An affine scaling method using a class of differential barrier functions

TL;DR

This paper introduces a new affine scaling interior point algorithm for linear programming that leverages a broad class of differential barrier functions, demonstrating robustness and efficiency over classical methods.

Contribution

The paper presents a novel affine scaling algorithm based on differential barrier functions, expanding the toolkit for linear programming optimization.

Findings

The proposed algorithm is robust and efficient.

Differential barrier functions offer new perspectives in linear optimization.

Comparison shows advantages over classical affine scaling methods.

Abstract

In this paper we address a practical aspect of differential barrier penalty functions in linear programming. In this respect we propose an affine scaling interior point algorithm based on a large classe of differential barrier functions. The comparison of the algorithm with a vesion of the classical affine scaling algorithm shows that the algorithm is robust and efficient. We thus show that differential barrier functions open up new perspectives in linear optimization.

Tables

Problem	r=0	r=0.1	r=0.2	r=0.3	r=0.4	r=0.5	r=0.6	r=0.7
25fv47	58	59	60	64	66	81	90	$\star \star$
80bau3b	80	79	77	76	78	75	91	$\star \star$
adlittle	34	34	33	33	34	37	40	57
afiro	25	23	23	21	21	20	21	22
agg	45	49	43	62	66	85	129	$\star \star$
agg2	41	41	42	46	48	54	69	$\star \star$
agg3	39	40	43	43	48	56	69	$\star \star$
bandm	46	52	53	60	72	96	143	$\star \star$
beaconfd	35	34	33	32	30	30	30	32
blend	40	41	40	43	44	37	44	49
bnl1	66	67	60	65	75	92	$\star \star$	$\star \star$
bnl2	75	87	91	99	120	148	$\star \star$	$\star \star$
boeing1	62	71	70	71	72	103	149	$\star \star$
boeing2	40	57	45	50	59	70	92	131
bore3d	67	72	80	86	95	121	147	$\star \star$
brandy	36	37	38	39	44	53	66	92
capri	$\star \star$	$\star \star$	$\star \star$	$\star \star$	49	$\star \star$	$\star \star$	$\star \star$
cycle	101	$\star \star$	90	107	118	136	165	$\star \star$
czprob	80	64	61	57	59	61	77	$\star \star$
d2q06c	71	69	69	72	75	103	134	$\star \star$
d6cube	117	87	85	72	68	70	74	82
degen2	44	43	43	37	42	44	49	62
degen3	53	51	50	47	51	49	64	91
dfl001	$\star \star$	$\star \star$	144	166	174	$\star \star$	$\star \star$	$\star \star$
e226	53	53	52	51	69	74	88	$\star \star$
etamacro	49	50	55	65	79	100	$\star \star$	$\star \star$
fffff800	61	56	55	58	68	62	140	212
finnis	78	82	80	85	$\star \star$	$\star \star$	$\star \star$	$\star \star$
fit1d	41	40	66	65	52	76	$\star \star$	$\star \star$
fit1p	31	32	32	41	48	50	73	$\star \star$
fit2d	55	45	43	42	44	58	$\star \star$	$\star \star$
fit2p	44	47	48	51	45	56	73	110
forplan	48	46	46	50	52	55	64	$\star \star$
ganges	27	26	34	29	32	36	44	56
gfrd-pnc	29	30	37	49	65	89	134	$\star \star$
greenbea	$\star \star$	${112}^{\ast }$	105	${99}^{\ast }$	${117}^{\ast }$	${132}^{\ast }$	${181}^{\ast }$	$\star \star$
greenbeb	85	81	84	90	98	112	$\star \star$	$\star \star$
grow7	97	87	83	82	81	79	72	67
grow15	113	97	101	94	88	81	71	66
grow22	112	109	103	89	82	72	63	60
israel	54	55	67	82	91	118	153	$\star \star$
kb2	38	42	36	40	45	61	65	$\star \star$
lotfi	46	46	48	53	58	66	88	106
maros	58	61	63	66	77	101	164	$\star \star$
maros-r7	30	29	29	29	30	32	37	48
modszk1	$\star \star$	58	55	59	58	75	80	$\star \star$
nesm	97	97	92	98	104	115	$\star \star$	$\star \star$
perold	$\star \star$	$\star \star$	116	146	198	288	$\star \star$	$\star \star$
pilot	100	$\star \star$	101	108	121	$\star \star$	$\star \star$	$\star \star$
pilot4	125	144	144	173	219	$\star \star$	$\star \star$	$\star \star$
pilot87	116	118	121	136	178	247	$\star \star$	$\star \star$
pilot_ja	$\star \star$	162	265	$\star \star$	$\star \star$	${227}^{\ast }$	$\star \star$	$\star \star$
pilotnov	47	53	58	70	87	129	$\star \star$	$\star \star$
pilot_we	$\star \star$	$\star \star$	$\star \star$	$\star \star$	${229}^{\ast }$	229	$\star \star$	$\star \star$
qap8	$39$	$36$	$36$	32	34	37	37	38
qap12	59	53	55	54	53	53	55	54
qap15	${219}^{\ast }$	${168}^{\ast }$	66	61	58	61	${191}^{\ast }$	68
recipe	38	38	37	38	38	33	31	48
sc105	26	27	27	28	26	29	29	32
sc205	35	37	31	34	35	40	47	63
sc50a	33	32	23	23	23	22	23	25
sc50b	23	23	22	21	21	21	20	21
scagr25	31	32	32	34	36	39	45	63
scagr7	37	29	31	31	32	42	39	52
scfxm1	44	44	45	48	53	61	83	118
scfxm2	46	47	51	54	62	74	95	148
scfxm3	45	45	50	54	64	81	102	146
scorpion	45	44	46	44	46	$\star \star$	$\star \star$	$\star \star$
scrs8	$\star \star$	$\star \star$	$\star \star$	$\star \star$	107	134	190	$\star \star$
scsd1	47	45	44	41	41	37	40	41
scsd6	53	47	46	43	41	41	32	34
scsd8	42	41	39	38	37	37	29	30
sctap1	49	49	51	51	55	52	62	73
sctap2	$52$	51	48	48	49	46	54	67
sctap3	56	49	47	46	48	50	59	73
seba	47	48	47	43	45	56	62	99
share1b	59	66	73	113	119	145	141	$\star \star$
share2b	37	39	38	38	29	33	33	50
shell	51	48	47	45	46	48	54	67
ship04l	52	50	49	47	46	37	37	42
ship04s	52	50	48	46	36	45	35	39
ship08l	49	48	48	45	36	37	34	38
ship08s	52	50	49	48	46	39	37	50
ship12l	51	49	48	46	44	45	49	$\star \star$
ship12s	49	48	46	46	40	40	49	53
sierra	52	51	51	54	68	86	141	$\star \star$
stair	46	42	34	45	41	46	59	99
standata	56	63	57	57	63	63	75	89
standgub	56	54	53	57	55	58	73	88
standmps	71	66	64	65	68	69	81	101
stocfor1	43	43	43	45	49	48	56	73
stocfor2	70	72	78	89	99	116	${162}^{\ast }$	${232}^{\ast }$
stocfor3	49	46	47	46	47	48	51	56
truss	49	46	47	46	47	48	51	56
tuff	50	47	45	44	53	52	57	75
vtp.base	46	46	50	58	73	94	90	$\star \star$
wood1p	$\star \star$	76	63	60	$\star \star$	$\star \star$	50	41
woodw	79	71	64	61	$\star \star$	$\star \star$	$\star \star$	$\star \star$

Equations

$$\min \{F(x):Ax=b,x\ge 0\}$$

$$\xi_{r}:x\mapsto\left\{\begin{array}[]{ll}\big{(}\sum x_{i}^{r}\big{)}^{1\over r}&\mbox{if }x\in[0,+\infty)^{n},\cr-\infty&\mbox{elsewhere,}\end{array}\right.$$

$$[0,+\infty {)}^n=\left\{x\in {\mathbb{R}}^n:{\xi }_r(x)\ge 0\right\}.$$

$$\min \{⟨c,x⟩:Ax=b,{\xi }_r(x)\ge 0\}.$$

$$g_{r}:x\mapsto\left\{\begin{array}[]{ll}-{1\over r}\big{(}\xi_{r}(x)\big{)}^{r}&\mbox{if }x\in[0,+\infty)^{n},\cr+\infty&\mbox{elsewhere,}\end{array}\right.$$

$$F_{r,\mu}(x)=\left\{\begin{array}[]{ll}\left\langle c,x\right\rangle+\mu g_{r}(x)&\mbox{if }x\in[0,+\infty)^{n},\cr+\infty&\mbox{elsewhere.}\end{array}\right.$$

$$\min \{\lambda :Ax+\lambda \left(b-A{x}^k\right)=b,x\in [0,+\infty {)}^n\text{mbox}and\lambda \in [0,+\infty )\}.$$

$$s=c-A^{t}y+w,w_{\mathcal{I}}=-U_{\mathcal{I}}^{-1}{X}_{\mathcal{I}}(c-A^{t}y{)}_{\mathcal{I}}\text{mbox}and{w}_{\overline{\mathcal{I}}}=0$$

$$\xi_{r}:x\mapsto\left\{\begin{array}[]{ll}\left(\sum x_{i}^{r}+\sum\limits_{i\in\cal U}(u_{i}-x_{i})^{r}\right)^{1\over r}&\mbox{if }x\geq 0\mbox{ and }u_{i}-x_{i}\geq 0,\ \forall i\in{\cal I},\cr-\infty&\mbox{elsewhere.}\end{array}\right.$$

$$\xi_{0}(x)=\displaystyle\left\{\begin{array}[]{ll}\left(\prod\limits_{i\in\{1,\cdots,n\}}x_{i}\prod\limits_{i\in{\cal I}}(u_{i}-x_{i})\right)^{1\over n+n_{\cal I}}&\mbox{if }x\in[0,+\infty)^{n}\cr-\infty&\mbox{elsewhere}.\end{array}\right.$$

$$\min \left\{\frac{1}{2}⟨{\nabla }^2{F}_{r,\mu }(x)d,d⟩+⟨\nabla F_{r,\mu }(x),d⟩:Ad=0\right\}.$$

$$\left\{\begin{array}[]{*{1}c@{\;=\;}c}\nabla^{2}F_{r,\mu}(x)d(\mu)+\nabla F_{r,\mu}(x)+A^{t}y&0\cr\hfill Ad(\mu)&0.\end{array}\right.$$

$$G_{ii}=\left\{\begin{array}[]{*{1}c@{\; \;}c}x_{i}^{r-1}&\mbox{ if }i\notin{\cal I},\\ x_{i}^{r-1}-(u_{i}-x_{i})^{r-1}&\mbox{ otherwise }\end{array}\right.$$

$$H_{ii}=\left\{\begin{array}[]{*{1}c@{\; \;}c}x_{i}^{r-2}&\mbox{ if }i\notin{\cal I},\\ x_{i}^{r-2}+(u_{i}-x_{i})^{r-2}&\mbox{ otherwise. }\end{array}\right.$$

$$P=I-H^{-{1\over 2}}A^{t}\big{(}AH^{-1}A^{t}\big{)}^{-1}AH^{-{1\over 2}},$$

$$d=\underset{\mu \downarrow 0}{\lim }\mu (1-r)d(\mu )=-H^{-\frac{1}{2}}P{H}^{-\frac{1}{2}}c$$

$$\min \left\{⟨\tilde{c},\tilde{x}⟩:\tilde{A}\tilde{x}=b,\tilde{x}\ge 0,{\tilde{x}}_i\le u_i,\forall i\in \mathcal{I}\right\},$$

$$\begin{array}[]{ll}{\tilde{P}}&=\left(\matrix{P-\delta\lambda^{2-r}H^{-1\over 2}A^{t}ww^{t}AH^{-1\over 2}&\delta\lambda^{1-{r\over 2}}H^{-1\over 2}A^{t}w\cr&\cr\delta\lambda^{1-{r\over 2}}w^{t}AH^{-1\over 2}&-\delta\cr}\right)\cr&\cr&=\left(\matrix{P&0\cr&\cr 0&0\cr}\right)-\delta\left(\matrix{\lambda^{1-{r\over 2}}H^{-1\over 2}A^{t}w\cr&\cr-1\cr}\right)\left(\matrix{\lambda^{1-{r\over 2}}w^{t}AH^{-1\over 2}&&-1\cr}\right).\end{array}$$

$$\min \left\{\lambda :Ax+\lambda \left(b-A{x}^k\right)=b,x\ge 0,x_i\le u_i,\forall i\in \mathcal{I},\lambda \ge 0\right\}.$$

$$d_x^k=-H^{-1}{A}^t{\left(A{H}^{-1}{A}^t\right)}^{-1}(b-A{x}^k)$$

$${\Vert X^{1-\frac{r}{2}}P{X}^{1-\frac{r}{2}}c\Vert }_2\le L(A,c)⟨c,X^{1-\frac{r}{2}}P{X}^{1-\frac{r}{2}}c⟩.$$

$${\Vert {\left(A{X}^2{A}^t\right)}^{-1}A{X}^{2}p\Vert }_2\le q(A){\Vert p\Vert }_2,$$

$$0<x_{i_{r^{\prime }}}^{1-r}|s_{i_{r^{\prime }}}|\le \Vert X^{1-r}s{\Vert }_{\infty }=x_{i_r}^{1-r}|s_{i_r}|(1)$$

$$0<x_{i_r}^{1-r^{\prime }}|s_{i_r}|\le \Vert X^{1-r^{\prime }}s{\Vert }_{\infty }=x_{i_{r^{\prime }}}^{1-r^{\prime }}|s_{i_{r^{\prime }}}|(2)$$

$$0<{\overline{L}}^a={\left(1-\frac{L(A,c{)}^{\frac{6-r}{2-r}}}{2^{\frac{3-r}{2-r}}{g}^{\frac{6-r}{2-r}}{n}_I^{\frac{7}{2}-\frac{r(1-r)}{2(2-r)}}}\right)}^a<1,\forall a>0.$$

$$\Vert \overline{s}{\Vert }_2\Vert x_I^k{\Vert }_2\ge ⟨\overline{s},x^k⟩=⟨c,x^k⟩-\overline{c}\ge \frac{1}{h}\Vert x^k-\overline{x}{\Vert }_2\ge \frac{1}{h}\Vert x_I^k{\Vert }_2.$$

$$F(x)=\ln\left(\displaystyle{\langle c,x\rangle-{\overline{c}}\over\tilde{\xi}_{r}(x)}\right)\mbox{, where }{\tilde{\xi}_{r}}:x\mapsto\left\{\begin{array}[]{cc}\left(\sum\limits_{i\in I}x_{i}^{r}\right)^{1\over r}&\mbox{if }x\geq 0,\cr-\infty&\mbox{elsewhere.}\end{array}\right.$$

$$F(x^{k+1})-F(x^k)\le -{\theta }^k\left({{\rm Y}}^k+(1-{{\rm Y}}^k)\frac{2-3\beta }{3(1-\beta )}\right)\Vert X_{k,I}^{\frac{r}{2}}{w}_I^k{\Vert }_2^2-(1+{{\rm Y}}^k)\frac{{\theta }^k}{\sum _{j\in I}{x^k}^r}{\delta }^k-{\theta }^k{\gamma }^k$$

$$\underset{y,s}{\max }\{{\xi }_{t,I}(s_I):A^{t}y+s=c,s_J=0,s\ge 0\}$$

$$\xi_{t,I}(x)=\left\{\begin{array}[]{ll}\left(\sum\limits_{i\in I}x_{i}^{t}\right)^{1\over t}&\mbox{if }x_{I}\geq 0\cr-\infty&\mbox{elsewhere,}\end{array}\right.$$

Full text

Key words: Barrier, concave gauge, differential barrier, interior point methods, linear programs, primal algorithm.

