Question

The Rover 6 is a new​ custom-designed sports car. An analysis of the task of building the Rover 6 reveals the following list.

This exercise contains only parts​ b, c, and d.

​b) The activities on the critical path are (1)

The total project completion time for custom designed Rover 6 car is

​(Round your response to one decimal​ place.)​

c) For Rover 6 to be completed 2 days​ earlier, one should (2)

d) For reducing the project​ duration, resources from (3) activities should be borrowed to speed activities on the speed activities on the (4) path.

Job Letter	Description	Normal Time​ (days)	Immediate​ Predecessor(s)
A	Start	0.00.0	minus−
B	Design	8.08.0	A
C	Order special accessories	0.10.1	B
D	Build frame	1.01.0	B
E	Build doors	1.01.0	B
F	Attach​ axles, wheels, gas tank	1.01.0	D
G	Build body shell	2.02.0	B
H	Build transmission and drivetrain	3.03.0	B
I	Fit doors to body shell	1.01.0	​G, E
J	Build engine	4.04.0	B
K	​Bench-test engine	2.02.0	J
L	Assemble chassis	1.01.0	​F, H, K
M	​Road-test chassis	0.50.5	L
N	Paint body	2.02.0	I
O	Install wiring	1.01.0	N
P	Install interior	1.51.5	N
Q	Accept delivery of special accessories	5.05.0	C
R	Mount body and accessories on chassis	1.01.0	​M, O,​ P, Q
S	Road test car	0.50.5	R
T	Attach exterior trim	1.01.0	S
U	Finish	0.00.0	T

(1)

A - B - D - F - L - M - R - S - T - U

A - B - C - Q - R - S - T - U

A - B - J - K - L - M - R - S - T - U

(2)

buy preassembled transmissions and drivetrains

speed delivery of special accessories by 3 days

speed the building of body shell

none of the above

(3)

non critical

critical

(4)

non critical

critical

Answer 1

The critical path of a project is composed off activities with zero total slack. The critical path(s) is/ are the longest path(s) in a project network. The slacks for each activity can be computed by carrying out forward pass and backward pass in the project network.

Forward Pass:

• ES of the starting activities = 0
• ES of all other activities = Max. (EF of their immediate predecessors)
• EF of an activity = Its ES + Its duration

Backward Pass:

• LF of ending activities = Max. (All the EFs)
• LF of all other activities = Min. (LS of their immediate successors)
• LS of an activity = Its LF - Its duration

Total slack (or, float) = LF - EF for each activity.

Job Letter	Normal Time​ (days)	Immediate​ Predecessor(s)	ES	EF	LS	LF	Slack
A	0	-	0	0	0	0	0
B	8	A	0	8	0	8	0
C	0.1	B	8	8.1	10.4	10.5	2.4
D	1	B	8	9	12	13	4
E	1	B	8	9	10	11	2
F	1	D	9	10	13	14	4
G	2	B	8	10	9	11	1
H	3	B	8	11	11	14	3
I	1	​G, E	10	11	11	12	1
J	4	B	8	12	8	12	0
K	2	J	12	14	12	14	0
L	1	​F, H, K	14	15	14	15	0
M	0.5	L	15	15.5	15	15.5	0
N	2	I	11	13	12	14	1
O	1	N	13	14	14.5	15.5	1.5
P	1.5	N	13	14.5	14	15.5	1
Q	5	C	8.1	13.1	10.5	15.5	2.4
R	1	​M, O,​ P, Q	15.5	16.5	15.5	16.5	0
S	0.5	R	16.5	17	16.5	17	0
T	1	S	17	18	17	18	0
U	0	T	18	18	18	18	0

(b)

The critical path is composed of activities with zero slack. So, the critical path is:

A - B - J - K - L - M - R - S - T - U

(c)

The given three alternatives are activity H, Q, and G and none of them are on the critical path. Shortening them will not result in the early completion of the project. So, the correct answer should be none of the above.

(d)

For reducing the project​ duration, resources from the non-critical activities should be borrowed to speed activities on the speed up activities on the critical path.