SDA 8: The Analytic Hierarchy
Process
The importances of the criteria could be
approximated by the AHP using pairwise
comparisons:
• T. L. Saaty, The Analytic Hierarchy Process: Planning, Priority Setting, Resource Allocation, McGraw-Hill, New
York, 1980.
Suppose that the value function has the
form
1
v(y) =
q
wiyi
i=1
If wi = 0, the corresponding outcome yi
can be deleted from consideration. Thus,
we shall assume that wi > 0, i = 1, 2, . . . , q.
Define the weight ratio by
wi
wij = .
wj
Note that, for any i, j, k indexes
−1,
wij = wji
wij = wik wkj .
2
Define the matrix of weight ratios as W =
[wij ]q×q :

w1 w1
w w
 1 2
w w
 2 2

 w1 w2

. . . . .

 wq wq
w1 w2

w1
w1
...
w3
wq 

w2
w2

...

w3
wq 

. . . . . . . .

wq
wq 
...
w3
wq
A matrix W is called consistent if its com−1,
ponents satisfy the equalities wij = wji
wij = wik wkj for any i, j and k.
3
Observe that:
• Since each row of W is a multiple of
the first row, the rank of W is one, and
thus there is onlu one nonzero eigenvalue which is q.
This due to the fact that wii = 1 and
that the sum of all eigenvalues is equal
q
to the trace of W (i.e. i=1 wii = q).
• We can easily chack that
W w = qw
therefore w must be the eigenvector of
W corresponding to the maximum eigenvalue q.
4
As a living system, human perception and
judgment are subject to change when the
information inputs or psychological states
of the decsion maker change.
A fixed weight vector is difficult to find.
Saaty proposed the following to overcome
this difficulty:
Estimate or elicit the weight ratio wij by
aij and let A = [aij ]q×q be the matrix of
components {aij }.
Note that as each wij > 0, we expect and
shall assume that all aij > 0.
5
−1, Saaty sugFurthermore, as wij = wji
gested that in practice, only aij , j > i
need to be assessed.
Since A is found as an approximate for
W , when the consistency conditions are
almost satisfied for A, one would expect
that the normalized eigenvector corresponding to the maximum eigenvector of A, denoted by λmax, will also be close to w.
Theorem 1. The maximum eigenvalue, λmax,
of A is a positive real number.
Let ŵ be the normalized eigenvector corresponding to λmax of A. Then ŵi > 0 for
all 1 ≤ i ≤ q.
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Theorem 2. The maximum eigenvalue of
A satisfies the inequality
λmax ≥ q.
Assume we have q objectives and we want
to construct a scale, rating these objectives as to their importance with respect
to the decision, as seen by the analyst.
We ask the decision maker to compare the
objectives in paired comparisons.
If we are comparing objective i with objective j, we assign the values aij and aji
as follows:
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• aij = a−1
ji
• If objective i is more important than
objective j then aij gets assigned a number as follows:
Note that the above observation is valid
for any matrix which is consistent.
8
Intensity of
relative importance
Definition
1
equal importance
3
weak importance
(of one over the other)
5
strong importance
7
demonstrated importance
over the other
9
absolute importance
2,4,6,8
intermediate
values between
Saaty’s scale of relative importances.
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Example 1. Let us consider the following
matrix


1 9 7
1

1


1


A = 9
5
1

5 1
7
To find λmax we solve
det[A − λI] = 0
that is,
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

1−λ
9
7
 1

1


1
−
λ


det  9
5 
 1

5
1−λ
7
= (1 − λ)3 − 3(1 − λ) + 9/35 + 35/9 = 0
The maximum solution is
λmax = 3.21.
After normalization we get
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ŵ1 = 0.77, ŵ2 = 0.05, ŵ3 = 0.17.
We illustrate Saaty’s method on a job selection problem (3 alternatives compared
on 6 criteria)
overall satisfaction with the job
level 1: focus
criteria
research
level 3: alternatives
growth
benefits
A
colleauges
B
Choice of job.
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location
C
reputation
The question asked was, which of a given
pair of criteria is seen as contributing more
to overall satisfaction with a job and what
is the intensity or strength of the difference?
res. growth benefits coll. location reputation priority
research
1
1
1
4
1
1/2
0.16
growth
1
1
2
4
1
1/2
0.19
benefits
1
1/2
1
5
3
1/2
0.19
colleaug.
1/4
1/4
1/5
1
1/3
1/3
0.05
location
1
1
1/3
3
1
1
0.12
reputation
2
2
2
3
1
1
0.30
Pairwise comparison matrix of criteria.
The relative weights of criteria (priorities)
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can be computed as normalized geometric
means of the rows (which are very close
to the eigenvector corresponding to the largest
eigenvalue of the matrix)
The geometric means are computed as
m1 =
1 × 1 × 1 × 4 × 1 × 1/2
6
1 × 1 × 2 × 4 × 1 × 1/2
m2 =
6
m3 =
1 × 1/2 × 1 × 5 × 3 × 1/2
6
14
m4 =
6
1/4 × 1/4 × 1/5 × 1 × 1/3 × 1/3
1 × 1 × 1/3 × 3 × 1 × 1
m5 =
6
m6 =
√
6
2×2×2×3×1×1
So, the relative weight (priority) of the
criterion research is obtained as
m1
p1 =
m1 + m2 + m3 + m4 + m5 + m6
15
Then we compare the alternatives on each
of the criteria
research A B
A
B
C
1 1/4 1/2
4 1 3
2 1/3 1
growth A B
A
B
C
C priority
C priority
1 1/4 1/2
4 1 3
2 1/3 1
16
0.14
0.63
0.24
0.10
0.33
0.57
benefits A B C priority
A
B
C
1 3 1/3
1/3 1
3 1 1
colleaug. A
A
B
C
B C priority
1 1/3 5
3 1 7
1/5 1/7 1
17
location A
A
B
C
1 1 7
1 1 7
1/7 1/7 1
reputat. A
A
B
C
B C priority
B C priority
1 7 9
1/7 1 5
1/9 1/5 1
18
0.77
0.17
0.05
We obtain




0.14
0.10




0.16 ×  0.63  + 0.19 ×  0.33  + · · ·
0.24
0.57


0.77


+0.30 ×  0.17 
0.05


0.40
A


=  0.34  B
0.26
C
So job A should be selected as the best
alternative.
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