On a Dual Direct Cosine Simplex Type Algorithm and Its Computational Behavior

Table 2

The solution of Example 1 by the two-phase method.

Iteration							R1	R2		R.H.S
0 Phase1	Z′	0	0	0	0	0	−1	−1	0	0
		3	1	0	0	1	0	0	0	3
	R1	3	1	−1	0	0	1	0	0	3
	R2	4	3	0	−1	0	0	1	0	6
		1	2	0	0	0	0	0	1	4
1 Phase1	Z′	7	4	−1	−1	0	0	0	0	9
		3	1	0	0	1	0	0	0	3
	R1	3	1	−1	0	0	1	0	0	3
	R2	4	3	0	−1	0	0	1	0	6
		1	2	0	0	0	0	0	1	4
2 Phase1	Z′	0	1.67	−1	−1	−2.33	0	0	0	2
		1	0.33	0	0	0.33	0	0	0	1
	R1	0	0	−1	0	−1	1	0	0	0
	R2	0	1.67	0	−1	−1.33	0	1	0	2
		0	0	0	0	−0.33	0	0	1	3
3 Phase1	Z′	0	0	−1	0	−1	0	−1	0	0
		1	0	0	0.2	0.6	1	−0.2	0	0.6
	R1	0	0	−1	0	−1	0	0	0	0
		0	1	0	−0.6	−0.8	0	0.6	0	1.2
		0	0	0	1	1	0	−1	1	1
4 Phase2	Z′	0	0	0	0.2	1.6	Blocked	Blocked	0	3.6
		1	0	0	0.2	0.6	0	−0.2	0	0.6
	R1	0	0	−1	0	−1	1	0	0	0
		0	1	0	−0.6	−0.8	0	0.6	0	1.2
		0	0	0	1	1	0	−1	1	1
5 Phase 2	Z′	0	0	−1.6	0.2	0	Blocked	Blocked	0	3.6
		1	0	−0.6	0.2	0	0	−0.2	0	0.6
		0	0	1	0	1	1	0	0	0
		0	1	0.8	0.6	0	0	0.6	0	1.2
		0	0	−1	1	0	0	−1	1	1
6 Phase 2	Z′	0	0	−1.4	0	0	Blocked	Blocked	−0.2	3.4
		1	0	−0.4	0	0	0	−0.2	−0.2	0.4
		0	0	1	0	1	1	0	0	0
		0	1	0.2	1	0	0	0.6	0.6	1.8
		0	0	−1	0	0	0	−1	1	1