> GCE Further Mathematics B (MEI) Y433/01: Modelling with algorithms Advanced GCE Mark Scheme for Autumn 2021. RATED A_+

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Mark Scheme October 2021 Question Answer Marks AOs Guidance 1 (a) Bin 1: 5 16 12 10 M1 1.1 First six values placed correctly (the values Bin 2: 15 21 5 3 in bold) – so the 10 in the correct bin ... Bin 3: 17 6 13 5 Bin 4: 24 A1 1.1 cao [2] 1 (b) e.g. M1 1.1 At least two full bins (= 45) Bin 1: 24 21 Bin 2: Bin 3: Bin 4: 16 17 12 15 6 13 10 5 55 3 A1 1.1 cao (three full bins with 17 units in the nonfull bin) [2] 22 (a) (b) M1 A1 A1 3.1b 3.1a 1.1 Activity on arc, single start vertex Precedences correct for A, B, C, D, G, H Directions may be implied Durations not necessary Single finish Precedences correct for E, F, I, J, K Directions may be implied Durations not necessary All three dummies correct and no extras All arcs directed [3] M1 ft M1 ft A1 [3] 3.1b 1.1 1.1 Network must have at least one burst and at least one merge, other than start and finish Forward pass, increasing, allow 1 blank Backward pass, decreasing, allow 1 blank Forward pass and backward pass correctY433/01 Mark Scheme October 2021 Question Answer Marks AOs Guidance 2 (c) Minimum completion time is 31 (hours) B1ft [1] 2.2a Follow through their network 2 (d) Interfering float for H is (22 – 8) – (21 – 8) = 1 (hour) B1ft [1] 3.4 Follow through using their early and late event times at the beginning and end of H 2 (e) Total float for E is 21 – 11 – 6 (= 4) and Total float for G is 21 – 8 – x (= 13 – x) 13 – x < 2×4 or 13 – x ≤ 8 5 ≤ x <13 M1 * M1dep* A1 [3] 1.1 2.1 2.2a Correct calculations of the total float for their E and G Using the given information to set up an inequality for x cao 3 (a) (i) The sum of the vertex orders equals the number of arc endings Each arc has two ends so the sumber of arc endings is twice the number of arcs So the sum of the vertex orders is twice the number of arcs, which is even B1 2.1 States or uses the result that the sum of the order of the vertices is equal to twice the number of arcs Alternative method Let a graph have e edges and n nodes (vertices), let di represent n the order of the ith node so ∑di = 2e , which is even i=1 B1 [1] 3 (a) (ii) The sum of the orders of all the even vertices will be an even number so the sum of the order of the odd vertices must be an even number too Hence a graph must have an even number of vertices of odd order So no graph has an odd number of odd vertices B1 2.2a Correctly explains why a graph cannot have an odd number of vertices with odd order (or must have an even number of vertices with odd order) Must refer to even vertices as well as od [Show More]

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