A company produces 2 products in a week. Let $x_i$ denote the number of units of product $i$ to produce. Each product requires liters of Chemical X to make. Info is given below: Others:
Company has goals:
I have to formulate as a goal programming problem and then solve it graphically. The answer is given as $5P_2 + 140P_3$ A hint is given: Show that minimising Goal 2 is equivalent to minimising $200  x_1  x_2$. My formulation Company has goals: Goal 1: Minimum Revenue of 7000. $$100x_1 + 150x_2 \ge 7000$$ $$ \to 100x_1 + 150x_2 + d_1^  d_1^+ = 7000$$ $ \ $ Goal 2: 40 units each. $$x_1 \ge 40$$ $$x_2 \ge 40$$ $$ \to x_1 + d_2^  d_2^+ = 40$$ $$ \to x_2 + d_3^  d_3^+ = 40$$ $ \ $ Goal 3: At most 120 hours of labour. $$4x_1 + 3x_2 \le 120$$ $$ \to 4x_1 + 3x_2 + d_4^  d_4^+ = 120$$ $ \ $ Company has other constraints:
Is that right? In which part does or should the 'prohibitively expensive to dispose' come? I was thinking to include deviational variables in the 50L constraint but there is no goal for that. Maybe $d_2^+$ and $d_3^+$ have their own constraints? Trying to answer it:
What is the correct answer?
I think it is true because: $$2x_1 \ge 80 \to 2x_1 + 2d_2^  2d_2^+ = 80$$ $$3x_2 \ge 120 \to 3x_1 + 3d_3^  3d_3^+ = 120$$ $$\to 2x_1 + 3x_2 + 2d_2^  2d_2^+ + 3d_3^  3d_3^+ = 200$$ $$\to 2d_2^ + 3d_3^ = 200  2x_1  3x_2 + 2d_2^+ + 3d_3^+$$ Thus, minimising $d_2^$ and $d_3^$ means minimising $200  2x_1  3x_2$ because we would have $d_2^+ = d_3^+ = 0$ Is that right?
Having less lines to draw? Like rewriting Goal 2 constraints as $$200  2x_1  3x_2 \le 0$$ ? I think I would rather draw two perpendicular lines than draw that. *I used LiPS, and I used the code:
