Aufgaben:Exercise 5.6Z: Gilbert-Elliott Model - Revision history

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 *Furthermore, the mean error probability (with given $p_{\rm G}$ and $p_{\rm B}$) can also be represented as follows:
-:$$p_{\rm M} = \frac{p_{\rm G} \cdot {\rm Pr}(\rm
+:$$p_{\rm M} = \frac{p_{\rm G} \cdot {\rm Pr}(\rmG\hspace{0.05cm}|\hspace{0.05cm} B)+ p_{\rm B} \cdot {\rm Pr}(\rmB\hspace{0.05cm}|\hspace{0.05cm} G)}{{\rm Pr}(\rmG\hspace{0.05cm}|\hspace{0.05cm} B) + {\rm Pr}(\rmB\hspace{0.05cm}|\hspace{0.05cm} G)} =  \frac{0.001 \cdot {\rmPr}(\rm G\hspace{0.05cm}|\hspace{0.05cm} B)+ 0.1 \cdot {\rmPr}(\rm B\hspace{0.05cm}|\hspace{0.05cm} G)}{{\rm Pr}(\rmG\hspace{0.05cm}|\hspace{0.05cm} B) + {\rm Pr}(\rmB\hspace{0.05cm}|\hspace{0.05cm} G)}\hspace{0.05cm}.$$*With $\rm Pr(B|G) = 0.01$ and with $\rm Pr(G|B) = 0.1$, respectively, the following equations are obtained::$${\rm Pr}(\rm B\hspace{0.05cm}|\hspace{0.05cm} G) =0.01\hspace{-0.15cm}:\hspace{0.2cm} {\it p}_{\rm M} \hspace{-0.1cm} \ = \\hspace{-0.1cm}  \frac{0.001 \cdot {\rm Pr}(\rmG\hspace{0.05cm}|\hspace{0.05cm} B)+ 0.001 }{{\rm Pr}(\rmG\hspace{0.05cm}|\hspace{0.05cm} B) +0.01}\hspace{0.05cm},\hspace{0.5cm}{\rm Pr}(\rm G\hspace{0.05cm}|\hspace{0.05cm} B) =0.1\hspace{-0.15cm}:\hspace{0.2cm}{\it p}_{\rm M} \hspace{-0.1cm} \ = \\hspace{-0.1cm} \frac{0.0001 + 0.1 \cdot {\rm Pr}(\rmB\hspace{0.05cm}|\hspace{0.05cm} G)}{0.1 +{\rm Pr}(\rmG\hspace{0.05cm}|\hspace{0.05cm} B) }\hspace{0.05cm}.$$
 *From the last equation it can be seen that with no $\rm Pr(G|B)$ value the result $p_{\rm M} = 0.005$ is possible.

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	'''(2)'''  The GE model is a stationary Markov chain.		'''(2)'''  The GE model is a stationary Markov chain.
	*For the probability that it is in the state "GOOD", taking into account the result of subtask '''(1)''':		*For the probability that it is in the state "GOOD", taking into account the result of subtask '''(1)''':
	:$$w_{\rm G} = {\rm Pr}(\rm G\hspace{0.05cm}\|\hspace{0.05cm} G) \~~cdot~~		:$$w_{\rm G} = {\rm Pr}(\rm G\hspace{0.05cm}\|\hspace{0.05cm} G) \cdotw_{\rm G} + {\rm Pr}(\rm G\hspace{0.05cm}\|\hspace{0.05cm} B) \cdotw_{\rm B}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} {\rm Pr}(\rmB\hspace{0.05cm}\|\hspace{0.05cm} G) \cdot w_{\rm G} = {\rm Pr}(\rmG\hspace{0.05cm}\|\hspace{0.05cm} B) \cdot w_{\rm B}\hspace{0.05cm}.$$*Further, $w_{\rm B} = 1 \, –w_{\rm G}$::$${\rm Pr}(\rmB\hspace{0.05cm}\|\hspace{0.05cm} G) \cdot w_{\rm G} + {\rm Pr}(\rmG\hspace{0.05cm}\|\hspace{0.05cm} B) \cdot w_{\rm G} = {\rm Pr}(\rmG\hspace{0.05cm}\|\hspace{0.05cm} B)$$:$$\Rightarrow \hspace{0.3cm} w_{\rm G} = \frac{{\rm Pr}(\rmG\hspace{0.05cm}\|\hspace{0.05cm} B)}{{\rm Pr}(\rmG\hspace{0.05cm}\|\hspace{0.05cm} B) + {\rm Pr}(\rmB\hspace{0.05cm}\|\hspace{0.05cm} G)} = \frac{0.1}{0.1 + 0.01}\hspace{0.15cm}\underline {\approx 0.909} \hspace{0.05cm},\hspace{0.2cm} w_{\rm B} = 1 -w_{\rm G }\hspace{0.15cm}\underline {\approx 0.091}\hspace{0.05cm}.$$
	w_{\rm G} + {\rm Pr}(\rm G\hspace{0.05cm}\|\hspace{0.05cm} B) \~~cdot~~
	w_{\rm B}\hspace{0.3cm}
	\Rightarrow \hspace{0.3cm} {\rm Pr}(\rm
	B\hspace{0.05cm}\|\hspace{0.05cm} G) \cdot w_{\rm G} = {\rm Pr}(\rm
	G\hspace{0.05cm}\|\hspace{0.05cm} B) \cdot w_{\rm B}
	\hspace{0.05cm}.$$*Further, $w_{\rm B} = 1 \, –w_{\rm G}$::$${\rm Pr}(\rmB\hspace{0.05cm}\|\hspace{0.05cm} G) \cdot w_{\rm G} + {\rm Pr}(\rmG\hspace{0.05cm}\|\hspace{0.05cm} B) \cdot w_{\rm G} = {\rm Pr}(\rmG\hspace{0.05cm}\|\hspace{0.05cm} B)$$:$$\Rightarrow \hspace{0.3cm} w_{\rm G} = \frac{{\rm Pr}(\rm
	G\hspace{0.05cm}\|\hspace{0.05cm} B)}{{\rm Pr}(\rm
	G\hspace{0.05cm}\|\hspace{0.05cm} B) + {\rm Pr}(\rm
	B\hspace{0.05cm}\|\hspace{0.05cm} G)} = \frac{0.1}{0.1 + 0.01}
	\hspace{0.15cm}\underline {\approx 0.909} \hspace{0.05cm},\hspace{0.2cm} w_{\rm B} = 1 -
	w_{\rm G }\hspace{0.15cm}\underline {\approx 0.091}\hspace{0.05cm}.$$


