Generalized Newton Method for a Kind of Complementarity Problem

Table 3

𝑥0=(1,1)^𝑇
Step (𝑑𝑘 was computed by (12))	Ψ(𝑥)
16	9.441173879087828𝑒−005
𝑥0=(1,1)^𝑇
Step (𝑑𝑘 was computed by (35))	Ψ(𝑥)
22	9.162651686955847𝑒−005
𝑥0=(1,1)^𝑇
Step (𝑑𝑘 was computed by (36))	Ψ(𝑥)
42	9.397247497289207𝑒−005
𝑥0=(10,10)^𝑇
Step (𝑑𝑘 was computed by (12))	Ψ(𝑥)
18	8.575708886859879𝑒−005
𝑥0=(10,10)^𝑇
Step (𝑑𝑘 was computed by (35))	Ψ(𝑥)
21	9.313859056403473𝑒−005
𝑥0=(10,10)^𝑇
Step (𝑑𝑘 was computed by (36))	Ψ(𝑥)
24	9.637003553941703𝑒−005