ANALYTICAL

HEIRARCHY PROCESS

Theory

with a Case

AHP Process is explained

with two methods namely Normalization and Eigen vector Method and error

rectification is explained for global weights calculations. Same is

explained with an example as well.

DEEPAK AGGARWAL

1/29/2018

Analytical Hierarchy Process

INTRODUCTION

The AHP process was first developed

and introduced by Thomas L Saaty as early as in 1970 which afterwards was used

and researched exhaustively.

As per Thomas L Saaty, the

Analytical Hierarchy process can be defined as is

“Structured technique for

organizing and analysing complex decisions, based on mathematics and

psychology.”

Basically it is a decision making

method. It can be used at number of areas like government, purchasing goods or

services or making decisions based on perception like going for a college.

AHP need not necessarily give the

correct or accurate decision but can be made the most consistent one with the

data based on ratings or perception. It helps individuals or organisations to

set priorities where data based on ideas, feelings etc. can be quantified.

AHP PROCESS

AHP can be used in number of steps:

A. Organizing

the parameters of criteria and alternative:

1.

Setting of Goal or

Objective: In order to approach the solution

of the problem, the first step includes freezing of agenda or output intended

to become the outcome like selection of college or purchasing a house etc.

2.

Setting of Criterion or

Criteria: The next step involves setting of

the criteria based on ideas, feelings or ratio data collected through

questionnaire

3.

Setting of Sub Criterion or Criteria: The

above step can be further de-clustered into Sub criterion for the objective set

to give in depth analysis of the decision making.

4.

Alternatives: The

final Process is to evaluate the alternatives available among which the choices

have to be made.

5.

Tree Structure:

Finally arrange all the above steps into a tree like structure to evaluate the

global preferences of all the alternatives.

B. Collect

data for comparison matrix:

1. Only

two alternatives / criterion can be compared at a time.

2. The

pair can be compared using fundamental scale (1 to 9)

3. It

can measure both tangible and intangible criterions.

4. The

criterions can be scaled in following manner:

5:

Extremely favourable à For Alternative “A”

4: Very Strongly Favourable

3: Strongly Favourable

2: Slightly Favourable

1: Equally Favourable

_______________________________________

2: Slightly Favourable

3: Strongly Favourable

4: Very Strongly Favourable

5:

Extremely favourable à For Alternative “B”

From the above example if choose let’s say 4 for B

alternative then preference of B over A is 4 OR Ratio of B: A = 4. In the

similar manner other data can be collected through questionnaire or any other

method.

C. Develop

the comparison matrix:

While preparing comparison matrix,

arrange all alternatives both on Rows and columns and while calculating

preference Numerator to be kept as Row Alternative and Denominator to be kept

as column alternative.

For example if the weight-age of

each alternative is S1, S2……Sn, then the Matrix cell will look like “S1/S2” for

1st row and 2nd column and so on. The resultant will be a

square matrix as per the following details:

Alternative 1

Alternative 2

Alternative 3

Alternative 1

S1/S1

S1/S2

S1/S3

Alternative 2

S2/S1

S2/S2

S2/S3

Alternative 3

S3/S1

S3/S2

S3/S3

OR

Alternative 1

Alternative 2

Alternative 3

Alternative 1

1

S1/S2

S1/S3

Alternative 2

S2/S1

1

S2/S3

Alternative 3

S3/S1

S3/S2

1

With

weight-age of Alternative 1 à S1

Alternative

2 à S2

Alternative 3 à S3

Note:

1.

If there are n

alternatives then comparison matrix can be made using nC2 formula which gives

number of comparisons required for it.

2.

Also it can be noticed

that diagonal elements are always unity as it’s the ratio of same alternatives.

3.

It can also be noticed

that elements which are diagonally opposite to each other are inverse of each

other i.e.

aij

= (1/aji)

Continuing to

it,

S1/S2

= 1 / (S2/S1)

S1/S3

= 1 / (S3/S1)

S2/S3

= 1 / (S3/S2)

D. Problem

Solving:

There are two ways in

which these problems can be solved to prioritise the decisions:

a. Normalization

Method

b. Eigen

Vector Method

a. Normalization

method:

In this method the

above square matrix is first normalized column wise using the following

formula:

After normalization of

the matrix, the sum of column elements will be unity.

