A polynomial time infeasible interior-point arc-search algorithm for convex optimization

Abstract

This paper proposes an infeasible interior-point algorithm for the convex optimization problem using arc-search techniques. The proposed algorithm simultaneously selects the centering parameter and the step size, aiming at optimizing the performance in every iteration. Analytic formulas for the arc-search are provided to make the arc-search method very efficient. The convergence of the algorithm is proved and a polynomial bound of the algorithm is established. The preliminary numerical test results indicate that the algorithm is efficient and effective

Appendix: Selection of the centering parameter \(\sigma _{k}\) and step size \(\alpha _{k}\)

Although the method of selecting $\alpha _{k}$ described in Sects. 3 and 4 assures that the algorithm converges in polynomial iteration, but this selection is very conservative. A better method is to simultaneously select centering parameter $\sigma _{k}$ and step size $\alpha _{k}$ to maximize the step size in every iteration. The merit of this holistic strategy is proved in theory (Yang 2021), and has been demonstrated in computational experiments (Yang 2017, 2018). The same strategy is proposed in Step 5 of Algorithm 3.1, but there is no details provided there. In this “Appendix”, we discuss how this strategy is implemented. Although the formulas in this “Appendix” are similar to the ones in Yang (2018), they are different. To avoid the confusion and implementation errors, we would like to list them in this “Appendix”. For the sake of completeness, we also provide the proofs even though they follow the same ideas of Yang (2018).

Let the current iterate be $\mathbf{v}^{k}=(\mathbf{x}^{k},\mathbf{y}^{k},\mathbf{w}^{k},\mathbf{s}^{k},\mathbf{z}^{k})$, $({\dot{\mathbf{x}}},{\dot{\mathbf{y}}},{\dot{\mathbf{w}}},{\dot{\mathbf{s}}},{\dot{\mathbf{z}}})$ be computed by solving (13), $(\mathbf{p}_{\mathbf{x}},\mathbf{p}_{\mathbf{y}},\mathbf{p}_{\mathbf{w}},\mathbf{p}_{\mathbf{s}},\mathbf{p}_{\mathbf{z}})$ be computed by solving (17) and $(\mathbf{q}_{\mathbf{x}},\mathbf{q}_{\mathbf{y}},\mathbf{q}_{\mathbf{w}},\mathbf{q}_{\mathbf{s}},\mathbf{q}_{\mathbf{z}})$ be computed by solving (18), $\phi _{k}$ and $\psi _{k}$ be computed by using (23) and (24). An intuition based on Propositions 3.1 and 3.3 is that the step size $\alpha _{k}$ should be chosen as large as possible provided that Condition (C4), (26) and (27) hold. Given $\mathbf{v}^{k}=(\mathbf{x}^{k},\mathbf{y}^{k},\mathbf{w}^{k},\mathbf{s}^{k},\mathbf{z}^{k})$, $({\dot{\mathbf{x}}},{\dot{\mathbf{y}}},{\dot{\mathbf{w}}},{\dot{\mathbf{s}}},{\dot{\mathbf{z}}})$, $(\mathbf{p}_{\mathbf{x}},\mathbf{p}_{\mathbf{y}},\mathbf{p}_{\mathbf{w}},\mathbf{p}_{\mathbf{s}},\mathbf{p}_{\mathbf{z}})$, $(\mathbf{q}_{\mathbf{x}},\mathbf{q}_{\mathbf{y}},\mathbf{q}_{\mathbf{w}},\mathbf{q}_{\mathbf{s}},\mathbf{q}_{\mathbf{z}})$, $\phi _{k}$ and $\psi _{k}$, similar to the derivation of Yang (2017), the largest ${\tilde{\alpha }}$ that meet conditions (26) and (27) can be expressed as a function of $\sigma _{k}$. For each $i \in \lbrace 1,\ldots , n \rbrace$, given $\sigma$, we can select the largest $\alpha _{s_{i}}$ such that for any $\alpha \in [0, \alpha _{s_{i}}]$, the ith inequality of (26) holds, and the largest $\alpha _{z_{i}}$ such that for any $\alpha \in [0, \alpha _{z_{i}}]$ the ith inequality of (27) holds. We then define

$$\begin{aligned}&{\alpha ^s}=\displaystyle \min _{i \in \lbrace 1,\ldots , n \rbrace } \lbrace \alpha _{s_{i}}\rbrace , \end{aligned}$$

