Optimization Letters (2020) 14:1855–1867
https://doi.org/10.1007/s11590-019-01484-z
ORIGINAL PAPER
Globally solving extended trust region subproblems
with two intersecting cuts
Zhibin Deng1,2
· Cheng Lu3 · Ye Tian4 · Jian Luo5
Received: 21 May 2018 / Accepted: 19 September 2019 / Published online: 24 September 2019
© Springer-Verlag GmbH Germany, part of Springer Nature 2019
Abstract
In this paper, we propose an algorithm to globally solve the extended trust-region
subproblem with two linear intersecting cuts. Based on the tightness of the linear cuts
at optimal solution, we can resolve the original problem by solving either a traditional
trust-region subproblem over a sphere or a semidefinite programming relaxation with
second-order cone constraints. Two examples from literature and numerical experiment on randomly generated instances are used to demonstrate how the proposed
algorithm works.
Keywords Extended trust region subproblem · Intersecting cuts ·
Semidefinite programming relaxation · Second-order cone constraints
1 Introduction
Consider the extended trust-region subproblem (ETRS) with linear cuts as follows:
min x T Qx + 2q T x
s.t. x ≤ 1,
B
(ETRSm )
Jian Luo
luojian546@hotmail.com
1
School of Economics and Management, University of Chinese Academy of Sciences,
Beijing 100190, China
2
Key Laboratory of Big Data Mining and Knowledge Management, Chinese Academy of
Sciences, Beijing 100190, China
3
School of Economics and Management, North China Electric Power University, Beijing 102206,
China
4
School of Business Administration, Southwestern University of Finance and Economics,
Chengdu 611130, China
5
School of Management Science and Engineering, Dongbei University of Finance and Economics,
Dalian 116025, China
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a1T x ≤ b1 ,
..
.
amT x ≤ bm ,
where x ∈ n is the decision variable, x is the Euclidean norm of x, Q is an
n × n symmetric real matrix, q, a1 , . . . , am ∈ n , and b1 , . . . , bm are constants. In
particular, we assume Q is not positive semidefinite. Otherwise, (ETRSm ) is a tractable
convex program, which can be solved in polynomial time to any given precision by
interior-point methods [1]. Problem (ETRSm ) arises from the trust-region method for
constrained optimization problems (ref. [18]).
When m = 1, i.e., there is only one linear cut in (ETRSm ), Sturm and Zhang [14]
showed that (ETRSm ) can be reformulated to an equivalent semi-definite programming (SDP) problem with second-order cone (SOC) constraints. Therefore, (ETRSm )
is polynomial-time solvable within any given precision for m = 1. Locatelli also
established the exactness conditions for an SDP relaxations of this problem in [8]. For
large-scale instances, Salahi and Taati proposed a fast eigenvalue approach in [13].
For m ≥ 3, Jeyakumar and Li [6] show that the classical SDP relaxation is tight
as long as a dimensional condition holds. Burer and Yang proved in [3] that the SDP
relaxation with additional SOC reformulation (SDPR-SOCR) is exact when any two
of the linear cuts are nonintersecting in the unit ball. For solving (ETRSm ) in a general
case, we refer the readers [4,9] and [10] for conic approximation methods. We point
out that the general (ETRSm ) is NP-hard since it includes the copositivity detection
problem as a special case [12].
For m = 2, Ye and Zhang [16] prove that SDPR-SOCR is tight if one of the two
linear cuts is tight at optimal solution and the SDPR-SOCR and its dual problem have
complementary optimal solutions. Burer and Anstreicher [2] then prove that SDPRSOCR is also tight when the two linear cuts are parallel. Jin et al. [7] further prove that
the conclusion still holds when the two linear cuts are not parallel but nonintersecting
in a second-order cone.
In this paper, we are interested in globally solving (ETRSm ) with two intersecting
linear cuts in the unit ball. A counterexample in [2] shows that a positive gap exists
between the original problem and its SDPR-SOCR in the intersecting case. Yuan et al.
