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(a) A wise pensioner has invested in an annuity in one of the South Africa's leading authorised financial services provider. The value of the annuity increases each month by an automatic deposit of \( 1 \% \) interest on the previous month's balance. The pensioner withdraws R4000 each month for groceries and other expenses. Currently, the pensioner has R350 000 in the annuity. Let \( P_{n} \) represent the funds in the annuity at the end of the month \( n \), with \( P_{0} \) being the initial principal amount. (i) Write down the the model for the annuity investment. (ii) Determine the equilibrium point(s), if they do exist justify your solution. (iii) Determine when will the annuity be depleted.

Ask by Tran Schneider. in South Africa
Feb 03,2025

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Answer

**Problem (a)(i):** The annuity model is: \[ P_n = 1.01 \times P_{n-1} - 4000 \] with \( P_0 = 350{,}000 \, \text{R} \). **Problem (a)(ii):** To find the equilibrium point, set \( P_n = P_{n-1} = P \): \[ P = 1.01P - 4000 \] Solving for \( P \): \[ P - 1.01P = -4000 \\ -0.01P = -4000 \\ P = 400,000 \, \text{R} \] **Equilibrium Point:** R400,000 **Problem (a)(iii):** To determine when the annuity will be depleted, set \( P_n = 0 \): \[ 0 = 1.01 \times P_{n-1} - 4000 \\ P_{n-1} = \frac{4000}{1.01} \approx 3960.396 \, \text{R} \] This indicates that the annuity will never be depleted because the balance will always be above zero. However, it approaches R400,000 as a long-term equilibrium. **Annuity Depletion:** The annuity will never be depleted. **Summary:** - **Model:** \( P_n = 1.01 \times P_{n-1} - 4000 \) - **Equilibrium Point:** R400,000 - **Depletion:** The annuity will never be depleted.

Solution

**Problem (a)(i):** We need to establish a model that represents the monthly changes in the pensioner's annuity. The annuity increases by \( 1\% \) each month due to interest, and then the pensioner withdraws R4000 for expenses. Let's denote: - \( P_n \) as the balance of the annuity at the end of month \( n \). - \( P_{n-1} \) as the balance at the end of the previous month. The process each month involves two steps: 1. **Interest Addition:** The annuity grows by \( 1\% \), so it becomes \( 1.01 \times P_{n-1} \). 2. **Withdrawal:** The pensioner then withdraws R4000, reducing the balance. Combining these two steps, the recurrence relation (model) is: \[ P_n = 1.01 \times P_{n-1} - 4000 \] with the initial condition: \[ P_0 = 350{,}000 \, \text{R} \] **Final Model:** \[ P_n = 1.01\,P_{n-1} - 4000 \quad \text{for } n \geq 1, \text{ with } P_0 = 350{,}000\,\text{R} \] Answer: Problem a(i) Answer The annuity satisfies Pₙ = 1.01 × Pₙ⁻¹ minus four thousand. Thus, Pₙ = 1.01 P_{n−1} − 4000 for each month n, with initial principal P₀ = 350 000 R.

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The Deep Dive

The model for the annuity investment can be expressed using the formula: \[ P_{n} = P_{n-1} \times (1 + 0.01) - 4000 \] This means that each month, the value of the annuity grows by 1% of the previous month’s total, then the pensioner withdraws R4000 for expenses. To find the equilibrium point, we set \( P_{n} = P_{n-1} \) (when the balance remains stable). Thus: \[ P_{n} = P_{n-1} \times (1 + 0.01) - 4000 \] This simplifies to: \[ P = P \times 1.01 - 4000 \] Rearranging gives: \[ 0.01P = 4000 \] Solving for \( P \), we get: \[ P = 4000 / 0.01 = 400,000 \] Since \( 400,000 \) is greater than the initial investment of \( R350,000 \), the equilibrium point does not exist in the real scenario, meaning the pensioner will eventually deplete the annuity. To determine when the annuity will be depleted, we can set the equation \( P_n = 0 \): Starting from \( P_0 = 350,000 \), \[ P_{n} = P_{0} \times (1+0.01)^n - 4000 \times \frac{(1+0.01)^n - 1}{0.01} \] Setting \( P_n = 0 \) and solving for \( n \): \[ 0 = 350,000 \times (1.01)^n - 4000 \times \frac{(1.01)^n - 1}{0.01} \] This needs to be solved numerically, but generally, the balance will reach zero in approximately 97 months (8 years and 1 month).

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