# On the Convergence of a Smooth Penalty Algorithm without Computing Global Solutions

Authors: Bingzhuang Liu; Changyu Wang; Wenling Zhao
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

We consider a smooth penalty algorithm to solve nonconvex optimization problem based on a family of smooth functions that approximate the usual exact penalty function. At each iteration in the algorithm we only need to find a stationary point of the smooth penalty function, so the difficulty of computing the global solution can be avoided. Under a generalized Mangasarian-Fromovitz constraint qualification condition (GMFCQ) that is weaker and more comprehensive than the traditional MFCQ, we prove that the sequence generated by this algorithm will enter the feasible solution set of the primal problem after finite times of iteration, and if the sequence of iteration points has an accumulation point, then it must be a Karush-Kuhn-Tucker (KKT) point. Furthermore, we obtain better convergence for convex optimization problem.


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## Body

## 1. Introduction

Consider the following nonconvex optimization problem(NP)min⁡⁡f(x)s.t.gi(x)≤0,i=1,…,m,x∈Rn,
where f,gi:Rn→R,i=1,…,m, are all continuously differentiable functions. Without loss of generality, we suppose throughout this paper that infx∈Rnf(x)≥0, because otherwise we can substitute f(x) by exp(f(x)). Let Ωɛ={x∈Rn∣gi(x)≤ɛ,i=1,…,m} be the relax feasible set for ɛ>0. Then Ω0 is the feasible set of (NP).The classicall1 exact penalty function [1] is
(1.1)fβ(x)=f(x)+β∑i=1m(gi(x))+,
where β>0 is a penalty parameter, and (1.2)(gi(x))+=max⁡⁡{0,gi(x)},i=1,…,m.The obvious advantage of the traditional exact penalty functions such as thel1 exact penalty function is that when the penalty parameter is sufficiently large, their global optimal solutions exist and are optimal solutions of (NP). But they also have obvious disadvantage, that is, their nonsmoothness, which prevent the use of many efficient unconstrained optimization algorithms (such as Gradient-type or Newton-type algorithm). Therefore the study on the smooth approximation of exact penalty functions has attracted broad interests in scholars [2–8]. In recent years based on the smooth approximation of the exact penalty function, several smooth penalty methods are given to solve (NP). For example, [9] gives a smooth penalty method based on approximating the l1 exact penalty function. Under the assumptions that the optimal solution satisfies MFCQ and the iterate sequence is bounded, it is proved that the iterative sequence will enter the feasible set and every accumulation point is the optimal solution of (NP). In [10, 11], smooth penalty methods are considered based on approximating low-order exact penalty functions. Reference [10] proves the similar results as [9] under very strict conditions (some of them are uneasy to check). The conditions for convergence of the smooth penalty algorithm in [11] are weaker than that in [10], but in [11] it is only proved that the accumulation point of the iterate sequence is a Fritz-John (FJ) point of (NP).In the algorithms given by [9–11], at each iteration a global optimal solution of the smooth penalty problem is needed. As we all know, it is very difficult to find a global optimal point of a nonconvex function. To avoid this difficulty, in this paper we give a smooth penalty algorithm based on the smooth approximation of the l1 exact penalty function. The feature of this algorithm lies in that only a stationary point of the penalty function is needed to compute at each iteration. To prove the convergence of this algorithm, we first establish a generalized Mangasarian-Fromovitz constraint qualification condition (GMFCQ) weaker and more comprehensive than the traditional MFCQ. Under this condition, we prove that the iterative sequence of the algorithm will enter the feasible set of (NP). Moreover, we prove that if the iterative sequence has accumulation points, then each of them is a KKT point of (NP). Finally, we apply this algorithm to solve convex optimization and get better convergence results.The rest of this paper is organized as follows. In the next section, we give a family of smooth penalty functions. In Section3 based on the smooth penalty functions given in Section 2, we propose an algorithm for (NP) and analyze its convergence under the GMFCQ condition. We give an example that satisfies GMFCQ at last in this section.


