1
$\begingroup$

enter image description here

Some additional information:

In the next season the harvesting amount is estimated at 900 for farm A, 1200, 1500, 1800 for farm B,C and D respectively.

In this scenario I'm asked to minimize the maintenance costs. Unfortunately I can't solve it.

Denoting $s_i$ as the harvesting machine; senior 1,2,3


I've made the obvious constraint: $$s_1,s_2,s_3 \geq 0$$

But I'm not sure how to handle the fact there are 6 machines of each type and how to involve the estimated amount....

I want to minimize these costs: $$16s_1+13s_2+15s_3$$ $$20s_1+29s_2+25s_3$$ $$40s_1+38s_2+42s_3$$ $$37s_1+49s_2+45s_3$$

Can someone help me out?

Thanks in advance!

Edit: and what is in this case parameter/variable? I seem to mix them up quite often...

$\endgroup$
4
$\begingroup$

I can help you with the constraints.

You can't have more than 6 of any machine:

$$0 \leq s_1,s_2,s_3 \leq 6$$

The machines must be able to harvest at least the estimated amount:

$$250s_1+200s_2+230s_3 \geq 900$$ $$270s_1+350s_2+300s_3 \geq 1200$$ $$490s_1+470s_2+520s_3 \geq 1500$$ $$460s_1+630s_2+550s_3 \geq 1800$$


It seems that a better way to approach this problem is to use a different variable for each machine on each farm. I am going to use the simplex algorithm. So, if $w$ is our total cost, we will be trying to minimize the equation:

$$w = 16x_1+13x_2+15s_3+20x_4+29x_5+25x_6+40x_7+38x_8+42x_9+37x_{10}+49x_{11}+45x_{12}$$

Where $x_1$ is machine type 1 on farm A, $x_2$ is machine type 2 on farm A, etc. By adding "surplus" variables, we can get rid of the inequalities we had above, so:

$$250x_1+200x_2+230x_3-s_1 = 900$$ $$270x_4+350x_5+300x_6-s_2 = 1200$$ $$490x_7+470x_8+520x_9-s_3 = 1500$$ $$460x_{10}+630x_{11}+550x_{12}-s_4 = 1800$$

And since we still can't use the same machine on more than one farm at a time, we have to restate our other constraint, this time with "slack" variables:

$$x_1+x_4+x_7+x_{10}+s_4 = 6$$ $$x_2+x_5+x_8+x_{11}+s_5 = 6$$ $$x_3+x_6+x_9+x_{12}+s_6 = 6$$

We use our original equation with all of the constraints to produce our tableau. It's a big matrix; I'm sorry if it doesn't render correctly.

1   0   0   1   0   0   1   0   0   1   0   0   1  0  0  0  0  0  0  0  6
0   1   0   0   1   0   0   1   0   0   1   0   0  1  0  0  0  0  0  0  6
0   0   1   0   0   1   0   0   1   0   0   1   0  0  1  0  0  0  0  0  6
250 200 230 0   0   0   0   0   0   0   0   0   0  0  0  -1 0  0  0  0  900
0   0   0   270 350 300 0   0   0   0   0   0   0  0  0  0  -1 0  0  0  1200
0   0   0   0   0   0   490 470 520 0   0   0   0  0  0  0  0  -1 0  0  1500
0   0   0   0   0   0   0   0   0   460 630 550 0  0  0  0  0  0  -1 0  1800
16  13  15  20  29  25  40  38  42  37  49  45  0  0  0  0  0  0  0  1  0

By using one of several online calculators, we get the optimal solution $w = \frac{5306}{13}$ when: $x_1 = \frac{14}{9}, x_2 = {23}{9}, x_3 = 0, x_4 = \frac{40}{9}, x_5 = 0, x_6 = 0, x_7= 0, x_8 = 0, x_9 = \frac{75}{26}, x_{10} = 0, x_{11} = \frac{20}{7}, x_{12} = 0$

However, in reality we can't have fractions of a machine, so this answer will have to be further refined by using a cutting plane technique (you can't necessarily just round up). Unfortunately, I did not find a calculator that does this.

$\endgroup$
  • $\begingroup$ I tried to solve for these constraints but found values that don't fit the 1st constraint. I find something like $s_1=8,....$ so must be doing something wrong by just solving for these inequalities...? $\endgroup$ – Bob Mar 7 '13 at 11:43
  • $\begingroup$ Are you able to answer my edit? I guess I'll check the actual solving later with my teacher then ;) $\endgroup$ – Bob Mar 7 '13 at 20:22
  • $\begingroup$ @Bob The solution I found might be overkill, depending on what your teacher had in mind. $\endgroup$ – Alex Mar 7 '13 at 21:02
  • $\begingroup$ wow i think this indeed is overkill haha. Do you think you can explain me which of the items are variable, and which are a parameter? $\endgroup$ – Bob Mar 7 '13 at 21:05
  • $\begingroup$ No, I'm not sure what the distinction is in this context. $\endgroup$ – Alex Mar 7 '13 at 21:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.