Oliver Yu © 2020
QUANTITATIVE BUSINESS ANALYSIS
Analytic Hierarchy Process
An Improved Approach for Quantifying Values and Selecting the Best choice:
Oliver Yu, Ph.D.
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OUTLINE
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SIMPLE HIERARCHICAL AND SYSTEMATIC PROCESS FOR SELECTING BEST CHOICE
A simple hierarchical and systematic process for selecting the best choice is the previously-discussed Multifactor Evaluation, shown here with another example:
Hierarchical Step Hypothetical Example
1. State overall decision Finding the best computer for the decision- maker
2. Specify key values and Affordability (A), Quality (Q), and Style (S).
determine the relative The decision-maker qualitatively assigns
weight Wj of each value j High (H), Medium (M), and Low (L)
by simultaneous direct weights respectively to A, Q, and S.
comparisons of the values
3. Identify major choices High-end, Middle, and Low-end computers
and rate the Choice k Affordability Quality Style
with respective to each choice _____ Initial Ratings (Rjk)______
Value j by simultaneous direct High-end L H H
comparisons Rjk as shown in Middle M M M
table at right Low-end H L L
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SIMPLE HIERARCHICAL AND SYSTEMATIC PROCESS (concluded)
Hierarchical Step Hypothetical Example
4. Quantify the weights It is not unreasonable to assign
and ratings H=9, M=6, and L=3, or normalized:
H=9/(9+6+3)=0.5, M= 0.33, and L=0.17
5. Determine the Total Score The numerical example is given below
of each choice through with normalized ratings in parentheses
the weighted average of
the ratings
Value j Afford. Quality Style
Weight Wj 0.50 0.33 0.17 Total Score (Sj WjRjk)
Choice k Quantitative Ratings (Rjk)_
High-end L=3 (0.17) H (0.50) H (0.50) 0.50(0.17)+0.33(0.50)+0.17(0.50) = 0.33
Middle M=6 (0.33) M (0.33) M (0.33) 0.50(0.33)+0.33(0.33)+0.17(0.33) = 0.33
Low-end H=9 (0.50) L (0.17) L (0.17) 0.50(0.50)+0.33(0.17)+0.17(0.17) = 0.33
The best choice is the one with the highest total score; however, this approach often results in total scores that are not very differentiated.
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SIMPLE HIERARCHICAL AND SYSTEMATIC PROCESS – MAJOR SHORTCOMINGS
In this process, the weights of the values or the ratings of choices with respect to a given value are evaluated by simultaneous direct comparison. The comparison usually starts qualitatively then refines quantitatively.
As shown in the previous example, the decision-maker compares the weights of the three values first qualitatively through the ordinal scale of H, M, and L, and then quantitatively as 9, 6, and 3. A similar approach is used for rating the choices with respect to each value
The qualitative scale can be refined, such as HH, HM, and HL. The quantitative scale can also be changed to 1 to 5, or 1 to 100, or others.
Major Shortcomings:
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ANALYTIC HIERARCHY PROCESS (AHP):
Psychological & Mathematical Foundations
Analytic Hierarchy Process (AHP) was developed by Prof. Tom Saaty of University of Pittsburgh in 1970s by observing the following psychological and mathematical characteristics in making comparisons:
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THE PAIRWISE COMPARISON SCALE AND MATRIX
At each hierarchy, the factors, are values, subvalues, or choices in the decision process, and their weights or ratings represent the relative degrees of preference, desirability, or importance of these factors.
Let wjk be the decision-maker’s relative preference of factor j over that of factor k. A pairwise comparison scale of wjk for AHP is defined as follows:
wjk = 1 Factor j is equally preferred to factor k
3 Factor j is moderately preferred to factor k
5 Factor j is strongly preferred to factor k
7 Factor j is very strongly preferred to factor k
9 Factor j is extremely preferred to factor k
Even numbers between 2 and 8 are used for in-between preferences.
Clearly, wjj = 1 and wkj = 1/wjk
Thus, for n factors, the decision-maker needs to make n(n-1)/2 comparisons.
A = [wjk ] is then the pairwise comparison matrix.
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ESTIMATING RELATIVE PREFERENCES OF FACTORS BY ROW AVERAGING
For n comparable factors, V1, V2, …, Vn, in a hierarchy of factors, we can create a nxn pairwise comparison matrix [wik] by defining wik, the element on the j-th row and k-th column of the matrix, to be the comparative or relative degree of preference of Vj to that of Vk for a decision-maker, in that, if wik > 1, then Vj is more important than Vk, and vice versa.
