1 Introduction

Analytic hierarchy process (AHP) is a scaling method for priorities in the hierarchical structure [1, 2]. When people (decision makers) make decisions, they should often treat multiple alternatives. Decision makers’ main task is to define the priorities of alternatives.

Pairwise comparison is one of the most tangible ways to prioritize them. Let \(C_i, i=1,\ldots ,n\) be alternatives to be prioritized. One defines the value \(a_{ij}\) in pairs between \(C_i\) and \(C_j\). The value represents the priority of \(C_i\) as compared to \(C_j\). In the context of AHP, one usually uses its value by Saaty’s discrete scale from 1 to 9 and their reciprocals. It is based on the verbal expressions in Table 1 [1].

Table 1 Saaty’s discrete scale
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For instance, suppose that one wishes to determine his/her preference for fruits. In comparison of apples (\(C_i\)) and oranges (\(C_j\)), if he/she prefers apples to oranges moderately, \(a_{ij}\) is assigned to 3 and \(a_{ji}\) to 1/3. Logically the value \(a_{ji}\) in pairs between \(C_j\) and \(C_i\) should be \(1/a_{ij}\). After all pairwise comparisons, one obtains the following so-called pairwise comparison matrix:

$$\begin{aligned} A= \begin{pmatrix} 1 &{} a_{12} &{} \cdots &{} a_{1n} \\ 1/a_{12} &{} 1 &{} \cdots &{} a_{2n} \\ \vdots &{} \vdots &{} \ddots &{} \vdots \\ 1/a_{1n} &{} 1/a_{2n} &{} \cdots &{} 1 \end{pmatrix}. \end{aligned}$$

In AHP, employing a pairwise comparison matrix enables the treatment of subjective judgment.

In the context of AHP, evaluating the principal (maximum magnitude) eigenvalues of pairwise comparison matrices is a key ingredient. In AHP, solving the following linear system is called by the eigenvector method:

$$\begin{aligned} Aw=\lambda _\mathrm {max} w, \end{aligned}$$

where \(\lambda _\mathrm {max}\) is the maximum magnitude eigenvalue and w is the associated eigenvector. The maximum magnitude eigenvalue defines the first and most popular consistency index of pairwise comparison matrices [2, 3]. The associated eigenvector is utilized as a priority vector (weight vector) [1, 2, 4, 5]. From the viewpoint of mathematical programming, the theoretical foundation of the eigenvector method was established by [6, 7]. In this note, we focus on third-order pairwise comparison matrix. The third-order pairwise comparison matrix has importance in AHP. The reason is that it is a source of inconsistency. There are several research articles devoted to third-order pairwise comparison matrix; see [8,9,10] and references therein. In this note, we treat inconsistent third-order pairwise comparison matrix A. Inconsistency is characterized by \(\det A > 0\).

It is clear that the existence of the real-number root of the characteristic equation because the polynomial to be considered is odd order. In Sect. 2, we will specify the area where the real-number root exists. The root is shown to be the maximum magnitude eigenvalue among all eigenvalues of the pairwise comparison matrix.

The analytic form of the maximum magnitude eigenvalue is known; see [9] and references therein. Nevertheless, in this note, we shed light on Newton’s method to obtain the maximum magnitude eigenvalue. We will see mathematical nature of Newton’s method for third-order comparison matrix stands out. In Sect. 3, we will construct a sequence by the usual Newton’s method. We will take an initial value to be 3, which is equal to the order of the matrix. We rely on the favorable properties of inconsistent third-order pairwise comparison matrix; uniqueness of the real-number root and \(\det A > 0\). We will theoretically prove that the sequence generated by Newton’s method converges to the maximum magnitude eigenvalue. We also confirm the property of convergence by Newton’s method by computational experiment in Sect. 4. Utilization of Newton’s method is expected to extend the result to possible application for fourth or more order pairwise comparison matrix.

