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Chapter 3
The AHP and SMART
3.1
Introduction
In the previous chapter we discussed the parts from which methods are constructed in order to select
one particular item from among many. Two of these methods, with variations, are the subject of
this chapter.
The best-known method is probably Thomas L Saaty’s [7] analytical hierarchical process (AHP).
This method is based on Saaty’s eigenvector method for calculating weights from an
A
matrix,
but any other method for calculating weights may also be used. However, the AHP is susceptible
to several serious errors, inter alia two types of rank reversal.
(These problems are discussed in
Chapter 4.)
F A Lootsma [6] proposed an alternative, namely the geometric (or multiplicative)
AHP, which is increasingly accepted as an alternative to the AHP because it is not subject to rank
reversal.
The most attractive method for large practical problems is SMART, the abbreviation of a somewhat
forced name, the Simple Multi-Attribute Review Technique. SMART also uses a method to calculate
weights, and although the earliest versions proposed quite primitive methods, methods based on a
A
matrix have become the norm. SMART may also be seen as a methodology that is independent
of the method used to determine weights.
When one item has to be selected from many, we usually find one item that is the strongest in
a particular area whereas another item is better in another area.
This immediately tells us that
there is more than one criterion or objective. All these methods allocate weights to all the different
criteria, and values are assigned to the different items under each criterion. The criteria weights are
then utilised to weigh the values assigned to the items under the different criteria before they are
added together. Put differently: The criteria weights are used to calculate a weighted average of the
values allocated to an item under the different criteria.
In Example 2.2 of the previous chapter, values are allocated to six criteria. Assume that these values
are encountered in an actual problem and that the figures presented in the last four columns present
the performance of four items under the six criteria. Higher values are better. In this instance the
values may be seen as percentages (or scores out of one hundred).
The question is: Which item
should be selected?
51
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CHAPTER 3.
THE AHP AND SMART
Scores for
Scores for
Scores for
Scores for
Criterion
Item 1
Item 2
Item 3
Item 4
1
85
70
65
70
2
55
90
70
85
3
35
60
85
60
4
68
65
80
55
5
65
60
30
60
6
70
65
45
65
The only correct answer is probably: It all depends!
However, something that can be deduced from a direct comparison of the items is that Item 4 is
outperformed by Item 2 in every instance. The scores given to Item 2 for each criterion are either
the same or higher than those allocated to Item 4. Item 2
dominates
Item 4 and the latter can be
removed from the list of candidates because it offers nothing that Item 2 cannot deliver to the same
extent or better.
Definition 3.1.
Item A is dominated by item B if B’s performance equals or exceeds A’s in every area (or for every
criterion).
In the table above item performance was presented as columns (or vectors) and the decision maker
had no clear way of ranking the items. The bottom row of the table below presents a single value
that summarises the column above it and this facilitates item comparison. Inability to compare the
items in the above table has been eliminated by the weighted average.
Criterion
Weight
Scores for
Scores for
Scores for
i
w
i
Item 1
Item 2
Item 3
1
0,15
85
70
65
2
0,18
55
90
70
3
0,25
35
60
85
4
0,21
68
65
80
5
0,10
65
60
30
6
0,11
70
65
45
Weighted
average
59,88
68,50
68,35
The table can be produced by any one of the methods if the allocated scores were discounted. The
principle of using a weighted average on the basis of criteria weights remains the same.
The value functions of SMART and AHP are weighted arithmetic means of their scoring functions,
while the value function of the multiplicative AHP is a weighted geometric mean of its scoring
functions.
SMART is discussed first because it is the simplest method and because its value function is so
visible – which is not the case with AHP. It is therefore easier to understand AHP once you have
understood SMART.
52
CHAPTER 3.
THE AHP AND SMART
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3.2
SMART
SMART (Von Winterfeldt and Edwards [9]) makes use of a scoring function
f
i
to allocate values
between 0 and 100 to items on behalf of criterion
i
. The value function
z
is then
z
=
∑
w
i
f
i
, a
typical weighted average of the values provided by the scoring functions.
