Progressively Interactive Evolutionary Multi-Objective Optimization ...

4.5.2 ProblemDS1
This problemhas 2K variableswith K real-valuedvariableseachfor lower and upper
levels. Sinceinthis studyweconsider bileveloptimizationproblemshaving m = M =
2,thevectors u, tand sallhaveasingleelement. Following threeparametricfunctions
areused forDS1:
Step1: Here, one of the upper level variables y1 is chosen as u. The mapping ΦU is
chosen as follows: v1 = (1 + r) − cos(πy1) and v2 = (1 + r) − sin(πy1) (r is a
user-supplied constant). Depending on the value of y1, the v1-v2 point lies on the
quarterofacircleofradiusoneandcenterat (1 + r, 1 + r)inthe F-space,asshown
inFigure 4.
Step2: For every (v1, v2) point on the F-space,
the following one-dimensional
(t) envelope
is chosen: v1(y1) = −r cos t and v2(y1) = −r sin t , where
γ π
2
y1
t ∈ [0, y1] and γ (=1) is a constant. The envelope is a quarter of a circle of radius
r and center located at (v1, v2) point. Although for each (v1, v2) point, the entire
envelope (quarter circle)is a non-dominated front, when all (v1, v2) points are
considered, only one point from each envelope qualifies to be on the upper level
Pareto-optimal front. Thus, the Pareto-optimal front for the upper level problem
lies on the quarter of a circle having radius (1 + r) and center at (1 + r, 1 + r) on
the F-space,as indicatedinthe figure.
Step3: Each U1-U2 pointisthenmappedtoa(f ∗ 1 , f ∗ 2 )pointbythefollowing mapping:
f ∗ 1 = s2 , f ∗ 2 = (s − y1) 2 ,where s = t is assumed.
Step4: We donot use any function lj here.
Step5: But, weuse Ej = (yj − j−1
2 )2 for j = 2, . . . , K inthe upperlevelproblem. This
willensurethat yj = (j − 1)/2(for j = 2, . . . , K)willcorrespondtotheupperlevel
Pareto-optimalfront.
Step6: Different lj and Lj termsareusedwith (K−1)remainingupperandlowerlevel
variablessuchthatalllowerlevelPareto-optimalsolutions mustsatisfy xi = yi for
i = 2, . . . , K.
The complete DS1problemis givenbelow:
Minimize F(y, x) =
⎛
(1 + r − cos(απy1)) +
⎜
⎝
K j−1
j=2 (yj −
+τ K i=2 (xi − yi)2
− r cos
(1 + r − sin(απy1)) + K j−1
j=2 (yj
−
+τ K
i=2 (xi − yi)2 − r sin
2 )2
γ π x1
2 y1
γ π
2
x1
y1
2 )2
−K ≤ xi ≤ K, for i = 1, . . . , K,
1 ≤ y1 ≤ 4, −K ≤ yj ≤ K, j = 2, . . . , K.
γ π
2
⎞
⎟
⎠ ,
subjectto (x) ∈ argmin f(x) =
⎧
(x) ⎛
⎪⎨
x
⎜
⎝
⎪⎩
2 1 + K i=2 (xi − yi)2
+ ⎞⎫
⎪⎬
K
π
i=2 10(1 − cos( (xi − yi))) ⎟
K ⎟
K
i=1 (xi − yi)2
⎠ ,
⎪⎭
y1
+ K π
i=2 10| sin( (xi − yi)|
K
(7)
For this test problem, we suggest K = 10 (overall 20 variables), r = 0.1, α = 1, γ = 1,
and τ = 1. Since τ = 1 is used, for every lower level Pareto-optimal point, xi = yi for
i = 2, . . . , K andboth lj and Lj terms arezero,thereby making anagreement between
this relationship betweenoptimal values of xi and yi variablesinboth levels.
Aninterestingscenariohappenswhen τ = −1isset. Ifany xisnotPareto-optimal
to a lower level problem (meaning xi = yi for i = 2, . . . , K), a positive quantity is
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