The Analytic Network Process

Abstract

Analysis to break down a problem into its constituent components to study their behavior has been the major tool of scientific inquiry to test hypotheses and solve problems. It has proven to be extremely successful in dealing with the world of matter and energy. It has enabled man to land on the moon, to harness the energy of the atom, to master global communication, to invent the computer and to produce tens of thousands of useful and not so useful things. But it has not been so effective in the world of man.

Appendix

Proof of Theorem 5:

• 1. The condition of unrestricted domain is automatically satisfied because the aggregation procedure is defined on all pairwise reciprocal profiles without restriction.

• 2. Given ${\mathbf{P}} \in \mathfrak{P}^{m}$ such that for any pair $A_{i} {\text{ and }}A_{j}$ of $\mathfrak{A}$, $P_{k} (A_{i} ,A_{j} ) > 1$ for all $k \in \mathfrak{M}$, then by definition,

  $$ f({\mathbf{P}})(A_{i} ,A_{j} ) = \left[ {\prod\limits_{k = 1}^{m} {P_{k} (A_{i} ,A_{j} )} } \right]^{\frac{1}{m}} > 1 $$

  and f satisfies pairwise unanimity.

• 3. Let P and Q be two reciprocal pairwise profiles. Let $a_{ijk} \equiv P_{k} (A_{i} ,A_{j} )$ and $b_{ijk} \equiv Q_{k} (A_{i} ,A_{j} )$. Let ${\mathbf{A}} = \{ A_{i} ,A_{j} ,A_{l} \} \subset \mathfrak{A}$. If P| _A  = Q| _A , then $a_{ijk} = b_{ijk}$, $a_{ilk} = b_{ilk}$ and $a_{jlk} = b_{jlk}$, for all k, and f(P)| _A  = f(Q)| _A follows.

• 4. In general, the geometric mean of a set of (positive) numbers is not equal to any one of these numbers, and hence, it is non-dictatorial.

Proof of Theorem 6:

Let ${\mathbf{P}} = \left( {P_{k} (A_{i} ,A_{j} ) \equiv \frac{{w_{ik} }}{{w_{jk} }},k = 1,2, \ldots ,m} \right)$ be a consistent pairwise profile. We have

$\begin{aligned} \sigma_{P} ({\mathbf{P}}) = (\omega_{2} \circ f_{P} )({\mathbf{P}}) = \omega_{2} (P) = \omega_{2} \left( {P(A_{i} ,A_{j} ) = \frac{{\prod\limits_{k = 1}^{m} {w_{ik}^{1/m} } }}{{\prod\limits_{k = 1}^{m} {w_{jk}^{1/m} } }}} \right) \hfill \\ = \left( {\prod\limits_{k = 1}^{m} {w_{ik}^{1/m} } ,i = 1, \ldots ,n} \right) = w. \hfill \\ \end{aligned}$.

Also, $\sigma_{W} ({\mathbf{P}}) = (f_{W} \circ \omega_{1} )({\mathbf{P}}) = f_{W} (\omega_{1} (P_{1} ), \ldots ,\omega_{1} (P_{m} )) = f_{W} (w_{1} , \ldots ,w_{m} )$where $\omega_{1} \left( {P_{k} } \right) = w_{k} = (w_{1k} , \ldots ,w_{nk} )^{T}$ and we have $\sigma_{W} ({\mathbf{P}}) = \left( {\prod\limits_{k = 1}^{m} {w_{1k}^{1/m} } , \ldots ,\prod\limits_{k = 1}^{m} {w_{nk}^{1/m} } } \right)^{T} = w$ and hence, $\sigma_{P} = \sigma_{W}$ on consistent pairwise profiles. In addition, because we have shown that the geometric mean satisfies Arrow’s conditions, it follows that $\sigma_{P} = \sigma_{W}$ also satisfies these conditions.

