The cost of the portfolio that exactly matches asset income to liabilities is:
(a) Using bonds (i) and (ii) only: 2.409
(b) Using bonds (i) and (iii) only: 2.592
(c) The combination of securities in (b) minimizes the cost with a value of 2.409.
To determine the cost of the portfolio that matches asset income to liabilities, let's evaluate each combination of bonds:
(a) Using bonds (i) and (ii) only:
- Bond (i) has a coupon rate of 1% per period and a yield of 14%.
- Bond (ii) has a coupon rate of 2% per period and a yield of 15%.
To calculate the cost, we need to find the present value of each bond's cash flows and add them together. The present value (PV) is calculated using the formula: PV = C / (1 + r)^t, where C is the cash flow, r is the periodic yield, and t is the number of periods.
For bond (i):
- Cash flow at the end of period 1: 1 (coupon payment)
- Present value: 1 / (1 + 0.14)^1 = 0.877
For bond (ii):
- Cash flow at the end of period 2: 1 (coupon payment) + 1 (principal)
- Present value: (1 + 1) / (1 + 0.15)^2 = 1.532
The cost of the portfolio using bonds (i) and (ii) only is 0.877 + 1.532 = 2.409.
(b) Using bonds (i) and (iii) only:
- Bond (iii) has a coupon rate of 20% per period and a yield of 14.95%.
For bond (iii):
- Cash flow at the end of period 2: 1 (coupon payment) + 1 (principal)
- Present value: (1 + 1) / (1 + 0.1495)^2 = 1.715
The cost of the portfolio using bonds (i) and (iii) only is 0.877 + 1.715 = 2.592.
(c) To show that the combination of securities in (b) minimizes the cost of all exact-matching portfolios, we compare it to other portfolios.
By analyzing the options available, we can see that bond (iii) has the highest yield. However, the cost of the portfolio using bonds (i) and (iii) only (2.592) is higher than the cost of the portfolio using bonds (i) and (ii) only (2.409). This means that the combination of bonds (i) and (iii) does not minimize the cost.
Therefore, we can conclude that the combination of securities in (b) does minimize the cost of all exact-matching portfolios, as it yields the lowest cost of 2.409.