Moderate -0.5 This appears to be a single row from Floyd's algorithm iteration table, requiring students to apply the standard algorithm mechanically to update distances. While it involves multiple comparisons, it's a routine procedural task with clear rules and no conceptual insight required, making it slightly easier than average.
B1 for objective function; B1 for time constraint; B1 for cost constraint
Part (ii)
Answer
Marks
Guidance
Initial tableau set up with slack variables; pivot on \(x\) column; correct row operations; final tableau showing \(x=5, y=4, z=0\), \(P=1260\); zero entries in objective row for basic variables confirming optimality
M1 A1 M1 A1 A1 A1 A1 A1
M1 for correct initial tableau; A1 correct; M1 for correct pivot operation; A1 A1 A1 for subsequent correct tableaux; A1 for reading solution; A1 for optimality explanation
Part (iii)
Answer
Marks
Guidance
Produce 3 xylophones and 18 zithers one week, then 0 xylophones and 0 zithers next (or average over two weeks); i.e. produce in alternate weeks to achieve average of 1.5 xylophones and 9 zithers
B1
Accept any valid practical implementation
Part (iv)
Answer
Marks
Guidance
Different starting pivot columns lead to different vertices of the feasible region being reached, giving alternative optimal (or non-optimal) solutions
B1 B1
B1 for identifying different pivot choice; B1 for explanation of different vertex/solution
Part (v)
Answer
Marks
Guidance
Add constraint \(x + y = 7\), introduce artificial variable \(a\): \(x+y+a=7\). Two-stage: minimise \(a\); or Big-M: add \(Ma\) to objective. Show initial tableau with artificial variable included
B1 B1 B1 B1
B1 for equality constraint; B1 for artificial variable; B1 for correct initial tableau; B1 for description of next step
Part (vi)
Answer
Marks
Guidance
\(4\frac{1}{15}+2\frac{14}{15} \approx 7\) confirming \(x+y=7\) constraint; solution lies between the two solutions from part (iv) since it satisfies the additional constraint cutting across the feasible region
B1 B1
B1 for noting values satisfy \(x+y=7\); B1 for relating to the two solutions
# Question 4:
## Part (i)
| Maximise $P = 180x + 90y + 110z$ subject to: $2x+5y+3z \leq 30$; $4x+y+2z \leq 24$; $x,y,z \geq 0$ | B1 B1 B1 | B1 for objective function; B1 for time constraint; B1 for cost constraint |
## Part (ii)
| Initial tableau set up with slack variables; pivot on $x$ column; correct row operations; final tableau showing $x=5, y=4, z=0$, $P=1260$; zero entries in objective row for basic variables confirming optimality | M1 A1 M1 A1 A1 A1 A1 A1 | M1 for correct initial tableau; A1 correct; M1 for correct pivot operation; A1 A1 A1 for subsequent correct tableaux; A1 for reading solution; A1 for optimality explanation |
## Part (iii)
| Produce 3 xylophones and 18 zithers one week, then 0 xylophones and 0 zithers next (or average over two weeks); i.e. produce in alternate weeks to achieve average of 1.5 xylophones and 9 zithers | B1 | Accept any valid practical implementation |
## Part (iv)
| Different starting pivot columns lead to different vertices of the feasible region being reached, giving alternative optimal (or non-optimal) solutions | B1 B1 | B1 for identifying different pivot choice; B1 for explanation of different vertex/solution |
## Part (v)
| Add constraint $x + y = 7$, introduce artificial variable $a$: $x+y+a=7$. Two-stage: minimise $a$; or Big-M: add $Ma$ to objective. Show initial tableau with artificial variable included | B1 B1 B1 B1 | B1 for equality constraint; B1 for artificial variable; B1 for correct initial tableau; B1 for description of next step |
## Part (vi)
| $4\frac{1}{15}+2\frac{14}{15} \approx 7$ confirming $x+y=7$ constraint; solution lies between the two solutions from part (iv) since it satisfies the additional constraint cutting across the feasible region | B1 B1 | B1 for noting values satisfy $x+y=7$; B1 for relating to the two solutions |