Present Value: Continuous Compounding - Business Calculus

1. A perpetual income stream pays $\$A$ dollars continuously per year, with the first payment starting immediately. If the continuous discount rate is $r$, and after $n$ years the payment rate increases to $\$kA$ dollars per year forever, what is the present value of this income stream?
   Answer: $\$\frac{A}{r} + \frac{A(k-1)e^{-rn}}{r}$ The present value consists of two parts: (1) A/r for the perpetual stream at rate A, and (2) the present value of the increase from A to kA starting at time n. The increase is (k-1)A per year starting at time n, which has present value (k-1)A/r discounted back n years: (k-1)A·$e^{-rn}$/r. Total PV = A/r + A(k-1)$e^{-rn}$/r. Choice B double-counts the base payment. Choice C has incorrect algebraic form. Choice D has wrong sign and structure.
2. An asset provides a continuous income stream at a constant rate $C$. An analyst calculates its present value over $T$ years, $PV_T$, and its present value over $2T$ years, $PV_{2T}$. It is found that $PV_{2T}$ is exactly 50% larger than $PV_T$. Based on this information, what is the value of the discount factor $e^{-rT}$?
   Answer: $0.50$ The present value of a constant income stream $C$ over a period of length $L$ is $PV_L = \int_0^L C e^{-rt} dt = \frac{C}{r}(1 - e^{-rL})$. We are given that $PV_{2T} = 1.5 \cdot PV_T$. Substituting the formula: $\frac{C}{r}(1 - e^{-r(2T)}) = 1.5 \cdot \frac{C}{r}(1 - e^{-rT})$. The $\frac{C}{r}$ terms cancel. Let $x = e^{-rT}$. Then $e^{-2rT} = (e^{-rT})^2 = x^2$. The equation becomes $1 - x^2 = 1.5(1 - x)$. Since $1-x^2 = (1-x)(1+x)$, we have $(1-x)(1+x) = 1.5(1-x)$. Since $T>0$ and $r>0$, $x=e^{-rT}$ cannot be 1, so we can divide by $(1-x)$. This leaves $1+x = 1.5$, which gives $x = 0.5$. Therefore, $e^{-rT} = 0.5$.
3. A new tech startup projects its revenue will be generated at a rate of $f(t) = 50 + 10t$ thousand dollars per year, where $t$ is the number of years from now. Assuming an annual interest rate of 4% compounded continuously, what is the present value of this income stream over the first 10 years, rounded to the nearest thousand dollars?
   Answer: $797,000 The present value (PV) is calculated by the integral $PV = \int_{0}^{T} f(t)e^{-rt} dt$. Here, $f(t) = 50 + 10t$, $r = 0.04$, and $T = 10$. The integral is $PV = \int_{0}^{10} (50 + 10t)e^{-0.04t} dt$. This requires integration by parts. Let $u = 50 + 10t$ and $dv = e^{-0.04t} dt$. Then $du = 10 dt$ and $v = -25e^{-0.04t}$. The antiderivative is $(-7500 - 250t)e^{-0.04t}$. Evaluating from $t=0$ to $t=10$: $[(-7500 - 250(10))e^{-0.04(10)}] - [(-7500 - 0)e^{0}] = -10000e^{-0.4} + 7500 \approx -10000(0.67032) + 7500 = -6703.2 + 7500 = 796.8$. In thousands of dollars, this is $796,800, which rounds to $797,000.
4. Two investment options are being compared. Option A provides $\$40,000$ per year continuously for $6$ years. Option B provides $\$20,000$ per year continuously for $6$ years, followed by $\$60,000$ per year continuously for another $6$ years. Using a $5\%$ continuous discount rate, what is the difference in present values (Option B minus Option A)?
   Answer: $\$89,247$ When comparing investment streams with continuous cash flows, you need to calculate the present value of each option using continuous discounting. This involves integrating the cash flow function multiplied by the discount factor $e^{-rt}$. For Option A, the present value is $\int_0^6 40000e^{-0.05t} dt = 40000 \cdot \frac{1-e^{-0.3}}{0.05} = \$209,424$ Option B has two phases: $\int_0^6 20000e^{-0.05t} dt + \int_6^{12} 60000e^{-0.05t} dt$ The first integral equals $20000 \cdot \frac{1-e^{-0.3}}{0.05} = \$104,712$ For the second integral, factor out $e^{-0.3}$: $60000e^{-0.3} \int_0^6 e^{-0.05u} du = 60000e^{-0.3} \cdot \frac{1-e^{-0.3}}{0.05} = \$193,959$ Option B's total present value is $\$104,712 + \$193,959 = \$298,671$ The difference (Option B minus Option A) is $\$298,671 - \$209,424 = \$89,247$, confirming answer C. Answer A ($82,914) likely results from calculation errors in the exponential terms. Answer B ($76,183) probably comes from incorrectly applying discrete rather than continuous discounting formulas. Answer D ($95,672) suggests errors in handling the two-phase integration for Option B, possibly from incorrect bounds or discount factor application. Study tip: For continuous cash flow problems, remember that the present value formula $\int_a^b Ce^{-rt} dt = \frac{C}{r}(e^{-ra} - e^{-rb})$ is your foundation. Always check that your exponential calculations are precise, as small errors compound significantly.