AMS Subject Classifications: 90C05, 90C51, 49M30, 49N15

1 Introduction

In this paper we present an algorithm based on a family of penalty functions introduced in [1]. Contrary to the classical logarithmic barrier function, these functions are not necessarily barriers, since they can be well defined on the positive orthant including its boundary. But they are differentially barriers. In fact, these functions generalize the notion of barrier functions since (Proposition 17 of [1]) a barrier function is in particular a differential barrier one. We recall that (Definition 1 of [1]) a function $F$ is said to be a differential barrier on the positive orthant ${\cal P}=[0,+\infty)^{n}$ if $F$ is differentiable on $(0,+\infty)^{n}$ and $\limsup\limits_{{x\to x^{\prime}}\atop{x>0}}||\nabla F(x)||=+\infty$, for every $x^{\prime}$ being on the boundary of ${\cal P}$. So $\nabla F$ plays the role of a barrier. Also, the fact that a method based on the minimization of a penalty function is of interior points type is closely related to the following property.

Proposition 1

(Proposition 18 of [1])

Let $F$ be a convex, lower semi-continuous and differential barrier function on ${\cal P}$. Then every optimal solution ${\overline{x}}$ to the problem

[TABLE]

is an interior point of the positive orthant.

Through the example of the concave gauge functions111A background on concave gauge functions is given in [1] and a complete description is done in [2] we will consider, we will show the important role that penalty functions of the differential barrier type can play as an alternative to the classical logarithmic barrier function. In this respect we consider the familiy of differential barrier functions builded from the following concave gauge functions:

[TABLE]

where $r$ is taken arbitrary in $(0,1)$. To be more precise, let us consider the linear program given by

$\hfill\min\{\left\langle c,x\right\rangle:Ax=b,\ x\in[0,+\infty)^{n}\ \}\hfill(LP)$

where $A$ is an $m\times n$ matrix of rank $m$, $c$, $x\in\mathbb{R}^{n}$ and $b\in\mathbb{R}^{m}$. By definition of a concave gauge function the positive cone can be expressed as

[TABLE]

Hence the original linear program can be equivalently rewritten as

[TABLE]

Applying the approach developed in [1], we propose to penalize the constraint $\xi_{r}(x)\geq 0$ by the functions

[TABLE]

So the nonlinear optimization problem approximating the linear program222We recall that the idea to approximate a linear program by a nonlinear optimization problem is du originally to Courant [3] in 1941 with a penalty function of exterior type and later to Frisch [4], in 1955, when he introduced the logarithmic barrier function which is an interior penalty one. We recall also that the notion of interior penalty operators were introduced by Auslender [5] in 1976 to generalize the concept of barrier functions. is as follow

$\hfill\min\{F_{r,\mu}(x):\ Ax=b\}\hfill(P_{\mu,r})_{\mu>0}$

where

[TABLE]

It is easy to see that $F_{r,\mu}$ is a differential barrier function and then \Big{(}Proposition 1\Big{)} the optimal solution of $(P_{\mu,r})$ belongs to the interior of the positive orthant.

The algorithm we build, called galpv4, is of primal type and uses an affine scaling approach333An affine scaling algorithm was originally proposed by Dikin in 1967 [6]. It was rediscovered by several researchers such as barnes [7] and Vanderbei et al [8] after Karmarkar [9] proposed his famous projective scaling algorithm.. It consists of two combined phases. The first one improves the feasibility of the current point and the second brings the point closer to an optimal solution. At each iteration, this requires the computation of two directions. The direction $d^{k}$, bringing a current point $x^{k}$ closer to the optimal solutions’ set is obtained as follows. We compute at $x^{k}$ the Newton direction $d^{k}(\mu)$ for the problem $(P_{\mu,r})$. Vector $d^{k}$ is then the part of the expression of $d^{k}(\mu)$, independent of parameter $\mu$ and satisfies $\delta_{\mu}d^{k}(\mu)=d^{k}+O(\mu)$, where $\delta_{\mu}$ is a positive real function of $\mu$. That is $d^{k}=\lim\limits_{\mu\downarrow 0}\delta_{\mu}d^{k}(\mu)$. The direction $d^{\prime k}$ that improves the feasibility of the current point is obtained by using the same process with the linear program

[TABLE]

We show that the sequence $(x^{k})$ converges. Its limit is an interior point of the optimal solutions’ face of the linear program when $\beta\in\left(0,\displaystyle{2\over 3}\right)$, where $\beta$ is the factor of the maximal step size with respect to $x^{k}$ and $d^{k}$. Moreover, by calculating $d^{k}$, the algorithm generates a sequence $(y^{k},s^{k})$ that converges to the $\xi^{\oplus}_{r}$-analytic center of the dual optimal solutions’ face, where $\xi_{r}^{\oplus}$ is the polar concave gauge function (see [2]) of $\xi_{r}$.

In Proposition 2 of Section 2, we show that galpv4 includs the classical affine scaling approach by setting $r=0$. In this respect, we compare the algorithm’s performances between different values of $r\in[0,1)$ through numerical experiments using the familiar netlib test set [10].

The paper is organized as follows. In section 2 we present the algorithm, the computation of an affine scaling direction and how to find approximately a relative interior feasible solution. Section 3 deals with the convergence results and a stopping criteria, followed by numerical results and comments in Section 4. Finally, we close the paper by some concluding remarks in Section 5.

2 Presentation of a primal affine scaling method

In order to take account of possible bound constraints, we consider in all the following, the linear program

$\hfill\min\{\left\langle c,x\right\rangle:Ax=b,\ x\geq 0,\ x_{i}\leq u_{i}\ i\in{\cal I}\}\hfill(LPB)$

where ${\cal I}$ is a subset to $\{1,2,\cdots,n\}$ and $u\in\mathbb{R}^{n}$ is given such that $u_{i}>0$ if $i\in{\cal I}$ and $u_{i}=+\infty$ if not. It is easy to see that the dual problem of $(LPB)$ is

$\hfill\max\{\langle b,y\rangle-\langle u_{\cal I},w_{\cal I}\rangle:\ A^{t}y+s-w=c,\ w_{\overline{\cal I}}=0,s,w\geq 0\}\hfill(LDB)$

where $\overline{\cal I}=\{1,2,\cdots,n\}-{\cal I}$. Moreover if $(x,y,s,w)$ is a primal-dual optimal solution then it is easy to see from the KKT optimality conditions that

[TABLE]

where $U_{\cal I}=diag\left(u_{\cal I}\right)$ and $X_{\cal I}=diag\left(x_{\cal I}\right)$. We assume that there exists a relative interior feasible solution to $(LPB)$ and that the minimum is finite. Hence the optimal solutions’ set of $(LPB)$ and of $(LDB)$ are non empty, and there is no duality gap. Moreover the set of optimal solutions to $(LDB)$ is compact. Now taking account of the slack variables $u_{i}-x_{i}$, we adapt the definition of $\xi_{r}$ as

[TABLE]

The following algorithm, called galpv4, uses an approach based on a version of the classical affine scaling algorithm presented in [7, 8, 11].

$\underline{Algorithm}$ galpv4

$\underline{Initialization}$

Construct a starting point $x$ as described just bellow and choose $r\in[0,1)$.

Compute $y$, $w$, $s$ according to (8), (9) and (10) respectively.

Compute the expected relative duality gap $Rgap$ according to (11)

Set the feasibility measure $Rf\leftarrow\displaystyle{\|Ax-b\|_{\infty}\over\|b\|_{\infty}+1}$

Choose $\epsilon>0$ a stopping rule parameter.

While $\min(Rf,Rgap)>\epsilon$ do

Compute $d_{x}$ the feasible direction according to (7)

Compute $d$ the descent direction according to (6)

Set $t_{max}\leftarrow\min\left(\min\limits_{d_{x_{i}}<0}-\displaystyle{x_{i}\over d_{x_{i}}},\min\limits_{{d_{x_{i}}>0},{i\in{\cal I}}}-\displaystyle{u_{i}-x_{i}\over d_{x_{i}}},1\right)$

If $Rf>\epsilon$ then $t\leftarrow 0.95t_{max}$ else set $t\leftarrow 0.65t_{max}$

Set $x\leftarrow x+td_{x}$

Set $t_{max}\leftarrow\min\left(\min\limits_{d_{i}<0}-\displaystyle{x_{i}\over d_{i}},\min\limits_{{d_{i}>0},{i\in{\cal I}}}-\displaystyle{u_{i}-x_{i}\over d_{i}}\right)$

If $Rf>\epsilon$ then $t\leftarrow 0.65t_{max}$ else set $t\leftarrow 0.95t_{max}$

Update $x\leftarrow x+td$

Update $y$, $w$, $s$ according to (8), (9) and (10) respectively.

Update the expected relative duality gap $Rgap$ according to (11)

Update $Rf\leftarrow\displaystyle{\|Ax-b\|_{\infty}\over\|b\|_{\infty}+1}$

End while

Let us describe now how to construct, empirically, a starting point. In fact we construct two starting points $x^{1}$ and $x^{2}$. The first one is defined as follow. $\mbox{For }j=1,2,..,n$, $x_{j}^{1}=\min\left(\displaystyle{n\over\left\|A_{j.}\right\|},0.9u_{j}\right)$ if $c_{j}<0$ and $x_{j}^{1}=\min\left(\displaystyle{n\over\left\|A_{j.}\right\|},0.1u_{j}\right)$ otherwise, where $A_{j.}$ is the $j^{th}$ column of matrix $A$. The second one is defined as in the routine $pcinit.f$ of the software HOPDM of Gondzio [12]. Our starting point $x$ is chosing as follow. If $\min x^{2}_{i}>\min\limits_{j}x_{j}^{1}$ or $\min\limits_{j}x_{j}^{1}<1$ then we set $x=x^{2}$ else we set $x=x^{1}$.

Note that the algorithm can be extended to the case $r=0$. It is justified by the following proposition.

Proposition 2

Set $n_{\cal I}=card({\cal I})$ and define $\xi_{0}$ as

[TABLE]

For any $r\in(-\infty,0)\cup(0,1)$ we set ${\tilde{\xi}_{r}}={1\over(n+n_{\cal I})^{1\over r}}\xi_{r}$. Then for every $x\in(0,+\infty)^{n}$,

(i)* $\lim\limits_{r\uparrow 0}{\tilde{\xi}_{r}}(x)=\lim\limits_{r\downarrow 0}{\tilde{\xi}_{r}}(x)={1\over n+n_{\cal I}}\xi_{0}(x)$,*

(ii)* $\lim\limits_{r\uparrow 0}\nabla{\tilde{\xi}_{r}}(x)=\lim\limits_{r\downarrow 0}\nabla{\tilde{\xi}_{r}}(x)={1\over n+n_{\cal I}}\nabla\xi_{0}(x)$,*

(iii)* $\lim\limits_{r\uparrow 0}\nabla^{2}{\tilde{\xi}_{r}}(x)=\lim\limits_{r\downarrow 0}\nabla^{2}{\tilde{\xi}_{r}}(x)={1\over n+n_{\cal I}}\nabla^{2}\xi_{0}(x)$.*

Proof. (i). Without loss of generality we can assume that ${\cal I}=\emptyset$. Let $x\in(0,+\infty)^{n}$ and set $\psi_{1}(r)=\ln({1\over n}\Sigma x_{i}^{r})\hbox{ and }\psi_{2}(r)=r.$ We have $\lim\limits_{r\to 0}\psi_{1}(r)=\lim\limits_{r\to 0}\psi_{2}(r)=0\hbox{ and }\psi^{\prime}_{2}(r)=1\not=0.$ Then by the classical Hôpital theorem $\lim\limits_{r\to 0}\ln{\tilde{\xi}_{r}}(x)=\displaystyle\lim\limits_{r\to 0}{\psi_{1}(r)\over\psi_{2}(r)}=\displaystyle\lim\limits_{r\to 0}{\psi^{\prime}_{1}(r)\over\psi^{\prime}_{2}(r)},$ but $\psi^{\prime}_{1}(r)=\displaystyle{\sum\limits_{i=1}^{n}x_{i}^{r}\ln x_{i}\over\sum\limits_{i=1}^{n}x_{i}^{r}}.$ The result follows.

