QUANTITATIVE BUSINESS ANALYSIS

Oliver Yu © 2020

QUANTITATIVE BUSINESS ANALYSIS

Analytic Hierarchy Process

 

An Improved Approach for Quantifying Values and Selecting the Best choice:

 

Oliver Yu, Ph.D.

 

AHP-*

Oliver Yu © 2020

OUTLINE

    • Simple hierarchical structure for selecting the best choice

 

    • The multi-factor evaluation approach and its major shortcomings

 

    • Analytic Hierarchy Process (AHP): Psychological and mathematical foundations

 

    • The pairwise comparison scale and matrix

 

    • Estimating the relative weights of factors from the matrix

 

    • Checking the consistency of a pairwise comparison matrix

 

    • Making a pairwise comparison matrix totally consistent

 

    • A summary example of AHP

 

    • Strengths, weaknesses, and major applications of AHP

 

  • Homework assignments 2, 3, and 4

AHP-*

Oliver Yu © 2020

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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Oliver Yu © 2020

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.

AHP-*

Oliver Yu © 2020

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:

 

  • Humans usually have only 1.5 digit of precision.
  • It is generally difficult to differentiate more than 2 factors, especially when they have weights or ratings relatively close to one another.
  • There is often instability in the comparisons in that the weights or ratings may shift when the comparisons are repeated.
  • There is no way to determine whether the comparisons are internally consistent with one another.

AHP-*

Oliver Yu © 2020

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:

 

  • The hierarchical structure is an effective way for a person to systematically break down a large problem into smaller and smaller sub-problems or factors.
  • However, for the breakdown process to be efficient, at each hierarchy, the factors should be easily comparable in weights or ratings; i.e., their weights or ratings should be within an order of magnitude of one another.
  • At each hierarchy, humans can differentiate the factors best by comparing two, i.e., pairwise, rather than more than two at a given time.
  • Since the comparisons are within an order of magnitude of one another, a scale of 1 through 9 is appropriate.
  • Pairwise comparisons of the weights or ratings of factors at a given hierarchy form a matrix that can be studied mathematically with precision and consistency by using matrix theory.

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Oliver Yu © 2020

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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Oliver Yu © 2020

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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Oliver Yu © 2020

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

  • Dividing each element in the matrix by its respective column sum
  • Averaging the ratios of each row, and the row averages are the relative preferences

 

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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Oliver Yu © 2020

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.

 

  • max can be precisely determined mathematically. However, it can also be approximated as follows:

 

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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Oliver Yu © 2020

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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Oliver Yu © 2020

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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Oliver Yu © 2020

MAKING A MATRIX TOTALLY CONSISTENT – Systematic Procedure

  • Let C be the group of pairwise comparisons that have been accepted as valid and wjk be the next comparison on the confidence ranking list to be evaluated for validity.
  • If there exist wjr, wsk, and wsr that are already in C for some r and s, (i.e., wjk and these exsiting 3 members of C form the 4 corners of a rectangle or square) then wjk must be equal or made equal to wjr(wsk/wsr) to ensure column proportionality before it can be a valid new member of C.
  • On the other hand, if such members of C do not exist, then wjk is automatically valid and can be included in C without modification because no challenges from members of C.
  • Proceed until all comparisons are members of C.

The two highest ranked non-diagonal comparisons are automatically valid.

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Oliver Yu © 2020

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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Oliver Yu © 2020

A SUMMARY EXAMPLE OF AHP

Selecting the Best Computer for the Decision-Maker

 

  • First determine the weights of the values

 

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.

AHP-*

Oliver Yu © 2020

STRENGTHS, WEAKNESSES, AND MAJOR APPLICATIONS OF AHP

Strengths:

  • Intuitive, simple to use, and easily programmable
  • Numerically stable, i.e., not susceptible to small errors
  • Widely applied throughout the world

 

Weaknesses:

  • No time dynamics or feedbacks
  • In rating the choices, one may need to compare each choice with a hypothetical ideal to assure all choices are up to minimum standard.
  • For resource allocation applications, no sophisticated relationships between choices and preferences can be developed or applied

 

Major Applications:

  • Relative preference estimation
  • Resource allocation in proportion to preference
  • Probability forecasting

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Oliver Yu © 2020

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.

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