Critical Path Analysis

Topic

From an AS paper

Edexcel

Edexcel · Old spec

A2 June 2025 Q6

EdexcelCurrent spec13 marksCritical Path Analysis

Figure 3: activity network with activities A(2), B(4), C(3), D(7), E(4), F(5), G(5), H(6), I(10), J(2), K(6), L(10), M(7) and three dummies — Figure 3

A project is modelled by the activity network shown in Figure 3. The activities are represented by the arcs. The number in brackets on each arc gives the time, in hours, to complete the corresponding activity. The project is to be completed in the shortest possible time.

(a) Complete Table 1 in the answer book to show the immediately preceding activities for each activity. (2)

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

Table 1

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

Diagram 1: the network of Figure 3 with an empty box at each event for the early event time (top) and late event time (bottom) — Diagram 1

Grid 1: blank grid with a time axis from 0 to 24 — Grid 1

Figure 4: resource histogram, number of workers against time in hours: 4 from 0 to 2, 5 from 2 to 3, 8 from 3 to 4, 6 from 4 to 7, 8 from 7 to 8, 9 from 8 to 9, 7 from 9 to 12, 6 from 12 to 13, 5 from 13 to 15, 4 from 15 to 17, 2 from 17 to 20 — Figure 4

The resource histogram in Figure 4 shows the number of workers required when each activity starts at its earliest possible time. When an activity is started it must be completed without interruption. Each activity requires at least one worker.

(d) By considering which activities are happening each hour, and by working backwards from the minimum project completion time, complete Table 2 in the answer book to show the number of workers needed for each activity. (3)

Activity	Number of Workers	Activity	Number of Workers
A		G
B		H
C		I
D		J
E		K
F		L
		M

Table 2

Mark scheme (a)

Scheme

Marks

Activity	IPA	Activity	IPA
A	-	G	A, B, E
B	-	H	D, G
C	-	I	A, B, E
D	A	J	H, I
E	C	K	A, B, E
F	C	L	A, B, E, F
		M	A, B, E, F

B1
B1

1.1b
1.1b

(2)

Notes

a1B1: Five correct rows (not including A, B and C)

a2B1: All rows correct (accept blanks for A, B and C)

activity network with early and late event times: start 0, 0; end of A 2, 5; end of C 3, 3; node after B and E 7, 7; end of D and G 12, 12; end of H and I 18, 18; end of F 8, 10; end of J and K and L 20, 20; finish 20, 20 — Mark scheme (b)

Scheme	Marks	AO
	M1 A1 M1 A1	1.1b 1.1b 1.1b 1.1b
	(4)

Gantt chart: critical activities C 0-3, E 3-7, G 7-12, H 12-18, J 18-20 on the top row; A 0-2 float to 5, B 0-4 float to 7, D 2-9 float to 12, F 3-8 float to 10, I 7-17 float to 18, K 7-13 float to 20, L 8-18 float to 20, M 8-15 float to 20 — Mark scheme (c)

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

Activity	Start	Late End	Float	Critical	Start	End
A	0	5	3	C	0	3
B	0	7	3	E	3	7
D	2	12	3	G	7	12
F	3	10	2	H	12	18
I	7	18	1	J	18	20
K	7	20	7
L	8	20	2
M	8	20	5

AS June 2025 Q2

EdexcelAS paperCurrent spec8 marksCritical Path Analysis

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

(a) Explain how you can deduce from the precedence table that at least one dummy will be needed when drawing the activity network. Your explanation should refer to specific activities. (1)

(b) Draw the activity network described in the precedence table, using activity on arc. Your activity network must contain the minimum number of dummies only. (5)

Each activity in the precedence table takes 3 hours to complete.

One of the activities now needs to be chosen to be extended by an hour. When the change is made the minimum completion time must not be affected.

(d) List the activities that could be chosen. (1)

Mark scheme (a)

Scheme	Marks	AO
e.g. Activity D (or F or G) is preceded by activity B only, but activity E is preceded by both activity A and activity B.	B1	2.4
	(1)

Notes

a1B1: e.g. Reference to D (or F or G) is preceded by B only, E depends on both A and B (oe).
Reference to J is preceded by E only, K depends on E and F and/or I
L and M have the same start node and finish node

activity network with dummies from the end of B to the end of A, from the end of E to the end of F, and from the end of M to the end of L — Mark scheme (b)

Scheme	Marks	AO
	M1 A1 (ABCDE) A1 (FGHJ) A1 (IKLM) A1	1.1b 1.1b 1.1b 1.1b 1.1b
	(5)

Scheme	Marks	AO
15 (hours)	B1	3.1b
	(1)

Scheme	Marks	AO
A C E F G J L M	B1	2.2a
	(1)
	(8 marks)

A2 June 2024 Q6

EdexcelCurrent spec11 marksCritical Path Analysis

6. The precedence table below shows the 12 activities required to complete a project.

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

(a) Draw the activity network described in the precedence table, using activity on arc.
Your activity network must contain the minimum number of dummies only. (5)

Each of the activities shown in the precedence table requires one worker. The project is to be completed in the minimum possible time.

Figure 3: schedule for three workers on a time axis from 0 to 20: worker 1 C 0-7, E 7-12, J 12-15, K 15-20; worker 2 A 0-4, D 4-8, G 8-12, H 12-14, I 14-18; worker 3 B 0-5, idle 5-7, F 7-9, idle 9-12, L 12-15 — Figure 3

Figure 3 shows a schedule for the project using three workers.

(b)

(i) State the critical path for the network.

(ii) State the minimum completion time for the project.

(iii) Calculate the total float on activity B.

(iv) Calculate the total float on activity G. (4)

Immediately after the start of the project, it is found that the duration of activity I, as shown in Figure 3, is incorrect. In fact, activity I will take 8 hours.
The durations of all the other activities remain as shown in Figure 3.

(c) Determine whether the project can still be completed in the minimum completion time using only three workers when the duration of activity I is 8 hours.
Your answer must make specific reference to workers, times and activities. (2)

activity network: A, B, C from the start; dummies from the end of A and from the end of C to the end of B; D from the end of A, G from the end of C, E and F from the end of B; D and E meet, then H, I and J; a dummy from the end of F/G node to the end of J for K; L from the end of F and G; a dummy between the ends of H and I — Mark scheme (a)

Scheme	Marks	AO
e.g.	M1 A1 A1 A1 A1	2.1 1.1b 1.1b 1.1b 1.1b
	(5)

Mark scheme (b)

Scheme	Marks	AO
(i) Critical path: C – E – J – K	B1	1.1b
(ii) Minimum completion time: 20	B1	1.1b
(iii) Total float on activity B is 2	B1	1.1b
(iv) Total float on activity G is 4	B1	3.1b
	(4)

Notes

(i) B1: CAO (critical path CEJK – in this order)

(ii) B1: CAO (20)

(iii) B1: CAO (2 – total float on B)

(iv) B1: CAO (4 – total float on G)

revised schedule for workers 2 and 3: A, D, G, H then L from 14; B, idle, F, idle, then I from 12 to 20 — Mark scheme (c)

Scheme	Marks	AO
e.g. (One worker continues to do the critical activities (in time 20)) Second worker completes A, D, G and H as originally planned but now completes activity L in the interval 14 to 20 (but must have started L by time 17 at the latest due to its duration of 3) Third worker completes B and F as planned but now completes activity I exactly in the time interval 12 to 20 so yes, the project can be completed by 3 workers	B2, 1, 0	2.4 2.4
e.g. Yes – can be completed by 3 workers
	(2)
	(11 marks)

AS June 2024 Q2

EdexcelAS paperCurrent spec8 marksCritical Path Analysis

2. A company manages an awards evening.

The table below lists the activities required to set up the room for the evening, and their immediately preceding activities. Each activity requires exactly one person.

Activity	Immediately preceding activities
A	-
B	A
C	A
D	C
E	C
F	B, D, E
G	E
H	B
J	H, F, G

Figure 1 shows a partially completed activity network used to model the project. Each activity is represented by an arc.

Figure 1: partial activity network with A from the start node, B and C from the end of A, and D from the end of C — Figure 1

(a) Add the remaining five activities to Diagram 1 in the answer book to complete the activity network, using exactly two dummies. (3)

[Diagram 1 in the answer book is a copy of Figure 1.]

In addition to setting up the room, the company must prepare the meals for the guests. Figure 2 shows the activity network for preparing the main courses. The numbers in brackets represent the time, in minutes, to complete each task.

Figure 2: activity network with activities L(17), M(18), N(11), X(11), P(9), Q(12), S(7), R(15), T(12), W(5), U(23), V(20), Y(12) and three dummies — Figure 2

(b) Complete Diagram 2 in the answer book to show the early event times and the late event times for the activity network shown in Figure 2. (3)

Diagram 2: the network of Figure 2 with an empty box at each event for the early event time (top) and late event time (bottom) — Diagram 2

(d) Given that the main courses need to be ready to be served (with all activities completed) at 8 pm, state the latest time that activity R can start. (1)

completed activity network: dummy from the end of B and dummy from the end of E both lead to the start of F; H from the end of B; G from the end of E; H, F and G lead to the start of J — Mark scheme (a)

Scheme	Marks	AO
	M1 A1 (E,F,H) A1	1.1b 1.1b 1.1b
	(3)

early and late event times: start 0/0; end of L 17/17; start of Q 17/17; end of M 35/37; end of Q 29/29; end of U 23/28; end of S 36/36; end of N 46/48; start of Y 48/48; finish 60/60 — Mark scheme (b)

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

Scheme	Marks	AO
The critical activities are L, Q, S, T and Y	B1	1.1b
	(1)

Scheme	Marks	AO
7:33pm	B1	3.2a
	(1)
	(8 marks)

A2 June 2023 Q6

EdexcelCurrent spec9 marksCritical Path Analysis

6. The precedence table below shows the twelve activities required to complete a project.

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

(a) Draw the activity network described in the precedence table, using activity on arc. Your activity network must contain the minimum number of dummies only. (5)

Figure 6: partially completed cascade chart on a time axis from 0 to 26: B 0-6, E 6-12, I 12-19, L 19-24 on the top row; A 0-5 with float to 6; C 0-7 with float to 15; D 5-9 with float to 12; G 7-10 with float to 18; H 16-22 with float to 24 — Figure 6

Figure 6 shows a partially completed cascade chart for the project. The non-critical activities F, J and K are not shown in Figure 6.

