MPV_APPE 2016
Dominated and non-dominated variants (dominance analysis)
(model example)
Local police department wants to purchase new cars. Following table contains data about several
considered models. Decide which models are dominated, determine basal and ideal variants and full
solution set (the models that should be considered for further choice).
Model Acceleration Top speed Fuel consum. Trunk size Price
Forman 17 141 8.1 450 220
Felicia combi 16 148 7.6 450 250
Lada 1500 15 153 7.6 480 210
Trabant 30 110 8.1 380 180
With increasing amount of available options and evaluated criteria in multi criteria evaluation it can
get very difficult to orientate. One way to simplify it is to reduce the available set of options, let’s call
it a full solution, in which we do not consider further those options that are not relevant (dominated
ones) – the options that practically cannot be chosen over some other available option.
Irrelevant variant is in this case the one to which there exists at least one other variant that is not
worse in any of the considered criteria while being better in at least one criterion. Such variant is
then considered as a dominated variant, and is being dominated by all other variants that fulfill the
condition of being better in at least one criterion while not being worse in any other.
Available variants in most cases do not dominate each other, meaning that one is better in some
criteria, while worse in other. Then they are considered to be non-dominated.
Sometimes it might help to determine theoretically worst and best variant. The worst variant is the
one with the worst available values from the set, it is called basal variant (B), and contains basal
values. On contrary, with the best available values from the set we get the ideal variant (I) with the
ideal values of criteria.
Solution: Basal variant has acceleration of 30, top speed of 110, fuel consumption 8.1, trunk size 380
and costs 250. Ideal variant has acceleration of 15, top speed of 153, fuel consumption 7.6, trunk size
480 and costs 180. Forman and Felicia are dominated (by Lada), Lada and Trabant are not
dominated. For further evaluation we would consider Lada and Trabant (full solution).
*In case of more complex problem you can use available tools, like SANNA from PSE.
MPV_APPE 2016
Transformation of minimizing criteria to maximizing
(model example)
Police department wants to purchase new cars. Following table contains data about considered
models. Some of the criteria are minimizing. Transform all to maximizing criteria.
Model Acceleration Top speed Fuel consum. Trunk size Price
Forman 17 141 8.1 450 220
Felicia combi 16 148 7.6 450 250
Lada 1500 15 153 7.6 480 210
Trabant 30 110 8.1 380 180
In practice of multicriteria evaluation we often encounter situation where some criteria are desired
to be maximized (like output level), while other minimized (like price). Transformation to the one
type can reduce possibility of making a mistake due to such difference and can be also useful later.
We can transform the values of minimizing criteria to maximizing using the following transformation:
( ) ( ) ( )
Where y(max) means transformed value from min criterion to max criterion, B(min) means basal
value of given min criterion (in such case the highest value of such criterion), and y(min) means
original value of min criterion.
*in case we have a fixed available interval of values, like using grades 1-5, we use 5 as basal value
independently from the fact that none of the evaluated variants actually got grade 5 in the criterion.
Solution:
Model T-Acceleration Top speed T-Fuel consum. Trunk size T-Price
Forman 13 141 0 450 30
Felicia combi 14 148 0,5 450 0
Lada 1500 15 153 0,5 480 40
Trabant 0 110 0 380 70
MPV_APPE 2016
WSA – weight sum approach
(model example)
Police department wants to purchase new cars. Following table contains data about considered
models. Use WSA to select the best variant (weight of criteria are 30%, 10%, 30%, 30%).
Model Acceleration Top speed Fuel consum. Price
Octavia 9 200 6.8 410
Rapid 10 190 6.5 360
Fabia 11 180 6.3 330
WSA means that individual evaluated criteria are assigned with certain weights, that represent their
level of importance in the final evaluation. Significantly worse parameters in one less important
criterion therefore do not mean that the variant will automatically not be selected, if it has better
parameters in more important criteria. For correct use we need to transform original values to the
appropriate form. We transform values to the same type and then normalize them, so we would
have comparable values. Final score for each variant is then scalar product of normalized values of
criteria and their weights.
