Compound Interest

?

Percentage Accurate: 28.6% → 95.8%
Time: 23.2s
Precision: binary64
Cost: 35276

?

\[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
\[\begin{array}{l} t_0 := {\left(1 + \frac{i}{n}\right)}^{n}\\ t_1 := \frac{t_0 + -1}{\frac{i}{n}}\\ \mathbf{if}\;t_1 \leq -2000:\\ \;\;\;\;n \cdot \frac{t_0 \cdot 100 + -100}{i}\\ \mathbf{elif}\;t_1 \leq 0:\\ \;\;\;\;100 \cdot \frac{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}{\frac{i}{n}}\\ \mathbf{elif}\;t_1 \leq \infty:\\ \;\;\;\;\frac{n}{\frac{i}{\mathsf{fma}\left(100, t_0, -100\right)}}\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100\\ \end{array} \]
(FPCore (i n)
 :precision binary64
 (* 100.0 (/ (- (pow (+ 1.0 (/ i n)) n) 1.0) (/ i n))))
(FPCore (i n)
 :precision binary64
 (let* ((t_0 (pow (+ 1.0 (/ i n)) n)) (t_1 (/ (+ t_0 -1.0) (/ i n))))
   (if (<= t_1 -2000.0)
     (* n (/ (+ (* t_0 100.0) -100.0) i))
     (if (<= t_1 0.0)
       (* 100.0 (/ (expm1 (* n (log1p (/ i n)))) (/ i n)))
       (if (<= t_1 INFINITY)
         (/ n (/ i (fma 100.0 t_0 -100.0)))
         (* n 100.0))))))
double code(double i, double n) {
	return 100.0 * ((pow((1.0 + (i / n)), n) - 1.0) / (i / n));
}
double code(double i, double n) {
	double t_0 = pow((1.0 + (i / n)), n);
	double t_1 = (t_0 + -1.0) / (i / n);
	double tmp;
	if (t_1 <= -2000.0) {
		tmp = n * (((t_0 * 100.0) + -100.0) / i);
	} else if (t_1 <= 0.0) {
		tmp = 100.0 * (expm1((n * log1p((i / n)))) / (i / n));
	} else if (t_1 <= ((double) INFINITY)) {
		tmp = n / (i / fma(100.0, t_0, -100.0));
	} else {
		tmp = n * 100.0;
	}
	return tmp;
}
function code(i, n)
	return Float64(100.0 * Float64(Float64((Float64(1.0 + Float64(i / n)) ^ n) - 1.0) / Float64(i / n)))
end
function code(i, n)
	t_0 = Float64(1.0 + Float64(i / n)) ^ n
	t_1 = Float64(Float64(t_0 + -1.0) / Float64(i / n))
	tmp = 0.0
	if (t_1 <= -2000.0)
		tmp = Float64(n * Float64(Float64(Float64(t_0 * 100.0) + -100.0) / i));
	elseif (t_1 <= 0.0)
		tmp = Float64(100.0 * Float64(expm1(Float64(n * log1p(Float64(i / n)))) / Float64(i / n)));
	elseif (t_1 <= Inf)
		tmp = Float64(n / Float64(i / fma(100.0, t_0, -100.0)));
	else
		tmp = Float64(n * 100.0);
	end
	return tmp
end
code[i_, n_] := N[(100.0 * N[(N[(N[Power[N[(1.0 + N[(i / n), $MachinePrecision]), $MachinePrecision], n], $MachinePrecision] - 1.0), $MachinePrecision] / N[(i / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]
code[i_, n_] := Block[{t$95$0 = N[Power[N[(1.0 + N[(i / n), $MachinePrecision]), $MachinePrecision], n], $MachinePrecision]}, Block[{t$95$1 = N[(N[(t$95$0 + -1.0), $MachinePrecision] / N[(i / n), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[t$95$1, -2000.0], N[(n * N[(N[(N[(t$95$0 * 100.0), $MachinePrecision] + -100.0), $MachinePrecision] / i), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$1, 0.0], N[(100.0 * N[(N[(Exp[N[(n * N[Log[1 + N[(i / n), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]] - 1), $MachinePrecision] / N[(i / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$1, Infinity], N[(n / N[(i / N[(100.0 * t$95$0 + -100.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(n * 100.0), $MachinePrecision]]]]]]
100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}}
\begin{array}{l}
t_0 := {\left(1 + \frac{i}{n}\right)}^{n}\\
t_1 := \frac{t_0 + -1}{\frac{i}{n}}\\
\mathbf{if}\;t_1 \leq -2000:\\
\;\;\;\;n \cdot \frac{t_0 \cdot 100 + -100}{i}\\

