max $$z=4x_1+5x_2$$ s.t $$3x_1+x_2\leq27$$ $$x_1+x_2=12$$ $$3x_1+2x_2\geq 30$$ $$x_1,x_2\geq 0$$

To start the process we find an initial solution, so we add a slack variables:

$$3x_1+x_2+x_3 = 27$$ $$x_1+x_2+x_4=12$$ $$3x_1+2x_2-x_5= 30$$ $$x_1,x_2, x_3,x_4,x_5\geq 0$$

or should it be

$$3x_1+x_2+x_3 = 27$$ $$x_1+x_2+x_4=12$$ $$3x_1+2x_2-x_5+x_6= 30$$ $$x_1,x_2, x_3,x_4,x_5,x_6\geq 0$$

  • $\begingroup$ I have got $$x_1=x_2=6$$ $\endgroup$ Commented May 26, 2019 at 15:04

1 Answer 1


To set up an LP in augmented form, to prepare it for either the two-phase method or the big-$M$, method:

  • Each $\le$ constraint gets a slack variable
  • Each $=$ constraint gets an artificial variable
  • Each $\ge$ constraint gets both a surplus variable and an artificial variable

So, your second set of constraints is correct.

  • $x_3$ is a slack variable
  • $x_4$ is an artificial variable
  • $x_5$ is a surplus variable and $x_6$ is an artificial variable.

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