16: BOTH COMPARISON IN THE EXPERT METHOD

Lecture questions:

1. Pairwise object matching

2. Double pairwise matching

of objects

3. Ranking in the method

pairwise matching

1. Pairwise object matching

Expert assessment in pairwise comparison of

viewed objects carried out if the number of objects

even. The preference of the expert is expressed by

Preferred object numbers in the appropriate column

mapping tables as shown for example for six

objects in the table. ten.

Table 10

Results of pairwise comparison of objects by an expert

Object number ->

amount

expertise

one

2

3

four

five

6

preferences

i-th object, Ni

one

X

one

one

one

five

one

four

2

X

2

2

five

2

3

3

X

3

five

3

2

four

X

five

four

one

five

X

five

five

6

X

0

The maximum possible number of preferences for any of

considered objects, obtained from one of the experts,

equals N

 m 1, where t is the number of evaluated objects.

max

Frequency of these preferences f i is as private from de

N

N

i

i

N i on N max, i.e.

F





i

N

max

m

one

Using the data table. 10, we get N max = 6 - 1 = 5, and

Tots of preferences given by an expert are equal to:

four

3

2

F

one





0.8

;

F

2





0.6;

F

3





0.4;

five

five

five

one

five

0

F





0.2;

F



1;

F





0

four

five

6

five

five

five

The total number of judgments of one expert, associated

with the number of objects of expertise t, are found from the ratio

(

1)

C

 mm

2

6

(6 1)

With six objects of expertise

C



15

2

Indicator of object i determined by one expert

or weight compared with other objects are calculated

according to the formula:

n



Q

i,

j

i1

g



,

i

n

m



Q

i,

j

i

1, j

1

where n is the number of experts;

t - the number of indicators evaluated;

Qi, j is the weight coefficient of the jth index in ranks (points), which

I gave the i- th expert.

Transformed to view:

, n

F

i

Qi



m

,

i

1, j

1

C

where n is the number of experts in the group;

i

F

- frequency of preference of objects;

C - the number of possible judgments of one expert

Let the number of experts in a group be five and their estimates

about f i . are summarized in table. eleven.

Table 11

Frequency of object preferences given by experts

Object Preference Frequencies

Expert numbers

F1

F2

F3

F4

F5

F6

one

0.8

0.6

0.4

0.2

1.0

0

2

0.7

0.7

0.4

0.3

0.9

0.1

3

0.8

0.5

0.5

0.3

1.0

0.1

four

0.9

0.5

0.6

0.2

0.8

0

five

0.8

0.5

0.5

0.2

0.9

0

Total  F ij

4.0

2.8

2.4

1.2

4.5

0.2

In this case, the results of the examination by definition

performance indicators are as follows:

four

2.8

2.4

Q





0.27

;

Q





0.18;

Q



0.16;

one

2

3

15

15

15

one,

2

4.5

0.2

Q





0.08;

Q





0.3;

Q





0.01

four

five

6

15

15

15

Find the sum of the weight values:

m

Q

0.270.180,160,080.30,011.0



i

i

1

This result indicates that the indicators are estimated

experts are fairly accurate. Therefore, it is obvious that the

bevy ranked number of objects considered by their

teli has the form:

№ 6 <№ 4 <№ 3 <№ 2 <№ 1 <№ 5.

2. Double pairwise mapping of objects

If the sum of the weights is significantly different

from 1, in order to increase the reliability of the assessment,

second object mapping using free

part of the pairwise mapping table. In this case, the repeated

delivery is done in a chaotic manner. In this case

each pair of objects is mapped twice. Such a complete or

double mapping of objects significantly reduces the chance

Significant errors of expert estimates. Consequently, the double

The treatment is more reliable than once.

Let, after double comparison and establishment of

respects obtained the results of evaluations of one expert

put in table. 12.

Table 12

Results of double pairwise comparison of objects by an expert

Object number ->

amount

expertise

one

2

3

four

five

6

preferences

i-th object, Ni

one

X

one

one

one

five

one

7

2

one

X

2

2

five

2

6

3

3

2

X

3

five

3

3

four

one

2

four

X

five

four

3.5

five

five

five

five

four

X

five

eight

6

one

2

3

0

five

X

0.5

Note. If the objects being mapped are the same, the inter-

then it is denoted by the number 0, but both objects are given by

0.5 preference.

Possible greatest number of preferences one

object equals

N



2 ( m 1),

and the frequency of preference

max

N

N

i

i

F





where N i is the number of preferences of the i th object

i

N

2 ( m



one)

max

that, N max - the greatest number of preferences.

According to table 12 we find that when N max = 10

7

6

3

F





0.7;

F





0.6;

F



0.3;

one

2

3

ten

ten

ten

3.5

eight

0.5

F





0.35

;

F





0.8

;

F





0.05

four

five

6

ten

ten

ten

Indicators of the estimated objects are found by the formula:

, n

F

i

Qi



m

where n is the number of experts in the group

Provided that in the case of double pairing

The number of possible judgments of one expert is equal to

C = t (t - 1). In our example, C = 6 (6 - 1) = thirty.

Therefore, the "average" indicators of the evaluated objects

these are:

The results obtained are given

estimates of actual, real pairwise comparisons

considered objects.

The sum of all indicators is:

m



Q

i

23, 23

, 2

,

, 12

, 27



0,002



0.922

i

1

Ranked number of objects compiled by estimates

first expert, such:

Q6

If, for example, the remaining four experts gave ratings

the same as listed in table. 11, then in the table. 13 will be changed

compared with the table. 11, only the first line.

Table 13

Object Preference Frequency

Object Preference Frequencies

Expert numbers

F1

F2

F3

F4

F5

F6

one

0.7

0.6

0.3

0.35

0.8

0.05

2

0.7

0.7

0.4

0.3

0.9

0.1

3

0.8

0.5

0.5

0.3

1.0

0.1

four

0.9

0.5

0.6

0.2

0.8

0

five

0.8

0.5

0.5

0.2

0.9

0

Total  F ij

3.9

2.8

2.3

1.35

4.4

0.25

The final result of the examination of all experts, calculated

formula based:

, n

F

i



m

,

Qi

i

1, j

1

C

where n is the number of experts in the group

In this example it will be like this:

3.9

2.8

2.3

Q





0.26

;

Q





0.19;

Q





0.15;

one

2

3

15

15

15

1.35

4.4

0.25

Q





0.09

;

Q





0.29

;

Q





0.02

four

five

6

15

15

15

The sum of all indicators of weight or significance (quality

equal to:

m



Q

i

0.260,190,150,090,290,021

i

1

Consequently, the ranked number according to the examination

has the form:

Q6

Thus, they receive the results of the examination in case of double

Mr. pairwise comparison of evaluated objects.

3. Ranking in the pair-wise comparison method

As can be seen from the above example, the method of

method is constantly implemented in the process of applying the method

pairwise matching