AS October 2020 Q2

EdexcelAS paperCurrent spec14 marksCritical Path Analysis

Figure 1: activity network with activities A(5), B(3), C(4) from the start, D(6), E(8), F(5), G(3), H(7), I(8), J(9), K(6), and four dummies — 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, in hours, to complete the corresponding activity. Each activity requires one worker. The project is to be completed in the shortest possible time.

(a) Complete the precedence table in the answer book. (2)

Activity	Immediately preceding activities
A
B
C
D
E
F
G
H
I
J
K

(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

(c)

(i) State the minimum project completion time.

(ii) List the critical activities. (2)

(d) Calculate the maximum number of hours by which activity H could be delayed without affecting the shortest possible completion time of the project. You must make the numbers used in your calculation clear. (1)

(e) Calculate a lower bound for the number of workers needed to complete the project in the minimum time. You must show your working. (2)

(f) Draw a cascade chart for this project on Grid 1 in the answer book. (3)

Grid 1: blank grid with a time axis from 0 to 26 — Grid 1

(g) Using the answer to (f), explain why it is not possible to complete the project in the shortest possible time using the number of workers found in (e). (1)

Mark scheme (a)

Scheme

Marks

Activity	Immediately preceding activity
A	-
B	-
C	-
D	A
E	A
F	A, B
G	A, B, C
H	A, B, C
I	D
J	D, E, F, G, H
K	H

B1
B1

1.1b
1.1b

(2)

Notes

B1: 5 non-empty rows correct (so any 5 of the rows for activities D to K correct)

B1: All 11 rows correct

Diagram 1 completed: early/late event times 0/0 at the start; 5/5 at the end of A; 5/6 at the end of B; 5/6 at the end of C; 11/13 at the end of D; 13/13 at the end of E, F and G; 12/13 at the end of H; 22/22 at the finish — Mark scheme (b)

Scheme	Marks	AO
	M1 A1 A1	2.1 1.1b 1.1b
	(3)

Mark scheme (c)

Scheme	Marks	AO
(i) Minimum project completion time is 22 hours	B1ft	1.1b
(ii) Critical activities are A, E and J	B1	1.1b
	(2)

Notes

(c)(i) B1ft: CAO following through their completed top boxes from (b)

(c)(ii) B1: CAO (A, E and J only)

Mark scheme (d)

Scheme	Marks	AO
H could be delayed by 13 – 5 – 7 = 1 hour	B1ft	3.4
	(1)

Notes

B1ft: Correct calculation for their activity H (from (b)) – must see all 3 numbers (so just 13 – 12 = 1 is B0)

Mark scheme (e)

Scheme	Marks	AO
\(\dfrac{5 + 3 + 4 + \ldots + 8 + 9 + 6}{22}\)	M1	1.1b
= 2.909... so a lower bound of 3 workers	A1	2.2a
	(2)

Notes

M1: (55 to 73 inclusive) / their duration (their answers to (b) and (c)(i) must be consistent)

A1: Correct deduction of lower bound from a correct calculation – answer of 3 with no working scores no marks in this part

Cascade chart: A 0-5, E 5-13, J 13-22 on the critical row; B 0-3 float to 6; C 0-4 float to 6; D 5-11 float to 13; F 5-10 float to 13; G 5-8 float to 13; H 5-12 float to 13; I 11-19 float to 22; K 12-18 float to 22 — Mark scheme (f)

Scheme	Marks	AO
	M1 A1 A1	2.1 1.1b 1.1b
	(3)

Mark scheme (g)

Scheme	Marks	AO
e.g. between times 5 and 13 activities E, D, F, G and H must all be happening. The total time to complete these five activities is 29 hours and 29/8 > 3 so it is not possible to complete with the lower bound of 3 workers e.g. at time 8.5 activities E, D, F and H must be happening so not possible to complete with only 3 workers	B1	3.4
	(1)
	(14 marks)

Notes

B1: Correct reasoning that it is not possible to complete the project with only 3 workers – candidates must refer to both times and activities for this mark (as an indication that they have used (f))