In this case,

? max =
(2.0154+0.2648+0.7306) = 3.0108

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Now consistency index
can be found out by above mentioned formula as

CI
= (? max-n)/ (n-1) = (3.0108-3)/(3-1) = 0.0054

RI = 1.98*(N-2)/N =
1.98*1/3 = 0.66

CR = CI/RI = 0.0082

As
we have discussed we accept consistency ratio value
up to 0.1. Here, CR value is 0.0082, which is under acceptable range. Therefore,
we can assume that our judgments matrix is reasonably consistent so we may
continue the process of decision-making using AHP.

Step-7 Local Priorities for Alternatives

We
have to find out local priorities of alternatives following the same procedure
as we did for local priorities of criteria’s.

We
have three criteria’s and two alternatives; hence, three matrixes will be
formed for both alternative to find out weightage of alternative A and B for
coal, water and market.

For
this purpose, we do a pair wise comparison, of all the alternatives, with
respect to each criterion.

In
our example,

1.      Compare
Project A and B with respect to availability of coal criterion.

2.      Compare
Project A and B with respect to availability of water criterion.

3.      Compare
Project A and B with respect to availability of market criterion.

We
use the same Saaty’s pair wise comparison scale to give weightage to
alternatives.

Let’s
assume that we obtained following three responses-

Comparison
with respect to coal : X Coal

Alternatives

Project A

Project B

Project A

1

8

Project B

0.125

1

Comparison
with respect to water : X water

Alternatives

Project A

Project B

Project A

1

7

Project B

0.1429

1

Comparison
with respect to water : X Market

Alternatives

Project A

Project B

Project A

1

0.25

Project B

4.00

1

Step-8 Local weightage of Alternatives

To
find out the weightage of alternatives, same procedures can be followed and
repeated as done for finding out local weightage for criteria’s. Normalized
matrix can be written as –

X’
coal –

Alternatives

Project A

Project B

Project A

0.889

0.889

Project B

0.111

0.111

W Coal A – 0.889

W Coal B- 0.111

X’
Water-

Alternatives

Project A

Project B

Project A

0.875

0.875

Project B

0.125

0.125

W water A – 0.875

W water B- 0.125

X’
market-

Alternatives

Project A

Project B

Project A

0.200

0.200

Project B

0.800

0.800

W market A – 0.2

W market B- 0.8

As
we can see, the columns of all three consistency matrix is identical.
Therefore, we can say that all three matrix are consistent. There is no further
need of finding consistency ratios for validation.

Step-9 Global weights

In this step we find out the overall priorities of each
alternative.

0.3421

0.0882

0.6687

0.8

0.2

0.875

0.125

0.889

0.111

Global
weights can be found out my multiplying local weights for each alternatives.

Global
weights for project A = (0.6687*.889+ 0.0882*0.875+ 0.3421*0.2) = 0.72

Global weights for
project B = (0.6687*0.111+ 0.0882*0.125+ 0.3421*0.8) = 0.28

Step-9 Analysis of result and decision making

For
NTPC project A (0.72) is more feasible compared to project B (0.28) given the
preference of each criteria’s availability of coal, water and market.

EIGEN
VECTOR METHOD

There
is an alternate method to solve analytic hierarchy process problems called
Eigen vector method.

This
method is similar to the additive normalization method in the sense that all
initial steps are same in both the method. It differs only in the process of
calculation of eigen values or weightage of criteria’s and alternatives.

Followings
are the steps to solve analytic hierarchy process problem through eigen vector
method-

Step1:
Find out the comparison matrix

Step2:
Prepare N*N identity matrix, where N is the size of comparison matrix.

Step3:
Find out determinant of matrix (A-?I), where ? is eigen value.  Find out the maximum value of eigen value ?
for which determinant of matrix (A-?I) is zero, using goal seek (what –if
analysis) feature of Excel.

Step-4: Use
value of ? max to check the consistency of matrix as we did in additive
normalization method.

To
elaborate and understand concepts behind each step, we’ll again consider one
example. Since we have taken one example of individual decision making. Lets
take example of group decision.

Group Decision

Here
unlike last example, priorities of criteria’s and that of alternatives are
finding out by putting the questions and options in front of group of
respondents.

Lets
take one example where goal is to take the decision to buy a steam generator
(boiler) for 500 MW NTPC plant. We have again two alternatives supplier for it

1.      M/s
BHEL

2.      M/s
Doosan

To
choose one alternative NTPC is considering various criteria’s such as

a)      Safety
of boiler

b)      Cost
of boiler

c)      Efficiency
of boiler

To
take the decision of such type a better approach is to put the questionnaire in
front of subject experts and take the decisions.

Comparison
matrix can be prepared once we have answers of all questionnaires.

For
this project following questionnaires has been prepared-

Questionnaires-

A power company has
to purchase steam generator for its plant. Company is considering various
criterions to take decisions of choosing best supplier (Between M/s BHEL and
M/s Doosan). Comparison is being done for exactly same type and capacity of
boiler. Please fill the questionnaires as per your priorities of one over other
from scale 1 to 9.

1.
Preference of safety of boiler over cost
of boiler.

2.
Preference of safety of boiler over efficiency
of boiler.

3.
Preference of cost of boiler over
efficiency of boiler.

4.
If only efficiency is considered, How
much M/S BHEL is preferred over M/s Doosan ?

5.
If only safety is considered, How much
M/S BHEL is preferred over M/s Doosan ?

6.
If only cost is considered, How much M/S
BHEL is preferred over M/s Doosan ?

Question 1,2,3: What
is relative importance of one criterion over other for selection of steam
generator?