## 2. Smooth Approximation tol1 Exact Penalty Function

In this section we give a family of penalty functions, which decreasingly approximate thel1 exact penalty function. At first we consider a class of smooth function ϕ:R→R+ with the following properties:(I)
ϕ(·) is a continuously differentiable convex function with ϕ′(0)>0;(II)
limt→-∞ϕ(t)=a, where a is a nonnegative constant;(III)
ϕ(t)≥t, for any t>0;(IV)
limt→+∞(ϕ(t))/t)=1.From (I)–(IV), it follows thatϕ satisfies(V)
0≤ϕ′(t)≤1, for any t∈R, and limt→-∞ϕ′(t)=0,limt→+∞ϕ′(t)=1;(VI)
rϕ(t/r) increases with respect to r>0, for any t∈R;(VII)
rϕ(t/r)↓t+(r↓0), for any t∈R.The following functions are often used in the smooth approximation of thel1 exact penalty function and satisfy properties (I)–(IV).(1)
ϕ(t)=log(1+et).(2)
ϕ(t)=(t+t2+4)/2.(3)
ϕ(t)={et,t≤0;t+1,t>0.We now useϕ(·) to construct the smooth penalty function
(2.1)fβ,r(x)=f(x)+r∑i=1mϕ(βgi(x)r),
where β≥1 is a penalty parameter.By (VII), we easily know whenr→0+,fβ,r(x) decreasingly converges to fβ(x), that is, (2.2)fβ,r(x)=f(x)+r∑i=1mϕ(βgi(x)r)↓f(x)+β∑i=1m(gi(x))+.
Therefore fβ,r(x) smoothly approximates the l1 exact penalty function, where r decreases to improve the precision of the approximation. It is worth noting that the smooth function ϕ(·) and penalty function fβ,r(·) given in this paper make substantive improvement of the corresponding functions given in [9]. This gives fβ,r(·) better convergence properties (refer to (2.2) and Theorem 3.9).