[17] recently derived the sufficient and necessary condition for the tightness of SDPRSOCR. However, when the condition fails, no solution method is given for solving
(ETRSm ). This motives us to investigate the global resolution method for this problem
in a general case. In this paper, we solve problem (ETRSm ) for m = 2 by utilizing its
SDPR-SOCR. When the SDPR-SOCR is not tight, we only need to solve a traditional
trust region subproblem. When the SDPR-SOCR is tight, we solve the corresponding
semidefinite programming and obtain the global solution of original problem by the
rank-one decomposition.
n and S n denote the set of
The following notations are adopted in this paper. S n , S+
++
all n × n real symmetric matrices, the set of n × n real positive semidefinite matrices
and the set of n × n real positive definite matrices, respectively. For A, B ∈ S,
A • B = trace(A T B) is the matrix inner-product between matrices A and B. In
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Globally solving extended trust region subproblems with…
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and On denote the n-dimensional identity and
zero matrix, respectively. For an ndimensional vector x = (x1 , ..., xn ), x = x12 + · · · + xn2 denotes the Euclidean
norm of x. For an n × n matrix A, A is the 2-norm of A. S OC(n) denotes the
n-dimensional
standard second order cone, that is, x = (x1 , ..., xn ) ∈ S OC(n) if
x22 + · · · + xn2 ≤ x1 . Moreover, x ∈ ∂ S OC(n) if x is on the boundary
of S OC(n), i.e., x22 + · · · + xn2 = x1 . Finally, v(P) is the optimal value of an
optimization problem (P).
and only if
2 Literature review
In this paper, we consider the special case m = 2 of (ETRSm ):
min x T Qx + 2q T x
(ETRS2 )
s.t. x ≤ 1,
a1T x ≤ b1 ,
a2T x ≤ b2 .
It is easy to see that (ETRS2 ) is equivalent to the following optimization problem [2]:
min x T Qx + 2q T x
s.t. x ≤ 1,
(QP)
(a1T x − b1 )(a2T x − b2 ) ≥ 0,
|b1 − a1T x|x ≤ b1 − a1T x,
|b2 − a2T x|x ≤ b2 − a2T x.
Obviously, (QP) is a variant of (ETRS2 ) by forming two SOC constraints from the
two linear cuts. The SDP relaxation of (QP) is
min
s.t.
M0 • Y
M1 • Y ≤ 0,
(SP)
M2 • Y = d1T Y d2 ≥ 0,
Y d1 ∈ S OC(n + 1),
Y d2 ∈ S OC(n + 1),
E • Y = 1,
n+1
Y ∈ S+
,
where
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0 qT
−1 0T
, M2 =
, M1 =
M0 =
q Q
0 In
−b1
−b2
1
, d2 =
and E =
d1 =
a1
a2
0
1
(d1 d2T + d2 d1T ),
2
0T
.
On
The dual problem of (SP) is
max
s.t.
y0
1
1
n+1
Z = M0 − y0 E + y1 M1 − y2 M2 − (u 1 d1T + d1 u 1T ) − (u 2 d2T + d2 u 2T ) ∈ S+
,
2
2
(DSP)
y0 ∈ , y1 ≥ 0, y2 ≥ 0, u 1 ∈ S OC(n + 1), u 2 ∈ S OC(n + 1).
By the definition of M1 , we can choose y0 and y1 large enough such that (DSP) is
strictly feasible. That is the Slater condition always holds for (DSP). The following
definition, given in [17], is used to characterize the tightness of (SP).