Please compare the LHS and RHS criteria and circle your

1-Equal
3- Moderate 5-Strong 7-Very Strong 9- Extreme

Safety of Boiler

9 8 7 6 5 4 3 2 1 2 3 4 5 6 7 8 9

Cost of Boiler

Safety of Boiler

9 8 7 6 5 4 3 2 1 2 3 4 5 6 7 8 9

Efficiency of Boiler

Cost of Boiler

9 8 7 6 5 4 3 2 1 2 3 4 5 6 7 8 9

Efficiency of Boiler

Question 4: If only
efficiency of boiler is considered, which supplier is preferable?

1-Equal
3- Moderate 5-Strong 7-Very Strong 9- Extreme

BHEL

9 8 7 6 5 4 3 2 1 2 3 4 5 6 7 8 9

DOOSAN

Question 5: If only
safety of boiler is considered, which supplier is preferable?

1-Equal
3- Moderate 5-Strong 7-Very Strong 9- Extreme

BHEL

9 8 7 6 5 4 3 2 1 2 3 4 5 6 7 8 9

DOOSAN

Question 6: If only
cost of boiler is considered, which supplier is preferable?

1-Equal
3- Moderate 5-Strong 7-Very Strong 9- Extreme

BHEL

9 8 7 6 5 4 3 2 1 2 3 4 5 6 7 8 9

DOOSAN

Explanation:

1-
If you
choose 1, when comparing BHEL with DOOSAN that means you give equal importance
to both.

2-
If you choose 9 towards right side (Towards
Doosan) that means you prefer Doosan extremely higher than BHEL on given
questions.

3-
If you choose 9 towards left side (Towards BHEL)
that means you prefer BHEL extremely higher than DOOSAN on given questions.

Following is the outcome of the survey received from 10
industry expert in matrix form.

Q1

Q2

Q3

Q4

Q5

Q6

RES1

9

7

0.5

1

1

5

RES2

8

8

0.2

0.33

3

4

RES3

9

5

0.5

0.25

2

2

RES4

4

6

1

0.13

0.25

6

RES5

8

5

2

2

0.2

3

RES6

9

7

0.5

8

0.5

0.25

RES7

6

8

0.25

4

0.33

0.5

RES8

8

7

3

3

0.5

1

RES9

7

9

4

2

4

1

RES10

9

9

1

0.11

8

0.5

To analyze further through Eigen value method, we’ll have to
take geometric means of responses.

Q1

Q2

Q3

Q4

Q5

Q6

7.502

6.960

0.827

0.922

0.978

1.463

Using these results now we can form Comparison matrix for 3
criteria’s as following-

Step1: Comparison Matrix:

Safety of Boiler

Cost of Boiler

Efficiency of Boiler

Safety of Boiler

1

7.5

6.96

Cost of Boiler

0.133

1

0.83

Efficiency of Boiler

0.144

1.205

1

Step2: Prepare Identity matrix

Identity matrix I (3*3) is given as  –

1

0

0

0

1

0

0

0

1

Step3: (A-?I) Matrix

(A-?I) Matrix is
prepared for a variable unknown value of ?.
The value of ? is calculated
such that determinant of the matrix becomes zero. We use goal use tool in excel
to calculate value of ?.

(A-?I) Matrix

-2.00

7.50

6.96

? max

3.001385

0.13

-2.00

0.83

Deter

0.00

0.14

1.20

-2.001385353

Step4: Consistency check

Similar to additive normalization method, we can check for
the consistency.

Here Consistency Index (CI)=
0.000693

Random Consistency Index (RI) = 0.66

Consistency ratio = CI/RI = 0.00105

Since consistency ratio is much lower than acceptable range
of 0.1, we can use it for further analysis.

Step4: Weights using linear programming

Safety of Boiler

Cost of Boiler

Efficiency of Boiler

W1

W2

W3

CONSTRAINTS

0.7827

0.1006

0.1167

LHS

SIGN

RHS

(A-? maxI) Matrix

-2.00

7.50

6.96

0.00

=

0

0.13

-2.00

0.83

0.00

=

0

0.14

1.20

-2.00

0.00

=

0

Objective Function

1

Weights are found out using linear programming.

Objective function is taken-
w1+w2+w3 =1

Constraints are – (A-? maxI)*W
=0

Weights obtained clearly show that the experts give
maximum priority to safety than efficiency of boiler than cost of boiler.

Note- Similar procedure can be done to find out
local weights of alternatives. Global weights can be easily found out by
multiplying local weights following the same procedure as already explained in

Evaluation
of  Error in AHP

If Wi, where i= 1,2,3,….n are actual
weights of priorities then element of comparison matrix (Aij) can be written as

Aij = (Wi / Wj )

If error ? is present, then (Wi / Wj ) = A ij* ?ij

?ij
= (Wi / Wj )* A ji

Steps to find out error in the AHP.

Step-1: Prepare
the comparison matrix from respondent’s priorities. For example- following table
is being prepared from the people response for the priorities of four
criteria’s for purchase of new car.

Brand

Cost

Efficiency

service

Brand

1

0.2

0.167

0.125

Cost

5

1

3

0.5

Efficiency

6

0.333

1

2

service

8

2

0.5

1

Step-2: Find out
transpose matrix Aji

Aji =

Brand

Cost

Efficiency

service

Brand

1

5

6

8

Cost

0.2

1

0.333

2

Efficiency

0.167

3

1

0.5

service

0.125

0.5

2

1

Weights can be found out by using eigen value method.

Calculated weights are as follows-

W1

W2

W3

W4

0.044

0.328

0.290

0.337

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