find interest rate - not your average problem
- Thread starter trasheroo
- Start date
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- Apr 12, 2005
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Basic Principles will save you EVERY TIME! However, in this case, there is no direct solution. It will require an iterative process.
"Joe wants to invest $150 every month for 10 years. At the end of that time, he would like to have $25000. At what annual interest rate, compounded monthly, does Joe need to invest to reach his goal?"
i = The desired annual rate, compounded monthly.
j = i/12 the equivalent monthly rate.
r = 1+j the monthly accumulation factor
That's all we need. Now build it. It's easier if you start from the end. The last payment is not accumulated at all. It is listed first.
\(\displaystyle 150 + 150r + 150r^{2} + 150r^{3} + ... + 150r^{119} = 25000\)
Now add them all up.
\(\displaystyle 150\dfrac{1 - r^{120}}{1-r} = 25000\)
This is where we run into a problem. You CANNOT solve for r directly. Break out your favorite numerical solution (Newton-Raphson, or whatever), and track it down. I managed r = 1.0052584098394929
Thus:
j = 0.0052584098394929
i = 0.0631009181
That was fun!!!
"Joe wants to invest $150 every month for 10 years. At the end of that time, he would like to have $25000. At what annual interest rate, compounded monthly, does Joe need to invest to reach his goal?"
i = The desired annual rate, compounded monthly.
j = i/12 the equivalent monthly rate.
r = 1+j the monthly accumulation factor
That's all we need. Now build it. It's easier if you start from the end. The last payment is not accumulated at all. It is listed first.
\(\displaystyle 150 + 150r + 150r^{2} + 150r^{3} + ... + 150r^{119} = 25000\)
Now add them all up.
\(\displaystyle 150\dfrac{1 - r^{120}}{1-r} = 25000\)
This is where we run into a problem. You CANNOT solve for r directly. Break out your favorite numerical solution (Newton-Raphson, or whatever), and track it down. I managed r = 1.0052584098394929
Thus:
j = 0.0052584098394929
i = 0.0631009181
That was fun!!!