Aufgaben:Exercise 2.10: Reed-Solomon Error Detection - Revision history

Hwang at 15:27, 23 January 2023

2023-01-23T15:27:53Z

Guenter at 15:27, 11 October 2022

2022-10-11T15:27:40Z

Guenter at 14:17, 11 October 2022

2022-10-11T14:17:38Z

Guenter at 13:33, 11 October 2022

2022-10-11T13:33:56Z

Guenter at 13:32, 11 October 2022

2022-10-11T13:32:40Z

Guenter: Guenter moved page Aufgaben:Exercise 2.10: Reed-Solomon Fault Detection to Aufgaben:Exercise 2.10: Reed-Solomon Error Detection

2022-10-05T15:09:51Z

Guenter moved page Exercise 2.10: Reed-Solomon Fault Detection to Exercise 2.10: Reed-Solomon Error Detection

Noah at 17:57, 2 September 2022

2022-09-02T17:57:46Z

Noah at 17:55, 2 September 2022

2022-09-02T17:55:45Z

Noah at 17:25, 2 September 2022

2022-09-02T17:25:07Z

Noah at 17:24, 2 September 2022

2022-09-02T17:24:43Z

@@ Line 11: / Line 11: @@
 :$$\underline {y} \notin C_{\rm RS} = \{ \underline {c}_{\hspace{0.05cm}0}, \hspace{0.05cm}\text{...} \hspace{0.1cm}, \underline {c}_i,  \hspace{0.05cm}\text{...} \hspace{0.1cm},  \underline {c}_{\hspace{0.05cm}n -1} \}
   \hspace{0.05cm}. $$
-* However,&nbsp; if the corrupted received word&nbsp; $(\underline{y} &ne; \underline{c})$&nbsp; is a valid code word&nbsp; $(\underline{y\in C_{\rm RS}})$,&nbsp; the decoding process will leave the corrupted block undetected.
+* However,&nbsp; if the falsified received word&nbsp; $(\underline{y} &ne; \underline{c})$&nbsp; is a valid code word&nbsp; $(\underline{y\in C_{\rm RS}})$,&nbsp; the decoding process will leave the falsified block undetected.
@@ Line 66: / Line 66: @@
 $W_3 \ = \ ${ 245 }
-{What is the probability that an incorrect block remains undetected? Let the corruption probability of each symbol be&nbsp; $\varepsilon = 0.1$.
+{What is the probability that an incorrect block remains undetected? Let the falsification probability of each symbol be&nbsp; $\varepsilon = 0.1$.
 |type="{}"}
 $\rm RSC \, (7, \, 3, \, 5) \text{:} \hspace{0.4cm} Pr(block error) \ = \ ${ 0.263 3% } $\ \cdot 10^{-5}$

@@ Line 11: / Line 11: @@
 :$$\underline {y} \notin C_{\rm RS} = \{ \underline {c}_{\hspace{0.05cm}0}, \hspace{0.05cm}\text{...} \hspace{0.1cm}, \underline {c}_i,  \hspace{0.05cm}\text{...} \hspace{0.1cm},  \underline {c}_{\hspace{0.05cm}n -1} \}
   \hspace{0.05cm}. $$
-* However,&nbsp; if the corrupted received word&nbsp; $(\underline{y} &ne; \underline{c})$&nbsp; is a valid code word,&nbsp; &nbsp; the decoding process will leave the corrupted block undetected.
+* However,&nbsp; if the corrupted received word&nbsp; $(\underline{y} &ne; \underline{c})$&nbsp; is a valid code word&nbsp; $(\underline{y\in C_{\rm RS}})$,&nbsp; the decoding process will leave the corrupted block undetected.