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	'''(4)'''  According to the general equation on the specification section, for $k > 0$:		'''(4)'''  According to the general equation on the specification section, for $k > 0$:
	:$$\varphi_{e}(k) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} p_{\rm M}^2 +		:$$\varphi_{e}(k) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} p_{\rm M}^2 +(p_{\rm B} - p_{\rm M}) \cdot (p_{\rm M} - p_{\rm G}) \cdot[1 - {\rm Pr}(\rmB\hspace{0.05cm}\|\hspace{0.05cm} G )- {\rm Pr}(\rmG\hspace{0.05cm}\|\hspace{0.05cm} B )]^{\it k} = 10^{-4} + 0.09 \cdot 0.009\cdot 0.89^{\it k} = 10^{-4} \cdot \left ( 1 + 8.1 \cdot0.89^{\it k} \right )\hspace{0.05cm}.$$
	(p_{\rm B} - p_{\rm M}) \cdot (p_{\rm M} - p_{\rm G}) \cdot
	[1 - {\rm Pr}(\rm
	B\hspace{0.05cm}\|\hspace{0.05cm} G )- {\rm Pr}(\rm
	G\hspace{0.05cm}\|\hspace{0.05cm} B )]^{\it k} = 10^{-4} + 0.09 \cdot 0.009
	\cdot 0.89^{\it k} = 10^{-4} \cdot \left ( 1 + 8.1 \~~cdot~~
	0.89^{\it k} \right )\hspace{0.05cm}.$$
	:$$\Rightarrow \hspace{0.3cm}\varphi_{e}(k = 1 ) \hspace{-0.1cm} \ =\ \hspace{-0.1cm} 10^{-4} \cdot \left ( 1 + 8.1 \cdot 0.89^{ 1}\right ) \hspace{0.15cm}\underline {= 8.209 \cdot 10^{-4}} \hspace{0.05cm},$$		:$$\Rightarrow \hspace{0.3cm}\varphi_{e}(k = 1 ) \hspace{-0.1cm} \ =\ \hspace{-0.1cm} 10^{-4} \cdot \left ( 1 + 8.1 \cdot 0.89^{ 1}\right ) \hspace{0.15cm}\underline {= 8.209 \cdot 10^{-4}} \hspace{0.05cm},$$
	:$$\Rightarrow \hspace{0.3cm}\varphi_{e}(k = 2 ) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 10^{-4}\cdot \left ( 1 + 8.1 \cdot 0.89^{ 2}\right )\hspace{0.15cm}\underline { = 7.416 \cdot 10^{-4}} \hspace{0.05cm},$$		:$$\Rightarrow \hspace{0.3cm}\varphi_{e}(k = 2 ) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 10^{-4}\cdot \left ( 1 + 8.1 \cdot 0.89^{ 2}\right )\hspace{0.15cm}\underline { = 7.416 \cdot 10^{-4}} \hspace{0.05cm},$$

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	B\hspace{0.05cm}\|\hspace{0.05cm} G) \cdot w_{\rm G} = {\rm Pr}(\rm		B\hspace{0.05cm}\|\hspace{0.05cm} G) \cdot w_{\rm G} = {\rm Pr}(\rm
	G\hspace{0.05cm}\|\hspace{0.05cm} B) \cdot w_{\rm B}		G\hspace{0.05cm}\|\hspace{0.05cm} B) \cdot w_{\rm B}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$*Further, $w_{\rm B} = 1 \, –w_{\rm G}$::$${\rm Pr}(\rmB\hspace{0.05cm}\|\hspace{0.05cm} G) \cdot w_{\rm G} + {\rm Pr}(\rmG\hspace{0.05cm}\|\hspace{0.05cm} B) \cdot w_{\rm G} = {\rm Pr}(\rmG\hspace{0.05cm}\|\hspace{0.05cm} B)$$:$$\Rightarrow \hspace{0.3cm} w_{\rm G} = \frac{{\rm Pr}(\rm

	*Further, $w_{\rm B} = 1 \, –w_{\rm G}$:
	:$${\rm Pr}(\rmB\hspace{0.05cm}\|\hspace{0.05cm} G) \cdot w_{\rm G} + {\rm Pr}(\rmG\hspace{0.05cm}\|\hspace{0.05cm} B) \cdot w_{\rm G} = {\rm Pr}(\rmG\hspace{0.05cm}\|\hspace{0.05cm} B)$$
	:$$\Rightarrow \hspace{0.3cm} w_{\rm G} = \frac{{\rm Pr}(\rm
	G\hspace{0.05cm}\|\hspace{0.05cm} B)}{{\rm Pr}(\rm		G\hspace{0.05cm}\|\hspace{0.05cm} B)}{{\rm Pr}(\rm
	G\hspace{0.05cm}\|\hspace{0.05cm} B) + {\rm Pr}(\rm		G\hspace{0.05cm}\|\hspace{0.05cm} B) + {\rm Pr}(\rm
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	Pr}(\rm B\hspace{0.05cm}\|\hspace{0.05cm} G)}{{\rm Pr}(\rm		Pr}(\rm B\hspace{0.05cm}\|\hspace{0.05cm} G)}{{\rm Pr}(\rm
	G\hspace{0.05cm}\|\hspace{0.05cm} B) + {\rm Pr}(\rm		G\hspace{0.05cm}\|\hspace{0.05cm} B) + {\rm Pr}(\rm
	B\hspace{0.05cm}\|\hspace{0.05cm} G)}\hspace{0.05cm}.$$		B\hspace{0.05cm}\|\hspace{0.05cm} G)}\hspace{0.05cm}.$$*With $\rm Pr(B\|G) = 0.01$ and with $\rm Pr(G\|B) = 0.1$, respectively, the following equations are obtained::$${\rm Pr}(\rm B\hspace{0.05cm}\|\hspace{0.05cm} G) =

