MHB How do I decide this total cost and how do I make a minimum estimate?

[BTerry]

Principle: 10K
Interest: 10% compounded annually
Time: Over five years

What is the total cost of all of this factored in?

Could I get the formula so I can use it to apply to a real problem?

 

[MarkFL]

Is this for a loan repayment? If so, consider the following, which can easily be changed to an annual payment:

Let $P$ = monthly payment, $A$ = amount borrowed, $i$ = monthly interest rate, and $n$ = the number of payments.

Also, let $D_n$ be the debt amount after payment $n$.

Consider the recursion:

(1) [math]D_{n}=(1+i)D_{n-1}-P[/math]

(2) [math]D_{n+1}=(1+i)D_{n}-P[/math]

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

[math]D_{n+1}=(2+i)D_{n}-(1+i)D_{n-1}[/math]

whose associated auxiliary equation is:

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

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

Thus, the closed-form for our recursion is:

[math]D_n=k_1(1+i)^n+k_2[/math]

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

[math]D_0=k_1+k_2=A[/math]

[math]D_1=k_1(1+i)+k_2=(1+i)A-P[/math]

Solving this system, we find:

[math]k_1=\frac{Ai-P}{i},\,k_2=\frac{P}{i}[/math] and so we have:

[math]D_n=\left(\frac{Ai-P}{i} \right)(1+i)^n+\left(\frac{P}{i} \right)=\frac{(Ai-P)(1+i)^n+P}{i}[/math]

Now, equating this to zero, we can solve for $P$:

[math]\frac{(Ai-P)(1+i)^n+P}{i}=0[/math]

[math](Ai-P)(1+i)^n+P=0[/math]

[math](P-Ai)(1+i)^n=P[/math]

[math]P\left((1+i)^n-1 \right)=Ai(1+i)^n[/math]

[math]P=\frac{Ai(1+i)^n}{(1+i)^n-1}[/math]

[math]P=\frac{Ai}{1-(1+i)^{-n}}[/math]

 

[Wilmer]

After solving for P (as shown by Mark), the cost is simply: P*n - A

In other words: what you paid back less what you got.