Question

4.16. consider the following problem.
maximize ( z = 2x_1 - x_4 - 5x_5 + 2x_7 )
subject to
( x_1 + x_3 - 2x_5 - x_7 = 12 )
( -x_1 + x_2 + x_4 + 3x_5 - x_7 = 6 )
( 2x_1 - 2x_4 + 6x_5 + x_6 + 4x_7 = 18 )
( quad quad quad quad quad quad quad x_j geq 0, quad \text{for all } j )
the tableau of table 4.38 represents the current basic solution.

table 4.38

	( z )	( x_1 )	( x_2 )	( x_3 )	( x_4 )	( x_5 )	( x_6 )	( x_7 )	rhs
0	0	3	1	1	( b )	1	0	( f )
0	0	1	0	0	( c )	( \frac{1}{4} )	1	( g )
0	1	1	1	0	( d )	( \frac{1}{4} )	0	( h )

(a) identify the basic variables and the basis inverse corresponding to the given tableau.
(b) determine the values of the unknowns in the tableau.

Explanation:

Step1: Identify Basic Variables

In a simplex tableau, basic variables correspond to columns with a single 1 and the rest 0s (pivot columns). Looking at the tableau:
- Column \(x_2\): Has 1, 3, 1, 1? Wait, no, let's check each column. Wait, the tableau rows: first row is \(z\)-row, then three constraint rows. Wait, the tableau structure: rows are \(z\), then row 1 (first constraint), row 2 (second), row 3 (third). Columns: \(z, x_1, x_2, x_3, x_4, x_5, x_6, x_7, \text{RHS}\).

Wait, basic variables are from columns with a single 1 (unit vector) in the constraint rows (excluding \(z\)-row). Let's check each \(x_j\) column in constraint rows (rows 2,3,4? Wait, the tableau has 4 rows: \(z\), then three constraint rows? Wait the given tableau:

Row 1 (z-row): \(1, 0, 1, 1, 0, a, 1, 0, e\)

Row 2: \(0, 0, 3, 1, 1, b, 1, 0, f\)

Row 3: \(0, 0, 1, 0, 0, c, \frac{1}{4}, 1, g\)

Row 4: \(0, 1, 1, 1, 0, d, \frac{1}{4}, 0, h\)

Wait, no, maybe the constraint rows are three (since three constraints). Let's re-express:

Constraints:

1. \(x_1 + x_3 - 2x_5 - x_7 = 12\) (row 1, non-z)
2. \(-x_1 + x_2 + x_4 + 3x_5 - x_7 = 6\) (row 2, non-z)
3. \(2x_1 - 2x_4 + 6x_5 + x_6 + 4x_7 = 18\) (row 3, non-z)

In the tableau, the columns for basic variables should have a 1 in one constraint row and 0s in others. Let's check each column:

- \(x_1\): column entries (z-row: 0, row1: 0, row2: 0, row3: 1) → so row3 has 1, others 0? Wait the tableau given:

Wait the user's tableau:

First row (z): \(1, 0, 1, 1, 0, a, 1, 0, e\)

Second row: \(0, 0, 3, 1, 1, b, 1, 0, f\)

Third row: \(0, 0, 1, 0, 0, c, \frac{1}{4}, 1, g\)

Fourth row: \(0, 1, 1, 1, 0, d, \frac{1}{4}, 0, h\)

Ah, fourth row (third constraint row) has \(x_1\) entry 1, others (row1, row2) have 0. So \(x_1\) is basic? Wait no, row1 (first constraint) has \(x_1\) entry 0, row2 (second) 0, row3 (third) 1. So \(x_1\) is in row3.

\(x_2\): row1 (z-row) 1, row2: 3, row3: 1, row4: 1? No, z-row is first, then row2 (second constraint) 3, row3 (third) 1, row4 (fourth? No, three constraints, so three rows below z. Wait maybe the tableau has z-row, then row1 (constraint1), row2 (constraint2), row3 (constraint3). So:

z-row: \(1, 0, 1, 1, 0, a, 1, 0, e\)

row1 (constraint1): \(0, 0, 3, 1, 1, b, 1, 0, f\)

row2 (constraint2): \(0, 0, 1, 0, 0, c, \frac{1}{4}, 1, g\)

row3 (constraint3): \(0, 1, 1, 1, 0, d, \frac{1}{4}, 0, h\)

Now, basic variables: columns with a single 1 in the constraint rows (row1, row2, row3).

