1.
Introduction
Bilevel programming problem has increasingly been addressed in the literature, both from the theoretical and computational points of view [14]. This model has been widely applied to decentralized planning problems involving a decision progress with a hierarchical structure. It is characterized by the existence of two optimization problems in which the constraint region of the first-level problem is implicitly determined by another optimization problem. The NBLP problem is hard to solve. In fact, the problem has been proved to be NP-hard [8]. However, the NBLP problem is used so extensively in resource allocation, finance budget, price control, transaction network etc. [1,7,28,29,39] that many researches have been devoted to this field, which leads to a rapid development in theories and algorithms. For the detailed expositions, the reader may consult [21,33].
In this paper we will consider the following NBLP problem
where t∈ℜn1 and y∈ℜn2. The functions fu:ℜn1+n2→ℜ, fl:ℜn1+n2→ℜ, gu:ℜn1+n2→ℜm1, and gl:ℜn1+n2→ℜm2 are assumed to be at least twice continuously differentiable function.
There are several approaches have proposed to solve problem 1.1, see [2,3,25,35,40]. One of these approaches and used in this paper, is converted the original two level problems to a single level one by replacing the lower level optimization problem with its Karush-Kuhn-Tucker (KKT) conditions, see [24,41]. By KKT optimality conditions for the lower-level problem, then we can reduce the NBLP problem 1.1 to one-level programming problem. This problem is non-convex and non-differentiable, moreover the regularity assumptions which are needed to successfully handle smooth optimization problems are never satisfied and it is not good to use our approach. So, we add slack variables for inequalities constraints in problem 1.1.
By adding slack variables su∈ℜm1 and sl∈ℜm2 to the upper inequality constraint gu(t,y) and the lower inequality constraint gl(t,y) respectively, then NBLP problem 1.1 can be written as follows
The above NBLP problem can be simplified as follows
where ˜gu(t,y,su)=gu(t,y)+su∈ℜm1, ˜gl(t,y,sl)=gl(t,y)+sl∈ℜm2, and s=(su,sl)T∈ℜm1+m2.
Applying KKT conditions only on the lower-level problem without the constraint s≥0, then we can reduce the NBLP problem 1.2 to the following smooth SONP problem:
where μl∈ℜm2 is a Lagrange multiplier vector associated with equality constraint ˜gl(t,y,sl), see [5].
Using problem 1.3, to overcome the difficulty that problem 1.1 does not satisfy any regularity assumptions, which are needed for successfully handling smooth optimization problems, and pave the way for using the proposed approach to solve problem 1.1. To simply our discussion, we introduce the following notations. x=(t,y,s)T∈ℜn, n=n1+n2+m1+m2, h(x)∈ℜm represents the vector of equality constraints such that m=m1+m2+n2. Then problem 1.3 can be written as follows
where v∈{ℜ⋃{−∞}}n, w∈{ℜ⋃{+∞}}n, and v<w.
Various approaches have been proposed to solve the SONP problem 1.4, see [5,9,10,11,15,16,17,18,19]. In this paper, we use Newton's interior point method with Das scaling matrix [12] to solve problem 1.4. Newton's method converges quadratically to a stationary point under reasonable assumptions if the starting point sufficiently closed to the stationary point. It may not converge if the starting point is far away from the stationary point. To guarantee convergence from any starting point, a trust-region strategy is used. The trust-region strategy can induce strongly global convergence, which is very important method for solving SONP problem and is more robust when it deals with rounding errors. It does not require the objective function of the model be convex or the Hessian of the objective function must be positive definite. Also, some criteria are used to test the trial step is acceptable or no. If it is not acceptable, then the subproblem must be resolved with a reduced the trust-region radius. For the detailed expositions, the reader may consult [4,17,20,21,22,23,30,32,36,42,43,45,46].
A reduced hessian technique is used in this paper to overcome some difficulties in trust-region subproblem. This technique was suggested by [6,37] and used by [19,20].
In this paper, we use the symbol, fuk≡fu(xk), hk≡h(xk), Pk≡P(xk) ℓk≡ℓ(xk,λk), ∇xℓk≡∇xℓ(xk,λk), and so on to denote the function value at a particular point. Finally, all norms are l2-norms.
