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

Table 3

Numerical results for Example 3′.

ST	𝑐=10^2		𝑐=10^3
	IT	SOL	IT	SOL
(−1,−1,−1)^𝑇	5	(−0.0952,0.0952,0.4471)^𝑇	4	(−0.0946,0.0944,0.4474)^𝑇
(0,0,0)^𝑇	13	(−0.0953,0.0953,0.4471)^𝑇	10	(−0.0948,0.0946,0.4472)^𝑇
(1,1,1)^𝑇	5	(−0.0953,0.0952,0.4471)^𝑇	4	(−0.0947,0.0946,0.4476)^𝑇
(0,1,0)^𝑇	5	(−0.0952,0.0952,0.4471)^𝑇	5	(−0.0950,0.0949,0.4472)^𝑇