\mathbf{elif}\;t_1 \leq 0:\\
\;\;\;\;100 \cdot \frac{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}{\frac{i}{n}}\\

\mathbf{elif}\;t_1 \leq \infty:\\
\;\;\;\;\frac{n}{\frac{i}{\mathsf{fma}\left(100, t_0, -100\right)}}\\

\mathbf{else}:\\
\;\;\;\;n \cdot 100\\


\end{array}

Local Percentage Accuracy?

The average percentage accuracy by input value. Horizontal axis shows value of an input variable; the variable is choosen in the title. Vertical axis is accuracy; higher is better. Red represent the original program, while blue represents Herbie's suggestion. These can be toggled with buttons below the plot. The line is an average while dots represent individual samples.

Herbie found 19 alternatives:

AlternativeAccuracySpeedup

Accuracy vs Speed

The accuracy (vertical axis) and speed (horizontal axis) of each of Herbie's proposed alternatives. Up and to the right is better. Each dot represents an alternative program; the red square represents the initial program.

Bogosity?

Bogosity

Target

Original28.6%
Target34.2%
Herbie95.8%
\[100 \cdot \frac{e^{n \cdot \begin{array}{l} \mathbf{if}\;1 + \frac{i}{n} = 1:\\ \;\;\;\;\frac{i}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{i}{n} \cdot \log \left(1 + \frac{i}{n}\right)}{\left(\frac{i}{n} + 1\right) - 1}\\ \end{array}} - 1}{\frac{i}{n}} \]

Derivation?

  1. Split input into 4 regimes
  2. if (/.f64 (-.f64 (pow.f64 (+.f64 1 (/.f64 i n)) n) 1) (/.f64 i n)) < -2e3

    1. Initial program 99.7%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
    2. Simplified100.0%

      \[\leadsto \color{blue}{n \cdot \frac{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, -100\right)}{i}} \]
      Step-by-step derivation

      [Start]99.7

      \[ 100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]

      associate-/r/ [=>]99.8

      \[ 100 \cdot \color{blue}{\left(\frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i} \cdot n\right)} \]

      associate-*r* [=>]100.0

      \[ \color{blue}{\left(100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i}\right) \cdot n} \]

      *-commutative [=>]100.0

      \[ \color{blue}{n \cdot \left(100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i}\right)} \]

      associate-*r/ [=>]100.0

      \[ n \cdot \color{blue}{\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{i}} \]

      sub-neg [=>]100.0

      \[ n \cdot \frac{100 \cdot \color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n} + \left(-1\right)\right)}}{i} \]

      distribute-lft-in [=>]100.0

      \[ n \cdot \frac{\color{blue}{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + 100 \cdot \left(-1\right)}}{i} \]

      fma-def [=>]100.0

      \[ n \cdot \frac{\color{blue}{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, 100 \cdot \left(-1\right)\right)}}{i} \]

      metadata-eval [=>]100.0

      \[ n \cdot \frac{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, 100 \cdot \color{blue}{-1}\right)}{i} \]

      metadata-eval [=>]100.0

      \[ n \cdot \frac{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, \color{blue}{-100}\right)}{i} \]
    3. Applied egg-rr100.0%

      \[\leadsto n \cdot \frac{\color{blue}{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + -100}}{i} \]
      Step-by-step derivation

