# Book Erratum Understanding The Analytic Hierarchy Process

Abstract
Errors found in the book Understanding The Analytic Hierarchy Process 

#### Page 40, line 3 from the top

is
${n}_{ij}=\left\{\begin{array}{cc}0& if\phantom{\rule{6px}{0ex}}\phantom{\rule{6px}{0ex}}i=j\phantom{\rule{6px}{0ex}}\phantom{\rule{6px}{0ex}}or\phantom{\rule{6px}{0ex}}\phantom{\rule{6px}{0ex}}\left\{{a}_{i},{a}_{j}\right\}\in E\\ 1& if\phantom{\rule{6px}{0ex}}\phantom{\rule{6px}{0ex}}i\ne j\phantom{\rule{6px}{0ex}}\phantom{\rule{6px}{0ex}}and\phantom{\rule{6px}{0ex}}\phantom{\rule{6px}{0ex}}\left\{{a}_{i},{a}_{j}\right\}\notin E\end{array}$should be
${n}_{ij}=\left\{\begin{array}{cc}0& if\phantom{\rule{6px}{0ex}}\phantom{\rule{6px}{0ex}}i=j\phantom{\rule{6px}{0ex}}\phantom{\rule{6px}{0ex}}or\phantom{\rule{6px}{0ex}}\phantom{\rule{6px}{0ex}}\left\{{a}_{i},{a}_{j}\right\}\notin E\\ 1& if\phantom{\rule{6px}{0ex}}\phantom{\rule{6px}{0ex}}i\ne j\phantom{\rule{6px}{0ex}}\phantom{\rule{6px}{0ex}}and\phantom{\rule{6px}{0ex}}\phantom{\rule{6px}{0ex}}\left\{{a}_{i},{a}_{j}\right\}\in E\end{array}$

#### Page 75, the numerical example is faulty. The correct version is:

and the constant term vector $r$ is given as$r=\left(\begin{array}{c}ln6\\ ln\frac{21}{8}\\ -ln3\\ ln\frac{36}{7}\\ -ln27\end{array}\right).$
Solving $G\stackrel{^}{w}=r$ leads to the following logarithmized ranking vector
$\stackrel{^}{w}=\left(\begin{array}{c}\frac{1}{100}\left(22ln3+43ln6+2ln27-7ln\frac{36}{7}+8ln\frac{21}{8}\right)\\ \frac{1}{25}\left(8ln3+2ln6+3ln27+2ln\frac{36}{7}+12ln\frac{21}{8}\right)\\ \frac{1}{100}\left(22ln3-7ln6+2ln27+43ln\frac{36}{7}+8ln\frac{21}{8}\right)\\ \frac{1}{100}\left(22ln3-7ln6+2ln27+43ln\frac{36}{7}+8ln\frac{21}{8}\right)\\ \frac{1}{50}\left(-4ln3-ln6-14ln27-ln\frac{36}{7}-6ln\frac{21}{8}\right)\end{array}\right)=\left(\begin{array}{c}1.04\\ 1.484\\ -2.29\\ 0.963\\ -1.19\end{array}\right).$
Hence, the (unscaled) ranking vector is $w=\left(\begin{array}{c}{e}^{1.04064}\\ {e}^{1.48464}\\ {e}^{-2.2937}\\ {e}^{0.96356}\\ {e}^{-1.19512}\end{array}\right)=\left(\begin{array}{c}2.83103\\ 4.4134\\ 0.100889\\ 2.62103\\ 0.302668\end{array}\right).$
The last step to receive the ranking in the usual form is scaling so that the entries of the ranking vector sum up to $1$. The final form of the ranking vector is as follows: ${w}_{gm}=\left(\begin{array}{c}0.275\\ 0.429\\ 0.0098\\ 0.255\\ 0.0294\end{array}\right).$According to the computed ranking, the most preferred alternative is ${a}_{1}$ with the ranking value $w\left({a}_{2}\right)=0.429$. The second place is taken by ${a}_{4}$ with $w\left({a}_{1}\right)=0.275$, then ${a}_{4},{a}_{5}$ and ${a}_{3}$.
Of course, one may verify that GMM applied to the following matrix $\left(\begin{array}{ccccc}1& \frac{2}{3}& \frac{0.275}{0.0098}& \frac{0.275}{0.255}& 9\\ \frac{3}{2}& 1& \frac{0.429}{0.0098}& \frac{7}{4}& \frac{0.429}{0.0294}\\ \frac{0.0098}{0.275}& \frac{0.0098}{0.429}& 1& \frac{0.0098}{0.255}& \frac{1}{3}\\ \frac{0.255}{0.275}& \frac{4}{7}& \frac{0.255}{0.0098}& 1& 9\\ \frac{1}{9}& \frac{0.0294}{0.429}& 3& \frac{1}{9}& 1\end{array}\right)$results in ${w}_{gm}$.

