A New Noninterior Continuation Method for Solving a System of Equalities and Inequalities

Table 4

Numerical results for Example 4′.

ST	𝑐=10^2		𝑐=10^3
	IT	SOL	IT	SOL
(0,0,0)^𝑇	18	(0.5769,0.4787,99.9981)^𝑇	22	(0.7944,0.3344,100.0022)^𝑇
(0,0,−1)^𝑇	18	(0.3516,0.7573,100.0028)^𝑇	14	(1.2045,0.0456,100.0051)^𝑇
(1,0,1)^𝑇	19	(0.8154,0.4056,99.9981)^𝑇	9	(1.2619,−1.2051,99.9994)^𝑇
(0,0,1)^𝑇	17	(0.6092,0.4122,99.9992)^𝑇	13	(1.0040,−1.0030,100.0533)^𝑇