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Chapter 5: Problem 50

The capitalized cost, \(c,\) of an asset over its lifetime is the total of the initial cost and the present value of all maintenance expenses that will occur in the future. It is computed with the formula $$ c=c_{0}+\int_{0}^{L} m(t) e^{-k t} d t $$ where \(c_{0}\) is the initial cost of the asset, \(L\) is the lifetime (in years), \(k\) is the interest rate (compounded continuously), and \(m(t)\) is the annual cost of maintenance. Find the capitalized cost under each set of assumptions. $$ \begin{array}{l} c_{0}=\$ 600,000, k=4 \% \\ m(t)=\$ 40,000+\$ 1000 e^{0.01 t}, L=40 \end{array} $$

Short Answer

Expert verified
The capitalized cost is approximately $915,217.44.

Step by step solution

01

Analyze the Given Problem

We need to find the capitalized cost of an asset over its lifetime. The formula to compute this is \( c = c_0 + \int_0^L m(t) e^{-kt} dt \). Given values are: \( c_0 = \\(600,000, k=0.04, L=40, \) and \( m(t) = \\)40,000 + \$1,000 e^{0.01t} \).
02

Substitute Values into the Formula

Substitute the given values into the formula for \( c \). This gives us:\[c = 600,000 + \int_0^{40} (40,000 + 1,000 e^{0.01t}) e^{-0.04t} dt\]
03

Simplify the Maintenance Function

Split the integral into two separate integrals:\[\int_0^{40} 40,000 e^{-0.04t} dt + \int_0^{40} 1,000 e^{0.01t} e^{-0.04t} dt\] This simplifies to:\[\int_0^{40} 40,000 e^{-0.04t} dt + \int_0^{40} 1,000 e^{-0.03t} dt\]
04

Solve the First Integral

Calculate \( \int_0^{40} 40,000 e^{-0.04t} dt \):\[40,000 \int_0^{40} e^{-0.04t} dt = 40,000 \left[ -\frac{e^{-0.04t}}{0.04} \right]_0^{40}\]Solve the definite integral:\[= 40,000 \times \left(-\frac{1}{0.04}\right) (e^{-0.04 \times 40} - 1)\]\[= -1,000,000(e^{-1.6} - 1)\]
05

Solve the Second Integral

Calculate \( \int_0^{40} 1,000 e^{-0.03t} dt \):\[1,000 \int_0^{40} e^{-0.03t} dt = 1,000 \left[-\frac{e^{-0.03t}}{0.03}\right]_0^{40}\]Solve the definite integral:\[= 1,000 \times \left(-\frac{1}{0.03}\right) (e^{-0.03 \times 40} - 1)\]\[= -33,333.33(e^{-1.2} - 1)\]
06

Calculate the Capitalized Cost

Sum the obtained values from the integrals with the initial cost:\[c = 600,000 - 1,000,000(e^{-1.6} - 1) - 33,333.33(e^{-1.2} - 1)\] Use a calculator to find the exponential terms and compute the total. Using values:\[e^{-1.6} \approx 0.2019 \quad \text{and} \quad e^{-1.2} \approx 0.3012\]The calculated cost is:\[c \approx 600,000 - 1,000,000(0.2019 - 1) - 33,333.33(0.3012 - 1)\]

