Difference between revisions of "Aufgaben:Exercise 1.5Z: Probabilities of Default"
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− | {{quiz-Header|Buchseite= | + | {{quiz-Header|Buchseite=Theory_of_Stochastic_Signals/Statistical_Dependence_and_Independence}} |
− | [[File:P_ID87__Sto_Z_1_5.png|right|frame| | + | [[File:P_ID87__Sto_Z_1_5.png|right|frame|Functional circuit diagram of a device with two identical parts $T_1$ and $T_2$]] |
− | + | A sub-device 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 $p_{\rm B}$. |
− | * | + | *Sub-device $T_1$ functions only if all $n$ components are functional. |
− | + | To increase reliability, important assemblies are often duplicated. The overall device $G$ can be described in terms of set theory as follows: | |
− | $$ G = T_1 \cup T_2.$$ | + | :$$ G = T_1 \cup T_2.$$ |
− | + | This means: The overall device $G$ is already operational if at least one of the two identical sub-devices $(T_1$ or $T_2)$ is functional. | |
Line 19: | Line 19: | ||
− | + | Hints: | |
− | * | + | *The exercise belongs to the chapter [[Theory_of_Stochastic_Signals/Statistical_Dependence_and_Independence|Statistical dependence and independence]]. |
− | * | + | *The topic of this chapter is illustrated with examples in the (German language) learning video |
− | :[[Statistische_Abhängigkeit_und_Unabhängigkeit_(Lernvideo)|Statistische Abhängigkeit und Unabhängigkeit]]. | + | ::[[Statistische_Abhängigkeit_und_Unabhängigkeit_(Lernvideo)|Statistische Abhängigkeit und Unabhängigkeit]] $\Rightarrow$ "Statistical dependence and independence". |
+ | *Explanations: | ||
+ | #Component (German: "Bauteil" ⇒ subscript "B") | ||
+ | # Sub-device (German: "Teilgerät" ⇒ subscript "T") | ||
+ | #Overall device (German: "Gesamtgerät" ⇒ subscript "G") | ||
− | === | + | ===Questions=== |
<quiz display=simple> | <quiz display=simple> | ||
− | { | + | {The default probability $p_{\rm G}$ of the overall device must not be greater than $0.04\%$. <br>How large may then the default probabilities $p_{\rm T}$ of the two identical sub-devices existing in parallel be at most? |
|type="{}"} | |type="{}"} | ||
$p_\text{T, max} \ = \ $ { 2 3% } $ \ \%$ | $p_\text{T, max} \ = \ $ { 2 3% } $ \ \%$ | ||
− | { | + | {Let the default probability of all components be $\underline{p_{\rm B} = 0.1}$. Let each sub-device consist of $n = 3$ components. <br>Calculate the probability $p_{\rm T}$ exactly that a sub-device defaults. |
|type="{}"} | |type="{}"} | ||
− | $ | + | $p_{\rm T} \ = \ $ { 27.1 3% } $ \ \%$ |
− | { | + | {What value is obtained for $\underline{p_{\rm B} = 0.01}$? In what form can you approximate $p_{\rm T}$ for small values of $p_{\rm B}$ ? |
|type="{}"} | |type="{}"} | ||
− | $ | + | $p_{\rm T} \ = \ $ { 2.97 3% } $ \ \%$ |
− | { | + | {Now apply $p_{\rm B} = 0.4\%$ for the default probability of all components. What is the maximum number of components the sub-device can contain? $p_{\rm T} ≤ 2\%$ is to hold. |
|type="{}"} | |type="{}"} | ||
$n \ = \ $ { 5 3% } | $n \ = \ $ { 5 3% } | ||
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</quiz> | </quiz> | ||
− | === | + | ===Solution=== |
{{ML-Kopf}} | {{ML-Kopf}} | ||
− | '''(1)''' | + | '''(1)''' Since the two sub-devices default independently, set-theoretically holds: |
− | :$$\rm Pr( | + | :$${\rm Pr(}G \text{ drops out)} = {\rm Pr(}T_1 \text{ drops out)} \cdot {\rm Pr(}T_2 \text{ drops out)}. $$ |
− | + | *Moreover, since the sub-devices $T_1$ and $T_2$ are identical in construction, they default with the same probability $p_{\rm T}.$ It follows that: | |
− | :$$p_{\rm G} = \it p_{\rm T}^{\rm 2} \hspace{0.5cm} \ | + | :$$p_{\rm G} = \it p_{\rm T}^{\rm 2} \hspace{0.5cm} \Rightarrow \hspace{0.5cm} \rm \it p_{\rm T,\hspace{0.1cm}max}= \sqrt{\it p_{\rm G}} \le \rm\sqrt{0.0004} \hspace{0.15cm}\underline {= 2\%}.$$ |
