interest problem

Question

suppose that you save for retirement by contributing the same amount each month from your 23rd birthday until your 65th birthday in an account that pays a steady 4% annual interest compounded monthly. How much will be in your fund at age 65 if you save 100.00 a month?

Answer 1

Using Excel, we can use the Future Value "FV" formula..

Rate... devide annual rate by 12 periods

Nper... multiply by 12 periods

Pmt... as a negative since cash outflow

Type... 1 if payments are in the beginning of the month; skip if at the end

 

The result is close to Aaron's. The small difference is likely due to rounding.

 

 

Answer 2

Well...I don't know your background, and so I'm not sure whether you would like just a formula here or a derivation of a formula. I'll give a derivation using difference quotients (which you can look up if you'd like). First, some assumptions:

• You deposit money at the beginning of the month
• Interest is calculated at the end of the month

Finally, let t = number of months since your 23rd birthday    and x_t = total amt of money in account after t months.

 

Then, we want x_t to satisfy:      x_t+1 = (1+.04/12) * x_t + 100     with  x₀ = 100.

 

For simplicity, I will simplify:    1+.04/12 = 1 + .01/3 = 1 + 1/300 = 301/300.

 

So we want x_t to satisfy:  x_t+1 = 301/300 * x_t + 100 with x₀ = 100.

 

Again, I'm not sure about your background here, but for a difference equation of this type, we can guess that the solution will be of the form:

 

x_t = x_h + x_p

 

Where x_h is the homogeneous solution to the difference equation (i.e.  x_h+1 = 301/300 * x_h ) and x_p is the particular solution (i.e.  x_p+1 = 301/300 * x_p +100). 

 

Now, since x_h satisfies such a simple difference equation, we see that it needs to be of the form: 

 

xh = c₀ * (301/300)^t      where c₀ is an unknown constant

 

Then, since the inhomogeneous part of the original difference equation was the 100, we can guess that x_p is just a constant function, or that x_p = c₁ for c₁ unknown. But we want x_p to satisfy:

 

x_p+1 = 301/300 x_p + 100            so we want:

 

c₁ = 301/300 c₁ + 100    which, after some algebra, means that:

 

c₁ = -30000

 

So now we know our answer is of the form:

 

x_t = c₀ * (301/300)^t - 30000        with x₀ = 100.

 

Since we want x₀ = 100, we substitute those values in to see what c₀ needs to be:

 

c₀ * (301/300)⁰ - 30000 = 100      becomes:       c₀ - 30000 = 100      which means that:

 

c₀ = 30100

 

So our final equation is:

 

x_t = 30100 * (301/300)^t - 30000

 

As a check, after 1 month, my formula says that you will have:

 

x₁ = 30100 * 301/300 - 30000 = $200.33     which is what you would get from the direct calculation:

 

100*301/300 + 100 = $200.33

 

So anyway, now that we have a formula, we just plug in the month that we want. Since we want 65 - 23 = 42 years to pass, we want 42*12 = 504 months to pass. So we plug that in:

 

x₅₀₄ = 30100 * (301/300)⁵⁰⁴ - 30000 

= $131,052.66

 

And that should be your answer. If you use an online calculator, such as the one at:

http://www.crown.org/tools/calculators/Savings_MonthlyDeposit.aspx

You get the same result. Hopefully I've helped somewhat.