A Ramp Cash Flow Function with Continuous Compounding. Consider a situation in which money flows according to the diagram shown below (Figure 4.13). Here, a cash flow increases uniformly over two years to accumulate $10,000. (a) What is the expression for the present worth at 10% interest

Question

A Ramp Cash Flow Function with Continuous Compounding

Consider a situation in which money flows according to the diagram shown below (Figure 4.13). Here, a cash flow increases uniformly over two years to accumulate $10,000.
(a) What is the expression for the present worth at 10% interest compounded continuously?
(b) Suppose that the rate of operating expenditure is essentially constant at $2,000 per year. Determine the present worth of operating profit over a life of 2 years.

Accepted Answer

Given: C = $10,000 over 2 years, r = 6% per year, compounded continuously, N = 2 years, and the rate of operating expenditure = $2,000 per year.
Find: (a) P of revenue; (b) P of operating profit.

(a) The ramp cash flow function in Figure 4.13 shows the total amount to be received over two years, (i.e., the area), but the slope of the ramp function is not given. Therefore, our first task is to determine the slope G. Since the area of the ramp function is given, we can find the unknown slope of the function as

[latex]\int_{0}^{2}{} Gt\cdot dt=\$10,000[/latex]

[latex]G\frac{t^{2} }{2} \mid ^{2}_{0} =2G=\$10,000[/latex]

G = $5,000

Substituting these values back into the formula in Table 4.4 yields

[latex]P= \int_{0}^{2}{5,000t\cdot e^{-0.06t} dt }[/latex]

[latex]= \frac{5,000}{0.06^{2} } \left(1-e^{-0.06\left(2\right) } \right) -\frac{5,000}{0.06 }\left(2e^{-0.06\left(2\right) } \right)[/latex]

= $9,234.88

(b) The present worth of operating profit over a life of two years is

[latex]P= \int_{0}^{2}{} \left(5000t+2000\right) e^{-0.06t }dt[/latex]

[latex]= \$9,234.88 - 2000 \int_{0}^{2}{} e^{-0.06t }dt[/latex]

[latex]= \$9,234.88 - 2000 \left[\frac{e^{0.06\left(2\right)-1 } }{0.06e^{0.06\left(2\right) } } \right][/latex]

= $9,234.88 - $3,769.32

= $5,465.56

Question 4.11: A Ramp Cash Flow Function with Continuous Compounding. Consi...

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