# Showing an inequality with interest rates and compounding

Suppose that your grandmother is receiving Social Security at the start of next year and has to choose between two plans for how to receive her payments. Divide each year into $k$ payment periods ($k$ periodic compounding). From the start of next year, she plans to save over $n+1$ years the proceeds from her Social Security into an account that compounds $k$ times per year at annual interest rate $r$.

From both plans, the last payment is made at the start of the last period during the $(n+1)$st year.

Plan $A$ we take the first year to be the current year so plan $A$ begins at the start of next year (the second year). She receives the amount $A$ per period starting at the end of the first period of next year.

I have shown that the total amount she would accrue by the end of the $(n+1)$st year under plan $A$ is $$\tag{*} FV_A \equiv A \left[{(1 + r/k)^{(n+1)k} - (1 + r/k) \over r/k}\right]$$

For plan $B$, she receives amount $B$ per period with the plan starting at the end of the $q$th year, so the first payment is at the end of the first period of the $(q+1)$st year. So I have similarly shown that the amount she would have accrued by the end of the $(n+1)$st year is $$\tag{**} FV_B \equiv B \left[{(1 + r/k)^{[(n+1)-q]k} - (1 + r/k) \over r/k}\right]$$

My Question

Using those two FV amounts, and assuming that plan $B$ pays more per period than plan $A$, (but starts later) i.e. $B > A$, assuming also that the interest rate is above a certain threshold: $$r > k[(B/A)^{1/q} - 1]$$ I must show that there is no $n$ such that the amounts accrued under both plans are equal by the end of the $(n+1)$st year.

I must then show that plan $A$ is superior to plan $B$ i.e. $FV_A > FV_B$.

I know that I need only use the $FV$ values I derived above, but what assumptions should I make? I am working myself in circles doing this problem.

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You want to show $FV_A>FV_B$ assuming $B>A$ and $r>k[(B/A)^{1/q}-1]$. Let us write $R:=r/k$. Then your two assumptions give:

$$(1+R)^{q}>\frac{B}{A}>1$$

Using this form, write:

$$(1+R)^{(n+1)k}=(1+R)^{[(n+1)-q]k}(1+r)^{qk}>(1+R)^{[(n+1)-q]k}\left(\frac{B}{A}\right)^{qk}$$

Then:

$$FV_A=A\left[\frac{(1+R)^{(n+1)k}-(1+R)}{R}\right]>A\left[\frac{(1+R)^{[(n+1)-q]k}\left(\frac{B}{A}\right)^{qk}-(1+R)}{R}\right]$$

Now, multiply the right-hand-side by $1=\left(\frac{B}{A}\right)\left(\frac{A}{B}\right)$ to get:

$$FV_A>B\left[\frac{(1+R)^{[(n+1)-q]k}\left(\frac{B}{A}\right)^{qk-1}-(A/B)(1+R)}{R}\right]$$

And I'm assuming that $q,k\geq 1$ so that $(B/A)^{qk-1}\geq 1$ and $(A/B)<1$ which gives you:

$FV_A>FV_B$

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Great stuff, thanks very much – Moderat Sep 18 '12 at 1:53