A Test Set of Semi-Infinite Programs∗
Alexander Mitsos†, revised by Hatim Djelassi
‡
Process Systems Engineering (AVT.SVT), RWTH Aachen University, Aachen, Germany
February 26, 2016 (first published August 26, 2009)
Keywords
SIP ; NLP ; Slater point ; nonconvex ; global optimization
1
Test Set
Test problems for semi-infinite programs (SIPs) without convexity assumptions are presented
1.1
Problems based on Watson [9]
For consistency purposes the problem labels of Watson [9] are used.
•
X = [−1000, 1000]2
Y = [0, 1]
f (x) = x21 /3 + x22 + x1 /2
g(x, y) = (1 − x21 y 2 )2 − x1 y 2 − x22 + x2 .
(2)
∗ Financial support from the Deutsche Forschungsgemeinschaft (German Research Association) through grant GSC 111 is
gratefully acknowledged.
† [email protected]
‡ [email protected].
•
X = [−1000, 1000]3
Y = [0, 1]
f (x) = exp(x1 ) + exp(x2 ) + exp(x3 )
g(x, y) = 1/(1 + y 2 ) − x1 − x2 y − x3 y 2 .
(5)
•
X = [−1000, 1000]2
Y = [0, 1]
f (x) = (x1 − 2 x2 + 5 x22 − x22 x2 − 13)2 + (x1 − 14x2 + x22 + x32 − 29)2
g(x, y) = x21 + 2 x2 y 2 + exp(x1 + x2 ) − exp(y).
(6)
Note that in [1] the exponent is missing in the first term of the objective function.
•
X = [−1000, 1000]3
Y = [0, 1]2
f (x) = x21 + x22 + x23
g(x, y) = x1 (y1 + y22 + 1) + x2 (y1 y2 − y22 ) + x3 (y1 y2 + y22 + y2 ) + 1.
(7)
•
X = [−1000, 1000]6
Y = [0, 1]2
f (x) = x1 + x2 /2 + x3 /2 + x4 /3 + x5 /4 + x6 /3.
g(x, y) = exp(y12 + y22 ) − x1 − x2 y1 − x3 y2 − x4 y12 − x5 y1 y2 − x6 y22 .
(8)
Note that presumably in Watson’s collection [9] the coefficient of x4 in the objective function is
mistyped. This is suggested by the optimal solution value reported in [9] and by the symmetry of
the problem with respect to the variables x4 and x6 .
•
X = [−1000, 1000]6
Y = [−1, 1]2
f (x) = −4 x1 − 2/3 (x4 + x6 )
g(x, y) = x1 + x2 y1 + x3 y2 + x4 y12 + x5 y1 y2 + x6 y22 − 3 − (y12 − y22 )2 .
(9)
Note that the last term was rewritten from −(y1 − y2 )2 (y1 + y2 )2 since bilinear products result in weak
relaxations, and higher CPU requirements for the solution of the NLPs.
•
X = [0, 1] × [−1000, 1000]
Y = [−1, 1]
f (x) = x2
g(x, y) = −(x1 − y)2 − x2 .
(H)
This problem was first proposed without bounds on x2 as Example 4.1 in [7]. The bounds were
subsequently added in [2].
•
X = [−1000, 1000]2
Y = [−1, 1]
f (x) = x2
g(x, y) = 2 x21 y 2 − y 4 + x21 − x2 .
1.2
New Continuous Problems
The following two problems are based on problem 4 by Watson [9]
(N)
1.
X = [−1000, 1000]3
Y = [0, 1]
f (x) = x1 + x2 /2. + x3 /3.
g(x, y) = exp(y − 1) − x1 − x2 y − x3 y 2 .
(4 3)
2.
X = [−1000, 1000]6
Y = [0, 1]
f (x) = x1 + x2 /2. + x3 /3 + x4 /4 + x5 /5 + x6 /6.
g(x, y) = exp(y − 1) − x1 − x2 y − x3 y 2 − x4 y 3 − x5 y 4 − x6 y 5 .
(4 6)
The following problem is based on the Shekel Function in [5]
1.