Definition 1 Let Ŷ and ( ŷ0 , ŷ1 , ŷ2 , û 1 , û 2 ) be a pair of optimal solutions of (SP) and
(DSP). If the following conditions hold
– rank(Ŷ ) = 3, rank( Ẑ ) = n − 2;
– yˆ1 > 0 or M1 • Ŷ = 0;
– yˆ2 = 0 or M2 • Ŷ > 0;
– û 1 = 0, û 2 = 0 and Ŷ d1 ∦ Ŷ d2 ;
where Ẑ = M0 − ŷ0 E + ŷ1 M1 − ŷ2 M2 − 21 (û 1 d1T + d1 û 1T ) − 21 (û 2 d2T + d2 û 2T ), and
“∦” indicates that the vector Ŷ d1 is not parallel to vector Ŷ d2 , then we say this solution
pair has Property I. Furthermore, if (SP) and (DSP) have one pair of optimal solutions
satisfying Property I, then we say (SP) and (DSP) have Property I.
The following theorem in [17] states that Property I is the sufficient and necessary
condition for the tightness of (SP).
Theorem 1 (Theorem 2.7 in [17] ) Suppose that (ETRS2 ) satisfies the Slater condition.
Then v(S P) < v(Q P) = v(ETRS2 ) if and only if (SP) and (DSP) have Property I.
A direct corollary from this theorem is that when the two linear cuts are nonintersecting
in the unit ball, then Property I will never hold and there is no duality gap between (SP)
and (ETRS2 ). Hence, a positive gap can exist only when the two linear cuts intersecting
in the unit ball. That is, the two planes a1T x = b1 and a2T x = b2 intersect in the open
unit ball {x ∈ n |x < 1}. Yuan et al. did not provide a solution method for this case
in [17]. The main contribution of this paper is to solve this “hard” case. In the sequel,
we discuss the solution method depending on the number of active constraints at the
optimal solution.
3 Global resolution method of (ETRS2 )
Denote the feasible region of (ETRS2 ) as Ω = {x ∈ n |x ≤ 1, a1T x ≤ b1 , a2T x ≤
b2 }. Due to the nonconvexity of objective function, the optimal solution is achieved
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Globally solving extended trust region subproblems with…
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on the boundary of Ω. That is, none, one or two linear constraints are active at the
optimal solution. In this section, we only consider the case that there is a positive gap
between (ETRS2 ) and (SP). That is, (SP) and (DSP) have Property I. We discuss the
solution method according to the active constraints at optimal solution of (ETRS2 ).
3.1 No linear cut is active
When neither of the two linear cuts is active at the optimal solution of (ETRS2 ), it
follows that the optimal solution is actually a local minimizer of the following classical
trust region subproblem
min
x T Qx + 2q T x
(TRS)
s.t. x = 1.
Especially, when (SP) is not tight, this local minimizer is not a global solution of
(TRS). We state the result as the following theorem.
Theorem 2 Assume x̂ is an optimal solution of (ETRS2 ), and (SP) and (DSP) have
Property I. If (a1T x̂ − b1 )(a2T x̂ − b2 ) > 0, then x̂ is a local-nonglobal minimizer of
(TRS).
Proof Since neither of the two linear cuts is active at x̂, the unit ball constraint x ≤ 1
has to be active. This shows that x̂ is a local minimizer of (TRS). If x̂ is also a global
T
1 1
minimizer of (TRS), then Ŷ =
is an optimal solution of the following SDP
x̂ x̂
relaxation of (TRS) [14]
min
s.t.
M0 • Y
(STRS)
M1 • Y ≤ 0,
E • Y = 1,
n+1
Y ∈ S+
.
Note that (SP) is a tighter problem than (STRS), thus M0 • Ŷ ≤ v(SP). Also, Ŷ is a
feasible solution of (SP) due to the feasibility of x̂ to (ETRS2 ), thus M0 • Ŷ ≥ v(SP). In
all, we have Ŷ is an optimal solution of (SP) and v(SP) = v(ETRS2 ). This contradicts
with the condition that (SP) and (DSP) have Property I, under which there is a positive
gap between v(SP) and v(ETRS2 ) according to Theorem 1.
Theorem 2 shows that we need to find the local-nonglobal minimizers of (TRS)
when the optimal solution of (ETRS2 ) is not active at the linear cuts. Fortunately, there
is only one such local-nonglobal minimizer.
