Evaluating differing perpetuity payments

393 Views Asked by At

A perpetuity pays 1900 dollars on January 1 of 1980, 1982, 1984, ..., and pays X dollars on January 1 of 1981, 1983, 1985, ... If the present value on January 1, 1975 is 26500 dollars, and the effective rate of interest is 8.4 percent, what is X?

Can someone direct me in the right direction... the formula I am using is 26500=(100*(1/d)*v^5)+(1/d)*v^6*x where d=(0.084*2)/1+0.084*2 and i=0.084. After solving this equation for x i am putting it back into the present value for annuity due formula with n=5, but i am not getting the correct answer for x.

Thanks

1

There are 1 best solutions below

0
On

The idea is to write out the present value of the cash flow: $$PV = 1900v^5 + Xv^6 + 1900v^7 + Xv^8 + \cdots,$$ where $v = (1+i)^{-1}$ is the present value discount factor for an effective annual interest rate of $i = 0.084$. The reason why the cash flow begins at the fifth year ($v^5$) is because the valuation time is 1/1/1975, whereas the first payment is due to be made on 1/1/1980, five years later.

Now, the rest is mathematics: $$PV = 1900(v^5 + v^7 + v^9 + \cdots) + X(v^6 + v^8 + v^{10} + \cdots).$$ We can then factor out the highest common factor of each series: $$PV = 1900v^5(1 + v^2 + v^4 + \cdots) + Xv^6(1 + v^2 + v^4 + \cdots),$$ and notice how this itself can be factored: $$PV = (1900+Xv)v^5(1 + v^2 + v^4 + \cdots).$$ The infinite series factor is an infinite geometric series with common ratio $v^2$, thus has the value $$\sum_{t=0}^\infty (v^2)^t = \frac{1}{1 - v^2} = \frac{1}{1-(1+i)^{-2}} \approx 6.71246.$$ The other terms are then easy to compute, and then you just solve for $X$ for the given present value $PV$.

In actuarial notation, you would express the above cash flow as $$PV = 1900v^5 \ddot a_{\overline{\infty}\rceil j} + Xv^6 \ddot a_{\overline{\infty}\rceil j} = (1900+Xv)v^5 {\ddot a}_{\overline{\infty}\rceil j},$$ where $j$ is the effective two-year rate of interest satisfying $$1+j = (1+i)^2,$$ namely $j = 0.175056$.