Now if all the columns

are identical then the normalized matrix is consistent without any error and

the weights of each alternative are individual row value for that particular

alternative. For example, if normalized matrix comes out to be:

Alternative 1

Alternative 2

Alternative 3

Weights

Alternative 1

0.5

0.5

0.5

0.5

Alternative 2

0.4

0.4

0.4

0.4

Alternative 3

0.1

0.1

0.1

0.1

However, if the columns

are not identical then average of each column element of a row is taken to get

the weight of that particular row or alternative. And consistency error has to

be calculated using the following formulas:

For ?max calculations

use the following formula:

Now if the consistency ratio Cr

< 0.1, then the weights are consistent else correction is required which we
will see in Eigen vector method.
b. Eigen
Vector Method:
In
this method, instead of normalising the square matrix, we create a unity matrix
and multiply it with some value ? (keep it blank initially, we will solve and
try to find its value) to create a matrix "A-?I" and we have to make its
determinant zero to solve for the value of ? using the following formula:
To calculate the determinant of the
above equation, use "Mdeterm" in excel, it will give some value of determinant
of above equation.
Now for solving the above equation,
·
First set ? to any
higher value say 20, and then run the If analysis with goal seek.
·
To solve, set the goal
of making determinant to value zero by changing the values of ?.
·
Then run it.
We
will be able to get the value of ? which will be "?max" for
calculation of consistency ratio. Now, to solve for the weights of the
alternatives make n equations using following formulas:
And
make objective function as:
Using solver for above n equations and an objective function find individual
weights of all alternatives.
E. Error Calculation and rectification:
If
the matrix and weights are consistent and there is no error then the following
equation will hold true:
If not, then following equation will hold
true:
Where,
?ij
is the error associated with it.
If we solve the above equation we get:
So we have to take transpose of square matrix and multiply it with / matrix. In the result following analysis to be
done:
·
In this matrix if check
for the maximum value and if it lies above the unity diagonal then find out
corresponding wi / wj value
and substitute it the original square matrix in the same/corresponding cell and
again run the process for calculation of consistency ratio till it becomes less
than 0.1
·
If the maximum value
lies below the unity diagonal, then take corresponding mirror value (across
diagonal) which will be the minimum value of matrix and find out corresponding
wi / wj value and
substitute it the original square matrix in the same/corresponding cell and
again run the process for calculation of consistency ratio till it becomes less
than 0.1
This methodology gives you the
weights of all the corresponding alternatives. And to select the global
priority, multiply each corresponding weight of criterion and sub criterion and
sum for different products. This is explained further in detail with example.
Problem
Ram has to buy
a new house for himself and his dependent family. He came across three
different choices of houses which he thinks are good enough to be bought. And while
short listing these three houses he chose following criteria:
a.
Location
b.
Amenities
c.
Cost/resale
d.
Safety/Security
He asked to
his friend for advice who was studying analytical hierarchy process in his
college. He agreed to help him and gave him following questionnaire:
Rate circle, the following with a
score from 1-5 with 5 as most favourable and 1 as least favourable:
Q1: Location Amenities
Q2: Location Cost
Q3: Location Security
Q4: Amenities Cost
Q5: Amenities Security
Q6: Cost Security
Q7: Location A Location
B
Q8: Location A Location
C
Q9: Location B Location
C
Q10: Amenities
A Amenities B
Q11: Amenities
A Amenities C
Q12: Amenities
B Amenities C
Q13: Cost A Cost
B
Q14: Cost A Cost
C
Q15: Cost B Cost
C
Q17: Security
A Security
B
Q18: Security
A Security
C
Q19: Security
B Security
C
The response
to all of the questions were collected and used for analysis purpose using
Analytical Hierarchy Process.
SOLUTION
The following
matrixes were obtained from above questionnaire:
X =
Location
Amenities
Cost
Security
Location
1
4
2
1
Amenities
0.25
1
0.25
1
Cost
0.5
4
1
0.5
Security
1
1
2
1
XL =
A
B
C
A
1
0.5
0.25
B
2
1
0.5
C
4
2
1
XA =
A
B
C
A
1
2
4
B
0.5
1
2
C
0.25
0.5
1
XC =
A
B
C
A
1
0.5
2
B
2
1
4
C
0.5
0.25
1
XS =
A
B
C
A
1
4
2
B
0.25
1
0.5
C
0.5
2
1
Now using
Normalisation method 1st we get following matrixes as:
X =
Location
Amenities
Cost
Security
Weights
Location
0.36
0.40
0.38
0.29
0.36
Amenities
0.09
0.10
0.05
0.29
0.13
Cost
0.18
0.40
0.19
0.14
0.23
Security
0.36
0.10
0.38
0.29
0.28
Here, the
