I was helping a comrade how to answer this engineering economics question. It goes like this:
Find the present value of installment payments of \$1000 now, \$2000 at the end of the first year, \$3000 at the end of second year, \$4000 at the end of the third year, \$5000 at the end of the fourth year, if money is worth 10% compounded annually.
My work
I thought the present worth is the future worth of money 4 years from now.
Using the basic formula $F = P(1+i)^n$, where $F$ is future worth, $P$ is the principal, $i$ is the interest rate, and $n$ is the number of years, we can answer the question using this approach:
When I pay now: $$\$1000$$
When 1 year has elapsed, I pay $\$2000$ and my $\$1000$ was earning already. $$\$2000 + \$1000(1+0.1)^1 = \$3100$$
When 2 years has elapsed, I pay $\$3000$ and my $\$2000$ and $\$1000$ was earning already. $$\$3000 + \$2000(1+0.1)^1 + \$1000(1+0.1)^2 = \$6410$$
When 3 years has elapsed, I pay $\$4000$ and my $\$3000$, $\$2000$ and $\$1000$ was earning already. $$\$4000 + \$3000(1+0.1)^1 + \$2000(1+0.1)^2 + \$1000(1+0.1)^3 = \$11051$$
When 4 years has elapsed, I pay $\$5000$ and my $\$4000$, $\$3000$, $\$2000$ and $\$1000$ was earning already. $$\$5000 + \$4000(1+0.1)^1 + \$3000(1+0.1)^2 + \$2000(1+0.1)^3 + \$1000(1+0.1)^4 = \$17156.10$$
The present value of installment payments, therefore, would be: $$\$1000 + \$3100 +\$6410 +\$11051 +\$17156.10 = \$38717.10$$
Is my answer correct? Is it any good?
Edit
Someone said that I just found the future value, not the present one. With that in mind, to get the present value, do this:
When I pay now: $$\$1000$$
When 1 year has elapsed, I pay $\$2000$ and my $\$1000$ present value has changed already. $$\$2000 + \frac{\$1000}{(1+0.1)^1} = \$2909.09$$
When 2 years has elapsed, I pay $\$3000$ and my $\$2000$ and $\$1000$ present value has changed already. $$\$3000 + \frac{\$2000}{(1+0.1)^1} + \frac{\$1000}{(1+0.1)^2} = \$5644.63$$
When 3 years has elapsed, I pay $\$4000$ and my $\$3000$, $\$2000$ and $\$1000$ present value has changed already. $$\$4000 + \frac{\$3000}{(1+0.1)^1} + \frac{\$2000}{(1+0.1)^2} + \frac{\$1000}{(1+0.1)^3} = \$9131.48$$
When 4 years has elapsed, I pay $\$5000$ and my $\$4000$, $\$3000$, $\$2000$ and $\$1000$ present value has changed already. $$\$5000 + \frac{\$4000}{(1+0.1)^1} + \frac{\$3000}{(1+0.1)^2} + \frac{\$2000}{(1+0.1)^3} + \frac{\$1000}{(1+0.1)^4} = \$13301.34$$
The present value of installment payments, therefore, would be: $$\$1000 + \$2909.09 +\$5644.63 + \$9131.48 + $\$1301.34$= \$31986.54$$
Is my answer now correct? Is it any good?
No, you want the present value. What you have computed is the future value.
The present value can be thought of as the equivalent amount of money that would be paid up front as a lump sum, which has the same time value of money as the cash flow in question.
For example, if I say that $2000$ will be paid to you one year from now, and the effective annual rate of interest is $i = 0.1$, the present value of this payment is the amount which, if held by you over the same amount of time, would equal $2000$ at the end of one year. That is to say, $$PV(1+i) = 2000,$$ or $$PV = \frac{2000}{1.1} \approx 1818.18.$$ It should be clear that for a positive interest rate, the longer the deferral of the payment, the less its present value, since you have to wait longer to receive it. So for example, if I said that the same $2000$ will be paid to you in $10$ years from now, the present value is $2000/(1.1)^{10} \approx 771.09$. That decrease in value is a reflection of the fact that you could take $771.09$ now and assuming a constant effective annual interest rate of $10\%$ per year, it would grow to $2000$ at the end of $10$ years if the interest is compounded annually.
With this understanding, try recomputing the present value of your cash flow.