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Chapter 6: Problem 44

GE NERAL: Permanent Endowments Show that the size of the permanent endowment needed to generate an annual \(C\) dollars forever at interest rate \(r\) compounded continuously is \(C / r\) dollars.

Short Answer

Expert verified
The permanent endowment needed is \( \frac{C}{r} \).

Step by step solution

01

Understand the Concept of a Permanent Endowment

A permanent endowment aims to provide a set amount, \( C \) dollars, annually indefinitely. This requires calculating a principal that will yield \( C \) dollars a year forever when invested at a continuous interest rate \( r \).
02

Formula for Continuous Compounding

The formula for the amount \( A \) in continuous compounding is given by \( A = Pe^{rt} \), where \( P \) is the principal, \( r \) is the interest rate, and \( t \) is the time. This formula helps understand how money grows continuously over time.
03

Determine the Endowment's Annual Yield

To maintain \( C \) dollars annually from an endowment, an equivalent amount must be derived annually at the interest rate \( r \). For continuous compounding, the interest from principal \( P \) in one year is \( Pr \). Thus, the needed yield is \( Pr = C \).
04

Solve for the Principal \( P \)

From \( Pr = C \), solving for \( P \) gives \( P = \frac{C}{r} \). This represents the permanent endowment required to generate \( C \) dollars annually at continuous compound interest.
05

Conclusion on the Endowment Size

So, the endowment size necessary to generate a constant annual amount \( C \) at a continuous interest rate \( r \) is \( \frac{C}{r} \). This supports the endowment sustainability indefinitely.

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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 concept where the interest on an investment is calculated and added to the principal continuously, rather than at specific intervals like monthly or annually. This means that the investment grows infinitely faster as it earns "interest on interest" at every possible moment.

There are a few key points about continuous compounding that make it unique:
  • The formula used is: \[A = Pe^{rt}\]where:
    • \(A\) is the amount of money accumulated after time \(t\),
    • \(P\) is the principal amount (initial investment),
    • \(r\) is the annual interest rate (in decimal), and
    • \(t\) is the time the money is invested for, in years.
  • In this formula, \(e\) is the base of natural logarithms, approximately equal to 2.71828.
  • Continuous compounding results in a larger total amount compared to periodic compounding.
This method showcases how even small amounts of principal investment, with sufficient rate and time, can grow significantly. Understanding this formula is crucial for those interested in how investments can maximize returns over time.
Interest Rate
The interest rate is the percentage at which your investment grows over a certain period. It plays a vital role in calculating returns for investments and endowments.

A deeper understanding of interest rates involves:
  • It represents the cost of borrowing money or the reward for saving money.
  • When compounded continuously, it transforms how quickly money grows, beyond regular compounding periods.
  • Informs decision-making about financial products, investments, and savings strategies.
In the case of endowments, the interest rate indicates how much the original investment can yield annually. For example, if an endowment is set with a continuous interest rate of \(r\), this means that each dollar invested returns an annual interest of \(r\) dollars, which is crucial for calculating the necessary endowment size.
Principal Calculation
Principal calculation is a fundamental step in understanding how much you need to initially fund an endowment or investment to reach your financial goals.

When it comes to permanent endowments, you determine the principal by defining the amount you wish to draw each year, \(C\). Using the continuous interest rate, \(r\), you can use the relationship between these factors through the formula:
  • \[Pr = C\]
Solving for \(P\) gives:
  • \[P = \frac{C}{r}\]
Highlights of this calculation are:
  • \(P\) is the initial investment necessary to secure a continuous yield of \(C\) dollars annually.
  • This ensures sustainability, allowing the endowment to provide a permanent annual benefit.
  • Shows how an apparently small interest rate requires a correspondingly larger principal to maintain a desired income.
Understanding principal calculation is essential in assessing the feasibility and sustainability of financial goals through investments.

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Most popular questions from this chapter

Evaluate each definite integral using integration by parts. (Leave answers in exact form.) $$ \int_{0}^{4} z(z-4)^{6} d z $$

$$ \text { Use integration by parts to find each integral. } $$ $$ \int \frac{\ln t}{t^{2}} d t $$

$$ \text { Use integration by parts to find each integral. } $$ $$ \int \frac{\ln (x+1)}{\sqrt{x+1}} d x $$

In the reservoir model, the heart is viewed as a balloon that swells as it fills with blood (during a period called the systole), and then at time \(t_{0}\) it shuts a valve and contracts to force the blood out (the diastole). Let \(p(t)\) represent the pressure in the heart at time \(t\) a. During the diastole, which lasts from \(t_{0}\) to time \(T, p(t)\) satisfies the differential equation $$\frac{d p}{d t}=-\frac{K}{R} p$$ Find the general solution \(p(t)\) of this differential equation. ( \(K\) and \(R\) are positive constants determined, respectively, by the strength of the heart and the resistance of the arteries. The differential equation states that as the heart contracts, the pressure decreases \((d p / d t\) is negative) in proportion to itself.) b. Find the particular solution that satisfies the condition \(p\left(t_{0}\right)=p_{0} . \quad\left(p_{0}\right.\) is a constant representing the pressure at the transition time \(t_{0}\).) c. During the systole, which lasts from time 0 to time \(t_{0}\), the pressure \(p(t)\) satisfies the differential equation $$\frac{d p}{d t}=K I_{0}-\frac{K}{R} p$$ Find the general solution of this differential equation. \(\left(I_{0}\right.\) is a positive constant representing the constant rate of blood flow into the heart while it is expanding.) [Hint: Use the same \(u\) -substitution technique that was used in Example 7.] d. Find the particular solution that satisfies the condition \(p\left(t_{0}\right)=p_{0}\). e. In parts (b) and (d) you found the formulas for the pressure \(p(t)\) during the diastole \(\left(t_{0} \leq t \leq T\right)\) and the systole \(\left(0 \leq t \leq t_{0}\right)\). Since the heart behaves in a cyclic fashion, these functions must satisfy \(p(T)=p(0)\). Equate the solutions at these times (use the correct formula for each time) to derive the important relationship $$ R=\frac{p_{0}}{I_{0}} \frac{1-e^{-K T / R}}{1-e^{-K t_{0} / R}} $$

Derive each formula by using integration by parts on the left-hand side. (Assume \(n>0 .\) ) $$ \int(\ln x)^{n} d x=x(\ln x)^{n}-n \int(\ln x)^{n-1} d x $$
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