Network Analysis PERT and CPM Questions and Answers

Q: Activities A to K are required to complete a project. The time estimates and the immediate predecessors of these activities are given in the table. If the project is to be completed in the minimum possible time, the latest finish time for the activity G is ____________ hours.ActivityTime (hours)Immediate predecessorsA2-B3-C2-D4AE5BF4BG3CH10D, EI5FJ8GK3H, I, J

The latest finish time for G is 10 hours.Forward pass (earliest finish): A 2, B 3, C 2; D 6, E 8, F 7, G 5; H starts after D and E, so finishes at 8 + 10 = 18; I 12; J 13; K starts after H, I and J, at 18, and finishes at 21. The project takes 21 hours along the critical path B-E-H-K.Backward pass: K must finish by 21, so it starts by 18; J must therefore finish by 18 and start by \(18 - 8 = 10\). G is J's only predecessor, so G must finish by 10 hours.5 is G's earliest finish, 8 confuses it with J's duration, and 9 has no basis. G has 5 hours of slack, so it is not critical.

Q: The critical path for the project is

The critical path is a - d - f - h.List every path through the network and its duration:a - b - e - g - h: 3 + 4 + 2 + 6 + 2 = 17 daysa - c - g - h: 3 + 5 + 6 + 2 = 16 daysa - d - f - h: 3 + 4 + 9 + 2 = 18 daysThe longest path, 18 days, is the critical path and fixes the project duration.a - b - e - g - h and a - c - g - h are real paths with slack of 1 and 2 days. a - b - c - f - h is not a path at all: c does not follow b, and f follows d, not c.

Q: Gantt chart is used to solve the :

A Gantt chart is used for job sequencing problems.A Gantt chart draws each job as a bar on a time axis, one row per machine or activity, so it shows when each job starts and finishes and where machines sit idle. In operations research it is the standard way to display and check a sequencing solution, for example Johnson's rule for n jobs on two machines, and to read off the total elapsed time.Inventory problems are solved with EOQ models and stock-level diagrams, and replacement problems with cost-per-period tables, neither of which uses a time chart of job schedules, so 'all of the above' is too broad.

Q: Compute the maximal flow through the following network :

The maximal flow is 70 units.The two arcs leaving the source, S-1 (40) and S-2 (30), form a cut of capacity 70, so no flow can exceed 70. A flow of 70 can be achieved:S to 1: 40, sent on as 30 to node 3, 5 to node 4 and 5 to node 2.S to 2: 30, plus 5 from node 1, so 35 goes on to node 4.Node 4 receives 5 + 35 = 40: 30 goes to N and 10 to node 3 on the undirected arc 3-4 (capacity 25).Node 3 receives 30 + 10 = 40 and sends it to N (capacity 50).The flow into N is 40 + 30 = 70, equal to the cut, so by the max-flow min-cut theorem it is maximal. 75 exceeds the cut; 60 and 65 stop short of a flow that is achievable.

Q: In critical path analysis, CPM is :

CPM is deterministic in nature.The critical path method assumes each activity has a single known duration and finds the longest path through the network, which fixes the project time. It is activity-oriented and deterministic.PERT, by contrast, treats activity times as random (optimistic, most likely and pessimistic estimates) and is event-oriented, so 'event oriented' describes PERT rather than CPM. 'Dynamic in nature' does not describe either.

Which of the following is deterministic model?

A.Only PERT
B.Only CPM
C.Both CPM and PERT but in separate scenarios
D.Neither CPM nor PERT

Exam Year(s)

DSSSB - 2023,

Answer: B

CPM is a deterministic model, as the key states. In the critical path method the durations of activities are known and fixed, so there is no uncertainty, while PERT uses three time estimates (optimistic, most likely and pessimistic) for each activity, and so it is a probabilistic model. The options that name PERT or both, or neither, are wrong.

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Activities A to K are required to complete a project. The time estimates and the immediate predecessors of these activities are given in the table. If the project is to be completed in the minimum possible time, the latest finish time for the activity G is ____________ hours.

Activity	Time (hours)	Immediate predecessors
A	2	-
B	3	-
C	2	-
D	4	A
E	5	B
F	4	B
G	3	C
H	10	D, E
I	5	F
J	8	G
K	3	H, I, J

A.5
B.10
C.8
D.9

Exam Year(s)

GATE ME - 2022,

Answer: B

The latest finish time for G is 10 hours.

Forward pass (earliest finish): A 2, B 3, C 2; D 6, E 8, F 7, G 5; H starts after D and E, so finishes at 8 + 10 = 18; I 12; J 13; K starts after H, I and J, at 18, and finishes at 21. The project takes 21 hours along the critical path B-E-H-K.