The time taken to complete each activity is given in hours and the project is to be completed in the minimum possible time.

(b) State the critical activities. (1)

Given that the total float of activity F is 2 hours,

The duration of activity J is \(x\) hours, and the duration of activity K is \(y\) hours, where \(x \gt 0\) and \(y \gt 0\)

(d)

(i) State, in terms of \(y\), the maximum possible total float for activity K.

(ii) State, in terms of \(x\) and \(y\), the total float for activity J. (2)

activity network: A, B and C from the start; a dummy from the end of A to the end of B; D from the end of A and E from the end of B; a dummy from the end of B down to the end of C, where G starts; D and E meet, then F, I and J start there; F ends at the end of G, where H starts; dummies from the end of F/G to the end of J and from the end of I to the end of J; K from the end of J; L from the end of I; H, K and L end at the finish — Mark scheme (a)

Scheme	Marks	AO
	M1 A1 A1 A1 A1	2.1 1.1b 1.1b 1.1b 1.1b
	(5)

Activity	A	B	C	D	E	F	G	H	I	J	K	L
IPA	-	-	-	A	A, B	D, E	A, B, C	F, G	D, E	D, E	F, G, I, J	I

Scheme	Marks	AO
Critical activities: B, E, I and L	B1	2.2a
	(1)

Scheme	Marks	AO
(Earliest start for F is 12 and latest finish is 18 therefore if the total float is 2) the duration of F is 4 (hours)	B1	2.2a
	(1)

Mark scheme (d)

Scheme	Marks	AO
(i) Maximum possible total float for activity K is \(5 - y\)	B1	3.1b
(ii) Total float for activity J is \(12 - x - y\)	B1	2.2a
	(2)
	(9 marks)

Notes

(d)(i)

B1: CAO (\(5 - y\))

(d)(ii)

B1: CAO (\(12 - x - y\))

A2 June 2023 Q2

EdexcelCurrent spec7 marksCritical Path Analysis

Figure 3: activity network: A(3) and C(4) and D(7) leave the start; B(5) follows A; C ends at the same event as B; a dummy runs from the end of D to the end of B and C; E(1) and F(2) leave that event; H(1) follows E; G(4) follows D; F, G and H end at the finish — Figure 3

The network in Figure 3 shows the activities that need to be undertaken to complete a project. Each activity is represented by an arc and the duration, in hours, of the corresponding activity is shown in brackets.

(a)

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

(ii) State the minimum completion time of the project. (3)

The table below lists the number of workers required for each activity in the project.

Activity	Number of workers
A	2
B	1
C	2
D	2
E	3
F	2
G	1
H	3

Each worker is able to do any of the activities. Once an activity is started it must be completed without interruption. It is given that each activity begins at its earliest possible start time.

(b)

(i) On Grid 1 in the answer book, draw a resource histogram to show the number of workers required at each time.

(ii) Hence state the time interval(s) when six workers are required. (4)

Grid 1: blank grid, number of workers from 0 to 8 against time in hours from 0 to 14 — Grid 1

Diagram 1 completed: start 0, 0; end of A 3, 4; end of D 7, 7; end of B and C 8, 9; end of E 9, 10; finish 11, 11 — Mark scheme (a)

Scheme	Marks	AO
(i)	M1 A1	1.1b 1.1b
(ii) 11 (hours)	A1ft	2.2a
	(3)

resource histogram: 6 workers from 0 to 3, 5 from 3 to 4, 3 from 4 to 7, 2 from 7 to 8, 6 from 8 to 10, 1 from 10 to 11 — Mark scheme (b)

Scheme	Marks	AO
(i)	M1 A1 A1	1.1b 1.1b 1.1b
(ii) Six workers are required in the time intervals 0 – 3 and 8 – 10	A1	2.2a
	(4)
	(7 marks)

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)

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

A2 June 2022 Q5

EdexcelCurrent spec14 marksCritical Path Analysis

Figure 2: activity network with early event times: A(4), B(5) and C(7) leave the start (0); A ends at an event with early time 4, from which D(x) leaves and a dummy runs to the event at 11; C ends at an event with early time 7, from which E(4) and F(x) leave; B and E end at the event with early time 11, from which G(3) and H(5) leave; D and G end at the event with early time 14, from which J(7) leaves and a dummy runs to the event at 16; F and H end at the event with early time 16, from which K(6) leaves; J and K end at the finish (22); the late event time boxes are empty — Figure 2

The network in Figure 2 shows the activities that need to be completed for a project. Each activity is represented by an arc and the duration of the activity, in days, is shown in brackets. The early event times are shown in Figure 2.

(a) Complete Table 1 in the answer book to show the immediately preceding activities for each activity. (2)

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

Table 1

It is given that \(4 \lt x \leqslant m\)

(b) State the largest possible integer value of \(m\). (1)

(c)

(i) Complete Diagram 1 in the answer book to show the late event times.

(ii) State the activities that must be critical. (3)

[Diagram 1 in the answer book is a copy of Figure 2.]

(d) Calculate the total float for activity G. (1)

The resource histogram in Figure 3 shows the number of workers required when each activity starts at its earliest possible time. The histogram also shows which activities happen at each time.

Figure 3: resource histogram, number of workers against time: 0 to 4, A (3 workers), B (2) and C (1), total 6; 4 to 5, B (2), C (1) and D (2), total 5; 5 to 7, C (1) and D (2), total 3; 7 to 10, E (2), D (2) and F (1), total 5; 10 to 11, E (2) and F (1), total 3; 11 to 13, G (2), H (2) and F (1), total 5; 13 to 14, G (2) and H (2), total 4; 14 to 16, H (2) and J (1), total 3; 16 to 21, J (1) and K (3), total 4; 21 to 22, K (3), total 3 — Figure 3

(e) Complete Table 2 in the answer book to show the number of workers required for each activity of the project. (2)

Activity	Number of workers	Activity	Number of workers
A		F
B		G
C		H
D		J
E		K

Table 2

(f) Draw a Gantt chart on Grid 1 in the answer book to represent the activity network. (5)

Grid 1: blank grid with time from 0 to 22 — Grid 1

Mark scheme (a)

Scheme

Marks

Activity	IPA	Activity	IPA
A	-	F	C
B	-	G	A, B, E
C	-	H	A, B, E
D	A	J	D, G
E	C	K	D, F, G, H

B1
B1

1.1b
1.1b

(2)

Notes

(a) B1: Four correct rows (not including the rows for A, B and C)

B1: All rows correct (accept blanks or dashes (etc.) for A, B and C but any letters placed in these three rows scores B0)

Mark scheme (b)

Scheme	Marks	AO
\((m =)\ 9\)	B1	2.2a
	(1)

Notes

(b) B1: cao for the value of \(m\) – allow \(4 \lt x \leqslant 9\) or \(x \leqslant 9\) but not \(x = 9\) or \(m \leqslant 9\) unless stating the correct value too

Diagram 1 completed (early, late): start 0, 0; end of A 4, 15 − x; end of C 7, 7; end of B and E 11, 11; end of D and G 14, 15; end of F and H 16, 16; finish 22, 22 — Mark scheme (c)

Scheme	Marks	AO
(i)	M1 A1	1.1b 1.1b
(ii) C, E, H and K must be critical	B1	2.2a
	(3)

Scheme	Marks	AO
Total float for activity G is 15 – 11 – 3 = 1	B1	2.2a
	(1)

cascade (Gantt) chart: C 0–7, E 7–11, H 11–16, K 16–22 on the critical row; A 0–4 with float to 9; B 0–5 with float to 11; D 4–10 with float to 15; F 7–13 with float to 16; G 11–14 with float to 15; J 14–21 with float to 22; floats shaded — Mark scheme (f)

Scheme	Marks	AO
\(x = 6\)	B1	3.1b
	M1 A1 A1 A1	2.1 1.1b 1.1b 1.1b
	(5)
	(14 marks)

Activity	Duration + Float
A	0 to 4 F: 4 to 9
B	0 to 5 F: 5 to 11
C	0 to 7 Critical
D	4 to 10 F: 10 to 15
E	7 to 11 Critical
F	7 to 13 F: 13 to 16
G	11 to 14 F: 14 to 15
H	11 to 16 Critical
J	14 to 21 F: 21 to 22
K	16 to 22 Critical

AS June 2022 Q2

EdexcelAS paperCurrent spec8 marksCritical Path Analysis

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

(a) Draw the activity network described in the precedence table above, using activity on arc. Your activity network must contain the minimum number of dummies only. (5)

(b) Explain why it is necessary to draw a dummy from the end of activity A. (1)

Every activity shown in the precedence table has the same duration.