(ii) and (iii). Using (i) and the expressions of $\nabla{\tilde{\xi}_{r}}$ and $\nabla^{2}{\tilde{\xi}_{r}}$, it is easy to see that $\lim\limits_{r\to 0}\nabla{\tilde{\xi}_{r}}(x)=\nabla{1\over n}\xi_{0}(x)\hbox{ and }\lim\limits_{r\to 0}\nabla^{2}{\tilde{\xi}_{r}}(x)=\nabla^{2}{1\over n}\xi_{0}(x).$ ∎

2.1 Finding a descent direction

Let $x$ be a relative interior feasible point to $(LPB)$, $\mu>0$ and $r\in(0,1)$. The Newton direction at $x$ to the penalized problem $(P_{\mu,r})$ is obtained by solving the minimization problem

[TABLE]

Using the KKT optimality conditions, the problem amounts to finding $d(\mu)\in\mathbb{R}^{n}$ and $y\in\mathbb{R}^{m}$ solutions to the system of linear equations

[TABLE]

We have $\nabla F_{r,\mu}(x)=c-\mu Ge\mbox{ and }\nabla^{2}F_{r,\mu}(x)=\mu(1-r)H$ where $e^{t}=(1,1,..,1)\in\mathbb{R}^{n}$, $G$ and $H$ are diagonal matrices defined respectively by

[TABLE]

and

[TABLE]

Then setting

[TABLE]

the projection matrix on the kernel of $AH^{-{1\over 2}}$, system (2) reduces to $PH^{-{1\over 2}}\nabla F_{r,\mu}(x)+\mu(1-r)H^{1\over 2}d(\mu)=0$ and then $\mu(1-r)d(\mu)=-H^{-{1\over 2}}PH^{-{1\over 2}}\nabla F_{r,\mu}(x)=-H^{-{1\over 2}}PH^{-{1\over 2}}c+\mu H^{-{1\over 2}}PH^{-{1\over 2}}Ge.$ Since $\mu(1-r)>0$ we can so take as an affine scaling direction at $x$ to the linear program vector $d$ given by

[TABLE]

Observe that since $\langle c,d\rangle=-\|PH^{-{1\over 2}}c\|_{2}^{2}<0$, $d$ is a descent direction to the linear program at every point to $\mathbb{R}^{n}$.

Remark: To improve the quality of the direction $d$, in order to maintain a good feasibility to the current point, we can compute in addition the direction $H^{-{1\over 2}}PH^{{1\over 2}}d$ which can be used instead of $d$. Which in fact amounts to projecting a second time the direction $H^{{1\over 2}}d$ onto the vector subspace $\ker AH^{-{1\over 2}}$. Of course, theoretically the two directions are identical, but numerically there is a significant difference. However the computation of $H^{-{1\over 2}}PH^{{1\over 2}}d$ has some extra cost in number of operations. Therefore we use the technic only when the relative duality gap is less than 0.001 or the current number of iterations exceeds 20.

2.2 Finding a feasible solution

It is well known that an approximate relative interior feasible solution to $(LPB)$ can be obtained by solving a linear problem of the form

$\hfill\min\left\{\lambda:Ax+\lambda(b-Ax^{0})=b,\ x\geq 0,\ x_{i}\leq u_{i},\ \forall i\in{\cal I},\ \lambda\geq 0\right\},\hfill(FLP)$

where $x^{0}$ is a point arbitrarily chosen in $(0,+\infty)^{n}$. Write $(FLP)$ as

[TABLE]

where {\tilde{x}}=\left(\begin{array}[]{l}x\cr\lambda\end{array}\right),\ {\tilde{c}}=\left(\begin{array}[]{l}0_{\mathbb{R}^{n}}\cr 1\end{array}\right)\mbox{ and }{\tilde{A}}=\left(\matrix{A&b-Ax^{0}\cr}\right). Then using (6), the affine scaling direction ${\tilde{d}}$ with respect to ${\tilde{x}}$ is given by ${\tilde{d}}=-{\tilde{H}}^{-{1\over 2}}{\tilde{P}}{\tilde{H}}^{-{1\over 2}}{\tilde{c}}$ where {\tilde{H}}=\left(\matrix{H&0\cr 0&\lambda^{r-2}\cr}\right)\mbox{ and }{\tilde{P}}=I-{\tilde{H}}^{-{1\over 2}}{\tilde{A}}^{t}\left({\tilde{A}}{\tilde{H}}^{-1}{\tilde{A}}^{t}\right)^{-1}{\tilde{A}}{\tilde{H}}^{-{1\over 2}}.

But the matrix ${\tilde{A}}{\tilde{H}}^{-1}{\tilde{A}}^{t}$ will be generally dense when there is one dense column in ${\tilde{A}}$. Column $b-Ax^{0}$, in most cases, is dense. So for large-scale applications, we split such column from the others. We proceed as follow. Set $v=b-Ax^{0}$. Then ${\tilde{A}}{\tilde{H}}^{-1}{\tilde{A}}^{t}=AH^{-1}A^{t}+\lambda^{2-r}vv^{t}.$ Using the Sherman-Morrison formula we have $\left({\tilde{A}}{\tilde{H}}^{-1}{\tilde{A}}^{t}\right)^{-1}=\left(AH^{-1}A^{t}\right)^{-1}+\delta\lambda^{2-r}ww^{t},$ where $w=\left(AH^{-1}A^{t}\right)^{-1}v$ and $\delta=\displaystyle{-1\over 1+\lambda^{2-r}\left\langle w,v\right\rangle}.$ So

[TABLE]

It follows that {\tilde{d}}=-\delta\lambda^{2-r}\left(\matrix{H^{-1}A^{t}w\cr\cr&\cr-1\cr}\right). Since $-\delta\lambda^{2-r}>0$ the search directions with respect to $x$ and $\lambda$ can be expressed respectively as $d_{x}=-H^{-1}A^{t}\left(AH^{-1}A^{t}\right)^{-1}(b-Ax^{0})\mbox{ and }d_{\lambda}=-1.$ But numerical experiments show that as iterations go, the constraint $Ax+\lambda\left(b-Ax^{0}\right)=b$ is less and less satisfied. This is due to the rounding off errors generated by the projection onto $\ker\tilde{A}$ at each iteration and thus creating a snowball effect. To work around this problem, we proceed as follows: Let $x^{k}$ be a current point in the feasibility searching phase. Then \left(\begin{array}[]{l}x^{k}\cr 1\end{array}\right) is a feasible point of problem

[TABLE]

In this case the search direction with respect to $x^{k}$ is

[TABLE]

It follows that the point \left(\begin{array}[]{l}x^{k+1}\cr\lambda^{k+1}\end{array}\right)=\left(\begin{array}[]{l}x^{k}\cr 1\end{array}\right)+t^{k}\left(\begin{array}[]{l}{d_{x}^{k}}\cr-1\end{array}\right) for a step size $t^{k}$ suitably chosen, does not suffer from the snowball effect mentioned above.

Remark: To compute $\left(AH^{-1}A^{t}\right)^{-1}\left(b-Ax\right)$ and $PH^{-{1\over 2}}c$ we solve for $w$ and $\Delta$ by Cholesky factorization the linear systems $AH^{-1}A^{t}w=b-Ax$ and $\left(AH^{-1}A^{t}\right)\Delta=H^{-{1\over 2}}c$ and then we compute $PH^{-{1\over 2}}c=H^{-{1\over 2}}c-H^{-{1\over 2}}A^{t}\Delta$.

3 Convergence, dual solution and stopping criteria

Without loss of generality we can assume in this section that ${\cal I}=\emptyset$. In this case $H=X^{r-2}\mbox{,}\ P=I-X^{1-{r\over 2}}A^{t}(AX^{2-r}A^{t})^{-1}AX^{1-{r\over 2}},$ where $X=diag(x)$ and $x\in(0,+\infty)^{n}$. To simplify we assume that the starting point $x^{0}$ is a relative interior feasible solution to the linear program. So we consider $(x^{k})_{k\in\mathbb{N}}$ the sequence defined by $x^{k+1}=x^{k}+\beta t^{k}_{max}d^{k},$ where $\beta\in(0,1)$ and $t^{k}_{max}$ is the maximum step length with respect to $x^{k}$ and $d^{k}=-X_{k}^{1-{r\over 2}}PX_{k}^{1-{r\over 2}}c=-X_{k}^{2-r}c+X_{k}^{2-r}A^{t}(AX_{k}^{2-r}A^{t})^{-1}AX_{k}^{2-r}c$. Set $y^{k}=\left(AX_{k}^{2-r}A^{t}\right)^{-1}AX_{k}^{2-r}c\mbox{ and }s^{k}=c-A^{t}y^{k}.$ Here is our main result.

Theorem 3.1

Assume that $\displaystyle 0<\beta<{2\over 3}$. Then $\left(x^{k},y^{k},s^{k}\right)_{k\in\mathbb{N}}$ converges to $\left(\overline{x},{\overline{y}},{\overline{s}}\right)$, where $\left({\overline{y}},{\overline{s}}\right)$ is the $\xi_{t}$-analytic center to the dual optimal face of the linear program, $t$ is such that $\displaystyle{1\over t}+{1\over r}=1$ and ${\overline{x}}$ belongs to the relative interior of the primal optimal face of the linear program.

Before giving the proof of the theorem, we first establish some preliminary results.

Proposition 3

$\sum\limits_{k=0}\limits^{\infty}\beta t^{k}_{max}\left\|PX_{k}^{1-{r\over 2}}c\right\|_{2}^{2}$* is a converging series.*

Proof. We have $\left\langle c,x^{k+1}\right\rangle-\left\langle c,x^{k}\right\rangle=\beta t^{k}_{max}\left\langle c,d^{k}\right\rangle=-\beta t^{k}_{max}\left\|PX_{k}^{1-{r\over 2}}c\right\|_{2}^{2}.$ The sequence $\left(\left\langle c,x^{k}\right\rangle\right)_{k\in\mathbb{N}}$ is then decreasing. Since we assumed that the optimal value of the linear program is finite, the sequence is bounded and then converges. Set ${\overline{c}}$ its limit. Then we have $\sum\limits_{k=0}\limits^{\infty}\beta t^{k}_{max}\left\|PX_{k}^{1-{r\over 2}}c\right\|_{2}^{2}=\left\langle c,x^{0}\right\rangle-{\overline{c}}<+\infty.$ The result then follows. ∎

Now let us recall an important result. It was proved by Monteiro et al. [13], Saigal [11], Tseng and Luo [14] and Tsuchiya [15].

Theorem 3.2

There exists a constant $L(A,c)>0$ such that every optimal solution ${\overline{w}}$ to the following ellipsoidal problem

$\hfill\max\left\{\left\langle c,w\right\rangle:\ Aw=0,\ \left\|X^{-1}w\right\|_{2}^{2}\leq 1\right\}\hfill(EP)$**

satisfies $\left\|{\overline{w}}\right\|_{2}\leq L(A,c)\left\langle c,{\overline{w}}\right\rangle$.

Corollary 1

Let $x\in(0,+\infty)^{n}$. Then $X^{1-{r\over 2}}PX^{1-{r\over 2}}c$ satisfies

[TABLE]

Proof. First observe that $\displaystyle X^{1-{r\over 2}}PX^{1-{r\over 2}}c\over\displaystyle\left\|PX^{1-{r\over 2}}c\right\|_{2}$ can be viewed as the optimal solution to the following ellipsoidal problem

$\hfill\max\left\{\left\langle c,w\right\rangle:\ Aw=0,\ \left\|X^{{r\over 2}-1}w\right\|_{2}^{2}\leq 1\right\}\hfill(EP_{r})$

Hence using Theorem 3.2 by considering ${\tilde{X}}=X^{1-{r\over 2}}$ instead of $X$, the result follows. ∎

Proposition 4

$(x^{k})_{k\in\mathbb{N}}$* is a converging sequence, say to ${\overline{x}}$. Furthermore, for each $k\in\mathbb{N}$, $\left\|x^{k}-{\overline{x}}\right\|_{2}\leq h\left(\left\langle c,x^{k}\right\rangle-{\overline{c}}\right),$ where $\displaystyle h={1\over L(A,c)}$.*

Proof. By Corollary 1 we have $\left\langle c,x^{k}\right\rangle-\left\langle c,x^{k+1}\right\rangle=-\beta t^{k}_{max}\left\langle c,d^{k}\right\rangle\geq L(A,c)\left\|\beta t^{k}_{max}d^{k}\right\|_{2}=L(A,c)\left\|x^{k+1}-x^{k}\right\|_{2}.$ It follows that $+\infty>\left\langle c,x^{0}\right\rangle-{\overline{c}}=\sum\limits_{0\leq k<+\infty}\left\langle c,x^{k}-x^{k+1}\right\rangle\geq L(A,c)\sum\limits_{0\leq k<+\infty}\left\|x^{k+1}-x^{k}\right\|_{2}.$ Thus $(x^{k})_{k\in\mathbb{N}}$ converges. Now using again Corollary 1 we have $\displaystyle\left\langle c,x^{k}\right\rangle-{\overline{c}}=\sum\limits_{j=0}^{\atop\infty}\left\langle c,x^{k+j}-x^{k+j+1}\right\rangle\geq{1\over h}\sum\limits_{j=0}^{\atop\infty}\left\|x^{k+j}-x^{k+j+1}\right\|_{2}\geq\displaystyle{1\over h}\left\|\sum\limits_{j=0}^{\atop\infty}x^{k+j}-x^{k+j+1}\right\|_{2}={1\over h}\left\|x^{k}-{\overline{x}}\right\|_{2}.$ The result follows. ∎

Now we recall the next theorem proved by Dikin [16]. A proof can also be found in [11, 17, 18, 19].