− | |||
− | |||
− | |||
− | |||
− | '''( | + | '''(2)''' This result is easier to determine using the complementary event: |
+ | :$$\rm Pr(\it T_{\rm 1}\hspace{0.1cm}\rm functions) = \rm Pr(\it B_{\rm 1} \hspace{0.1cm}\rm functions \cap \it B_{\rm 2} \hspace{0.1cm} \rm functions \cap \it B_{\rm 3}\hspace{0.1cm} \rm functions).$$ | ||
+ | :$$\Rightarrow 1- p_{\rm T}= (1-p_{\rm B})^{3} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} | ||
+ | 1-p_{\rm T}=(0.9)^3= 0.729 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} p_{\rm T}\hspace{0.15cm}\underline {= 0.271 = 27.1\%}.$$ | ||
− | '''(4)''' | + | |
− | $$0.996^{\it n}\ge 0.98 \hspace{0.5cm} \rm\Rightarrow \hspace{0.5cm} \it n\le\rm\frac{log(0.98)}{log(0.996)} = 5.0406\hspace{0.15cm}\underline { \approx 5}.$$ | + | '''(3)''' With $p_{\rm B} = 0.01$, we obtain $p_{\rm T}\hspace{0.15cm}\underline {= 2.97\%}.$ |
+ | *In general, the approximation is: $p_{\rm T} \approx n \cdot p_{\rm B}\; (= 3\%)$. | ||
+ | |||
+ | |||
+ | |||
+ | '''(4)''' With the approximation of the last subtask ⇒ $\underline{n = 5}$ follows directly. | ||
+ | *For larger $p_{\rm B}$, one would have to proceed as follows: | ||
+ | :$$0.996^{\it n}\ge 0.98 \hspace{0.5cm} \rm\Rightarrow \hspace{0.5cm} \it n\le\rm\frac{log(0.98)}{log(0.996)} = 5.0406\hspace{0.15cm}\underline { \approx 5}.$$ | ||
{{ML-Fuß}} | {{ML-Fuß}} | ||
− | [[Category: | + | [[Category:Theory of Stochastic Signals: Exercises|^1.3 Statistical Dependence/Independence^]] |
Latest revision as of 13:26, 1 December 2021
A sub-device 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 $p_{\rm B}$.
- Sub-device $T_1$ functions only if all $n$ components are functional.
To increase reliability, important assemblies are often duplicated. The overall device $G$ can be described in terms of set theory as follows:
- $$ G = T_1 \cup T_2.$$
This means: The overall device $G$ is already operational if at least one of the two identical sub-devices $(T_1$ 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".
- Explanations:
- Component (German: "Bauteil" ⇒ subscript "B")
- Sub-device (German: "Teilgerät" ⇒ subscript "T")
- Overall device (German: "Gesamtgerät" ⇒ subscript "G")
Questions
Solution
(1) Since the two sub-devices default independently, set-theoretically holds:
- $${\rm Pr(}G \text{ drops out)} = {\rm Pr(}T_1 \text{ drops out)} \cdot {\rm Pr(}T_2 \text{ drops out)}. $$
- Moreover, since the sub-devices $T_1$ and $T_2$ are identical in construction, they default with the same probability $p_{\rm T}.$ It follows that:
- $$p_{\rm G} = \it p_{\rm T}^{\rm 2} \hspace{0.5cm} \Rightarrow \hspace{0.5cm} \rm \it p_{\rm T,\hspace{0.1cm}max}= \sqrt{\it p_{\rm G}} \le \rm\sqrt{0.0004} \hspace{0.15cm}\underline {= 2\%}.$$
(2) This result is easier to determine using the complementary event:
- $$\rm Pr(\it T_{\rm 1}\hspace{0.1cm}\rm functions) = \rm Pr(\it B_{\rm 1} \hspace{0.1cm}\rm functions \cap \it B_{\rm 2} \hspace{0.1cm} \rm functions \cap \it B_{\rm 3}\hspace{0.1cm} \rm functions).$$
- $$\Rightarrow 1- p_{\rm T}= (1-p_{\rm B})^{3} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 1-p_{\rm T}=(0.9)^3= 0.729 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} p_{\rm T}\hspace{0.15cm}\underline {= 0.271 = 27.1\%}.$$
(3) With $p_{\rm B} = 0.01$, we obtain $p_{\rm T}\hspace{0.15cm}\underline {= 2.97\%}.$
- In general, the approximation is: $p_{\rm T} \approx n \cdot p_{\rm B}\; (= 3\%)$.
(4) With the approximation of the last subtask ⇒ $\underline{n = 5}$ follows directly.
- For larger $p_{\rm B}$, one would have to proceed as follows:
- $$0.996^{\it n}\ge 0.98 \hspace{0.5cm} \rm\Rightarrow \hspace{0.5cm} \it n\le\rm\frac{log(0.98)}{log(0.996)} = 5.0406\hspace{0.15cm}\underline { \approx 5}.$$