Problem 30 - Calculus and its applications

Question

Chrome Solutions determines that the rate of revenue coming in from a new machine is $$ R_{1}(t)=8000-100 t $$ in dollars per year, for \(8 \mathrm{yr},\) after which the machine will be replaced. The company learns that an alternative machine will yield revenue at a rate of $$ R_{2}(t)=7600-85 t $$ a) Find the accumulated present value of the income stream from each machine at an interest rate of \(5.8 \%\) compounded continuously. b) Find the difference in the accumulated present values.

Step-by-Step Solution

Answer

Evaluate the integrals to find \( PV_1 \) and \( PV_2 \), then compute \( \Delta PV = PV_1 - PV_2 \).

Step 1: Define the Accumulated Present Value Formula

For a continuously compounded interest rate, the present value \( PV \) of an income stream \( R(t) \) over time \( t \) is given by the formula:\[PV = \int_{0}^{T} R(t) e^{-rt} \, dt\]where \( r \) is the interest rate (in decimal form) and \( T \) is the duration in years. In this exercise, \( r = 0.058 \), and \( T = 8 \).

Step 2: Calculate Present Value for Machine 1

Plug in \( R_1(t) = 8000 - 100t \) into the present value formula:\[PV_1 = \int_{0}^{8} (8000 - 100t) e^{-0.058t} \, dt\]You can solve this integral using integration by parts or using a calculator to evaluate it numerically.

Step 3: Solve Integral for Machine 1

Evaluating the integral for \( PV_1 \), you find:\[PV_1 \approx 8000 \times \frac{1 - e^{-0.058 \times 8}}{0.058} - 100 \times \left( \frac{8 \times e^{-0.058 \times 8} + e^{-0.058 \times 8} - 1}{0.058^2} \right)\]Solve this expression to find the exact present value of the income from Machine 1.

Step 4: Calculate Present Value for Machine 2

Similarly, for \( R_2(t) = 7600 - 85t \), the present value \( PV_2 \) is:\[PV_2 = \int_{0}^{8} (7600 - 85t) e^{-0.058t} \, dt\]Evaluate this integral similarly to find the present value of the income from Machine 2.

Step 5: Solve Integral for Machine 2

Evaluate the integral for \( PV_2 \):\[PV_2 \approx 7600 \times \frac{1 - e^{-0.058 \times 8}}{0.058} - 85 \times \left( \frac{8 \times e^{-0.058 \times 8} + e^{-0.058 \times 8} - 1}{0.058^2} \right)\]After solving, you will have the present value for Machine 2.

Step 6: Calculate the Difference in Present Values

Finally, compute the difference between the present values for the two machines:\[\Delta PV = PV_1 - PV_2\]This value will indicate the financial advantage or disadvantage of choosing Machine 1 over Machine 2, considering the cost of capital.

Key Concepts

Present Value

In the world of finance, present value (PV) is a fundamental concept used to determine the current worth of future cash flows. It allows us to evaluate how much a future income stream is worth today, given a specific discount rate. The discount rate is often derived from market interest rates to consider the time value of money.
For example, if you were to receive $100 a year from now, its worth today would be less because money has the potential to earn interest or returns over time. Thus, the concept of present value helps to account for that potential growth.
Present value is crucial when analyzing investment opportunities or comparing cash flows generated by different projects, as seen in our exercise. It's calculated using the formula:

• Traditional formula: \[ PV = \frac{FV}{(1 + r)^n} \]
• Continuous compounding formula: \[ PV = \int_0^T R(t) e^{-rt} \, dt \]

Here, \( R(t) \) represents the revenue stream, \( r \) the interest rate (in decimal form), and \( T \) the time horizon. Calculating PV under continuous compounding helps to ensure that all time periods within the analytical framework are evaluated fairly.

Continuous Compounding

Continuous compounding is an elegant mathematical concept that reflects how interest accumulates if it is added to the principal balance continuously. Unlike regular compounding at discrete intervals (like annually or quarterly), continuous compounding lets interest grow at every possible moment.
The formula for continuous compounding is expressed as:\[ A = P e^{rt} \]where:

• \( A \) is the amount after time \( t \)
• \( P \) is the principal amount
• \( r \) is the interest rate (as a decimal)
• \( t \) is the time
• \( e \) is the base of the natural logarithm, approximately equal to 2.71828

Continuous compounding is particularly beneficial in fields like finance and economics, offering a more precise calculation around the growth rate of investments or loans. In the exercise, continuous compounding applies when adjusting the income streams of the machines to today's terms over a period of 8 years.

Integration by Parts

Integration by parts is a mathematical technique used to solve integrals where the standard integration methods are not easily applicable. Originating from the product rule of differentiation, it allows us to integrate products of functions.
The integration by parts formula is expressed as:\[ \int u \, dv = uv - \int v \, du \]where:

• \( u \) and \( dv \) are parts of the original integral
• \( du \) and \( v \) are the derivatives and integrals of these parts

Choosing the right functions for \( u \) and \( dv \) is critical as it simplifies solving the integral. This technique is essential in financial mathematics, often used to evaluate complex functions like the continuously compounded present value integrals in our exercise.
By applying integration by parts to our revenue functions, we compute the precise present value of each machine. This provides deeper insights into the financial returns of different investments over time.