Theorem 3.3

For every $x>0$ and for every $p\in\mathbb{R}^{n}$, we have

[TABLE]

where $q(A)$ is a constant only function of $A$.

Proposition 5

The sequences $(y^{k})$ and $(s^{k})$ are bounded.

Proof According to Theorem 3.3, for every $x>0$ and for every $p\in\mathbb{R}^{n}$, we have $\|y^{k}\|_{2}=\left\|\left(AX_{k}^{2-r}A^{t}\right)^{-1}AX_{k}^{2-r}c\right\|_{2}=\left\|\left(A\left(X_{k}^{1-{r\over 2}}\right)^{2}A^{t}\right)^{-1}A\left(X_{k}^{1-{r\over 2}}\right)^{2}c\right\|_{2}\leq q(A)\left\|c\right\|_{2}$ and then $\|s^{k}\|_{2}=\|c-A^{t}y^{k}\|_{2}\leq(1+q(A)\|A\|_{2})\|c\|_{2}$. The result then follows. ∎

Let us now consider the following notation. Given $x\in(0,+\infty)^{n}$ and $s\in\mathbb{R}^{n}$ we set $I_{r}(x,s)=\{i:\ x_{i}^{1-r}|s_{i}|=\|X^{1-r}s\|_{\infty}\}$.

Lemma 1

Let $(x,s)\in(0,+\infty)^{n}\times\mathbb{R}^{n}$ be such that $Xs\not=0$. One has for every $(r,r^{\prime})\in[0,1]^{2}$, if $r<r^{\prime}$ then $x_{i_{r}}\geq x_{i_{r^{\prime}}}$ and $|s_{i_{r}}|\leq|s_{i_{r^{\prime}}}|$, $\forall(i_{r},i_{r^{\prime}})\in I_{r}(x,s)\times I_{r^{\prime}}(x,s)$.

Proof. We have

[TABLE]

and

[TABLE]

Multiplying side by side $(1)$ and $(2)$ one has $0<x_{i_{r^{\prime}}}^{1-r}x_{i_{r}}^{1-r^{\prime}}|s_{i_{r}}||s_{i_{r^{\prime}}}|\leq x_{i_{r}}^{1-r}x_{i_{r^{\prime}}}^{1-r^{\prime}}|s_{i_{r^{\prime}}}||s_{i_{r}}|$. That is $x_{i_{r^{\prime}}}^{r^{\prime}-r}\leq x_{i_{r}}^{r^{\prime}-r}$ and then $x_{i_{r^{\prime}}}\leq x_{i_{r}}$. Now using $(2)$ one has $0<x_{i_{r^{\prime}}}^{1-r^{\prime}}|s_{i_{r}}|\leq x_{i_{r}}^{1-r^{\prime}}|s_{i_{r}}|\leq\|X^{1-r^{\prime}}s\|_{\infty}=x_{i_{r^{\prime}}}^{1-r^{\prime}}|s_{i_{r^{\prime}}}|$. The result then follows. ∎

Define $I=\{i:\ {\overline{x}}_{i}=0\},\ J=\{i:\ {\overline{x}}_{i}>0\}\mbox{ and }n_{I}=card(I).$

Lemma 2

There is ${\tilde{h}}>0$ such that $\|x^{k}_{J}-{\overline{x}}_{J}\|_{2}\leq{\tilde{h}}\|x^{k}_{I}\|_{2},\ \forall k\in\mathbb{N}.$

Proof. Let $({\overline{y}},{\overline{s}})$ be an accumulation point of $(y^{k},s^{k})$. The existence of $({\overline{y}},{\overline{s}})$ is ensured by Proposition 5. Using Proposition 4 we have $\|x^{k}-{\overline{x}}\|_{2}^{2}=\|x^{k}_{I}\|_{2}^{2}+\|x^{k}_{J}-{\overline{x}}_{J}\|^{2}_{2}\leq h^{2}\langle c,x^{k}-{\overline{x}}\rangle^{2}=h^{2}\langle{\overline{s}},x^{k}-{\overline{x}}\rangle^{2}=h^{2}\langle{\overline{s}}_{I},x^{k}_{I}\rangle^{2}\leq h^{2}\|{\overline{s}}_{I}\|^{2}_{2}\|x^{k}_{I}\|_{2}^{2}$ and then $\|x^{k}_{J}-{\overline{x}}_{J}\|_{2}^{2}\leq(h^{2}\|{\overline{s}}_{I}\|^{2}_{2}-1)\|x^{k}_{I}\|_{2}^{2}.$ Thus $h^{2}\|{\overline{s}}_{I}\|^{2}_{2}-1$ is necessarily nonnegative. The result then follows by setting $\tilde{h}=\sqrt{h^{2}\|{\overline{s}}_{I}\|^{2}_{2}-1}$. ∎

N.B: The fact that $h=\displaystyle{1\over L(A,c)}$, $h^{2}\|{\overline{s}}_{I}\|^{2}_{2}-1\geq 0$ also means that $\|\overline{s}\|_{2}\geq L(A,c)$, for every $\overline{s}$ be an accumulation point of $(s^{k})$.

In all the following we set $g=\sup\limits_{k\in\mathbb{N}}\|s^{k}\|_{\infty}$, ${\overline{M}}=\sup\limits_{k}\|x^{k}\|_{\infty}<+\infty$ and $\underline{M}=\inf\limits_{k}\min\limits_{j\in J}x_{j}^{k}$. Note that since $\lim\limits_{k\uparrow+\infty}x_{J}^{k}={\overline{x}_{J}}>0$, $\underline{M}>0$ and that according to Proposition 5 $g<+\infty$.

Proposition 6

Let $\beta\in(0,1)$. Then there exists $K\in\mathbb{N}$ such that

i)* $\forall k\geq K$, $\forall r\in(0,1)$, $\forall i_{r}\in I_{r}(x^{k},s^{k})$, $x_{i_{r}}^{k}=O(\|x_{I}^{k}\|_{\infty})$ and $s_{i_{r}}^{k}=O(1)$. Furthermore $\|X_{k}^{1-r}s^{k}\|_{\infty}=\|X_{k,I}^{1-r}s_{I}^{k}\|_{\infty}=O\left(\|x_{I}\|^{1-r}_{\infty}\right)$ and there exists a constant $\hat{C}$ such that $\left\|s_{J}^{k}\right\|_{2}\leq\hat{C}\|x_{I}\|_{\infty}^{2-r}$.*

ii)* $\forall k\geq K$, $\left\langle c,x^{k+1}\right\rangle-{\overline{c}}\leq{\overline{L}}(\langle c,x^{k}\rangle-{\overline{c}})$, where $\displaystyle{\overline{L}}=1-\beta{L(A,c)^{6-r\over 2-r}\over 2^{3-r\over 2-r}g^{6-2r\over 2-r}n_{I}^{{7\over 2}-{r(1-r)\over 2(2-r)}}}$.*

iii)* $\sum\limits_{k=0}^{\atop\infty}\|x^{k}_{I}\|_{\infty}^{a}<+\infty,\ \forall a>0.$*

iv)* $\langle x^{k}_{I},s^{k}_{I}\rangle=O(\|x_{I}^{k}\|_{\infty})$.*

v)* $\langle c,x^{k}\rangle-\overline{c}=O(\|x^{k}_{I}\|_{\infty})$ and $\|x^{k}-\overline{x}\|_{2}=O(\|x^{k}_{I}\|_{\infty})$.*

Proof. i) and ii) We have $\left\langle c,x^{k+1}\right\rangle-{\overline{c}}=\left\langle c,x^{k}\right\rangle-{\overline{c}}-\beta t_{max}\left\|PX_{k}^{1-{r\over 2}}c\right\|_{2}^{2}=\left\langle c,x^{k}\right\rangle-{\overline{c}}-\beta t_{max}\left\|X_{k}^{1-{r\over 2}}s^{k}\right\|_{2}^{2}\leq\left\langle c,x^{k}\right\rangle-{\overline{c}}-\beta t_{max}\left\|X_{k}^{1-{r\over 2}}s^{k}\right\|^{2}_{\infty}.$ Now $\displaystyle t^{k}_{max}=\min\left\{-{x^{k}_{i}\over d^{k}_{i}}:\ d^{k}_{i}<0\right\}\geq{1\over\|X_{k}^{-1}d^{k}\|_{\infty}}={1\over\|X_{k}^{1-r}s^{k}\|_{\infty}}$. Then $\displaystyle\left\langle c,x^{k+1}\right\rangle-{\overline{c}}\leq\left\langle c,x^{k}\right\rangle-{\overline{c}}-\beta{\left\|X_{k}^{1-{r\over 2}}s^{k}\right\|^{2}_{\infty}\over\left\|X_{k}^{1-r}s^{k}\right\|_{\infty}}.$ Let $\left(i_{r\over 2},i_{r}\right)\in I_{r\over 2}(x^{k},s^{k})\times I_{r}(x^{k},s^{k})$. From Lemma 1 one has $x^{k}_{i_{r\over 2}}\geq x^{k}_{i_{r}}$ and then

$\hfill\displaystyle\left\langle c,x^{k+1}\right\rangle-{\overline{c}}\leq\left\langle c,x^{k}\right\rangle-{\overline{c}}-\beta x^{k}_{i_{r\over 2}}{\left|s^{k}_{i_{r\over 2}}\right|^{2}\over\left|s^{k}_{i_{r}}\right|}\hfill(1)$

Using Proposition 4 , the fact that $\displaystyle X_{k}^{1-{r\over 2}}PX_{k}^{1-{r\over 2}}c\over\displaystyle\left\|PX_{k}^{1-{r\over 2}}c\right\|_{2}$ is the optimal solution to $(EP_{r})$ and $\displaystyle x^{k}-{\overline{x}}\over\displaystyle\left\|X_{k}^{{r\over 2}-1}\left(x^{k}-{\overline{x}}\right)\right\|_{2}$ is a feasible solution of $(EP_{r})$, $L(A,c)\displaystyle{\|x^{k}-{\overline{x}}\|_{2}\over\left\|X_{k}^{{r\over 2}-1}(x^{k}-{\overline{x}})\right\|_{2}}\leq{\left\langle c,x^{k}-{\overline{x}}\right\rangle\over\left\|X_{k}^{{r\over 2}-1}(x^{k}-{\overline{x}})\right\|_{2}}\leq{\left\langle c,X_{k}^{1-{r\over 2}}PX_{k}^{1-{r\over 2}}c\right\rangle\over\left\|PX_{k}^{1-{r\over 2}}c\right\|_{2}}=\left\|PX_{k}^{1-{r\over 2}}c\right\|_{2}=\|X_{k}^{1-{r\over 2}}s^{k}\|_{2}.$ It follows that $\displaystyle L(A,c)^{2}{\|x^{k}_{I}\|_{2}^{2}+\|x^{k}_{J}-{\overline{x}}_{J}\|_{2}^{2}\over\left\|{X_{k}}_{I}^{r\over 2}e_{I}\right\|_{2}^{2}+\left\|{X_{k}}_{J}^{{r\over 2}-1}(x^{k}_{J}-{\overline{x}}_{J})\right\|_{2}^{2}}\leq\|X_{k}^{1-{r\over 2}}s^{k}\|_{2}^{2}.$ Now using Lemma 2, the fact that $r\in(0,1)$ and the fact that $\lim\limits_{k\to\infty}x^{k}_{I}=0$, for $k$ being large enough one has $\left\|{X_{k}}_{J}^{{r\over 2}-1}(x^{k}_{J}-{\overline{x}}_{J})\right\|_{2}^{2}\leq\underline{M}^{r-2}\tilde{h}^{2}\|x_{I}^{k}\|_{2}^{2}=\underline{M}^{r-2}\tilde{h}^{2}\|{X_{k}}_{I}^{1-{r\over 2}}{X_{k}}_{I}^{r\over 2}e_{I}\|_{2}^{2}\leq\underline{M}^{r-2}\tilde{h}^{2}\|x_{I}^{k}\|^{2-r}_{\infty}\left\|{X_{k}}_{I}^{r\over 2}e_{I}\right\|_{2}^{2}\leq\left\|{X_{k}}_{I}^{r\over 2}e_{I}\right\|_{2}^{2}$, where $e_{I}$ is the vector of $\mathbb{R}^{n_{I}}$ whose components are equal to 1. According to iii) of Proposition 4.3 in [2] one has $\left(\sum\limits_{i\in I}{x^{k}_{i}}^{r}\right)^{2\over r}n_{I}^{1-{2\over r}}=\left(\sum\limits_{i\in I}{({x^{k}_{i}}^{2})}^{r\over 2}\right)^{2\over r}n_{I}^{1-{2\over r}}=\xi_{{r\over 2},I}(x_{I}^{k})\xi_{{r\over r-2},I}(e_{I})\leq\langle x_{I}^{k},e_{I}\rangle=\sum\limits_{i\in I}{x^{k}_{i}}^{2}=\|x^{k}_{I}\|^{2}_{2},$ where $\xi_{{r\over 2},I}$ and $\xi_{{r\over r-2},I}$ are the concave gauge functions respectively defined by \xi_{{r\over 2},I}(z)=\left\{\begin{array}[]{ll}\left(\sum\limits_{i\in I}z_{i}^{r\over 2}\right)^{2\over r}&\mbox{if }z\in[0,+\infty)^{n_{I}},\cr-\infty&\mbox{elsewhere}\end{array}\right. and \xi_{{r\over r-2},I}(z)=\left\{\begin{array}[]{ll}\left(\sum\limits_{i\in I}z_{i}^{r\over r-2}\right)^{r-2\over r}&\mbox{if }z\in(0,+\infty)^{n_{I}},\cr 0&\mbox{if }z\in\partial[0,+\infty)^{n_{I}},\cr-\infty&\mbox{elsewhere.}\end{array}\right. Here $\partial[0,+\infty)^{n_{I}}$ denotes the boundary of $[0,+\infty)^{n_{I}}$. That is $\displaystyle\sum\limits_{i\in I}{x_{i}^{k}}^{r}=\|{X_{k}}_{I}^{r\over 2}e_{I}\|_{2}^{2}=\sum\limits_{i\in I}{x^{k}_{i}}^{r}\leq n_{I}^{1-{r\over 2}}\|x^{k}_{I}\|^{r}_{2}.$ Hence $\displaystyle{{L(A,c)}^{2}\over 2n_{I}^{1-{r\over 2}}}\|x_{I}\|_{2}^{2-r}\leq L(A,c)^{2}{\|x^{k}_{I}\|_{2}^{2}+\|x^{k}_{J}-{\overline{x}}_{J}\|_{2}^{2}\over\left\|{X_{k}}_{I}^{r\over 2}e_{I}\right\|_{2}^{2}+\left\|{X_{k}}_{J}^{{r\over 2}-1}(x^{k}_{J}-{\overline{x}}_{J})\right\|_{2}^{2}}\leq\|X_{k}^{1-{r\over 2}}s^{k}\|_{2}^{2}$ and then