The task is therefore to prepare a scoring function for each criterion, and it should preferably convert
an objectively measurable quantity (e.g. distance in kilometres, or income in rands per year) into a
score out of one hundred. Not all criteria can be measured objectively (e.g. appearance or status)
and generally depend on human judgement. SMART requires scoring functions for these criteria as
well, converting the initial assessment to a score between 0 and 100. The example illustrates both
possibilities.
Example 3.1.
A man of thirty-four who has two children of seven and nine years old is looking for a house in a price
category that would reflect the fact that both he and his wife are rapidly advancing in their chosen
careers. They have already inspected a dozen houses and then decide to compile a SMART model.
The first task is to determine the criteria, and they decide to concentrate on five characteristics:
distance from the house to the local primary school and to the wife’s place of work, age, appearance,
and the area of the house.
They determine the weights of the criteria as
0
,
15
(school),
0
,
12
(work),
0
,
21
(age),
0
,
22
(appear-
ance), and
0
,
30
(area).
Scoring functions are prepared, and the two distances (to school and to the workplace) are represented
by the functions below where
x
is the distance in kilometers.
f
School
(
x
) =
f
Work
(
x
) =
⎧
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎩
80 + 20
x
with
0
≤
x
≤
1
100
with
1
< x
≤
3
115
−
5
x
with
3
< x
≤
23
0
with
x >
23
The distances are scored in the same way. However, this is not necessary and the graphs may differ.
The graph is also piece-wise linear, which makes expression in algebraic form much easier.
Once
again this is not necessary, and the values may just as well be read from the graph itself.
The score for the
age
of the house is presented in the table below.
Age
Age in years
Score
0
+
to
1
70
1
+
to
2
100
2
+
to
5
80
5
+
to
8
60
8
+
to
50
0
50
+
100
The score for the
appearance
of the house is presented in the following table:
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CHAPTER 3.
THE AHP AND SMART
Appearance
Description
Score
Will impress yuppies
100
Will impress boss
80
Will impress parents
40
An ordinary house
0
The scoring function for the
surface
area,
x
, is presented by the following:
f
Area
(
x
) =
⎧
⎪
⎪
⎨
⎪
⎪
⎩
0
with
0
≤
x
≤
200
0
,
2
x
with
200
< x
≤
500
100
with
x >
500
The five functions above are not all in an accustomed form and the one that has to provide a score
under
appearance
is a function of a subjective input.
If the couple in the example were to select
the description that is consistent with their value system, the table would nevertheless allocate the
desired score under this criterion.
Allow
f
School
,
f
Work
,
f
Age
,
f
Appearance
,
f
Area
to represent the five scoring functions.
The value function would then be:
z
= 0
,
15
f
School
+ 0
,
12
f
Work
+ 0
,
21
f
Age
+ 0
,
22
f
Appearance
+ 0
,
30
f
Area
.
The details of the next three houses they view are presented in the table, and their values have to be
calculated in accordance with the value function
z
.
Characteristic
House A
House B
House C
School (km)
4
1
12
Work (km)
9
10
1
Age (years)
0
6
69
Appearance (category)
yuppies
yuppies
boss
Area (m
2
)
230
420
390
The next table presents the values of the scoring functions in the columns below the houses, with the
weighted averages at the bottom – the values assigned to the houses by the value function above.
Characteristic
Weight
House A
House B
House C
School (score)
0
,
15
95
100
55
Work (score)
0
,
12
70
65
100
Age (score)
0
,
21
70
60
100
Appearance (score)
0
,
22
100
100
80
Area (score)
0
,
30
46
84
78
Value
73
,
15
82
,
60
82
,
25
54
CHAPTER 3.
THE AHP AND SMART
DSC3704
The ranking for SMART is
B > C > A
.
The value for B is just a little higher than the value for C. Before B is chosen, we have to carefully
consider the reasons for B’s high value.
An error in the scoring function or an overenthusiastic
weight can sometimes be the underlying reason. A sensitivity analysis of each case would be a good
idea.