5. A company wants to pre-fund a marketing campaign that will cost $50,000 per year, paid as a continuous stream. They have allocated a budget of $400,000, which is the present value of the total campaign cost. If the company can earn 5% interest, compounded continuously, on their funds, for how many years $T$ can they fund the campaign with this budget?
   Answer: 10.2 years The present value is given by $PV = \int_{0}^{T} C e^{-rt} dt$. We have $PV=400000$, $C=50000$, and $r=0.05$. We need to solve for $T$. $400000 = \int_{0}^{T} 50000e^{-0.05t} dt = 50000[\frac{e^{-0.05t}}{-0.05}]_0^T = -1000000[e^{-0.05T} - 1]$. Dividing by -1,000,000 gives $-0.4 = e^{-0.05T} - 1$, so $0.6 = e^{-0.05T}$. Taking the natural logarithm of both sides: $\ln(0.6) = -0.05T$. Since $\ln(0.6) \approx -0.5108$, we have $-0.5108 = -0.05T$. Solving for $T$ gives $T = \frac{-0.5108}{-0.05} \approx 10.216$ years.
6. A business investment generates cash flows according to $\$C(t) = 30,000e^{-0.02t}$ dollars per year for $12$ years. If an investor requires a $8\%$ continuous return, but inflation is expected to be $2\%$ continuously compounded, what is the real present value of this investment?
   Answer: $\$243,890$ When you encounter present value problems with both required returns and inflation, you need to calculate the real present value using the real discount rate. The real discount rate accounts for both your required return and inflation's erosive effect on purchasing power. First, find the real discount rate. With an 8% required return and 2% inflation (both continuously compounded), the real rate is $r_{real} = 0.08 + 0.02 = 0.10$ or 10%. You add these rates because you need extra return to compensate for inflation on top of your base requirement. Next, calculate the present value using the continuous cash flow formula: $PV = \int_0^{12} 30,000e^{-0.02t} \cdot e^{-0.10t} \, dt$ Combining the exponentials: $PV = 30,000 \int_0^{12} e^{-0.12t} \, dt$ Evaluating: $PV = 30,000 \left[\frac{e^{-0.12t}}{-0.12}\right]_0^{12} = 30,000 \cdot \frac{1 - e^{-1.44}}{0.12} = \$243,890$ Choice A ($221,156) likely uses an incorrect real rate calculation, possibly subtracting inflation from the required return. Choice B ($198,765) appears to use too high a discount rate, perhaps incorrectly combining the rates. Choice C ($267,432) might represent using only the 8% required return without properly accounting for inflation. Study tip: Remember that real discount rates require adding the required return and inflation rate when both are continuously compounded. This reflects the need for returns above inflation to maintain purchasing power.
7. An oil well's production rate is $\$P(t) = 50,000e^{-0.08t}$ barrels per year, and oil sells for $80 per barrel. The well will be productive for$20$years. If the required rate of return is$9\%$compounded continuously, and extraction costs are$25 per barrel, what is the net present value of the well?
   Answer: $\$12,647,059$ This problem tests your understanding of net present value (NPV) calculations with continuous cash flows and exponential decay. When evaluating long-term investments like oil wells, you need to account for both changing production rates and the time value of money. Start by finding the net revenue per barrel: $80 - $25 = $55. The annual cash flow function becomes $C(t) = 55 \times 50,000e^{-0.08t} = 2,750,000e^{-0.08t}$. To find NPV, you must discount this continuous cash flow stream at the required 9% rate: $NPV = \int_0^{20} 2,750,000e^{-0.08t} \cdot e^{-0.09t} dt = \int_0^{20} 2,750,000e^{-0.17t} dt$ Evaluating this integral: $NPV = 2,750,000 \times \frac{1-e^{-0.17 \times 20}}{0.17} = 2,750,000 \times \frac{1-e^{-3.4}}{0.17} \approx 12,647,059$ Answer C ($12,647,059) correctly applies both the exponential production decline and continuous discounting. Answer A ($10,394,118) likely uses an incorrect discount rate or production function. Answer B ($15,823,824) probably fails to account for extraction costs or uses simple rather than continuous compounding. Answer D ($14,235,647) might calculate gross rather than net cash flows or use an inappropriate integration method. Remember: NPV problems with continuous cash flows require careful attention to three elements: the correct net cash flow function, proper discounting for the time value of money, and accurate integration over the specified time period.
