AS June 2023 Q2

EdexcelAS paperCurrent spec10 marksCritical Path Analysis

Figure 1: activity network with events 1 to 8: A(6) 1-2, B(10) 1-4, C(7) 1-3, D(7) 2-5, E(8) 2-4, dummy 3-4, F(11) 3-7, G(5) 4-5, H(9) 4-6, J(6) 4-7, I(2) 5-6, L(5) 6-8, K(9) 7-8 — Figure 1

A project is modelled by the activity network shown in Figure 1. The activities are represented by the arcs. The number in brackets on each arc gives the time required, in hours, to complete the corresponding activity. The numbers in circles are the event numbers. Each activity requires one worker, and the project is to be completed in the shortest possible time.

(a) Explain the significance of the dummy activity from event 3 to event 4 (1)

(b) Complete Diagram 1 in the answer book to show the early event times and the late event times. (3)

Diagram 1: the network of Figure 1 with an empty box at each event for the early event time (top) and late event time (bottom) — Diagram 1

(d) Calculate a lower bound for the number of workers needed to complete the project in the minimum time. You must show your working. (1)

(e) Draw a Gantt chart for this project on Grid 1 in the answer book. (4)

Grid 1: blank grid with a time axis from 0 to 32 — Grid 1

Mark scheme (a)

Scheme	Marks	AO
The dummy from event 3 to event 4 is required as activity F depends only on activity C, but activities G, H and J depend on activities C, B and E	B1	2.4
	(1)

Notes

B1: Correct reasoning for the dummy activity – must mention activities F and C (twice or clearly implied twice), at least one of B/E, and at least one of G/H/J (for example, ‘F relies on C, but G relies on C and E’)

early/late event times: event 1 0/0, event 2 6/6, event 3 7/9, event 4 14/14, event 5 19/22, event 6 23/24, event 7 20/20, event 8 29/29 — Mark scheme (b)

Scheme	Marks	AO
	M1 A1 A1	1.1b 1.1b 1.1b
	(3)

Scheme	Marks	AO
Critical activities are A, E, J and K	B1	1.1b
	(1)

Mark scheme (d)

Scheme	Marks	AO
\(\dfrac{6+10+7+7+8+11+5+9+2+6+9+5}{29} = \dfrac{85}{29} = 2.931\ldots\) so a lower bound of 3 workers	B1ft	2.2a
	(1)

Notes

B1ft: Correct deduction of lower bound from a correct calculation for their minimum project completion time from (b) (so if correct in (b), must be 85/29). The follow through is on their 29 only (so no follow through for incorrectly adding up the duration of all the activities). An answer of 3 with no working scores no marks. All working seen must be correct. As a minimum must either see \(\dfrac{6+10+7+7+8+11+5+9+2+6+9+5}{29}\) or \(\dfrac{85}{29}\) or an awrt 2.9 (not from incorrect working) followed by 3

Gantt (cascade) chart: A 0-6, E 6-14, J 14-20, K 20-29 on the critical row; B 0-10 float to 14; C 0-7 float to 9; D 6-13 float to 22; F 7-18 float to 20; G 14-19 float to 22; H 14-23 float to 24; I 19-21 float to 24; L 23-28 float to 29 — Mark scheme (e)

Scheme	Marks	AO
	M1 A1 A1 A1	2.1 1.1b 1.1b 1.1b
	(4)
	(10 marks)

Activity	Duration + Float
A	0 to 6 Critical
B	0 to 10 F: 10 to 14
C	0 to 7 F; 7 to 9
D	6 to 13 F: 13 to 22
E	6 to 14 Critical
F	7 to 18 F: 18 to 20
G	14 to 19 F: 19 to 22
H	14 to 23 F: 23 to 24
I	19 to 21 F: 21 to 24
J	14 to 20 Critical
K	20 to 29 Critical
L	23 to 28 F: 28 to 29