Problem 15 - Applied Calculus

Question

Your company is considering buying new production machinery. You want to know how long it will take for the machinery to pay for itself; that is, you want to find the length of time over which the present value of the profit generated by the new machinery equals the cost of the machinery. The new machinery costs \(\$ 130,000\) and earns profit at the continuous rate of \(\$ 80,000\) per year. Use an interest rate of \(8.5 \%\) per year compounded continuously.

Step-by-Step Solution

Answer

It will take approximately 1.74 years for the machinery to pay for itself.

Step 1: Identify the financial terms

The cost of the new machinery is \( C = \\(130,000 \). The profit it generates is at a continuous rate of \( P = \\)80,000 \) per year, and the annual interest rate is \( r = 8.5\% = 0.085 \). We need to find the time \( t \) when the present value of the profit equals the cost of the machinery.

Step 2: Present value of continuous profit stream

The present value \( PV \) of a continuous profit stream of rate \( P \) over time \( t \) with continuous compounding at rate \( r \) is given by the formula: \[ PV = \int_{0}^{t} P e^{-rt} dt = \frac{P}{r} (1 - e^{-rt}) \].

Step 3: Set the equation to solve for time

We need the present value of profits to equal the cost, hence: \[ \frac{P}{r} (1 - e^{-rt}) = C \]. Substitute the known values: \[ \frac{80,000}{0.085} (1 - e^{-0.085t}) = 130,000 \].

Step 4: Simplify the equation

Calculate the fraction: \( \frac{80,000}{0.085} = 941,176.47 \). Therefore, \[ 941,176.47 (1 - e^{-0.085t}) = 130,000 \].

Step 5: Solve for the exponential term

Divide both sides by \( 941,176.47 \): \[ 1 - e^{-0.085t} = \frac{130,000}{941,176.47} \]. Calculate the right-hand side: \( \frac{130,000}{941,176.47} \approx 0.138 \). Thus, \[ 1 - e^{-0.085t} \approx 0.138 \].

Step 6: Isolate the exponential and solve for \( t \)

Rearrange the equation: \( e^{-0.085t} = 1 - 0.138 = 0.862 \). Take the natural logarithm on both sides to solve for \( t \): \[ -0.085t = \ln(0.862) \].

Step 7: Compute the time \( t \)

Calculate the natural logarithm: \( \ln(0.862) \approx -0.148 \). Thus, \(-0.085t = -0.148\). Solve for \( t \) by dividing: \[ t = \frac{-0.148}{-0.085} \approx 1.74 \].

Step 8: Conclusion: Time needed for payback

It will take approximately 1.74 years for the machinery to pay for itself through the profit it generates.

Key Concepts

Continuous Compounding

Continuous compounding is a concept from finance where the calculation of interest happens in an infinitely small intervals.
In contrast to regular compounding, which happens monthly or annually, continuous compounding assumes that interest is being added continuously without pause. This allows for the calculation of a more accurate interest amount over time.
Key aspects of continuous compounding include:

• The formula used is: \[ A = Pe^{rt} \]where,
  • \(A\) is the amount of money accumulated after n years, including interest,
  • \(P\) is the principal amount,
  • \(r\) is the annual interest rate (decimal), and
  • \(t\) is the time the money is invested or borrowed for, in years.
• Interest is calculated using exponential functions, leveraging Euler's number (\(e\)), which is approximately 2.718.

Continuous compounding is beneficial when calculating present value because it provides a more precise view of the investment's returns over time, especially in cases like the machinery purchase where returns are steady and calculable.

Exponential Growth

Exponential growth describes a situation where the rate of growth is proportional to the size of the current amount.
This creates a rapid increase as the quantity multiplies over time, rather than increasing at a constant rate.
In context, it often refers to situations where there's a continuously increasing effect—whether in finance, population growth, or other phenomena.In finance, exponential growth can be visualized clearly:

• Continuous compounding is a primary example of exponential growth.
• Value grows exponentially with time since the interest earned also earns interest, effectively compounding over an infinite number of periods.
• Formula used includes the exponential constant \(e\), making it ideal for calculating processes involving growth over time, like continuous profit streams.

The machinery in your exercise generates profits at a continuous rate, resembling exponential growth since each year adds a bit more due to the effect of compounding. Understanding exponential growth is crucial for predicting long-term returns and assessing how quickly investments can mature.

Payback Period

Payback period is a straightforward financial metric that determines how long it takes for an investment to "pay back" its initial cost.
It measures the time required for the cash inflows to repay the original investment.
This metric is particularly useful for budgeting and assessing project viability. Here are some important points about the payback period:

• It does not account for the time value of money directly, but it is a quick way to assess investment risk.
• Shorter payback periods are preferred as they reduce exposure to risk and uncertainty.
• It's calculated by assessing the point at which cumulative cash flows equal the initial investment.
• For continuous earnings—like the machinery—it involves computing the present value of future earnings at each point in time.

In your exercise, the payback period was calculated using continuous compounding for precise results, ensuring you understand when the real profitability of your investment begins according to present value terms.