@@ Line 25: / Line 25: @@
 Furthermore,&nbsp; let it hold:
-* Each symbol is corrupted into another symbol with probability&nbsp; $\varepsilon_{\rm S} = 0.1$&nbsp; and correctly transferred with probability&nbsp; $1 - \varepsilon_{\rm S} = 0.9$.
+* Each symbol is falsified into another symbol with probability&nbsp; $\varepsilon_{\rm S} = 0.1$&nbsp; and correctly transferred with probability&nbsp; $1 - \varepsilon_{\rm S} = 0.9$.
 * For the distance spectrum of a Reed&ndash;Solomon code of length&nbsp; $n$&nbsp; holds with&nbsp; $d = d_{\rm min}$:
 :$$W_i =   {n \choose i} \cdot \sum_{j = 0}^{i-d}\hspace{0.15cm}(-1)^j \cdot {i \choose j} \cdot \big  [\hspace{0.03cm}q^{i\hspace{0.03cm}-\hspace{0.03cm}j\hspace{0.03cm}-\hspace{0.03cm}d\hspace{0.03cm}+\hspace{0.03cm}1}-1 \hspace{0.03cm} \big ]\hspace{0.05cm}.$$
@@ Line 84: / Line 84: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; The equation describing the weights $W_i$ is (the minimum distance $d_{\rm min}$ is abbreviated here as $d$):
+'''(1)'''&nbsp; The equation describing the weights&nbsp; $W_i$&nbsp; is&nbsp; $($the minimum distance&nbsp; $d_{\rm min}$&nbsp; is abbreviated here as&nbsp; $d)$:
 :$$W_i =   {n \choose i} \cdot \sum_{j = 0}^{i-d}\hspace{0.15cm}(-1)^j \cdot {i \choose j} \cdot \big  [\hspace{0.03cm}q^{i\hspace{0.03cm}-\hspace{0.03cm}j\hspace{0.03cm}-\hspace{0.03cm}d\hspace{0.03cm}+\hspace{0.03cm}1}-1 \hspace{0.03cm} \big ]\hspace{0.05cm}.$$
-Because of the minimum distance $d_{\min} = 5$, $W_3 \ \underline{= 0}$ and $W_4 \ \underline{= 0}$. The other weights result in
+*Because of the minimum distance&nbsp; $d_{\min} = 5$,&nbsp; $W_3 \ \underline{= 0}$&nbsp; and&nbsp; $W_4 \ \underline{= 0}$.&nbsp; The other weights result in
 :$$W_5 =   {7 \choose 5} \cdot (8^1 - 1) = \frac {7 \cdot 6 \cdot 5 \cdot4 \cdot 3}{1 \cdot 2 \cdot 3 \cdot4 \cdot 5} \cdot 7 = 21 \cdot 7
 \hspace{0.15cm}\underline {= 147}\hspace{0.05cm},$$
@@ Line 97: / Line 97: @@
-'''(2)'''&nbsp; Analogous to subtask (1), we obtain:
+'''(2)'''&nbsp; Analogous to subtask&nbsp; '''(1)''',&nbsp; we obtain:
 :$$W_3 =   {7 \choose 3} \cdot (8^1 - 1) = \frac {7 \cdot 6 \cdot 5 }{1 \cdot 2 \cdot 3 } \cdot 7 = 35 \cdot 7
 \hspace{0.15cm}\underline {= 245}\hspace{0.05cm}.$$
-'''(3)'''&nbsp; The corruption probability of a symbol is given by $\varepsilon_{\rm S} = 0.1$.
+'''(3)'''&nbsp; The falsification probability of a symbol is given by&nbsp; $\varepsilon_{\rm S} = 0.1$.
-:Then, for the probability that in a codeword with $n = 7$ code symbols.
+:Then,&nbsp; for the probability that in a code word with&nbsp; $n = 7$&nbsp; symbols
-* exactly 3 symbols are corrupted:
+* exactly three symbols are falsified:
 :$$p_3 =  0.1^3  \cdot 0.9^4 = 0.6561  \cdot 10^{-3}
 \hspace{0.05cm},$$
-* exactly 4 symbols are corrupted:
+* exactly four symbols are falsified:
 :$$p_4 =  0.1^4  \cdot 0.9^3 = 0.729  \cdot 10^{-4}
 \hspace{0.05cm},$$
