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$\left [ x,2 \right ]$ means to round $x$ to $2$ decimal places. For examples, $\left [ \pi,2 \right ]=3.14, \left [ \sqrt{9},2 \right ]=3.00,\left [ \sqrt{5},2 \right ]=2.24$

$\left \lfloor x,2 \right \rfloor$ means to round $x$ down to $2$ decimal places. For examples, $\left \lfloor 1.978,2 \right \rfloor=1.97,\left \lfloor 5.019,2 \right \rfloor=5.01,\left \lfloor 8.999,2 \right \rfloor=8.99$


How to find the minimum value of $\left [ s,2 \right ]$ such that

$$n\left [ b,2 \right ]+\left [ 0.00155n\left [ b,2 \right ],2 \right ]+\left \lfloor 0.15 \left [ 0.00155n\left [ b,2 \right ],2 \right ],2 \right \rfloor$$ $$\le n\left [ s,2 \right ]-\left [ 0.00155n\left [ s,2 \right ],2 \right ]-\left \lfloor 0.15 \left [ 0.00155n\left [ s,2 \right ],2 \right ],2 \right \rfloor$$

Where $n$ is a natural number, $b,s$ are positive real numbers.


Clarification with an example: (This example is taken from a real scenario):

$n$ is the number of units to be bought, say $350$

$\left [ b,2 \right ]$ is the buying (without commission and tax) price of one unit, say $28.62$

Commission is calculated as $0.00155 \times 350 \times 28.62 = 15.52635$ but rounding that to the nears $2$ decimal places, that is $15.53$

Tax is $15 \text{%}$ of that commission $=0.15 \times 15.53 = 2.3295$ but rounding that down to $2$ decimal places, that is $2.32$

So, this costs $350 \times 28.62 + 15.53 +2.32=10034.85$

So, the question is about finding the selling price $\left [ s,2 \right ]$ so that we have no losses.

By trial and error, $\left [ s,2 \right ]=28.73$


I think, it only looks complicated but actually easy for some of you. Please help me, your help would be appreciated. Thanks!

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  • $\begingroup$ I would first simplify the $2$-decimal-place rounding functions to common integer rounding functions: $$\lfloor x, 2\rfloor = \frac{\lfloor 100x\rfloor}{100}$$ Then many $100$'s will cancel out, and the inequality will be in terms of $100b$ and $100s$, i.e. prices in cents. $\endgroup$
    – peterwhy
    Dec 2 '20 at 12:25
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With $B=100b$ and $S=100s$ (which are already integers without rounding!), multiply your inequality with $100$ to arrive at $$nB+\left [ 0.00155nB \right ]+\left \lfloor 0.15 \left [ 0.00155nB\right ] \right \rfloor\\ \le nS-\left [ 0.00155nS \right ]-\left \lfloor 0.15 \left [ 0.00155nS \right ] \right \rfloor$$ In math, we are happier with the floor function, so rewrite as $$nB+\lfloor 0.00155nB +0.5\rfloor+\bigl \lfloor 0.15 \lfloor 0.00155nB+0.5\rfloor \bigr \rfloor\\ \le nS-\lfloor 0.00155nS +0.5\rfloor -\bigl \lfloor 0.15 \lfloor 0.00155nS +0.5\rfloor \bigr \rfloor$$ Now note that $\lfloor x\rfloor = x-\theta$ for some $\theta$ (depending on $x$) with $0\le \theta<1$. This allows us to write $$nB+ 0.00155nB +0.5-\theta_1+ 0.15 ( 0.00155nB+0.5) -0.15\theta_2 -\theta_3\\ \le nS-0.00155nS -0.5+\theta_4 - 0.15( 0.00155nS +0.5)+0.15\theta_5+\theta_6$$ and rearrange as $$ 1.0017825nB +0.575-2.15\theta_7\le 0.9982175 nS-0.575+2.15\theta_8,$$ or $$S\ge B+\frac{1426}{399287} B +\frac{460000}{399287n}-\frac{1720000\theta_9}{399287n}\approx B+\frac{1426}{399287} B + (1.152-4.3\theta_9)\frac1n$$ It seems natural to let $S$ be the right hand side rounded up, but we don't know $\theta_9$. Hence with
$$ S_0=B+\left\lceil\frac{1426 B-1260000}{399287} \right\rceil$$ we can simply try if $S=S_0$ fulfills the original inequality. If it doesn't, try $S=S_0+1$ and after that $S=S_0+2$ and so on. Note that for $n>4$, we will never have to check $S_0+2$, for $n>2$, we will never have to check $S_0+3$, and even for $n=1$, we need not check $S_0+5$ or above.

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