$\hfill\displaystyle{{L(A,c)}^{2}\over n_{I}^{1-{r\over 2}}}\|x_{I}\|_{2}^{2-r}\leq 2n_{I}\|X_{k}^{1-{r\over 2}}s^{k}\|_{\infty}\hfill(2)$

Now using Corollary 1 one has

$\underline{M}^{2-r}\|s^{k}_{J}\|_{2}\leq\underline{M}^{1-{r\over 2}}\left\|X_{k,J}^{1-{r\over 2}}s_{J}^{k}\right\|_{2}\leq\left(\min\limits_{j\in J}x_{j}^{k}\right)^{1-{r\over 2}}\left\|X_{k,J}^{1-{r\over 2}}s_{J}^{k}\right\|_{2}$

$\leq\left\|\left(X_{k}^{1-{r\over 2}}PX_{k}^{1-{r\over 2}}c\right)_{J}\right\|_{2}\leq\left\|X_{k}^{1-{r\over 2}}PX_{k}^{1-{r\over 2}}c\right\|_{2}\leq L(A,c)\left\langle c,X_{k}^{1-{r\over 2}}PX_{k}^{1-{r\over 2}}c\right\rangle$

$=L(A,c)\left\langle{\overline{s}}_{I},X_{k,I}^{1-{r\over 2}}X_{k,I}^{1-{r\over 2}}s^{k}_{I}\right\rangle\leq L(A,c)\left\|{\overline{s}}_{I}\right\|_{2}\left\|X_{k,I}^{1-{r\over 2}}X_{k,I}^{1-{r\over 2}}s^{k}_{I}\right\|_{2}$

$\leq L(A,c)\sqrt{n_{I}}\|\overline{s}_{I}\|_{\infty}\|x_{I}^{k}\|^{1-{r\over 2}}_{\infty}\left\|X_{k,I}^{1-{r\over 2}}s^{k}_{I}\right\|_{2}\leq L(A,c)\left\|{\overline{s}}_{I}\right\|_{\infty}n_{I}g\|x^{k}_{I}\|_{\infty}^{2-r}$.

So we get on the one hand $\|s^{k}_{J}\|_{2}\leq\hat{C}\|x^{k}_{I}\|^{2-r}$, where $\hat{C}=\displaystyle{L(A,c)\left\|{\overline{s}}_{I}\right\|_{\infty}n_{I}g\over\underline{M}^{2-r}}$, and

$\hfill\left\|X_{k,J}^{1-{r\over 2}}s_{J}^{k}\right\|_{2}\leq\displaystyle{L(A,c)\sqrt{n_{I}}\|\overline{s}_{I}\|_{\infty}\over\underline{M}^{1-{r\over 2}}}\|x_{I}^{k}\|^{1-{r\over 2}}_{\infty}\left\|X_{k,I}^{1-{r\over 2}}s^{k}_{I}\right\|_{2}\hfill(2bis)$

on the other hand. Since $\lim\limits_{k\uparrow\infty}x_{I}^{k}=0$, by $(2bis)$ we have necessarily $\left\|X_{k}^{1-{r\over 2}}s^{k}\right\|_{\infty}=\left\|X_{k,I}^{1-{r\over 2}}s^{k}_{I}\right\|_{\infty}={x_{i_{r\over 2}}^{k}}^{1-{r\over 2}}s_{i_{r\over 2}}^{k}$, for $k$ large enough. Now $\left\|X_{k,J}^{1-{r}}s_{J}^{k}\right\|_{2}\leq\underline{M}^{-{r\over 2}}\left\|X_{k,J}^{1-{r\over 2}}s_{J}^{k}\right\|_{2}$ and $\left\|X_{k,I}^{1-{r\over 2}}s^{k}_{I}\right\|_{2}=\left\|X_{k,I}^{{r\over 2}}X_{k,I}^{1-r}s^{k}_{I}\right\|_{2}\leq\|x_{I}^{k}\|_{\infty}^{r\over 2}\left\|X_{k,I}^{1-r}s^{k}_{I}\right\|_{2}$. Then using $(29bis)$ we get $\left\|X_{k,J}^{1-r}s_{J}^{k}\right\|_{2}\leq\displaystyle{L(A,c)\sqrt{n_{I}}\|\overline{s}_{I}\|_{\infty}\over\underline{M}}\|x_{I}^{k}\|_{\infty}\left\|X_{k,I}^{1-{r}}s^{k}_{I}\right\|_{2}.$ Hence using again the fact that $\lim\limits_{k\uparrow\infty}x_{I}^{k}=0$ we get $\left\|X_{k}^{1-{r}}s^{k}\right\|_{\infty}=\left\|X_{k,I}^{1-{r}}s^{k}_{I}\right\|_{\infty}$ for $k$ large enough.

Turn back now to (2). Then when $k$ is large enough we have

$\displaystyle{{L(A,c)}^{2}\over n_{I}^{1-{r\over 2}}}\|x^{k}_{I}\|_{2}^{2-r}\leq 2\|X_{k,I}^{1-{r\over 2}}s_{I}^{k}\|_{2}^{2}\leq 2n_{I}\|X_{k,I}^{1-{r\over 2}}s_{I}^{k}\|_{\infty}^{2}$

$\hfill=2n_{I}\left({x^{k}_{i_{r\over 2}}}^{1-{r\over 2}}s_{i_{r\over 2}}^{k}\right)^{2}\leq 2n_{I}\left\|x^{k}_{I}\right\|_{2}^{2-r}\left|s^{k}_{i_{r\over 2}}\right|^{2}\hfill(3)$

and

$\displaystyle{{L(A,c)}^{2}\over n_{I}^{1-{r\over 2}}}\|x^{k}_{I}\|_{2}^{2-r}\leq 2n_{I}\left({x^{k}_{i_{r\over 2}}}^{1-{r\over 2}}s_{i_{r\over 2}}^{k}\right)^{2}\leq 2n_{I}\left({x^{k}_{i_{r\over 2}}}\right)^{2-r}\left\|s^{k}_{I}\right\|_{\infty}^{2}.\hfill(4)$

Using (3) and Lemma 1 it follows that

$\hfill\displaystyle\left|s^{k}_{i_{r}}\right|\geq\left|s^{k}_{i_{r\over 2}}\right|\geq{{L(A,c)}\over\sqrt{2}n_{I}^{1-{r\over 4}}}\hfill(5)$

and then $\left|s^{k}_{i_{r}}\right|=O(1)$. Now using (4), (5) and the fact that $+\infty>g=\sup\limits_{k\in\mathbb{N}}\|s^{k}\|_{\infty}$, we get

$\hfill\left\|x_{I}^{k}\right\|_{2}\geq\displaystyle x^{k}_{i_{r\over 2}}\geq{L(A,c)^{2\over 2-r}\over 2^{1\over 2-r}g^{2\over 2-r}n_{I}^{1+{r\over 4-2r}}}\left\|x_{I}^{k}\right\|_{2}\hfill(6)$

and then $x^{k}_{i_{r\over 2}}=O\left(\left\|x_{I}^{k}\right\|_{2}\right)$ and then $i_{r\over 2}\in I$. Now $x_{i_{r}}^{k^{1-r}}O(1)=x_{i_{r}}^{k^{1-r}}\left|s^{k}_{i_{r}}\right|=\max\limits_{i}\left(x_{i}^{k^{1-r}}\left|s^{k}_{i}\right|\right)\geq x_{i_{r\over 2}}^{k^{1-r}}\left|s^{k}_{i_{r\over 2}}\right|=O\left(\left\|x_{I}^{k}\right\|_{2}\right)^{1-r}$. It follows that $\|X^{1-r}_{k,I}s^{k}_{I}\|_{\infty}=O(\|x^{k}_{I}\|^{1-r}_{\infty})$ and $x_{i_{r}}^{k}=O\left(\left\|x_{I}^{k}\right\|_{2}\right)$, witch implies that $i_{r}\in I$.

Now using (5), (6) and the fact that $|s_{i_{r}}^{k}|\leq\|s^{k}\|_{\infty}\leq g$ we get $\displaystyle x^{k}_{i_{r\over 2}}{\left|s^{k}_{i_{r\over 2}}\right|^{2}\over\left|s^{k}_{i_{r}}\right|}\geq{L(A,c)^{6-r\over 2-r}\over 2^{3-r\over 2-r}g^{4-r\over 2-r}n_{I}^{3-{r(1-r)\over 2(2-r)}}}\left\|x_{I}^{k}\right\|_{2}.$ But $\langle c,x^{k}\rangle-{\overline{x}}=\langle{\overline{s}},x^{k}-{\overline{x}}\rangle=\langle{\overline{s}}_{I},x^{k}_{I}\rangle\leq\|{\overline{s}}_{I}\|_{2}\|x^{k}_{I}\|_{2}\leq\sqrt{n_{I}}g\|x^{k}_{I}\|_{2}.$ It follows that $\displaystyle x^{k}_{i_{r\over 2}}{\left|s^{k}_{i_{r\over 2}}\right|^{2}\over\left|s^{k}_{i_{r}}\right|}\geq{L(A,c)^{6-r\over 2-r}\over 2^{3-r\over 2-r}g^{6-2r\over 2-r}n_{I}^{{7\over 2}-{r(1-r)\over 2(2-r)}}}(\langle c,x^{k}\rangle-\overline{c})$ and then by (1), $\left\langle c,x^{k+1}\right\rangle-{\overline{c}}\leq{\overline{L}}(\langle c,x^{k}\rangle-{\overline{c}}).$

iii) By ii) and Proposition 4, $\displaystyle\|x^{k}_{I}\|_{\infty}\leq\left\|x^{k}-{\overline{x}}\right\|_{2}\leq H\left(\left\langle c,x^{k}\right\rangle-{\overline{c}}\right)\leq H\overline{L}^{k-K}\left(\left\langle c,x^{K}\right\rangle-{\overline{c}}\right),\ \forall k\geq K.$ The result follows since

[TABLE]

iv) We have $L(A,c)\|x^{k}_{I}\|_{2}\leq L(A,c)\|x^{k}-\overline{x}\|_{2}\leq\langle c,x^{k}\rangle-\overline{c}=\langle s^{k},x^{k}-\overline{x}\rangle=\langle s^{k}_{I},x^{k}_{I}\rangle+\langle s^{k}_{J},x^{k}_{J}-\overline{x}_{J}\rangle=\langle\overline{s}_{I},x^{k}_{I}\rangle\leq\|\overline{s}_{I}\|_{2}\|x^{k}_{I}\|_{2}$. The result then follows from i).

v) Let $({\overline{y}},{\overline{s}})$ be an accumulation point to $(y^{k},s^{k})$. We have $\langle c,x^{k}\rangle-{\overline{c}}=\langle A^{t}{\overline{y}}+{\overline{s}},x^{k}-{\overline{x}}\rangle=\langle{\overline{s}}_{I},x^{k}_{I}\rangle$. Using Proposition 4 and the Cauchy-Schwartz inequality we get

[TABLE]

The result then follows. ∎

Now we establish some technical results given by Saigal [11] in the classical case.

