Relative error and percentage relative error

3 Relative error and percentage relative error

Two other measures of error can be obtained from a knowledge of the expression for the absolute error. As mentioned earlier, the relative error in $f$ is $\frac{δ f}{f}$ and the percentage relative error is $(\frac{δ f}{f} \times 100) %$ . Suppose that $f (x, y) = x^{2} + y^{2} + x y$ then

\begin{array}{rcl} δ f & ≃ & \frac{\partial f}{\partial x} δ x + \frac{\partial f}{\partial y} δ y \\ = & (2 x + y) δ x + (2 y + x) δ y \end{array}

The relative error is

\begin{array}{rcl} \frac{δ f}{f} & ≃ & \frac{1}{f} \frac{\partial f}{\partial x} δ x + \frac{1}{f} \frac{\partial f}{\partial y} δ y \\ = & \frac{(2 x + y)}{x^{2} + y^{2} + x y} δ x + \frac{(2 y + x)}{x^{2} + y^{2} + x y} δ y \end{array}

The actual value of the relative error can be obtained if the actual errors of the independent variables are known and the values of $x$ and $y$ at the point of interest. In the special case where the function is a combination of powers of the input variables then there is a short cut to finding the relative error in the value of the function. For example, if $f (x, y, u) = \frac{x^{2} y^{4}}{u^{3}}$ then

$\frac{\partial f}{\partial x} = \frac{2 x y^{4}}{u^{3}}, \frac{\partial f}{\partial y} = \frac{4 x^{2} y^{3}}{u^{3}}, \frac{\partial f}{\partial u} = - \frac{3 x^{2} y^{4}}{u^{4}}$

Hence

$δ f ≃ \frac{2 x y^{4}}{u^{3}} δ x + \frac{4 x^{2} y^{3}}{u^{3}} δ y - \frac{3 x^{2} y^{4}}{u^{4}} δ u$

Finally,

$\frac{δ f}{f} ≃ \frac{2 x y^{4}}{u^{3}} \times \frac{u^{3}}{x^{2} y^{4}} δ x + \frac{4 x^{2} y^{3}}{u^{3}} \times \frac{u^{3}}{x^{2} y^{4}} δ y - \frac{3 x^{2} y^{4}}{u^{4}} \times \frac{u^{3}}{x^{2} y^{4}} δ u$

Cancelling down the fractions,

$\frac{δ f}{f} ≃ 2 \frac{δ x}{x} + 4 \frac{δ y}{y} - 3 \frac{δ u}{u} (1)$

so that

rel. error in $f ≃ 2 \times$ (rel. error in $x$ )+ $4 \times$ (rel. error in $y$ )- $3 \times$ (rel. error in $u$ ).

Note that if we write

$f (x, y, u) = x^{2} y^{4} u^{- 3}$

we see that the coefficients of the relative errors on the right-hand side of $(1)$ are the powers of the appropriate variable.

To find the percentage relative error we simply multiply the relative error by $100$ .

Task!

If $f = \frac{x^{3} y}{u^{2}}$ and $x, y, u$ are subject to percentage relative errors of $1 %, - 1 %$ and $2 %$ respectively find the approximate percentage relative error in $f$ .

First find $\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}$ and $\frac{\partial f}{\partial u}$ :

$\frac{\partial f}{\partial x} = \frac{3 x^{2} y}{u^{2}}, \frac{\partial f}{\partial y} = \frac{x^{3}}{u^{2}}, \frac{\partial f}{\partial u} = - \frac{2 x^{3} y}{u^{3}}$ .

Now write down an expression for $δ f$

$δ f ≃ \frac{3 x^{2} y}{u^{2}} δ x + \frac{x^{3}}{u^{2}} δ y - \frac{2 x^{3} y}{u^{3}} δ u$

Hence write down an expression for the percentage relative error in $f$ :

$\frac{δ f}{f} \times 100 ≃ \frac{3 x^{2} y}{u^{2}} \times \frac{u^{2}}{x^{3} y} δ x \times 100 + \frac{x^{3}}{u^{2}} \times \frac{u^{2}}{x^{3} y} δ y \times 100 - \frac{2 x^{3} y}{u^{3}} \times \frac{u^{2}}{x^{3} y} δ u \times 100$

Finally, calculate the value of the percentage relative error:

\begin{array}{rcl} \frac{δ f}{f} \times 100 & ≃ & 3 \frac{δ x}{x} \times 100 + \frac{δ y}{y} \times 100 - 2 \frac{δ u}{u} \times 100 \\ = & 3 (1) - 1 - 2 (2) = - 2 % \end{array}

Note that $f = x^{3} y u^{- 2}$ .

3 Relative error and percentage relative error

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