Let's break down the problem: - **Maureen purchases a laddered portfolio of bonds for \$20,918.** - **Each bond has face value \(X\) and coupon rate 6.8% (paid semiannually).** - **There are 10 bonds, each maturing (redeemable) at the end of years 1 through 10.** - **Yield on the portfolio is 8% annual effective rate.** - **Find \(X\) (the face value of each bond).** Let's proceed step by step. --- ## 1. **Coupon payments** - Each bond has annual coupons of \( 6.8\% \) of face value \(X\), but paid semiannually. - Therefore, each coupon = \(\frac{6.8\%}{2} \times X = 3.4\% \times X\), paid twice per year. --- ## 2. **Portfolio cash flows** For bond maturing at end of year \(k\) (\(k=1\) to \(10\)): - Pays coupon every 6 months (semiannual) for \(2k\) periods. - Pays principal (\(X\)) at the end of period \(2k\). So the cash flows for the entire portfolio are: - At each semiannual period (\(t=1\) to \(20\)) the coupons are paid by all *outstanding* bonds. - At periods \(t=2,4,...,20\) (end of each year), the principal from one bond is paid (since one bond matures each year). Let’s write the semiannual periods as \(t=1\) to \(20\) (where \(t=1\) is 6 months, \(t=2\) is 1 year, ..., \(t=20\) is 10 years). At time \(t\): - The number of bonds outstanding = \(10 - \left\lfloor \frac{t-1}{2} \right\rfloor\) - Each outstanding bond pays coupon \(0.034X\) But let's structure the total coupon at each period: - For \(t=1,2,...,20\), the number of bonds outstanding = \(10-(t-1)//2\) - So at \(t=1\), 10 bonds pay coupon: \(10 \times 0.034X = 0.34X\) - At \(t=2\), 10 bonds pay coupon: \(0.34X\) - At \(t=3\), 9 bonds remain, pay coupon: \(0.306X\) - At \(t=4\), 9 bonds pay: \(0.306X\) - etc. But at the end of each year (i.e., \(t=2,4,...,20\)), one bond matures and pays principal \(X\). So at \(t=2n\) (end of year \(n\)): - Coupon: \( (10-n+1) \times 0.034X \) - Principal: \(X\) --- ## 3. **Discounting Cash Flows** Yield = 8% annual effective = \(i_{eff} = 0.08\) The equivalent semiannual rate \(i_{SA}\) is: \[ 1 + i_{eff} = (1 + i_{SA})^2 \implies 1.08 = (1 + i_{SA})^2 \] \[ 1 + i_{SA} = \sqrt{1.08} \implies i_{SA} = \sqrt{1.08} - 1 \approx 0.039230 \] --- ## 4. **Present Value Equation** Let’s sum up the present value of all the cash flows and set equal to the purchase price: \[ PV = \sum_{k=1}^{10} PV\text{(bond maturing in year $k$)} \] Each bond: - Pays coupon \(0.034X\) at times \(t=1\) to \(2k\) - Pays principal \(X\) at time \(t=2k\) So, for the bond maturing at year \(k\): \[ PV_k = 0.034X \cdot v^1 + 0.034X \cdot v^2 + \ldots + 0.034X \cdot v^{2k} + X \cdot v^{2k} \] \[ = 0.034X \cdot (v^1 + v^2 + \ldots + v^{2k}) + X \cdot v^{2k} \] \[ = 0.034X \cdot \frac{v - v^{2k+1}}{1-v} + X \cdot v^{2k} \] But let's use the sum of a geometric series