Lemma 1 (Theorem 3.1 in [11]) There is at most one local-nonglobal minimizer x̂ of
(TRS).
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Algorithm 1 Computing the local-nonglobal minimizer of (TRS)
Input: Data Q and q.
Output: A local-nonglobal minimizer of (TRS) if there is any.
1: Spectral decompose Q = PΛP T , where P = ( p1 , ..., pn ) is an orthogonal matrix consisting of
eigenvectors p1 , ..., pn of Q and Λ = Diag(λ1 , ..., λn ), λ1 ≤ λ2 ≤ · · · ≤ λn , is a diagonal matrix
L
consisting of the eigenvalues λ1 , ..., λn of Q. Set k = 0, μU
k = −λ1 , μk = −λ2 and μk = (−λ1 −
2
c
n
i
λ2 )/2. Let ci = piT q, i = 1, ..., n, and φ(μ) = i=1
2.
(λi +μ)
2: If φ(μk ) = 1 and φ (μk ) ≥ 0, stop.
U
U
L
3: If φ(μk ) ≤ 1 or φ (μk ) ≤ 0, set μk+1
= μk and μU
k+1 = μk . Otherwise, set μk+1 = μk and
L
= μkL .
μk+1
φ(μk )−1
L , μU ], declare that
4: If φ(μk ) > 1 and φ (μk ) = 0 or φ(μk ) > 1 and μkN = μk − φ (μ
∈
/ [μk+1
k+1
k)
the local-nonglobal minimizer does not exist and stop.
5: Compute the largest root R(μk ) of the quadratic equation φ(μk ) + φ (μk )(μ − μk ) + 21 φ (μk )(μ −
L +0.01(μU −μ L ), μ L +0.99(μU −μ L )], set μ
μk )2 = 1. If R(μk ) ∈ [μk+1
k+1 = R(μk ).
k+1
k+1
k+1
k+1
k+1
L
Otherwise, set μk+1 = (μk+1
+ μU
k+1 )/2.
6: Set k = k + 1. Go to Step 2.
Martínez [11] proposed a line-search algorithm to find the unique local-nonglobal
minimizer. For the completeness, we state the algorithm as Algorithm 1.
Algorithm 1 generates a sequence of μk ’s, which approximate the multiplier of the
local-nonglobal minimizer, together with a sequence of lower bounds μkL and μU
k .
=
−λ
,
and
μ
=
(−λ
−
λ
)/2.
In
each
iteration,
The initial state, μ0L = −λ2 , μU
1
0
2
1
0
we try to find the largest root of equation φ(μ) = 1 for μ ∈ (−λ2 , −λ1 ). If we find
such a solution, we stop in Step 2. Otherwise, the algorithm either searches in a new
L , μU ) as in Step 3 or declares that the non-global minimizer does not
interval (μk+1
k+1
exist as in Step 4. μk is then updated by solving a quadratic approximation of φ(μk )
in Step 5.
The global convergence of Algorithm 1 is described by the following theorem.
Theorem 3 Consider the sequence μk , k = 0, 1, ..., generated by Algorithm 1. If the
algorithm does not stop in a finite number of iterations, then there exists an μ∗ ∈
[−λ2 , −λ1 ) such that
∗
lim μk = lim μkL = lim μU
k =μ .
k→∞
k→∞
k→∞
(i) In the case that μ∗ = −λ2 , there does not exist an μ ∈ (−λ2 , −λ1 ) such that
φ(μ) = 1 and φ (μ) ≥ 0.
(ii) In the case that μ∗ ∈ (−λ2 , −λ1 ), then φ(μ∗ ) = 1 and φ (μ∗ ) ≥ 0.
