Answer to Question #127822 in Financial Math for Nur farah Hanim Bt Shukri

Question

Liabilities of RM100 each are due at the ends of periods 1 and 2. There are three securities available to produce asset income to cover these liabilities, as follows:

i) A bond due at the end of a period 1 with coupons at rate 0.01 per period, valued at a periodic yield of 7%.
ii) A bond due at the end of the period 2 with coupon rate 0.02 per period, valued at a periodic yield of 7.5%.
iii) A bond due at the end of period 2 with coupon rate 0.2 per period, valued at a periodic yield of 7.75%.
Determine the cost of the portfolio that exactly-matches asset income to liabilities due using:

a) bonds i) and ii) only.

b) bonds i) and iii) only.
(c) Which combination minimizes the cost of exact-matching portfolios made up of a combination of the three securities?

Answer 1

Equation to match assets and liabilities

100 = (1.01)x +(0.02)y+(0.2)z

100 = (1.02)y + (1.2)z

Solving the equations:

Solving for z

100 = (1.02)y + (1.2)z

100 = (1.01)x +(0.02)y+(0.2)z

Solving for y

z = $\frac{(100 – 1.02y)}{1.2}$

z = 83.33333 – 0.85y

100 = (1.01)x +(0.02)y+(0.2)( 83.33333 – 0.85y)

100 = 1.01x +0.02y+16.66667– 0.17y

Solving for x

83.33333 = 1.01x - 0.15y

1.01x = 83.33333 + 0.15y

x = 82.50825 + 0.148515y

Solving for y

x = 82.50825 + 0.148515y

0.148515y = x-82.50825

y = 6.733333x - 555.5556

z = 83.33333 – 0.85y

0.85y = 83.33333-z

y = 98.03922 - 1.176471z

Cost of combination

The total price of the bonds will be given by:            

a) Bonds i) and ii) only.

C = $[82.50825 + 0.148515y]*1.01*1.07^{-1} + [98.03922 - 1.176471z]*0.02*1.075^{-2}$

C = $[82.50825 + 0.148515y]*0.943925 + [98.03922 - 1.176471z]*0.017226$

C = 79.57049 + 0.140187y - 0.020266z

Since z = 83.33333 – 0.85y

C = $79.57049 + 0.140187y - 0.020266*(83.33333 – 0.85y)$

C = 79.57049 + 0.140187y - 1.688833 + 0.017226y

C = 77.88165 + 0.157413y

 

b) Bonds i) and iii) only.

C = $[82.50825 + 0.148515y]*1.01*1.07^{-1} + [83.33333 – 0.85y]*0.2*1.0775^{-2}$

C = $[82.50825 + 0.148515y]*0.943925 + [83.33333 – 0.85y]*0.172264$

C = 92.23698-0.00624y

(c) Combination that minimizes the cost of exact-matching portfolios

∆C = [Bonds i) and ii) only] – [Bonds i) and iii) only]

∆C = 77.88165 + 0.157413y – (92.23698-0.00624y)

∆C = -14.3553 + 0.163653y

∆C = [Bonds i) and iii) only] - [Bonds i) and ii) only]

∆C = [92.23698-0.00624y] – [77.88165 + 0.157413y]

∆C = 14.35533 -0.16365y

The total cost of Bonds i) and ii) is less more than that of the combination of Bonds i) and iii) since the cost difference (∆C = 14.35533 -0.16365y) obtained by [Bonds i) and iii) only] - [Bonds i) and ii) only] Is positive implying that combination of Bonds i) and iii) only is higher. Thus the combination that minimizes the cost of exact matching portfolios is that of Bonds i) and ii) only.