activity network: A, B, C and D start at the source; E from the end of A; dummies from the end of A and from the end of C to the start of F; G and H from the end of C; dummy from the end of E to the start of J; F and G lead to the start of J; D and H lead to the start of K; I, J and K end at the finish — Mark scheme (a)

Mark scheme (b)

Scheme	Marks	AO
e.g. Activity E is preceded by activity A only, but activity F is preceded by activity B (and/or C) as well as activity A	B1	2.4
	(1)

Notes

B1: Reference to E depends on A only, while F depends on A and B (and/or C). So must mention activities A, E, F and at least one of B and C

Mark scheme (c)

Scheme	Marks	AO
Activity D as this is the only activity on a path from start to finish of which contains only two activities. All other activities appear on at least one longer path	M1 A1	3.1b 3.4
	(2)
	(8 marks)

Notes

M1: cao - Activity D only – if more than one activity stated then M0

A1: Correct reasoning. Explain that the path/route through D is the only one containing two activities or that all other routes/paths have 3 activities. Or mention activities C, H and K and that C and H together take ‘longer’ to finish than D

A2 October 2021 Q7

EdexcelCurrent spec8 marksCritical Path Analysis

Figure 5: partial activity network: A, B and C leave the start; D, E and F leave the end of A; F ends at the end of B; a dummy runs from there to the end of C; G leaves the end of B and F; H leaves the end of C and the dummy; E, G and H end at the same event; D ends at its own event — Figure 5

Figure 5 shows a partially completed activity network for a project that consists of 14 activities.

(a) Complete the precedence table in the answer book for the 8 activities in Figure 5. (2)

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

The precedence table for the remaining 6 activities is given below.

Activity	Immediately preceding activities
I	D, E, G, H
J	D, E, G, H
K	E, G, H
L	I, J, K
M	J, K
N	J, K

(b) Complete the activity network in the answer book for the project. Your completed activity network must contain only the minimum number of dummies. (4)

[The activity network in the answer book is a copy of Figure 5.]

Given that all 14 activities have the same duration,

Mark scheme (a)

Scheme

Marks

Activity	Immediately preceding activities
A	-
B	-
C	-
D	A
E	A
F	A
G	B, F
H	B, C, F

B1
B1

1.1b
1.1b

(2)

Notes

(a) B1: Either row G or H correct

B1: All rows correct (condone blanks in A, B and C rows)

completion of the network: D ends at an event from which I and J leave; E, G and H end at an event from which K leaves, with a dummy from it up to the end of D; J and K end at the same event, from which N leaves; a dummy runs from there up to a middle event, from which M leaves, and a second dummy from the middle event up to the end of I, from which L leaves; L, M and N end at the finish — Mark scheme (b)

Mark scheme (c)

Scheme	Marks	AO
If all activities have the same duration then any critical path must contain 5 activities. All paths that pass-through D have only 4 activities and so therefore D cannot be critical.	B1 B1	2.4 2.4
	(2)
	(8 marks)

Notes

(c) B1: Explains that all critical paths must contain 5 activities (oe method e.g., attempting a forward and backward pass with each activity having the same duration)

B1: cao that D cannot be critical with mention of all paths through D only contain 4 activities (oe method e.g., showing that the total float on activity D is not zero)

SCB1: stating or implying that D has a float of 1 (oe) by considering a forward pass (which may or may not be done mathematically) up to at least activity D

A2 October 2021 Q2

EdexcelCurrent spec9 marksCritical Path Analysis

Figure 2: activity network: A(4), B(6) and C(10) leave the start; D(2) and E(4) follow A; B and E end at the same event, from which a dummy runs up to the end of D, and H(7), I(2) and F(5) leave; G(6) follows the end of D and the dummy; C and F end at the same event, from which a dummy runs to the end of I, and L(7) and M(5) leave; J(7) follows G and H; K(6) follows the end of I and the dummy; J, K and L end at the finish; a dummy runs from the end of M to the finish — Figure 2

A project is modelled by the activity network shown in Figure 2. The activities are represented by the arcs. The number in brackets on each arc gives the time, in hours, to complete the corresponding activity.

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

Diagram 1: the network of Figure 2 with an empty box at each event for the early event time (top) and late event time (bottom) — Diagram 1

Each activity requires one worker and the project must be completed in the shortest possible time using as few workers as possible.

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

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

Diagram 1 completed (early, late): start 0, 0; end of A 4, 4; end of D 8, 9; end of B and E 8, 8; end of G and H 15, 15; end of I 13, 16; end of C and F 13, 15; end of M 18, 22; finish 22, 22 — Mark scheme (a)

Mark scheme (b)

Scheme	Marks	AO
\(\dfrac{71}{22} = \ldots\)	M1	1.1b
…= 3.22… therefore 4 workers	A1	2.2a
	(2)

Notes

(b) M1: Attempt to find the lower bound (71 \(\pm\) 10) / their completion time (a value of 3.2… seen with no working can imply this mark)

A1: cso - correct calculation seen or 3.2 followed by 4. An answer of 4 with no working scores M0A0

schedule on 4 workers: worker 1 A 0–4, E 4–8, H 8–15, J 15–22; worker 2 B 0–6, D 6–8, G 8–14, I 14–16, K 16–22; worker 3 C 0–10, F 10–15, L 15–22; worker 4 M 15–20 — Mark scheme (c)

A2 October 2020 Q2

EdexcelCurrent spec15 marksCritical Path Analysis

Figure 1: activity network: A(6), B(7) and C(5) leave the start; D(4) and E(5) follow A; F(3) and G(4) follow B; a dummy from the end of B to the end of C; H(3) follows that event; I(4) runs from the end of E and F up to the end of D; K(5) and L(6) follow; J(6) follows G and H; a dummy from the end of K to the finish; L and J end at the finish — Figure 1

The network in Figure 1 shows the activities that need to be undertaken to complete a project. Each activity is represented by an arc and the duration, in hours, of the corresponding activity is shown in brackets.

(a) Explain why each of the dummy activities is required. (2)

(b) Complete the table in the answer book to show the immediately preceding activities for each activity. (2)

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

(c)

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

(ii) State the minimum completion time for the project.

(iii) State the critical activities. (6)

Each activity requires one worker. Each worker is able to do any of the activities. Once an activity is started it must be completed without interruption.

(d) On Grid 1 in the answer book, draw a resource histogram to show the number of workers required at each time when each activity begins at its earliest possible start time. (3)

Grid 1: blank grid, number of workers from 0 to 6 against time in hours from 0 to 23 — Grid 1

(e) Determine whether or not the project can be completed in the minimum possible time using fewer workers than the number indicated by the resource histogram in (d). You must justify your answer with reference to the resource histogram and the completed Diagram 1. (2)

Mark scheme (a)

Scheme	Marks	AO
The dummy at the end of activity B is required as F (and G) are dependent on activity B only, but activity H is dependent on both activities B and C	B1	2.4
The dummy at the end of activity K is required as two activities cannot start at the same event and finish at the same event	B1	2.4
	(2)

Notes

(a) B1: Correct explanation for precedence dummy (must mention B, C, H and one of F or G)

B1: Correct explanation for uniqueness dummy

Mark scheme (b)

Scheme

Marks

Activity	IPA	Activity	IPA	Activity	IPA
A	-	E	A	I	E, F
B	-	F	B	J	G, H
C	-	G	B	K	D, I
D	A	H	B, C	L	D, I

B1
B1

1.1b
1.1b

(2)

Notes

(b) B1: Six correct rows (not including A, B and C)

B1: All rows correct (accept blanks for A, B and C)

Diagram 1 completed (early, late): start 0, 0; end of A 6, 6; end of B 7, 8; end of C 7, 12; end of E and F 11, 11; end of D and I 15, 15; end of G and H 11, 15; end of K 20, 21; finish 21, 21 — Mark scheme (c)

resource histogram: 3 workers (A, B, C) from 0 to 5, 2 (A, B) from 5 to 6, 3 (B, D, E) from 6 to 7, 5 (D, E, F, G, H) from 7 to 10, 2 (E, G) from 10 to 11, 2 (I, J) from 11 to 15, 3 (J, K, L) from 15 to 17, 2 (K, L) from 17 to 20, 1 (L) from 20 to 21 — Mark scheme (d)

Mark scheme (e)

Scheme	Marks	AO
Currently five workers are required between time 7 and 10 and so one of the non-critical activities D, F, G or H would have to be delayed and start after time 10	M1	2.4
e.g. Activity H could be delayed and start at time 10 (as it has sufficient total float and can finish as late as time 15) and so the project can be completed with fewer workers than the number indicated by the resource histogram as J could be delayed too and start at time 15	A1	2.2a
	(2)
	(15 marks)

Notes

(e) M1: Explanation involving the need to delay just one of the non-critical activities (must mention one of D, F, G or H) to start after time 10 (oe) – follow through their histogram

A1: Dependent on a correct histogram and correct answer to (c)(i). Correct deduction that it is possible to complete with fewer workers e.g. delay H to start at 10 therefore delay J to start at its late time (or 15) – A0 if mention of delaying activity F

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)

(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)

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)

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))

A2 June 2019 Q5

EdexcelCurrent spec6 marksCritical Path Analysis

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

(a) Draw the activity network described in the precedence table above, using activity on arc. Your activity network must contain only the minimum number of dummies. (5)

Given that all the activities shown in the precedence table have the same duration,

(b) state the critical path for the network. (1)

activity network: A, B and C leave the start; G and D leave the end of A; B and D meet at a central event, with a dummy into it from the end of C; F and H leave the central event; a dummy runs from the central event to the end of G; I leaves that event and a further dummy leads to J; E follows C and ends with H; K follows; F, I, J and K end at the finish — Mark scheme (a)

A2 June 2019 Q4

EdexcelCurrent spec9 marksCritical Path Analysis

Figure 3: activity network with activities A(8), B(6), C(7), D(11), E(5), F(8), G(10), H(x), I(6), J(4), K(6), L(7), M(6) and four dummies, with an empty early/late event time box at each event; the late event time at the end of H is 25 — Figure 3

The network in Figure 3 shows the activities that need to be undertaken to complete a project. Each activity is represented by an arc and the duration of the activity, in days, is shown in brackets. The early event times and late event times are to be shown at each vertex and one late event time has been completed for you.