A valid question in this case is what the smallest change,
θ
, would be in a weight to increase the
value of C above that of B.
C would benefit most from an increase in the weight for
age
because the score for C exceeds that of
B most in the case of this characteristic. Indeed, if this weight were to increase by
θ
, the value for C
would increase by
40
θ
more than the value for B would increase. The gap is therefore reduced by
40
θ
.
In the same manner a reduction in the score for
school
would decrease the value of B quite quickly
relative to C’s value, namely
45
θ
. The extent to which the gap would decrease when the weight for
age
were to increase by
θ
and that for
school
were to decrease by the same amount, is therefore
85
θ
and the gap to be closed is
0
,
35
.
The equation
85
θ
= 0
,
35
gives
θ
= 0
,
004118
, a very small figure. The choice between B and C is
therefore quite sensitive to changes in the weights. Moreover, the weights are obtained in a process
that is subject to quite extensive errors. For practical purposes, the two are equal.
Activity 3.1.
Briefly describe the scoring functions for:
1. Distance from the school.
2. Area of the house.
Activity 3.2.
Calculate the values of the houses described in the table
Characteristic
House X
House Y
House Z
School (km)
6
5
12
Work (km)
3
10
15
Age (years)
1
10
65
Appearance (category)
Ordinary
Yuppies
Parents
Area (m
2
)
530
520
300
under the value function
z
= 0
,
20
f
School
+ 0
,
17
f
Work
+ 0
,
16
f
Age
+ 0
,
22
f
Appearance
+ 0
,
25
f
Area
.
Activity 3.3.
The outcome of the previous activity places Y before X, which is in turn before Z. What would be
the smallest weights change to give Z and Y the same values?
Use the solution of Activity 3.2.
55
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CHAPTER 3.
THE AHP AND SMART
3.3
AHP
The AHP differs from SMART only in the scoring functions.
The scoring functions of AHP are
not prepared beforehand.
The scoring takes place under each criterion by comparing the various
competitors pair-wise and by using the methods discussed in Chapter 2 to solve the resultant A
matrix.
(The AHP, as it was originally described, uses the eigenvector method to determine the
preference vectors.) A preference vector is determined under each criterion, and when the value of
the first item is calculated, the first component of each preference vector is used as the score for that
criterion.
The AHP cannot evaluate a competitor on its own; all of them are compared, a preference vector is
derived from the comparison, and the components of the preference vector represent the scores of the
competitors. Because the components of a preference vector add up to one (normalisation), it means
that the more competitors there are, the lower would be the score of each.
The values allocated
to each competitor by the value function consequently also depend on the number of competitors.
In principle this is not a problem because the competitors are only compared with each other, but
problems can occur as can be seen from the example.
Example 3.2.
Look again at the above example where different houses have to be compared with each other. Assume
the same weights were determined for the criteria, then the value function would again be
z
= 0
,
15
f
School
+ 0
,
12
f
Work
+ 0
,
21
f
Age
+ 0
,
22
f
Appearance
+ 0
,
30
f
Area
.
Also assume that there is an external value process that assigns values to the three houses (as in the
table). This same table has been used in the example and is selected to compare the outcome of the
AHP with that of SMART.
Characteristic
House A
House B
House C
School (score)
95
100
55
Work (score)
70
65
100
Age (score)
70
60
100
Appearance (score)
100
100
80
Area (score)
46
84
78
The
A
matrices are compiled by using the values in the table in place of the
u
i
.
For the first characteristic,
A
is
A
School
=
⎡
⎣
1
,
00
95
100
95
55
1
,
00
100
55
1
,
00
⎤
⎦
and its preference vector is
w
School
=
⎡
⎣
0
,
3800
0
,
4000
0
,
2200
⎤
⎦
.
