Compound Interest

Percentage Accurate: 27.7% → 27.7%
Time: 24.1s
Alternatives: 5
Speedup: 1.0×

Specification

?
\[\begin{array}{l} \\ 100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \end{array} \]
(FPCore (i n)
 :precision binary64
 (* 100.0 (/ (- (pow (+ 1.0 (/ i n)) n) 1.0) (/ i n))))
double code(double i, double n) {
	return 100.0 * ((pow((1.0 + (i / n)), n) - 1.0) / (i / n));
}

real(8) function code(i, n)
    real(8), intent (in) :: i
    real(8), intent (in) :: n
    code = 100.0d0 * ((((1.0d0 + (i / n)) ** n) - 1.0d0) / (i / n))
end function

public static double code(double i, double n) {
	return 100.0 * ((Math.pow((1.0 + (i / n)), n) - 1.0) / (i / n));
}

def code(i, n):
	return 100.0 * ((math.pow((1.0 + (i / n)), n) - 1.0) / (i / n))

function code(i, n)
	return Float64(100.0 * Float64(Float64((Float64(1.0 + Float64(i / n)) ^ n) - 1.0) / Float64(i / n)))
end

function tmp = code(i, n)
	tmp = 100.0 * ((((1.0 + (i / n)) ^ n) - 1.0) / (i / n));
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]

\begin{array}{l}

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

Sampling outcomes in binary64 precision:

Local Percentage Accuracy vs ?

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.

Accuracy vs Speed?

Herbie found 5 alternatives:

AlternativeAccuracySpeedup
The accuracy (vertical axis) and speed (horizontal axis) of each alternatives. Up and to the right is better. The red square shows the initial program, and each blue circle shows an alternative.The line shows the best available speed-accuracy tradeoffs.

Initial Program: 27.7% accurate, 1.0× speedup?

\[\begin{array}{l} \\ 100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \end{array} \]
(FPCore (i n)
 :precision binary64
 (* 100.0 (/ (- (pow (+ 1.0 (/ i n)) n) 1.0) (/ i n))))
double code(double i, double n) {
	return 100.0 * ((pow((1.0 + (i / n)), n) - 1.0) / (i / n));
}

real(8) function code(i, n)
    real(8), intent (in) :: i
    real(8), intent (in) :: n
    code = 100.0d0 * ((((1.0d0 + (i / n)) ** n) - 1.0d0) / (i / n))
end function

public static double code(double i, double n) {
	return 100.0 * ((Math.pow((1.0 + (i / n)), n) - 1.0) / (i / n));
}

def code(i, n):
	return 100.0 * ((math.pow((1.0 + (i / n)), n) - 1.0) / (i / n))

function code(i, n)
	return Float64(100.0 * Float64(Float64((Float64(1.0 + Float64(i / n)) ^ n) - 1.0) / Float64(i / n)))
end

function tmp = code(i, n)
	tmp = 100.0 * ((((1.0 + (i / n)) ^ n) - 1.0) / (i / n));
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]

\begin{array}{l}

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

Alternative 1: 27.7% accurate, 1.0× speedup?

\[\begin{array}{l} \\ 100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \end{array} \]
(FPCore (i n)
 :precision binary64
 (* 100.0 (/ (- (pow (+ 1.0 (/ i n)) n) 1.0) (/ i n))))
double code(double i, double n) {
	return 100.0 * ((pow((1.0 + (i / n)), n) - 1.0) / (i / n));
}

real(8) function code(i, n)
    real(8), intent (in) :: i
    real(8), intent (in) :: n
    code = 100.0d0 * ((((1.0d0 + (i / n)) ** n) - 1.0d0) / (i / n))
end function

public static double code(double i, double n) {
	return 100.0 * ((Math.pow((1.0 + (i / n)), n) - 1.0) / (i / n));
}

def code(i, n):
	return 100.0 * ((math.pow((1.0 + (i / n)), n) - 1.0) / (i / n))

function code(i, n)
	return Float64(100.0 * Float64(Float64((Float64(1.0 + Float64(i / n)) ^ n) - 1.0) / Float64(i / n)))
end

function tmp = code(i, n)
	tmp = 100.0 * ((((1.0 + (i / n)) ^ n) - 1.0) / (i / n));
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]

\begin{array}{l}

\\
100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}}
\end{array}
Derivation

  1. Initial program 26.7%

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

Alternative 2: 27.7% accurate, 1.0× speedup?