Backward pass: K must finish by 21, so it starts by 18; J must therefore finish by 18 and start by \(18 - 8 = 10\). G is J's only predecessor, so G must finish by 10 hours.

5 is G's earliest finish, 8 confuses it with J's duration, and 9 has no basis. G has 5 hours of slack, so it is not critical.

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Direction for [ Question No: 3 To 4 ] :

Statement for Linked Answer Questions:

For a particular project, eight activities are to be carried out. Their relationships with other activities and expected durations are mentioned in the table below.

Activity	Predecessors	Duration (days)
a	-	3
b	a	4
c	a	5
d	a	4
e	b	2
f	d	9
g	c, e	6
h	f, g	2

The critical path for the project is

A.a - b - e - g - h
B.a - c - g - h
C.a - d - f - h
D.a - b - c - f - h

Exam Year(s)

GATE ME - 2012,

Answer: C

The critical path is a - d - f - h.

List every path through the network and its duration:

a - b - e - g - h: 3 + 4 + 2 + 6 + 2 = 17 days
a - c - g - h: 3 + 5 + 6 + 2 = 16 days
a - d - f - h: 3 + 4 + 9 + 2 = 18 days

The longest path, 18 days, is the critical path and fixes the project duration.

a - b - e - g - h and a - c - g - h are real paths with slack of 1 and 2 days. a - b - c - f - h is not a path at all: c does not follow b, and f follows d, not c.

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If the duration of activity f alone is changed from 9 to 10 days, then the

A.critical path remains the same and the total duration to complete the project changes to 19 days.
B.critical path and the total duration to complete the project remain the same.
C.critical path changes but the total duration to complete the project remains the same.
D.critical path changes and the total duration to complete the project changes to 17 days.

Exam Year(s)

GATE ME - 2012,

Answer: A

The critical path remains the same and the total duration changes to 19 days.

Activity f lies on the critical path a - d - f - h, so lengthening it by one day lengthens that path to 3 + 4 + 10 + 2 = 19 days. The other paths (17 and 16 days) are unchanged and still shorter, so the critical path is still a - d - f - h.

Any change to a critical activity changes the project duration, so 'total duration remains the same' is wrong. The critical path would change only if another path became the longest, and 17 days is the length of a non-critical path, not the new project duration.

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Gantt chart is used to solve the :

A.Job sequencing problems
B.Inventory problems
C.Replacement problems
D.All of the above

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: A

A Gantt chart is used for job sequencing problems.

A Gantt chart draws each job as a bar on a time axis, one row per machine or activity, so it shows when each job starts and finishes and where machines sit idle. In operations research it is the standard way to display and check a sequencing solution, for example Johnson's rule for n jobs on two machines, and to read off the total elapsed time.

Inventory problems are solved with EOQ models and stock-level diagrams, and replacement problems with cost-per-period tables, neither of which uses a time chart of job schedules, so 'all of the above' is too broad.

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Compute the maximal flow through the following network :

Network from source S to sink N through nodes 1 to 4: arcs S-1 40, S-2 30, 1-3 30, 1-2 5, 1-4 5, 2-4 50, 3-4 25, 3-N 50 and 4-N 30

A.70 units
B.65 units
C.60 units
D.75 units

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: A

The maximal flow is 70 units.

The two arcs leaving the source, S-1 (40) and S-2 (30), form a cut of capacity 70, so no flow can exceed 70. A flow of 70 can be achieved:

S to 1: 40, sent on as 30 to node 3, 5 to node 4 and 5 to node 2.
S to 2: 30, plus 5 from node 1, so 35 goes on to node 4.
Node 4 receives 5 + 35 = 40: 30 goes to N and 10 to node 3 on the undirected arc 3-4 (capacity 25).
Node 3 receives 30 + 10 = 40 and sends it to N (capacity 50).

The flow into N is 40 + 30 = 70, equal to the cut, so by the max-flow min-cut theorem it is maximal. 75 exceeds the cut; 60 and 65 stop short of a flow that is achievable.

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In critical path analysis, CPM is :

A.event oriented
B.deterministic in nature
C.dynamic in nature
D.none of the above

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: B

CPM is deterministic in nature.

The critical path method assumes each activity has a single known duration and finds the longest path through the network, which fixes the project time. It is activity-oriented and deterministic.

PERT, by contrast, treats activity times as random (optimistic, most likely and pessimistic estimates) and is event-oriented, so 'event oriented' describes PERT rather than CPM. 'Dynamic in nature' does not describe either.

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Network Analysis PERT and CPM Questions and Answers

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