columns are not identical and there is some inconsistency. The consistency
ratio comes out to be:
n=
4.00
CI=
0.16
?MAX =
4.48
RI=
0.99
CR=
0.16
Which is
greater than 0.1. Hence Eigen Matrix will be used for error rectification. We
will do it in later part of the solution. Now other matrixes:
XL =
A
B
C
Weights
A
0.14
0.14
0.14
0.14
B
0.29
0.29
0.29
0.29
C
0.57
0.57
0.57
0.57
Location
Preferences of A, B & C and it is consistent so no need of error
calculation.
XA =
A
B
C
Weights
A
0.57
0.57
0.57
0.57
B
0.29
0.29
0.29
0.29
C
0.14
0.14
0.14
0.14
Amenities
Preferences of A, B & C and it is consistent so no need of error
calculation.
XC =
A
B
C
Weights
A
0.29
0.29
0.29
0.29
B
0.57
0.57
0.57
0.57
C
0.14
0.14
0.14
0.14
Cost
Preferences of A, B & C and it is consistent so no need of error
calculation.
XS =
A
B
C
Weights
A
0.57
0.57
0.57
0.57
B
0.14
0.14
0.14
0.14
C
0.29
0.29
0.29
0.29
Security
Preferences of A, B & C and it is consistent so no need of error
calculation.
Hence weights
of A,B & C with respect to Location, Amenities, Cost and Security is
calculated using normalization method and it is consistent since all the
columns are identical in each matrix. Therefore in these weights there is no
need to calculate error as it is nil.
However, while
calculating weights of Location, Amenities, Cost and Security, the columns are
not identical and there is consistence error of 0.16 which is greater than 0.1.
Hence, error rectification is required in it for further process.
To rectify
error, we need to first calculate weights using Eigen vector method:
X =
Location
Amenities
Cost
Security
Location
1.00
4.00
2.00
1.00
Amenities
0.25
1.00
0.25
1.00
Cost
0.50
4.00
1.00
0.50
Security
1.00
1.00
2.00
1.00
Form a unit
matrix for further calculation,
I =
1.00
0.00
0.00
0.00
0.00
1.00
0.00
0.00
0.00
0.00
1.00
0.00
0.00
0.00
0.00
1.00
Now using
blank values of
?=
Determinant =
Make following
matrix:
X-?I
Location
Amenities
Cost
Security
Location
1.00
4.00
2.00
1.00
Amenities
0.25
1.00
0.25
1.00
Cost
0.50
4.00
1.00
0.50
Security
1.00
1.00
2.00
1.00
Substitute a value of ?= 20 let's say and run
what if analysis with goal seek of determinant
setting to zero by changing ?. We get,
?=
4.45
Determinant =
0.00
And,
X-?I
Location
Amenities
Cost
Security
Location
-3.45
4.00
2.00
1.00
Amenities
0.25
-3.45
0.25
1.00
Cost
0.50
4.00
-3.45
0.50
Security
1.00
1.00
2.00
-3.45
For weights
calculations, first make empty weights cells like:
Location
Amenities
Cost
Security
And make 4
equations/constraints using formula:
And set the
objective function as per the following formula:
Now, run the
solver with these constraints and objective with linear function and non
negative variables. We will get following results:
Location
Amenities
Cost
Security
Objective
0.36
0.12
0.24
0.28
1.00
Now calculate
the consistency ratio using following formulas:
We get the
following results:
n=
4.00
CI=
0.15
?MAX =
4.45
RI=
0.99
CR=
0.15
The consistency
ratio is 0.15 which is greater than 0.1. Hence error rectification is required.
For error
rectification we first take transpose of original matrix:
XTRANSPOSE
1.00
0.25
0.50
1.00
4.00
1.00
4.00
1.00
2.00
0.25
1.00
2.00
1.00
1.00
0.50
1.00
Now using the
following formula we calculate the error matrix:
It comes out
to be:
0.36
0.12
0.24
0.28
0.36
1.00
0.73
0.76
1.30
0.12
1.37
1.00
2.10
0.45
0.24
1.31
0.48
1.00
1.70
0.28
0.77
2.24
0.59
1.00
Here, the value
of 2.24 is highest. But it is below the diagonal. Hence the corresponding mirror value across diagonal
is taken which is 0.45
Now the Wi/Wj
value of that cell is 0.45 which is needed to be replaced in original matrix
which comes out to be:
X =
Location
Amenities
Cost
Security
Location
1.00
4.00
2.00
1.00
Amenities
0.25
1.00
0.25
0.45
Cost
0.50
4.00
1.00
0.50
Security
1.00
1.00
2.00
1.00
If we again
solve it with Goal Seek and then solver and calculate the final consistency
ratio it comes out to be following values:
?=
4.07
Determinant =
0.00
n=
4.00
CI=
0.02
?MAX =
4.07
RI=
0.99
CR=
0.02
Location
Amenities
Cost
Security
Objective
0.37
0.09
0.23
0.30
1.00
Now the final
consistency ratio is 0.02. Hence it is consistent with above weights.
For the final
preference we have to prepare the chart once again:
0.37 0.09 0.23 0.3
0.29 0.57 0.29 0.57 0.14 0.57 0.14 0.29
0.14 0.57
0.29 0.14
Now the global
weights of each of the house A, B and C is calculated by multiplying
alternative weight of category and category weight and summing all these
products to get the total weight-age of that particular alternative.
WA
= (0.14)*(0.37) + (0.57)*(0.09) + (0.29)*(0.23) + (0.57)*(0.3)
= 0.341
Similarly,
WB
= (0.29)*(0.37) + (0.29)*(0.09) + (0.57)*(0.23) + (0.14)*(0.3)
= 0.307
And,
WC
= (0.57)*(0.37) + (0.14)*(0.09) + (0.14)*(0.23) + (0.29)*(0.3)
= 0.343
Hence weight-age of house C is
highest. So the friend recommend ram to go for house C as it is the best suited
according to his preferences with an error of 2% approximately.