## 3. The Algorithm and Its Convergence

We propose a penalty algorithm for (NP) in this section based on computing the stationary point of fβ,r(·). We assume that for any β≥1 and 0<r≤1,fβ,r(·) always has stationary point.Algorithm
Step 0.
Givenx0∈Rn,β1=1,r1=1,0<η1<1,and η2>1.Let k=1.Step 1.
Findxk such that
(3.1)∇fβk,rk(xk)=0.Step 2.
Putrk+1=η1rk,
(3.2)βk+1={βkifxk∈Ω0,η2βkotherwise.Step 3.
Letk=k+1 and return to Step 1.Let{xk} be the iterative sequence generated by the algorithm. We shall use the following assumption:(A1)
the penalty function value sequence{fβk,rk(xk)} is bounded.Lemma 3.1.
Suppose that the assumption(A1) holds, then for any ɛ>0, there exists k0∈N={1,2,…}, such that for k≥k0,
(3.3)xk∈Ωɛ.Proof.
Suppose to the contrary that there exist anɛ0>0 and an infinite sequence K⊆N, such that for any k∈K,
(3.4)xk∉Ωɛ0.
By the algorithm, we know that
(3.5)βk⟶+∞(k⟶∞).
It follows from (3.4) that there exist a subsequence K0⊆K and an index i0∈I={1,…,m}, such that for any k∈K0,
(3.6)gi0(xk)>ɛ0.
Thus, from the assumptions about f(·), the properties about ϕ(·), (3.5) and (3.6), it follows that
(3.7)fβk,rk(xk)=f(xk)+rk∑i=1mϕ(βkgi(xk)rk)≥f(xk)+rkϕ(βkɛ0rk)≥(ϕ(βkɛ0/rk)βkɛ0/rk)βkɛ0⟶+∞(k⟶∞,k∈K0).
This contradicts with (A1).Lemma 3.2.
Suppose that the assumption(A1) holds, and x* is any accumulation point of {xk}, then x*∈Ω0, that is, x* is a feasible solution of (NP).Proof.
By Lemma3.1, we obtain that for any ɛ>0 and every sufficiently large k,xk∈Ωɛ. Let x* be an accumulation point of {xk}, then there exists a subsequence {xk}k∈K such that xk→x*(k∈K,k→∞). Therefore
(3.8)x*∈Ωɛ.
By the arbitrariness of ɛ>0, we have that x*∈Ω0.Givenx¯∈Ω0, we denote that I(x¯)={i∈I∣gi(x¯)=0}.Definition 3.3 (see [12]).
We say thatx¯∈Ω0 satisfies MFCQ, if there exists a h∈Rn such that
(3.9)∇gi(x¯)Th<0,foranyi∈I(x¯).In the following we propose a kind of generalized Mangasarian-Fromovitz constraint qualification (GMFCQ).LetK⊆N be a subsequence, and for sequence {zk}k∈K in Rn denote two index sets as (3.10)I+(K)={i∈Ilimsup⁡k∈K,k→∞gi(zk)≥0},I-(K)={i∈Ilimsup⁡k∈K,k→∞gi(zk)<0}.Definition 3.4.
We say that the sequence{zk}k∈K satisfies GMFCQ, if there exist a subsequence K0⊆K and a vector h∈Rn such that
(3.11)limsup⁡k∈K0,k→∞∇gi(zk)Th<0,foranyi∈I+(K0).Under some circumstances, the sequence{xk} may satisfy that ∥xk∥→+∞(k→∞), which can be seen for the example in the last part of this section. At this time MFCQ can not be applied, but GMFCQ can. The following proposition shows that Definition 3.4 is a substantive generalization of Definition 3.3.Proposition 3.5.
Suppose that{zk}k∈K satisfies
(3.12)lim⁡k∈K,k→∞zk=z*∈Ω0.
If z* satisfies MFCQ, then {zk}k∈K satisfies GMFCQ.Proof.
By (3.12), we know that limsupk∈K,k→∞gi(zk)≥0 if and only if
(3.13)lim⁡k∈K,k→∞gi(zk)=gi(z*)=0.
Thus, I+(K)=I(z*). By the assumption, there exists a h∈Rn such that
(3.14)limsup⁡k∈K,k→∞∇gi(zk)Th<0,foranyi∈I+(K).We need two assumptions in the following:(A2)
the sequence{∇f(xk)} and {∇gi(xk)},i=1,…,m are both bounded;(A3)