      [Start]100.0

      \[ n \cdot \frac{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, -100\right)}{i} \]

      fma-udef [=>]100.0

      \[ n \cdot \frac{\color{blue}{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + -100}}{i} \]

    if -2e3 < (/.f64 (-.f64 (pow.f64 (+.f64 1 (/.f64 i n)) n) 1) (/.f64 i n)) < 0.0

    1. Initial program 25.2%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
    2. Applied egg-rr99.7%

      \[\leadsto 100 \cdot \frac{\color{blue}{0 + \mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}}{\frac{i}{n}} \]
      Step-by-step derivation

      [Start]25.2

      \[ 100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]

      add-log-exp [=>]25.2

      \[ 100 \cdot \frac{\color{blue}{\log \left(e^{{\left(1 + \frac{i}{n}\right)}^{n} - 1}\right)}}{\frac{i}{n}} \]

      *-un-lft-identity [=>]25.2

      \[ 100 \cdot \frac{\log \color{blue}{\left(1 \cdot e^{{\left(1 + \frac{i}{n}\right)}^{n} - 1}\right)}}{\frac{i}{n}} \]

      log-prod [=>]25.2

      \[ 100 \cdot \frac{\color{blue}{\log 1 + \log \left(e^{{\left(1 + \frac{i}{n}\right)}^{n} - 1}\right)}}{\frac{i}{n}} \]

      metadata-eval [=>]25.2

      \[ 100 \cdot \frac{\color{blue}{0} + \log \left(e^{{\left(1 + \frac{i}{n}\right)}^{n} - 1}\right)}{\frac{i}{n}} \]

      add-log-exp [<=]25.2

      \[ 100 \cdot \frac{0 + \color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}}{\frac{i}{n}} \]

      pow-to-exp [=>]25.2

      \[ 100 \cdot \frac{0 + \left(\color{blue}{e^{\log \left(1 + \frac{i}{n}\right) \cdot n}} - 1\right)}{\frac{i}{n}} \]

      expm1-def [=>]36.7

      \[ 100 \cdot \frac{0 + \color{blue}{\mathsf{expm1}\left(\log \left(1 + \frac{i}{n}\right) \cdot n\right)}}{\frac{i}{n}} \]

      *-commutative [=>]36.7

      \[ 100 \cdot \frac{0 + \mathsf{expm1}\left(\color{blue}{n \cdot \log \left(1 + \frac{i}{n}\right)}\right)}{\frac{i}{n}} \]

      log1p-udef [<=]99.7

      \[ 100 \cdot \frac{0 + \mathsf{expm1}\left(n \cdot \color{blue}{\mathsf{log1p}\left(\frac{i}{n}\right)}\right)}{\frac{i}{n}} \]
    3. Simplified99.7%

      \[\leadsto 100 \cdot \frac{\color{blue}{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}}{\frac{i}{n}} \]
      Step-by-step derivation

      [Start]99.7

      \[ 100 \cdot \frac{0 + \mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}{\frac{i}{n}} \]

      +-lft-identity [=>]99.7

      \[ 100 \cdot \frac{\color{blue}{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}}{\frac{i}{n}} \]

    if 0.0 < (/.f64 (-.f64 (pow.f64 (+.f64 1 (/.f64 i n)) n) 1) (/.f64 i n)) < +inf.0

    1. Initial program 99.6%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
    2. Simplified99.6%

      \[\leadsto \color{blue}{n \cdot \frac{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, -100\right)}{i}} \]
      Step-by-step derivation

      [Start]99.6

      \[ 100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]

      associate-/r/ [=>]99.4

      \[ 100 \cdot \color{blue}{\left(\frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i} \cdot n\right)} \]

      associate-*r* [=>]99.6

      \[ \color{blue}{\left(100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i}\right) \cdot n} \]

