Article Contents
Article Contents

# Double projection algorithms for solving the split feasibility problems

• * Corresponding author: Jie Sun
This work was partially supported by National Science Foundation of China (Grants 71572113 and 11401322) and Australian Research Council (Grant DP160102819).
• We propose two new double projection algorithms for solving the split feasibility problem (SFP). Different from the extragradient projection algorithms, the proposed algorithms do not require fixed stepsize and do not employ the same projection region at different projection steps. We adopt flexible rules for selecting the stepsize and the projection region. The proposed algorithms are shown to be convergent under certain assumptions. Numerical experiments show that the proposed methods appear to be more efficient than the relaxed- CQ algorithm.

Mathematics Subject Classification: Primary: 37C25, 47H09; Secondary: 90C25.

 Citation:

• Table 1.  The numerical results of Example 5.1

 Initiative point CQ Algorithm with stepsize $\frac{1}{\rho(A^{T}A})$ Algorithm 3.1 Algorithm 4.1 $x^{0}=$ $k=269; s=0.063$ $k=54; s=0.101$ $k=28; s=0.077$ $(-5,-2,-10)^{T}$ $x^{\ast}=(0.5071,-1.8186,01.9072)^{T}$ $x^{\ast}=(0.8718; -1.6577; -1.4080)^{T}$ $x^{\ast}=(1.1595;-1.0082;0-1.0814)^{T}$ $x^{0}=$ $k=261; s =0.063$ $k=16; s=0.061$ $k=1; s=0.045$ $(-2,-1,-5)^{T}$ $x^{\ast}=(0.1098;-1.7655; -1.6134)^{T}$ $x^{\ast}=(0.6814;-1.4212; -1.0762)^{T}$ $x^{\ast}=(0.4734;-1.7714 ; -1.3758)^{T}$ $x^{0}=$ $k=6450; s =0.525$ $k=59; s=0.096$ $k=1; s=0.048$ $(-6,0,-1)^{T}$ $x^{\ast}=(-3.9899;-0.6144; 1.8062)^{T}$ $x^{\ast}=((-3.8898;-0.5850; 1.9604)^{T}$ $x^{\ast}=(-3.9302,-1.0861,1.9786)^{T}$

Table 2.  The numerical results of Example 5.2

 $M, N$ CQ Algorithm with stepsize $\frac{1}{\rho(A^{T}A}$ $t_{k}$ Algorithm3.1 Algorithm 4.1 $M=20, N=10$ $k=485, s =1.040$ 0.8 $k=274, s =0.312$ $k=210, s =0.270$ 1.0 $k=193, s =0.100$ $k=108, s =0.070$ 1.8 $k=103, s =0.067$ $k=64, s =0.021$ $M=100, N=90$ $k=3987, s =3.100$ 0.4 $k=1534, s =0.690$ $k=1244, s =0.530$ 1 $k=1074, s =0.500$ $k=630, s =0.261$ 1.6 $k=674, s =0.201$ $k=412, s =0.132$
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