(A1)

$$\begin{aligned}&{\alpha ^z}=\displaystyle \min _{i \in \lbrace 1,\ldots , n \rbrace } \lbrace \alpha _{z_{i}} \rbrace , \end{aligned}$$

(A2)

$$\begin{aligned}&{{\tilde{\alpha }}}=\min \lbrace \alpha ^{s}, \alpha ^{z} \rbrace , \end{aligned}$$

(A3)

where $\alpha _{s_{i}}$ and $\alpha _{z_{i}}$ can be obtained, using a similar argument as in (Yang 2017), in analytical forms represented by $\phi _{k}$, $\dot{s}_{i}$, $\ddot{s}_{i}=p_{s_{i}}\sigma +q_{s_{i}}$, $\psi _{k}$, $\dot{z}_{i}$, and $\ddot{z}_{i}=p_{z_{i}}\sigma +q_{z_{i}}$. First, from (26), we have

$$\begin{aligned} s_{i} +\ddot{s}_{i} - \phi _{k} \ge \dot{s}_{i}\sin (\alpha )+\ddot{s}_{i}\cos (\alpha ). \end{aligned}$$

(A4)

Case 1a ($\dot{s}_{i}=0$ and $p_{s_{i}}\sigma +q_{s_{i}} \ne 0$):

In this case, if $\ddot{s}_{i} \ge -(s_{i}-\phi _{k})$ and $\alpha \in \left[ 0, \frac{\pi }{2}\right]$, then $s_{i}(\alpha ) \ge \phi _{k}$ follows from (26). If $\ddot{s}_{i} \le -(s_{i}-\phi _{k})$ or $s_{i} +\ddot{s}_{i} - \phi _{k} \le 0$, to meet (A4), we must have $\cos (\alpha ) \ge \frac{x_{i} +\ddot{x}_{i}-\phi _{k}}{\ddot{x}_{i}}$, or, $\alpha \le \cos ^{-1}\left( \frac{x_{i} +\ddot{x}_{i}-\phi _{k}}{\ddot{x}_{i}} \right)$. Therefore,

$$\begin{aligned} \alpha _{s_{i}}(\sigma ) = \left\{ \begin{array}{ll} \frac{\pi }{2} &{} \text{ if }\, s_{i} +(p_{s_{i}}\sigma +q_{s_{i}})-\phi _{k} \ge 0 \\ \cos ^{-1}\left( \frac{s_{i} -\phi _{k} +p_{s_{i}}\sigma +q_{s_{i}}}{p_{s_{i}}\sigma +q_{s_{i}}} \right) &{} \text{ if } \,s_{i} +(p_{s_{i}}\sigma +q_{s_{i}})-\phi _{k} \le 0 . \end{array} \right. \end{aligned}$$

(A5)

Case 2a ($p_{s_{i}}\sigma +q_{s_{i}}=0$ and $\dot{s}_{i}\ne 0$):

In this case, if $\dot{s}_{i} \le s_{i} -\phi _{k}$ and $\alpha \in \left[ 0, \frac{\pi }{2}\right]$, then $s_{i}(\alpha ) \ge \phi _{k}$ follows from (26). If $\dot{s}_{i} \ge s_{i} -\phi _{k}$, to meet (A4), we must have $\sin (\alpha ) \le \frac{s_{i} -\phi _{k}}{\dot{s}_{i}}$, or $\alpha \le \sin ^{-1}\left( \frac{s_{i} -\phi _{k}}{\dot{s}_{i}} \right)$. Therefore,

$$\begin{aligned} \alpha _{s_{i}}(\sigma ) = \left\{ \begin{array}{ll} \frac{\pi }{2} &{} \quad \text{ if } \dot{s}_{i} \le s_{i} -\phi _{k} \\ \sin ^{-1}\left( \frac{s_{i}-\phi _{k} }{\dot{s}_{i}} \right) &{} \quad \text{ if } \dot{s}_{i} \ge s_{i} -\phi _{k} \end{array} \right. \end{aligned}$$

(A6)

Case 3a ($\dot{s}_{i}>0$ and $p_{s_{i}}\sigma +q_{s_{i}}>0$):

