D1 June 2014 (R) Q7

EdexcelOld spec11 marksCritical Path Analysis

Figure 5: activity network with activities A to K and two dummies — Figure 5

A project is modelled by the activity network shown in Figure 5. The activities are represented by the arcs. The number in brackets on each arc gives the time, in days, to complete the activity. Each activity requires one worker. The project is to be completed in the shortest possible time.

(a) Complete Diagram 1 in the answer book to show the early event times and late event times. (4)

(b) Calculate the total float for activity D. You must make the numbers you use in your calculation clear. (2)

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

The project is to be completed in the minimum time using as few workers as possible.

(d) Schedule the activities using Grid 1 in the answer book. (3)

Activity network with early and late event times — Mark scheme (a)

Scheme	Marks
	M1 A1
	M1 A1
	(4)

Mark scheme (b)

Scheme	Marks
Total float for D = 12 – 4 – 4 = 4	M1 A1
	(2)

Notes

b1M1: Correct calculation for their activity D seen – their three numbers correct. Final value must be non-negative.

b1A1: CAO – no ft on this mark. The answer of 4 (with no working) scores no marks.

Mark scheme (c)

Scheme	Marks
\(\dfrac{52}{22} \approx 2.36\) so 3 workers	M1 A1
	(2)

Notes

c1M1: Attempt to find lower bound: [42-62 / their finish time].

c1A1: CAO – correct calculation seen then 3. No working scores M0 A0.

Gantt-style schedule for 3 workers: B, G, H, K; A, D, I; C, E, F, J — Mark scheme (d)

Activity	Duration	Time interval	IPA
A	4	0 – 7	-
B	5	0 – 5	-
C	3	0 – 5	-
D	4	4 – 12	A
E	2	4 – 9	A
F	3	5 – 9	B
G	4	5 – 9	B, C
H	6	9 – 15	E, F, G
I	4	9 – 15	G
J	10	9 – 22	D, E, F
K	7	15 – 22	H, I

Scheme	Marks
	M1
	A1
	A1
	(3)
	(11 marks)