	*With $\rm Pr(B\|G) = 0.01$ and with $\rm Pr(G\|B) = 0.1$, respectively, the following equations are obtained:
	:$${\rm Pr}(\rm B\hspace{0.05cm}\|\hspace{0.05cm} G) =
	0.01\hspace{-0.15cm}:\hspace{0.2cm} {\it p}_{\rm M} \hspace{-0.1cm} \ = \		0.01\hspace{-0.15cm}:\hspace{0.2cm} {\it p}_{\rm M} \hspace{-0.1cm} \ = \
	\hspace{-0.1cm} \frac{0.001 \cdot {\rm Pr}(\rm		\hspace{-0.1cm} \frac{0.001 \cdot {\rm Pr}(\rm

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	[[File:P_ID1843__Dig_Z_5_6.png\|right\|frame\|Given Gilbert–Elliott model]]		[[File:P_ID1843__Dig_Z_5_6.png\|right\|frame\|Given Gilbert–Elliott model]]
	We consider the <i>burst error channel model</i>  according to  [https://en.wikipedia.org/wiki/Edgar_Gilbert "E.N. Gilbert"]  and E.O. Elliott (see sketch). For the transition probabilities let hold:		We consider the <i>burst error channel model</i>  according to  [https://en.wikipedia.org/wiki/Edgar_Gilbert "E.N. Gilbert"]  and E.O. Elliott (see sketch). For the transition probabilities let hold:
	:$${\rm Pr}(\rm		:$${\rm Pr}(\rmG\hspace{0.05cm}\|\hspace{0.05cm} B)= 0.1, \hspace{0.2cm} {\rmPr}(\rm B\hspace{0.05cm}\|\hspace{0.05cm} G) = 0.01\hspace{0.05cm}.$$
	G\hspace{0.05cm}\|\hspace{0.05cm} B)= 0.1, \hspace{0.2cm} {\rm
	Pr}(\rm B\hspace{0.05cm}\|\hspace{0.05cm} G) = 0.01\hspace{0.05cm}.$$

	The error probability in the state "GOOD" is  $p_{\rm G} = 0.1\%$. For the state "BAD" is  $p_{\rm B} = 10\%$.		The error probability in the state "GOOD" is  $p_{\rm G} = 0.1\%$. For the state "BAD" is  $p_{\rm B} = 10\%$.
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	* the state probabilities  $w_{\rm G} = \rm Pr(Z = G)$  and  $w_{\rm B} = \rm Pr(Z = B)$,		* the state probabilities  $w_{\rm G} = \rm Pr(Z = G)$  and  $w_{\rm B} = \rm Pr(Z = B)$,
	* the values of the correlation function, which for  $k > 0$  is given analytically as follows:		* the values of the correlation function, which for  $k > 0$  is given analytically as follows:
	:$$\varphi_{e}(k) = p_{\rm M}^2 + (p_{\rm B} -		:$$\varphi_{e}(k) = p_{\rm M}^2 + (p_{\rm B} -p_{\rm M}) \cdot (p_{\rm M} - p_{\rm G}) \cdot\big [1 - {\rm Pr}(\rmB\hspace{0.05cm}\|\hspace{0.05cm} G )- {\rm Pr}(\rmG\hspace{0.05cm}\|\hspace{0.05cm} B )\big ]^{\it k} \hspace{0.05cm}.$$
	p_{\rm M}) \cdot (p_{\rm M} - p_{\rm G}) \cdot
	\big [1 - {\rm Pr}(\rm
	B\hspace{0.05cm}\|\hspace{0.05cm} G )- {\rm Pr}(\rm
	G\hspace{0.05cm}\|\hspace{0.05cm} B )\big ]^{\it k} \hspace{0.05cm}.$$


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	*Further, $w_{\rm B} = 1 \, –w_{\rm G}$:		*Further, $w_{\rm B} = 1 \, –w_{\rm G}$:
	:$${\rm Pr}(\rm		:$${\rm Pr}(\rmB\hspace{0.05cm}\|\hspace{0.05cm} G) \cdot w_{\rm G} + {\rm Pr}(\rmG\hspace{0.05cm}\|\hspace{0.05cm} B) \cdot w_{\rm G} = {\rm Pr}(\rmG\hspace{0.05cm}\|\hspace{0.05cm} B)$$
	B\hspace{0.05cm}\|\hspace{0.05cm} G) \cdot w_{\rm G} + {\rm Pr}(\rm
	G\hspace{0.05cm}\|\hspace{0.05cm} B) \cdot w_{\rm G} = {\rm Pr}(\rm
	G\hspace{0.05cm}\|\hspace{0.05cm} B)$$
	:$$\Rightarrow \hspace{0.3cm} w_{\rm G} = \frac{{\rm Pr}(\rm		:$$\Rightarrow \hspace{0.3cm} w_{\rm G} = \frac{{\rm Pr}(\rm
	G\hspace{0.05cm}\|\hspace{0.05cm} B)}{{\rm Pr}(\rm		G\hspace{0.05cm}\|\hspace{0.05cm} B)}{{\rm Pr}(\rm
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	'''(3)'''  The mean error probability $p_{\rm M}$ is obtained from the error probabilities $p_{\rm G}$ and $p_{\rm B}$ weighted by $w_{\rm G}$ and $w_{\rm B}$:		'''(3)'''  The mean error probability $p_{\rm M}$ is obtained from the error probabilities $p_{\rm G}$ and $p_{\rm B}$ weighted by $w_{\rm G}$ and $w_{\rm B}$:
	:$$p_{\rm M} = w_{\rm G} \cdot p_{\rm G} + w_{\rm B} \cdot p_{\rm B}		:$$p_{\rm M} = w_{\rm G} \cdot p_{\rm G} + w_{\rm B} \cdot p_{\rm B}= \frac{10}{11} \cdot 10^{-3} + \frac{1}{11} \cdot 10^{-1}= \frac{10+100}{11} \cdot 10^{-3}\hspace{0.15cm}\underline { =1\%}\hspace{0.05cm}.$$
	= \frac{10}{11} \cdot 10^{-3} + \frac{1}{11} \cdot 10^{-1}= \frac{10+100}{11} \cdot 10^{-3}\hspace{0.15cm}\underline { =
	1\%}\hspace{0.05cm}.$$