Let $\dot{s}_{i}=\sqrt{\dot{s}_{i}^{2}+\ddot{s}_{i}^{2}}\cos (\beta )$, and $\ddot{s}_{i}=\sqrt{\dot{s}_{i}^{2}+\ddot{s}_{i}^{2}}\sin (\beta )$, (A4) can be rewritten as

$$\begin{aligned} s_{i} + \ddot{s}_{i} -\phi _{k} \ge \sqrt{\dot{s}_{i}^{2}+\ddot{s}_{i}^{2}} \sin (\alpha + \beta ), \end{aligned}$$

(A7)

where

$$\begin{aligned} \beta = \sin ^{-1} \left( \frac{\ddot{s}_{i} }{\sqrt{\dot{s}_{i}^{2}+\ddot{s}_{i}^{2}}} \right) = \sin ^{-1}\left( \frac{p_{s_{i}}\sigma +q_{s_{i}} }{\sqrt{\dot{s}_{i}^{2}+(p_{s_{i}}\sigma +q_{s_{i}})^{2}}} \right) . \end{aligned}$$

(A8)

If $\ddot{s}_{i} + s_{i} -\phi _{k}\ge \sqrt{\dot{s}_{i}^{2}+\ddot{s}_{i}^{2}}$ and $\alpha \in \left[ 0, \frac{\pi }{2}\right]$, then $s_{i}(\alpha ) \ge \phi _{k}$ follows from (26). If $\ddot{s}_{i} + s_{i} -\phi _{k} \le \sqrt{\dot{s}_{i}^{2}+\ddot{s}_{i}^{2}}$, to meet (A7), we must have $\sin (\alpha + \beta ) \le \frac{s_{i} + \ddot{s}_{i} -\phi _{k}}{\sqrt{\dot{s}_{i}^{2}+\ddot{s}_{i}^{2}}}$, or $\alpha + \beta \le \sin ^{-1}\left( \frac{s_{i} + \ddot{s}_{i} -\phi _{k}}{\sqrt{\dot{s}_{i}^{2}+\ddot{s}_{i}^{2}}} \right)$. Therefore,

$$\begin{aligned} \alpha _{s_{i}}(\sigma ) = \left\{ \begin{array}{ll} \frac{\pi }{2} &{} \text{ if }\, s_{i}-\phi _{k} + p_{s_{i}}\sigma +q_{s_{i}} \ge \sqrt{\dot{s}_{i}^{2}+(p_{s_{i}}\sigma +q_{s_{i}})^{2}}, \\ \sin ^{-1}\left( \frac{s_{i} -\phi _{k} + p_{s_{i}}\sigma +q_{s_{i}} }{\sqrt{\dot{s}_{i}^{2}+(p_{s_{i}}\sigma +q_{s_{i}})^{2}}} \right) - \beta &{} \text{ if }\, s_{i} -\phi _{k} + p_{s_{i}}\sigma +q_{s_{i}} \le \sqrt{\dot{s}_{i}^{2}+(p_{s_{i}}\sigma +q_{s_{i}})^{2}}. \end{array} \right. \end{aligned}$$

(A9)

Case 4a ($\dot{s}_{i}>0$ and $p_{s_{i}}\sigma +q_{s_{i}}<0$):

Let $\dot{s}_{i}=\sqrt{\dot{s}_{i}^{2}+\ddot{s}_{i}^{2}}\cos (\beta )$, and $\ddot{s}_{i}=-\sqrt{\dot{s}_{i}^{2}+\ddot{s}_{i}^{2}}\sin (\beta )$, (A4) can be rewritten as

$$\begin{aligned} s_{i} + \ddot{s}_{i} -\phi _{k} \ge \sqrt{\dot{s}_{i}^{2}+\ddot{s}_{i}^{2}} \sin (\alpha - \beta ), \end{aligned}$$

(A10)

where

$$\begin{aligned} \beta = \sin ^{-1} \left( \frac{-\ddot{s}_{i} }{\sqrt{\dot{s}_{i}^{2}+\ddot{s}_{i}^{2}}} \right) = \sin ^{-1}\left( \frac{-(p_{s_{i}}\sigma +q_{s_{i}} )}{\sqrt{\dot{s}_{i}^{2}+(p_{s_{i}}\sigma +q_{s_{i}})^{2}}} \right) . \end{aligned}$$

(A11)

