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I would like to produce $R$ items in the shortest amount of time possible. For the sake of a visual, call these items bottles of carrot juice.

Let $t_1$ be the time to purchase and set up a blender to produce units of carrot juice. However, the blenders work less efficiently over time, and importantly, we can only one run blender after another. We can characterize this by saying that, to produce $x$ units of juice with a single blender, it takes time:

$f(x) = a + b*x + c*x^2$

Where $(a,b,c)$ are positive real numbers.

Provided a value for the number of units of juice $T$ we wish to produce, is there a closed-form optimal solution for the number of blenders we should bother setting up?

To get started, let's express the time to produce the $R$ items given the number of blenders which must be run sequentially $Q$, as $Q*t_1 + \frac{R}{Q}f(\frac{R}{Q}) = Q*t_1 + a + b(\frac{R}{Q})+c(\frac{R}{Q})^2$.

In response to Joriki's comment - please note that the $Q*t_1$ term in the above expression is the time cost with setting up the $Q$ blenders. Please also note my error in originally writing $\frac{Q}{R}*t_1$ instead of the correct term $Q*t_1$ as the cost to set up the $Q$ blenders. This, I suspect, was the cause of the trouble.

MirrorEcafrus
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1 Answers1

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No, there is no optimal solution. The more blenders you set up, the less time it takes to produce a given amount of juice. This is both intuitively plausible and follows from your formula, which is a sum of terms proportional to inverse powers of $Q$. In order for there to be an optimal number of blenders, you'd need to introduce some countervailing objective, e.g. that you also care about the cost of the blenders, and combine these two objectives into a single objective function (e.g. in a weighted sum).

You speak of "units" and "items"; in case this is meant to imply granularity and that each item must be produced by a single machine, then you can't gain anything from setting up more than $R$ machines.

joriki
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  • I'm having a trouble understanding your answer? In my original problem statement I wrote: "Let $t_1$ be the time to purchase and set up a blender to produce units of carrot juice." If I set $t_1$ to be arbitrarily large, I can always make it more costly to set up an additional blender than to produce all of the units with a single blender. – MirrorEcafrus Jul 14 '13 at 05:17
  • I originally wrote the cost of setting up the $Q$ blenders as $\frac{R}{Q}t_1$ instead of $Qt_1$ because of a typo. I'm sorry about that. – MirrorEcafrus Jul 14 '13 at 05:24
  • @Mirror: Hmm -- now I don't understand your calculation. From the first term, it seems that there's a single person setting up these blenders, so the time it takes to set them up is $Q$ times the time it takes to set one of them up. But then shouldn't the blenders start producing at different times, each one as soon as you're done setting it up? And why is $f$ being multiplied by $R/Q$? – joriki Jul 15 '13 at 11:39
  • Yes, I mean that the blenders must be set up sequentially, so you have the $Q*t_1$ cost to set up $Q$ blenders. $f$ shouldn't be multiplied by $\frac{R}{Q}$ and this has been removed. As for the blenders starting at different times - that's a good point. Let's say that they need to be run successively where one the next one is turned on once the first has finished its task. – MirrorEcafrus Jul 16 '13 at 11:24
  • Ah, the reason why we have $\frac{R}{Q}f$ is because I was originally imagining that the blenders must be run sequentially. I have hopefully clarified the problem statement. – MirrorEcafrus Jul 16 '13 at 11:27