      *-commutative [=>]99.6

      \[ \color{blue}{n \cdot \left(100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i}\right)} \]

      associate-*r/ [=>]99.6

      \[ n \cdot \color{blue}{\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{i}} \]

      sub-neg [=>]99.6

      \[ n \cdot \frac{100 \cdot \color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n} + \left(-1\right)\right)}}{i} \]

      distribute-lft-in [=>]99.6

      \[ n \cdot \frac{\color{blue}{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + 100 \cdot \left(-1\right)}}{i} \]

      fma-def [=>]99.6

      \[ n \cdot \frac{\color{blue}{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, 100 \cdot \left(-1\right)\right)}}{i} \]

      metadata-eval [=>]99.6

      \[ n \cdot \frac{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, 100 \cdot \color{blue}{-1}\right)}{i} \]

      metadata-eval [=>]99.6

      \[ n \cdot \frac{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, \color{blue}{-100}\right)}{i} \]
    3. Applied egg-rr99.7%

      \[\leadsto \color{blue}{\frac{n}{\frac{i}{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, -100\right)}}} \]
      Step-by-step derivation

      [Start]99.6

      \[ n \cdot \frac{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, -100\right)}{i} \]

      clear-num [=>]99.6

      \[ n \cdot \color{blue}{\frac{1}{\frac{i}{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, -100\right)}}} \]

      un-div-inv [=>]99.7

      \[ \color{blue}{\frac{n}{\frac{i}{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, -100\right)}}} \]

    if +inf.0 < (/.f64 (-.f64 (pow.f64 (+.f64 1 (/.f64 i n)) n) 1) (/.f64 i n))

    1. Initial program 0.0%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
    2. Taylor expanded in i around 0 71.3%

      \[\leadsto \color{blue}{100 \cdot n} \]
    3. Simplified71.3%

      \[\leadsto \color{blue}{n \cdot 100} \]
      Step-by-step derivation

      [Start]71.3

      \[ 100 \cdot n \]

      *-commutative [=>]71.3

      \[ \color{blue}{n \cdot 100} \]
  3. Recombined 4 regimes into one program.
  4. Final simplification95.4%

    \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq -2000:\\ \;\;\;\;n \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + -100}{i}\\ \mathbf{elif}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq 0:\\ \;\;\;\;100 \cdot \frac{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}{\frac{i}{n}}\\ \mathbf{elif}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq \infty:\\ \;\;\;\;\frac{n}{\frac{i}{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, -100\right)}}\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100\\ \end{array} \]