Proposition 7

Let $(u^{k})$ the sequence defined by $u^{k}=\displaystyle{X_{k}^{r\over 2}PX_{k}^{1-{r\over 2}}c\over\left\langle c,x^{k}\right\rangle-{\overline{c}}}=\displaystyle{X_{k}s^{k}\over\left\langle c,x^{k}\right\rangle-{\overline{c}}}.$ Then we have:

i)* The sequence $(u^{k})$ is bounded.*

ii)* $\sum\limits_{k=0}^{\atop\infty}\left\|u_{J}^{k}\right\|^{2}<+\infty$.*

iii)* $\sum\limits_{k=0}^{\atop\infty}|\delta^{k}|<+\infty$, where $\delta^{k}=\left\langle e_{I},u^{k}_{I}\right\rangle-1$.*

Proof.

i) and ii) By Proposition 4, there is $h>0$ such that $\left\|x^{k}-{\overline{x}}\right\|\leq h\left(\left\langle c,x^{k}\right\rangle-{\overline{c}}\right).$ Then $\|u_{I}\|_{2}=\displaystyle{\|X_{k,I}s^{k}_{I}\|_{2}\over\langle c,x^{k}\rangle-{\overline{c}}}\leq h{\|X_{k,I}s^{k}_{I}\|_{2}\over\|x^{k}-{\overline{x}}\|_{2}}\leq h{\|x_{I}^{k}\|_{\infty}\|s^{k}_{I}\|_{2}\over\|x^{k}_{I}\|_{\infty}}=h\|s^{k}_{I}\|_{2}$ Hence $u_{I}^{k}$ is bounded according to Proposition 5. According to Proposition 6 $\|u^{k}_{J}\|_{2}=\displaystyle{\|X_{k,J}s^{k}_{J}\|_{2}\over\langle c,x^{k}\rangle-{\overline{c}}}\leq h\|x^{k}_{J}\|_{\infty}{\|s^{k}_{J}\|_{2}\over\|x^{k}-{\overline{x}}\|_{2}}\leq h\overline{M}{\|s^{k}_{J}\|_{2}\over\|x^{k}_{I}\|_{2}}\leq h\hat{C}\overline{M}\|x_{I}^{k}\|^{1-r}.$ Then i) and ii) follows by using Proposition 6.

iii) Set $SD=\left\{(y,s):\ A^{t}y+s=c,\ s_{J}=0\right\}$ the expected dual optimal solutions’ set. Let $\left({\hat{y}}^{k},{\hat{s}}^{k}\right)$ a solution to the problem $\min\left\{\left\|s^{k}-s\right\|_{2}:\ (y,s)\in SD\right\}.$ We have $\left\langle c,x^{k}\right\rangle-{\overline{c}}=\left\langle c,x^{k}-{\overline{x}}\right\rangle=\left\langle{\hat{s}}^{k}+A^{t}{\hat{y}}^{k},x^{k}-{\overline{x}}\right\rangle=\left\langle{\hat{s}}_{I}^{k},x_{I}^{k}\right\rangle.$ Then $\left|\left\langle e_{I},u_{I}^{k}\right\rangle-1\right|=\displaystyle{\left\langle x_{I}^{k},s_{I}^{k}\right\rangle-\left\langle{\hat{s}}_{I}^{k},x_{I}^{k}\right\rangle\over\left\langle c,x^{k}\right\rangle-{\overline{c}}}\leq{\left\|s_{I}^{k}-{\hat{s}}_{I}^{k}\right\|_{2}\left\|x_{I}^{k}\right\|_{2}\over\left\langle c,x^{k}\right\rangle-{\overline{c}}}.$ By Proposition 4 we have $\left\|x_{I}^{k}\right\|_{2}\leq\left\|x^{k}-{\overline{x}}\right\|_{2}\leq h\left(\left\langle c,x^{k}\right\rangle-{\overline{c}}\right).$ Hence $\left|\left\langle e_{I},u_{I}^{k}\right\rangle-1\right|_{2}\leq h\left\|s_{I}^{k}-{\hat{s}}_{I}^{k}\right\|_{2}.$ From Theorem 7 of [11], there is ${\hat{M}}$ such that $\left\|{\hat{s}}^{k}-s^{k}\right\|_{2}\leq{\hat{M}}\left\|s^{k}_{J}\right\|_{2}.$ Using Proposition 6 we get $\left\|s_{I}^{k}-{\hat{s}}_{I}^{k}\right\|_{2}\leq{\hat{M}}\left\|s^{k}_{J}\right\|_{2}\leq{\hat{C}}{\hat{M}}\|x_{I}\|_{\infty}^{2-r}.$ The result then follows by using iii) of Proposition 6. ∎

Now let us introduce the potential function $F$ defined as follow:

[TABLE]

The following Proposition holds.

Proposition 8

There is $\Delta\in\mathbb{R}$ such that for every $k\geq 0$, $F(x^{k})\geq\Delta>-\infty.$

Proof. By Theorem 3.1 and Proposition 4.3 of [2], ${\tilde{\xi}_{r}}(x^{k}){\tilde{\xi}_{t}}(e)\leq\langle x^{k}_{I},e_{I}\rangle$, where $t$ is such that $\displaystyle{1\over t}+{1\over r}=1$. Hence ${\tilde{\xi}_{r}}(x^{k})\leq\displaystyle{1\over n_{I}^{1\over t}}\sum\limits_{i\in I}x^{k}_{i}\leq{n^{1\over 2}\over n_{I}^{1\over t}}\left\|x_{I}^{k}\right\|_{2}=n^{{1\over r}-{1\over 2}}\left\|x_{I}^{k}\right\|_{2}=n^{{1\over r}-{1\over 2}}\left\|x^{k}-{\overline{x}}\right\|_{2}\leq n^{{1\over r}-{1\over 2}}h\left(\left\langle c,x^{k}\right\rangle-{\overline{c}}\right)$ and then $-\infty<\ln\left(\displaystyle{1\over n^{{1\over r}-{1\over 2}}h}\right)\leq F(x^{k})$. The result then follows. ∎

Proposition 9

Let $\displaystyle\beta\in\left(0,{2\over 3}\right)$. There exists $K\in\mathbb{N}$ such that $\forall k\geq K$

[TABLE]

where \Upsilon^{k}=\left\{\begin{array}[]{ll}1&\mbox{if }\max\limits_{i\in I}w_{i}^{k}\leq 0,\cr 0&\mbox{if }\max\limits_{i\in I}w_{i}^{k}>0,\end{array}\right. $\theta^{k}=\displaystyle{{\tilde{t}}^{k}\over 1-\displaystyle{1\over\sum\limits_{i\in I}{x_{i}^{k}}^{r}}{\tilde{t}}^{k}}$, ${\tilde{t}}^{k}=t^{k}\left(\left\langle c,x^{k}\right\rangle-{\overline{c}}\right)$, $t^{k}=\beta t^{k}_{\max}$, $\gamma^{k}=\left\|X_{k,I}^{-{r\over 2}}u_{J}^{k}\right\|^{2}$ and $\displaystyle w^{k}_{I}=X_{k,I}^{-r}u^{k}_{I}-{1\over\sum\limits_{i\in I}{x_{i}^{k}}^{r}}e$.

Proof. Let us proof at first that given $\beta\in(0,1)$, there is $K\in\mathbb{N}$ such that $\forall k\geq K$, $F\left(x^{k+1}\right)-F\left(x^{k}\right)\leq\ln\left(1-\theta^{k}\|X_{k,I}^{r\over 2}w_{I}^{k}\|-\displaystyle 2{\theta^{k}\over\sum\limits_{i\in I}{x^{k}_{i}}^{r}}\delta^{k}-\theta^{k}\gamma^{k}\right)-\displaystyle\sum\limits_{i\in I}\displaystyle{{x^{k}_{i}}^{r}\over\sum\limits_{i\in I}{x^{k}_{i}}^{r}}\ln\left(1-\theta^{k}w_{i}^{k}\right).$ We have $F(x^{k+1})-F(x^{k})=\ln\left(\displaystyle{\langle c,x^{k+1}\rangle-{\overline{c}}\over\langle c,x^{k}\rangle-{\overline{c}}}\right)-{1\over r}\ln\left(\displaystyle{\sum x_{i}^{{k+1}^{r}}\over\sum x_{i}^{{k}^{r}}}\right)$ , $\langle c,x^{k+1}\rangle-{\overline{c}}=\langle c,x^{k}\rangle-{\overline{c}}-t^{k}\langle X_{k}^{-r}X_{k}S^{k},X_{k}s^{k}\rangle$, $\displaystyle u^{k}={X_{k}s^{k}\over\langle c,x^{k}\rangle-{\overline{c}}}$ and ${\tilde{t}^{k}}=(\langle c,x^{k}\rangle-{\overline{c}})t^{k}$. Then $\displaystyle{\langle c,x^{k+1}\rangle-{\overline{c}}\over\langle c,x^{k}\rangle-{\overline{c}}}=1-{\tilde{t}^{k}}\langle X^{-r}_{k,I}u^{k},u^{k}\rangle=1-{\tilde{t}^{k}}\langle X^{-r}_{k,I}u^{k}_{I},u^{k}_{I}\rangle-\tilde{t^{k}}\gamma^{k}.$ Now $u_{I}^{k}=X_{k,I}^{r}w_{I}^{k}+{1\over\sum\limits_{i\in I}x_{i}^{k^{r}}}X_{k,I}^{r}e_{I}$, $X_{k,I}^{-r}u_{I}^{k}=w_{I}^{k}+{1\over\sum\limits_{i\in I}x_{i}^{k^{r}}}e_{I}$ and $\delta^{k}=\langle u^{k}_{I},e_{I}\rangle-1=\langle X^{r}_{k,I}w_{I}^{k},e_{I}\rangle$. Then $\displaystyle\langle X^{-r}_{k,I}u^{k}_{I},u^{k}_{I}\rangle=\langle X^{r}_{k,I}w^{k}_{I},w^{k}_{I}\rangle+{2\over\sum\limits_{i\in I}x_{i}^{k^{r}}}\delta^{k}+{1\over\sum\limits_{i\in I}x_{i}^{k^{r}}}$. It follows that

\begin{array}[]{ll}\displaystyle{\langle c,x^{k+1}\rangle-{\overline{c}}\over\langle c,x^{k}\rangle-{\overline{c}}}&=\displaystyle 1-{\tilde{t}^{k}}\langle X_{k,I}^{r}w_{I}^{k},w_{I}^{k}\rangle-{2{\tilde{t}^{k}}\over\sum\limits_{i\in I}x_{i}^{k^{r}}}\delta^{k}-{{\tilde{t}^{k}}\over\sum\limits_{i\in I}x_{i}^{k^{r}}}-{\tilde{t}^{k}}\gamma^{k}\cr&=\displaystyle\left(1-{{\tilde{t}^{k}}\over\sum\limits_{i\in I}x_{i}^{k^{r}}}\right)\left[1-\theta^{k}\langle X_{k,I}^{r}w_{I}^{k},w_{I}^{k}\rangle-{2\theta^{k}\over\sum\limits_{i\in I}x_{i}^{k^{r}}}\delta^{k}-\theta^{k}\gamma^{k}\right].\end{array}

Let us show now, for $\beta\in(0,1)$, $\displaystyle\theta^{k}={{\tilde{t}^{k}}\over 1-{{\tilde{t}^{k}}\over\sum\limits_{i\in I}x_{i}^{k^{r}}}}={\sum\limits_{i\in I}x_{i}^{k^{r}}{\tilde{t}^{k}}\over\sum\limits_{i\in I}x_{i}^{k^{r}}-{\tilde{t}^{k}}}>0$, for $k$ large enough. We have $\langle c,x^{k}\rangle-{\overline{c}}=\langle s^{k},x^{k}-\overline{x}\rangle=\langle s^{k}_{I},x^{k}_{I}\rangle+\langle s^{k}_{J},x^{k}_{J}-\overline{x}_{J}\rangle$.

\begin{array}[]{ll}\mbox{Then }\sum\limits_{i\in I}x_{i}^{k^{r}}-{\tilde{t}^{k}}&=\sum\limits_{i\in I}x_{i}^{k^{r}}-\displaystyle\beta{\langle c,x^{k}\rangle-{\overline{c}}\rangle\over\max\limits_{i\in I}(X_{k}^{1-r}s^{k})_{i}}\cr&=\displaystyle{\sum\limits_{i\in I}x_{i}^{k^{r}}\max\limits_{i\in I}(X_{k}^{1-r}s^{k})_{i}-\beta(\langle s^{k}_{I},x^{k}_{I}\rangle+\langle s^{k}_{J},x^{k}_{J}-\overline{x}_{J}\rangle)\over\max\limits_{i\in I}(X_{k}^{1-r}s^{k})_{i}}\cr&\geq\displaystyle{\sum\limits_{i\in I}x_{i}^{k^{r}}(x_{i}^{k^{1-r}}s^{k}_{i})-\beta\langle s^{k}_{I},x^{k}_{I}\rangle-\beta\langle s^{k}_{J},x^{k}_{J}-\overline{x}_{J}\rangle)\over\max\limits_{i\in I}(X_{k}^{1-r}s^{k})_{i}}\cr&=\displaystyle{\langle x_{I}^{k},s_{I}^{k}\rangle\over\max\limits_{i\in I}(X_{k}^{1-r}s^{k})_{i}}\left(1-\beta\displaystyle{\langle s_{J}^{k},x_{J}^{k}-\overline{x}_{J}\rangle\over\langle x_{I}^{k},s_{I}^{k}\rangle}-\beta\right)\cr&\geq\displaystyle{\langle x_{I}^{k},s_{I}^{k}\rangle\over\max\limits_{i\in I}(X_{k}^{1-r}s^{k})_{i}}\left(1-\beta\displaystyle{\|s_{J}^{k}\|_{2}\|x_{J}^{k}-\overline{x}_{J}\|_{2}\over\langle x_{I}^{k},s_{I}^{k}\rangle}-\beta\right)\cr\end{array}

Using the fact that $\lim\limits_{k\uparrow\infty}x^{k}-\overline{x}=0$, from Proposition 6 we get $\displaystyle{\|s_{J}^{k}\|_{2}\|x_{J}^{k}-\overline{x}_{J}\|_{2}\over\langle x_{I}^{k},s_{I}^{k}\rangle}=o(\|x_{I}^{k}\|^{1-r})$. Therefore $\theta^{k}>0$. Now $\displaystyle{1\over r}\ln\left(\displaystyle{\sum\limits_{i\in I}x_{i}^{k^{r+1}}\over\sum\limits_{i\in I}x_{i}^{k^{r}}}\right)={1\over r}\ln\left(\sum\limits_{i\in I}\displaystyle{x_{i}^{k^{r}}\over\sum\limits_{j\in I}x_{j}^{k^{r}}}(1-{\tilde{t}^{k}}x_{i}^{k^{-r}}u_{i}^{k})^{r}\right).$ Since the function $t\mapsto\ln t$, $t>0$, is concave, one has

\begin{array}[]{ll}\displaystyle{1\over r}\ln\left(\displaystyle{\sum\limits_{i\in I}x_{i}^{k^{r+1}}\over\sum\limits_{i\in I}x_{i}^{k^{r}}}\right)^{r}&\geq\sum\limits_{i\in I}\displaystyle{x_{i}^{k^{r}}\over\sum\limits_{j\in I}x_{j}^{k^{r}}}\ln\left(1-{\tilde{t}^{k}}x_{i}^{k^{-r}}u_{i}^{k}\right)\cr&=\sum\limits_{i\in I}\displaystyle{x_{i}^{k^{r}}\over\sum\limits_{j\in I}x_{j}^{k^{r}}}\ln\left(1-{{\tilde{t}^{k}}\over\sum\limits_{j\in I}x_{j}^{k^{r}}}-{\tilde{t}^{k}}w_{i}^{k}\right)\cr&=\displaystyle\ln\left(1-{{\tilde{t}^{k}}\over\sum\limits_{j\in I}x_{j}^{k^{r}}}\right)+\sum\limits_{i\in I}\displaystyle{x_{i}^{k^{r}}\over\sum\limits_{j\in I}x_{j}^{k^{r}}}\ln\left(1-\theta^{k}w_{i}^{k}\right).\end{array}