If $\ddot{s}_{i} + s_{i} -\phi _{k} \ge \sqrt{\dot{s}_{i}^{2}+\ddot{s}_{i}^{2}}$ and $\alpha \in \left[ 0, \frac{\pi }{2}\right]$, then $s_{i}(\alpha ) \ge \phi _{k}$ follows from (26). If $\ddot{s}_{i} + s_{i} -\phi _{k} \le \sqrt{\dot{s}_{i}^{2}+\ddot{s}_{i}^{2}}$, to meet (A10), we must have $\sin (\alpha - \beta ) \le \frac{s_{i} + \ddot{s}_{i}}{\sqrt{\dot{s}_{i}^{2}+\ddot{s}_{i}^{2}}}$, or $\alpha - \beta \le \sin ^{-1}\left( \frac{s_{i} + \ddot{s}_{i} }{\sqrt{\dot{s}_{i}^{2}+\ddot{s}_{i}^{2}}} \right)$. Therefore,

$$\begin{aligned} \alpha _{s_{i}}(\sigma ) = \left\{ \begin{array}{ll} \frac{\pi }{2} &{} \text{ if }\, s_{i} -\phi _{k} + p_{s_{i}}\sigma +q_{s_{i}} \ge \sqrt{\dot{s}_{i}^{2}+(p_{s_{i}}\sigma +q_{s_{i}})^{2}},\\ \sin ^{-1}\left( \frac{s_{i}-\phi _{k} + p_{s_{i}}\sigma +q_{s_{i}} }{\sqrt{\dot{s}_{i}^{2}+(p_{s_{i}}\sigma +q_{s_{i}})^{2}}} \right) + \beta &{} \text{ if }\, s_{i}-\phi _{k} + p_{s_{i}}\sigma +q_{s_{i}} \le \sqrt{\dot{s}_{i}^{2}+(p_{s_{i}}\sigma +q_{s_{i}})^{2}}. \end{array} \right. \end{aligned}$$

(A12)

Case 5a ($\dot{s}_{i}<0$ and $p_{s_{i}}\sigma +q_{s_{i}}<0$):

Let $\dot{s}_{i}=-\sqrt{\dot{s}_{i}^{2}+\ddot{s}_{i}^{2}}\cos (\beta )$, and $\ddot{s}_{i}=-\sqrt{\dot{s}_{i}^{2}+\ddot{s}_{i}^{2}}\sin (\beta )$, (A4) can be rewritten as

$$\begin{aligned} s_{i} + \ddot{s}_{i} -\phi _{k} \ge -\sqrt{\dot{s}_{i}^{2}+\ddot{s}_{i}^{2}} \sin (\alpha +\beta ), \end{aligned}$$

(A13)

where

$$\begin{aligned} \beta = \sin ^{-1} \left( \frac{-\ddot{s}_{i} }{\sqrt{\dot{s}_{i}^{2}+\ddot{s}_{i}^{2}}} \right) = \sin ^{-1}\left( \frac{-(p_{s_{i}}\sigma +q_{s_{i}}) }{\sqrt{\dot{s}_{i}^{2}+(p_{s_{i}}\sigma +q_{s_{i}})^{2}}} \right) . \end{aligned}$$

(A14)

If $\ddot{s}_{i} + s_{i} -\phi _{k} \ge 0$ and $\alpha \in \left[ 0, \frac{\pi }{2}\right]$, then $s_{i}(\alpha ) \ge \phi _{k}$ follows from (26). If $\ddot{s}_{i} + s_{i} -\phi _{k} \le 0$, to meet (A13), we must have $\sin (\alpha + \beta ) \ge \frac{-(s_{i} + \ddot{s}_{i} -\phi _{k})}{\sqrt{\dot{s}_{i}^{2}+\ddot{s}_{i}^{2}}}$, or $\alpha + \beta \le \pi - \sin ^{-1} \left( \frac{-(s_{i} + \ddot{s}_{i} -\phi _{k})}{\sqrt{\dot{s}_{i}^{2}+\ddot{s}_{i}^{2}}} \right)$. Therefore,

$$\begin{aligned} \alpha _{s_{i}}(\sigma ) = \left\{ \begin{array}{ll} \frac{\pi }{2} &{} \text{ if }\, s_{i}-\phi _{k} + p_{s_{i}}\sigma +q_{s_{i}}\ge 0, \\ \pi - \sin ^{-1} \left( \frac{-(s_{i} -\phi _{k} + p_{s_{i}}\sigma +q_{s_{i}}) }{\sqrt{\dot{s}_{i}^{2}+(p_{s_{i}}\sigma +q_{s_{i}})^{2}}} \right) - \beta &{} \text{ if }\, s_{i}-\phi _{k} + p_{s_{i}}\sigma +q_{s_{i}} \le 0. \end{array} \right. \end{aligned}$$