The rest of the paper is organized as follows. In section 2, we introduce detailed description for the proposed method to solve problem 1.4. Section 3 is devoted to analysis of the global convergence of the proposed algorithm. Section 4 contains implementation of the proposed algorithm and the results of test problems. Section 5 contains concluding remarks.
2.
An interior-point method with trust-region algorithm
In this section, firstly, we will consider the detailed description for the Newton's interior-point method with Das scaling matrix to solve SONP problem 1.4. Secondly, to guarantee convergence from any starting point, we will introduce the detailed description for trust-region strategy. Finally, we clarify main steps for general algorithm to solve NBLP 1.1.
2.1. Newton's method with scaling matrix
Motivated by the impressive computational performance of Newton's interior-point method for solving SONP problem 1.4, let
be a Lagrangian function associated with problem 1.4 without the constraints v≤x≤w, and let
be a Lagrangian function associated with problem 1.4 with the constraints v≤x≤w. The vectors λ∈ℜm, μv∈ℜn, and μw∈ℜn represent Lagrange multiplier vectors associated with the constraints h(x)=0, 0≤(x−v), and 0≤(w−x) respectively. Let ˆG={x:v≤x≤w} and int(ˆG)={x:v<x<w}.
The first-order necessary conditions for the point x∗ to be a local minimizer of problem 1.4 are the existence of multipliers λ∗∈ℜm, μv∗∈ℜn+, and μw∗∈ℜn+, such that (x∗,λ∗,μv∗,μw∗) satisfies
and for all i corresponding to x(i) with finite bound, we have
where ∇xℓ(x∗,λ∗)=∇fu(x∗)+∇h(x∗)λ∗.
The proposed algorithm here, like its predecessors in [12,18,19], starts at a point strictly feasible with respect to the bounds on the variables and produces iterates that are strictly feasible with respect to the bounds (i.e. 'in the interior'). Define a diagonal scaling matrix P(x)=diag(p(x) whose diagonal elements p(x) are given by
Using the matrix P(x), then (x∗,λ∗,μv∗,μw∗) satisfy the KKT conditions [2.3-2.7] if and only if
For more details about the proof, see [12].
Applying Newton's method on the nonlinear system [2.9-2.10], then we have
where θ(x) is a vector whose components are given by
For more details see [18].
In our method, the matrix P(x) must be nonsingular, so we restrict the point x∈int(ˆG). Multiplying both sides of equation 2.11 by P−1(x), then we have
Substituting Δx=P(x)d in the above two system, then we have
where H(x,λ)=∇2xℓ(x,λ) represents the Hessian of the Lagrange function 2.1 or an approximation to it. It is easy to see that the step generated by the above system coincides with the solution of the following quadratic programming subproblem
where B=P(x)H(x,λ)P(x)+diag(∇xℓ(x,λ))diag(θ(x)). This means that, the point (x∗,λ∗) that satisfies the KKT conditions for subproblem 2.16 will satisfy the KKT conditions for problem 1.4.
Although Newton's method converges quadratically to a stationary point under reasonable assumptions, it may not converge to a stationary point if the starting point is far away from the solution. To overcome this disadvantage and to guarantee convergence from any starting point, we use the trust-region technique.
2.2. Trust-region technique
Trust-region methods can induce strongly global convergence, which are very important methods for solving a smooth nonlinear programming problem and are more robust when they deal with rounding errors. It does not require the objective function of the model be convex. Also, it does not demand the Hessian of the objective function must positive definite.
The trust-region subproblem associated with problem 2.16 is
where δk>0 is the radius of the trust-region.
Subproblem 2.17 may be infeasible, because there may be no intersecting points between the constraint ‖d‖≤δk and hk+(Pk∇hk)Td=0 constraints. Even if they intersect, there is no warranty that this will continue true if δk is decreased. For more details see [13]. To overcome this difficulty, we use a reduced hessian technique. This technique was suggested by [6,37] and used by [19,20]. In this technique, the trial step d is decomposed into two orthogonal components: the normal component dn to improve feasibility and the tangential component dtk to improve optimality. Each of components is computed by solving unconstrained trust-region subproblem.