Hence $F(x^{k+1})-F(x^{k})\leq\ln\left[1-\theta^{k}\langle X_{k,I}^{r}w_{I}^{k},w_{I}^{k}\rangle-{2\theta^{k}\over\sum\limits_{i\in I}x_{i}^{k^{r}}}\delta^{k}-\theta^{k}\gamma^{k}\right]-\sum\limits_{i\in I}\displaystyle{x_{i}^{k^{r}}\over\sum\limits_{j\in I}x_{j}^{k^{r}}}\ln\left(1-\theta^{k}w_{i}^{k}\right).$ Assume now that $\max\limits_{i\in I}w_{i}^{k}>0$. Then using Lemma 8 of [11] or its proof we can easily see $\sum\limits_{i\in I}\displaystyle{x_{i}^{k^{r}}\over\sum\limits_{j\in I}x_{j}^{k^{r}}}\ln\left(1-\theta^{k}w_{i}^{k}\right)\geq\displaystyle-{\theta^{k}\over\sum\limits_{j\in I}x_{j}^{k^{r}}}\delta^{k}-\displaystyle{{\theta^{k}}^{2}\over 2\sum\limits_{j\in I}x_{j}^{k^{r}}}{\|X_{k,I}^{r\over 2}w_{I}^{k}\|^{2}\over 1-\theta^{k}\max\limits_{i\in I}w_{i}^{k}}$. Using in addition the fact that $\ln(1-a)\leq-a,\ \forall a<1$ we get $F(x^{k+1})-F(x^{k})\leq-\theta^{k}\left(1-\displaystyle{{\theta^{k}}\over 2\sum\limits_{j\in I}x_{j}^{k^{r}}}{1\over 1-\theta^{k}\max\limits_{i\in I}w_{i}^{k}}\right)\|X_{k,I}^{r\over 2}w_{I}^{k}\|_{2}^{2}-\displaystyle{{\theta^{k}}\over\sum\limits_{j\in I}x_{j}^{k^{r}}}\delta^{k}-\theta^{k}\gamma^{k}.$ Now

\begin{array}[]{ll}1-\displaystyle{{\theta^{k}}\over 2\sum\limits_{j\in I}x_{j}^{k^{r}}}{1\over 1-\theta^{k}\max\limits_{i\in I}w_{i}^{k}}&=1-\displaystyle{{\theta^{k}}\over 2\sum\limits_{j\in I}x_{j}^{k^{r}}}\displaystyle{1\over 1+\displaystyle{{\theta^{k}}\over\sum\limits_{j\in I}x_{j}^{k^{r}}}-\theta^{k}\max\limits_{i\in I}(X_{k}^{-r}u^{k})_{i}}\cr&=1-\displaystyle{1\over 2\sum\limits_{j\in I}x_{i}^{k^{r}}}\displaystyle{\theta^{k}\over 1+\displaystyle{{\theta^{k}}\over\sum\limits_{j\in I}x_{j}^{k^{r}}}}\displaystyle{1\over 1-\displaystyle{\theta^{k}\over 1+\displaystyle{{\theta^{k}}\over\sum\limits_{j\in I}x_{j}^{k^{r}}}}\max\limits_{i\in I}(X_{k}^{-r}u^{k})_{i}}.\end{array}

We have on the one hand $\theta^{k}=\displaystyle{{\tilde{t}^{k}}\over 1-\displaystyle{{\tilde{t}^{k}}\over\sum\limits_{j\in I}x_{j}^{k^{r}}}}$ and then ${\tilde{t}^{k}}=\displaystyle{\theta^{k}\over 1+\displaystyle{\theta^{k}\over\sum\limits_{j\in I}x_{j}^{k^{r}}}}$. On the other hand, ${\tilde{t}^{k}}=\beta\displaystyle{\langle c,x^{k}\rangle-{\overline{c}}\over\max\limits_{i\in I}(X_{k}^{1-r}s^{k})_{i}}=\displaystyle{\beta\over\max\limits_{i\in I}(X_{k}^{-r}u^{k})_{i}}$. It follows that

\begin{array}[]{ll}1-\displaystyle{{\theta^{k}}\over 2\sum\limits_{j\in I}x_{j}^{k^{r}}}{1\over 1-\theta^{k}\max\limits_{i\in I}w_{i}^{k}}&=1-\displaystyle{{\tilde{t}^{k}}\over 2(1-\beta)\sum\limits_{i\in I}x_{i}^{k^{r}}}\cr&=1-\displaystyle{\beta\over 2(1-\beta)}\displaystyle{\langle c,x^{k}\rangle-{\overline{c}}\over\sum\limits_{i\in I}\left[x_{i}^{k^{r}}\max\limits_{j\in I}(X_{k}^{1-r}s^{k})_{i}\right]}.\end{array}

Using the fact that $\sum\limits_{i\in I}\left[x_{i}^{k^{r}}\max\limits_{j\in I}(X_{k}^{1-r}s^{k})_{i}\right]\geq\sum\limits_{i\in I}{x_{i}^{k}}^{r}{x_{i}^{k}}^{1-r}s_{i}^{k}=\sum\limits_{i\in I}x_{i}^{k}s_{i}^{k}$ and the fact that $\langle c,x^{k}\rangle-{\overline{c}}=\langle s^{k},x^{k}-\overline{x}\rangle=\langle s^{k}_{I},x^{k}_{I}\rangle+\langle s_{J}^{k},x_{J}^{k}-\overline{x}_{J}\rangle$ we get $1-\displaystyle{\beta\over 2(1-\beta)}\displaystyle{\langle c,x^{k}\rangle-{\overline{c}}\over\sum\limits_{i\in I}\left[x_{i}^{k^{r}}\max\limits_{j\in I}(X_{k}^{1-r}s^{k})_{i}\right]}\geq 1-\displaystyle{\beta\over 2(1-\beta)}\left(1+\displaystyle{\langle x_{J}^{k},s_{J}^{k}\rangle\over\langle s_{I}^{k},x_{I}^{k}\rangle}\right)=\displaystyle{2-3\beta\over 2(1-\beta)}-{\beta\over 2(1-\beta)}{\langle s_{J}^{k},x^{k}_{J}-\overline{x}_{J}\rangle\over\langle x_{I}^{k},s^{k}_{I}\rangle}\geq\displaystyle{2-3\beta\over 2(1-\beta)}-{\beta\over 2(1-\beta)}{\|s_{J}^{k}\|_{2}\|x^{k}_{J}-\overline{x}_{J}\|_{2}\over\langle x_{I}^{k},s^{k}_{I}\rangle}=\displaystyle{2-3\beta\over 2(1-\beta)}+o(\|x_{I}^{k}\|^{1-r}_{2})\geq\displaystyle{2-3\beta\over 3(1-\beta)},$ for k being large enough. Hence $F(x^{k+1})-F(x^{k})\leq-\theta^{k}\displaystyle{2-3\beta\over 3(1-\beta)}\|X_{k,I}^{r\over 2}w_{I}^{k}\|_{2}^{2}-\displaystyle{\theta^{k}\over\sum\limits_{j\in I}{x^{k}}^{r}}\delta^{k}-\theta^{k}\gamma^{k}.$ Consider now the case $\max\limits_{i\in I}w_{i}^{k}\leq 0$. Then $\displaystyle\sum\limits_{i\in I}\displaystyle{{x^{k}_{i}}^{r}\over\sum\limits_{i\in I}{x^{k}_{i}}^{r}}\ln\left(1-\theta^{k}w_{i}^{k}\right)\geq 0$ and the result follows by using the fact that $\ln(1-a)\leq-a,\ \forall a<1$. ∎

Lemma 3

We have $\displaystyle{\theta^{k}\over\sum\limits_{j\in I}{x_{i}^{k}}^{r}}=\displaystyle{\displaystyle{{\tilde{t}^{k}}\over\sum\limits_{j\in I}{x_{i}^{k}}^{r}}\over 1-\displaystyle{{\tilde{t}^{k}}\over\sum\limits_{j\in I}{x_{i}^{k}}^{r}}}=O(1),$ For $k$ large enough.

Proof. We have $n_{I}g\|x^{k}_{I}\|_{\infty}\geq\sum\limits_{i\in I}{x_{i}^{k}}^{r}\max\limits_{i\in I}(X_{k}^{1-r}s^{k})_{i}\geq\sum\limits_{i\in I}{x_{i}^{k}}^{r}(X_{k}^{1-r}s^{k})_{i}=\langle x_{I}^{k},s^{k}_{I}\rangle$. By Proposition 6 $\langle x_{I}^{k},s^{k}_{I}\rangle=O(\|x^{k}_{I}\|_{\infty})$ and $\langle c,x^{k}\rangle-\overline{c}=O(\|x^{k}_{I}\|_{\infty})$. It follows that $\sum\limits_{i\in I}{x_{i}^{k}}^{r}\max\limits_{i\in I}(X_{k}^{1-r}s^{k})_{i}=O(\|x^{k}_{I}\|_{\infty})$ and then $\displaystyle{{\tilde{t}^{k}}\over\sum\limits_{i\in I}{x_{i}^{k}}^{r}}=\beta\displaystyle{\langle c,x^{k}\rangle-{\overline{c}}\over\sum\limits_{i\in I}{x_{i}^{k}}^{r}\max\limits_{i\in I}(X_{k}^{1-r}s^{k})_{i}}=O(1)$. Now $\langle c,x^{k}\rangle-{\overline{c}}=\langle s^{k},x^{k}-\overline{x}\rangle=\langle s^{k}_{I},x^{k}_{I}\rangle+\langle s^{k}_{J},x^{k}_{J}-\overline{x}_{J}\rangle\leq\langle s^{k}_{I},x^{k}_{I}\rangle+\|s^{k}_{J}\|_{2}\|x^{k}_{J}-\overline{x}_{J}\|_{2}$. Then using again Proposition 6 we get $\displaystyle{{\tilde{t}^{k}}\over\sum\limits_{i\in I}{x_{i}^{k}}^{r}}\leq\beta+\beta\displaystyle{\|s^{k}_{J}\|_{2}\|x^{k}_{J}-\overline{x}_{J}\|_{2}\over\langle x^{k}_{I},s^{k}_{I}\rangle}=\beta+o\left(\|x_{I}^{k}\|^{1-r}\right).$ But $\beta\in(0,1)$. The result then follows. ∎

Now as is mentioned in Theorem 1 of [1], given $t$ in $(-\infty,0)\cup(0,1)$, we say the $\xi_{t}$-dual-analytic center the unique optimal solution to the problem

[TABLE]

where

[TABLE]

if $t\in(0,1)$ and

[TABLE]

if $t\in(-\infty,0)$. The unicity of the solution is ensured by the strict quasi-concavity of $\xi_{t}$ and Lemma 1 of [1]. The KKT optimality conditions are then expressed as follow. There exist $(y,s)\in\mathbb{R}^{m}\times\mathbb{R}^{n}$ and $v\in\mathbb{R}^{n}$ satisfying the following conditions, $\nabla\xi_{t,I}(s_{I})=v_{I}$, $Av=0$, $A^{t}y+s=c$, $s_{J}=0$, $s\geq 0$.

Proof of Theorem 3.1.