Solution steps: Transformation formulae for normalizing maximizing criteria:
( )
Resp. transformation formulae for normalizing minimizing criteria:
( )
Using these transformation we get normalized matrix of values between 0 and 1 and then we
multiply the values with the weights. Normalized matrix looks like this:
Model N-Acceleration N-Top speed N-Fuel consum. N-Price
Octavia 1 1 0 0
Rapid 0.5 0.5 0.6 0.625
Fabia 0 0 1 1
Solution: Octavia gets 40%, Rapid 56.75% and Fabia 60%. Best model is Fabia.
MPV_APPE 2016
Scales and ranges – assigning points within a scale
(model example)
Region is deciding between projects of several water plants on different rivers. Three projects were
submitted, with criteria of building costs, running costs, output and safety level (range 0-10). You as
an expert should evaluate criteria of projects on a scale of 0-100 and choose the best.
River Building costs Running costs Output Safety
Bobrava 170 73 67 9
Ponávka 132 38 45 7
Želetavka 99 41 33 5
Method of scales requires ability of quantitative evaluation of given parameters within evaluated
criteria, meaning that the evaluator assigns values based on his expert opinion. Unlike with strictly
mathematical methods this allows the consideration of other factors as well, like the experience of
the evaluator, preferences or other aspects. The better the value of a parameter, the more points are
assigned. Thanks to that we do not have any more issues with min/max criteria and their
transformations or normalization.
On the other hand, the disadvantage of this method is the dependency on the subjective evaluation
of parameters. For reducing the risk of making incorrect decision, multiple independent expert
evaluation are often used. Final score is then a result of sum of individual evaluations, or their
weighted sum, if opinions of different experts are weighted differently.
Analogically we can assign different weight to different criteria based on their importance. Assigned
points for each parameter are then weighted and summed together afterwards. Such method is then
called scoring method.
Example of a possible solution, a subjective assignation of points to the parameters (0-100):
River Building costs Running costs Output Safety Sum
Bobrava 40 50 100 90 280
Ponávka 61 100 77 70 308
Želetavka 80 95 50 50 275
MPV_APPE 2016
Lexicographic method
(model example)
Region wants to build a bridge over the river Jevišovka. Estimated parameters of variants are in the
following table. For decision use lexicographic method with criteria preferences C→B→D→A and
requirements for criteria being A ≥ 440, B ≥ 6, C ≥ 7 a D ≤ 50.
Bridge A) Capacity B) Looks C) Place D) Costs
of victory (1) 406 7 8 50
Red (2) 444 8 8 39
of friendship (3) 505 4 9 44
of labor (4) 568 5 10 32
of proletariat (5) 541 8 6 52
Lexicographic method evaluated available variants sequentially based on the importance of
individual criteria and limiting requirements. In the first step we take the most important criterion
and discard the variants that do not meet the requirements for this criterion. In the second step we
continue analogically with the reduced set of remaining variants. The evaluation process is finished
when only one variant remains, and this variant is chosen as the best. In case we are left with
multiple variants even after the last criterion, we need to use some additional method for choosing a
compromise variant (for instance selecting one with the best parameter in the last evaluated
criterion).
The disadvantage of this method is that we are practically taking into account only the last evaluated
criterion and do not consider previous as long as the minimal requirements were met. Moreover, the
result can be notably biased by the criteria preferences selection. In cases with no non-dominated
variants, with “appropriate” preference order and minimal requirements it is sometimes possible to
secure any of the variants as the winning one.
Solution steps: In the first step we evaluate according to the C criterion – we discard variant 5,
remaining set is {1, 2, 3, 4}. In the second step we evaluate according to the B criterion – we discard
variants 3 and 4, remaining set is {1, 2}. In the third step we evaluate according to the D criterion –
we do not discard any variants, the set remains {1, 2}. In the fourth step we evaluate according to the
A criterion – we discard variant 1, remaining set is {2}.
Solution: We choose the Red bridge (2).