● How to estimate the normal component dnk
The normal component dnk is computed by solving the following trust-region subproblem
for some 0<ζ<1. To solve the subproblem 2.18, we use a conjugate gradient method which is introduced by [38] and used by [21], see algorithm 2.1 in [21]. It is very cheap if the problem is large-scale and the Hessian is indefinite. By using the conjugate gradient method, the normal predicted decrease obtained by dnk is greater than or equal to a fraction of the normal predicted decrease obtained by the Cauchy step dncpk. This means that
such that dncpk is defined as follows
where the parameter φncpk is given by
Once dnk is obtained, we will evaluate dtk=Zkˉdtk where Zk is the matrix whose columns form a basis for the null space of (Pk∇hk)T.
● How to estimate the tangential component dtk
To obtain the tangential component dtk, we use the conjugate gradient method [21] to solve the following trust-region subproblem
where ∇qk(Pkdnk)=Pk∇xℓk+Bkdnk and Δk=√δ2k−‖dnk‖2.
By using the conjugate gradient method, the tangential predicted decrease which is obtained by tangential step ˉdtk is greater than or equal to a fraction of the tangential predicted decrease obtained by a tangential Cauchy step ˉdtcpk. This means that
for some 0<ϑ2≤1 and ˉdtcpk is defined as follows
where the parameter φtcpk is given by
where ˉBk=ZTkBkZk.
● How to estimate a parameter γk
Once obtaining dtk, we set dk=dnk+dtk and xk+1=xk+Pkdk. To ensure xk+1∈int(ˆG), we need to evaluate the parameter γk. To do this, evaluate
and
Compute
Once the trial step γkPkdk is evaluated, it needs to be tested to decide whether it will be accepted or not. To do this, we need to a merit function which is ties the objective function and the constraints in such a way that progress in the merit function means progress in solving problem. In our method, we use the following merit function which is introduced by [26] and known as an augmented Lagrange function
where ℓ(x,λ) is defined in 2.1 and ρ>0 represents the penalty parameter.
● How to estimate λk+1
The Lagrange multiplier vector λk+1 will be estimated as follows
To test whether the point (xk+1,λk+1), will be accepted in the next iterate or no we need to define the following actual reduction Aredk and the predicted reduction Predk.
The actual reduction Aredk in the merit function 2.27 in moving from (xk,λk) to (xk+γkPkdk,λk+1) is defined as follows
Also we can write the actual reduction Aredk as follows,
where Δλk=(λk+1−λk).
The predicted reduction Predk is defined as follows
Since qk(γkPkdk)=ℓk+(Pk∇xℓk)Tγkdk+12γ2kdTkBkdk, then Predk can be written as follows,
● How to update the penalty parameter ρk
To ensure Predk is strictly positive, we use the following scheme to update the positive penalty parameter ρk
Algorithm 2.1.: (Updating the penalty parameter ρk)
Set ρk+1=ρk.
If
then set
End if.
● How to test the step γkPkdk and update δk
To decide the trial step γkPkdk will be accepted in the next iteration or no, we use the following algorithm.
Algorithm 2.2.: (Testing the step γkPkdk and updating δk)
Step 0. Choose 0<τ1<τ2<1, 0<β1<1<β2, and δmin≤δ0≤δmax.
Step 1. While AredkPredk<τ1 or Predk≤0.
Do not accept the step and set δk=β1‖dk‖.
Compute a new trial step.
End while.
Step 2. If τ1≤AredkPredk<τ2.
Accept the step: xk+1=xk+γkPkdk.
Set δk+1=max(δk,δmin).
End if.
Step 3. If AredkPredk≥τ2.
Accept the step: xk+1=xk+γkPkdk.
Set δk+1=min{δmax,max{δmin,β2δk}}.
End if.
Finally, the algorithm is stopped when either ‖ZTkPk∇xℓk‖+‖hk‖≤ε1, for some ε1>0 or ‖dk‖≤ε2 for some ε2>0.
Main steps of the trust-region algorithm for solving subproblem 2.17 are summarized in the following algorithm.
