# ANALYTICAL be defined as is “Structured technique for

ANALYTICAL
HEIRARCHY PROCESS

Theory
with a Case

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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

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.