Question

The time to failure T of a component has probability density f ( t ) as shown
(b) Derive the corresponding survivor function R ( t ) .
(c) Derive the corresponding failure rate function z ( t ) , and make a sketch of z(t)

Note: The f(t) is a valid pdf (so we can obtain c or the height of the triangle). Information are enough to solve this problem.

f(t) a -b a b Time t Fig. 2.27 Probability density (Problem 2.28)

Answer 1

  $f(t)=\left\{\begin{matrix} \frac{c-0}{a-(a-b)}[t-(a-b)] &\;\;\;t\in [a-b,a] \\ \frac{c-0}{a-(a+b)}[t-(a+b)] &\;\;\;t\in [a,a+b] \end{matrix}\right.$

$\Rightarrow f(t)=\left\{\begin{matrix} \;\;\frac{c}{b}[t-a+b] &\;\;\;t\in [a-b,a] \\ -\frac{c}{b}[t-a-b] &\;\;\;t\in [a,a+b] \end{matrix}\right.$

Now   $\int_{a-b}^{a+b}f(t)dt=1$ i.e. area under the curve of f(t) = 1 $\Rightarrow$ (1/2)[(a+b)-(a-b)]c = 1 $\Rightarrow$ bc = 1 $\Rightarrow$ c = 1/b

$\therefore f(t)=\left\{\begin{matrix} \;\;\frac{1}{b^2}(t-a+b) &\;\;\;t\in [a-b,a] \\ -\frac{1}{b^2}(t-a-b) &\;\;\;t\in [a,a+b] \end{matrix}\right.$

$Now\;\;F(t)=P(T\leqslant t)=\int_{a-b}^{t}f(t)dt=\left\{\begin{matrix} \int_{a-b}^{t}f(t)dt &\;\;\;x\in[a-b,a] \\ \int_{a-b}^{a}f(t)dt+\int_{a}^{a+b}f(t)dt & \;\;\;x\in[a,a+b] \end{matrix}\right.$

$=\left\{\begin{matrix} \frac{1}{b^2}\int_{a-b}^{t}(t-a+b)dt & \;\;\;t\in[a-b,a]\\ \frac{1}{b^2}\int_{a-b}^{a}(t-a+b)dt-\frac{1}{b^2}\int_{a}^{t}(t-a-b)dt &\;\;\;t\in[a,a+b] \end{matrix}\right.$

$=\left\{\begin{matrix} \;\;\frac{1}{2b^2}(t-a+b)^2 &\;\;\;t\in [a-b,a] \\\frac{1}{2} -\frac{1}{2b^2}(t-a)(t-a-2b) &\;\;\;t\in [a,a+b] \end{matrix}\right.$

  $=\left\{\begin{matrix} \;\;\frac{1}{2b^2}(t-a+b)^2 &\;\;\;t\in [a-b,a] \\\frac{1}{2} +\frac{1}{2b^2}(t-a)(a+2b-t) &\;\;\;t\in [a,a+b] \end{matrix}\right.$

(b)

$R(t)=1-F(t)=\left\{\begin{matrix} \;\;1-\frac{1}{2b^2}(t-a+b)^2 &\;\;\;t\in [a-b,a] \\\frac{1}{2} -\frac{1}{2b^2}(t-a)(a+2b-t) &\;\;\;t\in [a,a+b] \end{matrix}\right.$

(c)

$h(t)=\frac{f(t)}{R(t)}=\left\{\begin{matrix} \frac{\frac{1}{b^2}(t-a+b)}{1-\frac{1}{2b^2}(t-a+b)^2} & \;\;\;t\in[a-b,b]\\ \frac{-\frac{1}{b^2}(t-a-b)}{\frac{1}{2}-\frac{1}{2b^2}(t-a)(a+2b-t)} & \;\;\;t\in[a,a+b] \end{matrix}\right. =\left\{\begin{matrix} \frac{2(t-a+b)}{2b^2-(t-a+b)^2} & \;\;\;t\in[a-b,b]\\ \frac{-2(t-a-b)}{b^2-(t-a)(a+2b-t)} & \;\;\;t\in[a,a+b] \end{matrix}\right.$