**Let's solve the problem step by step:** Given: - Liability: 750,000 due in 4 years. - Assets: - X: cash flow \(A_X\) in 2 years - Y: 250,000 in 4 years - Z: \(A_Z\) in 5 years - Annual discount factor: \(v = 0.95\) - **Full immunization**: Present value and duration of assets must match liability. --- ### **Step 1: Present Value Equations** #### **Present Value of Liability** \[ PV_L = 750,000 \cdot v^4 \] #### **Present Value of Assets** \[ PV_A = A_X \cdot v^2 + 250,000 \cdot v^4 + A_Z \cdot v^5 \] Set present values equal: \[ A_X \cdot v^2 + 250,000 \cdot v^4 + A_Z \cdot v^5 = 750,000 \cdot v^4 \quad \cdots (1) \] --- ### **Step 2: Duration Equations** #### **Duration of Liability** \[ D_L = 4 \] \[ \text{Weighted duration of assets:} \] \[ \frac{2 A_X v^2 + 4 \cdot 250,000 v^4 + 5 A_Z v^5}{A_X v^2 + 250,000 v^4 + A_Z v^5} = 4 \] But for immunization (Macaulay duration), set the numerator of the duration term equal to the numerator for liability: \[ 2A_X v^2 + 4 \cdot 250,000 v^4 + 5A_Z v^5 = 4 \cdot 750,000 v^4 \quad \cdots (2) \] --- ### **Step 3: Substitute \(v = 0.95\) and \(v^2, v^4, v^5\):** \[ v^2 = 0.95^2 = 0.9025 \] \[ v^4 = 0.95^4 = (0.9025)^2 = 0.81450625 \] \[ v^5 = 0.95^5 = 0.7737809375 \] --- ### **Step 4: Plug Values into Equations** #### **Equation (1):** \[ A_X \cdot 0.9025 + 250,000 \cdot 0.81450625 + A_Z \cdot 0.7737809375 = 750,000 \cdot 0.81450625 \] \[ A_X \cdot 0.9025 + 203,626.5625 + A_Z \cdot 0.7737809375 = 610,879.6875 \] \[ A_X \cdot 0.9025 + A_Z \cdot 0.7737809375 = 610,879.6875 - 203,626.5625 \] \[ A_X \cdot 0.9025 + A_Z \cdot 0.7737809375 = 407,253.125 \] #### **Equation (2):** \[ 2A_X \cdot 0.9025 + 4 \cdot 250,000 \cdot 0.81450625 + 5A_Z \cdot 0.7737809375 = 4 \cdot 750,000 \cdot 0.81450625 \] \[ 2A_X \cdot 0.9025 + 1,016,312.25 + 5A_Z \cdot 0.7737809375 = 2,443,518.75 \] \[ 2A_X \cdot 0.9025 + 5A_Z \cdot 0.7737809375 = 2,443,518.75 - 1,016,312.25 \] \[ 2A_X \cdot 0.9025 + 5A_Z \cdot 0.7737809375 = 1,427,206.5 \] --- ### **Step 5: Solve the System of Equations** Let’s call: - \(a = 0.9025\) - \(b = 0.7737809375\) So: \[ a A_X + b A_Z = 407,253.125 \quad (i) \] \[ 2a A_X + 5b A_Z = 1,427,206.5 \quad (ii) \] Multiply equation (i) by 2: \[ 2a A_X + 2b A_Z = 814,506.25 \] Subtract from equation (ii): \[ (2a A_X + 5b A_Z) - (2a A_X + 2b A_Z) = 1,427,206.5 - 814,506.25 \] \[ 3b A_Z = 612,700.25 \] \[ A_Z = \frac{612,700.25}{3b} = \frac{612,700.25}{3 \times 0.7737809375} = \frac{612,700.25}{2.3213428125} \approx 263,914.6 \] Now substitute \(A_Z\) into equation (i): \[ a A_X + b \cdot 263,914.6 = 407,253.125 \] \[ a A_X = 407,253.125 - b \cdot 263,914.6 \] \[ a A_X = 407,253.125 - 0.7737809375 \cdot 263,914.6 \] \[ a A_X = 407,253.125 - 204,238.5 = 203,014.625 \] \[ A_X = \frac{203,014.625}{a} = \frac{203,014.625}{0.9025} \approx 225,000 \] --- ## **Final Answer** \[ \boxed{225,000} \] The correct choice is **D**.

