Student Name: Akhmedov Akbar


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Student Name:

Akhmedov Akbar

Student Number:

B1903046

Course:

Year 1

Module Code and Title:

Financial Techniques and Analysis(ASB-1117)

Module Leader:

Mr Lim Khai Seng

Module Tutor:

Mr Abdul Aziz Buriev

Assessment:

Main Exam

Due Date:

24 May 2021

Date Submitted:

24 May 2021

Weighting within Module:

50%



ALL MARKS ARE PROVISIONAL AND ARE SUBJECT TO CHANGE UNTIL

CONFIRMED BY THE EXTERNAL EXAM BOARD




Comments from Internal Examiner (Name):

Strengths:

Weaknesses:



Areas of Improvement:



Provisional Mark out of 100:



ALL MARKS ARE PROVISIONAL AND ARE SUBJECT TO CHANGE UNTIL

CONFIRMED BY THE EXTERNAL EXAM BOARD


Section A.

Question 1.

𝐹𝑉 = π‘ƒπ‘Ÿπ‘–π‘›π‘π‘–π‘π‘™π‘’ βˆ— (1 + )

𝑛

𝑑

1,002512𝑑 = 1,6



12𝑑 ln 1,0025 = ln 1,6

12𝑑 =

12𝑑 =

12𝑑 = 0,0053



𝑑 = 4,42 = 4 π‘¦π‘’π‘Žπ‘Ÿπ‘  π‘Žπ‘›π‘‘ 5 π‘šπ‘œπ‘›π‘‘β„Žπ‘  Question 2.

  1. 𝐹𝑉 = π‘ƒπ‘Ÿπ‘–π‘›π‘π‘–π‘π‘™π‘’

𝑛 4

Β£48,551.109



𝐴𝐸𝑅 %

  1. 𝐹𝑉 = π‘ƒπ‘Ÿπ‘–π‘›π‘π‘–π‘π‘™π‘’

𝑛 365

𝐴𝐸𝑅 %

Question 3.

i.

22,203.66 = 144,072S

𝑆 = Β£154.12 ii. 154,12 βˆ— 12 βˆ— 10 = Β£18,494.4

18,494.4 βˆ’ 15,000 = Β£3,494.4 Question 4.



  1. Project X.

𝑁𝑃𝑉

Project Y.



𝑁𝑃𝑉

For both projects, just one with the highest NPV should be approved. It is preferable if the company invests capital in Project Y.





  1. Project X.

𝐹𝑉 1 13,000,000

𝐼𝑅𝑅 = () βˆ’ 1 = ( ) βˆ’ 1 = 12.91%

𝑃𝑉 8,00,000

Project Y.

𝐹𝑉 1 32,000,000

𝐼𝑅𝑅 = () βˆ’ 1 = ( ) βˆ’ 1 = 9,82% 𝑃𝑉 22,000,000



  1. For this particular initiative, the conglomerate of companies has a greater return on the internal rate and hence a higher rate of profit.



  1. IRR reviews the amount of return productivity, while NPV represents the cash difference, which combines the current value of less cash flows with less export for a certain duration. It takes the central impression that a future dollar now costs less than a dollar. In sum, I strongly recommend businesses in their operations to use the NPV system.

Question 5.

𝐹𝑉 600 600



𝑃𝑉

𝑒 𝑒 𝑒

Question 6.

  1. Asset A.

𝑛

(π‘₯) = βˆ‘ 𝑃(π‘₯𝑖) π‘₯𝑖 = 0,4 βˆ— 50 + 0,3 βˆ— 80 + 0,3 βˆ— 120 = 80

𝑖=1 Asset B.

𝑛

(π‘₯) = βˆ‘ 𝑃(π‘₯𝑖) π‘₯𝑖 = 0,1 βˆ— 40 + 0,7 βˆ— 80 + 0,2 βˆ— 100 = 80



𝑖=1



  1. Asset A.

𝑆𝐷 (π‘₯𝑖 βˆ’ (π‘₯))2 = √0,4(50 βˆ’ 80)2 + 0,3(80 βˆ’ 80)2 + 0,3(120 βˆ’ 80)2

= 28,98


Asset B.

𝑆𝐷 (π‘₯𝑖 βˆ’ (π‘₯))2 = √0,1(40 βˆ’ 80)2 + 0,7(80 βˆ’ 80)2 + 0,2(100 βˆ’ 80)2

= 15,49




  1. 𝑝 = π‘ƒπ‘Ÿπ‘–π‘›π‘π‘–π‘π‘™π‘’πΈ(π‘₯) βˆ’ 1 βˆ’ π‘Ÿπ‘“ = 8074 βˆ’ 1 βˆ’ 0,04 = 4,11%



  1. A risk-averse investor would opt for asset B because the uncertainty is lower, which implies a near clustering of potential returns.