56
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THE AHP AND SMART
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The other
A
matrices are
A
Work
=
⎡
⎣
1
,
00
70
65
70
100
1
,
00
65
100
1
,
00
⎤
⎦
A
Age
=
⎡
⎣
1
,
00
70
60
70
100
1
,
00
60
100
1
,
00
⎤
⎦
A
Appearance
=
⎡
⎣
1
,
00
100
100
100
80
1
,
00
100
80
1
,
00
⎤
⎦
A
Area
=
⎡
⎣
1
,
00
46
84
46
78
1
,
00
84
78
1
,
00
⎤
⎦
and their preference vectors are
w
Work
=
⎡
⎣
0
,
2979
0
,
2766
0
,
4255
⎤
⎦
,
w
Age
=
⎡
⎣
0
,
3043
0
,
2609
0
,
4348
⎤
⎦
,
w
Appearance
=
⎡
⎣
0
,
3571
0
,
3571
0
,
2857
⎤
⎦
and
w
Area
=
⎡
⎣
0
,
2212
0
,
4038
0
,
3750
⎤
⎦
.
The
i
-th component of a preference vector is the score the scoring function assigns to the
i
-th item.
In this example it means that
f
School
provides the score 0,3800 to house A and that
f
Age
= 0
,
2609
for house B.
The value of house A is therefore
z
A
= 0
,
15
f
School
+ 0
,
12
f
Work
+ 0
,
21
f
Age
+ 0
,
22
f
Appearance
+ 0
,
30
f
Area
= 0
,
15(0
,
3800) + 0
,
12(0
,
2979) + 0
,
21(0
,
3043) + 0
,
22(0
,
3571) + 0
,
30(0
,
2212)
= 0
,
3016
.
The value of house B is
z
B
= 0
,
15(0
,
4000) + 0
,
12(0
,
2766) + 0
,
21(0
,
2609) + 0
,
22(0
,
3571) + 0
,
30(0
,
4038)
= 0
,
3477
and that of house C is
z
C
= 0
,
15(0
,
2200) + 0
,
12(0
,
4255) + 0
,
21(0
,
4348) + 0
,
22(0
,
2857) + 0
,
30(0
,
3750)
= 0
,
3507
.
The AHP ranking is
C > B > A
.
57
DSC3704
CHAPTER 3.
THE AHP AND SMART
Activity 3.4.
The values of competitors M
1
, M
2
, M
3
, M
4
have to be calculated with the AHP.
There are three criteria: A, B and C, and a
3
×
3
matrix of pair-wise ratios is obtained from a
group of decision-makers. After applying one of the solution methods from the previous chapter, the
preference vector is determined as
w
criteria
=
⎡
⎣
0
,
2080
0
,
4305
0
,
3615
⎤
⎦
.
Three
4
×
4
matrices are also prepared by the group and their preference vectors are found to be
w
A
=
⎡
⎢
⎢
⎣
0
,
1719
0
,
1990
0
,
3166
0
,
3125
⎤
⎥
⎥
⎦
,
w
B
=
⎡
⎢
⎢
⎣
0
,
3004
0
,
2909
0
,
2109
0
,
1978
⎤
⎥
⎥
⎦
,
and
w
C
=
⎡
⎢
⎢
⎣
0
,
2710
0
,
2571
0
,
2571
0
,
2148
⎤
⎥
⎥
⎦
.
Calculate the AHP score of each competitor.
3.4
Multiplicative AHP
Another way of approaching the value function is by asking: Why the weighted sum of the scores?
Why not the weighted product?
Why
z
from
z
=
∑
w
i
f
i
and not from
z
=
(
f
i
)
w
i
?
Lootsma
(1999) is strongly in favour of this (multiplicative or geometric) approach as an alternative to the
(normal additive) AHP because it naturally arises from the nature and structure of some problems.
The
w
i
are the criteria weights and the
f
i
values are calculated as for the normal AHP scores.
We again use the example of the person looking for a house.
Example 3.3.
You have already been introduced to the table below. It contains the weights of the five criteria in
the second column and the performance of the three houses in the columns below the houses.