Proof According to the update rule in Step 5, we have
L
L
L
U
L
μk+1 ∈ μk+1
+ 0.01(μU
−
μ
),
μ
+
0.99(μ
−
μ
)
k+1
k+1
k+1
k+1
k+1
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Globally solving extended trust region subproblems with…
1861
L
for all k = 0, 1, 2, .... Therefore, by the definition of μk+1
and μU
k+1 in Step 3, we
have
L
U
L
μU
k+1 − μk+1 ≤ 0.99(μk − μk )
for all k = 0, 1, 2, .... This implies that there exists an μ∗ ∈ [−λ2 , −λ1 ] such that
∗
lim μkL = lim μU
k = lim μk = μ .
k→∞
k→∞
k→∞
Assume that μ∗ = −λ1 . Then, μkL must have been updated infinite times in Step
3. According to the update rule in Step 3, we have φ(μk ) ≤ 1 or φ (μk ) ≤ 0
and limk→∞ μk = −λ1 for some subsequence. This contradicts the fact that
limμ→−λ1 φ(μ) = ∞. Therefore, μ∗ = −λ1 . Now let us prove (i) and (ii).
(i) In the case that μ∗ = −λ2 , we have limk→∞ μU
k = −λ2 . According to the update
rule in Step 3, there exists a subsequence of μk satisfying
lim μk = −λ2 , μk > −λ2 , φ(μk ) > 1, φ (μk ) > 0.
k→∞
For any μ ∈ (−λ2 , −λ1 ), there exists some k0 such that μ = [μk0 , μk0 +1 ]. Then,
by the convexity of function φ(μ) over the interval (−λ2 , −λ1 ), we have
φ(μ) ≥ φ(μk0 ) + φ (μk0 )(μ − μk0 ) > φ(μk0 ) > 1,
and
φ (μ) ≥ φ (μk0 ) > 0.
Therefore, (i) is proved.
∗
(ii) Since μ∗ = limk→∞ μU
k < −λ1 , we have, by updating rule in Step 3, φ(μ ) ≥ 1
L
∗
∗
∗
and φ (μ ) ≥ 0. Assume that φ(μ ) > 1. Since limk→∞ μk = μ , we have, by
updating rule in Step 3, φ(μ∗ ) ≤ 1 or φ (μ∗ ) ≤ 0. Under the assumption that
φ(μ∗ ) > 1, we must have φ (μ∗ ) ≤ 0. Combing this with previous inequality
k )−1)
/
φ (μ∗ ) ≥ 0 implies φ (μ∗ ) = 0. If this is the case, then μkN = μk − (φ(μ
φ (μk ) ∈
U
L
[μk+1 , μk+1 ] for some k large enough. This means that the algorithm would stop
in Step 4 in a finite number of iterations. Therefore, φ(μ∗ ) = 1 and (ii) is proved.
The complexity of Algorithm 1 is given by the following theorem.
Theorem 4 For a given instance of (ETRS2 ) with variable size n and any given tolerance > 0, the computational complexity of Algorithm 1 is O(n 3 + log(Q/)).
Proof In Step 1 of Algorithm 1, the complexity of spectral decomposition of Q is
L , μU ] is 0.98
O(n 3 ). For each iteration in Step 5, the length of new interval [μk+1
k+1
times of the length of previous interval [μkL , μU
k ]. Note the length of initial interval
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[μ0L , μU
0 ] ⊆ [−λ2 , −λ1 ] is less than 2Q. Thus, for a given tolerance > 0, it
takes O(log(Q/)) iterations to terminate. In each iteration, the computational
complexity for solving a quadratic equation is constant. This completes the proof. 3.2 At least one linear cut is active
Now we consider the case that at least one of the two linear cuts is active. In this
case, the complementarity constraint (a1T x − b1 )(a2T x − b2 ) = 0 holds at the optimal
solution. It follows that (ETRS2 ) is equivalent to the following problem:
min x T Qx + 2q T x
s.t. x ≤ 1,
(QRCP )
(a1T x − b1 )(a2T x − b2 ) = 0,
|b1 − a1T x|x ≤ b1 − a1T x,
|b2 − a2T x|x ≤ b2 − a2T x.
