# Solve $n * \lceil{\frac{R}{T}}\rceil - \lceil{\frac{R*n}{T} - \frac{x*n}{T}}\rceil = \frac{x * n}{C}$ for x and n, $n \in Z^+$ and $x,R,T,C \in R^+$

I want to find the minimum value of n which satisfy given equation.

(Also, not stated in title but given is that $C < T$)

So far, I have been able to find the following properties:

• $$\frac{x * n}{C} \in Z^+$$
• $$x \leqslant \frac{\lceil{\frac{R}{T}}\rceil - \frac{R}{T}}{\frac{1}{C} - \frac{1}{T}}$$

Reasoning for the first observation is that since left hand side of the equation in title is all integers, right hand side too needs to be integer

The second observation comes from the following:

$$\frac{R*n}{T} - \frac{x*n}{T} \leqslant \lceil{\frac{R*n}{T} - \frac{x*n}{T}}\rceil$$

So

$$n * \lceil{\frac{R}{T}}\rceil - \lceil{\frac{R*n}{T} - \frac{x*n}{T}}\rceil = \frac{x * n}{C} \leqslant n * \lceil{\frac{R}{T}}\rceil - (\frac{R*n}{T} - \frac{x*n}{T} ) \implies$$

$$x*n(\frac{1}{C} - \frac{1}{T}) \leqslant n *(\lceil{\frac{R}{T}}\rceil - \frac{R}{T})$$

But I haven't been able to find any explicit constraints on n. What is the minimum bound on n?

Also, I suspect but can't prove, that if there exists a value of n, then there is no upper bound on n.

For context, I have asked a similar (but not exactly the same) question here: Solve $\lceil x\rceil-\frac{\lceil nx\rceil}n\geqslant y$ for $n$, where $n \in\mathbb Z^+, x \in\mathbb R^+, y \in\mathbb R^+$

Also, I need to solve this equation in a computer program. So an iterative solution also works.

Trivial but perhaps not efficient: Iterate over $n=1,2,\dots$ until you get a solution. For each $n$, substitute it (so that the only variable is $x$) and try to solve equation for $x$. Note that the left side is decreasing in $x$ and right side is increasing in $x$. Something "primitive" like bisection search should work.
• Sorry, yes; missed a minus sign. If you can find an interval for $x$ that straddles the solution for fixed $n$ (left side less than right at one endpoint, greater than right at the other endpoint), you can still do bisection search. Commented Mar 6, 2018 at 19:27