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Test: CPM & PERT Level - 3 - Mechanical Engineering MCQ


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20 Questions MCQ Test - Test: CPM & PERT Level - 3

Test: CPM & PERT Level - 3 for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Test: CPM & PERT Level - 3 questions and answers have been prepared according to the Mechanical Engineering exam syllabus.The Test: CPM & PERT Level - 3 MCQs are made for Mechanical Engineering 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: CPM & PERT Level - 3 below.
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Test: CPM & PERT Level - 3 - Question 1
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For the project given below, the number of critical paths are

Detailed Solution for Test: CPM & PERT Level - 3 - Question 1

Paths:

1‐ 3‐ 5‐ 6 = 6 + 4 + 3 = 13∗

1‐ 2‐ 5‐ 6 = 8 + 2 + 3 = 13∗

1‐ 4‐ 6 = 7 + 5 = 12

*Answer can only contain numeric values
Test: CPM & PERT Level - 3 - Question 2
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While drawing precedence diagram for given table, no. of dummy activities required is ___________.


Detailed Solution for Test: CPM & PERT Level - 3 - Question 2

No. of Dummy activities required is 3

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*Answer can only contain numeric values
Test: CPM & PERT Level - 3 - Question 3
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Find the completion time of the project


Detailed Solution for Test: CPM & PERT Level - 3 - Question 3

Paths:

A‐ C‐ G = 4 + 2 + 7 = 13

A‐D‐E‐G = 4 + 5 + 6 + 7 = 22

A‐ D‐ F‐ H‐ I = 4 + 5 + 6 + 10 + 11 = 36∗

B‐D‐E‐G = 3 + 5 + 6 + 7 = 21

B‐ D‐ F‐ H‐ I = 3 + 5 + 6 + 10 + 11 = 35

Completion time = 36 weeks

*Answer can only contain numeric values
Test: CPM & PERT Level - 3 - Question 4
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A PERT network is shown below. Find the standard deviation of critical path (Upto 2 decimal places)

Detailed Solution for Test: CPM & PERT Level - 3 - Question 4

Paths:

1‐ 2‐ 3‐ 7 = 9 + 5 + 11 = 25∗

1‐ 5‐ 7 = 16 + 8 = 24

1‐ 4‐ 7 = 9 + 4 = 13

standard Deviation of critical path

= 2.45

*Answer can only contain numeric values
Test: CPM & PERT Level - 3 - Question 5
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A probabilistic project consist of the following activities

Expected completion time of the project is ______________ days.

Detailed Solution for Test: CPM & PERT Level - 3 - Question 5

Paths

A‐ C‐ D‐ G = 10 + 12 + 7 + 10 = 39

A‐ C‐ D‐ F = 10 + 12 + 7 + 13 = 42

A‐B‐D‐G = 10 + 16 + 7 + 10 = 43

A‐ B‐ E‐ F = 10 + 16 + 7 + 13 = 46

A‐ B‐ D‐ F = 10 + 16 + 7 + 13 = 46∗

Test: CPM & PERT Level - 3 - Question 6
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For a project following data is given:

The critical path and the independent float for the activity 6-8 is
Detailed Solution for Test: CPM & PERT Level - 3 - Question 6

IF + 5 = 24 − 19 = 5

IF = 0

Test: CPM & PERT Level - 3 - Question 7
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The maximum completion time of the project is
Detailed Solution for Test: CPM & PERT Level - 3 - Question 7

A‐C‐F‐H‐J = 4 +2 + 7+ 2+ 2 = 17

A‐C‐F‐I = 4+ 2+ 7 +2 = 15

A‐D‐E‐F‐H‐J = 4 +1 +3 + 7+ 4+ 2 = 21

A‐D‐E‐F‐I = 4+ 1+ 3+ 7 +2 = 17

A‐ D‐ G‐ H‐ J = 4 + 1 + 5 + 4 + 7 = 21

A‐ D‐ G‐ I = 4 + 1 + 5 + 2 = 12

B‐E‐F‐H‐J = 6 +3 + 7+ 4+ 2 = 22∗ = μ B‐E‐F‐I = 6+ 3+ 7 +2 = 18

B‐G‐H‐J = 6 +5 +4+2 = 17

B‐G‐I = 6+ 5+ 2 = 13

μ = 22

Maximum completion time = μ + 3σ = 22 + 3 × 6.24

= 40.72 days

Test: CPM & PERT Level - 3 - Question 8
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In the network diagram shown below, the numbers adjacent to the arcs indicates the resource required to complete an activity. The resources are specified in terms of number of units. Find the critical path duration and the activity with maximum total float.
Detailed Solution for Test: CPM & PERT Level - 3 - Question 8

