Exercise 1.5Z: Probabilities of Default

• Return to book

Functional circuit diagram of a device

A device part is composed of the components  $B_1, \ B_2,\ \text{...} \ , B_n$  where the respective functionality can be assumed to be independent of all other components.

• Assume that all components default with equal probability  $p_{\rm A}$ .
• Part  $T_1$  functions only if all  $n$  components are functional.

To increase reliability, important assemblies are often duplicated. The device  $G$  can thus be described in terms of set theory as follows:

$$ G = T_1 \cup T_2.$$

This means:   Device  $G$  is already operational if at least one of the two identical subassemblies  $(T_1$  or  $T_2)$  is functional.

Hints:

• The exercise belongs to the chapter  Statistical dependence and independence.

• The topic of this chapter is illustrated with examples in the  (German language)  learning video

Statistische Abhängigkeit und Unabhängigkeit   $\Rightarrow$   "Statistical dependence and independence".

Questions

1

The default probability  $p_{\rm G}$  of the total device must not be greater than  $0.04\%$.
How large may then the default probabilities  $p_{\rm T}$  of the two identical device parts existing in parallel be at most?

$p_\text{T, max} \ = \ $ $ \ \%$

2

Let the default probability of all components be  $\underline{p_{\rm A} = 0.1}$.  Let each subdevice consist of  $n = 3$  components.
Calculate the probability  $p_{\rm T}$  exactly that a subdevice defaults.

$p_{\rm T} \ = \ $ $ \ \%$

3

What value is obtained for  $\underline{p_{\rm A} = 0.01}$?  In what form can you approximate  $p_{\rm T}$  for small values of  $p_{\rm A}$ ?

$p_{\rm T} \ = \ $ $ \ \%$

4

Now apply  $p_{\rm A} = 0.4\%$ for the default probability of all components.   What is the maximum number of components the subdevice can contain  $p_{\rm T} ≤ 2\%$  is to hold?

$n \ = \ $

Solution

Solution

(1)  Since the two subdevices default independently, set-theoretically holds:

$$\rm Pr(\it G \rm \hspace{0.1cm}f\ddot{a}llt\hspace{0.1cm}aus) = Pr(\it T_{\rm 1}\rm \hspace{0.1cm} f\ddot{a}llt \hspace{0.1cm}aus) \cdot Pr(\it T_{\rm 2}\rm \hspace{0.1cm} f\ddot{a}llt \hspace{0.1cm}aus). $$

• Moreover, since subdevices  $T_1$  and  $T_2$  are identical in construction, they default with the same probability  $p_{\rm T}$ . It follows that:

$$p_{\rm G} = \it p_{\rm T}^{\rm 2} \hspace{0.5cm} \rm bzw. \hspace{0.5cm} \rm \it p_{\rm T,\hspace{0.1cm}max}= \sqrt{\it p_{\rm G}} \le \rm\sqrt{0.0004} \hspace{0.15cm}\underline {= 2\%}.$$

(2)  This result is easier to determine using the complementary event:

$$\rm Pr(\it T_{\rm 1}\hspace{0.1cm}\rm functions) = \rm Pr(\it B_{\rm 1} \hspace{0.1cm}\rm functions \cap \it B_{\rm 2} \hspace{0.1cm} \rm functions \cap \it B_{\rm 3}\hspace{0.1cm} \rm functions).$$

$$\Rightarrow 1- p_{\rm T}= (1-p_{\rm A})^{3} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 1-p_{\rm T}=(0.9)^3= 0.729 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} p_{\rm T}\hspace{0.15cm}\underline {= 0.271 = 27.1\%}.$$

(3)  With  $p_{\rm A} = 0.01$ , we obtain  $p_{\rm T}\hspace{0.15cm}\underline {= 2.97\%}.$

• In general, the approximation is:  $p_{\rm T} \approx n \cdot p_{\rm A}\; (= 3\%)$.

(4)  With the approximation of the last subtaks  $\underline{n = 5}$ follows directly.

• For larger  $p_{\rm A}$ , one would have to proceed as follows:

$$0.996^{\it n}\ge 0.98 \hspace{0.5cm} \rm\Rightarrow \hspace{0.5cm} \it n\le\rm\frac{log(0.98)}{log(0.996)} = 5.0406\hspace{0.15cm}\underline { \approx 5}.$$