Characteristic
Weight
House A
House B
House C
School (score)
0
,
15
95
100
55
Work (score)
0
,
12
70
65
100
Age (score)
0
,
21
70
60
100
Appearance (score)
0
,
22
100
100
80
Area (score)
0
,
30
46
84
78
In order to determine the value of A in terms of the multiplicative AHP, the AHP scores under each
of the criteria have to be calculated first.
Under
School
it gets
95
out of the total allocation of
95 + 100 + 55 = 250
, therefore
f
School
=
95
250
= 0
,
38
.
The scores under the other criteria are:
f
Work
=
70
235
= 0
,
2979
;
f
Age
=
70
230
= 0
,
3043
;
f
Appearance
=
100
280
= 0
,
3571
; and
f
Area
=
46
208
= 0
,
2212
.
58
CHAPTER 3.
THE AHP AND SMART
DSC3704
The value function of the multiplicative AHP allocates to A the value
z
A
with
z
A
= 0
,
3800
0
,
15
×
0
,
2979
0
,
12
×
0
,
3043
0
,
21
×
0
,
3571
0
,
22
×
0
,
2212
0
,
30
= 0
,
2954
.
Strictly speaking, this is not yet the final
z
A
because the AHP always demands normalisation. The
values
z
B
and
z
C
also have to be calculated and the three of them have to be normalised.
Activity 3.5.
Complete the process by calculating
z
B
and
z
C
and normalising the three weights.
Activity 3.6.
Which values does the value function of the multiplicative AHP assign to the four items of
Activity
3
.
4
?
3.5
Ranking issues
Consider again the rankings that were obtained for the three houses in Example 3.1 using SMART
and in Example 3.2 using AHP.
The ranking that was obtained for SMART is B (value 82
,
60), C (82
,
25), A (73
,
15). This is contrary
to the AHP ranking of C (value 0
,
3507), B (0
,
3477), A (0
,
3016) and an explanation has to be found.
The difference cannot be explained by the criteria weights because exactly the same value function
is used, and the only difference is in the scoring functions. The difference therefore has to lie in the
scoring functions.
The scoring functions of the AHP do not provide values out of one hundred as SMART’s scoring
functions do.
In itself this should not make any difference, because the relative quantities of the
values that give rise to their ranking are at issue here and not their absolute quantities. The scores
of the scoring functions of the AHP are weights – which means that they add up to one – and this
is where the problem lies. An example would clarify this point, but first note that it is sometimes
unnecessary to compile
A
matrices.
In this study guide and in the literature it is sometimes necessary to construct examples to explain
a concept. Values are then postulated for the competitors and from these the weights are directly
calculated by way of normalisation.
For example, the above example uses the scores from the
SMART exercise.
The
A
matrices were then compiled and solved to determine the preference
vectors. However, this is unnecessary. Consider for example the first characteristic (the distance to
school) of the three houses. The scores are 95, 100, 55 with a total of 250. Normalisation provides
95
250
;
100
250
;
55
250
and we again have
w
School
=
⎡
⎣
0
,
3800
0
,
4000
0
,
2200
⎤
⎦
.
Returning to the argument, assume there are five items and that their actual performance cannot
be improved. SMART recognises that they are all good by allocating 100 to all of them. The AHP
(which provides scores to all and where the scores add up to one), allocates 0
,
20 to each. Suppose
the same five items are all equally poor with respect to a subsequent criterion. The AHP will again
allocate 0
,
20 to all of them, merely because they recorded the same performance, whereas SMART
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CHAPTER 3.
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might allocate 35. This difference between SMART and the AHP can easily give rise to different
rankings.
Once again consider five items and two criteria.
Suppose they all perform quite well on the
first criterion, with an excellent score for the fifth item.
SMART’s scoring function for the first
criterion allocates values of say 76; 76; 76; 76; 96.
