The Iteration-Complexity Upper Bound for the Mizuno-Todd-Ye Predictor-Corrector Algorithm is Tight

Abstract

It is an open question whether there is an interior-point algorithm for linear optimization problems with a lower iteration-complexity than the classical bound $\mathcal {O}(\sqrt{n} \log (\frac{\mu _1}{\mu _0}))$. This paper provides a negative answer to that question for a variant of the Mizuno-Todd-Ye predictor-corrector algorithm. In fact, we prove that for any $\varepsilon >0$, there is a redundant Klee-Minty cube for which the aforementioned algorithm requires $n^{( \frac{1}{2}-\varepsilon )}$ iterations to reduce the barrier parameter by at least a constant. This is provably the first case of an adaptive step interior-point algorithm where the classical iteration-complexity upper bound is shown to be tight.

Appendix

Lemma 8.1

For large enough r, there is one-dimensional LO problem with $(r+1)$ constraints for which $\tau _1 \sqrt{r}\le \kappa (\mu )\le \tau _2 \sqrt{r}$ for any $\mu \in [\alpha _1,\alpha _2]$, where $\alpha _1=\frac{1}{r-\frac{\sqrt{r}}{4}}$ and $\alpha _2=\frac{1}{r-\sqrt{r}}$ for some constants $\tau _1,\tau _2 \ge 0$.

Proof

Consider the problem $\min \{ \ y : \ y \le 1 \ \mathrm {and}, \ y \ge 0 \ \mathrm {counted}\ r \ \mathrm {times} \}$. The construction is given in [15], p:551. Consider the interval $[\alpha _1,\alpha _2]$, where $\alpha _1=\frac{1}{r-\frac{\sqrt{r}}{4}}$ and $\alpha _2=\frac{1}{r-\sqrt{r}}$. Let $s_0(\mu )=1-y(\mu )$. Then it is shown in [15], p. 551 that, $\displaystyle \frac{\dot{s}_0(\mu )}{s_0(\mu )} \ge \displaystyle \frac{r^2}{3\sqrt{r}}$ on $[\alpha _1,\alpha _2]$. This implies $\displaystyle \frac{\mu \dot{s}_0(\mu )}{s_0(\mu )}=\varOmega (\sqrt{r})$ on $[\alpha _1,\alpha _2]$. Then, from Proposition 2.1 part 1., we have $\kappa (\mu )=\varOmega (\sqrt{r})$ for all $\mu \in [\alpha _1,\alpha _2]$. The proof is complete.    $\square$

Proposition 8.1

Consider the LO problems

$$\begin{aligned} \begin{array}{ccc} \begin{array}{crl} \min &{} (c^1)^T x &{}\\ \mathrm {s.t. } &{} A^1 x^1 &{} =b^1 \\ &{} x^1 &{} \ge 0, \\ \end{array} &{} \text { and } \ &{} \begin{array}{crl} \min &{}(c^2)^T x &{}\\ \mathrm {s.t. } &{} A^2 x^2 &{} =b^2 \\ &{} x^2 &{} \ge 0, \ \ \\ \end{array} \\ \end{array} \end{aligned}$$

(19)

with the corresponding $\kappa ^1(\mu )$ and $\kappa ^2(\mu )$ on the interval $[\mu _0,\mu _1]$. Then for the problem

$$\begin{aligned} \begin{array}{crl} \min &{}c^T x &{}\\ \mathrm {s.t. } &{} A x &{} =b \\ &{} x &{} \ge 0, \ \ \\ \end{array} \end{aligned}$$

(20)

with the corresponding $\overline{\kappa }(\mu )$ where $c=\left[ \begin{array}{c} c^1\\ c^2\\ \end{array} \right]$, $b=\left[ \begin{array}{c} b^1\\ b^2\\ \end{array} \right]$ and $A= \left[ \begin{array}{cc} A^1 &{} 0 \\ 0 &{} A^2 \\ \end{array} \right]$, on $[\mu _0,\mu _1]$, we have $\overline{\kappa }(\mu ) \ge \kappa ^i(\mu )$ for $i=1,2$.