-* exactly 5 symbols are corrupted:
+* exactly five symbols are falsified:
 :$$p_5 =  0.1^5  \cdot 0.9^2 = 0.81  \cdot 10^{-5}
 \hspace{0.05cm},$$
-* exactly 6 symbols are corrupted:
+* exactly six symbols are falsified:
 :$$p_6 =  0.1^6  \cdot 0.9 = 0.9  \cdot 10^{-6}\hspace{0.05cm},$$
-* all $n = 7$ symbols are corrupted:
+* all $n = 7$ symbols are falsified:
 :$$p_7 =  0.1^7  = 10^{-7}\hspace{0.05cm}.$$
-At $\rm RSC \, (7, \, 3, \, 5)_8$ the zero word can be corrupted by symbol errors into one of $q^k - 1 = 8^3 - 1 = 511$ other codewords.
+&rArr; &nbsp; At the&nbsp; $\rm RSC \, (7, \, 3, \, 5)_8$&nbsp; the zero word can be falsified by symbol errors into one of&nbsp; $q^k - 1 = 8^3 - 1 = 511$&nbsp; other code words.&nbsp; Thus,&nbsp; using the weight enumerating functions of subtask&nbsp; '''(1)''',&nbsp; we obtain:
+:$${\rm Pr}({\rm block\hspace{0.15cm}error}) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \frac {W_5 \cdot p_5 + W_6 \cdot p_6 + W_7 \cdot p_7}{511} =\frac {147 \cdot 0.81  \cdot 10^{-5} + 147 \cdot 0.9  \cdot 10^{-6} + 217 \cdot 10^{-7}}{511}
 \hspace{0.15cm}\underline {\approx  0.263  \cdot 10^{-5}} \hspace{0.05cm}.$$
-At $\rm RSC \, (7, \, 5, \, 3)_8$ must be averaged over $8^5 - 1 = 32767$ corruption probabilities because of $k = 5$.
+&rArr; &nbsp; At the&nbsp; $\rm RSC \, (7, \, 5, \, 3)_8$&nbsp; we have to average over&nbsp; $8^5 - 1 = 32767$&nbsp; falsification probabilities because of&nbsp; $k = 5$.&nbsp; With&nbsp; $W_3 = 245$&nbsp; from subtask&nbsp; '''(2)'''&nbsp; and the weights&nbsp;
+:$$W_4 = 1224, \ W_5 = 5586, \ W_6 = 12838, \ W_7 = 12873$$
-With $W_3 = 245$ from subtask (2) and the weights $W_4 = 1224, \ W_5 = 5586, \ W_6 = 12838, \ W_7 = 12873$ according to the specification sheet, we obtain for this:
+according to the specification sheet,&nbsp; we obtain for this:
-:$${\rm Pr}({\rm Block\hspace{0.15cm}error}) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \frac {W_3 \cdot p_3 + W_4 \cdot p_4 + W_5 \cdot p_5 + W_6 \cdot p_6 + W_7 \cdot p_7}{32767} =\frac {245 \cdot 0.656  \cdot 10^{-3} + \hspace{0.15cm}... \hspace{0.15cm}+ 12873 \cdot 10^{-7}}{32767}
+:$${\rm Pr}({\rm block\hspace{0.15cm}error}) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \frac {W_3 \cdot p_3 + W_4 \cdot p_4 + W_5 \cdot p_5 + W_6 \cdot p_6 + W_7 \cdot p_7}{32767} =\frac {245 \cdot 0.656  \cdot 10^{-3} + \hspace{0.15cm}... \hspace{0.15cm}+ 12873 \cdot 10^{-7}}{32767}
 \hspace{0.15cm}\underline {\approx  0.942  \cdot 10^{-5}} \hspace{0.05cm}. $$
-'''(4)'''&nbsp; Let only $d_{\rm min}$ (further abbreviated as $d$) be known and thus also $p_d = \varepsilon_{\rm S}^d \cdot (1 - \varepsilon_{\rm S})^{n-d}$. This is also the upper bound we are looking for:
+:* ${\rm RSC} \, (7, \, 5, \, 3)_8 \text{:} \hspace{0.33cm} {\rm Pr(upper \ bound)} = p_3 = \underline{65.6 \cdot 10^{-5}}$.
