.com/questions/2011787/linear-programming-excercise/')
print('*********************************')
print('Minimize: [1;2;3] * x')
print('Subject to:')
print( 'x1 + 3*x3 >= 3')
print( '-x1 + x2 + 3*x3 <= 4')
print( 'x1 + 2*x2 - 3*x3 <= -6')
print( 'x1 >= 0')
print( 'x2 >= 0')
print( 'x3 >= 0')
c = [1, 2, 3] # objective function
# Inequality constraint matrix. Note that first inequality is negated.
A = [[-1,0,-3], [-1, 1, 3], [1,2,-3]]
b = [-3, 4, -6]
x0_bnd = (0, None)
x1_bnd = (0, None)
x2_bnd = (0, None)
print('Solving...')
result = linprog(c, A_ub=A, b_ub=b, bounds=(x0_bnd, x1_bnd, x2_bnd),
options={"disp": True})
print('Result:')
print(result)
if __name__ == "__main__":
main()
The output is:
Minimize: [1;2;3] * x
Subject to:
x1 + 3*x3 >= 3
-x1 + x2 + 3*x3 <= 4
x1 + 2*x2 - 3*x3 <= -6
x1 >= 0
x2 >= 0
x3 >= 0
Solving...
Optimization failed. Unable to find a feasible starting point.
Result:
fun: 2.0
message: 'Optimization failed. Unable to find a feasible starting point.'
nit: 2
status: 2
success: False
x: nan
Unfortunately your problem has no solution. Let's look at your constraints,
\begin{align} x_1 + 3x_3 \ge 3 \\ -x_1 + x_2 + 3x_3 \le 4 \tag{1} \\ x_1 + 2x_2 - 3x_3 \le -6 \tag{2} \end{align}
Adding up (1) and (2) we get
$$ 3x_2 \le -2 \quad\Rightarrow x_2 \le -2/3 $$
Which clearly contradicts the constraint $x_2 \ge 0$