2 Existence of the Root

We consider the third-order pairwise comparison matrix:

$$\begin{aligned} A= \begin{pmatrix} 1 &{} a_{12} &{} a_{13} \\ 1/a_{12} &{} 1 &{} a_{23} \\ 1/a_{13} &{} 1/a_{23} &{} 1 \end{pmatrix}, \end{aligned}$$

where \(a_{ij} > 0\). Its characteristic polynomial has the following form [11]:

$$\begin{aligned} P_A({\lambda }) = {\lambda }^3-3{\lambda }^2-\det A. \end{aligned}$$

We can calculate the determinant of the pairwise comparison matrix as follows.

$$\begin{aligned} \det A&= \begin{vmatrix} 1&a_{12}&a_{13} \\ 1/a_{12}&1&a_{23} \\ 1/a_{13}&1/a_{23}&1 \end{vmatrix} \\&= \frac{a_{12}a_{23}}{a_{13}}+\frac{a_{13}}{a_{12}a_{23}}-2\ge 0. \end{aligned}$$

The last inequality follows from the relationship between the arithmetic mean and the geometric mean. The equality holds if and only if \(\dfrac{a_{12}a_{23}}{a_{13}}=\dfrac{a_{13}}{a_{12}a_{23}}\), which is equivalent to \(a_{12}a_{23}=a_{13}\). We say the matrix A is consistent if \(a_{12}a_{23}=a_{13}\). For the third-order pairwise comparison matrix, as we mentioned above, it is obvious that the consistency of A and \(\det A=0\) is equivalent. If A is consistent, the characteristic polynomial is \(P_A({\lambda })={\lambda }^3-3{\lambda }^2={\lambda }^2({\lambda }-3)\). In this case, the roots of the characteristic equation are \({\lambda }=3,0\) (0 is a multiple root).

Hereafter, we treat the case when A is inconsistent (i.e., \(\det A >0\)). Taking the derivatives of \(P_A({\lambda })\), we have \(P^{\,\prime }_A({\lambda })=3{\lambda }(\lambda -2)\). So \(P_A({\lambda })\) takes a local maximum at the point \({\lambda }=0\) and it takes a local minimum at the point \({\lambda }=2\). Moreover, for \({\lambda }>2\), \({P^{\,\prime }_A}({\lambda })=3{\lambda }(\lambda -2)>0\), so \(P_A(\lambda )\) is monotone increasing in this area. The vertical axis section is \(-\det A< 0\). We indicate the shapes of the graph of \(P_A({\lambda })\) in Fig. 1.

Fig. 1
figure 1

Graph of \(P_A(\lambda )\) in inconsistent case

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An easy calculation shows the following:

$$\begin{aligned}P_A(3) = &\;3^3-3\cdot 3^2 -\det A =-\det A <0, \\ P_A(3+\det A) =&\; (3+\det A)^3-3(3+\det A)^2-\det A \\=&\; 27+27\det A+9(\det A)^2+(\det A)^3 \\& -(27+18\det A +3(\det A)^2)-\det A \\=&\; 8\det A + 6(\det A)^2 +(\det A)^3 > 0. \end{aligned}$$

We immediately have the following existence theorem of a root for the characteristic equation.

Theorem 1

For the inconsistent pairwise comparison matrix A, the characteristic equation has a unique real-number root in the interval \((3,3+\det A)\). This root is the maximum magnitude eigenvalue.

Proof

By the intermediate value theorem (p.206 in [12]), the existence of a real-number root in the interval \((3,3+\det A)\) is clear. From the shape of the graph, this root is the unique real-number root. We denote it by \(\lambda ^*\). In general, other two conjugate complex-number roots exist, say \(a\pm bi\), \(b\ne 0\).

From the general theory of eigenvalues and the trace of the matrix [2], we have \({\lambda ^*}+(a+bi)+(a-bi)=3\). So we have

$$\begin{aligned} a=\frac{3-{\lambda ^*}}{2}. \end{aligned}$$

Since \(a+bi\) is the root of the characteristic equation, we have

$$\begin{aligned} (a+bi)^3-3(a+bi)^2-\det A=0. \end{aligned}$$

By expanding the left-hand-side of the formula, we have

$$\begin{aligned} \left\{ \ \begin{aligned} 0&= a^3-3ab^2-3a^2+3b^2-\det A,\\ 0&= 3a^2b-b^3-6ab. \end{aligned} \right. \end{aligned}$$

Since \(b\ne 0\), from the last formula, we have

$$\begin{aligned} 3a^2-b^2-6a=0. \end{aligned}$$

So we finally have

$$\begin{aligned} b^2=3a^2-6a. \end{aligned}$$

Thus we have

$$\begin{aligned} | \, a \pm bi \, |^2&= a^2+b^2 = 4a^2 - 6a \\&= 4\Bigl (\frac{3-{\lambda ^*}}{2}\Bigr )^2 -6\Bigl (\frac{3-{\lambda ^*}}{2}\Bigr ) \\&= {\lambda ^*}^2-3{\lambda ^*}<{\lambda ^*}^2. \end{aligned}$$

This means \(\lambda ^*\) is maximum magnitude.

