# American Institute of Mathematical Sciences

October  2018, 14(4): 1595-1615. doi: 10.3934/jimo.2018023

## The modified inertial relaxed CQ algorithm for solving the split feasibility problems

 1 Department of Mathematics, Faculty of Science, Chiang Mai University, Chiang Mai 50200, Thailand 2 School of Science, University of Phayao, Phayao 56000, Thailand

∗ Corresponding author: prasitch2008@yahoo.com (P. Cholamjiak)

Received  April 2017 Revised  August 2017 Published  January 2018

In this work, we propose a new version of inertial relaxed CQ algorithms for solving the split feasibility problems in the frameworks of Hilbert spaces. We then prove its strong convergence by using a viscosity approximation method under some weakened assumptions. To be more precisely, the computation on the norm of operators and the metric projections is relaxed. Finally, we provide numerical experiments to illustrate the convergence behavior and to show the effectiveness of the sequences constructed by the inertial technique.

Citation: Suthep Suantai, Nattawut Pholasa, Prasit Cholamjiak. The modified inertial relaxed CQ algorithm for solving the split feasibility problems. Journal of Industrial & Management Optimization, 2018, 14 (4) : 1595-1615. doi: 10.3934/jimo.2018023
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##### References:
Comparison of the iterations of Choice 1 in Example 1
Comparison of the iterations of Choice 2 in Example 1
Comparison of the iterations of Choice 3 in Example 1
Comparison of the iterations of Choice 4 in Example 1
Comparison of the iterations of Choice 1 in Example 2
Comparison of the iterations of Choice 2 in Example 2
Comparison of the iterations of Choice 3 in Example 2
Comparison of the iterations of Choice 4 in Example 2
Error ploting of Choice 1 in Example 1
Error ploting of Choice 2 in Example 1
Error ploting of Choice 3 in Example 1
Error ploting of Choice 4 in Example 1
Algorithm 3.1 with different cases of $\rho_n$ and different choices of $x_0$ and $x_1$
 Case 1 Case 2 Case 3 Case 4 Choice 1 No. of Iter. 11 8 5 4 cpu (Time) $0.003553$ $0.002377$ $0.002195$ $0.002075$ Choice 2 No. of Iter. 7 6 4 4 cpu (Time) $0.002799$ $0.002769$ $0.002357$ $0.002184$ Choice 3 No. of Iter. 12 9 6 4 cpu (Time) $0.003828$ $0.002602$ $0.002401$ $0.002142$ Choice 4 No. of Iter. 27 17 11 9 cpu (Time) $0.007181$ $0.00343$ $0.002612$ $0.002431$ The numerical experiments for each case of $\rho_{n}$ are shown in Figure 1-4, respectively.
 Case 1 Case 2 Case 3 Case 4 Choice 1 No. of Iter. 11 8 5 4 cpu (Time) $0.003553$ $0.002377$ $0.002195$ $0.002075$ Choice 2 No. of Iter. 7 6 4 4 cpu (Time) $0.002799$ $0.002769$ $0.002357$ $0.002184$ Choice 3 No. of Iter. 12 9 6 4 cpu (Time) $0.003828$ $0.002602$ $0.002401$ $0.002142$ Choice 4 No. of Iter. 27 17 11 9 cpu (Time) $0.007181$ $0.00343$ $0.002612$ $0.002431$ The numerical experiments for each case of $\rho_{n}$ are shown in Figure 1-4, respectively.
Algorithm 3.1 with different cases of $\rho_n$ and different choices of $x_0$ and $x_1$
 Case 1 Case 2 Case 3 Case 4 Choice 1 No. of Iter. 19 10 5 5 cpu (Time) $0.005632$ $0.003408$ $0.003223$ $0.002791$ Choice 2 No. of Iter. 18 10 6 6 cpu (Time) $0.00391$ $0.002683$ $0.002447$ $0.002381$ Choice 3 No. of Iter. 19 10 6 6 cpu (Time) $0.004233$ $0.003016$ $0.002601$ $0.002575$ Choice 4 No. of Iter. 13 7 6 6 cpu (Time) $0.004812$ $0.003559$ $0.002922$ $0.002412$ The numerical experiments are shown in Figure 5-8, respectively.
 Case 1 Case 2 Case 3 Case 4 Choice 1 No. of Iter. 19 10 5 5 cpu (Time) $0.005632$ $0.003408$ $0.003223$ $0.002791$ Choice 2 No. of Iter. 18 10 6 6 cpu (Time) $0.00391$ $0.002683$ $0.002447$ $0.002381$ Choice 3 No. of Iter. 19 10 6 6 cpu (Time) $0.004233$ $0.003016$ $0.002601$ $0.002575$ Choice 4 No. of Iter. 13 7 6 6 cpu (Time) $0.004812$ $0.003559$ $0.002922$ $0.002412$ The numerical experiments are shown in Figure 5-8, respectively.
Comparison of MIner-R-Iter, Iner-R-Iter and H-R-Iter in Example 1
 MIner-R-Iter Iner-R-Iter H-R-Iter Choice 1 $u=(0, -1, -5)^T$ No. of Iter. 6 33 223 $x_{0}=(2, 6, -3)^T$ cpu (Time) 0.000737 0.007677 0.064889 $x_{1}=(-2, -1, 8)^T$ Choice 2 $u=(2, 1, 0)^T$ No. of Iter. 4 26 378 $x_{0}=(3, 4, -1)^T$ cpu (Time) 0.000522 0.004861 0.137471 $x_{1}=(-5, -2, 1)^T$ Choice 3 $u=(5, -3, -1)^T$ No. of Iter. 9 29 140 $x_{0}=(2, 1, -1)^T$ cpu (Time) 0.001458 0.005175 0.026824 $x_{1}=(-5, 3, 5)^T$ Choice 4 $u=(-2, -1, 4)^T$ No. of Iter. 9 34 763 $x_{0}=(7.35, 1.75, -3.24)^T$ cpu (Time) 0.001481 0.008058 0.687214 $x_{1}=(-6.34, 0.42, 7.36)^T$ The error plotting of $E_n$ of MIner-R-Iter, Iner-R-Iter and H-R-Iter for each choice in Table 3 is shown in the following figures, respectively.
 MIner-R-Iter Iner-R-Iter H-R-Iter Choice 1 $u=(0, -1, -5)^T$ No. of Iter. 6 33 223 $x_{0}=(2, 6, -3)^T$ cpu (Time) 0.000737 0.007677 0.064889 $x_{1}=(-2, -1, 8)^T$ Choice 2 $u=(2, 1, 0)^T$ No. of Iter. 4 26 378 $x_{0}=(3, 4, -1)^T$ cpu (Time) 0.000522 0.004861 0.137471 $x_{1}=(-5, -2, 1)^T$ Choice 3 $u=(5, -3, -1)^T$ No. of Iter. 9 29 140 $x_{0}=(2, 1, -1)^T$ cpu (Time) 0.001458 0.005175 0.026824 $x_{1}=(-5, 3, 5)^T$ Choice 4 $u=(-2, -1, 4)^T$ No. of Iter. 9 34 763 $x_{0}=(7.35, 1.75, -3.24)^T$ cpu (Time) 0.001481 0.008058 0.687214 $x_{1}=(-6.34, 0.42, 7.36)^T$ The error plotting of $E_n$ of MIner-R-Iter, Iner-R-Iter and H-R-Iter for each choice in Table 3 is shown in the following figures, respectively.
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