- \(x_1\): row3 has 1, row1: 0, row2: 0 → unit vector? Wait row3: \(x_1\) is 1, others 0. So \(x_1\) is basic (row3).

- \(x_7\): row2 has 1 (row2, \(x_7\) column: 0? Wait row2: \(x_7\) column is 0? Wait row2: \(x_7\) entry is 0? Wait row1: \(x_7\) 0, row2: \(x_7\) 0, row3: \(x_7\) 0? No, wait the original constraints:

Constraint1: \(x_1 + x_3 - 2x_5 - x_7 = 12\) → so \(x_7\) coefficient -1.

Constraint2: \(-x_1 + x_2 + x_4 + 3x_5 - x_7 = 6\) → \(x_7\) coefficient -1.

Constraint3: \(2x_1 - 2x_4 + 6x_5 + x_6 + 4x_7 = 18\) → \(x_7\) coefficient 4.

In the tableau, \(x_7\) column: z-row 0, row1: 0, row2: 0, row3: 0? No, row2 (constraint2) has \(x_7\) entry 0? Wait row2: \(x_7\) is 0? Wait the tableau's \(x_7\) column: z-row 0, row1: 0, row2: 0, row3: 0? No, maybe I misread. Wait the tableau columns: \(x_7\) is the 8th column (after \(x_6\)). So:

z-row: \(x_7\) entry 0

row1 (constraint1): \(x_7\) entry 0

row2 (constraint2): \(x_7\) entry 0

row3 (constraint3): \(x_7\) entry 0? No, that can't be. Wait maybe the tableau is misformatted. Alternatively, basic variables are \(x_1, x_2, x_3, x_6\)? No, let's think differently.

In a simplex tableau, the basis inverse is the matrix formed by the columns of the basic variables in the constraint rows. Wait, maybe part (a) first: basic variables are those with a unit vector (single 1, others 0) in the constraint rows. Looking at the tableau:

- \(x_1\): column has 0, 0, 0, 1 (row3) → unit vector.

- \(x_2\): column has 1 (z-row), 3 (row1), 1 (row2), 1 (row3) → no, z-row is not a constraint row. Wait constraint rows are row1, row2, row3 (below z). So \(x_2\) in constraint rows: row1: 3, row2: 1, row3: 1 → not unit vector.

- \(x_3\): row1: 1, row2: 0, row3: 1 → no.

- \(x_4\): row1: 1, row2: 0, row3: 0 → row1: 1, row2: 0, row3: 0? Wait row1 (constraint1) \(x_4\) entry 1, row2: 0, row3: 0 → unit vector! So \(x_4\) is basic (row1).

- \(x_6\): row1: 1, row2: \(\frac{1}{4}\), row3: \(\frac{1}{4}\) → no. Wait row2 (constraint2) \(x_6\) entry \(\frac{1}{4}\), row3 (constraint3) \(x_6\) entry \(\frac{1}{4}\), row1: 1 → no.

- \(x_7\): row2: 1 (constraint2 \(x_7\) entry 0? Wait no, original constraint2: \(-x_7\), so in tableau, row2 (constraint2) \(x_7\) entry should be -1? Wait maybe the tableau is in canonical form, so maybe the basic variables are \(x_1, x_4, x_7, x_2\)? No, this is getting confusing. Maybe part (b) first: determine \(a, b, c, d, e, f, g, h\).

From the original constraints, we can express the tableau entries. Let's write the \(z\)-row: \(z - 2x_1 + x_4 + 5x_5 - 2x_7 = 0\) (since \(z = 2x_1 - x_4 -5x_5 + 2x_7\), so \(z - 2x_1 + x_4 + 5x_5 - 2x_7 = 0\)).

In the \(z\)-row, the coefficients of non-basic variables are the reduced costs. Wait, the \(z\)-row has:

\(z\)-row: \(1, 0, 1, 1, 0, a, 1, 0, e\)

So coefficient of \(x_2\) is 1, \(x_3\) is 1, \(x_5\) is \(a\), \(x_6\) is 1, \(x_7\) is 0.

From the \(z\)-row formula: \(z = 2x_1 - x_4 -5x_5 + 2x_7\), and in terms of basic variables, we can substitute basic variables from constraints.

First, find basic variables. Let's assume basic variables are \(x_1, x_2, x_4, x_6\)? No, let's use the constraint equations.

