MHB Number of payments to reach an investment goal

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To determine when an investment will grow from $20,000 to $50,000 with monthly contributions of $1,000 and compounded interest, the discussion outlines a formula involving future value, interest rate, and monthly investment. The user initially shares a formula for calculating the number of periods needed to reach a target from zero but seeks clarification on how to apply it with an existing balance. A recursive approach is proposed, leading to a closed-form solution for the account balance over time. The final formula allows for solving the number of periods required to achieve the investment goal. The conversation concludes with gratitude for the assistance received, highlighting the clarity provided on the topic.
ebroc
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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
 
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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:

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