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
...
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
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