I apologize if this is in the wrong category, I haven't needed to do math this hard in over a decade.
I'm trying to calculate finish times for 2 production lines running the same thing simultaneously using different run rates. The caveat is that one of these lines won't be starting the run for x amount of hours. So I have a certain amount of cases that one line will be running from constantly and the second line will start running from the same pool of cases in x hours. I'm looking for a finish time in hours.
Start by specifying run rates a and b in terms of units per hour, with a being the rate that the longer line runs.
Let t be total units to run
Let d be hours of delay.
Let h be hours that the longer line runs. this equals the finish time.
[MATH]da \ge t \implies h = \dfrac{t}{a}.[/MATH]
What does this mean? Suppose you need to run 500 items. The line that can run first can process 200 items per hour. The second line must start 3 hours later. So d = 3, t = 500, and a = 200.
Obviously, 3 * 200 = 600 > 500. Thus, the formula above is the one to use, and h = 500/200 = 2.5 hours. In 2.5 hours, the line that runs at 200 units per hour will process 500 units. You will be done before the second line can even start.
[MATH]da < t \implies t = ah + b(h - d) \implies h = \dfrac{t + bd}{a + b}.[/MATH]
What does this mean? Suppose again that a is 200 items per hour and d is the number hours of delay before the second line can start up. Let’s say the second line can process 500 items per hour and you need to run 1300 items.
Obviously, 3 * 200 = 600 < 1300, so we use the second formula.
(t + bd)/(a + b) = (1300 + 500 * 3)/(200 + 500) = 2800/700 = 4. Over 4 hours, the first line will produce 4*200 = 800 units. The second line will run only an hour to produce 500 units. 800+500=1300.
What probably had you stuck is the need for two different formulas to use in different circumstance and how to tell which to use when.
By the way, this is an algebra problem rather than a calculus problem.
