Inclass Quiz 2 Set A Solution PDF

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1

A UNSW Sydney CVEN3101 – Engineering Operations and Control In-class Quiz 1 (17th October 2019) Instructions: Time allowed – 40 minutes Total number of questions - 2 Total number of pages in this exam booklet - 6 pages Total marks available – 12 marks Marks available for each question are shown in the examination paper All answers must be written in ink and must be written in the space provided for each question. 7. No part of this paper is to be retained by the candidates. 8. Candidates may bring to the examination - UNSW approved calculator 9. A formula sheet is provided at the end of this paper. 10. Write your name and student id clearly in the space provided below. 1. 2. 3. 4. 5. 6.

Student ID: ____________________________________________________________________

Name: ________________________________________________________________________

2

QUESTION 1

[8 MARKS]

+5

5 5

C 20 15 20

-2

0 B 10 0 10 10 +1

10 D 11 3

24 J 24 8

22 22

13 14

-1

12 G 13 9

21 22

19 H 19 3

22 22

32 32

1@350

2@150

2@200 0 A 0 0 0 0

18 F 18 4

22 I 22 2

24 24

32 K 32 0

1@100 +2 11 E 11 8

19 19

1@80

Activity A B C D E F G H I J K

Normal Duration (days) 0 10 15 3 8 4 9 3 2 8 0

Crash Duration (days) 0 10 13 2 8 2 9 2 2 7 0

ES Activity EF LS Duration LF Normal Cost ($) 1000 2000 500 1000 500 200 150 300 500 -

Cost Slope ($/day) 200 100 150 80 350 -

For the network diagram shown above and using the information given in the table above, calculate the following: (a) Calculate the total project duration. (1.5 marks) [32 days] (b) Identify the critical path(s). (2 marks) [ ABCFIJK] and [ABEHIJK] (c) What is the minimum possible duration of the project? (3 marks) [30 days] (d) What is the Project direct cost corresponding to minimum project duration? (1.5 marks) [$6730] Show all steps.

32 32

3

You may like to use the space in the network given below for calculation. Indicate the purpose you are using the below network space for. Initial critical paths: [ABCFIJK] and [ABEHIJK] Project cost = $6150 Crash H 1@80 and F 1@150 Activity H on [ABEHIJK] has the least slope of $80, so it can be crashed. However, activities on both critical paths need to be crashed by the same number of days. On [ABCFIJK], activity F has the least slope, so F can be crashed by 1 day, matching activity H.

+5

5

C 20 15

-2

0 B 10 10

10 D 3

13

23 J 23 8

21

1@150

2@200 0 A 0 0

18 F 3

-1

12 G 9

21

19 H 2

21

31 31

1@350 21 I 2

23 23

31 K 31 0

31 31

1@100

+1

+2 11 E 8

19

Crashing H and F by one day makes all paths critical. New project cost = $6150 + $80 + $150 = $6380. Now we look for activities common to all critical paths with crashability. Activity J is such an activity which can be further crashed without imposing floats to any other activity.

+5

5

C 20 15

-2

0 B 10 0 10 10 +1

10 D 10 3

13 13

23 J 23 7

21

-1

12 G 12 9

21 21

19 H 19 2

21 21

21 I 21 2

23 23

1@100 11 E 11 8

19 19

30 30

1@150

2@200 0 A 0 0 0 0

18 F 3

+2

30 K 30 0

30 30

4

Calculation space (c) The minimum possible duration is 30 days. (d) New project cost = $6380 + $350 = $6730.

5

QUESTION 2

[4 MARKS]

The NPV of a project is given by the equation: 𝑁𝑃𝑉 = 5000 − 2𝑥 where, x is a random variable following uniform distribution with the range between $2000 and $2400, and between $2600 and $3000. Using Lehmer generator (𝑚 = 231 − 1 and 𝑎 = 75 with the seed value of 1), two random values are generated as shown: 282475249 and 1458777923. Answer the following questions: (a) Calculate the NPV values of the project using these 2 random values. Show complete steps. (3 marks) (b) If you are a Project Manager for this project, can you take any decision based on NPV values calculated in part (a) using two random values? Give reasoning. (1 mark) Calculation space Solution:

ℎ

2000

2400 2600 3000

ℎ(2400 − 2000) + ℎ(3000 − 2600) = 1 ℎ=

1 400+400

=

1 800

[0.5 marks]

𝑓(𝑥) = ℎ =

1 , 800

2000 ≤ 𝑥 ≤ 2400

𝑓(𝑥) = ℎ =

1 , 800

2600 ≤ 𝑥 ≤ 3000 [0.5 marks]

𝐹(𝑥) =

𝑥−2000 800

, 2000 ≤ 𝑥 ≤ 2400 [0.5 marks]

𝐹(𝑥) =

𝑥−2200 800

, 2600 ≤ 𝑥 ≤ 3000 [0.5 marks]

6

𝐹(𝑥) = 𝑢 → 𝑥 = 800𝑢 + 2000, 0 ≤ 𝑢 ≤ 0.5 𝐹(𝑥) = 𝑢 → 𝑥 = 800𝑢 + 2200, 0.5 < 𝑢 ≤ 1 [0.5 marks] Lehmer

u

x

NPV [0.5 marks]

282475249

0.132

2105.23

789.54

1458777923

0.679

2743.44

-486.87

Calculation space for Question 2 (b) Two discussion points: (i)

NPV has got one positive and one negative values. Thus using these 2 values, any decision cannot be taken since the return is greater than MARR in one case and less than MARR in another case. [0.5 marks]

(ii)

Also, any certain decision cannot be taken until a greater number of random values are generated. The expected NPV can then be calculated to take further decision. [0.5 marks]...