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:

enter image description here

Others:

  1. 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.

  2. Company can produce at most 75 units total.

Company has goals:

  1. Estimated Revenue of at least 7000/week

  2. Produce at least 40 units of each product (for the upcoming week).

  3. 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:

  1. At least 50L of Chemical X should be produced. $$2x_1 + x_2 \ge 50$$

  2. At most 75 units can be produced. $$x_1 + x_2 \le 75$$

  3. 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:

  1. 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?

  1. 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?

  1. 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's gravatar image

BCLC
8312
accept rate: 0%

closed 13 Apr '16, 01:42

Ehsan's gravatar image

Ehsan ♦
4.8k31122

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

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Asked: 10 Apr '16, 06:11

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Last updated: 13 Apr '16, 01:42

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