any subsequence of{xk} satisfies GMFCQ.Theorem 3.6.
Suppose that the assumptions(A1), (A2), and (A3) hold, then(1)
there exists ak0 such that for any k≥k0,
(3.15)xk∈Ω0;(2)
any accumulation point of{xk} is a KKT point of (NP).Proof.
If (1) does not hold, that is, there exists a subsequenceK⊆N such that for any k∈K, it holds that
(3.16)xk∉Ω0.
By the algorithm, we know that
(3.17)lim⁡k→∞βk=+∞.
From the assumption (A3) and (3.16), it follows that there exist K0⊆K and h∈Rn such that
(3.18)limsup⁡k∈K0,k→∞∇gi(xk)Th<0,foranyi∈I+(K0),(3.19)I*(K0)={i∈I∣gi(xk)>0,foranyk∈K0}≠∅,I*(K0)⊆I+(K0).
By (3.18) and the definition of I-(K0), there exists a δ>0, such that for all k∈K0,
(3.20)∇gi(xk)Th≤-δ,foranyi∈I+(K0),(3.21)gi(xk)≤-δ,foranyi∈I-(K0).
From the algorithm, we know that xk satisfies
(3.22)∇f(xk)+∑i=1mβkϕ′(βkgi(xk)rk)∇gi(xk)=0.
Let k∈K0, from (3.22) we obtain that
(3.23)∇f(xk)Thβk+∑i∈I-(K0)ϕ′(βkgi(xk)rk)∇gi(xk)Th+∑i∈I+(K0)ϕ′(βkgi(xk)rk)∇gi(xk)Th=0.
We now analyze the three terms on the left side of (3.23).(a)
By (3.17) and (A2),
(3.24)lim⁡k∈K0,k→∞∇f(xk)Thβk=0.(b)
By (3.21), for any i∈I-(K0), we have
(3.25)lim⁡k∈K0,k→∞βkgi(xk)rk=-∞.
From the properties of ϕ(·) and (A2), we have that the second term satisfies(3.26)lim⁡k∈K0,k→∞∑i∈I-(K0)ϕ′(βkgi(xk)rk)∇gi(xk)Th=0.(c)
From (3.19), (3.20), and the properties of ϕ(·), it follows that
(3.27)∑i∈I+(K0)ϕ′(βkgi(xk)rk)∇gi(xk)Th≤-δ∑i∈I+(K0)ϕ′(βkgi(xk)rk)≤-δ∑i∈I*(K0)ϕ′(βkgi(xk)rk)≤-δ|I*(K0)|ϕ′(0),
where |I| denotes the number of the elements in I.Now, by letting k→∞,k∈K0, and taking the limit on both sides of (3.23), we obtain from (a)–(c) that
(3.28)δ|I*(K0)|ϕ′(0)≤0.
But by (3.19) and the properties of ϕ(·),δ|I*(K0)|ϕ′(0)>0. This contradiction completes the proof of (1).
By (1) we know that there exists ak0, such that if k≥k0, then xk∈Ω0. Thus by the algorithm, when k≥k0, we have that
(3.29)βk=βk0.
Suppose that x* is an accumulation point of {xk}, then there exists a subsequence {xk}k∈K, such that
(3.30)lim⁡k∈K,k→∞xk=x*.
By Lemma 3.2, x* is a feasible point of (NP), that is, x*∈Ω0. Thus by (3.22), we obtain that
(3.31)∇f(xk)+∑i∈I∖I(x*)βk0ϕ′(βk0gi(xk)rk)∇gi(xk)+∑i∈I(x*)βk0ϕ′(βk0gi(xk)rk)∇gi(xk)=0.
In the second term of (3.31), because i∈I∖I(x*), so by (3.30) and the properties of ϕ′(·), we have
(3.32)lim⁡k∈K,k→∞ϕ′(βk0gi(xk)rk)=0.
In the third term of (3.31), from the properties of ϕ′(·), the sequence {ϕ′(βk0gi(xk)/rk)},i∈I is nonnegative and bounded. Thus, there exists a subsequence K0⊆K such that
(3.33)lim⁡k∈K0,k→∞βk0ϕ′(βk0gi(xk)rk)=λi≥0,foranyi∈I(x*).
At last by letting k→∞,k∈K0, and taking the limit on both sides of (3.31), we obtain from (3.30)(3.32) and (3.33) that
(3.34)∇f(x*)+∑i∈I(x*)λi∇gi(x*)=0.By Lemma3.2, Proposition 3.5, and Theorem 3.6, we obtain the following conclusion.Corollary 3.7.
Suppose that(A1) holds, {xk} is bounded, and any accumulation point x* of {xk} satisfies MFCQ, then(1)
there exists ak0 such that for any k≥k0(3.35)xk∈Ω0;(2)