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Key Concepts

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

Present Value
Present Value is an essential concept in finance used to determine the current worth of a sum of money that will be received in the future, considering a specific interest rate. It reflects how much future money is worth today.
Learn more about Present Value.
  • **Formula:**
    The present value (PV) can be calculated using the formula for continuous compounding: \[ PV = FV \times e^{-kt} \] where \( FV \) is the future value, \( e \) is the base of the natural logarithm, \( k \) is the interest rate, and \( t \) is the time in years.
  • **Application in Capitalized Cost:**
    In the context of capitalized cost, the present value is used to discount future maintenance expenses back to their value in today's terms. This ensures that all costs are comparable when determining the total lifetime expense of an asset.
    Understanding present value helps us grasp why future payments carry less weight than their present equivalents, especially when compounded over time.
    • Continuous Compounding
    Continuous Compounding refers to calculating interest on an account so that it鈥檚 compounded at an infinite number of times per year. This concept is crucial in finance to maximize the potential of investment returns.
    What is Continuous Compounding?
  • **Calculation with Continuous Compounding:**
    Continuous compounding modifies the traditional formula for compound interest to: \[ A = P \times e^{kt} \] where \( A \) is the amount of money accumulated after \( n \) years, including interest, \( P \) is the principal amount, \( k \) is the rate of interest, and \( t \) is the time period.
  • **Use in Capitalized Cost Formula:**
    When evaluating capitalized cost, the interest rate \( k \) is used to discount future maintenance costs. This continuous nature of compounding affects the final values significantly, making it crucial to apply it consistently to obtain an accurate cost assessment.
    Leveraging continuous compounding allows more precise calculations, aligning closely with real-world financial complexities.
    • Maintenance Expenses
    Maintenance Expenses represent the recurring costs required to operate and maintain an asset over its lifespan. These expenses can vary and are influenced by factors such as usage intensity, age of the asset, and prevailing economic conditions.
    Learn about Maintenance & Repairs
  • **Understanding Maintenance Cost Functions:**
    In the formula for the capitalized cost, \( m(t) \) denotes the maintenance cost function, which can fluctuate over time. For instance, in our original problem, \( m(t) \) includes both a fixed cost \( \\(40,000 \) and a time-dependent term \( \\)1,000 e^{0.01t} \), reflecting increasing costs over time.
  • **Incorporating Maintenance into Cost Calculations:**
    Analyzing maintenance expenses helps in accurate financial planning. In continuous compounding, integrating \( m(t) \) over the asset's lifetime allows us to determine the accumulated cost effect.
    By accounting for future costs today, stakeholders can make informed decisions on the viability and sustainability of asset management strategies.
    • Integral Calculus
    Integral Calculus is a branch of calculus focusing on the aggregation of quantities, such as areas under curves, effectively used in various engineering and economic applications.
    Discover more about Integral Calculus
  • **Role in Capitalized Cost Calculation:**
    The capitalized cost formula employs integral calculus to accumulate maintenance expenses, computed continuously over the asset's life span. It uses definite integrals to sum up these costs efficiently.
  • **Breaking Down the Integration Process:**
    In step-by-step solutions, integration plays a pivotal role in computing costs involving changing rates, like maintenance expenses. The combined techniques of splitting integrals and different exponent rules are often necessary to solve real problems.
    Properly applying these techniques ensures precise financial evaluations, enhancing our understanding of the long-term financial implications of asset procurement and maintenance. Integral calculus helps visualize and compute these complex costs over time, allowing for comprehensive asset management.

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    Most popular questions from this chapter

    The value of a share of stock of Leslie's Designs, Inc., is modeled by $$ \frac{d V}{d t}=k(L-V) $$ where \(V\) is the value of the stock, in dollars, after \(t\) months; \(k\) is a constant; \(L=\$ 24.81,\) the limiting value of the stock; and \(V(0)=20 .\) Find the solution of the differential equation in terms of \(t\) and \(k\).

    Let \(x\) be a continuous random variable that is normally distributed with mean \(\mu=22\) and standard deviation \(\sigma=5 .\) Using Table A, find the following. $$ P(18 \leq x \leq 26) $$

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    Let \(x\) be a continuous random variable that is normally distributed with mean \(\mu=22\) and standard deviation \(\sigma=5 .\) Using Table A, find the following. $$ P(19 \leq x \leq 25) $$

    Let \(x\) be a continuous random variable with a standard normal distribution. Using Table A, find each of the following. $$ P(1.35 \leq x \leq 1.45) $$
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