The total float of activity H is 7 days.

(a) Explain, with detailed reasoning, why \(x = 11\) (2)

(b) Determine the missing early event times and late event times, and hence complete Diagram 1 in your answer book. (3)

[Diagram 1 in the answer book is a copy of Figure 3.]

Each activity requires one worker and the project must be completed in the shortest possible time using as few workers as possible.

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

Grid 1: blank grid with a time axis from 0 to 34 — Grid 1

Mark scheme (a)

Scheme	Marks	AO
The early event time at the end of activity C is 7 (as no other activity leads into this event). Therefore the float on activity H is \(25 - 7 - x\)	B1	3.1a
The float on activity H is given as 7 and so therefore \(25 - 7 - x = 7\) implies that the value of \(x\) is equal to \(25 - 7 - 7 = 11\)	dB1	2.4
	(2)

Notes

(a) B1: correct reasoning for why the float on activity H is given by \(25 - 7 - x\), must mention that the early event time at the end of activity C is 7 or the early event time at the start of H is 7 and that the total float for H is therefore \(\underline{25 - 7 - x}\) (or \(25 - x - 7\) but not just \(18 - x\)) (no reason for why the early event time at the end of C is 7 is required)

dB1: correct explanation for why \(x = 11\) (dependent on previous B mark) – as a minimum must equate \(\underline{25 - 7 - x}\) to \(\underline{7}\) (allow \(18 - x = 7\) as they must have shown where the 18 comes from to get the first B mark) and hence \(\underline{x = 11}\)

SC B1B0: – for those who write or imply \(25 - 7 - x = 7\) (but not just \(18 - x = 7\)) and state \(x = 11\) without any mention of the early event time at the end of C or the total float of activity H. However, \(25 - 7 - 7 = 11\) only is no marks in this part

Diagram 1 completed (early, late): start 0, 0; end of A 8, 8; start of F 8, 17; end of C 7, 14; end of D 19, 19; end of I 25, 32; start of L 25, 25; end of H 18, 25; finish 32, 32; H(11) — Mark scheme (b)

schedule for four workers on Grid 1: A 0-8, D 8-19, K 19-25, L 25-32; B 0-6, idle 6-8, E 8-13, F 13-21, I 21-27; C 0-7, G 7-17, idle 17-19, J 19-23, M 23-29; H 7-18 — Mark scheme (d)

AS June 2019 Q3

EdexcelAS paperCurrent spec7 marksCritical Path Analysis

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

(a) Draw the activity network described in the precedence table above, using activity on arc. Your activity network must contain the minimum number of dummies. (5)

Every activity shown in the precedence table has the same duration.

(b) Explain why activity B cannot be critical. (1)

activity network: A and B from the start; C from the end of A to the end of B; D and E from the end of A; F and G from the end of B; dummies from the end of D and from the end of G into the end of E and F; H from the end of D; I and J from the end of E and F; K from the end of G; a final dummy joining the ends of H and I to the finish — Mark scheme (a)

Mark scheme (b)

Scheme	Marks	AO
Activity F (and/or G) requires activity B and the two activities A and C to be completed before F (and/or G) can begin. The time to complete A and C is double that of B and so B can be delayed waiting for A and C to be completed and so B is therefore not critical.	B1	2.4
	(1)

Notes

B1: CAO - some mention of the time required to complete A + C compared with B (for the next activity to begin (either F and/or G)) oe e.g. paths through B have a maximum length of 3 (non-dummy) activities and there is at least one path of length 4 which does not include B so B cannot be critical OR the late time for B must be the same as the late time for A + C which is twice the duration of B and therefore B is not critical. Give bod to responses that imply that B and C meet at the same event, but C is also dependent on A (the key point for awarding this mark is that activities A and C imply that B is not critical)

AS June 2018 Q3

EdexcelAS paperCurrent spec10 marksCritical Path Analysis

Activity	Time taken (days)	Immediately preceding activities
A	5	–
B	8	–
C	4	–
D	14	A
E	10	A
F	3	B, C, E
G	7	C
H	5	D, F, G
I	7	H
J	9	H

The table above shows the activities required for the completion of a building project. For each activity, the table shows the time it takes, in days, and the immediately preceding activities. Each activity requires one worker. The project is to be completed in the shortest possible time.

Figure 2: partially completed activity network with A(5), B(8) and C(4) from the start, and F(3), H(5) and I(7) in a chain after B — Figure 2

Figure 2 shows a partially completed activity network used to model the project. The activities are represented by the arcs and the number in brackets on each arc is the time taken, in days, to complete the corresponding activity.

(a) Add the missing activities and necessary dummies to Diagram 1 in the answer book. (3)

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

At the beginning of the project it is decided that activity G is no longer required.

(d) Explain what effect, if any, this will have on

(i) the shortest completion time of the project if activity G is no longer required,

(ii) the timing of the remaining activities. (3)

completed activity network with early/late event times: start 0/0; end of A 5/5; end of B 15/16; end of C 4/12; end of F 19/19; end of H 24/24; end of J 33/33; finish 33/33; E from the end of A to the end of B, dummy from the end of C to the end of B, D from the end of A and G from the end of C both to the end of F, and a dummy from the end of J to the finish — Mark scheme (a)

Mark scheme (d)

Scheme	Marks	AO
(i) No effect (as G is not one of the critical activities)	B1	2.4
(ii) Activity C is the only affected activity and it can now start 4 days later at time 12 (rather than at time 8) or finish as late as time 16	M1 A1	3.4 1.1b
	(3)
	(10 marks)

Notes

(d)(i) B1: Explanation that there is no effect on the completion time (as G is not critical)

(d)(ii) M1: Use their model to deduce that C is the (only) activity that is affected

A1: Correct answer that activity C (only – maybe implicit) can e.g. finish at time 16 or start at time 12 – for this mark the explanation must give a relevant time

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

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)

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

D1 June 2019 Q5

EdexcelOld spec7 marksCritical Path Analysis

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

(a) Draw the activity network described in the precedence table above, using activity on arc and exactly 4 dummies. (5)

(b) Explain why one of the activities I or J must be critical. (1)

It is given that activity C is a critical activity.

Activity on arc network for activities A to J with four dummies — Mark scheme (a)

Mark scheme (b)

Scheme	Marks
In the network there must be at least one critical path from the start node to the end node and as all paths must pass through either I or J (as these are the only two activities that go into the end node) so one of these must be critical	B1
	(1)

Notes

b1B1: CAO (must either mention that any critical path for the network would need to pass through either I or J

D1 June 2018 Q6

EdexcelOld spec11 marksCritical Path Analysis

Figure 4: activity network with activities A to N and dummies — Figure 4

A project is modelled by the activity network shown in Figure 4. The activities are represented by the arcs. The number in brackets on each arc gives the time, in days, to complete the corresponding 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 the late event times. (4)

(b) State the critical activities. (1)

(d) Use your cascade chart to determine the minimum number of workers needed to complete the project in the shortest possible time. You must make specific reference to times and activities. (You do not need to provide a schedule of the activities.) (2)

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

Cascade chart for activities A to N with floats shaded — Mark scheme (c)

Mark scheme (d)

Scheme	Marks
Minimum workers is 4 e.g. D, E, F and G	M1
together with \(11 \lt \text{time} \lt 12\)	A1
	(2)
	(11 marks)

Notes

d1M1: Either a statement with the correct number of workers (4) and the correct activities (D, E, F and G) with any numerical time stated or the correct number of workers (4) and a correct time

d1A1: A completely correct statement with details of both time and activities. Candidates only need to give a time within the correct interval of \(11 \lt \text{time} \lt 12\). Please note the strict inequalities for the time interval (e.g. implying a time of 11 is incorrect). Answers given as an interval of time are acceptable provided the time interval stated is correct for all its possible values (e.g. ‘time 11 – 12’ is A0, ‘in the interval 11 – 12’ is A0 but ‘betweeen 11 and 12’ is A1). Allow for example, ‘on day 12’ as equivalent to \(11 \lt \text{time} \lt 12\)

D1 June 2018 Q3

EdexcelOld spec7 marksCritical Path Analysis

(a) Draw the activity network described in the precedence table below, using activity on arc and exactly four dummies. (5)

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

Given that D is a critical activity,

(b) state which other activities must also be critical. (2)

Activity on arc network for activities A to K with four dummies — Mark scheme (a)

Mark scheme (b)

Scheme	Marks
Given that D is critical this implies that A and E are guaranteed to be critical	M1 A1
	(2)
	(7 marks)

Notes

b1M1: One correct activity with at most 4 activities stated (ignore any mention of D in this part)

b1A1: Both (A and E) correct and no others

D1 June 2017 Q6

EdexcelOld spec11 marksCritical Path Analysis

Figure 6: activity network with activities A to N and dummies — Figure 6

A project is modelled by the activity network shown in Figure 6. The activities are represented by the arcs. The number in brackets on each arc gives the time, in days, to complete the corresponding activity. Each activity requires exactly 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 the late event times. (4)

(b) Draw a Gantt chart for the project on the grid provided in the answer book. (4)

An additional activity, P, is now included in the activity network shown in Figure 6. Activity P is immediately preceded only by activity D. No activity is dependent on the completion of activity P.

