Question

Today is 1 January 2018. Mary is 40 years old today and she is planning to set up a university education fund for her 12 year old daughter Emily. The fund should be enough to cover Emily’s university tuition fee costs .

a. Assume that Emily will attend university at age 18 to complete a three-year bachelor degree. Mary estimates that current average university tuition fees are $25,000 per person per year and are growing at a rate of 3% p.a. (e.g., the estimated tuition fee for 2019 is $25, 000 × (1 + 3%)). Find Emily’s first year, second and third year tuition fee amount. Calculate the amount that Mary needs to deposit into the fund today. Assume the fund yield rate is j1 = 3.6% p.a. and the tuition fee is paid on an annual basis at the beginning of each study year. Round your answer to three decimal places.

b. Mary has a portfolio which consists of three securities (henceforth referred to as security A, security B and security C).

• Security A is a 180-day $50,000 bank bill which is purchased on its issue date 15 November 2017. The purchase yield rate is 4.1% p.a. (simple interest rate).

• Security B is a Treasury bond which matures on 1 January 2025. This Treasury bond has a coupon rate of j2 = 3.25% p.a. and a face value of $100. Mary purchased this Treasury bond on 23 September 2017. The purchase yield rate was j2 = 3.7% p.a.

• Security C is a Treasury bond which matures on 1 January 2023. This Treasury bond has a coupon rate of j2 = 4.75% p.a. and a face value of $100. Mary purchased this treasury bond on 28 December 2016. The purchase yield rate was j2 = 4.1% p.a. Calculate the purchase price of security A, security B and security C. Round your answer to three decimal places. The price of both security B and security C should be calculated using the RBA method.

c. Mary decides to sell all three securities today to pay the required deposit for the education fund. Security A can be sold at a yield rate of 3.9% p.a. (simple interest rate). Security B and security C can be sold at a yield rate of j2 = 3.75% p.a. Calculate the sale price of security A, security B and 1 ACST201 Financial Modelling Take Home Quiz 1 S1 2018 security C. Round your answer to three decimal places. Note that the sale of security B and security C occurs after the coupon payment.

d. Based on Mary’s perspective, draw three carefully labelled cash flow diagrams to represent the above financial transactions of security A, security B and security C, respectively.

Answer 1

1.a) Since Emily is currently 12 years old, she will attend university after 6 years. We need to accordingly calculate the tuition fees.

Future value = Present value x (1+r)ⁿ
Present value = Future value / (1+r)ⁿ
Where, “r” is interest rate and “n” is number of years.

Tuition fees in first year = $25,000 x (1.03)⁶ = $29,851.307
Tuition fees in second year = $25,000 x (1.03)⁷ = $30,746.847
Tuition fees in third year = $25,000 x (1.03)⁸ = $31,669.252

Since the fund yield rate is 3.6%, we would use this rate to determine the fund to be deposited today.

Present value of all tuition fees at Year 6 = [($29,851.307)/((1+(0.036*0))] + [($30,746.847)/(1+(0.036*1))] + [($31,669.252)/(1+(0.036*2)]
=> $29,851.307 + [($30,746.847)/(1.036)] + [($31,669.252)/(1.072)]
=> $29,851.307 + $29,678.423 + $29,542.213 = $89,071.943

Amount required to be deposited today = $89,071.943 / ((1+(0.036*6)) = $73,249.953

1.b) Purchase price of Security A would be equal to present value of $50,000 discounted at the yield rate.

Purchase price of Security A = $50,000 / [1+(0.041/(180/365)] = $49,009.077

Formula to be used for calculating price of Security B and C:

P =      the price per $100 face value.

v =       1/(1+i)

i =        the annual percentage yield to maturity divided by 200

f =        the number of days from the date of settlement to the next interest payment date

d =       the number of days in the half year ending on the next interest payment date

g =       the half-yearly rate of coupon payment per $100 face value

n =       the term in half years from the next interest-payment date to maturity

               (1-vⁿ) / i

Price of Security B:

i =        3.7/200 = 0.0185

v =       1/(1+i) = 0.981836033

f =        100

d =       181

g =       3.25/2 = 1.625

n =       14

(1-vⁿ) / i = 12.23503851

• Price = 0.981836033^(100/181) x [(1.625*(1+12.23503851) + (100*0.773651787)]
  => 0.989923511 x (21.50693759 + 77.36517875)
  = $97.876

So, the purchase price of Security b would be $97.876

Price of Security C:

i =        4.1/200 = 0.0205

v =       1/(1+i) = 0.979911808

f =        4

d =       181

g =       4.75/2 = 2.375

n =       10

(1-vⁿ) / i = 8.959131452

• Price = 0.979911808 ^(4/181) x [(2.375*(1+8.959131452) + (100*0.816337805)]
  => 0.999551643 x (23.6529372 + 81.63378052)
  = $105.239

So, the purchase price of Security C would be $105.239

1.c) Selling price of Security A:

Days to maturity = 180 days – 47 days = 133 days

Selling price of Security A = $50,000 / [1+(0.039/(133/365)] = $49,299.408

Selling price of Security B:

i =        3.75/200 = 0.01875

v =       1/(1+i) = 0.981595092

f =        1

d =       181

g =       3.25/2 = 1.625

n =       14

(1-vⁿ) / i = 12.21343583

• Price = 0.981595092 ^(1/181) x [(1.625*(1+12.21343583) + (100*0.770998078)]
  => 0.999897373 x (21.47183322 + 77.09980782)
  = $98.562

So, the selling price of Security B would be $98.562

Selling Price of Security C:

i =        3.75/200 = 0.01875

v =       1/(1+i) = 0.981595092

f =        1

d =       181

g =       4.75/2 = 2.375

n =       10

(1-vⁿ) / i = 9.041616934

• Price = 0.981595092 ^(1/181) x [(2.375*(1+9.041616934) + (100*0.830469682)]
  => 0.999897373 x (23.84884022 + 83.04696825)
  = $106.885

So, the selling price of Security C would be $106.885

d) Cash Flow Diagram