3 Convergence of Newton’s Method

Here we use the usual Newton’s method (Chapter 8 in [13]). We give an elementary proof of convergence. We set the initial value to be \({\lambda }_0=3\) and generate a sequence by the following iteration.

$$\begin{aligned} {\lambda }_0&= 3,\\ {\lambda }_{n+1}&= {\lambda }_n-\frac{P_A({{\lambda }_n})}{{P^{\,\prime }_A}({{\lambda }_n})}, \quad \text {for n} \ge 0. \end{aligned}$$

Denote the maximum magnitude eigenvalue guaranteed by Theorem 1 by \(\lambda _\mathrm {max}\) . So \(P_A(\lambda _{\text {max}})=0\) and \(3<\lambda _{\text {max}}<3+\det A\).

Lemma 1

For all \(n\ge 1\), we have \({\lambda }_n > \lambda _{\text {max}}\).

Proof

We prove it by induction. Set \(n=1\). \({\lambda }_1=3+\frac{\det A}{9}>3\) and \(P_A({\lambda }_{1})=\frac{2}{27}(\det A)^2+(\frac{\det A}{9})^3>0\). So we have \({\lambda }_1>\lambda _{\text {max}}\).

Assume \({\lambda }_n > \lambda _{\text {max}}\) is true. Then from this assumption, we can take \(\lambda _{\text {max}}\le {\lambda } < {\lambda }_n\). Since \({P^{\,\prime }_A}({\lambda })\) is monotone increasing for \({\lambda }>1\) (because \({P^{\,\prime \prime }_A}({\lambda }) = 6(\lambda -1)\)) and \(\lambda _{\text {max}}> 3\), we have

$$\begin{aligned} {P^{\,\prime }_A}({\lambda })<{P^{\,\prime }_A}({\lambda }_n) \quad \text {for} \lambda \in [\lambda _{\text {max}}, {\lambda }_n). \end{aligned}$$

Thus we have

$$\begin{aligned} \int _{{\lambda _{\text {max}}}}^{{\lambda }_n}{P^{\,\prime }_A}({\lambda })\,d{\lambda } < \int _{{\lambda _{\text {max}}}}^{{\lambda }_n}{P^{\,\prime }_A}({\lambda }_n)\,d{\lambda }. \end{aligned}$$

So we conclude

$$\begin{aligned} P_A({\lambda }_n)-P_A({\lambda _{\text {max}}}) < {P^{\,\prime }_A}({\lambda }_n)({\lambda }_n-{\lambda _{\text {max}}}). \end{aligned}$$

Taking account into \(P_A(\lambda _{\text {max}})=0\), we have

$$\begin{aligned} \lambda _{\text {max}}<{\lambda }_n-\frac{P_A({{\lambda }_n})}{{P^{\,\prime }_A}({{\lambda }_n})} ={\lambda }_{n+1}. \end{aligned}$$

Lemma 2

The sequence \(\{{\lambda }_n\}\) is monotone decreasing for \(n\ge 1\).

Proof

From Lemma 1, we have \(3<\lambda _{\text {max}}< {\lambda }_n\). Since \(P_A({\lambda })\) is monotone increasing for \({\lambda } > 3\), we have \(P_A({\lambda }_n)> P_A(\lambda _{\text {max}})=0\). Obviously \({P^{\,\prime }_A}({{\lambda }_n})>0\) and \(P_A({{\lambda }_n})>0\), so we have

$$\begin{aligned} {\lambda }_{n+1}={\lambda }_n-\frac{P_A({{\lambda }_n})}{{P^{\,\prime }_A}({{\lambda }_n})} <{\lambda }_{n}. \end{aligned}$$

From Lemmas 1 and 2, the sequence is monotone decreasing and bounded below. So it converges, see p.183 in [12].