-Since the weight $W_d$ was assumed to be unknown, it is set to the maximum possible value ($W_5 = 511$ or $W_3 = 32767$) so that the prefactors in the equations for subtask (3) disappear. This is the only way to ensure an upper bound.
+*Since the weight&nbsp; $W_d$&nbsp; was assumed to be unknown,&nbsp; it is set to the maximum possible value&nbsp; $(W_5 = 511$&nbsp; or&nbsp; $W_3 = 32767)$&nbsp; so that the prefactors in the equations for subtask&nbsp; '''(3)'''&nbsp; disappear.&nbsp; This is the only way to ensure an upper bound.
-In both cases, the upper bound is significantly higher than the results of subtask (3):
+*In both cases,&nbsp; the upper bound is significantly higher than the results of subtask&nbsp; '''(3)''':
-* $\rm RSC \, (7, \, 3, \, 5)_8 \text{:} \hspace{0.33cm} 0.810 \cdot 10^{-5}$ statt $0.263 \cdot 10^{-5}$ (Faktor approx. $3$),
+:* $\rm RSC \, (7, \, 3, \, 5)_8 \text{:} \hspace{0.33cm} 0.810 \cdot 10^{-5}$&nbsp; instead of &nbsp;$0.263 \cdot 10^{-5}$&nbsp; $($faktor approx.&nbsp; $3)$,
-'''(5)'''&nbsp; Using the abbreviation $d = d_{\rm min}$, one obtains for the lower bound:
+'''(5)'''&nbsp; Using the abbreviation&nbsp;  $d = d_{\rm min}$,&nbsp; one obtains for the lower bound:
-:$${\rm Pr}({\rm Lower\hspace{0.15cm} bound})= \frac{W_d \cdot p_d}{q^k -1}
+:$${\rm Pr}({\rm lower\hspace{0.15cm} bound})= \frac{W_d \cdot p_d}{q^k -1}
   \hspace{0.05cm}. $$
-* For the $\rm RSC \, (7, \, 3, \, 5)_8$, because of $W_d = W_5$ and $p_d = p_5$, the lower bound is about $11\%$ below the actual value $(0.263 \cdot 10^{-5})$:
+* For the&nbsp; $\rm RSC \, (7, \, 3, \, 5)_8$,&nbsp; because of&nbsp; $W_d = W_5$&nbsp;and&nbsp; $p_d = p_5$,&nbsp; the lower bound is about&nbsp; $11\%$&nbsp; below the actual value $(0.263 \cdot 10^{-5})$:
-:$${\rm Pr}({\rm Lower\hspace{0.15cm} bound}) = \frac{147 \cdot 0.81  \cdot 10^{-5}}{511}
+:$${\rm Pr}({\rm lower\hspace{0.15cm} bound}) = \frac{147 \cdot 0.81  \cdot 10^{-5}}{511}
   \hspace{0.15cm}\underline {\approx  0.233  \cdot 10^{-5}}$$
-* For the $\rm RSC \, (7, \, 5, \, 3)_8$ holds with $W_d = W_3$ and $p_d = p_3$, the lower bound here deviates from the actual value $(0.942 \cdot 10^{-5})$more strongly, because in this code the contributions of the higher weights $(W_4, \ W_5, \ W_6, \ W_7)$ are more relevant in relation to $W_3$:
+* For the&nbsp; $\rm RSC \, (7, \, 5, \, 3)_8$&nbsp; holds with&nbsp; $W_d = W_3$&nbsp; and&nbsp; $p_d = p_3$,&nbsp; the lower bound here deviates from the actual value&nbsp; $(0.942 \cdot 10^{-5})$&nbsp; more strongly,&nbsp; because in this code the contributions of the higher weights&nbsp; $(W_4, \ W_5, \ W_6, \ W_7)$&nbsp; are more relevant in relation to&nbsp; $W_3$:
-:$${\rm Pr}({\rm Lower\hspace{0.15cm} bound}) = \frac{245 \cdot 0.656  \cdot 10^{-3}}{32767}
+:$${\rm Pr}({\rm lower\hspace{0.15cm} bound}) = \frac{245 \cdot 0.656  \cdot 10^{-3}}{32767}
   \hspace{0.15cm}\underline {\approx  0.494  \cdot 10^{-5}}\hspace{0.05cm}.$$
 {{ML-Fuß}}

