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I am no math wiz at all.

I have an 8 day set iteration that I want to run as many times in a year as I can. I can run 2 at a time an I can start 2 more every 3rd day.

So day 1 I start 2 iterations, day 3 I start 2 more, Day 9 I start a 3rd, day 11 I start a 4th. How many can I run per year?

EDIT to clarify and add context-

I'll try to make it more of a word problem. I brew 10 gallons of beer at a time and I have 4 fermenters. Each batch ferments for 6 days. After the beer ferments it has to clear in another tank of which I have 2 and the beer has to sit in this clearing tank for 2 days.

so I have 4 beers in fermentation but I can only move 2 into clearing tanks for 2 days, then after that I can move the other two into the clearing tanks.

At the point where I empty a fermenter I want to refill it.

How many times can I do all this in a year, and whats the equation?

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    I don't want to worry about leap year.2012-11-01

1 Answers 1

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If you can only run $2$ at a time, how can you start more on day $3$? You have four running on day $4$. It sounds like you start $2$ on days $1,9,17,25 \ldots$, which are the days of the form $8k+1$. You have to start them before day $357$. What is the greatest $8k+1$ that is less than $357$?

Added: if you can have four batches in process but have only two clearing tanks, you should start two each on day $1, 3, 9, 11, \ldots , 8k+1, 8k+3.$ They finish on day $8k+8$ and $8k+10$. So find the highest $k$ that each series is done within the year. In this case, for each series it is $44$, so you can make $4 \cdot 44=176$ batches in a year.

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    @amWhy: Generally you want to start them as early as possible so they stop as soon as possible. It may not matter. If you were asking about 12 days, for example, you could start two processes any time up to the 5th day and they would finish by day 12. Earlier wouldn't help here. But waiting never gets more done.2012-11-02