Constraint1: \(x_1 + x_3 - 2x_5 - x_7 = 12\) → \(x_3 = 12 - x_1 + 2x_5 + x_7\)

Constraint2: \(-x_1 + x_2 + x_4 + 3x_5 - x_7 = 6\) → \(x_2 = 6 + x_1 - x_4 - 3x_5 + x_7\)

Constraint3: \(2x_1 - 2x_4 + 6x_5 + x_6 + 4x_7 = 18\) → \(x_6 = 18 - 2x_1 + 2x_4 - 6x_5 - 4x_7\)

Now substitute \(x_2, x_3, x_6\) into \(z\):

\(z = 2x_1 - x_4 -5x_5 + 2x_7\)

Substitute \(x_2\): no, \(z\) is in terms of non-basic variables. Wait, non-basic variables are \(x_3, x_5, x_7\) (assuming \(x_1, x_2, x_4, x_6\) are basic? No, let's check the tableau again.

In the tableau, the \(z\)-row has \(x_1\) coefficient 0, \(x_4\) coefficient 0 (since \(x_1\) and \(x_4\) are basic? Wait \(z\)-row \(x_1\) is 0, \(x_4\) is 0. So basic variables are \(x_1, x_2, x_4, x_6\)? No, \(x_1\) in \(z\)-row is 0, \(x_4\) is 0, so they are basic (their coefficients in \(z\)-row are 0).

So \(x_1, x_2, x_4, x_6\) are basic? Wait three constraints, so three basic variables. Oh! Three constraints, so three basic variables. So I made a mistake: three constraints, so three basic variables (since it's a linear program with three equations, so basis has three variables).

So three basic variables. Let's find which three columns have unit vectors in the three constraint rows (row1, row2, row3 below z).

Looking at the tableau:

Row1 (constraint1): \(0, 0, 3, 1, 1, b, 1, 0, f\) → columns: \(x_2=3, x_3=1, x_4=1, x_5=b, x_6=1, x_7=0\)

Row2 (constraint2): \(0, 0, 1, 0, 0, c, \frac{1}{4}, 1, g\) → columns: \(x_2=1, x_3=0, x_4=0, x_5=c, x_6=\frac{1}{4}, x_7=1\)

Row3 (constraint3): \(0, 1, 1, 1, 0, d, \frac{1}{4}, 0, h\) → columns: \(x_1=1, x_2=1, x_3=1, x_4=0, x_5=d, x_6=\frac{1}{4}, x_7=0\)

Ah! Now, row2 (constraint2) has \(x_7=1\) (unit vector), row3 (constraint3) has \(x_1=1\) (unit vector), row1 (constraint1) has \(x_3=1\) (unit vector)? Wait row1: \(x_3=1\), row2: \(x_3=0\), row3: \(x_3=1\) → no, row1 and row3 both have 1. So that's not a unit vector. Wait row2: \(x_7=1\), row3: \(x_1=1\), and row1: \(x_4=1\) (row1: \(x_4=1\), row2: \(x_4=0\), row3: \(x_4=0\)) → yes! \(x_4\) in row1: 1, row2: 0, row3: 0 → unit vector. \(x_7\) in row2: 1, row1: 0, row3: 0 → unit vector. \(x_1\) in row3: 1, row1: 0, row2: 0 → unit vector. So basic variables are \(x_1, x_4, x_7\) (three variables, three constraints).

Now, express non-basic variables (\(x_2, x_3, x_5, x_6\)) in terms of basic variables.

From constraint3 (row3): \(x_1 = h - x_2 - x_3 - \frac{1}{4}x_6\) (wait no, row3: \(x_1 + x_2 + x_3 + \frac{1}{4}x_6 = h\) (since \(x_4=0, x_5=d, x_7=0\)) → no, original constraint3: \(2x_1 - 2x_4 + 6x_5 + x_6 + 4x_7 = 18\). With \(x_4=0, x_7=0\) (basic variables \(x_1, x_4, x_7\)? No, \(x_4\) and \(x_7\) are basic, so \(x_4\) and \(x_7\) are basic, so their values are from RHS. Wait, basic variables' values are RHS of their rows.

Row1 (constraint1, basic variable \(x_4\)): \(x_3 + x_4 + b x_5 + x_6 = f\) (since \(x_1=0, x_2=3, x_7=0\)) → no, original constraint1: \(x_1 + x_3 - 2x_5 - x_7 = 12\). With \(x_1=0, x_7=0\) (basic variables \(x_4, x_7, x_1\)? \(x_1\) is basic (row3: \(x_1=1\)), so \(x_1 = h\) (RHS of row3). \(x_4 = f\) (RHS of row1), \(x_7 = g\) (RHS of row2).

From row3 (constraint3, \(x_1\) basic): \(x_1 + x_2 + x_3 + \frac{1}{4}x_6 = h\) → but original constraint3

Answer:

Step1: Identify Basic Variables

In a simplex tableau, basic variables correspond to columns with a single 1 and the rest 0s (pivot columns). Looking at the tableau:

• Column \(x_2\): Has 1, 3, 1, 1? Wait, no, let's check each column. Wait, the tableau rows: first row is \(z\)-row, then three constraint rows. Wait, the tableau structure: rows are \(z\), then row 1 (first constraint), row 2 (second), row 3 (third). Columns: \(z, x_1, x_2, x_3, x_4, x_5, x_6, x_7, \text{RHS}\).

Wait, basic variables are from columns with a single 1 (unit vector) in the constraint rows (excluding \(z\)-row). Let's check each \(x_j\) column in constraint rows (rows 2,3,4? Wait, the tableau has 4 rows: \(z\), then three constraint rows? Wait the given tableau:

Row 1 (z-row): \(1, 0, 1, 1, 0, a, 1, 0, e\)

Row 2: \(0, 0, 3, 1, 1, b, 1, 0, f\)

Row 3: \(0, 0, 1, 0, 0, c, \frac{1}{4}, 1, g\)

Row 4: \(0, 1, 1, 1, 0, d, \frac{1}{4}, 0, h\)

Wait, no, maybe the constraint rows are three (since three constraints). Let's re-express:

Constraints:

1. \(x_1 + x_3 - 2x_5 - x_7 = 12\) (row 1, non-z)
2. \(-x_1 + x_2 + x_4 + 3x_5 - x_7 = 6\) (row 2, non-z)
3. \(2x_1 - 2x_4 + 6x_5 + x_6 + 4x_7 = 18\) (row 3, non-z)

In the tableau, the columns for basic variables should have a 1 in one constraint row and 0s in others. Let's check each column:

• \(x_1\): column entries (z-row: 0, row1: 0, row2: 0, row3: 1) → so row3 has 1, others 0? Wait the tableau given:

Wait the user's tableau:

First row (z): \(1, 0, 1, 1, 0, a, 1, 0, e\)

Second row: \(0, 0, 3, 1, 1, b, 1, 0, f\)

Third row: \(0, 0, 1, 0, 0, c, \frac{1}{4}, 1, g\)

Fourth row: \(0, 1, 1, 1, 0, d, \frac{1}{4}, 0, h\)

Ah, fourth row (third constraint row) has \(x_1\) entry 1, others (row1, row2) have 0. So \(x_1\) is basic? Wait no, row1 (first constraint) has \(x_1\) entry 0, row2 (second) 0, row3 (third) 1. So \(x_1\) is in row3.

\(x_2\): row1 (z-row) 1, row2: 3, row3: 1, row4: 1? No, z-row is first, then row2 (second constraint) 3, row3 (third) 1, row4 (fourth? No, three constraints, so three rows below z. Wait maybe the tableau has z-row, then row1 (constraint1), row2 (constraint2), row3 (constraint3). So:

z-row: \(1, 0, 1, 1, 0, a, 1, 0, e\)

row1 (constraint1): \(0, 0, 3, 1, 1, b, 1, 0, f\)

row2 (constraint2): \(0, 0, 1, 0, 0, c, \frac{1}{4}, 1, g\)

row3 (constraint3): \(0, 1, 1, 1, 0, d, \frac{1}{4}, 0, h\)

Now, basic variables: columns with a single 1 in the constraint rows (row1, row2, row3).

• \(x_1\): row3 has 1, row1: 0, row2: 0 → unit vector? Wait row3: \(x_1\) is 1, others 0. So \(x_1\) is basic (row3).

• \(x_7\): row2 has 1 (row2, \(x_7\) column: 0? Wait row2: \(x_7\) column is 0? Wait row2: \(x_7\) entry is 0? Wait row1: \(x_7\) 0, row2: \(x_7\) 0, row3: \(x_7\) 0? No, wait the original constraints:

Constraint1: \(x_1 + x_3 - 2x_5 - x_7 = 12\) → so \(x_7\) coefficient -1.

Constraint2: \(-x_1 + x_2 + x_4 + 3x_5 - x_7 = 6\) → \(x_7\) coefficient -1.

Constraint3: \(2x_1 - 2x_4 + 6x_5 + x_6 + 4x_7 = 18\) → \(x_7\) coefficient 4.

In the tableau, \(x_7\) column: z-row 0, row1: 0, row2: 0, row3: 0? No, row2 (constraint2) has \(x_7\) entry 0? Wait row2: \(x_7\) is 0? Wait the tableau's \(x_7\) column: z-row 0, row1: 0, row2: 0, row3: 0? No, maybe I misread. Wait the tableau columns: \(x_7\) is the 8th column (after \(x_6\)). So:

z-row: \(x_7\) entry 0

row1 (constraint1): \(x_7\) entry 0

row2 (constraint2): \(x_7\) entry 0

row3 (constraint3): \(x_7\) entry 0? No, that can't be. Wait maybe the tableau is misformatted. Alternatively, basic variables are \(x_1, x_2, x_3, x_6\)? No, let's think differently.

In a simplex tableau, the basis inverse is the matrix formed by the columns of the basic variables in the constraint rows. Wait, maybe part (a) first: basic variables are those with a unit vector (single 1, others 0) in the constraint rows. Looking at the tableau:

• \(x_1\): column has 0, 0, 0, 1 (row3) → unit vector.

• \(x_2\): column has 1 (z-row), 3 (row1), 1 (row2), 1 (row3) → no, z-row is not a constraint row. Wait constraint rows are row1, row2, row3 (below z). So \(x_2\) in constraint rows: row1: 3, row2: 1, row3: 1 → not unit vector.

• \(x_3\): row1: 1, row2: 0, row3: 1 → no.

• \(x_4\): row1: 1, row2: 0, row3: 0 → row1: 1, row2: 0, row3: 0? Wait row1 (constraint1) \(x_4\) entry 1, row2: 0, row3: 0 → unit vector! So \(x_4\) is basic (row1).

• \(x_6\): row1: 1, row2: \(\frac{1}{4}\), row3: \(\frac{1}{4}\) → no. Wait row2 (constraint2) \(x_6\) entry \(\frac{1}{4}\), row3 (constraint3) \(x_6\) entry \(\frac{1}{4}\), row1: 1 → no.

• \(x_7\): row2: 1 (constraint2 \(x_7\) entry 0? Wait no, original constraint2: \(-x_7\), so in tableau, row2 (constraint2) \(x_7\) entry should be -1? Wait maybe the tableau is in canonical form, so maybe the basic variables are \(x_1, x_4, x_7, x_2\)? No, this is getting confusing. Maybe part (b) first: determine \(a, b, c, d, e, f, g, h\).

From the original constraints, we can express the tableau entries. Let's write the \(z\)-row: \(z - 2x_1 + x_4 + 5x_5 - 2x_7 = 0\) (since \(z = 2x_1 - x_4 -5x_5 + 2x_7\), so \(z - 2x_1 + x_4 + 5x_5 - 2x_7 = 0\)).

In the \(z\)-row, the coefficients of non-basic variables are the reduced costs. Wait, the \(z\)-row has:

\(z\)-row: \(1, 0, 1, 1, 0, a, 1, 0, e\)

So coefficient of \(x_2\) is 1, \(x_3\) is 1, \(x_5\) is \(a\), \(x_6\) is 1, \(x_7\) is 0.

From the \(z\)-row formula: \(z = 2x_1 - x_4 -5x_5 + 2x_7\), and in terms of basic variables, we can substitute basic variables from constraints.

First, find basic variables. Let's assume basic variables are \(x_1, x_2, x_4, x_6\)? No, let's use the constraint equations.

Constraint1: \(x_1 + x_3 - 2x_5 - x_7 = 12\) → \(x_3 = 12 - x_1 + 2x_5 + x_7\)

Constraint2: \(-x_1 + x_2 + x_4 + 3x_5 - x_7 = 6\) → \(x_2 = 6 + x_1 - x_4 - 3x_5 + x_7\)

Constraint3: \(2x_1 - 2x_4 + 6x_5 + x_6 + 4x_7 = 18\) → \(x_6 = 18 - 2x_1 + 2x_4 - 6x_5 - 4x_7\)

Now substitute \(x_2, x_3, x_6\) into \(z\):

\(z = 2x_1 - x_4 -5x_5 + 2x_7\)

Substitute \(x_2\): no, \(z\) is in terms of non-basic variables. Wait, non-basic variables are \(x_3, x_5, x_7\) (assuming \(x_1, x_2, x_4, x_6\) are basic? No, let's check the tableau again.