Normalisation provides the following weights:
0
,
190; 0
,
190; 0
,
190; 0
,
190; 0
,
240, and these are the AHP’s scores for the five items under the first
criterion. Further assume that SMART allocates values of 48; 38; 38; 38; 38 under the second crite-
rion. The weights to be used as scores for the AHP are then 0
,
240; 0
,
190; 0
,
190; 0
,
190; 0
,
190. Also
suppose that the two criteria carry the same weight. The value function of the AHP then allocates
the values 0
,
215; 0
,
190; 0
,
190; 0
,
190; 0
,
215 and the SMART values are 62; 57; 57; 57; 67.
The example could refer to five students who each writes two papers. The students who are com-
peting for the best scores are One and Five.
In the first paper, Five beats the other four by 20
points, but in the second paper One is the winner with 10 points more than the rest. The SMART
approach uses the average of the two papers and the 67 average of Five is the highest. This is not
unexpected, because he beat the other four by 20 points in the first paper whereas One earned 10
points more in the second paper. The AHP – where the points allocated by the scoring function
have to add up to one – shows 0
,
24 for both top students in the paper against 0
,
19 for the rest. The
fact that one student achieved 20 points more in one paper compared to the other’s 10 points more
in another paper is lost in the process.
The problem lies in normalisation. SMART provides 400 points (for the five competitors together)
for the first, and 200 for the second criterion.
The AHP always allocates a total of one.
In the
comprehensive criticism of AHP (which is discussed later in the guide), “normalisation at every
level” is repeatedly mentioned as the reason for and the cause of specific problems.
60
CHAPTER 3.
THE AHP AND SMART
DSC3704
3.6
Answers to activities
Activity 3.1.
1. The graph of the scoring function appears below.
D
i s t a n c e i n k m
S c h o o l ( a s w
e l l a s w
o r k )
S c o r e
2. The graph of the scoring function appears below.
A
r e a
A
r e a i n m
2
S c o r e
61
DSC3704
CHAPTER 3.
THE AHP AND SMART
Activity 3.2.
Characteristic
House X
House Y
House Z
School (km)
6
5
12
Work (km)
3
10
15
Age (years)
1
10
65
Appearance (category)
Ordinary
Yuppies
Parents
Area (m
2
)
530
520
300
Characteristic
Weight
House X
House Y
House Z
School (score)
0
,
20
85
90
55
Work (score)
0
,
17
100
65
40
Age (score)
0
,
16
70
0
100
Appearance (score)
0
,
22
0
100
40
Area (score)
0
,
25
100
100
60
Value
70
,
20
76
,
05
57
,
60
Activity 3.3.
The gap between the values for Y and Z is
18
,
45
and it can be closed by increasing the weight of a
characteristic so that Z exceeds Y and simultaneously decreasing to the same extent the weight of a
characteristic for which the opposite is true in order to keep the sum of the weights equal to one.
The two characteristics that show the highest score differences are selected to ensure the smallest
weight change.
The difference under
age
between Z and Y’s scores is the biggest, and for an increase of
θ
the gap
narrows to
100
θ
. Under
appearance
, the difference between the scores for Z and Y is the smallest
(most negative), and the gap narrows by a further
60
θ
. The overall narrowing of the gap is therefore
160
θ
and the
θ
that makes
160
θ
= 18
,
45
has to be found. The answer is
θ
= 0
,
1153
and this is the smallest change in weights (one weight up
and one down) that is necessary to equalise the values for Z and Y.
Activity 3.4.
The value function is
z
= 0
,
2080
f
A
+ 0
,
4305
f
B
+ 0
,
3615
f
C
.
The value of M
1
is therefore
z
1
= 0
,
2080
f
A
+ 0
,
4305
f
B
+ 0
,
3615
f
C
= 0
,
2080(0
,
1719) + 0
,
4305(0
,
3004) + 0
,
3615(0
,
2710)
= 0
,
2630
.
The calculations can be summarised in a table:
62
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CHAPTER 3.
THE AHP AND SMART
DSC3704
Criterion
Weight
M1
M2
M3
M4
A
0
,
2080
0
,
1719
0
,
1990
0
,
3166
0
,
3125
B
0
,
4305
0
,
3004
0
,
2909
0
,
2109
0
,
1978
C
0
,
3615
0
,
2710
0
,
2571
0
,
2571
0
,
2148
Value
0
,
2630
0
,
2596
0
,
2496
0
,
2278
Activity 3.5.