Similar to (SP), we can define the SDPR-SOCR of (QRCP ) as
min
s.t.
M0 • Y
(SPCP )
M1 • Y ≤ 0,
M2 • Y = d1T Y d2 = 0,
Y d1 ∈ S OC(n + 1),
Y d2 ∈ S OC(n + 1),
E • Y = 1,
n+1
Y ∈ S+
.
Note that the feasible region of (SPCP ) is the same as the one of (SP) except that
the second constraint is equality instead of inequality. The dual problem of (SPCP ),
denoted by (DSPCP ) is as follows.
max
s.t.
y0
1
1
n+1
Z = M0 − y0 E + y1 M1 − y2 M2 − (u 1 d1T + d1 u 1T ) − (u 2 d2T + d2 u 2T ) ∈ S+
,
2
2
(DSPCP )
y0 , y2 ∈ , y1 ≥ 0, u 1 ∈ S OC(n + 1), u 2 ∈ S OC(n + 1).
We have the following result from [16].
Theorem 5 (Theorem 2.7 in [16]) Suppose (SPCP ) and its dual problem (DSPCP )
have complementary optimal solutions. Then the optimal value of (SPCP ), which
equals to that of (DSPCP ) by strong duality, is identical to the optimal value of
(QRCP ). Moreover, an optimal solution of (QRCP ) can be obtained in polynomial
time, provided that the solution of (SPCP ) is available.
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Now, we are ready to present the algorithm to find the global optimal solution for
(ETRS2 ).
Algorithm 2 Computing the global minimizer of (ETRS2 ) with two intersecting cuts
Input: Data Q, q, a1 , b1 , a2 and b2 .
Output: A global minimizer x ∗ of (ETRS2 ).
∗ of (TRS). If x ∗ is feasible to
1: Use a traditional trust-region algorithm to obtain a global minimizer x G
G
∗ as the global minimizer of (ETRS ) and STOP. Otherwise, go to Step 2.
(ETRS2 ), output x G
2
∗
∗ is feasible,
2: Run Algorithm 1 to obtain a local-nonglobal minimizer x N G of (TRS) if there is any. If x N
G
∗ . Go to Step 3.
let x ∗ = x N
G
∗
3: Solve SDP problem (SPCP ) and its dual problem. Decompose the solution Y of (SPCP ) to obtain an
optimal solution x̂ ∗ of (QRCP ). If the objective value of x̂ ∗ is less than the one of x ∗ , let x ∗ = x̂ ∗ .
Two remarks are as follows.
Remark 1 There are a lot of methods for solving the traditional trust-region problem
(TRS). In the numerical experiment in Sect. 4, we adopted the linear-time algorithm
from [15] for its low computational complexity.
Remark 2 The method of obtaining an optimal solution x ∗ of (QRCP ) from an optimal
solution Y ∗ of (SPCP ) is the matrix rank-one decomposition. The idea is to construct
a rank-one feasible solution to (SPCP ), while keeping the complementarity conditions
are still satisfied, thus ensuring optimality. We refer the readers to [16] for more details.
The idea behind Algorithm 2 is straightforward. We first find the global minimizer
∗ of (TRS). If the global minimizer happens to be feasible to (ETRS ), then the
xG
2
∗ is also the global
two linear constraints in (ETRS2 ) are actually redundant, and x G
minimizer to (ETRS2 ). Otherwise, we move to Step 2 and find the only one localnonglobal minimizer x N∗ G by Algorithm 1 if these is any. If neither of the two linear
inequalities is active at the optimal solution of (ETRS2 ), then x N∗ G must be the global
minimizer of (ETRS2 ) following from Theorem 2. Otherwise, at least one of the two
linear inequalities is binding at the optimal solution, and we need to find the optimal
solution x̂ ∗ of problem (QRCP ), which is equivalent to its SDP relaxation (SPCP ).