Alternatives

Alternative 1
Accuracy94.0%
Cost29004
\[\begin{array}{l} t_0 := {\left(1 + \frac{i}{n}\right)}^{n}\\ t_1 := \frac{t_0 + -1}{\frac{i}{n}}\\ \mathbf{if}\;t_1 \leq -100000000000:\\ \;\;\;\;n \cdot \frac{t_0 \cdot 100 + -100}{i}\\ \mathbf{elif}\;t_1 \leq 0:\\ \;\;\;\;100 \cdot \left(\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right) \cdot \frac{n}{i}\right)\\ \mathbf{elif}\;t_1 \leq \infty:\\ \;\;\;\;t_1 \cdot 100\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100\\ \end{array} \]
Alternative 2
Accuracy94.9%
Cost29004
\[\begin{array}{l} t_0 := {\left(1 + \frac{i}{n}\right)}^{n}\\ t_1 := \frac{t_0 + -1}{\frac{i}{n}}\\ \mathbf{if}\;t_1 \leq -2000:\\ \;\;\;\;n \cdot \frac{t_0 \cdot 100 + -100}{i}\\ \mathbf{elif}\;t_1 \leq 0:\\ \;\;\;\;100 \cdot \left(n \cdot \frac{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}{i}\right)\\ \mathbf{elif}\;t_1 \leq \infty:\\ \;\;\;\;t_1 \cdot 100\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100\\ \end{array} \]
Alternative 3
Accuracy95.8%
Cost29004
\[\begin{array}{l} t_0 := {\left(1 + \frac{i}{n}\right)}^{n}\\ t_1 := \frac{t_0 + -1}{\frac{i}{n}}\\ \mathbf{if}\;t_1 \leq -2000:\\ \;\;\;\;n \cdot \frac{t_0 \cdot 100 + -100}{i}\\ \mathbf{elif}\;t_1 \leq 0:\\ \;\;\;\;100 \cdot \frac{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}{\frac{i}{n}}\\ \mathbf{elif}\;t_1 \leq \infty:\\ \;\;\;\;t_1 \cdot 100\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100\\ \end{array} \]
Alternative 4
Accuracy80.8%
Cost7632
\[\begin{array}{l} t_0 := 100 \cdot \frac{n}{\frac{i}{\mathsf{expm1}\left(i\right)}}\\ \mathbf{if}\;n \leq -3.9 \cdot 10^{-175}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;n \leq 1.96 \cdot 10^{-184}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \mathbf{elif}\;n \leq 4.5 \cdot 10^{-67}:\\ \;\;\;\;100 \cdot \frac{i}{\frac{i}{n}}\\ \mathbf{elif}\;n \leq 3.7 \cdot 10^{-50}:\\ \;\;\;\;n \cdot \left(\log \left(\frac{i}{n}\right) \cdot \left(100 \cdot \frac{n}{i}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \]
Alternative 5
Accuracy80.8%
Cost7632
\[\begin{array}{l} t_0 := 100 \cdot \frac{n}{\frac{i}{\mathsf{expm1}\left(i\right)}}\\ \mathbf{if}\;n \leq -9.6 \cdot 10^{-176}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;n \leq 2.65 \cdot 10^{-185}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \mathbf{elif}\;n \leq 4.5 \cdot 10^{-67}:\\ \;\;\;\;100 \cdot \frac{i}{\frac{i}{n}}\\ \mathbf{elif}\;n \leq 3.7 \cdot 10^{-50}:\\ \;\;\;\;100 \cdot \left(n \cdot \left(\frac{n}{i} \cdot \log \left(\frac{i}{n}\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \]
Alternative 6
Accuracy80.8%
Cost7632
\[\begin{array}{l} t_0 := 100 \cdot \frac{n}{\frac{i}{\mathsf{expm1}\left(i\right)}}\\ \mathbf{if}\;n \leq -1.65 \cdot 10^{-174}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;n \leq 1.25 \cdot 10^{-184}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \mathbf{elif}\;n \leq 4.5 \cdot 10^{-67}:\\ \;\;\;\;100 \cdot \frac{i}{\frac{i}{n}}\\ \mathbf{elif}\;n \leq 3.7 \cdot 10^{-50}:\\ \;\;\;\;100 \cdot \left(\frac{n}{\frac{i}{n}} \cdot \log \left(\frac{i}{n}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \]
Alternative 7
Accuracy73.5%
Cost7376
\[\begin{array}{l} t_0 := 100 \cdot \frac{\mathsf{expm1}\left(i\right)}{\frac{i}{n}}\\ \mathbf{if}\;i \leq -9.5 \cdot 10^{-64}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;i \leq -8.6 \cdot 10^{-229}:\\ \;\;\;\;\frac{1}{i} \cdot \frac{i \cdot 100}{\frac{1}{n}}\\ \mathbf{elif}\;i \leq 4 \cdot 10^{-36}:\\ \;\;\;\;n \cdot 100\\ \mathbf{elif}\;i \leq 4.5 \cdot 10^{+161}:\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \end{array} \]