(A15)

Case 6a ($\dot{s}_{i}<0$ and $p_{s_{i}}\sigma +q_{s_{i}}>0$):

$$\begin{aligned} \alpha _{s_{i}}(\sigma ) = \frac{\pi }{2}. \end{aligned}$$

(A16)

Case 7a ($\dot{s}_{i}=0$ and $p_{s_{i}}\sigma +q_{s_{i}}=0$):

$$\begin{aligned} \alpha _{s_{i}}(\sigma ) = \frac{\pi }{2}. \end{aligned}$$

(A17)

Using the same idea, we can obtain the similar formulas for $\alpha _{z_{i}}(\sigma )$.

Case 1b ($\dot{z}_{i}=0$, $p_{z_{i}}\sigma +q_{z_{i}} \ne 0$):

$$\begin{aligned} \alpha _{z_{i}}(\sigma ) = \left\{ \begin{array}{ll} \frac{\pi }{2} &{} \text{ if }\, z_{i}-\psi _{k} +p_{z_{i}}\sigma +q_{z_{i}} \ge 0, \\ \cos ^{-1}\left( \frac{z_{i} -\psi _{k} +p_{z_{i}}\sigma +q_{z_{i}}}{p_{z_{i}}\sigma +q_{z_{i}}} \right) &{} \text{ if }\,z_{i}-\psi _{k} +p_{z_{i}}\sigma +q_{z_{i}} \le 0 . \end{array} \right. \end{aligned}$$

(A18)

Case 2b ($p_{z_{i}}\sigma +q_{z_{i}}=0$ and $\dot{z}_{i} \ne 0$):

$$\begin{aligned} \alpha _{z_{i}}(\sigma ) = \left\{ \begin{array}{ll} \frac{\pi }{2} &{} \text{ if }\, \dot{z}_{i} \le z_{i} -\psi _{k}, \\ \sin ^{-1}\left( \frac{z_{i} -\psi _{k} }{\dot{z}_{i}} \right) &{} \text{ if }\, \dot{z}_{i} \ge z_{i} -\psi _{k}. \end{array} \right. \end{aligned}$$

(A19)

Case 3b ($\dot{z}_{i}>0$ and $p_{z_{i}}\sigma +q_{z_{i}}>0$):

Let

$$\begin{aligned}&\beta = \sin ^{-1}\left( \frac{p_{z_{i}}\sigma +q_{z_{i}}}{\sqrt{\dot{z}_{i}^{2}+(p_{z_{i}}\sigma +q_{z_{i}})^{2}}} \right) . \end{aligned}$$

(A20)

$$\begin{aligned}&\alpha _{z_{i}}(\sigma ) = \left\{ \begin{array}{ll} \frac{\pi }{2} &{} \text{ if }\, z_{i} -\psi _{k} + p_{z_{i}}\sigma +q_{z_{i}} \ge \sqrt{\dot{z}_{i}^{2}+(p_{z_{i}}\sigma +q_{z_{i}})^{2}}, \\ \sin ^{-1}\left( \frac{z_{i} -\psi _{k} + p_{z_{i}}\sigma +q_{z_{i}} }{\sqrt{\dot{z}_{i}^{2}+(p_{z_{i}}\sigma +q_{z_{i}})^{2}}} \right) - \beta &{} \text{ if }\, z_{i} -\psi _{k} + p_{z_{i}}\sigma +q_{z_{i}} < \sqrt{\dot{z}_{i}^{2}+(p_{z_{i}}\sigma +q_{z_{i}})^{2}}. \end{array} \right. \end{aligned}$$

(A21)

Case 4b ($\dot{z}_{i}>0$ and $p_{z_{i}}\sigma +q_{z_{i}}<0$):

Let

$$\begin{aligned}&\beta = \sin ^{-1}\left( \frac{-(p_{z_{i}}\sigma +q_{z_{i}})}{\sqrt{\dot{z}_{i}^{2}+(p_{z_{i}}\sigma +q_{z_{i}})^{2}}} \right) . \end{aligned}$$