Theorem 2

The sequence \(\{{\lambda }_n\}\) converge to \(\lambda _{\text {max}}\).

Proof

From Lemmas 1 and 2, the sequence is monotone decreasing and bounded below. So it converges to, say \(\hat{{\lambda }} \ge {\lambda _{\text {max}}}\). Taking the limit of the iteration

$$\begin{aligned} {\lambda }_{n+1}={\lambda }_n-\frac{P_A({{\lambda }_n})}{{P^{\,\prime }_A}({{\lambda }_n})}, \end{aligned}$$

we have

$$\begin{aligned} \hat{{\lambda }}=\hat{{\lambda }}-\frac{P_A({\hat{{\lambda }})}}{{P^{\,\prime }_A}(\hat{{\lambda }})} \end{aligned}$$

by the continuity of \(P_A({\lambda })\) and \({P^{\,\prime }_A}({{\lambda }})\). Since \(\hat{{\lambda }} \ge {\lambda _{\text {max}}}>3\), the inequality \({P^{\,\prime }_A}(\hat{{\lambda }})>0\) holds. So we have \(P_A(\hat{{\lambda }})=0\). Because the root of \(P_A({\lambda })=0\) is unique, \(\hat{{\lambda }}\) is identical to \({\lambda _{\text {max}}}\).

Now we state the rate of convergence.

Theorem 3

The sequence \(\{{\lambda }_n\}\) converge to \(\lambda _{\text {max}}\) quadratically.

Proof

For each iteration of \({\lambda }_n\), by the second-order Taylor’s theorem (p.254 in [12]), there exists \({\xi }_n \in (\lambda _{\text {max}}, {\lambda }_n)\), we have

$$\begin{aligned} 0&= P_A(\lambda _{\text {max}}) \\&= P_A({\lambda }_n)+{P^{\,\prime }_A}({{\lambda }_n})(\lambda _{\text {max}}-{\lambda }_n) +\dfrac{1}{2}{P^{\,\prime \prime }_A}({{\xi }_n})(\lambda _{\text {max}}-{\lambda }_n)^2. \end{aligned}$$

Hence we have

$$\begin{aligned} {\lambda }_n-\dfrac{P_A({\lambda }_n)}{{P^{\,\prime }_A}({{\lambda }_n})}-\lambda _{\text {max}}&= \dfrac{1}{2}\dfrac{{P^{\,\prime \prime }_A}({{\xi }_n})}{{P^{\,\prime }_A}({{\lambda }_n})}({\lambda }_n-\lambda _{\text {max}})^2, \\ {\lambda }_{n+1}-\lambda _{\text {max}}&= \dfrac{1}{2}\dfrac{{P^{\,\prime \prime }_A}({{\xi }_n})}{{P^{\,\prime }_A}({{\lambda }_n})}({\lambda }_n-\lambda _{\text {max}})^2\\&\approx \dfrac{1}{2}\dfrac{{P^{\,\prime \prime }_A}(\lambda _{\text {max}})}{{P^{\,\prime }_A}({\lambda _{\text {max}}})}({\lambda }_n-\lambda _{\text {max}})^2. \end{aligned}$$

By Theorem 1, we know \(\lambda _{\text {max}}> 3\), so \({P^{\,\prime }_A}(\lambda _{\text {max}})=3{\lambda _{\text {max}}}({\lambda _{\text {max}}}-2)>0\) and \({P^{\,\prime \prime }_A}({\lambda _{\text {max}}})=6({\lambda _{\text {max}}}-1)>0\). We also know \({\lambda }_{n+1}-\lambda _{\text {max}}>0\). Hence we have

$$\begin{aligned} {\lambda }_{n+1}-\lambda _{\text {max}}\approx \dfrac{1}{2}\dfrac{{P^{\,\prime \prime }_A}(\lambda _{\text {max}})}{{P^{\,\prime }_A}({\lambda _{\text {max}}})}({\lambda }_n-\lambda _{\text {max}})^2, \end{aligned}$$

which implies quadratic convergence.