Algorithm 2.3. (Trust-region algorithm)
Step 0. Starting with x0∈int(ˆG). Evaluate λ0, P0, and β0. Set ρ0=1 and c0=0.1.
Choose ε1>0, ε2>0, and set k=0.
Step 1. If ‖ZTkPk∇xℓk‖+‖hk‖≤ε1, then stop.
Step 2. (To compute dk)
a) Compute dnk by solving trust-region subproblem 2.18.
b) Compute ˉdtk by solving trust-region subproblem 2.22.
c) Set dk=dnk+Zkˉdtk.
Step 3. If ‖dk‖≤ε2, then stop.
Step 4. Compute γk using equation 2.26.
Step 5. Update λk+1 using subproblem 2.28.
Step 6. Update the penalty parameter using scheme 2.1.
Step 7. Test the step γkPkdk and update δk by using algorithm 2.2.
Step 8. Compute Pk+1 and αk+1 using definitions 2.8 and 2.13 respectively.
Step 9. Set k=k+1 and go to Step 1.
Main steps for solving NBLP problem1.1 are summarized in the following algorithm.
Algorithm 2.4. (Interior-point trust-region (IPTR) algorithm)
Step 1. Adding slack variables to inequalities in NBLP problem1.1 and convert it to problem 1.2.
Step 2. By KKT optimality conditions for the lower-level problem, NBLP problem 1.2 is equivalent to the one level problem 1.3 which can be written in the form 1.4.
Step 3. Using Newton's method and Das strategy to transform problem 1.4 to subproblem 2.16.
Step 4. Using trust-region algorithm 2.3 to solve subproblem 2.16.
The following section is devoted to global convergence analysis for IPTR algorithm 2.4.
3.
Global convergence theory
We state the general assumption under which the global convergence theory for IPTR algorithm 2.4 is proved.
3.1. A general assumptions
Let Ω be a convex subset of ℜn that contains all points xk∈int(ˆG) and (xk+γkPkdk)∈int(ˆG). On the set Ω we state the following general assumptions under which the global convergence theory of IPTR algorithm is proved
[GS1.] The functions fu(x), h(x)∈C2 for all x∈Ω.
[GS2.] The matrix Pk∇hk has full column rank.
[GS3.] All of fu(x), ∇fu(x), ∇2fu(x), h(x), ∇h(x), ∇2hi(x) for i=1,...,m and (Pk∇Hk)((Pk∇hk)T(Pk∇hk))−1 are uniformly bounded in Ω.
[GS4.] The sequence of Lagrange multiplier vectors {λk} is bounded.
[GS5.] The sequence of approximate Hessian matrices {Hk} is bounded.
An immediate consequence of the above general assumptions is that the existence of positive constant b1, such that
3.2. Technical lemmas
In this section, we introduce some important results which are needed in the subsequent proof.
The following lemma shows how accurate the definition of Predk is as an approximation to Aredk.
Lemma 3.1. Under assumptions GS1-GS5, there exists a positive constant K1, such that
Proof. From Equations 2.29, 2.30, and using the inequality of Cauchy-Schwarz, we have
for some ξ1 and ξ2∈(0,1). Using the general assumptions GS1−GS5 and 0<γk≤1, we have
where κ1, κ2, and κ3 are positive constants. Since ρk≥0, ‖dk‖≤δmax, and ‖hk‖ is uniformly bounded, then inequality 3.2 hold.