← Older revision		Revision as of 13:33, 11 October 2022
Line 46:		Line 46:
	*Reference is made in particular to the page  [[Channel_Coding/Definition_and_Properties_of_Reed-Solomon_Codes#Singleton_bound_and_minimum_distance\|"Singleton Bound and Minimum Distance"]].		*Reference is made in particular to the page  [[Channel_Coding/Definition_and_Properties_of_Reed-Solomon_Codes#Singleton_bound_and_minimum_distance\|"Singleton Bound and Minimum Distance"]].

−	* The weights  $W_i$ marked in red in the above graph are to be calculated.	+	* The weights  $W_i$  marked in red in the above graph are to be calculated.

@@ Line 1: / Line 1: @@
 {{quiz-Header|Buchseite=Channel_Coding/Definition_and_Properties_of_Reed-Solomon_Codes}}
-[[File: P_ID2524__KC_A_2_10.png|right|frame|Distance spectra of two different Reed-Solomon codes]]
+[[File: P_ID2524__KC_A_2_10.png|right|frame|Distance spectra of two <br>different Reed-Solomon codes]]
-For a linear block code, up to&nbsp; $e = d_{\rm min} - 1$&nbsp; errors can be detected. For all Reed&ndash;Solomon codes the minimum distance is
+For a linear block code,&nbsp; up to &nbsp; $e = d_{\rm min} - 1$ &nbsp; errors can be detected.&nbsp; For all Reed&ndash;Solomon codes the minimum distance is
 :$$d_{\rm min} = n-k+1  \hspace{0.05cm}.$$
 One must distinguish the following cases:
-* If no more than&nbsp; $e = n - k$&nbsp; symbol errors occur, the block is always detected as faulty.
+* If no more than&nbsp; $e = d_{\rm min} - 1= n - k$&nbsp; symbol errors occur,&nbsp; the block is always detected as faulty.
-* Error detection can still work even if more than&nbsp; $n - k$&nbsp; symbol errors occur, and this is when the received word is not a valid codeword of the Reed&ndash;Solomon code:
 :$$\underline {y} \notin C_{\rm RS} = \{ \underline {c}_{\hspace{0.05cm}0}, \hspace{0.05cm}\text{...} \hspace{0.1cm}, \underline {c}_i,  \hspace{0.05cm}\text{...} \hspace{0.1cm},  \underline {c}_{\hspace{0.05cm}n -1} \}
   \hspace{0.05cm}. $$
-* However, if the corrupted received word&nbsp; $(\underline{y} &ne; \underline{c})$&nbsp; is a valid codeword &nbsp; &#8658; &nbsp; $\underline{y}$, the decoding process will leave the corrupted block undetected.
+* However,&nbsp; if the corrupted received word&nbsp; $(\underline{y} &ne; \underline{c})$&nbsp; is a valid code word,&nbsp; &nbsp; the decoding process will leave the corrupted block undetected.
-We define as block error probability:
+We define as&nbsp; '''block error probability''':
-:$${\rm Pr}({\rm Block\hspace{0.15cm}error}) = {\rm Pr}(\underline {y} \ne \underline {c})
+:$${\rm Pr}({\rm block\hspace{0.15cm}error}) = {\rm Pr}(\underline {y} \ne \underline {c})
   \hspace{0.05cm}.$$
-In this exercise, this probability is to be determined for the following codes:
+In this exercise,&nbsp; this probability is to be determined for the following codes:
 * Reed&ndash;Solomon code&nbsp; $(7, \, 3, \, 5)_8 \ \Rightarrow \ d_{\rm min} = 5$,
 * Reed&ndash;Solomon code&nbsp; $(7, \, 5, \, 3)_8 \ \Rightarrow \ d_{\rm min} = 3$.
-Furthermore, let it hold:
+Furthermore,&nbsp; let it hold:
-* Each symbol is corrupted into another symbol with probability&nbsp; $\varepsilon_{\rm S} = 0.1$&nbsp; and correctly transferred with probability&nbsp; $1 - \varepsilon_{\rm S} = 0.9$&nbsp; .
+* Each symbol is corrupted into another symbol with probability&nbsp; $\varepsilon_{\rm S} = 0.1$&nbsp; and correctly transferred with probability&nbsp; $1 - \varepsilon_{\rm S} = 0.9$.
 * For the distance spectrum of a Reed&ndash;Solomon code of length&nbsp; $n$&nbsp; holds with&nbsp; $d = d_{\rm min}$:
 :$$W_i =   {n \choose i} \cdot \sum_{j = 0}^{i-d}\hspace{0.15cm}(-1)^j \cdot {i \choose j} \cdot \big  [\hspace{0.03cm}q^{i\hspace{0.03cm}-\hspace{0.03cm}j\hspace{0.03cm}-\hspace{0.03cm}d\hspace{0.03cm}+\hspace{0.03cm}1}-1 \hspace{0.03cm} \big ]\hspace{0.05cm}.$$
