# Present Value of a 10-Year Continuously Increasing Arithmetic Annuity **Given:** - Annual effective interest rate \( i = 8\% \) - Payment rate increases continuously at \( 400 \) per year - Annuity term: \( 10 \) years - Payment rate at time \( t \): \( 400t \) (starts at \( \) and increases linearly) ## Step 1: Express the Payment Function At any time \( t \) (in years), the payment rate is: \[ \text{Payment at time } t = 400t \] ## Step 2: Present Value Formula The present value (PV) is the integral of the discounted payments over the 10-year period: \[ PV = \int_{}^{10} 400t \cdot v^t \, dt \] where \( v^t = e^{-\delta t} \) and \( \delta \) is the force of interest. ### Find the Force of Interest \( \delta \): \[ 1 + i = e^{\delta} \implies \delta = \ln(1 + i) = \ln(1.08) \approx .076961 \] ## Step 3: Plug in Values \[ PV = \int_{}^{10} 400t \cdot e^{-.076961 t} \, dt \] ## Step 4: Solve the Integral The integral is of the form \( \int t e^{-a t} dt \), where \( a = .076961 \). ### The antiderivative is: \[ \int t e^{-a t} dt = -\frac{t}{a} e^{-a t} - \frac{1}{a^2} e^{-a t} + C \] ### Apply limits from to 10: \[ PV = 400 \left[ -\frac{t}{a} e^{-a t} - \frac{1}{a^2} e^{-a t} \right]_{}^{10} \] \[ = 400 \left( \left[ -\frac{10}{a} e^{-10a} - \frac{1}{a^2} e^{-10a} \right] - \left[ - \frac{1}{a^2} \cdot 1 \right] \right) \] \[ = 400 \left( -\frac{10}{a} e^{-10a} - \frac{1}{a^2} e^{-10a} + \frac{1}{a^2} \right) \] ## Step 5: Substitute \( a = .076961 \) Calculate \( e^{-10a} \): \[ 10a = .76961 \] \[ e^{-10a} = e^{-.76961} \approx .463 \] Now substitute: - \( a = .076961 \) - \( a^2 = .005927 \) - \( \frac{1}{a} = 12.991 \) - \( \frac{1}{a^2} = 168.77 \) \[ PV = 400 \left( -10 \times 12.991 \times .463 - 168.77 \times .463 + 168.77 \right) \] \[ = 400 \left( -129.91 \times .463 - 78.17 + 168.77 \right) \] Calculate \( -129.91 \times .463 = -60.22 \): \[ = 400 \left( -60.22 - 78.17 + 168.77 \right) \] \[ = 400 \times (30.38) \] \[ = 12,152 \] ## Step 6: Final Answer \[ \boxed{12,\!152} \] --- ### **Summary Table** | Parameter | Value | |-----------|-----------| | \( i \) | 8% | | \( \delta \) | .076961 | | Term | 10 years | | Increase rate | 400 per year | | Present Value | \( 12,\!152 \) | --- ## **Conclusion** The present value at an annual effective rate of 8% of this 10-year continuously increasing arithmetic annuity is **\$12,152**.

# Present Value of a 10-Year Continuously Increasing Arithmetic Annuity **Objective:** Calculate the present value of a 10-year continuously increasing arithmetic annuity with payments starting at 0 and increasing at a rate of 400 per year at an annual effective interest rate of 8%. ## Given Data - Annual effective interest rate: \( i = 8\% \) - Payment increase rate: \( 400 \) per year - Annuity duration: \( 10 \) years ## Step 1: Define the Payment Function The payment at time \( t \) (in years) is given by: \[ \text{Payment}(t) = 400t \] ## Step 2: Present Value Formula The present value (PV) can be expressed using the integral of the discounted payments over the annuity term: \[ PV = \int_0^{10} 400t \cdot v^t \, dt \] where \( v^t = e^{-\delta t} \) and \( \delta \) is the force of interest. ### Step 2.1: Calculate the Force of Interest \( \delta \) The force of interest is calculated as: \[ \delta = \ln(1 + i) = \ln(1.08) \approx 0.076961 \] ## Step 3: Substitute Values into the PV Formula Now substituting \( \delta \) into the present value formula: \[ PV = \int_0^{10} 400t \cdot e^{-0.076961 t} \, dt \] ## Step 4: Solve the Integral ### Step 4.1: Antiderivative Calculation The integral \( \int t e^{-at} dt \) has the antiderivative: \[ \int t e^{-at} dt = -\frac{t}{a} e^{-at} - \frac{1}{a^2} e^{-at} + C \] where \( a = 0.076961 \). ### Step 4.2: Apply Limits Calculating the definite integral from \( 0 \) to \( 10 \): \[ PV = 400 \left[-\frac{t}{a} e^{-at} - \frac{1}{a^2} e^{-at}\right]_0^{10} \] ### Step 4.3: Evaluate at the Limits Evaluating at \( t = 10 \): \[ PV = 400 \left[-\frac{10}{0.076961} e^{-10 \times 0.076961} - \frac{1}{(0.076961)^2} e^{-10 \times 0.076961}\right] \] Evaluating at \( t = 0 \) (where \( e^{0} = 1 \)): \[ + \frac{1}{(0.076961)^2} \] ### Step 4.4: Calculate \( e^{-10a} \) Calculate \( 10a \): \[ 10a = 0.76961 \quad \Rightarrow \quad e^{-10a} = e^{-0.76961} \approx 0.463 \] ### Step 4.5: Substitute Values Substituting \( a = 0.076961 \): - \( \frac{1}{a} \approx 12.991 \) - \( \frac{1}{a^2} \approx 168.77 \) Final substitution: \[ PV = 400 \left[-10 \times 12.991 \times 0.463 - 168.77 \times 0.463 + 168.77\right] \] ### Step 4.6: Simplify the Expression Calculating: \[ -10 \times 12.991 \times 0.463 \approx -60.22 \] \[ -168.77 \times 0.463 \approx -78.17 \] Combining: \[ PV = 400 \left[-60.22 - 78.17 + 168.77\right] \] \[ = 400 \times 30.38 \] ## Step 5: Final Calculation Calculating the final present value: \[ PV \approx 12152 \] ## Final Answer The present value at an annual effective rate of 8% of this 10-year continuously increasing arithmetic annuity is **\$12,152**.