(A22)

$$\begin{aligned}&\alpha _{z_{i}}(\sigma ) = \left\{ \begin{array}{ll} \frac{\pi }{2} &{} \text{ if }\,z_{i} -\psi _{k} + p_{z_{i}}\sigma +q_{z_{i}} \ge \sqrt{\dot{z}_{i}^{2}+(p_{z_{i}}\sigma +q_{z_{i}})^{2}}, \\ \sin ^{-1}\left( \frac{z_{i} -\psi _{k} + p_{z_{i}}\sigma +q_{z_{i}}}{\sqrt{\dot{z}_{i}^{2}+(p_{z_{i}}\sigma +q_{z_{i}})^{2}}} \right) + \beta &{} \text{ if }\, z_{i}-\psi _{k} + p_{z_{i}}\sigma +q_{z_{i}} \le \sqrt{\dot{z}_{i}^{2}+(p_{z_{i}}\sigma +q_{z_{i}})^{2}}. \end{array} \right. \end{aligned}$$

(A23)

Case 5b ($\dot{z}_{i}<0$ and $p_{z_{i}}\sigma +q_{z_{i}}<0$):

Let

$$\begin{aligned}&\beta = \sin ^{-1}\left( \frac{-(p_{z_{i}}\sigma +q_{z_{i}}) }{\sqrt{\dot{z}_{i}^{2}+(p_{z_{i}}\sigma +q_{z_{i}})^{2}}} \right) . \end{aligned}$$

(A24)

$$\begin{aligned}&\alpha _{z_{i}}(\sigma ) = \left\{ \begin{array}{ll} \frac{\pi }{2} &{} \text{ if }\, z_{i}-\psi _{k} + p_{z_{i}}\sigma +q_{z_{i}} \ge 0 \\ \pi - \sin ^{-1} \left( \frac{-(z_{i} -\psi _{k} + p_{z_{i}}\sigma +q_{z_{i}}) }{\sqrt{\dot{z}_{i}^{2}+(p_{z_{i}}\sigma +q_{z_{i}})^{2}}} \right) - \beta &{} \text{ if }\, z_{i} -\psi _{k} + p_{z_{i}}\sigma +q_{z_{i}} \le 0. \end{array} \right. \end{aligned}$$

(A25)

Case 6b ($\dot{z}_{i}<0$ and $p_{z_{i}}\sigma +q_{z_{i}}>0$):

$$\begin{aligned} \alpha _{z_{i}}(\sigma ) = \frac{\pi }{2}. \end{aligned}$$

(A26)

Case 7b ($\dot{z}_{i}=0$ and $p_{z_{i}}\sigma +q_{z_{i}}=0$):

$$\begin{aligned} \alpha _{z_{i}}(\sigma ) = \frac{\pi }{2}. \end{aligned}$$

(A27)

Using this analytic formulas, our strategy to reduce the duality gap is to simultaneously select $\alpha _{k}$ and $\sigma _{k}$ by an iterative method similar to the idea of Yang (2018). This is implemented as follows: in every iteration k, given fixed $\phi _{k}$, $\psi _{k}$, ${\dot{\mathbf{s}}}$, ${\dot{\mathbf{z}}}$, $\mathbf{p}_{\mathbf{s}}$, $\mathbf{p}_{\mathbf{z}}$, $\mathbf{q}_{\mathbf{s}}$ and $\mathbf{q}_{\mathbf{z}}$, several different values of $\sigma$ are tried to find the best $\sigma _{k}$ for the maximum of ${\tilde{\alpha }}$. Therefore, we will find a $\sigma _{k}$ which maximizes the step size ${\tilde{\alpha }}$, i.e.,

$$\begin{aligned} \displaystyle \max _{\sigma \in [\sigma _{\min },\sigma _{\max }]} \displaystyle \min _{i \in \{ 1, \ldots ,n\} } \{ \alpha _{s_{i}}(\sigma ), \alpha _{z_{i}}(\sigma ) \}, \end{aligned}$$