Let's carefully work through the problem step by step to find the value of \(A_y\). --- ### **Given Data:** - **Liability:** 750,000 due in 4 years - **Assets:** - **X:** Cash flow \(A_x\) in 2 years - **Y:** Cash flow 250,000 in 4 years - **Z:** Cash flow \(A_z\) in 5 years - **Discount factor (annual):** \(v = 0.95\) --- ### **Step 1: Present Value (PV) of liability** \[ PV_{liability} = 750,000 \times v^4 \] Calculate \(v^4\): \[ v^4 = 0.95^4 = (0.95)^4 \approx 0.8145 \] Thus, \[ PV_{liability} = 750,000 \times 0.8145 \approx 610,875 \] --- ### **Step 2: Present value of assets** \[ PV_{assets} = A_x \times v^2 + 250,000 \times v^4 + A_z \times v^5 \] Calculate \(v^2\) and \(v^5\): \[ v^2 = 0.95^2 = 0.9025 \] \[ v^5 = 0.95^5 \approx 0.7738 \] Now, the PV equation: \[ A_x \times 0.9025 + 250,000 \times 0.8145 + A_z \times 0.7738 = 610,875 \] Calculate \(250,000 \times 0.8145\): \[ 250,000 \times 0.8145 \approx 203,625 \] So, \[ A_x \times 0.9025 + A_z \times 0.7738 = 610,875 - 203,625 = 407,250 \] *(Equation 1)*: \[ 0.9025 A_x + 0.7738 A_z = 407,250 \] --- ### **Step 3: Duration matching** The Macaulay duration of the liability is 4 years. For the assets, the weighted average time: \[ \frac{2 \times A_x \times v^2 + 4 \times 250,000 \times v^4 + 5 \times A_z \times v^5}{A_x \times v^2 + 250,000 \times v^4 + A_z \times v^5} = 4 \] Numerator: \[ 2 \times A_x \times 0.9025 + 4 \times 203,625 + 5 \times A_z \times 0.7738 \] Calculate \(4 \times 203,625\): \[ 4 \times 203,625 = 814,500 \] Numerator: \[ 1.805 A_x + 814,500 + 3.869 A_z \] Denominator (PV total): \[ A_x \times 0.9025 + 203,625 + A_z \times 0.7738 \] Set the duration equal to 4: \[ \frac{1.805 A_x + 814,500 + 3.869 A_z}{A_x \times 0.9025 + 203,625 + A_z \times 0.7738} = 4 \] Cross-multiplied: \[ (1.805 A_x + 814,500 + 3.869 A_z) = 4 \times (A_x \times 0.9025 + 203,625 + A_z \times 0.7738) \] Calculate RHS: \[ 4 \times 0.9025 A_x = 3.61 A_x \] \[ 4 \times 203,625 = 814,500 \] \[ 4 \times 0.7738 A_z = 3.095 A_z \] So: \[ 1.805 A_x + 814,500 + 3.869 A_z = 3.61 A_x + 814,500 + 3.095 A_z \] Subtract RHS from LHS: \[ (1.805 A_x - 3.61 A_x) + (814,500 - 814,500) + (3.869 A_z - 3.095 A_z) = 0 \] Simplify: \[ -1.805 A_x + 0 + 0.774 A_z = 0 \] Rearranged: \[ 1.805 A_x = 0.774 A_z \] \[ A_x = \frac{0.774}{1.805} A_z \approx 0.429 A_z \] --- ### **Step 4: Solve for \(A_z\) using Equation (1)** Recall Equation (1): \[ 0.9025 A_x + 0.7738 A_z = 407,250 \] Substitute \(A_x \approx 0.429 A_z\): \[ 0.9025 \times 0.429 A_z + 0.7738 A_z = 407,250 \] Calculate: \[ 0.9025 \times 0.429 \approx 0.387 \] So: \[ 0.387 A_z + 0.7738 A_z = 407,250 \] Sum coefficients: \[ (0.387 + 0.7738) A_z = 407,250 \] \[ 1.1608 A_z = 407,250 \] \[ A_z = \frac{407,250}{1.1608} \approx 350,756 \] Now, \(A_y\) is given as 250,000, which matches the problem statement. --- ### **Final Answer:** **The value of \(A_y\) is approximately \(\boxed{250,000}\).** **Therefore, the correct choice is:** > **E) 250,000** --- ### **Summary:** - The asset cash flow \(A_y\) is **250,000**. - The calculations confirm that the asset Y cash flow is 250,000, matching the given option. **Answer:** \(\boxed{250,000}\)

A