X = [−1000, 1000]2
Y = [−10, 10]
f (x) = −
1
1
1
−
−
(x1 − 4)2 + (x2 − 4)2 + 0.1 (x1 − 1)2 + (x2 − 1)2 + 0.2 (x1 − 8)2 + (x2 − 8)2 + 0.2
g(x, y) = 0.1 − (x1 − y)2 − (x2 − y)2 .
(S)
The following problem is aimed at provoking a dense discretization in methods similar to [3].
1.
X = [0, 6]
Y = [2, 6]
f (x) = 10 − x
g(x, y) =
1.3
y2
+ x − y − 2.
1 + exp(−40(x − y))
New Problems Involving Integer Variables
The following examples involve also discrete variables
(DP)
1.
Z = {0, 1}3
Y = [0, 1]3
f (x) = −z1 + 2z2 /2. − 3z3 /3
(IP1)
g(x, z, y) = z1 y12 − y22 + z2 (y2 y1 − y3 ) + z3 (exp(y3 ) − 3).
2.
X = [−1000, 1000]2
Z = {0, 1}3
Y = [0, 1]
f (x) = x21 + exp(x2 ) + z1 x1 + 2 z2 x1 x2 − 3 z3
(MINLP1)
g(x, z, y) = x1 exp(y1 ) + x2 (y1 − y22 ) + z1 (y12 − y22 ) + z2 (y2 y1 − y3 ) + z3 (exp(y3 ) − 3).
3.
X = [−10, 10]2
Z = {0, 1}3
Y = [0, 1]
f (x) = x21 + exp(x2 ) + z1 x1 + 2 z2 x1 x2 + 3 z3
(MINLP2.)
g(x, z, y) = x1 exp(y1 ) + x2 (y1 − y22 ) + z1 (y12 − y22 ) + z2 (y2 y1 − y3 ) + z3 (exp(y3 ) − 3).
1.4
Design Centering
The goal of design centering is to place a body B(x) into a container C. Typically, some metric of the body,
e.g., volume, or area, is maximized. The constraint B(x) ⊂ C, leads in general to a GSIP, but in some cases
an equivalent SIP can be formulated.
The following problems are based on the design centering problem by Floudas and Stein [6, Section 5.2]
with C ⊂ R2 after some modifications to avoid the trigonometric functions used therein, which are not
currently supported by BARON [8]. The body is a disk, parameterized by the midpoint (x1 , x2 ) and the
radius x3 , i.e., B(x) = {z ∈ R2 : (z1 − x1 )2 + (z2 − x2 )2 ≤ x23 }. The container is described by two inequality
constraints.
Since C is simply connected, Floudas and Stein [6] use ∂B(x) ⊂ C, or equivalently the constraints
ci (z) ≤ 0,
∀z ∈ ∂B(x),
i ∈ {1, 2}.
However, the parameterization of ∂B(x) leads to trigonometric functions. Here, the following formulation is
used instead:
ci (x1 , x2 ) ≤ 0,
i ∈ {1, 2}
(y − x1 )2 + (wi (y) − x2 )2 ≥ x23 ,
∀y ∈ Y,
i ∈ {1, 2}
where Y ⊂ R is a large enough set and the intermediate wi (y) is calculated by ci (y, wi ) = 0. Obviously,
this calculation is not possible for any function ci , but it is for the functions of interest herein. The first
constraint ensures that the center of the circle is within the container. The second constraint mandates that
the boundary of the container is sufficiently far away from the center of the disk, or equivalently the radius
is sufficiently small.
The container C used by Floudas and Stein is described by two functions, cl (y) = 0.3 sin(π y1 ) − y2
(for the lower side) and cu = y12 + 0.3 y22 − 1 (for the upper side). The former contains a trigonometric
function, and therefore is replaced by (the simpler) x2 ≥ x3 . For the latter setting c2 = 0 directly gives
p
y2 = (1 − y12 )/0.3.
min −x3
x
q
s.t. x2 − (1 − x21 )/0.3 ≤ 0
p
2
(1 − y 2 )/0.3 − x2 ≤ 0,
x23 − (y − x1 )2 −
∀y ∈ [−1, 1]
−x2 + x3 ≤ 0
x ∈ [−1, 1] × [0, 2] × [0, 1].
Note the presence of regular constraints. Note also that
√
(D1 0 1)
1 − t2 is not differentiable at t = ±1 (the boundaries
of X and Y ).