(A28)

where $0 \le \sigma _{\min } < \sigma _{\max } \le 1$, $\alpha _{s_{i}}(\sigma )$ and $\alpha _{z_{i}}(\sigma )$ are calculated using (A5)–(A27) for $\sigma \in [\sigma _{\min },\sigma _{\max }]$. Problem (A28) has no regularity conditions involving derivatives. Golden section search for variable $\sigma$ (Ekefer 1953) seems to be an appropriate method for solving this problem. Noting the fact from (26) that $\alpha _{s_{i}}({\sigma })$ is a monotonic increasing function of $\sigma$ if $p_{s_{i}}>0$ and $\alpha _{s_{i}}({\sigma })$ is a monotonic decreasing function of $\sigma$ if $p_{s_{i}}<0$ [and similar properties hold for $\alpha _{z_{i}}(\sigma )$], we can use the condition

$$ \begin{aligned} \min \left\{ \displaystyle \min _{\{i \in p_{s_{i}}<0\}} \alpha _{s_{i}}(\sigma ), \displaystyle \min _{\{i \in p_{z_{i}}<0\}} \alpha _{z_{i}}(\sigma )\right\}> \min \left\{ \displaystyle \min _{\{i \in p_{s_{i}}>0\}} \alpha _{s_{i}}(\sigma ), \displaystyle \min _{\{i \in p_{z_{i}}>0\}} \alpha _{z_{i}}(\sigma ) \right\} , \end{aligned} $$

(A29)

and the following bisection search for variable $\sigma$ to solve (A28).

Algorithm 6.1

(bisection search devised for solving (A28))

Data: $(\dot{x},\dot{s})$, $(p_{x}, p_{s} )$, $(q_{x}, q_{s})$, $(x^{k},s^{k})$, $\phi _{k}$, and $\psi _{k}$.

Parameter: $\epsilon \in (0,1)$, $\sigma _{lb}=\sigma _{\min }$, $\sigma _{ub}=\sigma _{\max } \le 1$.

for iteration $k=0,1,2,\ldots$

Step 0::

If $\sigma _{ub}-\sigma _{lb} \le \epsilon$, set $\alpha =\displaystyle \min _{i \in \{ 1, \ldots ,n\} } \{ \alpha _{x_{i}}(\sigma ), \alpha _{s_{i}}(\sigma ) \}$, stop.

Step 1::

Set $\sigma =\sigma _{lb}+0.5(\sigma _{ub}-\sigma _{lb})$.

Step 2::

Calculate $\alpha _{x_{i}}(\sigma )$ and $\alpha _{s_{i}}(\sigma )$ using (A5)–(A27).

Step 3::

If (A29) holds, set $\sigma _{lb}=\sigma$, otherwise, set $\sigma _{ub}=\sigma$.

Step 4::

Set $k+1 \rightarrow k$. Go back to Step 1.

end (for) $\square$

This algorithm reduces interval length by 0.5 in every iteration while golden section method reduces interval length by 0.618. The bisection is more efficient.

In view of Proposition 3.3, if $({\dot{\mathbf{s}}}^{\text{T}} {\mathbf{p}}_{\mathbf{z}}+{\dot{\mathbf{z}}}^{\text{T}} {\mathbf{p}}_{\mathbf{s}}) < 0$, to minimize $\mu _{k+1}$, we should select $\sigma _{k} =0$. Therefore, Problem (A28) is reduced to solve a much simpler problem

$$\begin{aligned} {\tilde{\alpha }}= \min _{\alpha } b_u(\alpha ). \end{aligned}$$

(A30)

This is a one-dimensional unconstrained optimization problem that can be solved by many existing methods, such as golden section method. Given ${\tilde{\alpha }}$, we still need to find the largest $\alpha _{k} \in (0, {\tilde{\alpha }}]$ such that Condition (C4) holds. We summarize the algorithm described above as follows:

Algorithm 6.2

(bisection search devised for Step 5 of Algorithm 3.1) )

Data: $({\dot{\mathbf{x}}},{\dot{\mathbf{s}}})$, $(\mathbf{p}_{\mathbf{x}}, \mathbf{p}_{\mathbf{s}} )$, $(\mathbf{q}_{\mathbf{x}}, \mathbf{q}_{\mathbf{s}})$, $(\mathbf{x}^{k},\mathbf{s}^{k})$, $\phi _{k}$, and $\psi _{k}$.

Parameter: $\epsilon \in (0,1)$.

Step 1::

If $({\dot{\mathbf{s}}}^{\text{T}} {\mathbf{p}}_{\mathbf{z}}+{\dot{\mathbf{z}}}^{\text{T}} {\mathbf{p}}_{\mathbf{s}}) < 0$, set $\sigma _{k} =0$, solve (A30) to get ${\tilde{\alpha }}$.

Step 2::

Otherwise, call Algorithm 6.1 to get ${\tilde{\alpha }}$.

Step 3::

Find the largest $\alpha _{k} \in (0, {\tilde{\alpha }}]$ such that Condition (C4) holds. $\square$