for the coupons: \[ PV_k = 0.034X \cdot \frac{1 - v^{2k}}{1 - v} + X v^{2k} \] where \(v = \frac{1}{1+i_{SA}}\). But actually, the coupons are paid at times \(t=1\) to \(2k\), and the principal at \(t=2k\). So, for all 10 bonds: \[ PV = \sum_{k=1}^{10} \left[ 0.034X \cdot \frac{1 - v^{2k}}{1 - v} + X v^{2k} \right] \] But notice that if we sum the coupon payments over all bonds, at each time \(t\) (from 1 to 20), the number of outstanding bonds is \(10 - \lfloor (t-1)/2 \rfloor\), as previously stated. But it's easier to sum over the 10 bonds as above. --- ## 5. **Plug in the numbers** First, calculate \(i_{SA}\): \[ i_{SA} = \sqrt{1.08} - 1 \approx 0.039230 \] \[ v = \frac{1}{1 + i_{SA}} \approx \frac{1}{1.039230} \approx 0.962250 \] Now, compute: \[ PV = \sum_{k=1}^{10} \left[ 0.034X \cdot \frac{1 - v^{2k}}{1 - v} + X v^{2k} \right] = 20918 \] Let’s split this sum: \[ PV = 0.034X \cdot \frac{1}{1-v} \sum_{k=1}^{10} (1 - v^{2k}) + X \sum_{k=1}^{10} v^{2k} \] \[ = 0.034X \cdot \frac{1}{1-v} \left[ 10 - \sum_{k=1}^{10} v^{2k} \right] + X \sum_{k=1}^{10} v^{2k} \] \[ = 0.034X \cdot \frac{10}{1-v} - 0.034X \cdot \frac{1}{1-v} \sum_{k=1}^{10} v^{2k} + X \sum_{k=1}^{10} v^{2k} \] \[ = 0.034X \cdot \frac{10}{1-v} + \left[1 - 0.034 \cdot \frac{1}{1-v}\right] X \sum_{k=1}^{10} v^{2k} \] Let’s compute all these terms. ### Compute \(v\), \(1-v\), \(\frac{1}{1-v}\): \[ v = 0.962250 \\ 1-v = 0.037750 \\ \frac{1}{1-v} = 26.493 \] ### Compute \(\sum_{k=1}^{10} v^{2k}\): Note that \(v^{2k} = (v^2)^k\), so this is a geometric series with ratio \(r = v^2\): \[ v^2 = (0.96225)^2 = 0.926929 \] \[ \sum_{k=1}^{10} v^{2k} = v^2 + v^4 + \ldots + v^{20} = \frac{v^2(1 - (v^2)^{10})}{1 - v^2} \] Compute: \[ (v^2)^{10} = (0.926929)^{10} = e^{10 \ln(0.926929)} \approx e^{-0.7604} \approx 0.4676 \] \[ 1 - (v^2)^{10} = 1 - 0.4676 = 0.5324 \] \[ 1 - v^2 = 1 - 0.926929 = 0.073071 \] \[ \sum_{k=1}^{10} v^{2k} = \frac{0.926929 \times 0.5324}{0.073071} = \frac{0.4934}{0.073071} \approx 6.749 \] --- Now, plug into the \(PV\) equation: \[ PV = 0.034X \cdot 26.493 \cdot 10 + \left[1 - 0.034 \cdot 26.493\right] X \cdot 6.749 \] First term: \[ 0.034 \times 26.493 \times 10 = 0.034 \times 264.93 = 9.007 \] Second term: \[ 0.034 \times 26.493 = 0.9007 \] \[ 1 - 0.9007 = 0.0993 \] \[ 0.0993 \times 6.749 = 0.670 \] So, \[ PV = 9.007X + 0.670X = 9.677X \] Set equal to 20,918: \[ 9.677X = 20,918 \implies X = \frac{20,918}{9.677} \approx 2,161.65 \] --- ## **Final Answer** \[ \boxed{X \approx 2,162} \] (Rounded to the nearest dollar.) --- ### **Summary of Calculation Steps** 1. Find semiannual yield (\(i_{SA}\)): \(0.039230\) 2. Find \(v = 0.962250\), \(v^2 = 0.926929\) 3. Compute present value of all cash flows, sum as geometric series. 4. Solve for \(X\): \[ X \approx \$2,162 \]