8. The total profit $P(t)$ from a new product line, in millions of dollars, is projected to grow at a rate of $P'(t) = 2 - 0.2t$ for the first five years ($0 \le t \le 5$). Find the present value of this profit stream over the five-year period, assuming an interest rate of 5% compounded continuously.
   Answer: $6.73 million The present value is $PV = \int_{0}^{5} (2 - 0.2t)e^{-0.05t} dt$. Integration by parts ($u = 2 - 0.2t$, $dv = e^{-0.05t}dt$) gives the antiderivative $(40 + 4t)e^{-0.05t}$. Evaluating the definite integral: $[(40+4(5))e^{-0.05(5)}] - [(40+0)e^0] = 60e^{-0.25} - 40 \approx 60(0.7788) - 40 = 46.728 - 40 = 6.728$. So the present value is approximately $6.73 million.
9. A new automated system generates revenue at a constant rate of $250,000 per year. Its maintenance costs are modeled by the function $C(t) = 20,000 + 5,000t$, where $t$ is in years. The system has a useful life of 10 years. If the company uses a discount rate of 7% compounded continuously, what is the present value of the net income stream generated by the system?
   Answer: $1,497,000 The net income stream is $f(t) = \text{Revenue} - \text{Cost} = 250000 - (20000 + 5000t) = 230000 - 5000t$. We need to find the present value: $PV = \int_{0}^{10} (230000 - 5000t)e^{-0.07t} dt$. This can be solved by splitting the integral or using integration by parts. A simpler way is to find the PV of revenue and subtract the PV of costs. $PV_{rev} = \int_{0}^{10} 250000e^{-0.07t}dt \approx 1,800,058$. $PV_{cost} = \int_{0}^{10} (20000+5000t)e^{-0.07t}dt \approx 302,674$. The net present value is $1,800,058 - 302,674 = 1,497,384$, which is approximately $1,497,000.
10. A company is selling a patent that is expected to generate a continuous income stream modeled by $f(t) = 150,000e^{-0.04t}$ dollars per year, where $t$ is in years. This income is expected to continue indefinitely. What is the present value of this perpetual income stream, assuming an interest rate of 6% compounded continuously?
    Answer: $1,500,000 The present value of a perpetual stream is found using an improper integral: $PV = \int_{0}^{\infty} f(t)e^{-rt} dt$. Substituting the given functions: $PV = \int_{0}^{\infty} 150000e^{-0.04t} e^{-0.06t} dt = \int_{0}^{\infty} 150000e^{-0.10t} dt$. Evaluating the integral: $PV = [\frac{150000}{-0.10}e^{-0.10t}]_0^{\infty} = [-1500000e^{-0.10t}]_0^{\infty}$. As $t \to \infty$, $e^{-0.10t} \to 0$. So, the evaluation is $0 - ($-1500000e^0$) = 1,500,000.
11. A company's cash flow is modeled as $\$F(t) = 25,000e^{0.04t}$ dollars per year. If the company wants to find the present value of cash flows from year $3$ to year $8$ using a continuous discount rate of $7\%$, which integral correctly represents this calculation?
    Answer: $\$\int_3^8 25,000e^{0.04t} \cdot e^{-0.07t} dt$ For present value of cash flows from year 3 to year 8, we integrate the cash flow function F(t) multiplied by the discount factor $e^{-rt}$ from t=3 to t=8. This gives ∫₃⁸ 25,$000e^{0.04t}$·$e^{-0.07t}$dt. Choice B is mathematically equivalent but unnecessarily complex. Choice C incorrectly adjusts the discount factor time reference. Choice D incorrectly changes the integration limits and applies an extra discount factor.
12. A project requires an initial investment of $221,000. It is expected to generate a continuous income stream of $30,000 per year for 10 years. To the nearest percent, what is the annual interest rate $r$, compounded continuously, that makes the present value of the income stream equal to the initial investment?