The table below has been given. The first task is to calculate the weights for house B and house C
and the last column has been added for this reason.
Characteristic
House A
House B
House C
Total
School (score)
95
100
55
250
Work (score)
70
65
100
235
Age (score)
70
60
100
230
Appearance (score)
100
100
80
280
Area (score)
46
84
78
208
Normalisation requires each of the values allocated under a particular criterion to be divided by the
total of those values (the value on the right). For the sake of completeness, the figures for house A
are also shown.
Characteristic
Criterion
House A
House B
House C
Weight
(Weight)
(Weight)
(Weight)
School
0
,
15
0
,
3800
0
,
4000
0
,
2200
Work
0
,
12
0
,
2979
0
,
2766
0
,
4255
Age
0
,
21
0
,
3043
0
,
2609
0
,
4348
Appearance
0
,
22
0
,
3571
0
,
3571
0
,
2857
Area
0
,
30
0
,
2212
0
,
4038
0
,
3750
The value function of the multiplicative AHP allocated the value
z
A
to A with
z
A
= 0
,
3800
0
,
15
×
0
,
2979
0
,
12
×
0
,
3043
0
,
21
×
0
,
3571
0
,
22
×
0
,
2212
0
,
30
= 0
,
2954
.
The value of B is
z
B
= 0
,
4000
0
,
15
×
0
,
2766
0
,
12
×
0
,
2609
0
,
21
×
0
,
3571
0
,
22
×
0
,
4038
0
,
30
= 0
,
3422
and that of C is
z
C
= 0
,
2200
0
,
15
×
0
,
4255
0
,
12
×
0
,
4348
0
,
21
×
0
,
2857
0
,
22
×
0
,
3750
0
,
30
= 0
,
3415
.
These three values add up to
0
,
9791
and they are normalised by dividing each by this figure. According
to the multiplicative AHP, the values of houses A, B and C are then
0
,
3017
;
0
,
3495
;
0
,
3488
.
63
DSC3704
CHAPTER 3.
THE AHP AND SMART
Activity 3.6.
The weights vector of the three criteria A, B and C
w
Criteria
=
⎡
⎣
0
,
2080
0
,
4305
0
,
3615
⎤
⎦
has already been calculated, and the weights (acting as scores) of the four competitors (M
1
, M
2
, M
3
and M
4
) under the various criteria are presented by the preference vectors
w
A
=
⎡
⎢
⎢
⎣
0
,
1719
0
,
1990
0
,
3166
0
,
3125
⎤
⎥
⎥
⎦
,
w
B
=
⎡
⎢
⎢
⎣
0
,
3004
0
,
2909
0
,
2109
0
,
1978
⎤
⎥
⎥
⎦
,
and
w
C
=
⎡
⎢
⎢
⎣
0
,
2710
0
,
2571
0
,
2571
0
,
2148
⎤
⎥
⎥
⎦
which have also already been determined. The value function of the multiplicative AHP allocates the
following values to M
1
, M
2
, M
3
and M
4
.
z
1
= 0
,
1719
0
,
2080
×
0
,
3004
0
,
4305
×
0
,
2710
0
,
3615
= 0
,
2577
,
z
2
= 0
,
1990
0
,
2080
×
0
,
2909
0
,
4305
×
0
,
2571
0
,
3615
= 0
,
2571
,
z
3
= 0
,
3166
0
,
2080
×
0
,
2109
0
,
4305
×
0
,
2571
0
,
3615
= 0
,
2465
,
z
4
= 0
,
3125
0
,
2080
×
0
,
1978
0
,
4305
×
0
,
2148
0
,
3615
= 0
,
2241
.
After normalisation, the values of M
1
, M
2
, M
3
and M
4
are
0
,
2615
;
0
,
2609
;
0
,
2502
and
0
,
2274
.
64