Let's analyze the problem step by step to verify the value of \(A_y\). --- ### **Given Data:** - Liability: **750,000** due in **4 years** - Assets: - **X:** Cash flow \(A_x\) in **2 years** - **Y:** Cash flow **250,000** in **4 years** - **Z:** Cash flow \(A_z\) in **5 years** - Discount factor per year: \(v = 0.95\) --- ### **Objective:** Determine \(A_y\), which in this setup is the cash flow from Asset Y. From the options, the known value for Asset Y is **250,000**, matching the problem statement. --- ### **Step 1: Compute Present Value of Liability** \[ PV_{liability} = 750,000 \times v^4 \] Calculate \(v^4\): \[ v^4 = 0.95^4 \approx 0.8145 \] Therefore, \[ PV_{liability} = 750,000 \times 0.8145 \approx 610,875 \] --- ### **Step 2: Present Value of Assets** \[ PV_{assets} = A_x \times v^2 + 250,000 \times v^4 + A_z \times v^5 \] Calculate \(v^2\) and \(v^5\): \[ v^2 = 0.95^2 = 0.9025 \] \[ v^5 = 0.95^5 \approx 0.7738 \] Now, \[ PV_{assets} = A_x \times 0.9025 + 250,000 \times 0.8145 + A_z \times 0.7738 \] Calculate \(250,000 \times 0.8145 \approx 203,625\). Set the PV of assets equal to the PV of liability: \[ A_x \times 0.9025 + A_z \times 0.7738 = 610,875 - 203,625 = 407,250 \] *(Equation 1)*: \[ 0.9025 A_x + 0.7738 A_z = 407,250 \] --- ### **Step 3: Duration Matching** The goal is to ensure the duration of the assets equals the liability duration (which is 4 years). The Macaulay duration of the assets is: \[ \frac{2 \times A_x \times v^2 + 4 \times 250,000 \times v^4 + 5 \times A_z \times v^5}{A_x \times v^2 + 250,000 \times v^4 + A_z \times v^5} = 4 \] Calculate numerator: \[ 2 \times A_x \times 0.9025 + 4 \times 203,625 + 5 \times A_z \times 0.7738 \] \[ = 1.805 A_x + 814,500 + 3.869 A_z \] The denominator (total PV): \[ A_x \times 0.9025 + 203,625 + A_z \times 0.7738 \] Set the ratio equal to 4: \[ \frac{1.805 A_x + 814,500 + 3.869 A_z}{A_x \times 0.9025 + 203,625 + A_z \times 0.7738} = 4 \] Cross-multiplied: \[ 1.805 A_x + 814,500 + 3.869 A_z = 4 \times (A_x \times 0.9025 + 203,625 + A_z \times 0.7738) \] Calculate RHS: \[ = 3.61 A_x + 814,500 + 3.095 A_z \] Now, equate: \[ 1.805 A_x + 814,500 + 3.869 A_z = 3.61 A_x + 814,500 + 3.095 A_z \] Subtract RHS from LHS: \[ (1.805 A_x - 3.61 A_x) + (814,500 - 814,500) + (3.869 A_z - 3.095 A_z) = 0 \] Simplify: \[ -1.805 A_x + 0 + 0.774 A_z = 0 \] Rearranged: \[ 1.805 A_x = 0.774 A_z \] \[ A_x = \frac{0.774}{1.805} A_z \approx 0.429 A_z \] --- ### **Step 4: Solve for \(A_z\)** Substitute \(A_x \approx 0.429 A_z\) into Equation 1: \[ 0.9025 \times 0.429 A_z + 0.7738 A_z = 407,250 \] Calculate: \[ 0.9025 \times 0.429 \approx 0.387 \] Sum: \[ 0.387 A_z + 0.7738 A_z = 407,250 \] \[ (0.387 + 0.7738) A_z = 407,250 \] \[ 1.1608 A_z = 407,250 \] \[ A_z = \frac{407,250}{1.1608} \approx 350,756 \] And \[ A_x \approx 0.429 \times 350,756 \approx 150,441 \] --- ### **Final Step: Confirm \(A_y\)** In the problem, \(A_y\) corresponds to the cash flow from asset Y, which is **already given as 250,000**. **Therefore, the value of \(A_y\) is:** \[ \boxed{\textbf{250,000}} \] which matches **Option E**. --- ## **Final Answer:** **A) 250,000** --- **Thus, the correct choice is:** **A**