    Answer: 6% The present value formula is $PV = \int_{0}^{T} C e^{-rt} dt = \frac{C}{r}(1 - e^{-rT})$. We are given $PV=221000$, $C=30000$, and $T=10$. The equation to solve is $221000 = \frac{30000}{r}(1 - e^{-10r})$. Since this cannot be solved algebraically for $r$, we must test the given interest rates. For $r=0.06$ (6%), $PV = \frac{30000}{0.06}(1 - e^{-10(0.06)}) = 500000(1 - e^{-0.6}) \approx 500000(1 - 0.54881) = 225,595$. For $r=0.07$ (7%), $PV = \frac{30000}{0.07}(1 - e^{-10(0.07)}) \approx 428571(1 - e^{-0.7}) \approx 428571(1 - 0.49659) = 215,745$. The value $225,595 is closer to the target investment of $221,000 than $215,745 is. Thus, 6% is the best approximation.
13. An investment will generate a continuous cash flow of $\$C(t) = 10,000 + 2,000t$ dollars per year for $10$ years. If the discount rate is $5\%$ compounded continuously, what is the present value of the investment to the nearest thousand dollars?
    Answer: $\$142,000$ PV = ∫₀¹⁰ (10,000 + 2,000t)$e^{-0.05t}$dt = ∫₀¹⁰ 10,$000e^{-0.05t}$dt + ∫₀¹⁰ 2,$000te^{-0.05t}$dt. The first integral equals 10,000[$e^{-0.05t}$/(-0.05)]₀¹⁰ ≈ 78,732. The second requires integration by parts: 2,000∫$te^{-0.05t}$dt ≈ 63,212. Total PV ≈ $141,944 ≈ $142,000. Choice A ignores discounting. Choice C uses wrong discount rate. Choice D incorrectly applies integration by parts.
14. A lease agreement specifies rent payments of $\$R(t) = 36,000(1.03)^t$ dollars per year, paid continuously for $10$ years. However, due to a discount clause, the effective payment rate is $\$0.95R(t)$. If money is worth $6\%$ compounded continuously, what is the present value of the lease payments?
    Answer: $\$267,389$ Effective payment rate = 0.95 × 36,000$(1.03)^t$ = 34,200$(1.03)^t$ = 34,$200e^{ln(1.03)t}$ = 34,$200e^{0.02956t}$. PV = ∫₀¹⁰ 34,$200e^{0.02956t}$·$e^{-0.06t}$dt = ∫₀¹⁰ 34,$200e^{-0.03044t}$dt = 34,200[1-$e^{-0.3044}$]/0.03044 ≈ $267,389. Choice A ignores the 0.95 discount factor. Choice C uses discrete compounding conversion. Choice D uses ln(1.03) ≈ 0.03 instead of exact value 0.02956.
15. An investor is considering two projects. Project A generates revenue at a constant rate of $20,000 per year. Project B generates revenue at a rate of $f(t) = 15000e^{0.02t}$ dollars per year. Both projects run for 8 years. If the prevailing interest rate is 5% compounded continuously, which project has a higher present value, and by approximately how much?
    Answer: Project A has a higher present value, by approximately $25,187. First, calculate the present value of Project A: $PV_A = \int_{0}^{8} 20000e^{-0.05t} dt = 20000[\frac{e^{-0.05t}}{-0.05}]_0^8 = -400000[e^{-0.4} - 1] \approx -400000(0.67032 - 1) = 131,872$. Next, calculate the present value of Project B, combining the exponents: $PV_B = \int_{0}^{8} 15000e^{0.02t}e^{-0.05t} dt = \int_{0}^{8} 15000e^{-0.03t} dt = 15000[\frac{e^{-0.03t}}{-0.03}]_0^8 = -500000[e^{-0.24} - 1] \approx -500000(0.78663 - 1) = 106,685$. Comparing the two, Project A has a higher present value. The difference is $131,872 - 106,685 = 25,187.
16. Consider two perpetual income streams. Stream A provides a constant $C$ dollars per year. Stream B's initial rate is also $C$ dollars per year, but it decays at a rate $k$, such that its flow is $C e^{-kt}$. Both are discounted at the same continuous interest rate $r$, where $r > k > 0$. What is the ratio of the present value of Stream B to the present value of Stream A?
    Answer: $\frac{r}{r+k}$ The present value of perpetual Stream A is $PV_A = \int_0^\infty C e^{-rt} dt = \frac{C}{r}$. The present value of perpetual Stream B is $PV_B = \int_0^\infty (C e^{-kt}) e^{-rt} dt = \int_0^\infty C e^{-(r+k)t} dt = \frac{C}{r+k}$. The ratio of B to A is $\frac{PV_B}{PV_A} = \frac{C/(r+k)}{C/r} = \frac{C}{r+k} \cdot \frac{r}{C} = \frac{r}{r+k}$.

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