Proof

Let $\left( x^1(\mu ),y^1(\mu ),s^1(\mu )\right)$ and $\left( x^2(\mu ),y^2(\mu ),s^2(\mu )\right)$ be the central paths in (19). Then the term $\overline{\kappa }(\mu )$ for the combined problem (20) becomes $\overline{\kappa }(\mu )=\left| \left| [\mu \dot{x}^1 \dot{s}^1 , \mu \dot{x}^2 \dot{s}^2] \right| \right| ^{\frac{1}{2}} \ge \kappa ^i(\mu )$ for $i=1,2$.    $\square$

Proposition 8.2

Let $\eta >0$ and consider the central path ((2) and its $\kappa (\mu )$. Let $(\hat{A},\hat{b},\hat{c})$ be another problem instance, where $(\hat{A},\hat{b},\hat{c})=(A,\frac{b}{\eta },c)$ with its corresponding $\hat{\kappa }(\mu )$. Then, we have

$$\begin{aligned} \hat{\kappa }(\mu )=\kappa (\eta \mu ), \ \mu \in \left[ \frac{\mu _0}{\eta },\frac{\mu _1}{\eta } \right] . \end{aligned}$$

(21)

Proof

Using (2), it is straightforward to verify that the central path $(\hat{x}(\mu ),\hat{y}(\mu ),\hat{s}(\mu ))$ of the new problem satisfies $\hat{x}(\mu )= \displaystyle \frac{x(\eta \mu )}{\eta }$, $\hat{y}(\mu )= y(\eta \mu )$ and $\hat{s}(\mu )= s(\eta \mu )$. Using the definition of $\kappa (\mu )$, we get $\hat{\kappa }({\mu })=\kappa ({\eta \mu })$. Hence the claim follows.    $\square$

Lemma 8.2

Given an interval $[\mu _0,\mu _1]$ and a constant $\nu >0$, there exists an LO problem of size $n=\varTheta \left( \log (\frac{\mu _1}{\mu _0})\right)$ such that $\overline{\kappa }(\mu ) \ge \nu$ for all $\mu \in [\mu _0,\mu _1]$. The hidden constant in $n=\varTheta \left( \log (\frac{\mu _1}{\mu _0})\right)$ depends on $\nu$.

Proof

Let a constant $\nu >0$ and an interval $[\mu _0,\mu _1]$ be given. For the given $\nu >0$, by Lemma 8.1, there exists an LO problem with its $\kappa (\mu ) \ge \nu$ on an interval $\mu \in [\alpha _1,\alpha _2]$. By applying Proposition 8.2 for $\eta :=\frac{\alpha _1}{\left( \frac{\alpha _2}{\alpha _1}\right) ^i \mu _0}$ for $i=0,1,\dots ,k$, we find $(k-1)$ scaled LO problems with their corresponding $\kappa ^i(\mu )$, $i=0,1,\dots ,k-1$ such that $\kappa ^i(\mu )=\kappa (\eta \mu )$ on $\mu \in \left[ (\frac{\alpha _2}{\alpha _1})^i \mu _0, (\frac{\alpha _2}{\alpha _1})^{i+1} \mu _0 \right]$, for $i=0,1,\dots ,k-1$. Then by using Proposition 8.1, we can obtain a block diagonal LO problem with its $\overline{\kappa }(\mu ) \ge \kappa ^i(\mu ) \ge \nu$ for $i=0,1,\dots ,k-1$ for any $\mu \in \left[ \mu _0, \left( \frac{\alpha _2}{\alpha _1}\right) ^k \mu _0 \right]$. In order to have $\overline{\kappa }(\mu ) \ge \nu$ for any $\mu \in [\mu _0, \mu _1 ]$, it is then enough to have $\left( \frac{\alpha _2}{\alpha _1}\right) ^k \mu _0 \ge \mu _1$. This is true if and only if $k \log \left( \frac{\alpha _2}{\alpha _1}\right) \ge \log \left( \frac{\mu _1}{\mu _0} \right)$. Since by Lemma 8.1, the ratio $\frac{\alpha _2}{\alpha _1}$ is a constant depending only on the given $\nu$, the number of blocks k needed is $\varTheta \left( \log (\frac{\alpha _2}{\alpha _1}) \right)$. Also since the size of the LO problem with its $\kappa (\mu )$ is a constant only determined by $\nu$, the size of the problem is $n=\varTheta \left( \log (\frac{\mu _1}{\mu _0})\right)$ to achieve $\overline{\kappa }(\mu ) \ge \nu$ for all $\mu \in [\mu _0,\mu _1]$. This completes the proof.    $\square$