I'm having trouble with converting the last constraint of the following linear program to standard form.
My attempt:
I'm confused on how to convert the last constraint, given that it is a variable located between 2 values.
Would I have to convert this into 2 equations, with one equalling the lower bound and another equalling the upper bound?
For convention, let’s use the variable $s_m$ to denote the surplus required for the left-hand-side of a $\le$ constraint to achieve equality, and let us use the variable $e_m$ to denote the excess amount required to take away from the left-hand-side of a $\ge$ constraint to achieve equality. In addition, let’s use artificial variables $a_m$ to be used with the $=$ and $\ge$ constraints to allow for basic variables to exist from those constraints. Where $m$ is a positive integer that denotes the row the slack variables $s, e,$ and $a$ are located in. By this approach, we are then able to know which slack variable corresponds to what constraint, and their purpose, at any point in the tableau.
Thus, the standard form of the problem you provided will be of the form:
(Via the Big-M Method)
$$max \text{ } z -5x_1-4x_2 + Ma_3 + Ma_4=0$$ Subject to, $$x_1 + x_2 + s_1 = 8$$ $$x_1 + 2x_2 + s_2= 10$$ $$x_1 - e_3 + a_3 = 2$$ $$x_2 -e_4 + a_4 =3$$ $$x_2 + s_5 = 7$$ $$x_1, x_2, s_1, s_2, s_5, e_3, e_4, a_3, a_4 \ge 0$$
This is assuming that $x_1 \ge 0$, but for the last constraint of the provided problem $3 \le x_2 \le 7$, we can split this singular constraint into two separate constraints as depicted in the last two constraints shown above.
However, for Simplex purposes, this type of approach will bloat the tableau and make it slower to compute both by hand and by machine, thus it would be worthwhile to investigate how to do the Simplex method for variables with upper bounds
Another way to split the lower-bound of $3\le x_2$ and $2 \le x_1$ is to do lower-bound substitution of letting $x_2 = x_2’ + 3$ and $x_1 = x_1’ + 2$. Then, replace all instances of $x_2$ and $x_1$ in the above with this substitution such that:
$$max \text{ } z -5x_1’-4x_2’=22$$ Subject to, $$x_1’ + x_2’ + s_1 = 3$$ $$x_1’ + 2x_2’ + s_2= 2$$ $$x_2 + s_3 = 7$$ $$x_1’, x_2’, s_1, s_2, s_3, \ge 0$$
Then, we would proceed with the simplex method for upper-bounds.