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QM Assignment One (1)

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ASSIGNMENT 1
Quantitative Methods for Business
1. A football club is considering buying a player on 1 January for K5 million. The
player's wages will be K10,000 per month higher than those of the man he will
replace. The manager expects the purchase to generate a level increase in
attendances, which will yield an extra income in the first year of K50,000 from each
home match. The manager also expects the new player to increase the club's
chance of reaching the Cup Final in any one year from 5% to 20%. The extra amount
generated for club funds by an appearance in a Cup Final on 30 April is K1 million.
The club plays a home match on the second of each month throughout the year,
but all Cup matches are played away from home. Wages are paid at the end of each
month. Wages, ticket prices and the reward for reaching a Cup Final rise at 3% pa,
the increments taking place on 1 January.
If the player is purchased, the cost will be borrowed from a bank, which will charge
interest at 1% per month and will accept repayment at any time. The owner of the
club insists that any purchase should show a profit if the managers expectation are
borne out in practice.
(a) If the manager expects that he will keep the player for 3 1/2 years until he
retires, calculate the net present value of the cashow, in order to assess whether
or not the purchase should go ahead. (12)
(b) The purchase goes ahead. Attendances rise as expected, but the club does not
reach the Cup Final and 12 months after being bought the player is sold again. The
club owner calculates that he has made a pro_t of K379,490. Calculate the sale
price. (5)
2. A woman who has won a prize is o_ered a lump sum of K500,000 to invest now,
or K300,000 to invest at the end of this year and another K300,000 to invest at the
end of the following year. If all investments are assumed to earn 20% pa, which
should she choose if she intends to withdraw the money after
(a) 5 years,
(b) 3 years. (3)
3. A 90-day government bill was bought by an investor for a price of K94 per K100
nominal. After 30 days the investor sold the bill to a second investor for K96.50 per
K100 nominal. The second investor held the bill to maturity when it was redeemed
at par. Determine which investor obtained the higher annual e_ective rate of
return. (3)
4. Mr Banda is struggling to repay his loan of K300,000 with payments of K6,200
made monthly in arrears for 5 years.
(a) Calculate the flat rate of interest per annum. (2)
(b) Hence, or otherwise, calculate the APR of Mr Banda's loan. (3)
After exactly one year, a loan company offers to `help' Mr Banda by restructuring
his loan with new monthly payments of K3,000 made in arrears.
(c) Assuming the company charges the same APR as Mr Banda's original loan,
calculate the term of the new loan. (4)
(d) Calculate how much more interest in total Mr Banda will pay on his restructured
loan than on his original loan. (2)
5. A businessman wishes to borrow an amount of K1 million for a term of 5 years.
The agreed rate of interest is 6% per annum effective for the first 3 years, and 8%
per annum effective thereafter. Repayments on the loan are made annually in
arrears.
(a) Find the amount of the level annual repayment. (2)
(b) Draw up the loan schedule for the full five-year period. (3)
(c) Calculate what percentage of the loan has been repaid by the end of year 3. (1)
(d) Without doing any further calculations, explain how this percentage figure
would alter if the rate of interest had instead been 8% for the first three years and
6% thereafter. (2)
End of Assignment
SOLUTION:
(a) The expected changes in the cashflows in the first year the player is bought
are as follows:
(i)
extra wages of K10,000 per month, paid at the end of each month;
(ii)
extra ticket sales of K50,000 per month, paid at the start of each
month; and
(iii) extra winnings of (20%-5%) x K1,000,000 = K150,000
The increase in the net present value of these cashflows in Year 1 (working
in months) is as follows:
NPV₁ = -10,000a₁₂┐ + 50,000ä₁₂┐ + 150,000v⁴
1−(1+0.01)¯¹²
= -10,000(
0.01
1−1.01¯¹²
= -10,000(
0.01
1−(1+0.01)¯¹²
) + 50,000 ( 1−(1+0.01)¯¹ ) + 150,000(1 + 0.01)¯⁴
1−1.01¯¹²
) + 50,000 ( 1−1.01¯¹ ) + 150,000(1.01)¯⁴
= K599,977.69
The net present values of the cashflows in Years 2-9 (as at the beginning of
each year) will be the same, but increased by 3% (compound) each year. In
Year 4, it will only apply for the first 6 months so that NPV₄ = -10,000a₆┐ + 50,000ä₆┐ + 150,000v ⁴
1−1.01¯⁶
= -10,000(
0.01
1−1.01¯⁶
) + 50,000 (1−1.01¯¹) + 150,000(1.01)¯⁴
= K378,863.85
NPV = 599,977.69(1 + 1.03v12 + 1.032 v²⁴ ) + 378,863.85(1.03³v³⁶)
= 599,977.69(
= 599,977.69(
1−(1.03v12 )³
1−1.03v12
) + 378,863.85(1.03³)(1.01)¯³⁶
1−(1.03(1.01)¯¹²)³
1−1.03(1.01)¯¹²
= K1,939,050.89
) + 289,351.27
Since this is below the purchase price of K5,000,000, the purchase should not
go ahead.