The following lemma obviously that the normal predicted reduction at any iteration k, is at least equal to the decrease in the 2-norm of the linearized constrained by the Cauchy step
Lemma 3.2. Under assumptions GS1-GS5, there exists a constant K2>0, such that
Proof. Since dnk is obtained by approximating the solution of subproblem 2.18 using the conjugate gradient method [21], then the fraction of Cauchy decrease condition 2.19 is hold. We will consider two cases:
Firstly, if dncp=−δk‖Pk∇hkhk‖(Pk∇hkhk) and ‖δk‖(Pk∇hk)TPk∇hkhk‖2≤‖Pk∇hkhk‖3 then
Secondly, if dncp=−‖Pk∇hkhk‖2‖(Pk∇hk)TPk∇hkhk‖2(Pk∇hkhk) and δk‖(Pk∇hk)TPk∇hkhk‖2≥‖Pk∇hkhk‖3, then
Using assumption GS2, we have
Then, from the above inequality, inequalities 3.5, 3.6, and using assumption GS3, we have
From the above inequality and 2.19, we have
Since 0<γk≤1, then we have
From inequality 3.7 and the above inequality, we have
From inequalities 2.32 and 3.8 we have
Lemma 3.3. Under assumptions GS1-GS5, there exists a positive constant K3, such that
Proof. Since the conjugate gradient method is used to solve subproblem 2.22 to obtain an approximate solution for ˉdtk, then the fraction of Cauchy decrease condition 2.23 is hold and we will consider two cases:
Firstly, if ˉdtcpk=−Δk‖ZTk∇qk(Pkdnk)‖(ZTk∇qk(Pkdnk)) and Δk(ZTk∇qk(Pkdnk))TˉBkZTk∇qk(Pkdnk)≤‖ZTk∇qk(Pkdnk)‖3, then
Secondly, if ˉdtcpk=−‖ZTk∇qk(Pkdnk)‖2ZTk∇qk(Pkdnk))TˉBkZTk∇qk(Pkdnk)ZTk∇qk(Pkdnk) and Δk(ZTk∇qk(Pkdnk))TˉBkZTk∇qk(Pkdnk)≥‖ZTk∇qk(Pkdnk)‖3, then
From inequalities 3.10, 3.11, and using necessary assumptions, we have
From condition 2.23 and the above inequality, we have
Since 0<γk≤1, then we have
From the above inequality and inequality 3.12, we have
This completes the proof.
The following lemma is needed in many forthcoming lemmas. In what follows, we use implicitly that ∇hkdnk=∇hkdk.
Lemma 3.4. Under assumptions GS1-GS5, there exists a positive constant K4, such that
Proof. Since dnk is normal to the tangent space, then we have
Using the fact that ‖hk+(Pk∇hk)Tdk‖≤‖hk‖, we have
From above inequality and necessary assumptions, we have
Since
From the above inequality and inequality 3.14 and using the fact that ‖dnk‖≤δmax, we have
This completes the proof.
Lemma 3.5. Under assumptions GS1-GS5,
Proof. From Equation 2.31, we have
Using inequalities 3.9 and 3.13, we obtain the desired result.
The following lemma shows that, if ‖ZTkPk∇xℓk‖≥ε1 and ‖hk‖≤ηδki at any trial iteration ki, then the penalty parameter ρk is not increased.
Lemma 3.6. Under assumptions GS1−GS5, if ‖ZTkPk∇xℓk‖≥ε1 and ‖hk‖≤ηδki at any trial iteration ki, then there exists a positive constant K5, such that
where η is given by
Proof. Since ‖ZTkPk∇xℓk‖≥ε1 and ‖hk‖≤ηδki, and using inequalities 3.1 and 3.14, we have
But η≤ε12b1κ4δmax, hence
From inequality 3.14, assumption ‖hk‖≤ηδki, and η≤√32κ4, then we have ‖dnki‖≤κ4ηδki≤κ4√32κ4δki=√32δki. Since Δki=√δ2ki−‖dnki‖2, then Δki≥12δki. Hence, from inequalities 3.15 and 3.17, we have
But η≤K3ε18K4min{ε1b1δmax,1}, hence
The result follows if we take K5=K3ε18min{ε1b1δmax,1}.
The following lemma shows that, at any iteration k, we can find an acceptable step after finite number of trials, or equivalently, the condition Aredkj/Predkj≥τ1 will be satisfied for some finite j.
Lemma 3.7. Under assumptions GS1−GS5, if ‖hk‖>ε1, where ε1>0, then Aredkj/Predkj≥τ1 will be satisfied for some finite j.
Proof. From inequalities 3.2, 3.4, and assumption ‖hk‖>ε1, we have
If the trial step dkj gets rejected, then δkj becomes small and hence we have
That is the acceptance rule will be met after finite number of trials (i.e., for finite j) and this completes the proof.
