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

Table 2

Numerical results for Example 2′.

ST	𝑐=10^2		𝑐=10^3
	IT	SOL	IT	SOL
(0,0,0)^𝑇	12	(−0.8362,−0.8605,1.9566)^𝑇	10	(−0.8396,−0.8600,1.9558)^𝑇
(−1,−1,−1)^𝑇	12	(−0.8362,−0.8606,1.9565)^𝑇	11	(−0.8394,−0.8600,1.9558)^𝑇
(1,1,1)^𝑇	10	(−0.8364,−0.8605,1.9565)^𝑇	9	(−0.8446,−0.8592,1.9548)^𝑇
(0,1,0)^𝑇	11	(−0.8363,−0.8605,1.9565)^𝑇	13	(−0.8376,−0.8603,1.9562)^𝑇