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#1
03-25-2005, 12:22 PM
 kme Member Join Date: Mar 2004 Posts: 44
Question from How to Pass

This problem has been driving me nuts. I'm probably making a silly mistake.

At an effective rate of interest of 5%, the present value of a perpetuity-immediate is the same as the accumulated value of an n-year annuity-immediate. Find the present value of an n-year annuity-due.

#2
03-25-2005, 12:32 PM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 26,452

You should know the value of a perpetuity immediate at i = .05.
Call it X.

Then write the formula for the n-year immediate annuity, set it equal to X.
Multiply through by i (which equals 0.05).
Solve for (1+i)^n.

That gives you v^n.

Then you have everything you need to get the value of the annuity due. Get d from i; plug in the values of v^n and d.

Comment: n = 14.2067. Non-integral n is somewhat strange.
#3
03-25-2005, 12:37 PM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 26,452

Polls are still open for almost 30 minutes, (or 24 hours and 30 minutes, if the mole is to be trusted). I trail by one.
#4
03-25-2005, 12:43 PM
 kme Member Join Date: Mar 2004 Posts: 44

x=1/.05=20

20=((1+.05)^n)-1)/.05

1=((1.05)^n)-1

2=(1+i)^n=1.05^n

d=.05/1.05=.04762

AV(annuity due)=(((1+i)^n)-1)/d
=(2-1)/.04762
=1/.04762
=21
#5
03-25-2005, 12:49 PM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 26,452

Quote:
 Originally Posted by kme At an effective rate of interest of 5%, the present value of a perpetuity-immediate is the same as the accumulated value of an n-year annuity-immediate. Find the present value of an n-year annuity-due.
You are getting the accumulated value of the annuity due.
#6
03-25-2005, 12:50 PM
 kme Member Join Date: Mar 2004 Posts: 44

ACK! that's been my silly mistake all along!
#7
03-25-2005, 02:18 PM
 D Member Join Date: Jun 2004 Posts: 170

given a_inf = s_n, i=0.05

=> a_inf = 1/i = [(1+i)^n -1 ]/i = s_n
=>1 = (1+i)^n -1
=> (1+i)^n = 2
=>v^n = 1/2

now a_n(due) = (1+i)*a_n

=(1+i)*(1-v^n)/i
=(1.05)*(1-1/2)/0.05
=10.5
#8
03-25-2005, 03:09 PM
 Svak Member Join Date: Jun 2002 Location: Lost Posts: 150

$\huge
\begin{array}{l}
{\rm PV of perpetuity = s}_{\overline {{\rm n|}} 0.05} \\
\frac{1}{{0.05}} = \frac{{1.05^n - 1}}{{0.05}} \\
1 = 1.05^n - 1 \\
2 = 1.05^n \\
n = \frac{{\ln (2)}}{{\ln (1.05)}} = 14.0206 \\
{\ddot a}_{\overline {{\rm 14}{\rm .026|}} 0.05} {\rm = 1}{\rm .05}\left[ {\frac{{1 - 1.05^{ {\small - 14.206}} }}{{0.05}}} \right] = 10.5 \\
\end{array}
$
__________________

Last edited by Svak; 03-25-2005 at 03:55 PM..

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