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Say annual benefit is $120$. Taking an annual discount rate of 3.5%, the present discounted value of the benefit a year from now is $\frac{120}{1+0.035}=115.94$. Assume that benefit is accrued evenly throughout the year i.e. $10$ per month.

How do I go about calculating a monthly discount rate such that the sum of the discounted monthly benefits is $115.94$ after a year?

EDIT: To clarify, I would like $r$ such that $\sum_{n=1}^{12} \frac{10}{(1+r)^n} = 115.94$.

Generally, is there some way to convert from a yearly discount rate to a monthly one?

Hamza
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2 Answers2

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I would like $r$ such that $\sum_{n=1}^{12} \frac{10}{(1+r)^n}=115.94$

This can be managed with the partial sum of the geometric series. Let $q=1+r$ and $S_n=115.94$

$\sum\limits_{n=1}^{12} \frac{10}{(1+r)^n}$

$=S_n=10\cdot \left( \ \ \quad \ \frac1{q}+ \frac1{q^2}+ \frac1{q^3}+ \ldots + \frac1{q^{11}}+ \frac1{q^{12}}\right) \qquad (1)$

$q\cdot S_n=10\cdot\left(1+\frac1{q}+ \frac1{q^2}+ \frac1{q^3}+ \ldots + \frac1{q^{11}}\right)\qquad (2)$

Subtracting (2) from (1).

$S_n\cdot (1-q)=10\cdot \left(\frac1{q^{12}}-1 \right)$

$S_n\cdot (q^{12}-q^{13})=10\cdot \left(1-q^{12} \right)$

$-S_n\cdot q^{13}+(S_n+10)\cdot q^{12}-10=0$

This is a polynomial with a degree of 13. Unfortunately this equation cannot be solved algebraically. Thus you have to solve the equation numerically. For this purpose you can apply the Newton-Raphson method.

callculus42
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Well, you have 12 cash flows, each a month from the previous one. So your portfolio looks like 10 next month, and 10 the following month, etc. If you assume the same monthly discounting rate $r$, can you find the value of the portfolio $V(r)$ as a function of $r$?

HINT (Find the value for each payment and add them together)

Then, solve $V(r) = 115.94$.

gt6989b
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