Problem 112
Question
BUSINESS: Capital Value of an Asset The capital value of an asset (such as an oil well) that produces a continuous stream of income is the sum of the present value of all future earnings from the asset. Therefore, the capital value of an asset that produces income at the rate of \(r(t)\) dollars per year (at a continuous interest rate \(i\) ) is $$ \left(\begin{array}{c} \text { Capital } \\ \text { value } \end{array}\right)=\int_{0}^{T} r(t) e^{-i t} d t $$ where \(T\) is the expected life (in years) of the asset. Use the formula in the preceding instructions to find the capital value (at interest rate \(i=0.05)\) of a uranium mine that produces income at the rate of \(r(t)=560,000 t^{1 / 2}\) dollars per year for 20 years.
Step-by-Step Solution
VerifiedKey Concepts
Capital Value
The purpose of calculating capital value is to determine how much an asset is worth in today’s dollars. This is especially relevant for businesses that want to compare the profitability of different investment opportunities. The formula given by the exercise, \[\text{Capital value} = \int_{0}^{T} r(t) e^{-i t} d t \]addresses this by integrating the income function \(r(t)\) over time, from the start to the end of the asset’s useful life. This continuous stream of discounted future income values provides a single figure representing the asset's capital value, adjusted for interest rates.
Continuous Income Stream
Understanding continuous income streams is vital because it allows for a more complex and realistic valuation of business operations. By expressing the income as a function of time \(t\), businesses can better predict earnings and perform financial planning.
- The income stream is consistent and coincides with real-world scenarios where businesses earn revenue evenly over time.
- Calculating the present value of this stream helps in understanding how much it is worth today, especially under the influence of interest rates.
Integration Techniques
When evaluating such integrals, especially those involving products of polynomial and exponential functions like \(560,000 t^{1/2} e^{-0.05 t}\), specific integration techniques are often necessary:
- Integration by Parts: Useful for integrals of products of functions, helping to break down complex expressions.
- Numerical Methods: Ideal when an integral cannot be solved analytically. Tools such as computational software allow for the approximation of integral values.