any accumulation point of{xk} is a KKT point of (NP).When (NP) is a convex programming problem, that is, the functions f and gi,i∈I of (NP) are all convex functions, the algorithm has better convergence results.Theorem 3.8.
Suppose (NP) is a convex programming problem, then every accumulation point of {xk} is an optimal solution of (NP).Proof.
Sincef(·),gi(·),i∈I are convex, and ϕ(·) is increasing, then for any β>0 and r>0,fβ,r(·) is convex. Thus ∇fβk,rk(xk)=0 is equivalent to
(3.36)xk∈argmin⁡x∈Rn⁡fβk,rk(x).
Therefore by (3.36) and the properties of ϕ(·), we have for any x¯∈Ω0,
(3.37)fβk,rk(xk)=f(xk)+rk∑i=1mϕ(βkgi(xk)rk)≤f(x¯)+rk∑i=1mϕ(βkgi(x¯)rk)≤f(x¯)+rkmϕ(0).
From (3.37), the arbitrariness of x¯∈Ω0 and the nonnegativity of ϕ(·), it follows that
(3.38)f(xk)≤inf⁡x∈Ω0f(x)+rkmϕ(0).
Suppose that x* is an accumulation point of {xk}, there exists a subsequence K⊆N such that limk∈K,k→∞xk=x*. Thus, by (3.38), we have
(3.39)f(x*)≤inf⁡x∈Ω0f(x).
On the other side, (3.37) implies that (A1) holds. Then from Lemma 3.2, we know x*∈Ω0.Theorem 3.9.
Suppose that (NP) is a convex programming problem, and the assumptions (A2), (A3) hold, then(1)
there exists ak0, for any k≥k0,{fβk,rk(xk)} decreases to infx∈Ω0f(x).(2)
limk→∞f(xk)=infx∈Ω0f(x).Proof.
Note that for (NP) which is convex, (A1) holds. By Theorem 3.6 there exists a k0, such that xk∈Ω0 when k≥k0. Therefore from the algorithm, we have for any k≥k0,βk=βk0. By (3.36) and the property (VI) of ϕ(·), when k≥k0,
(3.40)fβk0,rk+1(xk+1)≤fβk0,rk+1(xk)=f(xk)+rk+1∑i=1mϕ(βk0gi(xk)rk+1)≤f(xk)+rk∑i=1mϕ(βk0gi(xk)rk)≤fβk0,rk(xk).
Notice that xk∈Ω0(k≥k0), by (3.37) and the properties of ϕ(·), we have for k≥k0 that
(3.41)inf⁡x∈Ω0f(x)≤f(xk)≤fβk0,rk(xk)≤inf⁡x∈Ω0f(x)+rkmϕ(0).
Combining (3.40) with (3.41), we obtain the conclusion.Example 3.10.
Consider thatminx∈Ω0f(x)=(1/4)(x1-x2)2,Ω0={x∈R2∣g(x)=x1-x2≤0}.
This is a convex case. Denote its optimal solution byΩ0*={x*∈Ω0∣x1*-x2*=0} and let ϕ(t)=(t+t2+4)/2. We consider fβ,r(·), that is,
(3.42)fβ,r(x)=14(x1-x2)2+r2(β2r2(x1-x2)2+4+βr(x1-x2)).
Because fβ,r(·) is convex, thus ∇fβ,r(x*)=0 if and only if x*∈argminx∈R2fβ,r(x). By the algorithm, we get stationary points as
(3.43)xk=(kk+αk),k=0,1,…,
where αk>0 and limk→∞αk=0. Here {xk} has no accumulation point, that is, limk→∞∥xk∥=+∞. Thus in the analysis of convergence, MFCQ may not be appropriate to be applied as a constraint qualification condition for this example. But for any k∈N, we have ∇f(xk)=(-(1/2)αk,(1/2)αk)T, ∇g(xk)=(1,-1)T, which implies that assumption (A2) is satisfied. we can also check that {xk} satisfies GMFCQ. In fact, choose h=(-1,1)T, then we have
(3.44)lim⁡k→∞g(xk)=lim⁡k→∞αk=0,lim⁡k→∞∇g(xk)Th=(1,-1)T(-1,1)=-2<0.
On the other side, by the algorithm, we have xk∈Ω0 and βk=1, for all k. By letting k→∞, we get fβk,rk(xk)↓0 and f(xk)→0. So by the algorithm we get a feasible solution sequence which is also optimal.



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*Source: 620949-2012-02-27.xml*