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

Table 4

The solution of Example 2 by the two-phase method.

Iteration								R1	R2	R3	R.H.S
0 Phase1	Z′	0	0	0	0	0	0	−1	−1	−1	0
	R1	1	20	200	−1	0	0	1	0	0	100
	R2	0	−1	−20	0	−1	0	0	1	0	10
	R3	0	0	(−1)	0	0	−1	0	0	1	1
1 Phase1	Z′	1	21	221	−1	−1	−1	0	0	0	111
	R1	1	20	200	−1	0	0	1	0	0	100
	R2	0	1	20	0	−1	0	0	1	0	10
	R3	0	0	1	0	0	−1	0	0	1	1
2 Phase1	Z′	−0.11	−1.10	0	0.11	−1	−1	−1.11	0	0	0.50
	x3	0.01	0.10	1	−0.01	0	0	0.01	0	0	0.50
	R2	−0.10	−1	0	(0.10)	−1	0	−0.10	1	0	0
	R3	−0.01	−0.10	0	0.01	0	−1	−0.01	0	1	0.50
3 Phase1	Z′	0	−0.05	0	0	0.05	−1	−1	−1.05	0	0.50
	x3	0	0.05	1	0	−0.05	0	0	0	0	0.50
	x4	−1	−10	0	1	−10	0	−1	10	0	0
	R3	0	−0.05	0	0	0.05	−1	0	−0.05	1	0.50
4 Phase1	Z′	0	0	0	0	0	0	−1	−1	−1	0
	x3	0	0	1	0	0	−1	0	0	1	1
	x4	−1	−20	0	1	0	−200	−1	0	200	100
	R3	0	−1	0	0	1	−20	0	−1	20	10
5 Phase 2	Z′	−1	−100	0	0	0	−10^4	Blocked	Blocked	Blocked	10000
	x3	0	0	1	0	0	−1	0	0	1	1
	x4	−1	−20	0	1	0	−200	−1	0	200	100
	x5	0	−1	0	0	1	−20	0	−1	20	10