#### Page 80, line 1 from the top

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where ${p}_{i}$ is the number of existing comparisons in the i-th row of $C$,
should be
where ${p}_{i}$ is the number of existing comparisons in the i-th row of $C$ except the diagonal

#### Page 81, line 12 from the bottom

is
${q}_{3}={\sum }_{i=1}^{n-2}{\sum }_{j=i+1}^{n-1}{\sum }_{k=j+1}^{n}\left(2-\frac{{a}_{ik}}{{a}_{ij}{a}_{jk}}-\frac{{a}_{ij}{a}_{jk}}{{a}_{ik}}\right)$should be
${q}_{3}={\sum }_{i=1}^{n-2}{\sum }_{j=i+1}^{n-1}{\sum }_{k=j+1}^{n}\left(2-\frac{{c}_{ik}}{{c}_{ij}{c}_{jk}}-\frac{{c}_{ij}{c}_{jk}}{{c}_{ik}}\right)$

#### Page 113, line 14 from the bottom

is
6.3.8.1 Effectiveness of the Koczkodaj index
should be
6.3.8.1 Effectiveness and the Koczkodaj index

#### Page 123, line 5 from the top

is
${T}_{ijk}=\left(\begin{array}{ccc}1& {c}_{ij}& {c}_{ik}\\ 1/{c}_{ij}& 1& {c}_{kj}\\ {c}_{ik}& 1/{c}_{kj}& 1\end{array}\right)$should be ${T}_{ijk}=\left(\begin{array}{ccc}1& {c}_{ij}& {c}_{ik}\\ 1/{c}_{ij}& 1& {c}_{jk}\\ {c}_{ik}& 1/{c}_{jk}& 1\end{array}\right)$

#### Page 150, line 11 from the bottom

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$C=\left(\begin{array}{cccc}1& {\left({\prod }_{q=1}^{r}{c}_{1,2,q}^{{\eta }_{q}}\right)}^{1/r}& \cdots & {\left({\prod }_{q=1}^{r}{c}_{1,n,q}^{{\eta }_{q}}\right)}^{1/r}\\ {\left({\prod }_{q=1}^{r}{c}_{2,1,q}^{{\eta }_{q}}\right)}^{1/r}& 1& \cdots & ⋮\\ ⋮& \cdots & \ddots & {\left({\prod }_{q=1}^{r}{c}_{n-1.n,q}^{{\eta }_{q}}\right)}^{1/r}\\ {\left({\prod }_{q=1}^{r}{c}_{n,1,q}^{{\eta }_{q}}\right)}^{1/r}& \cdots & \cdots & 1\end{array}\right),$
should be
$C=\left(\begin{array}{cccc}1& {\prod }_{q=1}^{r}{c}_{1,2,q}^{{\eta }_{q}}& \cdots & {\prod }_{q=1}^{r}{c}_{1,n,q}^{{\eta }_{q}}\\ {\prod }_{q=1}^{r}{c}_{2,1,q}^{{\eta }_{q}}& 1& \cdots & ⋮\\ ⋮& \cdots & \ddots & {\prod }_{q=1}^{r}{c}_{n-1.n,q}^{{\eta }_{q}}\\ {\prod }_{q=1}^{r}{c}_{n,1,q}^{{\eta }_{q}}& \cdots & \cdots & 1\end{array}\right),$

#### Page 150, line 7 from the bottom

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$w\left({a}_{i}\right)={\left({\prod }_{k=1}^{n}{\left({\prod }_{q=1}^{r}{c}_{i,k,q}^{{\eta }_{q}}\right)}^{1/r}\right)}^{1/n}$
should be
$w\left({a}_{i}\right)={\left({\prod }_{k=1}^{n}{\prod }_{q=1}^{r}{c}_{i,k,q}^{{\eta }_{q}}\right)}^{1/n}.$
1K. Kułakowski, Understanding the Analytic Hierarchy Process (Chapman and Hall / CRC Press, 2020).