/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 36 Find the present value of a \(\$... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the present value of a \(\$ 20,000\) payment to be made in 10 years. Assume an interest rate of \(3.2 \%\) per year compounded continuously.

Short Answer

Expert verified
The present value is approximately $14,522.

Step by step solution

01

Understanding the Problem

We need to find the present value (PV) of a future payment of $20,000 that is due in 10 years. The interest rate is 3.2% per annum, compounded continuously. This requires using the formula for continuous compounding.
02

Using the Present Value Formula for Continuous Compounding

The formula for calculating the present value when compounded continuously is:\[ PV = FV imes e^{-rt} \]where \( FV = 20,000 \), \( r = 0.032 \), and \( t = 10 \).
03

Substitute Given Values into the Formula

Replacing the given values in the formula, we have:\[ PV = 20000 imes e^{-0.032 imes 10} \]
04

Calculate the Exponential Term

First, calculate the value of the exponent:\(-0.032 imes 10 = -0.32\). Then calculate the exponential term:\( e^{-0.32} \).
05

Compute the Present Value

Substitute the value of \( e^{-0.32} \) into the formula to find the present value\[ PV \approx 20000 imes e^{-0.32} \approx 20000 imes 0.7261 \approx 14522 \].
06

Summarize the Result

The present value of the $20,000 payment due in 10 years, with a continuously compounded interest rate of 3.2%, is approximately $14,522.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Continuous Compounding
Continuous compounding is a method used to calculate how much an investment will grow over time, considering the most frequent compounding interval possible. Unlike regular compounding, which might occur annually, quarterly, or even daily, continuous compounding assumes that interest is added an infinite number of times per period. This makes it a powerful tool for exponential growth calculations.
To understand how continuous compounding works, think about your money making more money.
  • The smaller the compounding interval, the more often interest is calculated, resulting in more money at the end.
  • With continuous compounding, the compounding doesn't stop; it's like calculating interest between each heartbeat.
  • This method uses the natural exponential function, denoted as "e," to simplify calculations.
In the context of present value, continuous compounding allows us to find out what a future amount is worth today. Assessing it often involves breaking down the formula: \[ PV = FV \times e^{-rt} \] where "FV" is the future value, "r" is the interest rate, and "t" is the time in years.
Exponential Functions
Exponential functions play a crucial role in continuous compounding. They're a type of mathematical function characterized by a constant base raised to a variable exponent. The base "e" is approximately equal to 2.71828, known as Euler's number.
These functions are essential because they model growth and decay processes, like populations or bank balances, efficiently. In finance, exponential functions are used to calculate the growth of investments over time when interest is compounded continuously.
  • The key property here is that the function grows at a rate proportional to its current value, leading to exponential growth.
  • In our present value problem, the exponential term \(e^{-0.032 \times 10}\) acts as a scaling factor, reducing the future payment to its present value.
  • Understanding this function's properties helps make sense of how exponentially compounding interest shrinks the future value to something tangible today.
By grasping exponential functions, you will find it easier to understand not just financial formulas but also natural phenomena.
Interest Rates
Interest rates are the backbone of many financial calculations, serving as the cost of borrowing or the gain from lending in percentage terms. Continuous compounding uses a specific annual interest rate to determine how investments grow or shrink over time.
In our example, we have a rate of 3.2% per annum, a typical setting for moderate growth scenarios.
  • The rate determines the acceleration of your investment's or debt's value when combined with time.
  • This example specifically uses a 3.2% rate to find out the present value of a future payment.
  • The rate translates into a growth factor through the exponent \(-rt\) in the formula \(PV = FV \times e^{-rt}\).
Understanding interest rates is essential because even a minor change can lead to significant differences in final values, especially over long periods.
Thus, whether in investments or loans, knowing how to work with rates like 3.2% can greatly aid in financial planning.

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Most popular questions from this chapter

The following formulas give the populations of four different towns, \(A, B, C,\) and \(D,\) with \(t\) in years from now. $$\begin{array}{rl} P_{A}=600 e^{0.08 t} & P_{B}=1000 e^{-0.02 t} \\ P_{C}=1200 e^{0.03 t} & P_{D}=900 e^{0.12 t} \end{array}$$ (a) Which town is growing fastest (that is, has the largest percentage growth rate)? (b) Which town is the largest now? (c) Are any of the towns decreasing in size? If so, which one(s)?

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