A Two-Step Newton-Type Method for Solving System of Absolute Value Equations

Table 2

Numerical results for Example 4.

Method		1000	2000	3000	4000	5000	6000
NINA	IT	18	18	18	18	18	18
	RES	3.4947e − 07	4.9580e − 07	6.0786e − 07	7.0227e − 07	7.8541e − 07	8.6055e − 07
	ER
	R	1.8652e − 07	2.6468e − 07	3.2453e − 07	3.7494e − 07	4.1934e − 07	4.5946e − 07
	CP
	U	139.2571	879.3143	3083.7005	—	—	—
SOR-like-	IT	15	16	16	16	16	17
	RES	9.8477e − 07	6.3197e − 07	7.7428e − 07	8.9422e − 07	9.9988e − 07	2.0134e − 07
	ER
	R	4.4031e − 07	7.7428e − 07	1.2729e − 07	1.4701e − 07	1.6438e − 07	9.0038e − 08
	CP
	U	0.9999	7.0482	23.4849	54.02271	104.1137	178.4376
NTM	IT	2	2	2	2	2	2
	RES	1.1960e − 11	3.5203e − 11	6.3912e − 11	9.6775e − 11	1.4111e − 10	1.7906e − 10
	ER
	R	3.2890e − 12	9.3320e − 12	1.6375e − 11	2.5658e − 11	3.5328e − 11	4.4462e − 11
	CP
	U	0.7966	5.3339	18.0749	41.8621	80.8494	137.7330