Each activity still requires exactly one worker and the revised project is to be completed in the shortest possible time.

(d) Explain, briefly, whether or not the revised project can be completed in the same time as the original project if the duration of activity P is

(i) 10 days

(ii) 17 days (2)

Gantt chart for activities A to N with floats shaded — Mark scheme (b)

Mark scheme (d)

Scheme	Marks
The earliest that P can start is at time 9 (the time that the critical activity D must be finished by) – if the duration of P is 10 then the earliest P can finish is 19 < 24 and so the second project can be completed on time. If the duration of P is 17 days then the earliest P can finish is 26 > 24 and so the second project cannot be completed on time	B1 B1
	(2)
	(11 marks)

Notes

Both marks in (d) are dependent on the correct early event time (of 9) at the end of D and the correct late event time (of 24) at the end of N

d1B1: ‘yes’ (or clearly implied) and mention of 9 and 10 or 19 or a float (oe) of 5 - but not simply ‘it will finish before the end of the project’ (their answer must contain some form of mathematical argument)

d2B1: ‘no’ (or clearly implied) and mention of 9 and 17 or 26 or a delay (oe) of 2 – but not simply ‘it will finish late’ (their answer must contain some form of mathematical argument)

D1 June 2016 Q7

EdexcelOld spec12 marksCritical Path Analysis

Figure 6: activity network with early and late event times, including unknowns w, x, y and z — Figure 6

The network in Figure 6 shows the activities that need to be undertaken by a company to complete a project. Each activity is represented by an arc and the duration, in days, is shown in brackets. Each activity requires exactly one worker. The early event times and late event times are shown at each vertex.

Given that the total float on activity D is 1 day,

(a) find the values of \(w\), \(x\), \(y\) and \(z\). (3)

(b) On Diagram 1 in the answer book, draw a cascade (Gantt) chart for the project. (4)

(c) Use your cascade chart to determine a lower bound for the minimum number of workers needed to complete the project in the shortest possible time. You must make specific reference to times and activities. (2)

It is decided that the company may use up to 36 days to complete the project.

(d) On Diagram 2 in the answer book, construct a scheduling diagram to show how the project can be completed within 36 days using as few workers as possible. (3)

Mark scheme (a)

Scheme	Marks
\(w = 11,\ x = 21,\ y = 17,\ z = 4\)	B3, 2, 1, 0
	(3)

Notes

a1B1: Any two values correct (it must be clear which value corresponds to which letter)

a2B1: Any three values correct

a3B1: All four values correct

Mark scheme (c)

Scheme	Marks
At time 12.5, activities H, D, G, I and J must all be happening so 5 workers	M1 A1
	(2)

Notes

c1M1: A statement with the correct number of workers (5) and the correct activities (H, D, G, I and J) with any mention of time (need not be correct)

c1A1: A correct, complete statement with details of both time (\(12 \lt \text{time} \lt 13\)) and activities. Allow ‘on day 13’ or ‘during day 13’ as equivalent to this time interval but not ‘at day 13’ – note strict inequality for the time, for example, at time 12 is A0. Accept the time interval ‘\(12 \lt \text{time} \lt 13\)’ for this mark or a time that implies a time strictly between \(t = 12\) and \(t = 13\)

Scheduling diagram for 3 workers: A, E, G, I, N; B, D, J, M; C, F, H, K, L — Mark scheme (d)

D1 June 2016 Q2

EdexcelOld spec5 marksCritical Path Analysis

2. Draw the activity network described in the precedence table below, using activity on arc and exactly three dummies. (5)

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

Activity on arc network for activities A to K with three dummies — Mark scheme

D1 June 2015 Q7

EdexcelOld spec13 marksCritical Path Analysis

Activity	Time taken (days)	Immediately preceding activities
A	5	-
B	7	-
C	8	-
D	5	A
E	7	A
F	10	B, C
G	4	B, C
H	9	C
I	8	G, H
J	12	G, H
K	7	D
L	10	E, F, I, J

The table shows the activities required for the completion of a building project. For each activity the table shows the time taken, in days, and the immediately preceding activities. Each activity requires one worker. The project is to be completed in the shortest possible time.

Figure 6: partially completed activity network with activities A, B, C, D, G, H, J, K, L and a dummy — Figure 6

Figure 6 shows a partially completed activity network used to model the project. The activities are represented by the arcs and the numbers in brackets on the arcs are the times taken, in days, to complete each activity.

(a) Add activities, E, F and I, and exactly one dummy to Diagram 1 in the answer book. (3)

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

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

(d) Schedule the activities, using the minimum number of workers, so that the project is completed in the minimum time. (4)

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

Mark scheme (c)

Scheme	Marks
Lower bound = \(\frac{92}{39}\) =2.35… so 3 workers	M1 A1
	(2)

Notes

c1M1: Attempt to find lower bound: (a value in the interval [80 – 104] / their finish time) or (sum of the activities / their finish time) or (as a minimum) an awrt 2.4

c1A1: CSO – either a correct calculation seen or awrt 2.4 then 3. An answer of 3 with no working scores M0A0

Schedule for 3 workers: C, H, J, L; A, D, E, I; B, G, F, K — Mark scheme (d)

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)

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)

D1 June 2014 (R) Q6

EdexcelOld spec7 marksCritical Path Analysis

(i) Draw the activity network described in the precedence table below, using activity on arc and the minimum number of dummies.

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

(ii) Explain why each of your dummies is necessary.

Activity on arc network with events 1 to 9 and two dummies — Mark scheme (i)

Mark scheme (ii)

Scheme	Marks
1^st dummy – G depends on A only, but D depends on A and C.	B1
2^nd dummy – This is so that H and I will not share the same start and end events or so that H and I can be uniquely described in terms of their end events.	B1
	(7 marks)

Notes

1B1: CAO - all relevant activities must be referred to – so activities D, G, A and C must all be mentioned for this mark

2B1: CAO – please note that e.g. ‘so that activities can be defined uniquely’ is not sufficient to earn this mark. There must be mention of describing activities uniquely in terms of the event at each end. However, give bod on statements that imply that an activity begins at ends at the same event.

D1 June 2014 Q7

EdexcelOld spec14 marksCritical Path Analysis

(a) In the context of critical path analysis, define the term ‘total float’. (2)

Figure 3: activity network with events 1 to 9 and activities A to M — Figure 3

Figure 3 is the activity network for a building project. 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 exactly one worker. The project is to be completed in the shortest possible time.

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

(d) Calculate the maximum number of days by which activity G could be delayed without affecting the shortest possible completion time of the project. You must make the numbers used in your calculation clear. (2)

(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)

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

(f) Schedule the activities using Grid 1 in the answer book. (4)

Mark scheme (a)

Scheme	Marks
The total float \(F(i, j)\) of activity \((i, j)\) is defined to be \(F(i, j) = l_j - e_i -\) duration \((i, j)\), where \(e_i\) is the earliest time for event \(i\) and \(l_j\) is the latest time for event \(j\) (see note below)	B2,1,0
	(2)

Notes

a1B1 For the first mark: the idea that total float is ‘how long an activity can be delayed for’. Give bod.

a2B1 For both marks: A clear correct statement e.g. the total amount of time that an activity may be delayed from its early start without delaying the project finish time. The candidate must clearly demonstrate a knowledge that total float = latest finish – earliest start – duration of activity. Ignore comments that infer that total refers to the sum of the floats for all activities in an activity network. Note that B1B0 should be awarded for an answer that has the pertinent idea of ‘float’ (see a1B1 above) and B1B1 for a clear correct statement (see a2B1 above) – B0B1 cannot be awarded in this part.

Activity network with early and late event times: 1 0/0, 2 4/4, 3 11/11, 4 13/15, 5 15/17, 6 14/21, 7 21/21, 8 16/30, 9 30/30 — Mark scheme (b)

Mark scheme (d)

Scheme	Marks
G can be delayed by 21 – 11 – 3 = 7 (days)	M1 A1
	(2)

Notes

d1M1 Correct calculation for their activity G seen - all three numbers correct (ft). Final value must be non-negative.

d1A1 CAO (no follow through on this A mark). Answer of 7 with no working scores no marks in this part.

Mark scheme (e)

Scheme	Marks
\(\dfrac{69}{30} = 2.3\) so lower bound is 3 workers	M1 A1
	(2)

Notes

e1M1 Attempt to find lower bound [59 – 79 / their finish time]

e1A1 CAO – correct calculation seen then 3. [As 30/13 also gives 3, an answer of 3 with no working scores M0A0.]

Schedule on 3 workers: A, C, J, M; B, E, G, H, K; D, F, I, L — Mark scheme (f)

D1 June 2014 Q2

EdexcelOld spec7 marksCritical Path Analysis

(a) Draw the activity network described in the precedence table below, using activity on arc and exactly two dummies. (5)

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

(b) Explain why each of the two dummies is necessary. (2)

Activity network with events 1 to 9, activities A to K and two dummies — Mark scheme (a)

Mark scheme (b)

Scheme	Marks
1st dummy – A and B both must be able to be described uniquely in terms of the events at each end.	B1
2nd dummy – I depends on D only but J depends on D and G.	B1
	(2)
	(7 marks)

Notes

b1B1 CAO – with no incorrect terminology (e.g. event for activity) - please note that e.g. ‘so that activities can be defined uniquely’ is not sufficient to earn this mark. There must be a mention of describing activities uniquely in terms of the events at each end. However give bod on statements that imply that an activity begins and ends at the same event e.g. ‘so that activites do not have the same start and finish’ is sufficient for B1.

b2B1 CAO – all relevant activities must be referred to – so activities D, G, I and J must all be mentioned for this mark.