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	\cdot 0.89^{\it k} = 10^{-4} \cdot \left ( 1 + 8.1 \cdot		\cdot 0.89^{\it k} = 10^{-4} \cdot \left ( 1 + 8.1 \cdot
	0.89^{\it k} \right )\hspace{0.05cm}.$$		0.89^{\it k} \right )\hspace{0.05cm}.$$
	:$$\Rightarrow \hspace{0.3cm}\varphi_{e}(k = 1 ) \hspace{-0.1cm} \ =		:$$\Rightarrow \hspace{0.3cm}\varphi_{e}(k = 1 ) \hspace{-0.1cm} \ =\ \hspace{-0.1cm} 10^{-4} \cdot \left ( 1 + 8.1 \cdot 0.89^{ 1}\right ) \hspace{0.15cm}\underline {= 8.209 \cdot 10^{-4}} \hspace{0.05cm},$$
	\ \hspace{-0.1cm} 10^{-4} \cdot \left ( 1 + 8.1 \cdot 0.89^{ 1}		:$$\Rightarrow \hspace{0.3cm}\varphi_{e}(k = 2 ) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 10^{-4}\cdot \left ( 1 + 8.1 \cdot 0.89^{ 2}\right )\hspace{0.15cm}\underline { = 7.416 \cdot 10^{-4}} \hspace{0.05cm},$$
	\right ) \hspace{0.15cm}\underline {= 8.209 \cdot 10^{-4}} \hspace{0.05cm},$$		:$$\Rightarrow \hspace{0.3cm}\varphi_{e}(k = 5 ) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 10^{-4}\cdot \left ( 1 + 8.1 \cdot 0.89^{ 5}\right )\hspace{0.15cm}\underline {= 5.523 \cdot 10^{-4}} \hspace{0.05cm},$$
	:$$\Rightarrow \hspace{0.3cm}\varphi_{e}(k = 2 ) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 10^{-4}		:$$\Rightarrow \hspace{0.3cm}\varphi_{e}(k = 50 ) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 10^{-4}\cdot \left ( 1 + 8.1 \cdot 0.89^{ 50} \right ) \hspace{0.15cm}\underline {= 1.024 \cdot10^{-4}} \hspace{0.05cm}.$$
	\cdot \left ( 1 + 8.1 \cdot 0.89^{ 2}
	\right )\hspace{0.15cm}\underline { = 7.416 \cdot 10^{-4}} \hspace{0.05cm},$$
	:$$\Rightarrow \hspace{0.3cm}\varphi_{e}(k = 5 ) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 10^{-4}
	\cdot \left ( 1 + 8.1 \cdot 0.89^{ 5}
	\right )\hspace{0.15cm}\underline {= 5.523 \cdot 10^{-4}} \hspace{0.05cm},$$
	:$$\Rightarrow \hspace{0.3cm}\varphi_{e}(k = 50 ) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 10^{-4}
	\cdot \left ( 1 + 8.1 \cdot 0.89^{ 50} \right ) \hspace{0.15cm}\underline {= 1.024 \~~cdot~~
	10^{-4}} \hspace{0.05cm}.$$


	'''(5)'''  For each channel model, because of $e_{\nu} ∈ \{0, 1\}$:		'''(5)'''  For each channel model, because of $e_{\nu} ∈ \{0, 1\}$:
	:$$\varphi_{e}(k = 0 ) = {\rm E}[e_{\nu} ^2] = {\rm E}[e_{\nu} ] =		:$$\varphi_{e}(k = 0 ) = {\rm E}[e_{\nu} ^2] = {\rm E}[e_{\nu} ] =p_{\rm M} \hspace{0.05cm}.$$
	p_{\rm M} \hspace{0.05cm}.$$

	Using the result of subtask '''(3)''', $\varphi_e(k = 0) \ \underline {= 0.01}$ for the present case.		Using the result of subtask '''(3)''', $\varphi_e(k = 0) \ \underline {= 0.01}$ for the present case.
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	'''(6)'''  According to subtask '''(3)''' holds:		'''(6)'''  According to subtask '''(3)''' holds:
	:$$p_{\rm M} = {10}/{11} \cdot p_{\rm G} + {1}/{11} \~~cdot~~		:$$p_{\rm M} = {10}/{11} \cdot p_{\rm G} + {1}/{11} \cdotp_{\rm B} \hspace{0.05cm}.$$
	p_{\rm B} \hspace{0.05cm}.$$

	*With given $p_{\rm B} = 0.1$, even for $p_{\rm G} = 0$ (no error in state "G"), the average error probability is $p_{\rm M} \approx 0.009$.		*With given $p_{\rm B} = 0.1$, even for $p_{\rm G} = 0$ (no error in state "G"), the average error probability is $p_{\rm M} \approx 0.009$.
	*In contrast, with fixed $p_{\rm G} = 0.001$ the value $p_{\rm M} = 0.005$ is achievable:		*In contrast, with fixed $p_{\rm G} = 0.001$ the value $p_{\rm M} = 0.005$ is achievable:
	:$$0.005 = {10}/{11} \cdot 10^{-3} + {1}/{11} \cdot p_{\rm B}		:$$0.005 = {10}/{11} \cdot 10^{-3} + {1}/{11} \cdot p_{\rm B}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} p_{\rm B} \le 0.055 - 0.1= 4.5\%\hspace{0.05cm}.$$
	\hspace{0.3cm} \Rightarrow \hspace{0.3cm} p_{\rm B} \le 0.055 - 0.1
	= 4.5\%\hspace{0.05cm}.$$

