# 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.

πΉπ = πππππππππ β (1 + )

π

π‘

1,002512π‘ = 1,6

12π‘ ln 1,0025 = ln 1,6

12π‘ =

12π‘ =

12π‘ = 0,0053

π‘ = 4,42 = 4 π¦ππππ  πππ 5 ππππ‘βπ  Question 2.

1. πΉπ = πππππππππ

π 4

Β£48,551.109

π΄πΈπ %

1. πΉπ = πππππππππ

π 365

π΄πΈπ %

Question 3.

i.

22,203.66 = 144,072S

π = Β£154.12 ii. 154,12 β 12 β 10 = Β£18,494.4

18,494.4 β 15,000 = Β£3,494.4 Question 4.

1. Project X.

πππ

Project Y.

πππ

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.

πΉπ 1 13,000,000

πΌππ = () β 1 = ( ) β 1 = 12.91%

ππ 8,00,000

Project Y.

πΉπ 1 32,000,000

πΌππ = () β 1 = ( ) β 1 = 9,82% ππ 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.

πΉπ 600 600

ππ

π π π

Question 6.

1. Asset A.

π

(π₯) = β π(π₯π) π₯π = 0,4 β 50 + 0,3 β 80 + 0,3 β 120 = 80

π=1 Asset B.

π

(π₯) = β π(π₯π) π₯π = 0,1 β 40 + 0,7 β 80 + 0,2 β 100 = 80

π=1

1. Asset A.

ππ· (π₯π β (π₯))2 = β0,4(50 β 80)2 + 0,3(80 β 80)2 + 0,3(120 β 80)2

= 28,98

Asset B.

ππ· (π₯π β (π₯))2 = β0,1(40 β 80)2 + 0,7(80 β 80)2 + 0,2(100 β 80)2

= 15,49

1. π = ππππππππππΈ(π₯) β 1 β ππ = 8074 β 1 β 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₯2β1

1. π¦ =

π₯

1

π¦ = π₯ β

π₯

π¦ = π₯ β π₯β1

πβ² = 1 + π₯β2

πβ²β² = β2π₯β3

1. π¦ = π₯π2π₯+1

πβ² = π₯β² β π2π₯+1 + π₯ β (π2π₯+1)β² = π2π₯+1 + 4π₯π2π₯+1

πβ²β² = 4π2π₯+1 + (4π₯)β² β π2π₯+1 + 4π₯ β π2π₯+1 = 4π2π₯+1 + 4π2π₯+1 + 16π₯π2π₯+1

= 8π2π₯+1 + 16π₯π2π₯+1 = 8π2π₯+1(1 + 2π₯)

1. π¦

π¦

π¦

πβ²

πβ²β²

Question 2.

πππΈ

βπ ππππ 500

βπ ππππ€ 120

πππΈβ = β β β

βπ ππππ€

ππππ = 100

ππππ€ = 120

ππππ = 1000 β 5 β 100 = 500 ππππ€ = 1000 β 5 β 120 = 400

Question 3.

(π₯) = (4 β 0,0001π₯)

(π₯) = 4π₯ β 0,0001π₯2

πβ²(π₯) = 4 β 0,0002π₯

4 β 0,0002π₯ = 0

β0,0002π₯ = β4

π₯ = 20,000

(20,000) = 80,000 β 40,000 = 40,000

Question 4.

π¦ = β9π₯2 + 126π₯ β 45

πβ² = β18π₯ + 126

β18π₯ + 126 = 0

β18π₯ = β126

π₯ = 7

π¦ = β9 β 49 + 126 β 7 β 45 = 396

Turning point is (7;396)

πβ²β² = β18

If πβ²β² < 0 , is a max point.

Question 5.

1. π = 64,000 β 400π

400π = 64,000 β π

64,000 β π

π

π

π

π π2

1. ππ

2. ππ = βππ = ππβ² π2 β² 1 1 q

βπ

1. ππΆ ππΆβ² = (0,0025π2 + 15π)β² = 0,005π + 15

βπ

1. π

2. πβ² π

π = 14,500

1. πβ²β²

If β²β² < 0 , is a max point.

1. ππ = ππΆ

π = 14,500

1. ππβ² = 0

160 β q = 0

β π = β160

π = 32,000

1. π΅ππππ β ππ£ππ πππππ‘ ππ  ππ = ππΆ

160π β π2 = 0,0025π2 + 15π

400

160π β π2 β 0,0025π2 β 15π = 145π β 1 π2 = 0

400 200

π (145 β π) = 0

π1 = 0

145 β π = 0

β π = β145

π2 = 29,000