A‐ E‐ G = 13

A‐ D = 11

A‐ C‐ F = 14∗

B‐ F = 11

B, D bot are having maximum total float 14, B or 14, D both are correct

Test: CPM & PERT Level - 3 - Question 9
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Consider the following project

Which of the following combination of activities will be having independent float as zero?
Detailed Solution for Test: CPM & PERT Level - 3 - Question 9

The activities which are present in the critical path will have total float, free float and independent float values as zero.

Test: CPM & PERT Level - 3 - Question 10
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A network diagram is given below.

Due to some technical issues if activity ‘B’ is started at time 3 and activity C is started at time 7 the EST for E and F are
Detailed Solution for Test: CPM & PERT Level - 3 - Question 10

EFT )B = EST)B + duration = 3 + 6 = 9

EFT)C = 7 + 3 = 10 (max)

EST)E = (EST)F = 10

Test: CPM & PERT Level - 3 - Question 11
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Consider the following project

The minimum completion time of the project is ___________.
Detailed Solution for Test: CPM & PERT Level - 3 - Question 11

Paths are:

A‐B‐D‐H = 5 + 3 + 6 + 9 = 23

A‐C‐F‐H = 5 + 11 + 7 + 9 = 32∗

A‐C‐E‐G = 5 + 11 + 8 + 7 = 31

μ = 32

σ = 3.19

Minimum completion time = μ − 3σ

= 32 − 3 × 3.19

= 22.43

Test: CPM & PERT Level - 3 - Question 12
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What is the critical path?
Detailed Solution for Test: CPM & PERT Level - 3 - Question 12

Paths are:

A‐B‐C‐F‐H‐I‐J = 4 + 6 + 7 + 8 + 4 + 6 + 2 = 37∗

A‐B‐ C‐G‐H‐I‐J = 4 + 6 + 7 + 5 + 4 + 6 + 2 = 34

A‐B‐D‐I‐J = 4+ 6+ 9+ 6+ 2 = 27

A‐ B‐ E‐ I‐ J = 4 + 6 + 11 + 6 + 2 = 29

*Answer can only contain numeric values
Test: CPM & PERT Level - 3 - Question 13
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Expected completion time of the project is _______

Detailed Solution for Test: CPM & PERT Level - 3 - Question 13

1‐ 2‐ 3‐ 5‐ 7‐ 8‐ 10 = 19

1‐ 2‐ 3‐ 5‐ 7‐ 9‐ 10 = 24

1‐ 2‐ 4‐ 5‐ 7‐ 8‐ 10 = 19

1‐ 2‐ 4‐ 5‐ 7‐ 9‐ 10 = 24

1‐ 2‐ 4‐ 6‐ 7‐ 8‐ 10 = 23

1‐ 2‐ 4‐ 6‐ 7‐ 9‐ 10 = 22

*Answer can only contain numeric values
Test: CPM & PERT Level - 3 - Question 14
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Consider the following construction project.

If the indirect cost is Rs.20, find the optimal completion time of the project.

Detailed Solution for Test: CPM & PERT Level - 3 - Question 14

Paths are:

1‐ 2‐ 5 = 6 + 8 = 14

1‐ 2‐ 4‐ 5 = 6 + 5 + 2 = 13

1‐ 3‐ 4‐ 5 = 5 + 5 + 2 = 12

i) Total cost = Direct + Indirect Direct = 50 + 80 + 80 + 100 + 140 + 60 = 510

= 510 + 20 × 14

= 790 Rs

ii) Crash ′1‐ 2′activity by 2 day TC = 510 + 15 × 2 + 20 × 12

= 780

iii) Crash 2-5, 3-4 by 1 day TC = (510 + 15 × 2 + 100 × 1 + 20 × 1)+ 20 × 11

= 880

Cost is increasing so, we can stop Optimal duration is that when the total cost is minimum Optimal completion time = 12 days

*Answer can only contain numeric values
Test: CPM & PERT Level - 3 - Question 15
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An R & D project has following activities to be performed. The time estimates are given in the table.