D1 June 2013 (R) Q6

EdexcelOld spec7 marksCritical Path Analysis

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

(a) Draw the activity network described in the precedence table, using activity on arc and exactly two dummies. (5)

(b) Explain why each of the two dummies is necessary. (2)

Activity network with activities A to K and two dummies — Mark scheme (a)

Mark scheme (b)

Scheme	Marks
1st dummy – G depends on D only, F depends on C and D.	B1
2nd dummy – I and J must be expressed uniquely in terms of their end events.	B1
	(2)
	(7 marks)

Notes

b1B1 First dummy correctly described (C, D, F and G referred to)

b2B1 Second dummy correctly described (mention of ‘uniqueness’ alone is not sufficient for this mark).

D1 June 2013 (R) Q3

EdexcelOld spec12 marksCritical Path Analysis

Figure 3: activity network with activities A to M — Figure 3

A project is modelled by the activity network shown in Figure 3. 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 H. 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 shortest possible time. Show your calculation. (2)

Diagram 2 in the answer book shows a partly completed scheduling diagram for this project.

(d) Complete the scheduling diagram, using the minimum number of workers, so that the project is completed in the minimum time. (4)

Activity network with early and late event times: 0/0, 8/12, 9/9, 9/9, 17/17, 16/20, 26/26, 31/31, 37/37 — Mark scheme (a)

Mark scheme (c)

Scheme	Marks
93/37 = 2.5135 so 3 (workers)	M1 A1
	(2)

Notes

c1M1 Attempt to find lower bound. [82 – 104 / their finish time] accept awrt 2.5

c1A1 CAO – correct calculation seen or awrt 2.5, then 3. (Beware 37/13 gives 3 also, so 3 with no working gets M0A0.)

Scheduling diagram: Worker 1 C, F, I, K, M; Worker 2 A, E, H, J, L; Worker 3 B, G, D — Mark scheme (d)

D1 June 2013 Q7

EdexcelOld spec17 marksCritical Path Analysis

Figure 5: activity network with events 1 to 12 and activities A to S — Figure 5

[The sum of the duration of all activities is 172 days]

(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 M. You must make the numbers you use in your calculation clear. (2)

(i) A 2 day delay on the early start of activity P.

(ii) A 2 day delay on the early start of activity Q. (2)

(d) Calculate a lower bound for the number of workers needed to complete the project in the shortest possible time. (1)

Diagram 2 in the answer book shows a partly completed cascade chart for this project.

(e) Complete the cascade chart. (4)

(f) Use your cascade chart to determine a second lower bound on the number of workers needed to complete the project in the shortest possible time. You must make specific reference to times and activities. (2)

(g) State which of the two lower bounds found in (d) and (f) is better. Give a reason for your answer. (2)

Activity network with early and late event times: 1 0/0, 2 5/7, 3 7/7, 4 7/7, 5 15/18, 6 18/18, 7 26/31, 8 27/31, 9 28/28, 10 36/42, 11 42/42, 12 53/53 — Mark scheme (a)

Mark scheme (c)

Scheme	Marks
(c)(i) 2 day delay on P – no effect on the project completion date (float on P is 4)	B1
(c)(ii) 2 day delay on Q – project finishes 2 days late (Q is a critical activity)	B1
	(2)

Notes

c1B1 CAO

c2B1 CAO

Cascade chart: critical activities C, F, L, Q, S on the top row; other activities with floats shaded — Mark scheme (e)

Mark scheme (f)

Scheme	Marks
E.g. Activities H, I, J, K and L together with \(22 \lt \text{time} \lt 26\) stated. So 5 workers needed	M1 A1
	(2)

Notes

Examples for part (f):

Example 1: Activities H, I, J, K and L with \(22 \lt \text{time} \lt 26\) so 5 workers needed.

Example 2: At \(10 \lt \text{time} \lt 14\), F, D, E and H must be happening. Activity G must be happening \(7 \lt \text{time} \lt 18\) but its duration is 5 so it must also occur at some point in the interval \(10 \lt \text{time} \lt 14\) so 5 workers needed.

f1M1 Example 1: A statement with the correct number of workers (5) and the correct activities (H, I, J, K and L) with some mention of time, or
Example 2: A statement with the correct number of workers (5), the correct activities (F,D,E and H) with some mention of time and an indication that G must be happening with the other four activities at some point - give bod but e.g. ‘at time 11 F, D, E, G and H must be happening’ is M0). Scheduling the activities only scores M0.

f1A1 A correct, complete full statement with details of both time and activities. Candidates only need to give a time within the intervals stated.
Please note strict inequalities for the time. Allow e.g. on ‘day 23’ as equivalent to \(22 \lt \text{time} \lt 23\).

Mark scheme (g)

Scheme	Marks
The cascade gives a higher lower bound, so (f) is better.	M1 A1
	(2)
	(17 marks)

Notes

g1M1 Must have attempted both parts (d) and (f). Their higher lower bound chosen + attempt at a reason.
Allow for the M mark a reason which argues that e.g. the cascade chart gives a better lower bound (e.g. it takes into account exactly when activities must be taking place) or e.g. the calculation gives a better lower bound (e.g. as it takes into account the sum of all the activities) but without specifically answering the question of which of the two bounds is better. Give bod on an attempt at a reason.

g1A1 CAO plus a correct reason given. Acceptable reasons e.g. the cascade gives a larger value or the bound for the cascade shows that the project cannot be done with fewer workers, etc.

D1 January 2013 Q7

EdexcelOld spec16 marksCritical Path Analysis

Figure 7: activity network with events 1 to 9 and activities A to M — Figure 7

Figure 7 is the activity network relating to a building project. The activities are represented by the arcs. The number in brackets on each arc gives the time to complete the activity. Each activity requires one worker.

The project must be completed in the shortest possible time.

(a) Explain the reason for the dotted line from event 4 to event 6 as shown in Figure 7. (2)

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

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

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

(f) State the activities that must be happening at time 5.5 (1)

(g) Use your Gantt chart to determine the minimum number of workers needed to complete the project in the minimum time. You must justify your answer. (2)

Mark scheme (a)

Scheme	Marks
Activity K depends on activities E, F and B, but activity I depends on F and B only.	B2, 1, 0
	(2)

Notes

a1B1 K, I, E and at least one of B or F referred to. Correct statement but may be incomplete give bod here.

a2B1 Clear correct statement no bod (at least one of only B or F referred to can score this mark).

Activity network with early and late event times: 1 0/0, 2 9/9, 3 6/9, 4 16/16, 5 12/15, 6 16/17, 7 19/19, 8 14/18, 9 21/21 — Mark scheme (b)

Gantt chart: critical activities A, F, I, L on one row; non-critical activities B, C, D, E, G, H, J, K, M with floats shaded — Mark scheme (e)

Mark scheme (g)

Scheme	Marks
E.g. Activities F, B, C and G together with \(9 \lt \text{time} \lt 10\) stated	M1
So 4 workers are needed	A1
	(2)
	(16 marks)

Notes

g1M1 A statement with the correct number of workers and details of either time or activities correct. If no part of their statement is correct then allow M mark (only) on the ft with time and activities from their 13 activity, 9 float diagram. Scheduling the activities only or a lower bound calculation argument scores M0.

g1A1 A correct, complete full statement details of time and activities (The two options are F, B, C and G with \(9 \lt \text{time} \lt 10\) or F, C, G and H with \(10 \lt \text{time} \lt 11\)). Please note strict inequalities for the time. Allow e.g. on ‘day 10’ as equivalent to \(9 \lt \text{time} \lt 10\).

D1 June 2012 Q6

EdexcelOld spec14 marksCritical Path Analysis

Figure 5: activity network with events 1 to 8 and activities A to K — Figure 5

Figure 5 is the activity network relating to a development project. 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 the precedence table in the answer book. (2)

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

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

(e) Schedule the activities using the minimum number of workers so that the project is completed in the minimum time. (4)

Activity network with early and late event times: event 1 0/0, 2 5/18, 4 9/21, 3 14/14, 5 11/23, 6 14/25, 7 24/24, 8 28/28 — Mark scheme (b)

Mark scheme (d)

Scheme	Marks
\(\dfrac{62}{28} = 2.21\) so lower bound is 3 workers	M1 A1
	(2)

Notes

d1M1 Attempt to find lower bound. [52-72 / their finish time] accept awrt 2.2.

d1A1 CAO – correct calculation seen or awrt 2.2, then . [Beware 28/11 gives 3 also, so 3 with no working gets M0A0.]

Example schedule on 3 workers: C, H, K; A, E, D, I; B, F, G, J — Mark scheme (e)

B and E precede F and G; C, D and F precede J; D and F precede I — Mark scheme (e)

D1 January 2012 Q7

EdexcelOld spec16 marksCritical Path Analysis

Figure 7: activity network on events 1 to 8: A (6) 1–2, B (5) 1–4, C (7) 1–3, D (5) 2–6, E (6) 2–4, F (8) 3–4, G (4) 3–7, H (5) 3–5, dummy 4–6, dummy 5–7, I (6) 4–8, K (3) 6–8, J (4) 7–8 — Figure 7

A project is modelled by the activity network shown in Figure 7. The activities are represented by the arcs. The number in brackets on each arc gives the time required, in hours, to complete the activity. The numbers in circles are the event numbers. Each activity requires one worker.