4 Computational Experiment

We have a computational experiment to compare Newton’s method with other root-finding methods. The experiment consists of the following steps.

Step 1. Generate an inconsistent pairwise comparison matrix A of order 3, at random.

Step 2. Iterate the following three methods until the differences between \(\lambda _{\text {max}}\) and the obtained values become below the threshold and store the numbers of iterations.

  • Newton’s method with the initial value of \(\lambda _0=3\).

    $$\begin{aligned} {\lambda }_{n+1}={\lambda }_n-\frac{P_A({{\lambda }_n})}{{P^{\,\prime }_A}({{\lambda }_n})}, \qquad n=0,1,2.\dots . \end{aligned}$$

  • Secant methodFootnote 1 with initial values of \(\lambda _{-1}=3\) and \(\lambda _0=3+\det A\).

    $$\begin{aligned} {\lambda }_{n+1}={\lambda }_n-\frac{P_A({{\lambda }_n})({\lambda }_n-{\lambda }_{n-1})}{P_A({\lambda }_n)-P_A({\lambda }_{n-1})}, \qquad n=0,1,2.\dots . \end{aligned}$$

  • Bisection methodFootnote 2 with initial interval of \([\lambda _0^{\text {lower}}, \lambda _0^{\text {upper}}] = [3, 3+\det A]\).

    $$\begin{aligned} {\lambda }_{n+1}=\frac{{\lambda }_n^{\text {lower}}+{\lambda }_n^{\text {upper}}}{2}, \qquad n=0,1,2.\dots , \end{aligned}$$

    where

    $$\begin{aligned}{}[\lambda _{n+1}^{\text {lower}}, \lambda _{n+1}^{\text {upper}}] = [\lambda _{n+1}, \lambda _{n}^{\text {upper}}] \end{aligned}$$

    if \(P_A(\lambda _{n+1})\) and \(P_A(\lambda _{n}^{\text {lower}})\) have the same sign, and

    $$\begin{aligned}{}[\lambda _{n+1}^{\text {lower}}, \lambda _{n+1}^{\text {upper}}] = [\lambda _{n}^{\text {lower}}, \lambda _{n+1}], \end{aligned}$$

    if \(P_A(\lambda _{n+1})\) and \(P_A(\lambda _{n}^{\text {upper}})\) have the same sign, for \(n=0,1,2.\dots\).

Figure 2 shows the distributions of the numbers of iterations until convergence by each method, after repeating these steps 5,000 times. We set the threshold of convergence to \(10^{-12}\) here.

Fig. 2
figure 2

Distributions of the number of iterations until convergence

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The summary statistics of the 5,000 numbers of iterations until convergence by each method are shown in Table 2.

Table 2 Summary statistics of the distributions of three methods
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For example, for the pairwise comparison matrix of

$$\begin{aligned} A = \begin{pmatrix} 1 &{} 5 &{} 5 \\ 1/5 &{} 1 &{} 1/7 \\ 1/5 &{} 7 &{} 1 \end{pmatrix}, \end{aligned}$$

obtained values converge in 5, 8, and 41 iterations by Newton’s method, the secant method, and the bisection method, respectively. This case can be considered typical because these are the median of each method. The transitions of the values toward convergence for this matrix are shown in Fig. 3.

Fig. 3
figure 3

Transition toward convergence in a typical case

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This experiment confirms the speed of convergence of Newton’s method empirically.

5 Conclusion

In this note, we entirely use the favorable properties from which A is third order. About other properties concerning third-order pairwise comparison matrices, see [9]. The key contribution of this note is findings of the possible area where the root of the characteristic equation and to show the convergence of Newton’s method from the initial value \({\lambda }_0=3\). We also confirm the superiority of Newton’s method compared to other root-finding methods—the secant method and the bisection method. This is considered as a direct consequence of the quadratic convergence of Newton’s method.

For the matrices for fourth and more order dimensions, expanding this note is left to future research.

  • For fourth-order pairwise comparison matrix, there usually exist two roots of the characteristic equation. So, if we set the initial value \({\lambda }_0=4\), Newton’s method does not generally converge to \(\lambda _{\text {max}}\).

  • We need to study the appropriate initial value and the convergence property.