\[\begin{array}{l} \\ 100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \end{array} \]
(FPCore (i n)
 :precision binary64
 (* 100.0 (/ (+ (pow (+ 1.0 (/ i n)) n) -1.0) (/ i n))))
double code(double i, double n) {
	return 100.0 * ((pow((1.0 + (i / n)), n) + -1.0) / (i / n));
}

real(8) function code(i, n)
    real(8), intent (in) :: i
    real(8), intent (in) :: n
    code = 100.0d0 * ((((1.0d0 + (i / n)) ** n) + (-1.0d0)) / (i / n))
end function

public static double code(double i, double n) {
	return 100.0 * ((Math.pow((1.0 + (i / n)), n) + -1.0) / (i / n));
}

def code(i, n):
	return 100.0 * ((math.pow((1.0 + (i / n)), n) + -1.0) / (i / n))

function code(i, n)
	return Float64(100.0 * Float64(Float64((Float64(1.0 + Float64(i / n)) ^ n) + -1.0) / Float64(i / n)))
end

function tmp = code(i, n)
	tmp = 100.0 * ((((1.0 + (i / n)) ^ n) + -1.0) / (i / n));
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]

\begin{array}{l}

\\
100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}}
\end{array}
Derivation

  1. Initial program 26.7%

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

Alternative 3: 27.8% accurate, 1.0× speedup?

\[\begin{array}{l} \\ \frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{i} \cdot \left(n \cdot 100\right) \end{array} \]
(FPCore (i n)
 :precision binary64
 (* (/ (+ (pow (+ 1.0 (/ i n)) n) -1.0) i) (* n 100.0)))
double code(double i, double n) {
	return ((pow((1.0 + (i / n)), n) + -1.0) / i) * (n * 100.0);
}

real(8) function code(i, n)
    real(8), intent (in) :: i
    real(8), intent (in) :: n
    code = ((((1.0d0 + (i / n)) ** n) + (-1.0d0)) / i) * (n * 100.0d0)
end function

public static double code(double i, double n) {
	return ((Math.pow((1.0 + (i / n)), n) + -1.0) / i) * (n * 100.0);
}

def code(i, n):
	return ((math.pow((1.0 + (i / n)), n) + -1.0) / i) * (n * 100.0)

function code(i, n)
	return Float64(Float64(Float64((Float64(1.0 + Float64(i / n)) ^ n) + -1.0) / i) * Float64(n * 100.0))
end

function tmp = code(i, n)
	tmp = ((((1.0 + (i / n)) ^ n) + -1.0) / i) * (n * 100.0);
end

code[i_, n_] := N[(N[(N[(N[Power[N[(1.0 + N[(i / n), $MachinePrecision]), $MachinePrecision], n], $MachinePrecision] + -1.0), $MachinePrecision] / i), $MachinePrecision] * N[(n * 100.0), $MachinePrecision]), $MachinePrecision]

\begin{array}{l}

\\
\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{i} \cdot \left(n \cdot 100\right)
\end{array}
Derivation

  1. Initial program 27.0%

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

Alternative 4: 27.7% accurate, 0.5× speedup?

\[\begin{array}{l} \\ \frac{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, -100\right)}{\frac{i}{n}} \end{array} \]
(FPCore (i n)
 :precision binary64
 (/ (fma 100.0 (pow (+ 1.0 (/ i n)) n) -100.0) (/ i n)))
double code(double i, double n) {
	return fma(100.0, pow((1.0 + (i / n)), n), -100.0) / (i / n);
}

function code(i, n)
	return Float64(fma(100.0, (Float64(1.0 + Float64(i / n)) ^ n), -100.0) / Float64(i / n))
end

code[i_, n_] := N[(N[(100.0 * N[Power[N[(1.0 + N[(i / n), $MachinePrecision]), $MachinePrecision], n], $MachinePrecision] + -100.0), $MachinePrecision] / N[(i / n), $MachinePrecision]), $MachinePrecision]

\begin{array}{l}

\\
\frac{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, -100\right)}{\frac{i}{n}}
\end{array}
Derivation

  1. Initial program 26.7%

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

Alternative 5: 27.8% accurate, 0.5× speedup?

\[\begin{array}{l} \\ n \cdot \frac{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, -100\right)}{i} \end{array} \]
(FPCore (i n)
 :precision binary64
 (* n (/ (fma 100.0 (pow (+ 1.0 (/ i n)) n) -100.0) i)))
double code(double i, double n) {
	return n * (fma(100.0, pow((1.0 + (i / n)), n), -100.0) / i);
}

function code(i, n)
	return Float64(n * Float64(fma(100.0, (Float64(1.0 + Float64(i / n)) ^ n), -100.0) / i))
end

code[i_, n_] := N[(n * N[(N[(100.0 * N[Power[N[(1.0 + N[(i / n), $MachinePrecision]), $MachinePrecision], n], $MachinePrecision] + -100.0), $MachinePrecision] / i), $MachinePrecision]), $MachinePrecision]

\begin{array}{l}

\\
n \cdot \frac{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, -100\right)}{i}
\end{array}
Derivation