	*Furthermore, the mean error probability (with given $p_{\rm G}$ and $p_{\rm B}$) can also be represented as follows:		*Furthermore, the mean error probability (with given $p_{\rm G}$ and $p_{\rm B}$) can also be represented as follows:
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	*From the last equation it can be seen that with no $\rm Pr(G\|B)$ value the result $p_{\rm M} = 0.005$ is possible.		*From the last equation it can be seen that with no $\rm Pr(G\|B)$ value the result $p_{\rm M} = 0.005$ is possible.
	*On the other hand, with a smaller $\rm Pr(B\|G)$ the condition can be fulfilled:		*On the other hand, with a smaller $\rm Pr(B\|G)$ the condition can be fulfilled:
	:$$0.005 = \frac{0.0001 + 0.1 \cdot {\rm Pr}(\rm		:$$0.005 = \frac{0.0001 + 0.1 \cdot {\rm Pr}(\rmB\hspace{0.05cm}\|\hspace{0.05cm} G)}{0.1 +{\rm Pr}(\rmB\hspace{0.05cm}\|\hspace{0.05cm} G) } \hspace{0.3cm} \Rightarrow\hspace{0.3cm}{\rm Pr}(\rm B\hspace{0.05cm}\|\hspace{0.05cm} G) \le\frac{0.0004}{0.095} \approx 0.0042\hspace{0.05cm}.$$
	B\hspace{0.05cm}\|\hspace{0.05cm} G)}{0.1 +{\rm Pr}(\rm
	B\hspace{0.05cm}\|\hspace{0.05cm} G) } \hspace{0.3cm} \Rightarrow
	\hspace{0.3cm}{\rm Pr}(\rm B\hspace{0.05cm}\|\hspace{0.05cm} G) \le
	\frac{0.0004}{0.095} \approx 0.0042\hspace{0.05cm}.$$

	*<u>Solutions 2 and 4</u> are therefore correct.		*<u>Solutions 2 and 4</u> are therefore correct.

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	[[Category:Digital Signal Transmission: Exercises\|^5.3 Burst Error Channels^]]		[[Category:Digital Signal Transmission: Exercises\|^5.3 Burst Error Channels^]]
			[[de:5.6Z_GE-Modelleigenschaften]]

Hwang at 02:54, 19 September 2022

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	'''(6)'''  ~~Entsprechend der Teilaufgabe~~ '''(3)''' ~~gilt~~:		'''(6)'''  According to subtask '''(3)''' holds:
	:$$p_{\rm M} = {10}/{11} \cdot p_{\rm G} + {1}/{11} \cdot		:$$p_{\rm M} = {10}/{11} \cdot p_{\rm G} + {1}/{11} \cdot
	p_{\rm B} \hspace{0.05cm}.$$		p_{\rm B} \hspace{0.05cm}.$$

	*~~Bei vorgegebenem~~ $p_{\rm B} = 0.1$ ~~ergibt sich selbst für~~ $p_{\rm G} = 0$ (~~kein Fehler im Zustand~~ "G") ~~die mittlere Fehlerwahrscheinlichkeit zu~~ $p_{\rm M} \approx 0.009$.		*With given $p_{\rm B} = 0.1$, even for $p_{\rm G} = 0$ (no error in state "G"), the average error probability is $p_{\rm M} \approx 0.009$.
	*~~Dagegen ist mit festem~~ $p_{\rm G} = 0.001$ ~~der Wert~~ $p_{\rm M} = 0.005$ ~~erreichbar~~:		*In contrast, with fixed $p_{\rm G} = 0.001$ the value $p_{\rm M} = 0.005$ is achievable:
	:$$0.005 = {10}/{11} \cdot 10^{-3} + {1}/{11} \cdot p_{\rm B}		:$$0.005 = {10}/{11} \cdot 10^{-3} + {1}/{11} \cdot p_{\rm B}
	\hspace{0.3cm} \Rightarrow \hspace{0.3cm} p_{\rm B} \le 0.055 - 0.1		\hspace{0.3cm} \Rightarrow \hspace{0.3cm} p_{\rm B} \le 0.055 - 0.1
	= 4.5\%\hspace{0.05cm}.$$		= 4.5\%\hspace{0.05cm}.$$

	*~~Weiterhin kann die mittlere Fehlerwahrscheinlichkeit~~ (~~mit vorgegebenem~~ $p_{\rm G}$ ~~und~~ $p_{\rm B}$) ~~auch wie folgt dargestellt werden~~:		*Furthermore, the mean error probability (with given $p_{\rm G}$ and $p_{\rm B}$) can also be represented as follows:
	:$$p_{\rm M} = \frac{p_{\rm G} \cdot {\rm Pr}(\rm		:$$p_{\rm M} = \frac{p_{\rm G} \cdot {\rm Pr}(\rm
	G\hspace{0.05cm}\|\hspace{0.05cm} B)+ p_{\rm B} \cdot {\rm Pr}(\rm		G\hspace{0.05cm}\|\hspace{0.05cm} B)+ p_{\rm B} \cdot {\rm Pr}(\rm
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	B\hspace{0.05cm}\|\hspace{0.05cm} G)}\hspace{0.05cm}.$$		B\hspace{0.05cm}\|\hspace{0.05cm} G)}\hspace{0.05cm}.$$

	*~~Mit~~ $\rm Pr(B\|G) = 0.01$ ~~bzw. mit~~ $\rm Pr(G\|B) = 0.1$ ~~erhält man folgende Gleichungen~~:		*With $\rm Pr(B\|G) = 0.01$ and with $\rm Pr(G\|B) = 0.1$, respectively, the following equations are obtained:
	:$${\rm Pr}(\rm B\hspace{0.05cm}\|\hspace{0.05cm} G) =		:$${\rm Pr}(\rm B\hspace{0.05cm}\|\hspace{0.05cm} G) =
	0.01\hspace{-0.15cm}:\hspace{0.2cm} {\it p}_{\rm M} \hspace{-0.1cm} \ = \		0.01\hspace{-0.15cm}:\hspace{0.2cm} {\it p}_{\rm M} \hspace{-0.1cm} \ = \
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	G\hspace{0.05cm}\|\hspace{0.05cm} B) }\hspace{0.05cm}.$$		G\hspace{0.05cm}\|\hspace{0.05cm} B) }\hspace{0.05cm}.$$

	*~~Aus der letzten Gleichung ist zu erkennen, dass mit keinem~~ $\rm Pr(G\|B)$~~–Wert das Ergebnis~~ $p_{\rm M} = 0.005$ ~~möglich ist~~.		*From the last equation it can be seen that with no $\rm Pr(G\|B)$ value the result $p_{\rm M} = 0.005$ is possible.
	*~~Dagegen lässt sich durch ein kleineres~~ $\rm Pr(B\|G)$ ~~die Bedingung erfüllen~~:		*On the other hand, with a smaller $\rm Pr(B\|G)$ the condition can be fulfilled:
	:$$0.005 = \frac{0.0001 + 0.1 \cdot {\rm Pr}(\rm		:$$0.005 = \frac{0.0001 + 0.1 \cdot {\rm Pr}(\rm
	B\hspace{0.05cm}\|\hspace{0.05cm} G)}{0.1 +{\rm Pr}(\rm		B\hspace{0.05cm}\|\hspace{0.05cm} G)}{0.1 +{\rm Pr}(\rm
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	\frac{0.0004}{0.095} \approx 0.0042\hspace{0.05cm}.$$		\frac{0.0004}{0.095} \approx 0.0042\hspace{0.05cm}.$$