(a) Explain the significance of the dummy activity

(i) from event 4 to event 6,

(ii) from event 5 to event 7

(3)

(b) Calculate the early time and the late time for each event. Write these in the boxes in the answer book. (4)

(c) Calculate the total float on each of activities D and G. You must make the numbers you use in your calculations clear. (3)

(d) Calculate a lower bound for the minimum number of workers required to complete the project in the minimum time. (2)

(e) On the grid in your answer book, draw a cascade (Gantt) chart for this project. (4)

Mark scheme (a)

Scheme	Marks
(i) I depends on B, E and F only, K depends on B, E, F and D	B1 DB1
(ii) This is so that G and H will not share the same start and end events. So that G and H can be uniquely described in terms of their end events.	B1
	(3)

Notes

ai1B1: K, I, D and at least one of B, E, F referred to. Correct statement but maybe incomplete give bod here.

ai2DB1: Clear correct statement. No bod.

aii3B1: correct statement referring to either events or activities. (‘unique’ alone not enough)

Event times (early/late): 1: 0/0; 2: 6/9; 3: 7/7; 4: 15/15; 5: 12/17; 6: 15/18; 7: 12/17; 8: 21/21 — Mark scheme (b)

Mark scheme (c)

Scheme	Marks
Total float on D = 18 – 5 – 6 = 7 Total float on G = 17 – 4 – 7 = 6	M1 A1ft B1
	(3)

Notes

c1M1: Correct calculation seen once, all three numbers correct (ft).

c1A1ft: one float ( \(\geqslant\) 0) correct.

c1B1: Both floats correct (independent of working)

Cascade chart: critical C 0–7, F 7–15, I 15–21; A 0–6 with float; B 0–5 with float; D 6–11 with float; E 6–12 with float; G 7–11 with float; H 7–12 with float; J 12–16 with float; K 15–18 with float — Mark scheme (e)

D1 June 2011 Q7

EdexcelOld spec16 marksCritical Path Analysis

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

A project is modelled by the activity network shown in Figure 7. 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 the precedence table in the answer book. (3)

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

(d) On the grid in your answer book, draw a cascade (Gantt) chart for this project. (4)

(e) By considering the activities that must take place between time 7 and time 16, explain why it is not possible to complete this project with just 3 workers in the minimum time. (3)

Mark scheme (a)

Scheme

Marks

Activity	Proceeded by	Activity	Proceeded by	Activity	Proceeded by
(A)	(-)	E	A B	I	C D E
(B)	(-)	(F)	(B)	J	C D E
C	A B	(G)	(B)	K	F H I
(D)	(B)	H	C D	L	F G H I

B3,2,1,0

(3)

Notes

(a)1B1 Any two rows correct

2B1 Any four rows correct

3B1 All seven rows correct

Event times (early/late): 1: 0/0; 2: 7/8; 3: 7/7; 4: 12/12; 5: 12/12; 6: 16/16; 7: 16/16; 8: 22/22 — Mark scheme (b)

Cascade chart: critical B 0–7, D 7–12, H 12–16, L 16–22, J 12–22; A 0–4 float to 8; C 7–11 float to 12; E 7–9 float to 12; F 7–12 float to 16; G 7–11 float to 16; I 12–15 float to 16; K 16–21 float to 22 — Mark scheme (d)

Mark scheme (e)

Scheme	Marks
E.g. Between time 7 and 16, 3 workers could do 3 x 9 = 27 days work. Activities C, D, E, F, G, H, I and 4 days of J need to be done This totals 31 days work. So it is not possible to complete the project with three workers.	B3,2,1,0
OR If three workers are used three activities H, J and I need to happen at time 13.5, this reduces the float on F and G, meaning that at 10.5 D, C, F and G need to be happening. Our initial assumption is incorrect hence four workers are needed.
	(3)
	(16 marks)

Notes

(e)1B1 Attempt at explanation – one correct idea.

2B1 Good explanation, some imprecise or vague statements – give bod

3B1 Fully correct explanation. No bod needed

D1 January 2011 Q7

EdexcelOld spec16 marksCritical Path Analysis

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

The network in Figure 7 shows the activities that need to be undertaken to complete a maintenance project. The activities are represented by the arcs. The number in brackets on each arc gives the time, in days, to complete the activity. The numbers in circles are the events.

Each activity requires one worker. The project is to be completed in the shortest possible time.

(a) Complete the precedence table for this network in the answer book. (3)

(b) Explain why each of the following is necessary.

(i) The dummy from event 6 to event 7.

(ii) The dummy from event 8 to event 9.

(3)

(d) State the critical activities. (2)

(e) Calculate the total float on activity K. You must make the numbers used in your calculation clear. (2)

(f) Calculate a lower bound for the number of workers needed to complete the project in the minimum time. (2)

Mark scheme (b)

Scheme	Marks
Dummy from 6 to 7 needed because K and L depend on G H and I, but J depends on G and H only. Dummy from 8 to 9 needed because no two activities may share both the same start event number and the same finish event number.	B3,2,1,0
	(3)

Notes

1B1: first dummy (precedence) explained, maybe confused, be generous, give bod.

2B1: first dummy clearly explained – all relevant activities referred to. Must refer to K and/or L; H and/or G; I and J

3B1: second dummy (uniqueness) explained, maybe confused, be generous, give bod.

Event times (early/late): 1: 0/0; 2: 4/4; 3: 7/7; 4: 10/10; 5: 11/13; 6: 12/12; 7: 14/16; 8: 19/21; 9: 21/21 — Mark scheme (c)

Mark scheme (d)

Scheme	Marks
Critical activities: A C \(\left\{\begin{matrix}\text{F H}\\ \text{G}\end{matrix}\right\}\) J	B2,1,0
	(2)

Notes

1B1: Critical activities correct condone one omission or extra. SC allow ACGJ for B1 only

2B1: Critical activites cao

D1 June 2010 Q8

EdexcelOld spec11 marksCritical Path Analysis

Figure 7: activity network with activities A (4), B (2), C (3), D (5), E (6), F (9), G (10), H (2), I (4), J (5), K (3), L (6) and four dummies — Figure 7

(a) Complete Diagram 2 in the answer book to show the early and late event times. (4)

(b) State the critical activities. (1)

(d) Use your cascade chart to determine a lower bound for the number of workers needed.
You must justify your answer. (2)

Cascade chart: critical C, E, H, J, L along the top from 0 to 22; A 0–4 float to 6; B 0–2 float to 3; D 4–9 float to 11; F 3–12 float to 16; G 3–13 float to 16; I 9–13 float to 16; K 13–16 float to 22 — Mark scheme (c)

Mark scheme (d)

Scheme	Marks
4 workers needed e.g. at time 8 ½ (noon on day 9) activities E, D, F and G must be happening.	B2, 1, 0
	(2)
	(11 marks)

Notes

1B1: A correct statement, details of either time (7<time<9, 8<day<10), or activities, bod gets B1. Allow 1 B mark (only) on ft from their 12 activity, 7 float diagram.

2B1: A correct, complete full statement details of time and activities.

D1 January 2010 Q6

EdexcelOld spec15 marksCritical Path Analysis

Figure 5: activity network. A(3) 1 to 2; B(5) 1 to 3; dummy 2 to 3; C(4) 2 to 4; D(2) 2 to 5; E(3) 3 to 5; dummy 4 to 5; F(5) 4 to 6; G(6) 5 to 6; H(2) 6 to 7 — Figure 5

Figure 5 is the activity network relating to a building project. The number in brackets on each arc gives the time taken, in days, to complete the activity.

(a) Explain the significance of the dotted line from event 2 to event 3. (2)

(b) Complete the precedence table in the answer booklet. (3)

(d) Determine the critical activities and the length of the critical path. (2)

(e) On the grid in the answer booklet, draw a cascade (Gantt) chart for the project. (4)

Mark scheme (b)

Scheme

Marks

Activity	Immediately preceding activity
A	-
B	-
C	A
D	A
E	A,B
F	C (A)
G	C, D, E
H	F, G

B1
B1
B1

(3)

Notes

B1 To this point (A to D)

B1 For E & F, accepting correct “extra”

B1 Last two rows, correct only

D1 June 2009 Q8

EdexcelOld spec15 marksCritical Path Analysis

Figure 5: activity network with activities A to M and durations in days — Figure 5

A construction 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 2 in the answer book, showing the early and late event times. (4)

(b) State the critical activities. (1)

(d) On the grid provided, draw a cascade (Gantt) chart for this project. (4)

An inspector visits the project at 1pm on days 16 and 31 to check the progress of the work.

(e) Given that the project is on schedule, which activities must be happening on each of these days? (3)

D1 January 2009 Q8

EdexcelOld spec16 marksCritical Path Analysis

Figure 5: activity network with events 1 to 10 and activities A to N with durations in days — Figure 5

The network in Figure 5 shows the activities involved in a process. The activities are represented by the arcs. The number in brackets on each arc gives the time, in days, taken to complete the activity.