The number of days required to complete the project with a probability of 50% is __________

Detailed Solution for Test: CPM & PERT Level - 3 - Question 15

A‐ D‐ H = 6 + 2 + 10 = 18

A‐D‐G‐J = 6 +2 +5 +8 = 21

C‐ H = 9 + 10 = 19

C‐G‐J = 9+ 5+ 8 = 22

B‐ E‐ H = 4 + 7 + 10 = 21

B‐E‐G‐J = 4 + 7 + 5 + 8 = 24∗

B‐F‐I = 4+ 8+ 4 = 16

Expected completion time of the project itself gives the completion time of project with 50% probability.

μ = 24

Test: CPM & PERT Level - 3 - Question 16
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Convert the given AON network to that of AOA network The numbers of dummy activities in the network diagram are
Detailed Solution for Test: CPM & PERT Level - 3 - Question 16

No. of dummies = 1

*Answer can only contain numeric values
Test: CPM & PERT Level - 3 - Question 17
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A project consist of activities with the following data

The overhead cost per day is given as Rs.100. The minimum duration of the project with which minimum cost is associated ___________

Detailed Solution for Test: CPM & PERT Level - 3 - Question 17

Paths:

A‐ D = 8 + 10 = 18

B‐ C‐ D = 5 + 5 + 10 = 19 (critical path)

(i) Total cost = direct cost + Indirect cost

Direct cost = 100 + 200 + 150 + 300 = 750 TC = 750 + 100 × 20

= Rs. 2750

(ii) crash activity ‘B’ as it is having minimum cost slope by ‘1’ days TC

= [750 + (40 × 1) + 100 × 19

= Rs. 2690

(iii) crash activity ‘D’ by 1 day TC = (750 + 40 × 1 + 100 × 1) + 100 × 18

= 2690

No further possibility to crash minimum cost

= 2690 Duration

= 18 days

Test: CPM & PERT Level - 3 - Question 18
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Given that the direct fixed cost is Rs.4000. Find the optimal completion time of the project and the cost associated with it, if the overhead cost is Rs.260/day.
Detailed Solution for Test: CPM & PERT Level - 3 - Question 18

Paths are:

A‐ B‐ E‐ F‐ G‐ I = 3 + 3 + 5 + 6 + 4 + 10 = 31 30

A‐ B‐ E‐ F‐ H = 3 + 3 + 5 + 6 + 10 = 27 26

A‐ C‐ F‐ G‐ I = 3 + 7 + 6 + 4 + 10 = 30

A‐C‐F‐H = 3 + 7 + 6 + 13 = 29

A‐D‐F‐G‐ I = 3 + 9 + 6 + 4 + 10 = 32∗ 31 30

A‐D‐F‐H = 3 + 9 + 6 + 13 = 31 30 29

(i) TC = 4000 + 260 × 32 = 12,320

(ii) Crash activity ‘D’ by 1 day TC = (4000 + 250 × 1) + 260 × 31

= 12,310

(iii) Crash B, D each by 1 day TC = (4000 + 250 × 1 + 30 × 1 + 250× 1) + 260 × 30

= 12,330

Optimal time = 31

Cost = 12310

*Answer can only contain numeric values
Test: CPM & PERT Level - 3 - Question 19
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Consider the data of a project. If the indirect cost per week is Rs.160. find the total cost after crashing two weeks?

Detailed Solution for Test: CPM & PERT Level - 3 - Question 19

Draw a network diagram for given data and find the critical path.

Crash the critical activity 1-3 for 2 weeks.

∴ Total cost = Total normal cost +Indirect cost + slope cost − decrease in Indirect cost

= 5900 + 4000 + [50 × 2] − [160 × 2]

Total cost = 5900 + 4000 + 100 − 380 = Rs. 9680

*Answer can only contain numeric values
Test: CPM & PERT Level - 3 - Question 20
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A project consists of activities for which exact completion times are not known. The project consist of fair parallel paths for which the mean and variance values are given as

What is the probability of completing the project in 29 days?

Detailed Solution for Test: CPM & PERT Level - 3 - Question 20

Longer path is the path which is having maximum value of mean. It is the critical path.

Critical path duration = 32 days = μ variance = 9 = σ2

Standard deviation = σ = 3

z = −1

At z = −1; P = 16%

= 0.16

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