	*~~Richtig sind somit die~~ <u>~~Lösungsvorschläge~~ 2 ~~und~~ 4</u>.		*<u>Solutions 2 and 4</u> are therefore correct.
	{{ML-Fuß}}		{{ML-Fuß}}


	[[Category:Digital Signal Transmission: Exercises\|^5.3 Burst Error Channels^]]		[[Category:Digital Signal Transmission: Exercises\|^5.3 Burst Error Channels^]]

Hwang at 02:48, 19 September 2022

2022-09-19T02:48:33Z

Show changes

Guenter at 15:08, 9 September 2022

2022-09-09T15:08:35Z

← Older revision		Revision as of 17:08, 9 September 2022
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	[[Category:Digital Signal Transmission: Exercises\|^5.3 ~~Bündelfehlerkanäle~~^]]		[[Category:Digital Signal Transmission: Exercises\|^5.3 Burst Error Channels^]]

Guenter: Guenter moved page Aufgaben:Aufgabe 5.6Z: GE-Modelleigenschaften to Aufgaben:Exercise 5.6Z: Gilbert-Elliott Model

2022-08-30T10:10:26Z

Guenter moved page Aufgaben:Aufgabe 5.6Z: GE-Modelleigenschaften to Aufgaben:Exercise 5.6Z: Gilbert-Elliott Model

← Older revision		Revision as of 12:10, 30 August 2022
(No difference)

← Older revision		Revision as of 17:57, 16 March 2026
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	[[File:P_ID1843__Dig_Z_5_6.png\|right\|frame\|Given Gilbert–Elliott model]]		[[File:P_ID1843__Dig_Z_5_6.png\|right\|frame\|Given Gilbert–Elliott model]]
	We consider the <i>burst error channel model</i>  according to  [https://en.wikipedia.org/wiki/Edgar_Gilbert "E.N. Gilbert"]  and E.O. Elliott (see sketch). For the transition probabilities let hold:		We consider the <i>burst error channel model</i>  according to  [https://en.wikipedia.org/wiki/Edgar_Gilbert "E.N. Gilbert"]  and E.O. Elliott (see sketch). For the transition probabilities let hold:
	:$${\rm Pr}(\~~rmG~~\hspace{0.05cm}\|\hspace{0.05cm} B)= 0.1, \hspace{0.2cm} {\~~rmPr~~}(\rm B\hspace{0.05cm}\|\hspace{0.05cm} G) = 0.01\hspace{0.05cm}.$$		:$${\rm Pr}(\rm G\hspace{0.05cm}\|\hspace{0.05cm} B)= 0.1, \hspace{0.2cm} {\rm Pr}(\rm B\hspace{0.05cm}\|\hspace{0.05cm} G) = 0.01\hspace{0.05cm}.$$

	The error probability in the state "GOOD" is  $p_{\rm G} = 0.1\%$. For the state "BAD" is  $p_{\rm B} = 10\%$.		The error probability in the state "GOOD" is  $p_{\rm G} = 0.1\%$. For the state "BAD" is  $p_{\rm B} = 10\%$.
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	* the state probabilities  $w_{\rm G} = \rm Pr(Z = G)$  and  $w_{\rm B} = \rm Pr(Z = B)$,		* the state probabilities  $w_{\rm G} = \rm Pr(Z = G)$  and  $w_{\rm B} = \rm Pr(Z = B)$,
	* the values of the correlation function, which for  $k > 0$  is given analytically as follows:		* the values of the correlation function, which for  $k > 0$  is given analytically as follows:
	:$$\varphi_{e}(k) = p_{\rm M}^2 + (p_{\rm B} -p_{\rm M}) \cdot (p_{\rm M} - p_{\rm G}) \cdot\big [1 - {\rm Pr}(\~~rmB~~\hspace{0.05cm}\|\hspace{0.05cm} G )- {\rm Pr}(\~~rmG~~\hspace{0.05cm}\|\hspace{0.05cm} B )\big ]^{\it k} \hspace{0.05cm}.$$		:$$\varphi_{e}(k) = p_{\rm M}^2 + (p_{\rm B} -p_{\rm M}) \cdot (p_{\rm M} - p_{\rm G}) \cdot\big [1 - {\rm Pr}(\rm B\hspace{0.05cm}\|\hspace{0.05cm} G )- {\rm Pr}(\rm G\hspace{0.05cm}\|\hspace{0.05cm} B )\big ]^{\it k} \hspace{0.05cm}.$$


