# [closed] Formulate as goal programming problem to determine amount of units for production

 0 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 to buy at least 50 L of Chemical X. It is prohibitively expensive to dispose Chemical X. Hence, it must use as much Chemical X as possible. Company can produce at most 75 units total. Company has goals: Estimated Revenue of at least 7000/week Produce at least 40 units of each product (for the upcoming week). Use at most 120 hours of labour/week. 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: At least 50L of Chemical X should be produced. $$2x_1 + x_2 \ge 50$$ At most 75 units can be produced. $$x_1 + x_2 \le 75$$ Nonnegativity $$x_1, x_2, d_1^{\pm}, ..., d_4^{\pm} \ge 0$$ 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: Assuming my formulation is correct, I tried plugging this into a computer*, and I got $$\color{red}{10}P_2 + 140P_3$$ instead. What is the correct answer? Why is the hint true? 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? How does the hint help? 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: min: X3; min: 2X5 + 3X7; min: X10; Row1: 100X1 + 150X2 + X3 - X4 = 7000; Row2: X1 + X5 - X6 = 40; Row3: X2 + X7 - X8 = 40; Row4: 4X1 + 3X2 + X9 - X10 = 120; Row5: 2*X1 + X2 >= 50; Row6: X1 + X2 <= 75; asked 10 Apr '16, 06:11 BCLC 8●3●12 accept rate: 0% Ehsan ♦ 4.8k●3●11●22

### The question has been closed for the following reason "Duplicate Question" by Ehsan 13 Apr '16, 01:42

By Email:

Markdown Basics

• *italic* or _italic_
• **bold** or __bold__
• image?![alt text](/path/img.jpg "Title")
• numbered list: 1. Foo 2. Bar
• to add a line break simply add two spaces to where you would like the new line to be.
• basic HTML tags are also supported

Tags: