# Question 7:

## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| The pivot for this first iteration came from the $x$-column | B1 | AO1.1b |
| …as $x$ is now a basic variable | dB1 | AO2.5; dependent on B1 |

## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $x$-row: $4x + y - z \leq 40$ **or** $s_3$-row: $y + z \leq 26$ | B1 | AO3.4 |
| Eliminating $s_2$ from $s_1$ row using $x$ row: $-\frac{1}{2}y + \frac{3}{2}z + s_1 - 2\!\left(10 - x - \frac{1}{4}y + \frac{1}{4}z\right) = 30$ | M1 | AO2.1 |
| Eliminating $s_2$ from $P$ row using $x$ row: $P - \frac{1}{4}y - \frac{11}{4}z + 3\!\left(10 - x - \frac{1}{4}y + \frac{1}{4}z\right) = 30$ | M1 | AO1.1b |
| $P - 3x - y - 2z = 0$ **or** $2x + z + s_1 = 50$ | A1 | AO1.1b |
| LP: Maximise $P = 3x + y + 2z$, subject to $2x + z \leq 50$, $4x + y - z \leq 40$, $y + z \leq 26$, $(x, y, z \geq 0)$ | A1 | AO2.2a |

# Question 7:

## Part (a)

| Answer | Mark | Guidance |
|--------|------|----------|
| Correct column of $x$ stated | B1 | CAO |
| $x$ is now a basic variable; $x$ column has one 1 and rest 0, or $s_2$ row has been replaced with $x$ | dB1 | Dependent on first B1 |

## Part (b)

| Answer | Mark | Guidance |
|--------|------|----------|
| Either constraint for $x$-row or $s_3$-row correct (inequalities, not strict) | B1 | Allow non-integer coefficients |
| Eliminating $s_2$ from $s_1$ row equation using $x$ row | M1 | Correct constraint implies this mark |
| Eliminating $s_2$ from $P$ row using $x$ row | M1 | Correct objective implies this mark |
| $P - 3x - y - 2z = 0$ **or** $2x + z + s_1 = 50$ | A1 | Allow equivalent forms with non-integer coefficients, simplified to single term per variable |
| Correct LP formulation | A1 | Condone lack of 'maximise' and non-negative trivial constraints |

## Part (c)

| Answer | Mark | Guidance |
|--------|------|----------|
| Pivot row completely correct including change of b.v. | B1 | Accept correct recurring decimals in place of fractions |
| All values in one non-pivot row correct, or one of the 'non zero and one' columns ($y, s_1, s_2$ or Value) correct | M1 | Must have pivoted on correct value |
| All values correct including b.v. column | A1 | Ignore 'Row Ops' column |
| Correct row operations stated | B1 | Allow alternative row numbering; condone use of $R_1$ throughout |

**Second iteration tableau:**

| b.v. | $x$ | $y$ | $z$ | $s_1$ | $s_2$ | $s_3$ | Value | Row Ops |
|------|-----|-----|-----|--------|--------|--------|-------|---------|
| $z$ | $0$ | $-\frac{1}{3}$ | $1$ | $\frac{2}{3}$ | $-\frac{1}{3}$ | $0$ | $20$ | $\frac{2}{3}r1$ |
| $x$ | $1$ | $\frac{1}{6}$ | $0$ | $\frac{1}{6}$ | $\frac{1}{6}$ | $0$ | $15$ | $r2 + 0.25R1$ |
| $s_3$ | $0$ | $\frac{4}{3}$ | $0$ | $-\frac{2}{3}$ | $\frac{1}{3}$ | $1$ | $6$ | $r3 - R1$ |
| $P$ | $0$ | $-\frac{7}{6}$ | $0$ | $\frac{11}{6}$ | $-\frac{1}{6}$ | $0$ | $85$ | $r4 + 2.75R1$ |

**(4 marks)**

## Part (d)(i)

| Answer | Mark | Guidance |
|--------|------|----------|
| After second iteration optimal solution not found as profit row still contains negative values | dB1 | CAO; dependent on M mark in (c) and completed profit row; do not accept "not all values are positive" |

## Part (d)(ii)

| Answer | Mark | Guidance |
|--------|------|----------|
| $z = 20,\ x = 15,\ s_3 = 6$ | dB1ft | Follow through on their values; dependent on M mark in (c); all values must be positive |

**(2 marks)**

## Part (e)

| Answer | Mark | Guidance |
|--------|------|----------|
| If $y$ is a basic variable, the $y$ column should contain only one value of 1 (in the third row), therefore the entry of 1 in the profit row is incorrect (should be 0) | B1 | CAO; accept: there should not be a 1 in the objective row in the $y$ column, or there should not be two 1s in the $y$ column. If $s_2$ column also mentioned then B0 |

**(1 mark)**

**Total: 14 marks**