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

Table 1

Numerical results for Example 1′.

ST	𝑐=10^2		𝑐=10^3
	IT	SOL	IT	SOL
(0,0,0)^𝑇	8	(0.8771,0.6720,0.5725)^𝑇	6	(0.8697,0.6680,0.5741)^𝑇
(−1,−1,−1)^𝑇	6	(0.8771,0.6721,0.5726)^𝑇	5	(0.8664,0.6665,0.5730)^𝑇
(1,1,1)^𝑇	8	(0.8771,0.6720,0.5725)^𝑇	6	(0.8697,0.6680,0.5741)^𝑇
(1,0,1)^𝑇	8	(0.8766,0.6718,0.5726)^𝑇	9	(0.8702,0.6687,0.5737)^𝑇