Then, we compare the objective values of x N∗ G and x̂ ∗ , and set the global optimal
solution of problem (ETRS2 ) as the one with smaller objective value. Therefore, we
have established the correctness of Algorithm 2.
Theorem 6 Algorithm 2 correctly returns an optimal solution of problem (ETRS2 ) in
O(n 3 log(δ/)) for any given accuracy tolerance 0 < < 1, where δ = max{1, Q}.
Proof We analyze the complexity of each step in Algorithm 2. In Step 1, according
√
to [15], the complexity of solving traditional trust region subproblem is O(n 2 / ).
Theorem 4 shows that the complexity of Step 2 is O(n 3 + log(Q/)), and the
complexity of solving SDP problem (SPCP ) is O(n 3 log(1/)). In all, the complexity
of Algorithm 2 is O(n 3 log(δ/)) with δ = max{1, Q}.
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Z. Deng et al.
4 Numerical experiment
4.1 Numerical examples in the literature
In this subsection, we use two examples in the literature to demonstrate how the
proposed algorithms work.
Example 1 Consider the following example from [17]:
−100 10
45
0
−0.9
Q=
, q=
, a1 =
, a2 =
,
10 −61
−14
0.4
−0.6
b1 = −0.2 and b2 = 0.2. One can check that the optimal objective value is −12.5791
(ref. Example 3.1 in [17]). The SDPR-SOCR proposed in [17] gave an approximate
solution with a gap of 0.6106 to (ETRS2 ). This is due to the fact that the sufficient and
necessary condition for the exactness of second-order cone reformulation proposed
in [17] does not hold. In the next, we will show that the proposed Algorithm 2 can
successfully eliminate the gap, and return a global optimal solution of (ETRS2 ).
∗
In Step 1 of Algorithm 2, the global minimizer of problem (TRS) is x G =
−0.9636
, and Algorithm 1 returns the local-nonglobal minimizer of (TRS) x N∗ G =
0.2672
0.7950
. Note that x N∗ G is a feasible solution of (ETRS2 ) and its objective value
−0.6066
is −6.7590. By solving problem (SPCP ), we have
⎡
⎤
1.0000 0.9682 0.2500
Y ∗ = ⎣0.9682 0.9375 0.2421⎦
0.2500 0.2421 0.0625
0.9682
0.2500
with objective value −12.5791. By comparing the objective values of x N∗ G and x̂ ∗ ,
the global optimal solution of problem (ETRS2 ) is x̂ ∗ .
which is a rank-one matrix. Thus, the optimal solution of (QRCP ) is x̂ ∗ =
Example 2 Consider the following example from [2]:
⎡
⎤
⎡ ⎤
⎡ ⎤
⎡
⎤
2 3 12
7
1
−1
Q = ⎣ 3 −19 6 ⎦ , q = ⎣ 7 ⎦ , a1 = ⎣0⎦ , a2 = ⎣−6/5⎦ ,
12 6 0
9/2
0
0
b1 = 0 and b2 = 0.5. Using the global optimization software SCIP,
⎡ one can⎤verify
−0.8534
that the optimal value is −12.9420 and the optimal solution is x ∗ = ⎣ 0.2945 ⎦. The
0.4302
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Globally solving extended trust region subproblems with…
1865
⎡
⎤
−0.3552
SDPR-SOCR proposed in [2] found an approximate solution x̄ = ⎣ 0.3881 ⎦ with
−0.2119
objective value −13.8410, thus resulting in a relative gap of 7%.