According to Proposition 8, $\sum\limits_{k\geq 0}(F(x^{k+1})-F(x^{k}))$ converges and according to Proposition 7, $\sum\limits_{k\geq 0}\delta^{k}$ and $\sum\limits_{k\geq 0}\gamma^{k}$ converge too. Then using Proposition 9, $\sum\limits_{k\geq 0}\|X_{k,I}^{r\over 2}w_{I}^{k}\|_{2}^{2}$ converges. Hence

$\hfill\displaystyle\lim\limits_{k\to+\infty}{X_{k,I}^{r\over 2}\over\sum\limits_{i\in I}{x_{i}^{k}}^{r}}\left({\sum\limits_{i\in I}{x_{i}^{k}}^{r}\over\langle c,x^{k}\rangle-{\overline{c}}}X_{k,I}^{1-r}s^{k}_{I}-e\right)=0\hfill(1)$

For all $k\in\mathbb{N}$ we set $I(k)=\{i:\ \displaystyle{x_{i}^{k}\geq\|x^{k}_{I}\|_{\infty}^{2}}\}$. We shall prove that for some $K$ chosen large enough, $I(k)\subset I(k+1)$, $\forall k\geq K$. For $k\in\mathbb{N}$ and $i\in\{1,\cdots,n\}$ we set $\epsilon^{k}_{i}=\displaystyle{\sum\limits_{j\in I}{x_{j}^{k}}^{r}\over\langle c,x^{k}\rangle-{\overline{c}}}{x_{i}^{k}}^{1-r}s_{i}^{k}-1$. Let $\epsilon>0$ be small enough. Then by (1) there exists $K\in\mathbb{N}$ large enough such that $\forall k\geq K$, $|\epsilon_{i}^{k}|\leq\epsilon$, $\forall i\in I(k)$. Let $k\geq K$ and $i\in I(k)$. Since $K$ is assumed to be large we have necessarily from (1) $s_{i}^{k}>0$. Using in addition the fact that $\|s^{k}\|_{\infty}\leq g$ and Proposition 4 we get $gn_{I}\|x^{k}_{I}\|^{r}_{\infty}{x_{i}^{k}}^{1-r}s_{i}^{k}\geq\sum\limits_{j\in I}{x_{j}^{k}}^{r}{x_{i}^{k}}^{1-r}s_{i}^{k}=(\langle c,x^{k}\rangle-{\overline{c}})(1+\epsilon_{i}^{k})\geq L(A,c)\|x^{k}-{\overline{x}}\|_{2}(1+\epsilon^{k}_{i})\geq L(A,c)\|x^{k}_{I}\|_{\infty}(1-\epsilon).$ Hence $\|x^{k}_{I}\|_{\infty}\geq x_{i}^{k}\geq\displaystyle\left({L(A,c)(1-\epsilon)\over n_{I}g}\right)^{1\over 1-r}\|x^{k}_{I}\|_{\infty}$ and then $x_{i}^{k}=O(\|x_{I}^{k}\|_{\infty})$. Since in addition $x_{i}^{k+1}=\left(1-\beta\displaystyle{{x_{i}^{k}}^{1-r}s_{i}^{k}\over\max\limits_{j\in I}(X_{k}^{1-r}s^{k})_{j}}\right)x_{i}^{k}$ and $0<\displaystyle{{x_{i}^{k}}^{1-r}s_{i}^{k}\over\max\limits_{j\in I}(X_{k}^{1-r}s^{k})_{j}}\leq 1$ we have $(1-\beta)x_{i}^{k}\leq x_{i}^{k+1}\leq(1+\beta)x_{i}^{k}$ and then $x_{i}^{k+1}=O(x_{i}^{k})$. Now $\|x_{I}^{k+1}\|_{2}=\left\|x_{I}^{k}-\beta\displaystyle{{X_{k,I}}^{2-r}s_{I}^{k}\over\max\limits_{j\in I}(X_{k}^{1-r}s^{k})_{j}}\right\|_{2}\leq\|x_{I}^{k}\|_{2}+\|x_{I}^{k}\|_{2}\displaystyle{\|{X_{k,I}}^{1-r}s_{I}^{k}\|_{2}\over\max\limits_{j\in I}(X_{k}^{1-r}s^{k})_{j}}$ and $\sum\limits_{j\in I}{x_{j}^{k}}^{r}\|X_{k}^{1-r}s^{k}\|_{\infty}\geq\sum\limits_{j\in I}{x_{j}^{k}}^{r}\max\limits_{j\in I}{x_{j}^{k}}^{1-r}s_{j}^{k}\geq\sum\limits_{j\in I}{x_{j}^{k}}^{r}{x_{j}^{k}}^{1-r}s_{j}^{k}=\langle s_{I},x_{I}\rangle.$ It follows that $\|X_{k}^{1-r}s^{k}\|_{\infty}\geq\max\limits_{j\in I}{x_{j}^{k}}^{1-r}s_{j}^{k}\geq\displaystyle{\langle s_{I},x_{I}\rangle\over\sum\limits_{j\in I}{x_{j}^{k}}^{r}}.$ But (Proposition 6) $\|X_{k}^{1-r}s^{k}\|_{\infty}=O(\|x_{I}^{k}\|_{\infty}^{1-r})$, $\langle s^{k}_{I},x_{I}^{k}\rangle=O(\|x_{I}^{k}\|_{\infty})$ and $\sum\limits_{j\in I}{x_{j}^{k}}^{r}=O(\|x_{I}^{k}\|_{\infty}^{r})$. Then $\max\limits_{j\in I}{x_{j}^{k}}^{1-r}s_{j}^{k}=O(\|x_{I}^{k}\|_{\infty}^{1-r})$ and then $\displaystyle{\|{X_{k,I}}^{1-r}s_{I}^{k}\|_{2}\over\max\limits_{j\in I}(X_{k}^{1-r}s^{k})_{j}}=O(1).$ Hence there is $\varrho>0$ such that $\|x_{I}^{k+1}\|_{2}\leq\varrho\|x_{I}^{k}\|_{2}$. So $x_{i}^{k+1}=O(x_{i}^{k})=O(\|x_{I}^{k}\|_{2})\geq O(\|x_{I}^{k+1}\|_{2})\geq\|x_{I}^{k+1}\|_{2}^{2}$ and then $i\in I(k+1)$. Set now ${\hat{I}}=\cup{\atop{k\in\mathbb{N}}}I(k)$ and let us prove that in fact ${\hat{I}}=I$. Assume for contradiction that there is $i\in I-{\hat{I}}$. Then $\forall k\geq K$, $\displaystyle{x_{i}^{k+1}\over x_{i}^{k}}=1-\beta{{x_{i}^{k}}^{1-r}s_{i}^{k}\over\|X_{I,k}^{1-r}s^{k}_{I}\|_{\infty}}=1-\beta{{x_{i}^{k}}^{1-r}\over\|x^{k}_{I}\|^{2(1-r)}_{\infty}}{\|x_{I}^{k}\|_{\infty}^{1-r}\over\|X_{I,k}^{1-r}s^{k}_{I}\|_{\infty}}s_{i}^{k}\|x_{I}^{k}\|_{\infty}^{1-r}\geq 1-\beta{\|x_{I}^{k}\|_{\infty}^{1-r}\over\|X_{I,k}^{1-r}s^{k}_{I}\|_{\infty}}s_{i}^{k}\|x_{I}^{k}\|_{\infty}^{1-r}$. We know by Proposition 6 that $\|X_{I,k}^{1-r}s^{k}_{I}\|_{\infty}=O(\|x_{I}^{k}\|_{\infty}^{1-r})$. Using in addition the fact that $s^{k}$ is bounded it follows that $\displaystyle{x_{i}^{k+1}\over x_{i}^{k}}\geq 1+O(\|x_{I}^{k}\|_{\infty}^{1-r})$. Hence for all $K^{\prime}\geq K$, $\displaystyle{x_{i}^{K^{\prime}}\over x_{i}^{K}}\geq\prod\limits_{K\leq k\leq K^{\prime}}(1+O(\|x_{I}^{k}\|_{\infty}^{1-r}))=1+O\left(\sum\limits_{k=K}^{K^{\prime}}\|x_{I}^{k}\|_{\infty}^{1-r}\right)$. We know by Proposition 6 that $\sum\limits_{k\in\mathbb{N}}\|x_{I}^{k}\|_{\infty}^{1-r}$ is a converging serie. It follows that chosing $K$ large enough, $O\left(\sum\limits_{k=K}^{+\infty}\|x_{I}^{k}\|_{\infty}^{1-r}\right)\geq\displaystyle-{1\over 2}$. Then $0=\lim\limits_{K^{\prime}\to+\infty}\displaystyle{x_{i}^{K^{\prime}}\over x_{i}^{K}}\geq{1\over 2}$, which is absurde. Hence

$\hfill\lim\limits_{k\to+\infty}\epsilon^{k}=\lim\limits_{k\to+\infty}\displaystyle{\sum\limits_{i\in I}{x_{i}^{k}}^{r}\over\langle c,x^{k}\rangle-{\overline{c}}}X_{k,I}^{1-r}s^{k}_{I}-e_{I}=0\hfill(3)$

and then there is necessarily $\tau>0$ such that $\tau e_{I}<s_{I}^{k}$ for $k$ being large enough. Let now $({\tilde{y}},{\tilde{s}})$ be an accumulation point to $(y^{k},s^{k})$. Then we have ${\overline{X}}{\tilde{s}}=0,\ A{\overline{x}}=b,\ A{\tilde{y}}+{\tilde{s}}=c,\ {\overline{x}}\geq 0$ and ${\tilde{s}}\geq 0$. The KKT optimality conditions of $(LP)$ are then satisfied and then ${\overline{x}}$ is an $(LP)$ optimal solution and $({\tilde{y}},{\tilde{s}})$ is a dual optimal solution. Moreover since ${\overline{x}}_{I}=0,\ {\tilde{s}}_{I}>0,\ x_{J}>0$ and ${\tilde{s}}_{J}=0$ the strict complementary slackness condition holds. Let now $k$ be large enough. Then it is easy to see with the help of Proposition 6 that $(\langle c,x^{k}\rangle-{\overline{c}})\displaystyle{X_{k,I}^{r-1}\over\sum\limits_{i\in I}{x_{i}^{k}}^{r}}=O(1)$. It follows that $s_{I}^{k}=(\langle c,x^{k}\rangle-{\overline{c}})\displaystyle{X_{k,I}^{r-1}\over\sum\limits_{i\in I}{x_{i}^{k}}^{r}}(e_{I}+\epsilon^{k})=(\langle c,x^{k}\rangle-{\overline{c}})\displaystyle{X_{k,I}^{r-1}e_{I}\over\sum\limits_{i\in I}{x_{i}^{k}}^{r}}+\hat{O}(\epsilon)=\displaystyle{(\langle c,x^{k}\rangle-{\overline{c}})\over\xi_{r,I}(x_{I}^{k})}\nabla\xi_{r,I}(x_{I}^{k})+\hat{O}(\epsilon),$ where $\hat{O}({\epsilon})$ represents every function of $\epsilon$ satisfying $\lim\limits_{\epsilon\downarrow 0}\hat{O}({\epsilon})=0$. According to Theorem 3.1 of [2], we have $\xi_{t,I}(\nabla\xi_{r,I}(x_{I}^{k}))=1$ and then, by continuity of $\xi_{t,I}$ on $(0,+\infty)^{n_{I}}$, we get $\xi_{t,I}(s^{k}_{I})=\displaystyle{(\langle c,x^{k}\rangle-{\overline{c}})\over\xi_{r,I}(x_{I}^{k})}+\hat{O}(\epsilon)=O(1)$. Hence $\displaystyle{s_{I}^{k}\over\xi_{t,I}(s^{k}_{I})}=\nabla\xi_{r,I}(x_{I}^{k})+\hat{O}(\epsilon).$ Now it is easy to see from Theorem 3.1 of [2] that $\nabla\xi_{r,I}(\cdot)$ is positively homogeneous of degree 0. It follows that $\displaystyle{s_{I}^{k}\over\xi_{t,I}(s^{k}_{I})}=\nabla\xi_{r,I}\left({x_{I}^{k}\over\|x_{I}^{k}\|_{\infty}}\right)+\hat{O}(\epsilon).$ By (3) there is $\tau^{\prime}>0$ such that $\tau^{\prime}e_{I}\leq\displaystyle{x_{I}^{k}\over\|x_{I}^{k}\|_{\infty}}\leq e_{I}$ for $k$ being large enough. Then using in addition Lemma 2, $z^{k}=\displaystyle{x^{k}-\overline{x}\over\|x_{I}^{k}\|_{\infty}}$ is bounded. Let then $\overline{z}$ be a limit of a convergent subsequence of $(z^{k})$. Then $\displaystyle{\overline{s}_{I}\over\xi_{t,I}(\overline{s}_{I})}=\nabla\xi_{r,I}(\overline{z}_{I}).$ Using again Theorem 3.1 of [2] one has $\displaystyle{\overline{z}_{I}\over\xi_{r,I}(\overline{z}_{I})}=\nabla\xi_{t,I}(\overline{s}_{I}).$ But $Az^{k}=0$. It follows that $A\left(\displaystyle{\overline{z}\over\xi_{r,I}(\overline{z}_{I})}\right)=0$. Hence $\left(\overline{y},\overline{s},\displaystyle{\overline{z}\over\xi_{r,I}(\overline{z}_{I})}\right)$ satisfies the KKT optimality conditions for the problem

[TABLE]

The result then follows ∎

Turn back now to the case where ${\cal I}\not=\emptyset$. Then using (1) of Section 2 and adapting results of this section, the expected dual approximate optimal solution vectors $y,\ s$ and $w$, associated to a current point $x$ are

[TABLE]

[TABLE]

[TABLE]

where $U=diag(u)$. Hence the expected relative duality gap is

[TABLE]

4 Numerical results

The porpose of the following tests is to compare the algorithm performance according to different values of $r$ between 0 and 1. We have opted to consider the following values $r=0$ (the classical case), $r=0.1$, $r=0.2$, $r=0.3$, $r=0.4$, $r=0.5$, $r=0.6$ and $r=0.7$. We solved a large set of testing problems, taken from the familiar Netlib test set (GAY [10]). For values of $r$ exceeding 0.8 the algorithm showed lesser efficiency on most problems. Results obtained are shown in Table 1. Each row in the table contains the name of the problem and the number of iterations for the different values of $r$. The parameter of the stopping rule is $\epsilon=10^{-10}$. To read the mps-files, the specs-files and perform the symbolic Cholesky factorization we use rdmps1, rdmps2, rdspec, prepro and all dependencies written by Gondzio[12]. We use no presolving. Our numerical experiments were performed on a laptop hp ZBook (Processor: Intel core i7-4810 MQ, CPU $2.804\times 8$HZ - Operating system: Ubuntu linux). The code is written in GNU Fortran 95.

Table 1

$\star\star$ : Number of iterations exceeds 300.

$*$: Best optimal value obtained with $Rgap\in(10^{-8},10^{-10})$.

At first we can observe that for every problem there is at least one $r$ value for which the problem is solved. Also, we can see that most problems are solved for $r$ between 0 and 0.5. The maximum number of problems solved is reached for $r=0.2$, as shown in the graph below.

\psaxes[Ox=0,Oy=87,Dx=0.1,dx=1,Dy=1]-¿(0,87)(6,96) Percentage of solved problemsValues of $r$

5 Concluding remarks

The results are conclusive and show that differentially barriers penalty functions offer effective alternative to conventional logarithmic barrier function in linear programming.

As we point out in the introduction, we chose an algorithm of affine scaling type for the simplicity of its implementation. But the ”Predictor-corrector” method of Mehrotra [20] has proved highly efficient in the classical case ($r=0$). Our immediate goal is to adapt this method to these new penalty functions.