SOLUTION:
(b) If the player is sold for amount ‘X’, the equation value (ignoring any change
in the team’s chances of winning in the Cup Final, which do not affect the
actual profit) becomes –
379,490 = -5,000,000 – 10,000a₁₂┐ + 50,000ä₁₂┐ + xv¹²
379,490 = -5,000,000 – 10,000(
1−1.01¯¹²
0.01
1−1.01¯¹²
) + 50,000 ( 1−1.01¯¹ ) + xv¹²
379,490 = - 4,544,169.36 + xv¹²
xv¹² = 379,490 + 4,544,169.36
xv¹² = 4,923,659.36
X(1.01)¯¹²
(1.01)¯¹²
=
4,923,659.36
(1.01)¯¹²
X = 5,548,102.61
The Sale Price is, therefore, K5,548,102.61
QUESTION 2
SOLUTION:
She should choose the option that maximises the present value of the payments
(the time at which she intends to withdraw the money is irrelevant). Now the
present value of the two-payment option is as follows:
PV = 300,000v + 300,000v²
= 300,000(1.20)¯¹ + 300,000(1.20)¯²
= K458,333.33
Since this is less than K500,000, she should choose the lumpsum in either case.
QUESTION 3
SOLUTION:
For the first investor, 94(1+i)30/365 = 96.50, so that
i = (96.50/94) 365/30 – 1 = 37.62%
For the second investor, 96.50(1+i)60/365 = 100, so that
i = (100/96.50) 365/60 – 1 = 24.20%
Thus, the first investor obtained the higher annual effective rate of return, i.e.,
37.62%
QUESTION 4
SOLUTION:
(a) The flat rate of interest is given by:
F=
SOLUTION:
total interest paid
loan amount x term
=
5x12x6,200 – 300,000
3000, 000 x 5
=
4.80%
(b) The equation of the value is as follows:
300,000 = 12(6,200)a₅┐⁽¹²⁾
300,000 = 74,400 a₅┐⁽¹²⁾
74,400
74,400
a₅┐⁽¹²⁾ = 4.03
APR = 2 x F
Therefore APR = 9.60
= 6,200
1−(1+𝑖 )¯5
1
12
((1+𝑖)
= 6,200
− 1)
1−(1+0.09175 )¯5
1
12
((1+0.09175)
−
1)
= 300,003.20
Where the RHS is equal to 300,003.20 if i = 9.175% and 299,996.72 if i =
9.176%
Remark: Since APR ≈ 2 x F =9.60, we had to try a first guess of 9%
SOLUTION:
(c) L1 = 12(6,200) a₅┐⁽¹²⁾ @9.2% = 6,200
1−(1+0.092 )¯4
1
12
((1+0.092)
=6,200(
− 1)
1−1.092¯4
1.092
1
12
)
−1
= 249,939.23
Therefore, the loan outstanding after 1 year is K249,939.23
Thus, if n is the term of the new loan in years, then
249,939.23 = 12(3,000) an┐⁽¹²⁾ @9.2% = 3,000(
1−1.092¯n
1.092
249,939.23 = 3,000(
1
12
)
−1
1−1.092¯n
1.092
1
12
)
−1
Thus, you divide both sides by 3,000 which results in the following:
249,939.23 =
3,000
1 – (1.092)¯n
(1.092)1/12 – 1
83.313 =
1 – (1.092)¯n
(1.092)1/12 – 1
83.313(1.092)1/12 – 1 = 1 – (1.092)¯n
0.6133 = 1 – (1.092)¯n
(1.092)¯n = 1 – 0.6133
(1.092)¯n = 0.3867
- n In (1.092) =
- n (1.092)
In (0.3867)
In (1.092)
n = In (0.3867)
In (1.092)
n = 10.80
Therefore, the term of the new loan is 10.8 Years
SOLUTION:
(d) The original loan had four years remaining, therefore –
Total interest = 4x12x6,200 – 249,939.23
= 47,660.77
The new loan has a term of 10.8 years, thus –
Total interest = 10.8x12x3000 – 249,939.23
= 138,860.77
Therefore, Mr. Banda will pay 138,860.77 – 47,660.77 = K91,200.00 more
interest
QUESTION 5
SOLUTION:
(a) If the annual repayment is x, then
1,000,000 = Xa₃┐@6% + xv -3 a₂┐@8%
= x(
1−1.06¯³
0.06
1−1.08¯²
) + x(1.06)¯³ (
0.08
)
= 2.6730x + 1.4973x
1, 000, 000 = 4.1703x
x = 4.1703x
=
1,000,000
4.1703
x =
1,000,000
4.1703
x =
239,790.90
Therefore, the annual repayment is K239,790.90.
SOLUTION:
(b) We, therefore, have the following loan schedule:
Year
1
2
3
4
5
Loan Outstanding
before payment
1,000,000.00
820,209.10
629,630.75
427,617.69
222,036.20
Annual
payment
239,790.90
239,790.90
239,790.90
239,790.90
239,790.90
Interest
element
60,000
49,212.55
37,777.84
34,209.41
17,762.90
Capital element
179,790.90
190,578.35
202,013.05
205,581.48
222,028.00
Loan Outstanding
before payment
820,209.10
629,630.75
427,617.69
222,036.20
8.2
SOLUTION:
(c) Thus, the percentage of the loan that has been repaid by the end of Year 3 is
as follows:
=
1,000,000 – 427,617.69 x 100 = 57.24%
1,000,000
Therefore, the percentage of the loan repaid by the end of Year 3 is 57.24%
SOLUTION:
(d) If the interest rate would have been the other round, a lesser proportion of
the loan would have been repaid by the end of Year 3.
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