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enter image description here

Attempt:

First, I call $x$ be the number of shafts produced per year and $y$ the number of frames produced per year. We have that each machine works at most $4500$ hours. we can place all of our data in the following table:

enter image description here

Thus, we want to maximize $f(x,y) = x+y$ subject to

$$ 0.6 x + 0.8 y \leq 90 000 $$ $$ 0.3 x + 0 y \leq 22 500 $$ $$ 0.4 x + 0.6 y \leq 45 000 $$ $$ 0x + 0.2 y \leq 135 00 $$ $$ 0x + 0.3 y \leq 27 000 $$

and obviously $x \geq 0$ and $y \geq 0 $

Is this the correct formulation?

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  • $\begingroup$ Can a forging machine process a shaft and a frame at the same time? $\endgroup$ Commented Sep 3, 2018 at 22:02
  • $\begingroup$ no, I dont think so. $\endgroup$ Commented Sep 3, 2018 at 23:18
  • $\begingroup$ Since fractional shafts and frames are probably not desired, you should add $x \in \mathbb{Z}$ and $y \in \mathbb{Z}$. This, however, turns your LP into a MIP. $\endgroup$ Commented Sep 4, 2018 at 6:39
  • $\begingroup$ What is an MIP? But, as far as the modeling is concerned, am I correct? $\endgroup$ Commented Sep 4, 2018 at 6:44

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I suppose your model is not correct. Solving the model with an LP-solver yields the solution $x=75000$ and $y=25000$. I assume, however, that for every frame you also need a shaft. Hence, you need to add the following constraint:

$$x=y$$

Solving the problem with the additional constraint yields the solution $x=y=45000$. Please note that luckily the number of shafts/frames in the solution is an integer. This need not happen always. Thus, you should also add the constraints $x\in \mathbb{Z}$ and $y \in \mathbb{Z}$ in general, even though these were not needed here.

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  • $\begingroup$ the constraint is $x \geq y$ (not $x=y$) $\endgroup$ Commented Sep 6, 2018 at 17:41
  • $\begingroup$ @LinAlg Could you please elaborate on this? $\endgroup$ Commented Sep 6, 2018 at 17:44
  • $\begingroup$ The requirement "for every frame you also need a shaft" means that the number of shafts needs to be at least the number of frames (so $x \geq y$). Your solution probably remains optimal though. $\endgroup$ Commented Sep 6, 2018 at 18:07
  • $\begingroup$ @LinAlg In the original problem it did not say "for every frame you also need a shaft" but that the assembly consists "of a frame, a shaft, and a ball bearing". The other quote was by me and was due to that in the author's own problem formulation one would have obtained a higher number of frames than shafts. I still think that my constraint is correct in terms of the text in the author's post. It is just my wording that caused the confusion. $\endgroup$ Commented Sep 7, 2018 at 6:58

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