Let's break down the problem and solve for \( q \): ### **Given:** - **Bond term:** 10 years - **Face value:** $1,000 - **Purchase price:** $962 - **Redemption amount:** $1,150 (at maturity) - **Coupon schedule:** - **First 5 years:** Nominal rate 8% **annual**, paid semiannually \(\rightarrow\) each coupon is \( \frac{8\%}{2} \times \$1,000 = \$40 \). - **Last 5 years:** Nominal rate \( q\% \) (annual), paid semiannually \(\rightarrow\) each coupon is \( \frac{q\%}{2} \times \$1,000 \). - **Yield:** Annual effective rate \( i_{eff} = 7.35\% \). --- ## 1. **Find the semiannual yield \( i \) corresponding to \( 7.35\% \) effective annual** \[ (1 + i)^2 = 1.0735 \implies 1 + i = \sqrt{1.0735} \implies i = \sqrt{1.0735} - 1 \] \[ i = 1.036029 - 1 = 0.036029 \] \[ i \approx 3.6029\% \] --- ## 2. **Set up the present value equation** Let \( C_1 = \$40 \) (first 10 coupons), \( C_2 = \frac{q\%}{2} \times 1{,}000 = \$5q \) (last 10 coupons). There are 20 periods (10 years, semiannual). \[ 962 = \sum_{k=1}^{10} \frac{40}{(1+i)^k} + \sum_{k=11}^{20} \frac{5q}{(1+i)^k} + \frac{1{,}150}{(1+i)^{20}} \] --- ### **PV of first 10 coupons (\$40)** This is a 10-term ordinary annuity: \[ PV_1 = 40 \cdot \frac{1 - (1+i)^{-10}}{i} \] ### **PV of last 10 coupons (\$5q)** These start at \( k=11 \), so it's a 10-term annuity-immediate, discounted by 10 more periods: \[ PV_2 = 5q \cdot \frac{1 - (1+i)^{-10}}{i} \cdot (1+i)^{-10} \] Alternatively, write as: \[ PV_2 = \sum_{k=11}^{20} \frac{5q}{(1+i)^k} = 5q \cdot \sum_{k=11}^{20} (1+i)^{-k} \] But the annuity formula above is more convenient. ### **PV of redemption (\$1,150) at end of period 20** \[ PV_3 = \frac{1,150}{(1+i)^{20}} \] --- ## 3. **Combine into equation** \[ 962 = 40 \cdot \frac{1 - (1+i)^{-10}}{i} + 5q \cdot \frac{1 - (1+i)^{-10}}{i} \cdot (1+i)^{-10} + \frac{1,150}{(1+i)^{20}} \] --- ## 4. **Plug in values** - \( i = 0.036029 \) - \( (1+i)^{-10} = (1.036029)^{-10} \) - \( (1+i)^{-20} = (1.036029)^{-20} \) Calculate: ### \( (1.036029)^{10} \approx \) \[ \ln(1.036029) \approx 0.035371 \\ 10 \times 0.035371 = 0.35371 \\ e^{0.35371} \approx 1.4243 \] So, \[ (1.036029)^{-10} = \frac{1}{1.4243} \approx 0.70198 \] ### \( (1.036029)^{20} = (1.4243)^2 \approx 2.0306 \) So, \[ (1.036029)^{-20} = \frac{1}{2.0306} \approx 0.4926 \] --- ### **Now plug in the numbers:** - \( i = 0.036029 \) - \( (1+i)^{-10} \approx 0.70198 \) - \( (1+i)^{-20} \approx 0.4926 \) #### **First 10 coupons:** \[ A = 40 \times \frac{1 - 0.70198}{0.036029} = 40 \times \frac{0.29802}{0.036029} = 40 \times 8.2690 = 330.76 \] #### **Last 10 coupons:** \[ B = 5q \times \frac{1 - 0.70198}{0.036029} \times 0.70198 = 5q \times 8.2690 \times 0.70198 = 5q \times 5.803 \] #### **Redemption:** \[ C = \frac{1,150}{2.0306} = 566.70 \] --- ### **Combine:** \[ 962 = 330.76 + 5q \times 5.803 + 566.70 \] \[ 962 = 897.46 + 5q \times 5.803 \] \[ 962 - 897.46 = 5q \times 5.803 \] \[ 64.54 = 5q \times 5.803 \] \[ q = \frac{64.54}{5 \times 5.803} = \frac{64.54}{29.015} = 2.224 \] \[ q \approx 2.22\% \] --- ## **Final Answer** \[ \boxed{q = 2.22\%} \] So, the nominal annual coupon rate for the final five years is **2.22%**.