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I am working on a textbook problem, and I think I disagree with the solution. The problem is

(10.6) Bruce deposits 100 into a bank account. His account is credited interest at a nominal rate of interest of 4% convertible semiannually. At the same time, Peter deposits 100 into a separate account. Peter’s account is credited interest at a force of interest of δ. After 7.25 years, the value of each account is the same. Calculate δ.

The solution provided is $\delta = .0396$, which I can get by setting these two accumulation functions equal: $$ (1+\dfrac{.04}{2})^{2 \times 7.25}=e^{7.25 \delta} $$ The reason I feel this is incorrect is that the semiannual accumulation function should not permit a time value of $t=7.25$, which is between periods. Instead I feel that a value of $t=7$ should be used, being the end of the most recent period: $$ (1+\dfrac{.04}{2})^{2 \times 7}=e^{7.25 \delta} $$ This yields $\delta \approx 0.0382$.

ajnelson.alpha
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3 Answers3

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The same principal invested for the same period of time yields the same accumulated value. So over $1$ year we must have the equivalence $$ \left(1+\frac{i^{(2)}}{2}\right)^2=\mathrm e^\delta $$ and then $$ \delta=2\ln\left(1+\frac{i^{(2)}}{2}\right)\approx 0.0396 $$

alexjo
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  • Yes, but only if the 7.25 years cancel out. Can I use one-quarter year measurements when the interest compounds semianually, i.e. every one-half year? – ajnelson.alpha Feb 25 '18 at 01:55
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    @ajnelson.alpha No you can't use interest with non coherent compounding periods. But you can put, in this case, the $i^{(2)}$ and $\delta$ equivalent over the same period (one year or one semester). – alexjo Feb 25 '18 at 10:42
  • @ajnelson.alpha See here your problem (it's the n. 1) and here the solution (it's the n. 1) – alexjo Feb 25 '18 at 10:51
  • Thanks for that, did not realize this question was from the Sample FM Exam haha. I'm still confused about WHY we're allowed to use a non-coherent period in this case... but I will accept that this is the correct approach. – ajnelson.alpha Feb 25 '18 at 19:43
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I'm a little late to this one, but if you google the FM exam problem it shows up, and I think that the fundamental problem here still hasn't been addressed, so it's worth talking about.

The question says that after 7.25 years, the value of each account is the same. They very carefully did not mention the balance of the accounts. If they had, this would be a different problem, since one account credits interest every six months and the other does so constantly. The value of the account credited semiannually still increases constantly, since you are always getting closer to the next payout.

Here is a slightly different way to approach the problem: The value of both accounts is governed by exponential equations that start at the same place and grow at the same rate, so they must be equivalent. Since we're looking for an annual force of interest let's make everything annual. The effective annual rate is $$i = \left(1 + \frac{i^{(m)}}{m}\right)^{m} - 1 $$ so $i = .0404$. Then the force of interest given $i$ is $$\delta = \ln(1+i)$$ so $\delta = .0396$

Ian
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Usually you use the $\texttt{simple interest}$ for the remaining three months.

$$\left(1+\frac{0.04}{2}\right)^{2 \cdot 7}\cdot \left(1+\frac{0.04}4\right)=e^{7.25 \cdot \delta}$$

$$7.25 \cdot\delta=\ln\left[1.02^{14}\cdot 1.01 \right]$$

$$7.25 \cdot\delta=0.287187$$

$$\delta=\frac{0.287187}{7.25}=0.039612\approx 3.96\%$$

callculus42
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  • Actually you cannot change from compounding to simple interest.... – alexjo Feb 25 '18 at 10:43
  • @alexjo Sure, since the three months are less than half a year. If the bank would calculate with 7.5 years t bruce would get less money. There is no need to downvote. I don´t say that your solution is wrong, but mine isn´t as well. – callculus42 Feb 25 '18 at 11:23
  • I didn't downvote. Anyway the interest cannot be changed in the OP. This might be a practice in a bank but has nothing to do with the math. Using two different type of interest is completely wrong. In which book did you see it?? I've added also the solution of SOA...just for information – alexjo Feb 25 '18 at 11:53
  • What if we use $i^{(2)}=10%$ with 2.25 years? Your solution will be different... – alexjo Feb 25 '18 at 12:03