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	'''(2)'''  The GE model is a stationary Markov chain.		'''(2)'''  The GE model is a stationary Markov chain.
	*For the probability that it is in the state "GOOD", taking into account the result of subtask '''(1)''':		*For the probability that it is in the state "GOOD", taking into account the result of subtask '''(1)''':
	:$$w_{\rm G} = {\rm Pr}(\rm G\hspace{0.05cm}\|\hspace{0.05cm} G) \cdotw_{\rm G} + {\rm Pr}(\rm G\hspace{0.05cm}\|\hspace{0.05cm} B) \cdotw_{\rm B}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} {\rm Pr}(\~~rmB~~\hspace{0.05cm}\|\hspace{0.05cm} G) \cdot w_{\rm G} = {\rm Pr}(\~~rmG~~\hspace{0.05cm}\|\hspace{0.05cm} B) \cdot w_{\rm B}\hspace{0.05cm}.$$*Further, $w_{\rm B} = 1 \, –w_{\rm G}$::$${\rm Pr}(\~~rmB~~\hspace{0.05cm}\|\hspace{0.05cm} G) \cdot w_{\rm G} + {\rm Pr}(\~~rmG~~\hspace{0.05cm}\|\hspace{0.05cm} B) \cdot w_{\rm G} = {\rm Pr}(\~~rmG~~\hspace{0.05cm}\|\hspace{0.05cm} B)$$:$$\Rightarrow \hspace{0.3cm} w_{\rm G} = \frac{{\rm Pr}(\~~rmG~~\hspace{0.05cm}\|\hspace{0.05cm} B)}{{\rm Pr}(\~~rmG~~\hspace{0.05cm}\|\hspace{0.05cm} B) + {\rm Pr}(\~~rmB~~\hspace{0.05cm}\|\hspace{0.05cm} G)} = \frac{0.1}{0.1 + 0.01}\hspace{0.15cm}\underline {\approx 0.909} \hspace{0.05cm},\hspace{0.2cm} w_{\rm B} = 1 -w_{\rm G }\hspace{0.15cm}\underline {\approx 0.091}\hspace{0.05cm}.$$		:$$w_{\rm G} = {\rm Pr}(\rm G\hspace{0.05cm}\|\hspace{0.05cm} G) \cdotw_{\rm G} + {\rm Pr}(\rm G\hspace{0.05cm}\|\hspace{0.05cm} B) \cdotw_{\rm B}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} {\rm Pr}(\rm B\hspace{0.05cm}\|\hspace{0.05cm} G) \cdot w_{\rm G} = {\rm Pr}(\rm G\hspace{0.05cm}\|\hspace{0.05cm} B) \cdot w_{\rm B}\hspace{0.05cm}.$$*Further, $w_{\rm B} = 1 \, –w_{\rm G}$::$${\rm Pr}(\rm B\hspace{0.05cm}\|\hspace{0.05cm} G) \cdot w_{\rm G} + {\rm Pr}(\rm G\hspace{0.05cm}\|\hspace{0.05cm} B) \cdot w_{\rm G} = {\rm Pr}(\rm G\hspace{0.05cm}\|\hspace{0.05cm} B)$$:$$\Rightarrow \hspace{0.3cm} w_{\rm G} = \frac{{\rm Pr}(\rm G\hspace{0.05cm}\|\hspace{0.05cm} B)}{{\rm Pr}(\rm G\hspace{0.05cm}\|\hspace{0.05cm} B) + {\rm Pr}(\rm B\hspace{0.05cm}\|\hspace{0.05cm} G)} = \frac{0.1}{0.1 + 0.01}\hspace{0.15cm}\underline {\approx 0.909} \hspace{0.05cm},\hspace{0.2cm} w_{\rm B} = 1 -w_{\rm G }\hspace{0.15cm}\underline {\approx 0.091}\hspace{0.05cm}.$$