-Besides, two bounds for the block error probability will be considered and evaluated:
+Besides,&nbsp; two bounds for the block error probability will be considered and evaluated:
-* If the minimum distance alone is known, one can derive an <i>upper bound</i>&nbsp; from it. The weight enumerating factors&nbsp; $W_i$&nbsp; are to be chosen in such a way, that certainly (that is: &nbsp; with all constellations) applies:
+* If alone the minimum distance is known,&nbsp; one can derive an&nbsp; "upper bound"&nbsp; from it.&nbsp; The weight enumerating factors&nbsp; $W_i$&nbsp; are to be chosen in such a way,&nbsp; that certainly (that is: &nbsp; with all constellations)&nbsp; applies:
-:$${\rm Pr}({\rm Upper\hspace{0.15cm} bound}) \ge {\rm Pr}({\rm Block\hspace{0.15cm}error})
+:$${\rm Pr}({\rm upper\hspace{0.15cm} bound}) \ge {\rm Pr}({\rm block\hspace{0.15cm}error})
   \hspace{0.05cm}. $$
-* A <i>lower bound</i>&nbsp; additionally requires knowledge of the weight function&nbsp; $W_i$&nbsp; for&nbsp; $i = d_{\rm min}$. Thus, the following condition can be satisfied:
+* A&nbsp; "lower bound"&nbsp; additionally requires knowledge of the weight function&nbsp; $W_i$&nbsp; for&nbsp; $i = d_{\rm min}$.&nbsp; Thus,&nbsp; the following condition can be satisfied:
-:$${\rm Pr}({\rm Upper\hspace{0.15cm} bound}) \le {\rm Pr}({\rm Block\hspace{0.15cm}error})
+:$${\rm Pr}({\rm lower\hspace{0.15cm} bound}) \le {\rm Pr}({\rm block\hspace{0.15cm}error})
   \hspace{0.05cm}.$$
@@ Line 44: / Line 43: @@
 Hints:
 *The exercise belongs to the chapter&nbsp; [[Channel_Coding/Definition_and_Properties_of_Reed-Solomon_Codes|"Definition and Properties of Reed-Solomon Codes"]].
 *Reference is made in particular to the page&nbsp; [[Channel_Coding/Definition_and_Properties_of_Reed-Solomon_Codes#Singleton_bound_and_minimum_distance|"Singleton Bound and Minimum Distance"]].
-* To be calculated are the weights marked in red in the above graph&nbsp; $W_i$.
@@ Line 61: / Line 62: @@
 $W_7 \ = \ ${ 217 }
-{What is the missing weight of the&nbsp; $\rm RSC \, (7, \, 5, \, 3)_8$ in the graph?
+{What is the missing weight of the&nbsp; $\rm RSC \, (7, \, 5, \, 3)_8$&nbsp; in the graph?
 |type="{}"}
 $W_3 \ = \ ${ 245 }
-{What is the probability that an incorrect block remains undetected? Let the corruption probability of each symbol be $\varepsilon = 0.1$.
+{What is the probability that an incorrect block remains undetected? Let the corruption probability of each symbol be&nbsp; $\varepsilon = 0.1$.
 |type="{}"}
-$\rm RSC \, (7, \, 3, \, 5) \text{:} \hspace{0.4cm} Pr(Blockfehler) \ = \ ${ 0.263 3% } $\ \cdot 10^{-5}$
+$\rm RSC \, (7, \, 3, \, 5) \text{:} \hspace{0.4cm} Pr(block error) \ = \ ${ 0.263 3% } $\ \cdot 10^{-5}$
-$\rm RSC \, (7, \, 5, \, 3) \text{:} \hspace{0.4cm} Pr(Blockfehler) \ = \ ${ 0.942 3% } $\ \cdot 10^{-5}$
+$\rm RSC \, (7, \, 5, \, 3) \text{:} \hspace{0.4cm} Pr(block error) \ = \ ${ 0.942 3% } $\ \cdot 10^{-5}$
-{Calculate and evaluate for both codes the upper bound proposed in the specification&nbsp; $p_{\rm upper} = \rm Pr(Upper \ bound)$.
+{Calculate and evaluate for both codes the&nbsp; "upper bound"&nbsp; proposed in the specification:&nbsp; $p_{\rm upper} = \rm Pr(upper \ bound)$.
 |type="{}"}
 ${\rm RSC} \, (7, \, 3, \, 5) \text{:} \hspace{0.4cm} p_{\rm upper} \ = \ ${ 0.81 3% } $\ \cdot 10^{-5}$
 ${\rm RSC} \, (7, \, 5, \, 3) \text{:} \hspace{0.4cm} p_{\rm upper} \ = \ ${ 65.6 3% } $\ \cdot 10^{-5}$
-{Calculate and evaluate for both codes the lower bound proposed in the specification&nbsp; $p_{\rm lower} = \rm Pr(Lower \ bound)$.
+{Calculate and evaluate for both codes the&nbsp; "lower bound"&nbsp; proposed in the specification:&nbsp; $p_{\rm lower} = \rm Pr(lower \ bound)$.
 |type="{}"}
 ${\rm RSC} \, (7, \, 3, \, 5) \text{:} \hspace{0.4cm} p_{\rm lower} \ = \ ${ 0.233 3% } $\ \cdot 10^{-5}$