Lemma 3.8. Under assumptions GS1−GS5 and at any iteration k, if
at the jth trial step, then the step must be accepted.
Proof. Assume that inequality 3.18 holds and the step dkj is rejected. From the way of updating trust-region radius which is clarified in Algorithm 2.2 we have
From the above inequality and using inequalities 3.2, 3.4, and 3.18 we have
This is a contradiction with the assumption dkj was rejected. Hence the step must be accepted.
Lemma 3.9. Under assumptions GS1−GS5 and for all trail iterates j of any iteration k we have
where b2=supx∈Ω‖hk‖.
Proof. Consider any trial iterate kj, if j=1, then the step is accepted and hence
such that b2=supx∈Ω‖hk‖.
Now, if j>1, then there exists at least one rejected trial step. For all rejected trial steps, we have from Lemma 3.8,
for all i=1,2,...j−1. Since dik is rejected trial step, then from the way of updating the radius of trust-region, we have
From the above inequality and inequality 3.20, we obtain the desired results.
The following lemma show that the sequence of trust-region radii {δkj} is bounded away from zero if {‖hk‖} is bounded away from zero.
Lemma 3.10. Under assumptions GS1−GS5, if ‖hk‖≥ε1 where ε1>0, then there exists a constant K6>0 such that
for all trial iterates j of any iteration k.
Proof. From Lemma 3.9 and the condition ‖hk‖≥ε1, the proof follows directly by taking K6=min{δminb2,β1(1−τ1)K24K1,β1}ε1.
Lemma 3.11. Under assumptions GS1−GS5, there exists a subsequence {ki} of the iteration sequence at which ρk is increased such that at any trial steps j of any iteration k∈{ki}, we have
where K7 is a positive constant.
Proof. At any trial steps j of any iteration k, if ρkj is increased, then from equation 2.33, we have
Applying inequality 3.8 on the left hand side and inequalities 3.9, 3.13, and 3.14 on the right hand side, we have
From assumptions GS2, GS3, and using the fact that ‖dnkj‖≤δkj≤δmax, we have
From inequalities 3.19 and 3.23, there exists a constant K7>0 such that
at any trial steps j for any iteration k∈{ki}.
In the following lemma we will prove that the sequence {‖hk‖} is not bounded away from zero when {ρk} unbounded sequence.
Lemma 3.12. Under assumptions GS1−AS6, there exists a subsequence {ki} of the iteration sequence at which ρk is increased such that
Proof: From Lemma 3.11 and the assumption ρk is increased, we obtain the desired result.
In the following section, we prove the main global convergence results for IPTR algorithm 2.4.
3.3. Fundamental convergence theorem
In the following theorem we prove that the sequence of the iterates generated by algorithm 2.4 converges to the feasible set.
Theorem 3.1. Under assumptions GS1−GS5, the sequence of iterates generated by IPTR algorithm satisfies
Proof. The proof of this theorem is by contradiction, so we assume that lim supk→∞‖hk‖≥ε1 where ε1>0. This implies the existence an infinite subsequence of indices {kj} indexing iterates that satisfy ‖hk‖≥ε12, for all k∈{kj}. From Lemma 3.7, there exists a finite sequence of acceptable steps. Without lose of generality, we assume all members of the sequence {kj} are acceptable iterates. Now we will consider two cases:
Firstly, if the sequence of the penalty parameter {ρk} is unbounded, then there exists a subsequence {ki} of the iteration sequence at which ρk is increased. Using Lemma 3.12, we have limki→∞‖hki‖=0. Therefore, there are no common elements between {ki} and {kj} at iteration k which is sufficiently large.
From inequality 3.4 and Lemma 3.10, we have
for all k∈{kj}, such that ˉK6=ε12min{δminb2,β1(1−τ1)K22K1,β1}. Since
then from 3.25 we have
Hence
for all acceptable steps which are generated by IPTR algorithm 2.4. Let k∈{kj} be an element between the two elements kˆi and kˆi+1 which are consecutive elements of the sequence {ki}. From inequality 3.26, we have
Since the value of ρk is the same for all iterates kˆi,…,kˆi+1−1, we have
Since ρk→∞ as k→∞, and |ℓk| is bounded, we can write
for kˆi sufficiently large. But this leads to a contradiction with Lemma 3.12.