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	'''(4)'''  According to the general equation on the specification section, for $k > 0$:		'''(4)'''  According to the general equation on the specification section, for $k > 0$:
	:$$\varphi_{e}(k) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} p_{\rm M}^2 +(p_{\rm B} - p_{\rm M}) \cdot (p_{\rm M} - p_{\rm G}) \cdot[1 - {\rm Pr}(\~~rmB~~\hspace{0.05cm}\|\hspace{0.05cm} G )- {\rm Pr}(\~~rmG~~\hspace{0.05cm}\|\hspace{0.05cm} B )]^{\it k} = 10^{-4} + 0.09 \cdot 0.009\cdot 0.89^{\it k} = 10^{-4} \cdot \left ( 1 + 8.1 \cdot0.89^{\it k} \right )\hspace{0.05cm}.$$		:$$\varphi_{e}(k) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} p_{\rm M}^2 +(p_{\rm B} - p_{\rm M}) \cdot (p_{\rm M} - p_{\rm G}) \cdot[1 - {\rm Pr}(\rm B\hspace{0.05cm}\|\hspace{0.05cm} G )- {\rm Pr}(\rm G\hspace{0.05cm}\|\hspace{0.05cm} B )]^{\it k} = 10^{-4} + 0.09 \cdot 0.009\cdot 0.89^{\it k} = 10^{-4} \cdot \left ( 1 + 8.1 \cdot0.89^{\it k} \right )\hspace{0.05cm}.$$
	:$$\Rightarrow \hspace{0.3cm}\varphi_{e}(k = 1 ) \hspace{-0.1cm} \ =\ \hspace{-0.1cm} 10^{-4} \cdot \left ( 1 + 8.1 \cdot 0.89^{ 1}\right ) \hspace{0.15cm}\underline {= 8.209 \cdot 10^{-4}} \hspace{0.05cm},$$		:$$\Rightarrow \hspace{0.3cm}\varphi_{e}(k = 1 ) \hspace{-0.1cm} \ =\ \hspace{-0.1cm} 10^{-4} \cdot \left ( 1 + 8.1 \cdot 0.89^{ 1}\right ) \hspace{0.15cm}\underline {= 8.209 \cdot 10^{-4}} \hspace{0.05cm},$$
	:$$\Rightarrow \hspace{0.3cm}\varphi_{e}(k = 2 ) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 10^{-4}\cdot \left ( 1 + 8.1 \cdot 0.89^{ 2}\right )\hspace{0.15cm}\underline { = 7.416 \cdot 10^{-4}} \hspace{0.05cm},$$		:$$\Rightarrow \hspace{0.3cm}\varphi_{e}(k = 2 ) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 10^{-4}\cdot \left ( 1 + 8.1 \cdot 0.89^{ 2}\right )\hspace{0.15cm}\underline { = 7.416 \cdot 10^{-4}} \hspace{0.05cm},$$
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	*Furthermore, the mean error probability (with given $p_{\rm G}$ and $p_{\rm B}$) can also be represented as follows:		*Furthermore, the mean error probability (with given $p_{\rm G}$ and $p_{\rm B}$) can also be represented as follows:
	:$$p_{\rm M} = \frac{p_{\rm G} \cdot {\rm Pr}(\~~rmG~~\hspace{0.05cm}\|\hspace{0.05cm} B)+ p_{\rm B} \cdot {\rm Pr}(\~~rmB~~\hspace{0.05cm}\|\hspace{0.05cm} G)}{{\rm Pr}(\~~rmG~~\hspace{0.05cm}\|\hspace{0.05cm} B) + {\rm Pr}(\~~rmB~~\hspace{0.05cm}\|\hspace{0.05cm} G)} = \frac{0.001 \cdot {\~~rmPr~~}(\rm G\hspace{0.05cm}\|\hspace{0.05cm} B)+ 0.1 \cdot {\~~rmPr~~}(\rm B\hspace{0.05cm}\|\hspace{0.05cm} G)}{{\rm Pr}(\~~rmG~~\hspace{0.05cm}\|\hspace{0.05cm} B) + {\rm Pr}(\~~rmB~~\hspace{0.05cm}\|\hspace{0.05cm} G)}\hspace{0.05cm}.$$*With $\rm Pr(B\|G) = 0.01$ and with $\rm Pr(G\|B) = 0.1$, respectively, the following equations are obtained::$${\rm Pr}(\rm B\hspace{0.05cm}\|\hspace{0.05cm} G) =0.01\hspace{-0.15cm}:\hspace{0.2cm} {\it p}_{\rm M} \hspace{-0.1cm} \ = \\hspace{-0.1cm} \frac{0.001 \cdot {\rm Pr}(\~~rmG~~\hspace{0.05cm}\|\hspace{0.05cm} B)+ 0.001 }{{\rm Pr}(\~~rmG~~\hspace{0.05cm}\|\hspace{0.05cm} B) +0.01}\hspace{0.05cm},\hspace{0.5cm}{\rm Pr}(\rm G\hspace{0.05cm}\|\hspace{0.05cm} B) =0.1\hspace{-0.15cm}:\hspace{0.2cm}{\it p}_{\rm M} \hspace{-0.1cm} \ = \\hspace{-0.1cm} \frac{0.0001 + 0.1 \cdot {\rm Pr}(\~~rmB~~\hspace{0.05cm}\|\hspace{0.05cm} G)}{0.1 +{\rm Pr}(\~~rmG~~\hspace{0.05cm}\|\hspace{0.05cm} B) }\hspace{0.05cm}.$$		:$$p_{\rm M} = \frac{p_{\rm G} \cdot {\rm Pr}(\rm G\hspace{0.05cm}\|\hspace{0.05cm} B)+ p_{\rm B} \cdot {\rm Pr}(\rm B\hspace{0.05cm}\|\hspace{0.05cm} G)}{{\rm Pr}(\rm G\hspace{0.05cm}\|\hspace{0.05cm} B) + {\rm Pr}(\rm B\hspace{0.05cm}\|\hspace{0.05cm} G)} = \frac{0.001 \cdot {\rm Pr}(\rm G\hspace{0.05cm}\|\hspace{0.05cm} B)+ 0.1 \cdot {\rm Pr}(\rm B\hspace{0.05cm}\|\hspace{0.05cm} G)}{{\rm Pr}(\rm G\hspace{0.05cm}\|\hspace{0.05cm} B) + {\rm Pr}(\rm B\hspace{0.05cm}\|\hspace{0.05cm} G)}\hspace{0.05cm}.$$*With $\rm Pr(B\|G) = 0.01$ and with $\rm Pr(G\|B) = 0.1$, respectively, the following equations are obtained::$${\rm Pr}(\rm B\hspace{0.05cm}\|\hspace{0.05cm} G) =0.01\hspace{-0.15cm}:\hspace{0.2cm} {\it p}_{\rm M} \hspace{-0.1cm} \ = \\hspace{-0.1cm} \frac{0.001 \cdot {\rm Pr}(\rm G\hspace{0.05cm}\|\hspace{0.05cm} B)+ 0.001 }{{\rm Pr}(\rm G\hspace{0.05cm}\|\hspace{0.05cm} B) +0.01}\hspace{0.05cm},\hspace{0.5cm}{\rm Pr}(\rm G\hspace{0.05cm}\|\hspace{0.05cm} B) =0.1\hspace{-0.15cm}:\hspace{0.2cm}{\it p}_{\rm M} \hspace{-0.1cm} \ = \\hspace{-0.1cm} \frac{0.0001 + 0.1 \cdot {\rm Pr}(\rm B\hspace{0.05cm}\|\hspace{0.05cm} G)}{0.1 +{\rm Pr}(\rm G\hspace{0.05cm}\|\hspace{0.05cm} B) }\hspace{0.05cm}.$$

	*From the last equation it can be seen that with no $\rm Pr(G\|B)$ value the result $p_{\rm M} = 0.005$ is possible.		*From the last equation it can be seen that with no $\rm Pr(G\|B)$ value the result $p_{\rm M} = 0.005$ is possible.
	*On the other hand, with a smaller $\rm Pr(B\|G)$ the condition can be fulfilled:		*On the other hand, with a smaller $\rm Pr(B\|G)$ the condition can be fulfilled:
	:$$0.005 = \frac{0.0001 + 0.1 \cdot {\rm Pr}(\~~rmB~~\hspace{0.05cm}\|\hspace{0.05cm} G)}{0.1 +{\rm Pr}(\~~rmB~~\hspace{0.05cm}\|\hspace{0.05cm} G) } \hspace{0.3cm} \Rightarrow\hspace{0.3cm}{\rm Pr}(\rm B\hspace{0.05cm}\|\hspace{0.05cm} G) \le\frac{0.0004}{0.095} \approx 0.0042\hspace{0.05cm}.$$		:$$0.005 = \frac{0.0001 + 0.1 \cdot {\rm Pr}(\rm B\hspace{0.05cm}\|\hspace{0.05cm} G)}{0.1 +{\rm Pr}(\rm B\hspace{0.05cm}\|\hspace{0.05cm} G) } \hspace{0.3cm} \Rightarrow\hspace{0.3cm}{\rm Pr}(\rm B\hspace{0.05cm}\|\hspace{0.05cm} G) \le\frac{0.0004}{0.095} \approx 0.0042\hspace{0.05cm}.$$

	*<u>Solutions 2 and 4</u> are therefore correct.		*<u>Solutions 2 and 4</u> are therefore correct.
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	[[Category:Digital Signal Transmission: Exercises\|^5.3 Burst Error Channels^]]		[[Category:Digital Signal Transmission: Exercises\|^5.3 Burst Error Channels^]]
	[[de:5.~~6Z_GE~~-Modelleigenschaften]]		[[de:Aufgaben:Aufgabe 5.6Z: GE-Modelleigenschaften]]