2.1.6 Метод допустимого базиса

				↑
		b		x1		x2		x3		x4		x5		x6
	F	0		-1		4		0		0		0		0
			1/2		1/2		1/2		-1/2		0		0		0
←	ξ1	1		2		1		-1		0		0		0		1/2
			1/2		1/2		1/2		-1/2		0		0		0
	ξ2	2		-1		1		0		1		0		0		14/3
			1/2		1/2		1/2		-1/2		0		0		0
	ξ3	14		3		2		0		0		1		0		3
			-3/2		-3/2		-3/2		3/2		0		0		0
	ξ4	3		1		-1		0		0		0		1
			-1/2		-1/2		-1/2		1/2		0		0		0
	f	20		5		3		-1		1		1		1
			-5/2		-5/2		-5/2		5/2		0		0		0

								↑
		b		ξ1		x2		x3		x4		x5		x6
	F	1/2		1/2		9/2		-1/2		0		0		0
			5/2		-1/2		-3/2		1		0		0		1
	x1	1/2		1/2		1/2		-1/2		0		0		0
			5/2		-1/2		-3/2		1		0		0		1
	ξ2	5/2		1/2		3/2		-1/2		1		0		0
			5/2		-1/2		-3/2		1		0		0		1
	ξ3	25/2		-3/2		1/2		3/2		0		1		0		25/3
			-15/2		3/2		9/2		-3		0		0		-3
←	ξ4	5/2		-1/2		-3/2		1/2		0		0		1		5
			5		-1		-3		2		0		0		2
	f	35/2		-5/2		1/2		3/2		1		1		1
			-15/2		3/2		9/2		-3		0		0		-3

						↑
		b		ξ1		x2		ξ4		x4		x5		x6
	F	3		0		3		1		0		0		1
			-3		0		-3/5		9/5		0		-3/5		9/5
	x1	3		0		-1		1		0		0		1
			1		0		1/5		-3/5		0		1/5		-3/5
	ξ2	5		0		0		1		1		0		1
			0		0		0		0		0		0		0
←	ξ3	5		0		5		-3		0		1		-3		1
			1		0		1/5		-3/5		0		1/5		-3/5
	x3	5		-1		-3		2		0		0		2
			3		0		3/5		-9/5		0		3/5		-9/5
	f	10		-1		5		-3		1		1		-2
			-5		0		-1		3		0		-1		3

														↑
		b		ξ1		ξ3		ξ4		x4		x5		x6
	F	0		0		-3/5		14/5		0		-3/5		14/5
			-14		0		0		-14/5		-14/5		0		-14/5
	x1	4		0		1/2		2/5		0		1/5		2/5		10
			-2		0		0		-2/5		-2/5		0		-2/5
←	ξ2	5		0		0		1		1		0		1		5
			5		0		0		1		1		0		1
	x2	1		0		1/5		-3/5		0		1/5		-3/5
			3		0		0		3/5		3/5		0		3/5
	x3	8		-1		3/5		1/5		0		3/5		1/5		40
			-1		0		0		-1/5		-1/5		0		-1/5
	f	5		-1		-1		0		1		0		1
			-5		0		0		-1		-1		0		-1

		b		ξ1		ξ3		ξ4		x4		x5		ξ2
	F	-14		0		-3/5		0		-14/5		-3/5		-14/5
	x1	2		0		1/5		0		-2/5		1/5		-2/5
	x6	5		0		0		1		1		0		1
	x2	4		0		1/5		0		3/5		1/5		-3/5
	x3	7		-1		3/5		0		-1/5		3/5		-1/5
	f	0		-1		-1		-1		0		0		-1

	b	x4	x5
F	-14	-14/5	-3/5
x6	5	1	0
x2	4	3/5	1/5
x3	7	-1/5	3/5
x1	2	-2/5	1/5

Допустимое базисное оптимальное решение:

X= (2, 4, 7, 0, 0, 5)

F = -14