Secondly, if the sequence of the penalty parameters {ρk} is bounded, then there exists an integer ˉk such that for all k≥ˉk, we have ρk=ˉρ. Since all the iterates of {kj} are acceptable, then for any ˉk∈{kj} we have
From Lemma 3.10, inequality 3.4, we have for any ˜k∈{kj} and ˜k≥ˉk
such that K8=K2τ˜kˉρε14min{‖ε12δmax,1}. From inequalities 3.28 and 3.29, we have
This gives a contradiction with the fact that {Φk} is bounded below when {ρk} is bounded. Hence in both cases, we have a contradiction. Thus, the supposition is not correct and the theorem is proved.
Theorem 3.2. Under assumptions GS1−GS5, the algorithm is terminated because
Proof. Assume that IPTR algorithm 2.4 does not terminate and that some subsequences of {‖ZTkPk∇ℓk‖} convergence to zero, then the nontermination is immediately contradicted by Theorem 3.1.
Now assume that for ˉk sufficiently large, there exists an index ˜k>ˉk such that ‖ZTkPk∇ℓk‖≥ε1. Let {kj} be a subsequence of iterates that satisfy ‖hkj‖>ηδkj, then limkj→∞δkj=0 such that limkj→∞‖hkj‖=0. This implies the existence of an infinite sequence {kj} of rejected trial steps. But this leads to contradiction. To show this, we consider two cases:
Firstly, if the sequence of the penalty parameter {ρk} is unbounded, then from inequalities 3.3 and 3.4, we have
As ρkj→∞ and δkj→0, then |Aredkj−Predkj|Predkj→0. This means that for kj large enough, all trial steps ‖dkj‖ must be accepted. This leads to a contradiction, so δkj must be bounded away from zero in this case.
Secondly, if the sequence of the penalty parameter {ρk} is bounded, then there exists an integer ˉk such that for all k≥ˉk, ρk=ˉρ. Now, we discuss three cases:
1] If the previous step is accepted (j=1), then from the way of updating the trust-region radius in algorithm 2.2, we have δkj≥δmin. That is δkj is bounded away from zero in this case.
2] If j>1 and ‖hkr‖>ηδkr for all r=1,⋯,j−1. Then
such that all the trial steps on {kj} are rejected. From above inequality and inequalities 3.2 and 3.4 we have
Hence
But from the way of updating the radius of trust-region in algorithm 2.2, all the rejected trial steps satisfy δkr=β1‖dkr‖, hence
This means that δkr is bounded away from zero in this case.
3] If j>1 and ‖hkr‖>ηδkr does not hold for all r. Hence, there exists an integer i such that ‖hkr‖≤ηδkr for all r=1,⋯,i, and ‖hkr‖>ηδkr for all r=i+1,⋯,j−1. Since ‖hkr‖>ηδkr for all r=i+1,⋯,j−1, then as the above case we can prove δkr is bounded away from zero.
The case when ‖hkr‖≤ηδkr for all r=1,⋯,i, then for all rejected trial steps, we have
From inequality 3.2, Lemma 3.6, and the above inequality, we have
Hence
From the way of updating the radius of trust-region, we have for all rejected trial step
Hence, δkr is bounded away from zeros. This leads to a contradiction and then for kj sufficiently large, all the iterates satisfy ‖hk‖≤ηδkj.
For all successful steps and from the way of updating the radius of trust-region and Lemma 3.6, we have for all k∈{kj} and k≥ˉk
We proved in the above cases, that δkj is bounded away from zeros. Then Φk−Φk+1>0. This leads to a contradiction with the fact that {Φk} is bounded below when {ρk} is bounded. Hence in both cases, we have a contradiction. Thus, the supposition is not correct and the theorem is proved.
4.
Application
In this section, firstly the proposed algorithm IPTR is applied to the engineering application which is called two-echelon supply chain system with one manufacturer and one retailer.
