# Student Name: Akhmedov Akbar

 Sana 13.01.2022 Hajmi 283.65 Kb. #330295
Bog'liq
akbar 1117
akbar 1117

 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 = 𝑃𝑟𝑖𝑛𝑐𝑖𝑝𝑙𝑒 ∗ (1 + )

𝑛

𝑡

1,002512𝑡 = 1,6

12𝑡 ln 1,0025 = ln 1,6

12𝑡 =

12𝑡 =

12𝑡 = 0,0053

𝑡 = 4,42 = 4 𝑦𝑒𝑎𝑟𝑠 𝑎𝑛𝑑 5 𝑚𝑜𝑛𝑡ℎ𝑠 Question 2.

1. 𝐹𝑉 = 𝑃𝑟𝑖𝑛𝑐𝑖𝑝𝑙𝑒

𝑛 4

£48,551.109

𝐴𝐸𝑅 %

1. 𝐹𝑉 = 𝑃𝑟𝑖𝑛𝑐𝑖𝑝𝑙𝑒

𝑛 365

𝐴𝐸𝑅 %

Question 3.

i.

22,203.66 = 144,072S

𝑆 = £154.12 ii. 154,12 ∗ 12 ∗ 10 = £18,494.4

18,494.4 − 15,000 = £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 = () − 1 = ( ) − 1 = 12.91%

𝑃𝑉 8,00,000

Project Y.

𝐹𝑉 1 32,000,000

𝐼𝑅𝑅 = () − 1 = ( ) − 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

(𝑥) = ∑ 𝑃(𝑥𝑖) 𝑥𝑖 = 0,4 ∗ 50 + 0,3 ∗ 80 + 0,3 ∗ 120 = 80

𝑖=1 Asset B.

𝑛

(𝑥) = ∑ 𝑃(𝑥𝑖) 𝑥𝑖 = 0,1 ∗ 40 + 0,7 ∗ 80 + 0,2 ∗ 100 = 80

𝑖=1

1. Asset A.

𝑆𝐷 (𝑥𝑖 − (𝑥))2 = √0,4(50 − 80)2 + 0,3(80 − 80)2 + 0,3(120 − 80)2

= 28,98

Asset B.

𝑆𝐷 (𝑥𝑖 − (𝑥))2 = √0,1(40 − 80)2 + 0,7(80 − 80)2 + 0,2(100 − 80)2

= 15,49

1. 𝑝 = 𝑃𝑟𝑖𝑛𝑐𝑖𝑝𝑙𝑒𝐸(𝑥) − 1 − 𝑟𝑓 = 8074 − 1 − 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.

𝑥2−1

1. 𝑦 =

𝑥

1

𝑦 = 𝑥 −

𝑥

𝑦 = 𝑥 − 𝑥−1

𝑓= 1 + 𝑥−2

𝑓′′ = −2𝑥−3

1. 𝑦 = 𝑥𝑒2𝑥+1

𝑓′ = 𝑥′ ∗ 𝑒2𝑥+1 + 𝑥 ∗ (𝑒2𝑥+1)′ = 𝑒2𝑥+1 + 4𝑥𝑒2𝑥+1

𝑓′′ = 4𝑒2𝑥+1 + (4𝑥)′ ∗ 𝑒2𝑥+1 + 4𝑥 ∗ 𝑒2𝑥+1 = 4𝑒2𝑥+1 + 4𝑒2𝑥+1 + 16𝑥𝑒2𝑥+1

= 8𝑒2𝑥+1 + 16𝑥𝑒2𝑥+1 = 8𝑒2𝑥+1(1 + 2𝑥)

1. 𝑦

𝑦

𝑦

𝑓′

𝑓′′

Question 2.

𝑃𝑃𝐸

∆𝑝 𝑞𝑜𝑙𝑑 500

∆𝑞 𝑝𝑛𝑒𝑤 120

𝑃𝑃𝐸= − ∗ ∗

∆𝑝 𝑞𝑛𝑒𝑤

𝑝𝑜𝑙𝑑 = 100

𝑝𝑛𝑒𝑤 = 120

𝑞𝑜𝑙𝑑 = 1000 − 5 ∗ 100 = 500 𝑞𝑛𝑒𝑤 = 1000 − 5 ∗ 120 = 400

Question 3.

(𝑥) = (4 − 0,0001𝑥)

(𝑥) = 4𝑥 − 0,0001𝑥2

𝑓(𝑥) = 4 − 0,0002𝑥

4 − 0,0002𝑥 = 0

−0,0002𝑥 = −4

𝑥 = 20,000

(20,000) = 80,000 − 40,000 = 40,000

Question 4.

𝑦 = −9𝑥2 + 126𝑥 − 45

𝑓= −18𝑥 + 126

−18𝑥 + 126 = 0

−18𝑥 = −126

𝑥 = 7

𝑦 = −9 ∗ 49 + 126 ∗ 7 − 45 = 396

Turning point is (7;396)

𝑓′′ = −18

If 𝑓′′ < 0 , is a max point.

Question 5.

1. 𝑞 = 64,000 − 400𝑝

400𝑝 = 64,000 − 𝑞

64,000 − 𝑞

𝑝

𝑞

𝑝

𝑞 𝑞2

1. 𝑇𝑅

2. 𝑀𝑅 = ∆𝑇𝑅 = 𝑇𝑅′ 𝑞2 ′ 1 1 q

∆𝑞

1. 𝑀𝐶 𝑇𝐶= (0,0025𝑞2 + 15𝑞)= 0,005𝑞 + 15

∆𝑄

1. 𝜋

2. 𝜋 𝑞

𝑞 = 14,500

1. 𝜋′′

If ′′ < 0 , is a max point.

1. 𝑀𝑅 = 𝑀𝐶

𝑞 = 14,500

1. 𝑇𝑅= 0

160 − q = 0

− 𝑞 = −160

𝑞 = 32,000

1. 𝐵𝑟𝑒𝑎𝑘 − 𝑒𝑣𝑒𝑛 𝑝𝑜𝑖𝑛𝑡 𝑖𝑠 𝑇𝑅 = 𝑇𝐶

160𝑞 − 𝑞2 = 0,0025𝑞2 + 15𝑞

400

160𝑞 − 𝑞2 − 0,0025𝑞2 − 15𝑞 = 145𝑞 − 1 𝑞2 = 0

400 200

𝑞 (145 − 𝑞) = 0

𝑞1 = 0

145 − 𝑞 = 0

− 𝑞 = −145

𝑞2 = 29,000