# Clearing a manufacturing backlog

I have a simple problem that I think I may be over complicating, but I'm not certain. I work in a manufacturing environment as a process engineer and I'm setting up a calculator to determine how long it will take to clear our backlog from a certain step in our process.

There are two stages. We'll call them X and Y. Stage X can process a certain number of products a day. And stage Y can process a certain number of products a day, but not necessarily equal to stage X.

When stage X produces at a faster rate than stage Y, a backlog B is created.

To clear the backlog as it stands at any particular moment, it will simply take B/Y. However, while that backlog is being cleared, more cases are being added to it at a rate of X.

So to clear the current backlog and the backlog that is created while clearing the current backlog, it will take B/Y + ( (B/Y) * (X) ) / Y.

But... while you clear that additional backlog, the backlog is still growing. So we have ourselves an infinite series (I think). My questions is how can I simplify this into an equation that I can use programmatically (like in Excel or something)?

I would think that calculating the number of days required to clear the backlog would only produce a number less than infinity if X is less than Y...

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To clarify, $X$ and $Y$ are the rates at which the eponymous stages process widgets, correct? –  Drew Christianson Jun 5 '12 at 3:03
@DrewChristianson Correct –  mbeasley Jun 5 '12 at 11:22

Let's call $X$ the number of units processed by stage $\mathbb{X}$ in one day, $Y$ the number of units processed in stage $\mathbb{Y}$ in one day, and $B_t$ the backlog on day $t$. (I have used blackboard to distinguish numbers from names.) The quantities $X$ and $Y$ are rates or flows, while $B_t$ is a stock. If we start with some existing backlog, we'll call that $B_0$.

My initial way of thinking was slightly differently from yours. Each day, some of the backlog goes through $\mathbb{Y}$ but some more is added from $\mathbb{X}$. So, if we have $B_t$ units in backlog at the end of the $t^{th}$ day, on the $(t+1)^{st}$ day we have $B_{t+1} = B_t - Y + X$ units, since stage $\mathbb{Y}$ has processed $Y$ units from the backlog and stage $\mathbb{X}$ has added $X$ units to the backlog. This would be an Excel formula.

A closed expression for the number of units in backlog is: $B_t = B_0 + (X - Y)t$. If $X > Y$, then as your intuition suggests, the number of units in the backlog will grow forever. If $X < Y$, then the initial backlog will be cleared after $\frac{B_0}{Y - X}$ days.

From your approach, at any initial backlog $B_0$, if $\mathbb{X}$ is turned off, the number of days remaining to clear the backlog is $B_0/Y$ days. (This is also my answer if $X = 0$.)

If $\mathbb{X}$ is turned on, then to clear the initial backlog, we still need $B_0/Y$ days. In that time, as you note we have added $XB_0/Y$ units, which will take another $XB_0/Y^2$ days to clear. But during that time, we have added $X^2B_0/Y^2$ units. Here's where your thought was headed: the total time to clear the initial backlog $B_0$, if $X > 0$, is the infinite series $$\frac{B_0}{Y} + \frac{B_0X}{Y^2} + \frac{B_0X^2}{Y^3} + \cdots = \frac{B_0}{Y}\sum_{k = 0}^\infty \bigg(\frac{X}{Y}\bigg)^k.$$

The sum is a geometric series. As your intuition suggests (again correctly) if $X \geq Y$, the series diverges. If $X < Y$, then the series converges to $\frac{1}{1 - \frac{X}{Y}} = \frac{Y}{Y - X}$, so $$\frac{B_0}{Y}\sum \bigg(\frac{X}{Y}\bigg)^k = \frac{B_0}{Y}\frac{Y}{Y - X} = \frac{B_0}{Y - X}.$$
This confirms that the algebra we did earlier was probably correct, and that your line of thinking was productive. Hopefully this answers your question!

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This is really an excellent answer. Thank you for both the simple answer (which makes loads of sense now) but also the thorough explanation confirming my over complicated approach. Well done. –  mbeasley Jun 5 '12 at 12:46
@mbeasley Thank you :) I am glad to have helped. –  Neal Jun 6 '12 at 4:37