Arguably, Problem (D1 0 1) is significantly simpler than the original problem by Floudas and Stein.
Therefore, a second design centering problem is formulated with c2 (z) = |0.25 − (z1 − 0.5)2 | − z2 . The
resulting design problem is
min −x3
x
q
s.t. x2 − (1 − x21 )/0.3 ≤ 0
p
2
x23 − (y − x1 )2 −
(1 − y 2 )/0.3 − x2 ≤ 0,
∀y ∈ [−1, 1]
|0.25 − (x1 − 0.5)2 | − x2 ≤ 0
2
x23 − (y − x1 )2 − |0.25 − (y − 0.5)2 | − x2 ≤ 0,
∀y ∈ [−1, 1]
x ∈ [−1, 1] × [0, 2] × [0, 1].
(D2 0 1)
Note the presence of multiple semi-infinite constraints and regular constraints. Note also that |0.25−(t−0.5)2 |
is not differentiable at t = 0 (in the interior of X or Y ). To solve problems involving | · | in GAMS [4] the
keyword DNLP must be used instead of NLP.
Design problems with two discs are considered, resulting in twice as many variables and parameters.
Additionally it must be ensured that the discs don’t overlap which results into the constraint
(x4 − x1 )2 + (x5 − x4 )2 ≥ (x3 + x6 )2
An obvious strategy for upper bounding is to use the solution with one disc as an initial guess. However,
this is not exploited here.
min −x3 − x6
x
q
s.t. x2 − (1 − x21 )/0.3 ≤ 0
q
2
2
2
2
x3 − (y1 − x1 ) −
≤ 0,
(1 − y1 )/0.3 − x2
∀y1 ∈ [−1, 1]
x3 ≤ x2
q
(1 − x24 )/0.3 ≤ 0
2
q
≤ 0,
x26 − (y2 − x4 )2 −
(1 − y22 )/0.3 − x5
x5 −
∀y2 ∈ [−1, 1]
x6 ≤ x5
(x4 − x1 )2 + (x5 − x4 )2 ≥ (x3 + x6 )2
x3 ≥ x6
x3 + x6 ≤ 1.5
x ∈ [−1, 1] × [0, 2] × [0, 1] × [−1, 1] × [0, 2] × [0, 1].
(D1 0 2)
min −x3 − x6
x
q
s.t. x2 − (1 − x21 )/0.3 ≤ 0
q
2
2
2
2
x3 − (y1 − x1 ) −
≤ 0,
(1 − y1 )/0.3 − x2
∀y1 ∈ [−1, 1]
|0.25 − (x1 − 0.5)2 | − x2 ≤ 0
2
x23 − (y1 − x1 )2 − |0.25 − (y1 − 0.5)2 | − x2 ≤ 0,
q
x5 − (1 − x24 )/0.3 ≤ 0
q
2
x26 − (y2 − x4 )2 −
(1 − y22 )/0.3 − x5
≤ 0,
∀y2 ∈ [−1, 1]
|0.25 − (x4 − 0.5)2 | − x5 ≤ 0
2
x26 − (y2 − x4 )2 − |0.25 − (y2 − 0.5)2 | − x5 ≤ 0,
∀y2 ∈ [−1, 1]
∀y1 ∈ [−1, 1]
(x4 − x1 )2 + (x5 − x2 )2 ≥ (x3 + x6 )2
x3 ≥ x6
x3 + x6 ≤ 1
x ∈ [−1, 1] × [0, 2] × [0, 1] × [−1, 1] × [0, 2] × [0, 1].
(D2 0 2)
Finally, another variant of the design problems is considered where the sum of the areas of the discs is
maximized. For the case of one disc the problems are mathematically equivalent, but the new objective is
nonconvex and therefore potentially numerically more challenging. As expected the algorithm gives very
similar results for the two different objective functions. On the other hand, for the case of two discs the
objectives are different; the maximal area results in a slightly bigger difference between the radii of the two
discs.
min −x23
x
Constraints same as (D1 0 1)
(D1 1 1)
min −x23
x
Constraints same as (D2 0 1)
(D2 1 1)
min −x23 − x26
x
Constraints same as (D1 0 2)
(D1 1 2)
min −x23 − x26
x
Constraints same as (D2 0 2)
(D2 1 2)
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