D1 June 2019 Q7

EdexcelOld spec14 marksCritical Path Analysis

Figure 5: activity network for activities A to M with early and late event times at each node — Figure 5

The network in Figure 5 shows the activities that need to be undertaken in order to complete a project. Each activity is represented by an arc. The number in brackets is the duration of the activity in hours. The early event times and late event times are shown at each node. The project can be completed in 23 hours.

Given that the total float on activity G is 1 hour,

(a) find the values of \(w\), \(x\), \(y\) and \(z\). (4)

(b) Explain the purpose of the dummy activity that has a late event time of 16 (1)

This project is being completed by a company that has only two permanent workers available. The project must be completed in 23 hours and, in order to achieve this, the company is prepared to hire additional workers at a cost of £35 per hour payable only for the time that the workers are engaged in activities. The company wishes to minimise the money spent on additional workers. Any worker can undertake any activity and each activity requires only one worker. Once an activity has been started it must be completed without interruption and by the same worker.

(d) Explain why the company cannot complete the project in 23 hours using only their permanent workers. (1)

(e) Schedule the activities to workers on Grid 1 in the answer book so that the project is completed in 23 hours using the minimum number of workers and at minimum cost to the company. (3)

(f) Calculate the minimum extra cost to the company. You must make your working clear. (2)

Due to bad weather, activity H may take 7 hours to complete.

(g) Explain what affect this would have on the minimum time taken to complete the whole project. (You do not need to reschedule the project.) (2)

Mark scheme (a)

Scheme	Marks
\(w = 7,\ x = 8,\ y = 6,\ z = 7\)	B4,3,2,1,0
	(4)

Notes

a1B1: One value correct

a2B1: Two values correct

a3B1: Three values correct

a4B1: All four values correct (check carefully for answers written on the diagram rather than on the given answer lines)

Mark scheme (b)

Scheme	Marks
The dummy is required as J relies on D and E but M relies on D, E, F and I	B1
	(1)

Notes

b1B1: Correct answer regarding precedence – must mention J, M and one of D and E and one of F and I

Scheme	Marks
Critical activities: B, E, J and K	B1
	(1)

Mark scheme (d)

Scheme	Marks
e.g. \(\dfrac{53 + y + z}{23} = \dfrac{66}{23} = 2.869\ldots\) so at least three workers are required	B1
	(1)

Notes

d1B1: Correct calculation or argument with the correct values of \(y\) and \(z\). Other equivalent answers with regards to scheduling are acceptable (e.g. with only two workers the minimum completion time for the project is 33 hours or at time 2.5 activities A, B and C must be taking place (in situations like this detail of both activities and time must be given))

Schedule on three workers: B, E, J, K; C, H, F, M; A, D, G, I, L — Mark scheme (e)

Scheme	Marks
e.g.	M1 A1 A1
	(3)

Activity	Duration	Time interval	IPA
A	3	0 - 4	-
B	4	0 – 4	-
C	5	0 – 7	-
D	2	3 – 8	A
E	4	4 – 8	A, B
F	7	4 – 16	A, B
G	6	4 – 11	A, B
H	4	5 – 11	C
I	5	10 – 16	G, H
J	5	8 – 13	D, E
K	10	13 – 23	J
L	4	13 – 23	J
M	7	15 – 23	D, E, F, I

Mark scheme (f)

Scheme	Marks
\(35 \times \left(\sum \text{activities completed by additional worker(s)}\right)\)	M1
£700	A1
	(2)

Notes

f1M1: Correct calculation (so cost of one or more additional workers only) for their schedule – dependent on scheduling at least 12 activities in (e) – M0 if attempted cost of the two permanent workers is included

f1A1: CAO (their schedule must have had two workers working continuously from 0 to 23) – condone lack of units (but not incorrect units)

Mark scheme (g)

Scheme	Marks
H requiring 7 hours will delay the completion of the project because the total float on activity H is 2 hours and so the project will be delayed by 1 hour	M1 A1
	(2)
	14 marks

Notes

g1M1: Delayed together with some mention of time and/or float for activity H

g1A1: Project delayed by 1 hour (oe e.g. minimum completion time is now 24) - just mentioning that the total float for activity H is 2 (or that H cannot be completed on time) is A0. Give bod that ‘a delay of 1 hour’ is considering the entire project but A0 if clearly only talking about activities