Question

Capitalized Cost In Exercises 51 and 52, find the capitalized cost \(C\) of an asset \((a)\) for \(n=5\) years, \((b)\) for \(n=10\) years, and (c) forever. The capitalized cost is given by \(C=C_{0}+\int_{0}^{n} c(t) e^{-r t} d t\) where \(C_{0}\) is the original investment, \(t\) is the time in years, \(r\) is the annual interest rate compounded continuously, and \(c(t)\) is the annual cost of maintenance (measured in dollars). [Hint: For part (c), see Exercises \(35-38 .]\) $$ C_{0}=\$ 650,000, c(t)=25,000, r=10 \% $$

Short answer

The solution requires using the given formula to solve for the capitalized cost for 5 years, 10 years and an infinite time period. It involves performing integration using the provided annual cost of maintenance and the annual interest rate for each time period.

Step-by-step solution

Define the Given Variables

The original investment \(C_{0}\) is $650,000, the annual cost of maintenance \(c(t)\) is $25,000 and the annual interest rate \(r\) is 10 % or 0.1 in decimal.

Use the Formula for 5 Years

For 5 years, \(n=5\). Subsitute these values in the given formula to get \(C=650000 + \int_{0}^{5} 25000 e^{-0.1t} dt\). This integral can be solved using basic laws of integration.

Use the Formula for 10 Years

Similarly, for 10 years, \(n=10\). Replace these values in the given formula to get \(C=650000 + \int_{0}^{10} 25000 e^{-0.1t} dt\). This integral is solved in the same manner as the previous one.

Use the Formula for an Infinite Time

For an infinite time, \(n\) is infinity. So, the equation turns out to be \(C=650000 + \int_{0}^{\infty} 25000 e^{-0.1t} dt\). This integral, while a bit more complex due to the infinity, can be solved using similar laws of integration as before.

Key concepts

Continuous Compounding

Continuous compounding is a powerful concept in finance where an investment grows continuously at an interest rate. Unlike simple or compound interest, which compound at discrete intervals, continuous compounding assumes that the compounding process is constant and never stops.

Mathematically, we represent continuous compounding with the natural exponential function, denoted as \( e \). The formula for continuous compounding is \( A = Pe^{rt} \), where \( A \) is the amount of money accumulated after time \( t \), \( P \) is the principal amount, \( r \) is the annual interest rate, and \( t \) is time in years. This formula is derived from the limit of compound interest as the number of compounding periods per year approaches infinity.

Continuous compounding is especially relevant in the exercise as it allows us to calculate the present value of future maintenance costs accurately. By discounting these costs at a continuously compounded rate, the correct capitalized cost of an asset over time can be estimated.

Present Value of Maintenance Costs

The present value of maintenance costs refers to the current worth of all future maintenance expenses, discounted at a specific interest rate. In our exercise, maintenance costs are paid annually, and the interest rate is continuous. Thus, we use an integral with a discount factor to find the present value.

To calculate this, the exercise utilizes the formula \( C = C_{0} + \int_{0}^{n} c(t) e^{-rt} dt \) where \( C_{0} \) is the original investment made, \( c(t) \) is the annual cost of maintenance, \( r \) is the continuous interest rate, and \( n \) is the number of years. The exponential part \( e^{-rt} \) represents the continuous discounting factor that brings future costs back to their present value.

Why do we use this method? Because it reflects the reduced value of future money, acknowledging that a dollar today is worth more than a dollar tomorrow. This is the core tenet of the time value of money.

Integration in Economics

Integration is a fundamental operation in calculus that is widely applied in economics. It helps us sum up the infinite number of infinitesimally small contributions to a total quantity, like the area under a curve.

In the context of the capitalized cost exercise, integration calculates the entire area under the curve of maintenance costs over time. The function \( c(t) \) represents yearly maintenance costs, which are then integrated over a period of time while being discounted at a continuous rate represented by the term \( e^{-rt} \).

This process determines the entire value of all maintenance payments in today's dollars, also known as the present value. By integrating\( c(t) e^{-rt} \) from \( t = 0 \) to \( t = n \), economists can determine the financial impact of future obligations on present economic decisions, crucial for investment appraisal and capital budgeting.

Time Value of Money

The time value of money is a fundamental financial principle stating that money available today is worth more than the same amount in the future due to its potential earning capacity. It highlights the benefits of receiving money now rather than later.

This core concept underlies the practice of discounting in the exercise. By calculating the present value, we reflect the time value of money by reducing the worth of future maintenance costs to account for the interest that could have been earned if the money had been received today.

When calculating the capitalized cost of an asset, considering the time value of money through discounting ensures that we are accurately comparing costs and benefits that occur at different times. Essentially, it provides a common ground to make financially sound decisions over different time horizons.