By definition, if the comparisons are totally consistent with one another, then the columns of the matrix will be proportional to or multiples of one another.
However, a decision-maker’s perspectives from different factors are often not totally consistent. A simple smoothing process can be used to reduce the errors from these inconsistencies by first normalizing all values in the each column and then dividing the sum of the normalized values of each row by the number of columns, as shown by the formula below, to produce the row average Wj as the relative preference of the j-th factor among n factors in a given hierarchy:
Wj= Sk [wjk/(Ss wsk)]/n, s = 1,2,…,n; for j = 1,2,…,n
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ESTIMATING THE RELATIVE WEIGHTS –
AN EXAMPLE
For the 3 values in the previous example, it is reasonable for the decision-maker to have the following pairwise comparison matrix:
Affordability Quality Style
Affordability 1 4 9
Quality 1/4 1 3
Style 1/9 1/3 1
Then the relative weights of the values can be estimated by
For the above example, we have
Affordability Quality Style Row Average
Affordability 1/(1+1/4+1/9)=0.74 4/(4+1+1/3)=0.75 9/(9+3+1)=0.69 0.73
Quality (1/4)/(1+1/4+1/9)=0.18 1/(4+1+1/3)=0.19 3/(9+3+1)=0.23 0.20
Style (1/9)/(1+1/4+1/9)=0.08 (1/3)/(4+1+1/3)=0.06 1/(9+3+1)=0.08 0.07
These row averages are the relative weights of the 3 values, and in this way, they are more precise than the respective relative weights of 0.5, 0.33, and 0.17 obtained by the simultaneous direct comparison approach.
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CHECKING THE CONSISTENCY OF
A PAIRWISE COMPARISON MATRIX
Let l max be the largest eigenvalue of matrix [wik]. If the pairwise comparisons are totally consistent, then wjk= Wj/Wk for all j, k, and it can be shown that l max = n. Thus, the deviation of l max from n can be used to measure the degree of inconsistency among the pairwise comparisons. Since the deviation tends to increase with the number of factors n, Prof. Saaty has constructed a Consistency Index CI = |lmax – n|/(n-1) that normalizes the deviation. Clearly, a totally consistent matrix will have CI=0.
l max = [Sj (SkWkwjk)/Wj]/n
Based on empirical observations of a large number of comparison matrices, it is determined that if CI < 0.05*, then the matrix is sufficiently consistent; i.e., Wj’s are close to those of a similar matrix with CI=0, and the relative weights can then be used directly for decision-making purposes without modification. Otherwise, revisions of the pairwise comparisons will be required for the weights or ratings to be acceptable for use.
* This is a stronger criterion and can be used without the Consistency Ratios discussed in the technical article by R.W. Saaty.
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CHECKING THE CONSISTENCY OF A MATRIX – AN EXAMPLE
For the previous example of the pairwise comparison matrix for the 3 values:
Step 1. Multiply column i by the Row Average i and sum over each row
1 4 9 2.16
0.73x 1/4 +0.20x 1 +0.07x 3 = 0.59
1/9 1/3 1 0.22
Step 2. Divide element i of the resulting vector by Row Average i
2.16/0.73 = 2.96 0.59/0.20 = 2.96 0.22/0.07 = 3.14
Step 3. The average of the ratios in Step 2 is the best estimate for lmax
(2.96 + 2.96 + 3.14)/3 = 3.02
Step 4. Compute the consistency index CI = |lmax – n|/(n-1), where n is the number of factors in the hierarchy.
CI = (3.02 – 3)/2 = 0.01
For this example, CI < 0.05, so the level of consistency among the pair-wise comparisons is acceptable.
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MAKING A MATRIX TOTALLY CONSISTENT (New Materials not in the Module)
If CI > 0.05, then the matrix needs to be modified to be usable.
As mentioned previously, which can also be readily proven, the matrix will be totally consistent with CI=0 if the columns (or rows) are multiples of one another. Thus, we can modify an insufficiently consistent matrix into a totally consistent matrix by making the columns proportional to one another.
To achieve this proportionality rationally and systematically, the decision-maker needs to rank order the pairwise comparisons by the level of confidence in their validity, and modify comparisons of lower confidence to conform with those are at higher levels of confidence.
The most confident and valid comparisons are obviously the unity diagonal elements of the matrix as they are comparisons of the factors to themselves.
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MAKING A MATRIX TOTALLY CONSISTENT – Systematic Procedure
The two highest ranked non-diagonal comparisons are automatically valid.
