# Present Value of Perpetuity

Really really struggling with the following question:

Suppose you receive $$\44000$$ per year, but get a $$\1000$$ raise every $$k = 9$$ years. That is, you are paid $$\44000$$ in years $$1, 2, ..., k$$, and paid $$\44000 + \1000$$ in years $$k+1, k+2, ..., 2k$$, and paid $$\44000+\2000$$ in years $$2k+1, 2k+2, ..., 3k$$, and so on into perpetuity.  Given an interest rate $$2.7\%$$, what is the present value of these cash flows?  (to nearest $$\0.01$$) (Hint: work out a formula rather than just calculating numerically. Can you convert this question to the PV of an annuity of length 9 years?)

The answer should be $$\1,766,315.72$$.

I tried calculating annuity for $$n=9$$ first and get the annuity of $$\44,000+\1000$$ and $$\44,000+\2000$$ separately. Is that what I do? If yes, how?

Any help would be appreciated! Thanks.

• This can be broken down as a sum of 9 series, each of which is what's called an "arithmetico-geometric series" Commented Dec 23, 2020 at 20:46

The present value can be written as an infinite sum: \begin{align}PV \,&=\, \color{red}{44000} + \color{blue}{\frac{44000}{1.027}}+\color{green}{\frac{44000}{1.027^2}} +\color{purple}{\frac{44000}{1.027^3}} + \cdots + \frac{44000}{1.027^8}\\\\ &+\,\color{red}{\frac{45000}{1.027^9}} +\color{blue}{\frac{45000}{1.027^{10}}} +\color{green}{\frac{45000}{1.027^{11}}} +\color{purple}{\frac{45000}{1.027^{12}}}+ \cdots +\frac{45000}{1.027^{17}}\\\\ &+\,\color{red}{\frac{46000}{1.027^{18}}} +\color{blue}{\frac{46000}{1.027^{19}}} +\color{green}{\frac{46000}{1.027^{20}}} +\color{purple}{\frac{46000}{1.027^{21}}}+ \cdots +\frac{46000}{1.027^{26}}\\\\ &+ \,\,\,\, \cdots \\\\ \end{align}

We split this sum into $$9$$ sums, grouping terms as follows:

\begin{align} PV \,&=\, \color{red}{44000} + \color{red}{\frac{45000}{1.027^9}} + \color{red}{\frac{46000}{1.027^{18}}} + \cdots \\\\ &+ \, \color{blue}{\frac{44000}{1.027}} + \color{blue}{\frac{45000}{1.027^{10}}} + \color{blue}{\frac{46000}{1.027^{19}}} + \cdots \\\\ &+ \, \color{green}{\frac{44000}{1.027^2}} + \color{green}{\frac{45000}{1.027^{11}}} + \color{green}{\frac{46000}{1.027^{20}}} + \cdots \\\\ &+ \, \color{purple}{\frac{44000}{1.027^3}} + \color{purple}{\frac{45000}{1.027^{12}}} + \color{purple}{\frac{46000}{1.027^{21}}} + \cdots \\\\ &\phantom{+} \qquad \vdots \\\\ &+ \, \frac{44000}{1.027^8} + \frac{45000}{1.027^{17}}+\frac{46000}{1.027^{26}} + \cdots \\\\ \tag{1}\end{align}

Each of these series is what is known as an arithmetico-geometric series. In general, series of this form can be reduced to a geometric series by multiplying by the common ratio and subtracting the result. In one of my previous answers, I've described this reduction, for a simpler arithmetico-geometric series.

In this case, dividing both sides of $$(1)$$ by $$1.027^{9}$$ solves the problem:

\begin{align} \frac{PV}{1.027^9} \,&=\, \color{red}{\frac{44000}{1.027^9}} + \color{red}{\frac{45000}{1.027^{18}}} + \color{red}{\frac{46000}{1.027^{27}}} + \color{red}{\frac{47000}{1.027^{36}}} + \cdots \\\\ &+ \, \color{blue}{\frac{44000}{1.027^{10}}} + \color{blue}{\frac{45000}{1.027^{19}}} + \color{blue}{\frac{46000}{1.027^{28}}} + \color{blue}{\frac{47000}{1.027^{37}}} + \cdots \\\\ &+ \, \color{green}{\frac{44000}{1.027^{11}}} + \color{green}{\frac{45000}{1.027^{20}}} + \color{green}{\frac{46000}{1.027^{29}}} + \color{green}{\frac{47000}{1.027^{38}}} + \cdots \\\\ &+ \, \color{purple}{\frac{44000}{1.027^{12}}} + \color{purple}{\frac{45000}{1.027^{21}}} + \color{purple}{\frac{46000}{1.027^{30}}} + \color{purple}{\frac{47000}{1.027^{39}}} + \cdots \\\\ &\phantom{+} \qquad \vdots \\\\ &+ \, \frac{44000}{1.027^{17}} + \frac{45000}{1.027^{26}} + \frac{46000}{1.027^{35}}+\frac{47000}{1.027^{44}} + \cdots \\\\ \tag{2} \end{align}

Subtracting $$(2)$$ from $$(1)$$ gives:

\begin{align} PV - \left(\frac{PV}{1.027^9} \right) \, &= \, \color{red}{44000} + \left(\, \color{red}{\frac{1000}{1.027^9}} + \color{red}{\frac{1000}{1.027^{18}}} + \color{red}{\frac{1000}{1.027^{27}}} + \cdots \,\right)\\\\ &+\, \color{blue}{\frac{44000}{1.027}} + \left(\, \color{blue}{\frac{1000}{1.027^{10}}} + \color{blue}{\frac{1000}{1.027^{19}}} + \color{blue}{\frac{1000}{1.027^{28}}} + \cdots \,\right)\\\\ &+\, \color{green}{\frac{44000}{1.027^2}} + \left(\, \color{green}{\frac{1000}{1.027^{10}}} + \color{green}{\frac{1000}{1.027^{19}}} + \color{green}{\frac{1000}{1.027^{28}}} + \cdots \,\right)\\\\ &+\, \color{purple}{\frac{44000}{1.027^3}} + \left(\, \color{purple}{\frac{1000}{1.027^{10}}} + \color{purple}{\frac{1000}{1.027^{19}}} + \color{purple}{\frac{1000}{1.027^{28}}} + \cdots \,\right)\\\\ &\phantom{+} \qquad \vdots \\\\ &+ \frac{44000}{1.027^8} + \left(\,\frac{1000}{1.027^{17}} + \frac{1000}{1.027^{26}} + \frac{1000}{1.027^{35}} + \cdots \,\right)\\\\ \tag{3} \end{align}

Each of the nine series in parentheses on the right hand side of $$(3)$$ are infinite geometric series with common ratio $$1/1.027^9 \approx 0.786803$$. If the first term is $$a$$, and the common ratio is $$r$$, then the sum of each infinite geometric series is $$a + ar + ar^2 + \cdots = \frac{a}{1 - r}\tag{4}$$

Applying $$(4)$$ to each of the nine infinite geometric series on the right hand side of $$(3)$$ gives:

\begin{align} PV \left(1 - \frac{1}{1.027^9}\right) \, &=\, \left(\color{red}{44000} + \color{red}{\frac{1000000000000000000000000000000}{270966174450725282733813987}}\right)\\\\ &+\, \left(\color{blue}{44000} + \color{blue}{\frac{1000000000000000000000000000000}{270966174450725282733813987}}\right)\left(\color{blue}{\frac{1}{1.027}}\right)\\\\ &+\, \left(\color{green}{44000} + \color{green}{\frac{1000000000000000000000000000000}{270966174450725282733813987}}\right)\left(\color{green}{\frac{1}{1.027^2}}\right)\\\\ &+\, \left(\color{purple}{44000} + \color{purple}{\frac{1000000000000000000000000000000}{270966174450725282733813987}}\right)\left(\color{purple}{\frac{1}{1.027^3}}\right)\\\\ &\phantom{+} \qquad \vdots \\\\ &+ \, \left(44000 + \frac{1000000000000000000000000000000}{270966174450725282733813987}\right)\left(\frac{1}{1.027^8}\right)\\\\ \end{align}

This is, at last, equivalent to the present value of an annuity of length $$9$$ years.