The manufacturer purchases raw materials from the supplier first, then after the manufacturer's production and processing, the end products are sold to the retailer, this problem is formulated as bilevel models for joint pricing and lot-sizing decisions, see [34].
where ˜T=52; ˜Ps=4; ˜Tc=0.5; ˜Mc=1; ˜cm=˜cr=0.001; ˜Om=400; ˜Or=200. For more details about the above application and its notations, see [34].
We solve this model in case of the manufacturer is the leader, who makes the first decision, and the retailer is the follower. Our results, when applying Algorithm (2.4) is t1=5.8778, t2=6.002, t3=19710.195, y1=7.691, y2=2.6007, fu=431230, and fl=8548300, which is closed to whose reported in [34].
Secondly, we introduce an extensive variety of possible numeric bilevel nonlinear programming problems to clarify the effectiveness of our IPTR algorithm, since, Problems 1, 2, 6, 7, 13, and 14 have quadratic functions in both levels. Problems 3, 4, 5, 8, 9 all the inner level functions are convex and Problem 10 [27], at fixed x, the inner problem is convex. These problems are solved numerically with the help of algorithm (2.4) to clarify the effectiveness of that approach. For each test example, 10 independent runs with different initial starting point are performed to observe the consistency of the outcome. Statistical results of all examples are summarized in Table 1 which shows that the results found by the IPTR algorithm (2.4) are approximate or equal to those by the compared algorithms in the literature.
Table 1 also including the mean number of iterations (iter), the mean number of function evaluations (nfunc), the mean value of CPU time (CPUs) in seconds.
For comparison, we have included the corresponding results of the mean value of CPU time (CPUs) obtained by Method in [31](Table 2), [27](Table 3), and [44](Table 4) respectively. It is clear from the results that our approach is capable for treating nonlinear bilevel programming problems even the upper and the lower levels are convex or not and the computed results converge to the optimal solution which is similarly or approximate to the optimal that reported in literature. Finally, it is clear from the comparison between the solutions obtained using IPTR algorithm with literature, that IPTR is able to find the optimal solution of all problems by a small number of iterations, small number of function evaluations, and less time.
Problem 1 [31]:
Problem 2 [31]:
Problem 3 [31]:
Problem 4 [31]:
We offered the numerical results of our algorithm using MATLAB (R2013a)(8.2.0.701)64-bit(win64) and a starting point x0∈int(ˆG). The following parameter setting is used: δmin=10−3, δ0=max(‖scp0‖,δmin), δmax=103δ0, τ1=10−4, τ2=0.75, β1=0.5, β2=2, ˆε=0.01, ε1=10−8, and ε2=10−10.
Problem 5 [31]:
Problem 6 [31]:
Problem 7 [31]:
Problem 8 [31]:
Problem 9 [27]:
Problem 10 [27]:
Problem 11 [44]:
Problem 12 [27]:
Problem 13 [44]:
Problem 14 [44]:
Problem 15 [44]:
Problem 16 [44]:
5.
Concluding remarks
This paper presented a new technique for solving a nonlinear bilevel optimization problem based on using the slack variable with KKT condition to transform NBLP problem into an equivalent smooth SONP problem. A Newton's interior-point method with Das scaling matrix is utilized to solve the equivalent smooth SONP problem effectively. Newton's method is locally method, so a trust region technique is utilized to ensure global convergence from any starting point. On applying this methodology we overcome some known difficulties on treating such problems, as
● A trust-region technique can induce strongly global convergence, which is very important technique for solving a smooth optimization problems and is more robust when they deal with rounding errors
● Our approach used to transform Problem 1.3 which is not smooth to smooth problem
● Using the interior-point method guarantees the converges quadratically to a stationary point.
On the other hand, the global convergence theorems for the IPTR algorithm is presented and numerical results reflect the good behavior of our algorithm and computed results converge to the optimal solutions. Finally, it is clear from the comparison between the solutions obtained using IPTR algorithm with literature, that IPTR is able to find the optimal solution of all problems by a small number of iterations.
Acknowledgments
The authors would like to thank the anonymous referees for their valuable comments and suggestions which have helped to greatly improve this paper.
Conflict of interest
The authors declare that there is no conflict of interest in this paper.