Section B.

Question 1.

π‘₯2βˆ’1



  1. 𝑦 =

π‘₯

1

𝑦 = π‘₯ βˆ’



π‘₯

𝑦 = π‘₯ βˆ’ π‘₯βˆ’1

𝑓′ = 1 + π‘₯βˆ’2

𝑓′′ = βˆ’2π‘₯βˆ’3





  1. 𝑦 = π‘₯𝑒2π‘₯+1

𝑓′ = π‘₯β€² βˆ— 𝑒2π‘₯+1 + π‘₯ βˆ— (𝑒2π‘₯+1)β€² = 𝑒2π‘₯+1 + 4π‘₯𝑒2π‘₯+1

𝑓′′ = 4𝑒2π‘₯+1 + (4π‘₯)β€² βˆ— 𝑒2π‘₯+1 + 4π‘₯ βˆ— 𝑒2π‘₯+1 = 4𝑒2π‘₯+1 + 4𝑒2π‘₯+1 + 16π‘₯𝑒2π‘₯+1



= 8𝑒2π‘₯+1 + 16π‘₯𝑒2π‘₯+1 = 8𝑒2π‘₯+1(1 + 2π‘₯)



  1. 𝑦

𝑦

𝑦

𝑓′

𝑓′′

Question 2.

𝑃𝑃𝐸

βˆ†π‘ π‘žπ‘œπ‘™π‘‘ 500



βˆ†π‘ž 𝑝𝑛𝑒𝑀 120

π‘ƒπ‘ƒπΈβˆ† = βˆ’ βˆ— βˆ—

βˆ†π‘ π‘žπ‘›π‘’π‘€

π‘π‘œπ‘™π‘‘ = 100

𝑝𝑛𝑒𝑀 = 120

π‘žπ‘œπ‘™π‘‘ = 1000 βˆ’ 5 βˆ— 100 = 500 π‘žπ‘›π‘’π‘€ = 1000 βˆ’ 5 βˆ— 120 = 400

Question 3.

(π‘₯) = (4 βˆ’ 0,0001π‘₯)

(π‘₯) = 4π‘₯ βˆ’ 0,0001π‘₯2

𝑓′(π‘₯) = 4 βˆ’ 0,0002π‘₯

4 βˆ’ 0,0002π‘₯ = 0

βˆ’0,0002π‘₯ = βˆ’4

π‘₯ = 20,000

(20,000) = 80,000 βˆ’ 40,000 = 40,000



Question 4.

𝑦 = βˆ’9π‘₯2 + 126π‘₯ βˆ’ 45

𝑓′ = βˆ’18π‘₯ + 126

βˆ’18π‘₯ + 126 = 0

βˆ’18π‘₯ = βˆ’126

π‘₯ = 7


𝑦 = βˆ’9 βˆ— 49 + 126 βˆ— 7 βˆ’ 45 = 396

Turning point is (7;396)

𝑓′′ = βˆ’18

If 𝑓′′ < 0 , is a max point.



Question 5.

  1. π‘ž = 64,000 βˆ’ 400𝑝

400𝑝 = 64,000 βˆ’ π‘ž

64,000 βˆ’ π‘ž



𝑝

π‘ž

𝑝

π‘ž π‘ž2



  1. 𝑇𝑅

  2. 𝑀𝑅 = βˆ†π‘‡π‘… = 𝑇𝑅′ π‘ž2 β€² 1 1 q

βˆ†π‘ž

  1. 𝑀𝐢 𝑇𝐢′ = (0,0025π‘ž2 + 15π‘ž)β€² = 0,005π‘ž + 15

βˆ†π‘„

  1. πœ‹

  2. πœ‹β€² π‘ž

π‘ž = 14,500



  1. πœ‹β€²β€²

If β€²β€² < 0 , is a max point.

  1. 𝑀𝑅 = 𝑀𝐢

π‘ž = 14,500





  1. 𝑇𝑅′ = 0

160 βˆ’ q = 0

βˆ’ π‘ž = βˆ’160

π‘ž = 32,000



  1. π΅π‘Ÿπ‘’π‘Žπ‘˜ βˆ’ 𝑒𝑣𝑒𝑛 π‘π‘œπ‘–π‘›π‘‘ 𝑖𝑠 𝑇𝑅 = 𝑇𝐢

160π‘ž βˆ’ π‘ž2 = 0,0025π‘ž2 + 15π‘ž

400


160π‘ž βˆ’ π‘ž2 βˆ’ 0,0025π‘ž2 βˆ’ 15π‘ž = 145π‘ž βˆ’ 1 π‘ž2 = 0

400 200


π‘ž (145 βˆ’ π‘ž) = 0

π‘ž1 = 0

145 βˆ’ π‘ž = 0

βˆ’ π‘ž = βˆ’145



π‘ž2 = 29,000
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