  1. Initial program 27.0%

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

Developer target: 34.1% accurate, 0.5× speedup?

\[\begin{array}{l} \\ \begin{array}{l} t_0 := 1 + \frac{i}{n}\\ 100 \cdot \frac{e^{n \cdot \begin{array}{l} \mathbf{if}\;t_0 = 1:\\ \;\;\;\;\frac{i}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{i}{n} \cdot \log t_0}{\left(\frac{i}{n} + 1\right) - 1}\\ \end{array}} - 1}{\frac{i}{n}} \end{array} \end{array} \]
(FPCore (i n)
 :precision binary64
 (let* ((t_0 (+ 1.0 (/ i n))))
   (*
    100.0
    (/
     (-
      (exp
       (*
        n
        (if (== t_0 1.0)
          (/ i n)
          (/ (* (/ i n) (log t_0)) (- (+ (/ i n) 1.0) 1.0)))))
      1.0)
     (/ i n)))))
double code(double i, double n) {
	double t_0 = 1.0 + (i / n);
	double tmp;
	if (t_0 == 1.0) {
		tmp = i / n;
	} else {
		tmp = ((i / n) * log(t_0)) / (((i / n) + 1.0) - 1.0);
	}
	return 100.0 * ((exp((n * tmp)) - 1.0) / (i / n));
}

real(8) function code(i, n)
    real(8), intent (in) :: i
    real(8), intent (in) :: n
    real(8) :: t_0
    real(8) :: tmp
    t_0 = 1.0d0 + (i / n)
    if (t_0 == 1.0d0) then
        tmp = i / n
    else
        tmp = ((i / n) * log(t_0)) / (((i / n) + 1.0d0) - 1.0d0)
    end if
    code = 100.0d0 * ((exp((n * tmp)) - 1.0d0) / (i / n))
end function

public static double code(double i, double n) {
	double t_0 = 1.0 + (i / n);
	double tmp;
	if (t_0 == 1.0) {
		tmp = i / n;
	} else {
		tmp = ((i / n) * Math.log(t_0)) / (((i / n) + 1.0) - 1.0);
	}
	return 100.0 * ((Math.exp((n * tmp)) - 1.0) / (i / n));
}

def code(i, n):
	t_0 = 1.0 + (i / n)
	tmp = 0
	if t_0 == 1.0:
		tmp = i / n
	else:
		tmp = ((i / n) * math.log(t_0)) / (((i / n) + 1.0) - 1.0)
	return 100.0 * ((math.exp((n * tmp)) - 1.0) / (i / n))

function code(i, n)
	t_0 = Float64(1.0 + Float64(i / n))
	tmp = 0.0
	if (t_0 == 1.0)
		tmp = Float64(i / n);
	else
		tmp = Float64(Float64(Float64(i / n) * log(t_0)) / Float64(Float64(Float64(i / n) + 1.0) - 1.0));
	end
	return Float64(100.0 * Float64(Float64(exp(Float64(n * tmp)) - 1.0) / Float64(i / n)))
end

function tmp_2 = code(i, n)
	t_0 = 1.0 + (i / n);
	tmp = 0.0;
	if (t_0 == 1.0)
		tmp = i / n;
	else
		tmp = ((i / n) * log(t_0)) / (((i / n) + 1.0) - 1.0);
	end
	tmp_2 = 100.0 * ((exp((n * tmp)) - 1.0) / (i / n));
end

code[i_, n_] := Block[{t$95$0 = N[(1.0 + N[(i / n), $MachinePrecision]), $MachinePrecision]}, N[(100.0 * N[(N[(N[Exp[N[(n * If[Equal[t$95$0, 1.0], N[(i / n), $MachinePrecision], N[(N[(N[(i / n), $MachinePrecision] * N[Log[t$95$0], $MachinePrecision]), $MachinePrecision] / N[(N[(N[(i / n), $MachinePrecision] + 1.0), $MachinePrecision] - 1.0), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]], $MachinePrecision] - 1.0), $MachinePrecision] / N[(i / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

\begin{array}{l}

\\
\begin{array}{l}
t_0 := 1 + \frac{i}{n}\\
100 \cdot \frac{e^{n \cdot \begin{array}{l}
\mathbf{if}\;t_0 = 1:\\
\;\;\;\;\frac{i}{n}\\

\mathbf{else}:\\
\;\;\;\;\frac{\frac{i}{n} \cdot \log t_0}{\left(\frac{i}{n} + 1\right) - 1}\\


\end{array}} - 1}{\frac{i}{n}}
\end{array}
\end{array}

Reproduce

?
herbie shell --seed 2023276 
(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))))