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MAKING THE MATRIX CONSISTENT – AN EXAMPLE
For 4 factors A,B,C, and D, a decision-maker has the following pair-wise comparison matrix and rankings of the confidence of validity in the comparisons:
A B C D (1) wBD (i.e., comparison value of B vs D) Highest
A 1 5 1/2 1/4 (2) wAD
B 1 7 3 (3) wAB
C 1 2 (4) wCD
D 1 (5) wAC
(6) wBC Lowest
C will initially include the 4 diagonal elements. Since the top 2 comparisons, wBD=3 and wAD=1/4 do not form rectangles with any diagonal elements, they will be automatically included in C as valid. Now, because wAD, wBD, and wBB are already in C, the next comparison on the confidence list, wAB, must be equal or made equal to wAD(wBB/wBD) = (1/4)(1/3)=1/12* .
Following this procedure, we will finally modify the matrix into the following
A B C D
A 1 1/12 1/8 1/4
B 12 1 3/2 3
C 8 2/3 1 2
D 4 1/3 1/2 1
*: In making a matrix totally consistent, the modified comparison value can exceed the 1/9-9 scale; however, if the modified value is way out of this scale, decision-maker should review the appropriateness of the hierarchy and revise accordingly.
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A SUMMARY EXAMPLE OF AHP
Selecting the Best Computer for the Decision-Maker
Value Affordability Quality Style Row Av (Wj)
Affordability 1 (0.73) 4 (0.76) 9 (0.64) 0.73 lmax = 3.02
Quality 1/4 (0.18) 1 (0.19) 3 (0.29) 0.20
Style 1/9 (0.08) 1/3 (0.05) 1 (0.07) 0.07 CI = 0.01`
2. For each value, determine the ratings of the competing choices
Value Affordability (W1 = 0.73)
Choice High-end Middle Low-end Row Av (R1k)
HE 1 (0.09) 1/2 (0.08) 1/8 (0.09) 0.09 lmax = 3.01
M 2 (0.18) 1 (0.15) 1/5 (0.15) 0.16
LE 8 (0.73) 5 (0.77) 1 (0.75) 0.75 CI = 0.005
Value Quality (W2= 0.20)
Choice High-end Middle Low-end Row Av (R2k)
HE 1 (0.46) 1 (0.43) 6 (0.60) 0.50 lmax = 3.06
M 1 (0.46) 1 (0.43) 3 (0.30) 0.40
LE 1/6 (0.08) 1/3 (0.14) 1 (0.10) 0.11 CI = 0.03
Value Style (W3 = 0.07)
Choice High-end Middle Low-end Row Av (R3k)
HE 1 (0.69) 3 (0.72) 9 (0.56) 0.66 lmax = 3.05
M 1/3 (0.23) 1 (0.24) 6 (0.38) 0.28
LE 1/9 (0.08) 1/6 (0.04) 1 (0.06) 0.06 CI = 0.025
Numbers in parentheses are normalized for each column.
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A SUMMARY EXAMPLE OF AHP – concluded
3. Compute the total score of each choice by the formula SjWj Rjk
Choice/Rating High-end/Rj1 Middle/Rj2 Low-end/Rj3
Value and Weight Wj
1. Affordability 0.73 0.09 0.16 0.75
2. Quality 0.20 0.50 0.40 0.11
3. Style 0.07 0.66 0.28 0.06
Total Score Sj Wj Rjk 0.21 0.22 0.57
These total scores are much more differentiated than those based on simultaneous direct comparisons used in Multifactor Evaluation.
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STRENGTHS, WEAKNESSES, AND MAJOR APPLICATIONS OF AHP
Strengths:
Weaknesses:
Major Applications:
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HOMEWORK ASSIGNMENTS 2, 3 & 4
2. (10 points) Apply AHP to re-develop the weights for the three values: Affordability, Quality, and Style to HW1a. Compute CI for the pairwise comparison matrix; if it is greater than 0.05, then modify the matrix to make it totally consistent. Compare these weights with the normalized values of the original weights you have assigned in HW1a, and discuss the differences.
3. (30 points) Apply AHP to the remainder of HW1a and compare the resulting total scores of the 3 candidate computers with the normalized values of the scores you obtained in HW1a, and discuss the differences and any insights gained.
4. (20 points) For 4 choices A,B,C, and D, a decision-maker has the following pair-wise comparison matrix and rankings of the confidence in their validity:
A B C D (1) A vs. C Highest
A 1 1/2 1/3 1/4 (2) A vs. D
B 1 2 4 (3) C vs. D
C 1 1/2 (4) B vs. C
D 1 (5) A vs. B
(6) B vs. D Lowest
Use the rankings as the basis to make the matrix totally consistent.