Number of payments to reach an investment goal

[ebroc]

Hi

I'm trying to calculate at what point my investment will reach it's target. I'm fine doing this starting from zero using

number of periods = log(1+ ((FV * R)/P)) / log (1+R)

FV = Future Value
R = Rate
P = Monthly Investment

But what if I already have \$20000 in my account and want to know when it will reach \$50000 if I am saving \$1000 from now onwards. Assuming interest is compounded monthly.

Hope this is clear. Thanks for looking

Regards

E

 

[MarkFL]

Let $P$ = monthly investment, $A$ = existing investment, $R$ = monthly interest rate, and $n$ = the number of deposits.

Also, let $B_n$ be the account balance after payment $n$, and $B_0=A+P$ is the starting balance.

Consider the recursion:

[math]B_{n}=(1+R)B_{n-1}+P\tag{1}[/math]

[math]B_{n+1}=(1+R)B_{n}+P\tag{2}[/math]

Subtracting (1) from (2) yields the linear homogeneous recursion:

[math]B_{n+1}=(2+R)B_{n}-(1+R)B_{n-1}[/math]

whose associated auxiliary equation is:

[math]r^2-(2+R)r+(1+R)=0[/math]

[math](r-(1+R))(r-1)=0[/math]

Thus, the closed-form for our recursion is:

[math]B_n=k_1(1+R)^n+k_2[/math]

Using initial values, we may determine the coefficients $k_i$:

[math]k_1+k_2=B_0[/math]

[math]k_1(1+R)+k_2=(1+R)B_0+P[/math]

Solving this system, we find:

$$\left(k_1,k_2\right)=\left(B_0+\frac{P}{R},-\frac{P}{R}\right)$$

And so the solution is:

[math]B_n=\left(B_0+\frac{P}{R}\right)(1+R)^n-\frac{P}{R}[/math]

Solving for $n$, we obtain:

$$n=\frac{\ln\left(\dfrac{B_nR+P}{B_0R+P}\right)}{\ln(1+R)}$$

 

[jonah1]

Hi

I'm trying to calculate at what point my investment will reach it's target. I'm fine doing this starting from zero using

number of periods = log(1+ ((FV * R)/P)) / log (1+R)

FV = Future Value
R = Rate
P = Monthly Investment

But what if I already have \$20000 in my account and want to know when it will reach \$50000 if I am saving \$1000 from now onwards. Assuming interest is compounded monthly.

Hope this is clear. Thanks for looking

Regards

E

Compound interest formula plus FV of annuity formula (annuity due version, i.e. payments at beginning of period).
Thus,
$$50,000=200,000(1+i)^n+1,000(1+i)\frac{(1+i)^{n}-1}{i}$$

From your post, I take it that you can easily solve for n using logarithms.

 

[ebroc]

Thank you both so much for your help. Your answers not only answered my question, but also a couple I hadn't even thought of. :D

Kind regards

E