Proof of Theorem 7:

Let ${\mathbf{P}} = (P_{1} , \ldots ,P_{m} ) \in \mathfrak{P}_{\mathfrak{D}}^{m} \subset \mathfrak{P}^{m}$ be a reciprocal pairwise profile that satisfies pairwise dominance. Then, for any pair of alternatives i and j,$P_{k} (A_{i} ,A_{l} ) \ge P_{k} (A_{j} ,A_{l} )$, for all l and k, we have

$$ f_{P} ({\mathbf{P}})(A_{i} ,A_{l} ) = \left[ {\prod\limits_{k = 1}^{m} {P_{k} (A_{i} ,A_{l} )} } \right]^{\frac{1}{m}} \ge \left[ {\prod\limits_{k = 1}^{m} {P_{k} (A_{j} ,A_{l} )} } \right]^{\frac{1}{m}} = f_{P} ({\mathbf{P}})(A_{j} ,A_{l} ) $$

or

$$ P(A_{i} ,A_{l} ) \ge P(A_{j} ,A_{l} ){\text{ for all }}l $$

and the geometric mean satisfies strong unanimity.

Proof of Theorem 8:

Let ${\mathbf{P}} = (P_{1} , \ldots ,P_{m} )$ be a reciprocal cardinal profile. From Theorem 2, $\gamma_{P}$ satisfies strong unanimity, i.e., for every pair of alternatives (i,j), $f_{P} ({\mathbf{P}})(A_{i} ,A_{l} ) \ge f_{P} ({\mathbf{P}})(A_{j} ,A_{l} )$ for all l, or $a_{il} = P(A_{i} ,A_{l} ) \ge P(A_{j} ,A_{l} ) = a_{jl}$ for all l, and hence, by construction $\sum\limits_{l = 1}^{n} {a_{il} w(A_{l} )} \ge \sum\limits_{l = 1}^{n} {a_{jl} w(A_{l} )}$ or $w(A_{i} ) \ge w(A_{j} )$ and $\sigma_{P} |_{\mathfrak{D}}$ satisfies unanimity.

Let P and Q be two reciprocal pairwise profiles and let w and v the corresponding absolute cardinal relations, respectively. If for any subset A of $\mathfrak{A}$ with three elements P| _A  = Q| _A is true, then by Theorem 1 the geometric mean aggregation procedure satisfies pairwise independence from irrelevant alternatives and hence $f_{P} ({\mathbf{P}})|_{{\mathbf{A}}} = f_{P} ({\mathbf{Q}})|_{{\mathbf{A}}}$. Let $f_{P} ({\mathbf{P}}) = P$ and $f_{P} ({\mathbf{Q}}) = Q$, and let $w = \omega_{2} (P)$ and $v = \omega_{2} (Q)$. We have

$$ \begin{gathered} \sigma_{P} |_{\mathfrak{D}} ({\mathbf{P}})|_{{\mathbf{A}}} = (\omega_{2} \circ f_{P} )({\mathbf{P}})|_{{\mathbf{A}}} = \omega_{2} (f_{P} ({\mathbf{P}})|_{{\mathbf{A}}} ) \hfill \\ = \omega_{2} (f_{P} ({\mathbf{Q}})|_{{\mathbf{A}}} ) = (\omega_{2} \circ f_{P} )({\mathbf{Q}})|_{{\mathbf{A}}} = \sigma_{P} |_{\mathfrak{D}} ({\mathbf{Q}})|_{{\mathbf{A}}} \hfill \\ \end{gathered} $$

and thus $w|_{{\mathbf{A}}} = v|_{{\mathbf{A}}}$. ð

Let ${\mathbf{P}} = (P_{1} , \ldots ,P_{m} ) \in \mathfrak{P}_{\mathfrak{D}}^{m}$ and $w = \sigma_{P} ({\mathbf{P}}) \in \mathcal{W}$. $\sigma_{P} |_{\mathfrak{D}} :\mathfrak{P}^{m} \to \mathcal{W}$ is dictatorial if there is a member d of the group for which $\sigma_{P} ({\mathbf{P}}) = w_{d}$. For this to happen, the following must hold: $\sigma_{P} ({\mathbf{P}}) = \omega_{2} (f_{P} ({\mathbf{P}})) = \omega_{2} (P_{d} )$ and $w(A_{i} ) = w_{d} (A_{i} )$ for all i. But by Theorem 1, the geometric mean aggregation procedure $f_{P}$ is non-dictatorial, i.e.,$f_{P} ({\mathbf{P}}) \ne P_{d}$, from which we have $\sigma_{P} ({\mathbf{P}}) = \omega_{2} (f_{P} ({\mathbf{P}})) \ne \omega_{2} (P_{d} )$ and hence, $\sigma_{P} |_{\mathfrak{D}} :\mathfrak{P}^{m} \to \mathcal{W}$ is non-dictatorial.