Skill: Determine present value of an investment

Explanation and Free Practice Resources

Present value is the amount that must be invested now to obtain a specified future value, found by reversing compound growth: P=A/(1+r/n)ntP=A/(1+r/n)^{nt}, where rr is the annual interest rate, nn the number of compounding periods per year, and tt the time in years. The concept connects exponential growth with discounting and emphasizes that more frequent compounding, higher rates, or longer periods reduce the present amount required; continuous compounding, variable rates, and complex cash-flow streams are excluded.

Detailed Explanation: Determine present value of an investment

To find the amount needed now to reach a future value, reverse the compound-growth formula:

P=A(1+rn)ntP=\frac{A}{\left(1+\frac{r}{n}\right)^{nt}}

Here:

  • PP is the present value, or amount invested now.
  • AA is the future value.
  • rr is the annual interest rate written as a decimal.
  • nn is the number of compounding periods per year.
  • tt is the number of years.

Example

How much should be invested today to have 10{,}000in3yearsiftheaccountearnsin 3 years if the account earns6%$ interest compounded quarterly?

Step 1: Identify the values.

A=10,000,r=0.06,n=4,t=3A=10{,}000,\qquad r=0.06,\qquad n=4,\qquad t=3

The rate is 0.060.06 because 6%=0.066\%=0.06, and quarterly compounding means 44 times per year.

Step 2: Substitute into the formula.

P=10,000(1+0.064)4(3)P=\frac{10{,}000}{\left(1+\frac{0.06}{4}\right)^{4(3)}}

Step 3: Simplify the exponent and rate per period.

P=10,000(1.015)12P=\frac{10{,}000}{(1.015)^{12}}

There are 1212 total compounding periods because 4⋅3=124\cdot 3=12.

Step 4: Calculate.

P≈10,0001.1956P\approx \frac{10{,}000}{1.1956} P≈8,363.88P\approx 8{,}363.88

Therefore, about 8{,}363.88$ must be invested today.

The investment grows from 8{,}363.88toto$10{,}000$ over 3 years. In general, for a fixed future value, a higher interest rate or a longer investment period means a smaller amount is needed today.

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Exponential Function Solving - Compound Interest (Discrete) Equation to Starting Value


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