Accumulating Cash Flows Under Simple Interest

# Inconsistent results in multi-stage simple interest problems — what's the correct accumulation method?


# Inconsistent results in multi-stage simple interest problems — what's the correct accumulation method?












## Question on multi-stage simple interest problems

Textbook Source: Marcel Finan's A Basic Course in the Theory of Interest and Derivatives Markets

Notation: Let $\nu = \frac{1}{1+i}$ denote the discount factor, $A_0$ the principal, $A(t)$ the amount function, and $a(t)$ the accumulation function. These satisfy the relation $A(t) = A_0 \cdot a(t)$.

For the purpose of this question, assume that $T$ is the valuation time and let cash flows (deposits and withdrawals) $C_k$ occur at times $t_k\le T$.

Now, examine the questions from my textbook below:

Example 4.5

Suppose you make a deposit of \$100 at time $t = 0$. A year later, you make a withdrawal of \$50. Assume annual simple interest rate of 10%, what is the accumulated value at time $t = 2$ years?

Problem 6.17

Important note: for this problem, we are only concerned with Joe's accumulated value at $t=10$. The second part of the question involving Tina's deposits is largely irrelevant.

Joe deposits \$10 today and another \$30 in five years into a fund paying simple interest of 11% per year. Tina will make the same two deposits, but the \$10 will be deposited $n$ years from today and the \$30 will be deposited $2n$ years from today. Tina's deposits earn an annual effective rate of 9.15% . At the end of 10 years, the accumulated amount of Tina's deposits equals the accumulated amount of Joe's deposits. Calculate $n$.

The author has not yet introduced the concept of present value in the beginning chapters. However, while answering these questions, I found that it is pretty convenient to calculate the net present value at $t=0$ and then multiply by the accumulation function at $t=T$. The procedure for this is shown below:

$$A(T)\;=\; \sum_k C_k\,\frac{a(T)}{a(t_k)} \;=\; a(T)\sum_k C_k\,v(t_k)$$

This method works just fine with Example 4.5 and many other questions in the introductory chapters involving deposits and withdrawals at non-zero times $t_k$. However, this assumption was flipped on its head when it came to Problem 6.17. It appears that the author treats each stage in Problem 6.17 as the sum of individual accumulation functions multiplied by their respective principals, rather than a single accumulation function characterizing the behavior that can be multiplied to each of the present values at $t=0$. This procedure is shown below:

$$ A(T)\;=\; \sum_k C_k\,\bigl(1+i\,[T-t_k]\bigr) \;=\; a(T)\!\sum_k C_k \;-\; i\!\sum_k C_k\,t_k $$

The textbook's work under this approach looks like: $$ A(10)=10\,[1+0.11(10-0)]+30\,[1+0.11(10-5)] =10(2.1)+30(1.55)=21+46.5=\boxed{67.5}. $$ This implies that each deposit earns simple interest for its own holding period.

However, I stuck with my original method, using the following work:

$$ \text{PV at }0:\quad 10+\frac{30}{1+0.11\cdot5}=10+\frac{30}{1.55}\approx 10+19.3548=29.3548. $$ Accumulate to $T=10$ with $a(10)=1+0.11\cdot10=2.1$: $$ A(10)=2.1\times 29.3548\approx \boxed{61.645}\quad(\text{does not equal }67.5). $$

I am confused why the textbook did this problem in this way for simple interest. Are they not implicitly violating the below assumption under simple interest? $$ \frac{a(T)}{a(t_k)} \ne a(T-t_k)\quad \text{for simple interest, since}\quad \frac{1+iT}{1+it_k} \ne 1 + i(T - t_k). $$ The only way for this to work (in my mind) is if there were multiple accumulation functions for different time periods, and it were impossible to characterize the behavior of Joe's account under a single accumulation function. But then, how would I know, intuitively, whether this is the case for any given problem on the exam? It doesn't seem like the solution has anything to do with summing up the present values at $t=0$ like I did. So, how is this even possible? The textbook states that all deposits and withdrawals must use the accumulation factor formula (which is an implicit use of the sum of present values formula, since $a(T)$ gets distributed). The second method of computation for Problem 6.17 does not work with Example 4.5 either:

$$ \text{Using } A(T)=\sum_k C_k\bigl(1+i\,[T-t_k]\bigr): \\ A(2)=100\bigl(1+0.10\cdot2\bigr)-50\bigl(1+0.10\cdot(2-1)\bigr) =120-55=\boxed{65}. $$

$$ \text{Using } A(T)=\sum_k C_k\,\frac{a(T)}{a(t_k)}: \\ A(2)=100\cdot\frac{1.2}{1}-50\cdot\frac{1.2}{1.1} \approx 120-54 \approx \boxed{65.45}. $$

The second answer, which uses the first method of simple interest computation, is the correct answer according to the textbook. Clearly, this is inconsistent with the approach the author has chosen for Problem 6.17. Is this an errata in the textbook, or is there some nuance between the questions that make the computation different? I would really prefer a consistent way to perform these simple interest calculations that have multiple deposits/withdrawals at non-principal times.

## Answer by Attack68 (score 2, accepted)

https://quant.stackexchange.com/a/84104

> The author has not yet introduced the concept of present value in the beginning chapters

OK because that is not needed.

> Joe deposits \$10 today and another $30 in five years into a fund paying simple interest of 11% per year. What is the accumulated amount after 10 years.

$$10 * 1.11^{10} + 30 * 1.11^5 = 78.946 $$

> Tina will make the same two deposits, but the \$10 will be deposited n years from today and the $30 will be deposited 2n years from today. Tina's deposits earn an annual effective rate of 9.15% . At the end of 10 years, the accumulated amount of Tina's deposits equals the accumulated amount of Joe's deposits. Calculate n.

$$ 10 * 1.0915^{10-n} + 30 * 1.0915 ^{10 - 2n} = 78.946 $$

$n=1.285$ solved by iteration. I.e. Tina invests \$10 15.5months from today and then invests \$30 in 31months time. Clearly she has to deposit earlier than Joe because her interest rate is lower.

This is basic mathematics and isn't really suitable for Quant Finance forum.