The present value is:

$$PV \, = \, \frac{13271419491079374076175586444556000}{7316086710169582633812977649} \, \approx \, \ 1\text{,}814\text{,}005.22\,$$

This calculation assumes that the first payment occurs right now.

If, instead, the first payment of $$\44000$$ occurs one year from now, then every term is discounted by an additional factor of $$\frac{1}{1.027}$$. In that case, the present value is:

$$PV \, = \, \frac{12922511675831912440287815428000000}{7316086710169582633812977649} \, \approx \, \boxed{ \, \ 1\text{,}766\text{,}314.72\, }$$

This is an increasing perpetuity that has the following cash flow, in thousands: $$PV = 44 (v + v^2 + \cdots + v^9) + 45 v^{9} (v + v^2 + \cdots + v^9) + 46 v^{18} (v + v^2 + \cdots + v^9) + \cdots$$ where $$v = 1/(1+i)$$ is the annual present value discount factor. Note we are assuming that payments are made in arrears, so this is a perpetuity-immediate and the present value is calculated with respect to one year before the first payment of $$44$$. Clearly, we may perform a factorization: $$PV = (v + v^2 + \cdots + v^9)(44 + 45 v^{9} + 46 v^{18} + \cdots ),$$ and the first factor is simply $$\require{enclose} a_{\enclose{actuarial}{9}i} = \frac{1 - v^9}{i} \approx 7.89619.$$ The second factor is $$44(1 + v^9 + v^{18} + \cdots) + (v^9 + 2v^{18} + 3v^{27} + \cdots) = 44 \ddot a_{\enclose{actuarial}{\infty} j} + (Ia)_{\enclose{actuarial}{\infty} j},$$ where $$j = (1+i)^{9} - 1$$ is the effective nine-year rate of interest. Since a perpetuity-due has a present value of $$\ddot a_{\enclose{actuarial}{\infty} j} = 1 + \frac{1}{j},$$ and an increasing perpetuity-immediate has a present value of $$(Ia)_{\enclose{actuarial}{\infty} j} = \frac{1}{j} + \frac{1}{j^2},$$ the second factor in our equation of value is $$44(4.6905) + 17.3103 \approx 223.692.$$ Therefore the present value is $$PV \approx (7.89619) ( 223.692) = 1766.31472367$$ in units of thousands of dollars.

As an addendum, the formulas for the present values of the perpetuities can be found as follows. As usual, $$i$$ is some periodic interest rate and $$v = 1/(1+i)$$. We assume $$i > 0$$, otherwise the present value is infinite. Then $$0 < v < 1$$ and

First, $$\ddot a_{\enclose{actuarial}{\infty} i} = 1 + v + v^2 + \cdots = \frac{1}{1-v} = \frac{1}{1 - \frac{1}{1+i}} = \frac{1+i}{i} = 1 + \frac{1}{i},$$ from the formula for the sum of an infinite geometric series. Hence, $$a_{\enclose{actuarial}{\infty} i} = v + v^2 + v^3 + \cdots = \frac{1}{i},$$ which is the present value of a perpetuity-immediate. Finally, let $$x = (Ia)_{\enclose{actuarial}{\infty} i} = v + 2v^2 + 3v^3 + \cdots.$$ Then $$vx = v^2 + 2v^3 + 3v^4 + \cdots,$$ and $$vx + a_{\enclose{actuarial}{\infty} i} = (v^2 + 2v^3 + 3v^4 + \cdots) + (v + v^2 + v^3 + \cdots) = v + 2v^2 + 3v^3 + \cdots = x.$$ Solving for $$x$$ yields $$x = (Ia)_{\enclose{actuarial}{\infty} i} = a_{\enclose{actuarial}{\infty}i} \cdot \frac{1}{1-v} = (a_{\enclose{actuarial}{\infty}i}) (\ddot a_{\enclose{actuarial}{\infty}i}) = \frac{1}{i} + \frac{1}{i^2},$$ as claimed.

• I have never seen this method! +1 ... Looks like we both have the same final result, and both of us differ from what the asker wrote the answer "should" be, by about \$1 Commented Dec 23, 2020 at 23:47