Solving multiple ceil functions

Question

Hello,
I am working on an computer program and I am required to find the least possible value of $x$ for

$\lceil \dfrac a x \rceil + \lceil \dfrac b x \rceil + \lceil \dfrac c x \rceil + \lceil \dfrac d x \rceil = H $

where $x,a,b,c,d,H$ are all integers and the values of $a,b,c,d$ and $H$ are given.
I am supposed to find the least possible integral value for $x$ such that the value of $H$ that will be formed (after substituting the values) should be lesser than or equal to the value of $H$ in the question but not greater than $H$.
What I mean by this is that we need to find a minimum possible integral value for $x$ such that after plugging in the values of $a, b...$ the value of $H$ that we get is maximum possible and not greater than the proposed value of $H$ (See the examples for further understanding).

I saw this question on FLOOR function but it was about multiplication and following the method they used to solve it, I got,

$\lceil \dfrac a {n - \alpha} \rceil + \lceil \dfrac b {n - \alpha} \rceil + \lceil \dfrac c {n - \alpha} \rceil + \lceil \dfrac d {n - \alpha} \rceil = H $

where $n$ is the ceil of $x$ and $\alpha$ is the difference of $n$ and $x$. But this method has left me even more clueless on how to go about, since I'm pretty new to ceil and floor functions.

****** EXAMPLES ******

If the values are

$A = 3, B = 5, C = 6, D = 8, H = 6$

then the equation formed will be

$\lceil \dfrac 3 x \rceil + \lceil \dfrac 5 x \rceil + \lceil \dfrac 6 x \rceil + \lceil \dfrac 8 x \rceil = 6 $

for which the correct value for $x$ will be $5$ which is the lowest possible value since, if it is $6$ then the value of $H$ will be $5$ and if the value is $4$, then the value of $H$ will be $7$ which wont comply to the requirements of the value of $H$ and $x$.

But, If the values are

$A = 3, B = 5, C = 6, D = 9, H = 7$

then the equation formed will be

$\lceil \dfrac 3 x \rceil + \lceil \dfrac 5 x \rceil + \lceil \dfrac 6 x \rceil + \lceil \dfrac 9 x \rceil = 7 $

for which the correct integral value for $X$ will be $5$ since if $X = 4$ it would made $H = 8$ (Greater than the given value of $7$) and if $X = 6$ then it would make $H = 5$ (which would be smaller than the value for $x = 5$).

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What I thought is that for questions like the second example once we get an fractional value for $x$ (after mathematically solving it) we can floor[ $\lfloor x \rfloor$ ] it.

ANY HELP IS GREATLY APPRECIATED!!! (And if it's not possible to reduce it to an easier equation, a algorithm to computer-solve it is also OK :P )

Answer 1

So the question is $$ \text{maximize} \left(\left[\frac a x\right]+\left[\frac b x\right]+\left[\frac c x\right]+\left[\frac d x\right]\right) $$ under the contraint that $$ \left[\frac a x\right]+\left[\frac b x\right]+\left[\frac c x\right]+\left[\frac d x\right] \leq H $$ The map $$ x\to \left[\frac a x\right]+\left[\frac b x\right]+\left[\frac c x\right]+\left[\frac d x\right] $$ is decreasing with $x$, and its smallest non trivial value is obtained for $$ x_M=\max(a,b,c,d) $$ where it is equal to the multiplicity of that maximum (1,2,3, or 4).

Indeed, if $x>a$, then $\frac a x <1$, so $\left[\frac a x\right]=0$, and similarly if $x>a$, $x>c$, $x>c$ and $x>d$, that is, $x>\max(a,b,c,d)=x_M$, then $\left[\frac a x\right] + \left[\frac b x\right] + \left[\frac c x \right]+\left[\frac d x\right] =0$, and depending if $a,b,c,d$ are equal or not, $\left[\frac a {x_M}\right] + \left[\frac b {x_M}\right] + \left[\frac c {x_M}\right]+\left[\frac d {x_M}\right]=1,2,3$ or $4$.

The largest value is obtained for $x_m=1$ where it is equal to $a+b+c+d$.

So to maximize under the constraint, and choose the best possible value between $x_m$ and $X_M$, a simple algorithm is a dichotomy.

Start with $(x_m,x_M)$.

Set $x= [(x_m+x_M)/2] $ and $ y= \left[\frac a x\right] + \left[\frac b x\right] + \left[\frac c x \right]+\left[\frac d x\right] $.

If $y>H$, set $x_m=x$ and start again. If $y\leq H$, set $x_M=x$ and start again.

Stop when $x_M$ and $x_m$ are only 1 apart.

Answer 2

COMMENT.-Let $f(x)=\lceil a/ x \rceil + \lceil b/ x \rceil + \lceil c/ x \rceil + \lceil d/ x \rceil$ with $a,b,c,d\gt0$ . We have $$f(x)\to \infty\text{ when } x\to 0\\f(x)\to 4\text{ when } x\to \infty$$ Can you explain to me better what you want to calculate? I do not understand well. Thank you.