⎡
⎤
−0.2035
∗ = ⎣−0.9687⎦ of problem
In the other hand, the Step 1 of Algorithm 2 returns x G
0.1418
⎡
⎤
0.2769
(TRS), and the local-nonglobal minimizer of (TRS) is x N∗ G = ⎣ 0.2553 ⎦ in Step 2
−0.9264
of Algorithm 2. But neither of these two minimizers is feasible to (ETRS2 ). Then, in
Step 3 of Algorithm 2, the optimal solution of (SPCP )
⎡
1.0000
⎢−0.8534
∗
⎢
Y =⎣
0.2945
0.4301
−0.8534
0.7283
−0.2513
−0.3671
0.2945
−0.2513
0.0867
0.1267
⎤
0.4301
−0.3671⎥
⎥
0.1267 ⎦
0.1850
⎡
⎤
−0.8534
is a rank-one matrix. Therefore, the global minimizer of (ETRS2 ) is x ∗ = ⎣ 0.2945 ⎦
0.4301
with objective value −12.9420, which is exactly the same as the one obtained by the
global optimization solver SCIP.
4.2 Randomly generated instances
In this subsection, we test the proposed algorithm on randomly generated instances
of (ETRS2 ). The instances are generated according to the procedure in [17]. For the
matrix Q, all the upper triangular entries (including the diagonal entries) of Q are
generated uniform-randomly on the interval [−50, 50] (the lower part takes the values
by symmetry) and then, to ensure the “nonconvexity”, each of the diagonal entries is
renewed again by subtracting 60 except that the first entry is reset as zero. The vector
q is randomly generated in the open unit ball {q|q < 1}. The entries of two vectors
a1 and a2 are generated uniform-randomly on the interval [−1, 1], and then we set
b1 = a1T q and b2 = a2T q. Apparently, any instance generated by the above procedure
is always “nonconvex” and “intersecting”.
We generate 1000 instances for each problem size n = 2, 3, ..., 10. The tolerance
is set to 10−6 . The SDP problem (SPCP ) is solved by SeDuMi in CVX [5]. Problem
(ETRS2 ) is solved by the global optimization software SCIP in order to verify the
solution of Algorithm 2. We compare the objective value of the proposed algorithm
with the one of SDPR-SOCR in [17]. The results are summarized in Table 1. For
the table, the first column n is the size of instances, the second column “number for
SDPR-SOCR” denotes the number of the 1000 instances that there exist gaps between
the original problems and the SDPR-SOCR relaxations proposed in [17], and the
third column “number for Algorithm 2” denotes the number of 1000 instances whose
123
1866
Table 1 Gap existing for
numbers of 1000 random
instances
Z. Deng et al.
n
Number for
SDPR-SOCR in [17]
Number for Algorithm 2
2
51
0
3
6
0
4
4
0
5
4
0
6
2
0
7
2
0
8
1
0
9
1
0
10
2
0
objective values are different from the output of Algorithm 2. As shown in Table 1,
Algorithm 2 is able to find the global optimal solutions to all the instances while the
gap-existing SDPR-SOCR instances appear in every problem size n (although more
unlikely for larger n). The numerical results confirm that Algorithm 2 is indeed a
global resolution method for problem (ETRS2 ).
5 Conclusion
In this paper, we proposed an algorithm to globally solve the extended trust-region
subproblem with two intersecting cuts. We first used Algorithm 1 to check whether
there is any local-nonglobal minimizer lies on the unit sphere in the feasible region
of (ETRS2 ). Then, an SDP relaxation with SOC reformulation is solved to find an
optimal solution in the case that one of the two linear cuts is active. Two examples
in the literature and numerical experiment on randomly generated instances are given
to show that the proposed algorithm is able to find the global optimal solution of
(ETRS2 ). One interesting research direction is whether we can extend this algorithm
to solve the classical Celis–Dennis–Tapia (CDT) problem.
Acknowledgements The authors would like to thank the anonymous reviewers for their value comments that
help improve the quality of this work. Deng’s research has been supported by National Natural Science Foundation of China Grant #11501533 and University of Chinese Academy of Sciences Grant #Y95402MXX2.
Lu’s research has been supported by National Natural Science Foundation of China Grants #11701177 and
#11771243. Tian’s research has been supported by National Natural Science Foundation of China Grant
#71331004 and Fundamental Research Funds for the Central Universities #JBK1805005. Jian’s research
has been supported by National Natural Science Foundation of China Grants #71701035 and #71831003.
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