Alternative 8
Accuracy80.1%
Cost7113
\[\begin{array}{l} \mathbf{if}\;n \leq -8 \cdot 10^{-176} \lor \neg \left(n \leq 2.1 \cdot 10^{-152}\right):\\ \;\;\;\;100 \cdot \frac{n}{\frac{i}{\mathsf{expm1}\left(i\right)}}\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \end{array} \]
Alternative 9
Accuracy64.4%
Cost1992
\[\begin{array}{l} \mathbf{if}\;n \leq -7.8 \cdot 10^{-119}:\\ \;\;\;\;100 \cdot \left(n + \left(0.5 + i \cdot 0.16666666666666666\right) \cdot \left(i \cdot n\right)\right)\\ \mathbf{elif}\;n \leq 2.5 \cdot 10^{-151}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \left(n + n \cdot \left(i \cdot \left(i \cdot \left(\frac{0.3333333333333333}{n \cdot n} + \left(0.16666666666666666 + \frac{-0.5}{n}\right)\right) + \left(0.5 + \frac{-0.5}{n}\right)\right)\right)\right)\\ \end{array} \]
Alternative 10
Accuracy64.3%
Cost1097
\[\begin{array}{l} \mathbf{if}\;n \leq -4.2 \cdot 10^{-117} \lor \neg \left(n \leq 1.3 \cdot 10^{-152}\right):\\ \;\;\;\;100 \cdot \left(n + \left(0.5 + i \cdot 0.16666666666666666\right) \cdot \left(i \cdot n\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \end{array} \]
Alternative 11
Accuracy61.0%
Cost976
\[\begin{array}{l} t_0 := n \cdot \left(100 + i \cdot 50\right)\\ t_1 := 100 \cdot \frac{i}{\frac{i}{n}}\\ \mathbf{if}\;n \leq -7200000000000:\\ \;\;\;\;t_0\\ \mathbf{elif}\;n \leq -2.15 \cdot 10^{-265}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;n \leq 1.05 \cdot 10^{-281}:\\ \;\;\;\;50 \cdot \left(i \cdot n\right)\\ \mathbf{elif}\;n \leq 1.6 \cdot 10^{-49}:\\ \;\;\;\;t_1\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \]
Alternative 12
Accuracy61.4%
Cost968
\[\begin{array}{l} \mathbf{if}\;n \leq -1.22 \cdot 10^{-119}:\\ \;\;\;\;100 \cdot \left(n + n \cdot \left(i \cdot 0.5\right)\right)\\ \mathbf{elif}\;n \leq 4.4 \cdot 10^{-108}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;n \cdot \left(100 + i \cdot \left(50 + \frac{-50}{n}\right)\right)\\ \end{array} \]
Alternative 13
Accuracy56.1%
Cost713
\[\begin{array}{l} \mathbf{if}\;i \leq -7.2 \cdot 10^{-13} \lor \neg \left(i \leq 4 \cdot 10^{-36}\right):\\ \;\;\;\;100 \cdot \frac{i}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100\\ \end{array} \]
Alternative 14
Accuracy61.9%
Cost713
\[\begin{array}{l} \mathbf{if}\;n \leq -5.2 \cdot 10^{-119} \lor \neg \left(n \leq 1.05 \cdot 10^{-152}\right):\\ \;\;\;\;n \cdot \left(100 + i \cdot 50\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \end{array} \]
Alternative 15
Accuracy61.9%
Cost712
\[\begin{array}{l} \mathbf{if}\;n \leq -1.25 \cdot 10^{-119}:\\ \;\;\;\;100 \cdot \left(n + n \cdot \left(i \cdot 0.5\right)\right)\\ \mathbf{elif}\;n \leq 7.2 \cdot 10^{-150}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;n \cdot \left(100 + i \cdot 50\right)\\ \end{array} \]
Alternative 16
Accuracy53.9%
Cost452
\[\begin{array}{l} \mathbf{if}\;i \leq 9 \cdot 10^{+28}:\\ \;\;\;\;n \cdot 100\\ \mathbf{else}:\\ \;\;\;\;50 \cdot \left(i \cdot n\right)\\ \end{array} \]
Alternative 17
Accuracy2.8%
Cost192
\[i \cdot -50 \]
Alternative 18
Accuracy48.6%
Cost192
\[n \cdot 100 \]

Reproduce?

herbie shell --seed 2023161 
(FPCore (i n)
  :name "Compound Interest"
  :precision binary64

  :herbie-target
  (* 100.0 (/ (- (exp (* n (if (== (+ 1.0 (/ i n)) 1.0) (/ i n) (/ (* (/ i n) (log (+ 1.0 (/ i n)))) (- (+ (/ i n) 1.0) 1.0))))) 1.0) (/ i n)))

  (* 100.0 (/ (- (pow (+ 1.0 (/ i n)) n) 1.0) (/ i n))))