Let's break the problem step by step. ### Given: - Nominal annual interest rate: \( 7.5\% \) **convertible every five days** - There are 365 days in a year. We are to find the **sum** of: 1. The **force of interest** (\(\delta\)) 2. The **nominal annual discount rate compounded quarterly** (\(d^{(4)}\)) and **round the sum to four decimal places**. --- ## Step 1: Find the effective annual rate (\(i_{eff}\)) #### 1. **Number of conversion periods in a year:** \[ n = \frac{365}{5} = 73 \] #### 2. **Nominal rate convertible every 5 days:** Nominal rate per period = \( \frac{0.075}{73} \) So, the **effective annual rate**: \[ i_{eff} = \left(1 + \frac{0.075}{73}\right)^{73} - 1 \] Calculate: First, \( \frac{0.075}{73} \approx 0.00102739726 \) So, \[ i_{eff} = (1.00102739726)^{73} - 1 \] Calculate \( (1.00102739726)^{73} \): Let's use the natural logarithm for precision: \[ \ln(1.00102739726) \approx 0.001027 \] \[ 73 \times 0.001027 = 0.075 \] \[ e^{0.075} \approx 1.077884 \] So, \[ i_{eff} \approx 1.077884 - 1 = 0.077884 \] But let’s check with more precision: \[ (1.00102739726)^{73} = e^{73 \times \ln(1.00102739726)} \] \[ \ln(1.00102739726) \approx 0.00102687 \] \[ 73 \times 0.00102687 = 0.075162 \] \[ e^{0.075162} \approx 1.078143 \] \[ i_{eff} \approx 1.078143 - 1 = 0.078143 \] So, **effective annual rate:** \[ i_{eff} \approx 0.07814 \] --- ## Step 2: Find the **force of interest** (\(\delta\)) \[ \delta = \ln(1 + i_{eff}) \] Plug in \( i_{eff} \): \[ \delta = \ln(1 + 0.078143) = \ln(1.078143) \approx 0.075240 \] --- ## Step 3: Find the **nominal annual discount rate compounded quarterly** (\(d^{(4)}\)) Nominal annual **discount** rate compounded quarterly: Let \( d^{(4)} \) be the nominal annual discount rate, compounded quarterly. We know: - The **effective annual rate** is \( i_{eff} \). - The **quarterly discount rate** (\(d_{qu}\)) is such that: \[ (1 - d_{qu})^4 = 1 - i_{eff} \] But actually, the nominal rate is \( d^{(4)} = 4d_{qu} \). But more commonly, we relate the effective rate to the nominal discount rate by: \[ 1 + i_{eff} = \left(1 - \frac{d^{(4)}}{4}\right)^{-4} \] So, \[ \left(1 - \frac{d^{(4)}}{4}\right)^{-4} = 1 + i_{eff} \] \[ 1 - \frac{d^{(4)}}{4} = (1 + i_{eff})^{-1/4} \] \[ \frac{d^{(4)}}{4} = 1 - (1 + i_{eff})^{-1/4} \] \[ d^{(4)} = 4\left[1 - (1 + i_{eff})^{-1/4}\right] \] Plug in \( i_{eff} \): \[ d^{(4)} = 4\left[1 - (1.078143)^{-0.25}\right] \] Calculate \( (1.078143)^{-0.25} \): First, \( \ln(1.078143) = 0.07524 \) So, \[ -0.25 \times 0.07524 = -0.01881 \] \[ e^{-0.01881} \approx 0.98136 \] So, \[ d^{(4)} = 4 \times [1 - 0.98136] = 4 \times 0.01864 = 0.07456 \] --- ## Step 4: **Sum the two values** \[ \delta + d^{(4)} = 0.07524 + 0.07456 = 0.14980 \] --- ## **Final Answer:** \[ \boxed{0.1498} \] (Rounded to four decimal places.) --- ### **Summary Table** | Quantity | Value (rounded to 5 d.p.) | |----------|--------------------------| | \(\delta\) (force of interest) | 0.07524 | | \(d^{(4)}\) (nominal annual discount, quarterly) | 0.07456 | | **Sum** | **0.14980** | --- **The sum, rounded to four decimal places, is** \(\boxed{0.1498}\).