(a) Calculate the early time and the late time for each event, showing them on the diagram in the answer book. (4)

(b) Determine the critical activities and the length of the critical path. (3)

(d) On the grid in the answer book, draw a cascade (Gantt) chart for the process. (4)

Given that each task requires just one worker,

(e) use your cascade chart to determine the minimum number of workers required to complete the process in the minimum time. Explain your reasoning clearly. (2)

D1 January 2009 Q3

EdexcelOld spec7 marksCritical Path Analysis

(a) Draw the activity network described in this precedence table, using activity on arc and exactly two dummies. (5)

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

(b) Explain why each of the two dummies is necessary. (2)

Activity network with events 1 to 8: dummy from event 2 (end of B) to event 3 (end of C), and dummy from event 5 (end of G) to event 6 (end of H) — Mark scheme (a)

Mark scheme (b)

Scheme	Marks
1^st dummy – D depends on B only, but E and F depend on B and C	B1
2^nd dummy – G and H both must be able to be described uniquely in terms of the events at each end.	B1
	(2)
	(7 marks)

Notes

(b) 1B1: cao, but B, C, D, E and/or F referred to, generous

2B1: cao, but generous.

D1 June 2008 Q7

EdexcelOld spec14 marksCritical Path Analysis

Figure 6: activity network with activities A to Q, durations in days, and early and late event times including v, w, x, y and z — Figure 6

The network in Figure 6 shows the activities that need to be undertaken to complete a building project. Each activity is represented by an arc. The number in brackets is the duration of the activity in days. The early and late event times are shown at each vertex.

(a) Find the values of \(v\), \(w\), \(x\), \(y\) and \(z\). (3)

(b) List the critical activities. (1)

(d) Draw a cascade (Gantt) chart for the project. (4)

The engineer in charge of the project visits the site at midday on day 8 and sees that activity E has not yet been started.

(e) Determine if the project can still be completed on time. You must explain your answer. (2)

Given that each activity requires one worker and that the project must be completed in 35 days,

(f) use your cascade chart to determine a lower bound for the number of workers needed. You must justify your answer. (2)

Mark scheme (f)

Scheme	Marks
e.g At time 23 ½ activities L, I, J and N must be taking place At time 13 ½ or 14 ½ activities C, D, E and F must be taking place So 4 workers needed.	B2, 1, 0
	(2)
	(14 marks)

D1 January 2008 Q5

EdexcelOld spec7 marksCritical Path Analysis

(a) Draw the activity network described in this precedence table, using activity on arc and exactly two dummies. (4)

Activity	Immediately preceding activities
\(A\)	-
\(B\)	-
\(C\)	\(A\)
\(D\)	\(B\)
\(E\)	\(B\), \(C\)
\(F\)	\(B\), \(C\)

(b) Explain why each of the two dummies is necessary. (3)

Activity network: A from 1 to 2, B from 1 to 3, C from 2 to 4, dummy 1 from 3 to 4, D from 3 to 6, E from 4 to 5, F from 4 to 6, dummy 2 from 5 to 6 — Mark scheme (a)

Mark scheme (b)

Scheme	Marks
Dummy 1 is needed to show dependancy. E and F depend on C and B, but D depends on B only
Dummy 2 is needed so that each activity can be uniquely represented in terms of its events.	B3, 2, 1, 0
	(3)
	(7 marks)

Notes

(b) 1B1 Dummy 1 correctly justified – give bod

2B1 Dummy 2 correctly justified – give bod

3B1 A bonus for two good answers

D1 January 2008 Q4

EdexcelOld spec11 marksCritical Path Analysis

Figure 5: activity network with activities A to M, durations in hours, and some early and late event times — 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 hours, to complete the activity. Some of the early and late times for each event are shown.

(a) Calculate the missing early and late times and hence complete Diagram 1 in your answer book. (4)

(b) Calculate the total float on activities D, G and I. You must make your calculations clear. (4)

Each activity requires one worker.

(d) Calculate a lower bound for the number of workers needed to complete the project in the minimum time. (2)

Mark scheme (b)

Scheme	Marks
Total float on D = 18ft − 5 − 9 = 4ft G = 25 − 8 − 10 = 7 I = 25 − 20 − 3 = 2	M1 A2, 1ft, 0 B1
	(4)

Notes

(b) 1M1 Correct (ft) three numbers visible for at least one calculation.

1A1ft one correct value (ft on D)

1A1 2 correct values

1B1 3 correct values (even if no working seen)

D1 June 2007 Q6

EdexcelOld spec15 marksCritical Path Analysis

Figure 5: activity network with activities A to L, durations in brackets, and early and late event time boxes — Figure 5

The network in Figure 5 shows the activities that need to be undertaken to complete a project. Each activity is represented by an arc. The number in brackets is the duration of the activity in days. The early and late event times are to be shown at each vertex and some have been completed for you.

(a) Calculate the missing early and late times and hence complete Diagram 2 in your answer book. (3)

(b) List the two critical paths for this network. (2)

The sum of all the activity times is 110 days and each activity requires just one worker.

The project must be completed in the minimum time.

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

(e) List the activities that must be happening on day 20. (2)

(f) Comment on your answer to part (e) with regard to the lower bound you found in part (d). (1)

(g) Schedule the activities, using the minimum number of workers, so that the project is completed in 30 days. (3)

Mark scheme (c)

Scheme	Marks
Idea of ‘critical’ – zero float, no delay, immediate – if late project will finish late etc.	B1
Idea of ‘path’ – from start to end event + continuous – the event forming end of one activity forms the start of the next – sequence or series or link or run …	B1
	(2)

Notes

S.C. “longest path” gets 1st B1 only.

D1 January 2007 Q6

EdexcelOld spec16 marksCritical Path Analysis

Figure 5: activity network with events 1 to 10 and activities A(8) to M(7); dummies from 4 to 7 and from 6 to 8 — 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 hours, to complete the activity. The numbers in circles are the event numbers. Each activity requires one worker.

(a) Explain the purpose of the dotted line from event 6 to event 8. (1)

(b) Calculate the early time and late time for each event. Write these in the boxes in the answer book. (4)

(d) Determine the critical activities. (2)

(e) Given that the sum of all the times of the activities is 95 hours, 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) Given that workers may not share an activity, schedule the activities so that the process is completed in the shortest time using the minimum number of workers. (4)

D1 June 2006 Q5

EdexcelOld spec15 marksCritical Path Analysis

Figure 4: activity network with events 1 to 9 and activities A(10) to N(12) — Figure 4

An engineering project is modelled by the activity network shown in Figure 4. 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 time.

(a) Calculate the early time and late time for each event. Write these in boxes in Diagram 1 in the answer book. (4)

(b) State the critical activities. (1)

(d) On the grid in the answer book, draw a cascade (Gantt) chart for this project. (4)

The chief engineer visits the project on day 15 and day 25 to check the progress of the work.

Given that the project is on schedule,

(e) which activities must be happening on each of these two days? (3)

Mark scheme (c)

Scheme	Marks
Float on \(D = 21 - 5 - 14 = 2\)	B1ft
Float on \(F = 42 - 20 - 14 = 8\)	M1 A1ft
	(3)

Notes

B1ft cao ft from diagram

M1 method correct or ft correct answer (top ≤ bottom at both ends; must see appropriate working for M1)

A1ft cao ft from diagram

Cascade chart from 0 to 56 days with the critical activities A, C, G, I, M and H, K at the top and floats shaded — Mark scheme (d)

D1 January 2006 Q5

EdexcelOld spec15 marksCritical Path Analysis

Figure 5: activity network with activities A(4) to N(4) — Figure 5

(a) Calculate the early time and late time for each event, showing them on the diagram in the answer book. (4)

(b) Determine the critical activities and the length of the critical path. (2)

Each activity requires only one worker, and workers may not share an activity.

(d) Use your cascade chart to determine the minimum numbers of workers required to complete the process in the minimum time. Explain your reasoning clearly. (2)

(e) Schedule the activities, using the number of workers you found in part (d), so that the process is completed in the shortest time. (3)

D1 June 2005 Q4

EdexcelOld spec7 marksCritical Path Analysis

4. The precedence table shows the activities involved in a project.

Activity	Immediately preceding activities
\(A\)	–
\(B\)	–
\(C\)	–
\(D\)	\(A\)
\(E\)	\(A\)
\(F\)	\(B\)
\(G\)	\(B\)
\(H\)	\(C, D\)
\(I\)	\(E\)
\(J\)	\(F, H\)
\(K\)	\(G, J\)
\(L\)	\(G\)
\(M\)	\(L\)
\(N\)	\(L\)

(a) Draw the activity network for this project, using activity on arc and using two dummies. (4)

(b) Explain why each of the two dummies is necessary. (3)

Mark scheme (b)

Scheme	Marks
Reference to \(K, J, G\) and \(L\) – \(K\) depends on \(J\) and \(G\), but \(L\) depends on \(G\) only	B2,1,0
Both \(M\) and \(N\) must be uniquely represented in terms of events.	B1
	(3)
	(7 marks)

D1 January 2005 Q2

EdexcelOld spec7 marksCritical Path Analysis

2. The precedence table for activities involved in producing a computer game is shown below.

Activity	Must be preceded by
\(A\)	–
\(B\)	–
\(C\)	\(B\)
\(D\)	\(A, C\)
\(E\)	\(A\)
\(F\)	\(E\)
\(G\)	\(E\)
\(H\)	\(G\)
\(I\)	\(D, F\)
\(J\)	\(G, I\)
\(K\)	\(G, I\)
\(L\)	\(H, K\)

An activity on arc network is to be drawn to model this production process.

(a) Explain why it is necessary to use at least two dummies when drawing the activity network. (2)

(b) Draw the activity network using exactly two dummies. (5)