## Tuesday, February 14, 2012

$550 is too much for one person to spend on food. This works out to about$18 per day. I eat out about 60-80% of the time. I think that I may be able to cut this to $50 a week by eating in nearly exclusively. This will save me about$300 to $350 per month. That money can go toward my savings and investment. When I change my starting point for monthly investment to$700 from $400 it makes some significant changes to the time profile of my investment. For example I only need to raise my monthly investment by about$43 every month rather than $47. Additionally, it reduces the maximum amount I need to invest to$3275.48. This maximum investment also happens 8 months earlier after 134 months of investing. This will happen in March of 2023. This of course motivated me to get curious about different methods for figuring out how much I should contribute.

The current method I use, where my total contribution (my monthly savings plus interest from my investments) increases at a fixed rate every month might not be the easiest to achieve though it is one of the easiest to calculate. Another method which might make more sense is where my active investment increases by a fixed amount. This method is unfortunately more difficult to calculate, requiring a moderate familiarity with differential equations and some simplifying approximations. Suppose that we let the total net worth be f(t). The per month change in the net worth (the amount I invest in that month) is df(t)/dt (we’ll denote this as f’(t)). We’ll let the expected monthly investment income be I and g(t) will be our function for the amount I contribute personally. Then its off to the races!

$If(t)+g(t) = f'(t)$

Suppose that we let g(t) increase constantly thus,

$g(t)=kt+b$

In b will be my zeroth month of investment income which will be a variable that we control. k is the amount we’ll need to increase that investment by on a monthly basis, this variable we’ll calculate with the equation we develop.

$If(t)+kt+b = f'(t)$

To solve this inhomogeneous differential equation we’ll use the method of undetermined coefficients.

$f_p(t)=At+B$

$f_h(t)=c_1e^I^t$

$f(t)=f_h(t)+f_p(t)$

$f(t)=c_1e^I^t+At+B$

$f'(t)=Ic_1e^I^t+A$

$I(c_1e^I^t+At+B)+kt+b=Ic_1e^I^t+A$

$Ic_1e^I^t+IAt+IB+kt+b=Ic_1e^{It}+A$

Subtract the exponential from both sides.

$IAt+IB+kt+IB+b=A$

Now we can let t be very large so that all the non-t terms drop out of the equation:

$IAt+kt=0$

thus,

$IA=-k$

thus,

$A=-\frac{k}{I}$

and then we can let t be zero:

$A=IB+b$

thus,

$-k/I=IB+b$

solving for B:

$-(\frac{k}{I^2}+\frac{b}{I})=B$

therefore making the necessary substitutions,

$Ic_1e^{It}-kt-(\frac{k}{I}+b)+kt+b=Ic_1e^{It}-\frac{k}{I}$

Making all possible cancellations:

$0=0$

verifying the differential equation. We still have a couple undetermined coefficients, namely c1 and k. We know the initial amount of money I’m starting with and the final amount we wish to have, and when both events occur, this allows us to determine both c1 and k. I = 1/150 and b = $400.$f(0)=5850=c_1e^\frac{0}{150}-\frac{k}{\frac{1}{150}}0-(\frac{k}{\frac{1}{150}^2}+\frac{400}{\frac{1}{150}})f(198)=1,000,000=c_1e^\frac{198}{150}-\frac{198k}{\frac{1}{150}}-(\frac{k}{\frac{1}{150}^2}+\frac{400}{\frac{1}{150}})f(0)=5850=c_1-(1502k+400*150)0=c_1-(22500k+65850)0=c_1*3.7434-(52200k+1060000)$We’ll multiply the top equation by 3.7434.$0=3.7434c_1-3.7434(22500k+65850)0=3.7434c_1-84226.5k-246502.89$then we can set each as equal to each other.$84226.5k+246502.89=52200k+106000032026.5k=813497.11k = 25.4$This gives me all of the information I need in order to execute this strategy. If I increase my contribution by a set amount every month by 25.4 by the last month I need$5029.2 as a contribution. This is an annual contribution of about \$60,000. Keep in mind I don’t get to include any income from investments in this. In order to accomplish this plan I’ll have to be making more than twice as much and keep my costs way down. An easier solution probably has my contribution increasing not linearly, but only initially linearly, and at a decreasing rate. We’ll look at this next week.