Sites: Global Q&A | Wits | MathsGee Club | Joburg Libraries | StartUps | Zimbabwe | OER

MathsGee is Zero-Rated (You do not need data to access) on: Telkom |Dimension Data | Rain | MWEB

0 like 0 dislike
8 views
On 25 October 2008 a certain government issued a five-year index-linked bond. The bond paid coupons at a nominal rate of $3 \%$ per annum payable half-yearly in arrear and the bond was redeemed at par. The coupons and redemption payments were index-linked to a retail price index at the month of payment.

An investor, who is not subject to tax, bought R10,000 nominal of the bond on 26 October 2012. The investor held the bond until redemption.

\begin{array}{|c|c|c|}
\hline {\text { Retailprice index }} \\
\hline & \text { April } & \text { October } \\
\hline 2008 & \ldots & 149.2 \\
\hline & \ldots & \ldots \\
\hline 2012 & \ldots & 169.4 \\
\hline 2013 & 171.4 & 173.8 \\
\hline
\end{array}

i. Calculate all the payments received by the investor on 25 April 2013 and 25 October $2013 .$

ii. Calculate the purchase price that the investor paid on 26 October 2012 if the investor achieved an effective real yield of $3.5 \%$ per annum on this investment.
| 8 views

0 like 0 dislike
(i)
25 April 2013:
25 October 2013 :
$10,000 * 0.015 * \frac{171.4}{149.2}=172.3190349$
\begin{aligned} &10,000 * 0.015 * \frac{173.8}{149.2}=174.7319035 \\ &10,000 * \frac{173.8}{149.2}=1,1648.79357 \end{aligned}
(ii)
$\mathrm{Q}(0)=$ RPI on 25 October 2008 (issue date)
$Q\left(t_{0}\right)=$ RPI on 25 October 2012 (purchase date)
$\text { Price }=1.5 \times\left[\frac{Q\left(t_{1}\right)}{Q(0)}\right] \times\left[\frac{Q\left(t_{0}\right)}{Q\left(t_{1}\right)}\right] \times v_{3.55 \%}^{0.5}+1.5 \times\left[\frac{Q\left(t_{2}\right)}{Q(0)}\right] \times\left[\frac{Q\left(t_{0}\right)}{Q\left(t_{2}\right)}\right] \times v_{35 \%}^{1}+100 \times\left[\frac{Q\left(t_{2}\right)}{Q(0)}\right] \times\left[\frac{Q\left(t_{0}\right)}{Q\left(t_{2}\right)}\right] \times v_{35 \%}^{1}$
$=1.5 \times\left[\frac{Q\left(t_{0}\right)}{Q(0)}\right] \times v_{35 \%}^{05}+1.5 \times\left[\frac{Q\left(t_{0}\right)}{Q(0)}\right] \times v_{3.55}^{1}+100 \times\left[\frac{Q\left(t_{0}\right)}{Q(0)}\right] \times v_{3.5 \%}^{1}$
$=1.5 \times\left[\frac{169.4}{149.2}\right] \times v_{35 \%}^{0.5}+1.5 \times\left[\frac{169.4}{149.2}\right] \times v_{3.55}^{1}+100 \times\left[\frac{169.4}{149.2}\right] \times v_{35 \%}^{1}$
$=1.674039368+1.645490927+109.6993952=113.02$
by Platinum (91,198 points)
selected by

1 like 0 dislike
1 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
1 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
1 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
1 like 0 dislike
0 like 0 dislike
2 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
1 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
1 like 0 dislike
1 like 0 dislike
0 like 0 dislike
0 like 0 dislike