In the tableau, the \(z\)-row has \(x_1\) coefficient 0, \(x_4\) coefficient 0 (since \(x_1\) and \(x_4\) are basic? Wait \(z\)-row \(x_1\) is 0, \(x_4\) is 0. So basic variables are \(x_1, x_2, x_4, x_6\)? No, \(x_1\) in \(z\)-row is 0, \(x_4\) is 0, so they are basic (their coefficients in \(z\)-row are 0).

So \(x_1, x_2, x_4, x_6\) are basic? Wait three constraints, so three basic variables. Oh! Three constraints, so three basic variables. So I made a mistake: three constraints, so three basic variables (since it's a linear program with three equations, so basis has three variables).

So three basic variables. Let's find which three columns have unit vectors in the three constraint rows (row1, row2, row3 below z).

Looking at the tableau:

Row1 (constraint1): \(0, 0, 3, 1, 1, b, 1, 0, f\) → columns: \(x_2=3, x_3=1, x_4=1, x_5=b, x_6=1, x_7=0\)

Row2 (constraint2): \(0, 0, 1, 0, 0, c, \frac{1}{4}, 1, g\) → columns: \(x_2=1, x_3=0, x_4=0, x_5=c, x_6=\frac{1}{4}, x_7=1\)

Row3 (constraint3): \(0, 1, 1, 1, 0, d, \frac{1}{4}, 0, h\) → columns: \(x_1=1, x_2=1, x_3=1, x_4=0, x_5=d, x_6=\frac{1}{4}, x_7=0\)

Ah! Now, row2 (constraint2) has \(x_7=1\) (unit vector), row3 (constraint3) has \(x_1=1\) (unit vector), row1 (constraint1) has \(x_3=1\) (unit vector)? Wait row1: \(x_3=1\), row2: \(x_3=0\), row3: \(x_3=1\) → no, row1 and row3 both have 1. So that's not a unit vector. Wait row2: \(x_7=1\), row3: \(x_1=1\), and row1: \(x_4=1\) (row1: \(x_4=1\), row2: \(x_4=0\), row3: \(x_4=0\)) → yes! \(x_4\) in row1: 1, row2: 0, row3: 0 → unit vector. \(x_7\) in row2: 1, row1: 0, row3: 0 → unit vector. \(x_1\) in row3: 1, row1: 0, row2: 0 → unit vector. So basic variables are \(x_1, x_4, x_7\) (three variables, three constraints).

Now, express non-basic variables (\(x_2, x_3, x_5, x_6\)) in terms of basic variables.

From constraint3 (row3): \(x_1 = h - x_2 - x_3 - \frac{1}{4}x_6\) (wait no, row3: \(x_1 + x_2 + x_3 + \frac{1}{4}x_6 = h\) (since \(x_4=0, x_5=d, x_7=0\)) → no, original constraint3: \(2x_1 - 2x_4 + 6x_5 + x_6 + 4x_7 = 18\). With \(x_4=0, x_7=0\) (basic variables \(x_1, x_4, x_7\)? No, \(x_4\) and \(x_7\) are basic, so \(x_4\) and \(x_7\) are basic, so their values are from RHS. Wait, basic variables' values are RHS of their rows.

Row1 (constraint1, basic variable \(x_4\)): \(x_3 + x_4 + b x_5 + x_6 = f\) (since \(x_1=0, x_2=3, x_7=0\)) → no, original constraint1: \(x_1 + x_3 - 2x_5 - x_7 = 12\). With \(x_1=0, x_7=0\) (basic variables \(x_4, x_7, x_1\)? \(x_1\) is basic (row3: \(x_1=1\)), so \(x_1 = h\) (RHS of row3). \(x_4 = f\) (RHS of row1), \(x_7 = g\) (RHS of row2).

From row3 (constraint3, \(x_1\) basic): \(x_1 + x_2 + x_3 + \frac{1}{4}x_6 = h\) → but original constraint3