1. **Stating the problem:** You want to withdraw 100,000 pesos every 6 months for 3 years (which is 6 withdrawals), with an investment earning 12% interest compounded quarterly. Inflation is 5% annually, and you want to find the monthly deposit amount needed to achieve this. 2. **Understanding the problem:** - Withdrawals: 100,000 pesos every 6 months for 3 years means 6 withdrawals. - Interest rate: 12% annual, compounded quarterly means quarterly interest rate $i = \frac{0.12}{4} = 0.03$. - Inflation: 5% annually means the real value of money decreases, so withdrawals increase in nominal terms by inflation. - Monthly deposits: We want to find the fixed monthly deposit $D$ that will accumulate to cover these withdrawals. 3. **Adjusting withdrawals for inflation:** Each withdrawal increases by inflation every 6 months. Since inflation is 5% annually, the 6-month inflation rate is approximately $\left(1 + 0.05\right)^{0.5} - 1 = \sqrt{1.05} - 1 \approx 0.0247$ or 2.47% per 6 months. So, the $k$-th withdrawal amount is: $$W_k = 100000 \times (1.0247)^{k-1}$$ for $k=1,2,...,6$. 4. **Calculating the present value (PV) of withdrawals at the start:** The investment compounds quarterly, so the effective 6-month interest rate is: $$i_{6m} = (1 + 0.03)^2 - 1 = 1.03^2 - 1 = 1.0609 - 1 = 0.0609$$ The present value of all withdrawals is: $$PV = \sum_{k=1}^6 \frac{W_k}{(1 + i_{6m})^k} = \sum_{k=1}^6 \frac{100000 \times (1.0247)^{k-1}}{(1.0609)^k}$$ Calculate each term: - $k=1$: $\frac{100000 \times 1}{1.0609} = 94200.68$ - $k=2$: $\frac{100000 \times 1.0247}{1.0609^2} = \frac{102470}{1.1265} = 90944.54$ - $k=3$: $\frac{100000 \times 1.050}{1.0609^3} = \frac{105000}{1.195} = 87849.37$ - $k=4$: $\frac{100000 \times 1.076}{1.0609^4} = \frac{107600}{1.267} = 84900.47$ - $k=5$: $\frac{100000 \times 1.103}{1.0609^5} = \frac{110300}{1.342} = 82094.88$ - $k=6$: $\frac{100000 \times 1.130}{1.0609^6} = \frac{113000}{1.420} = 79329.58$ Sum: $$PV \approx 94200.68 + 90944.54 + 87849.37 + 84900.47 + 82094.88 + 79329.58 = 519319.52$$ 5. **Finding the monthly deposit $D$ to accumulate $PV$ in 6 years:** - Monthly interest rate: quarterly rate 3% means monthly rate $i_m = (1.03)^{1/3} - 1 \approx 0.009867$ (about 0.9867%). - Number of months: $6 \times 12 = 72$. The future value of an annuity (monthly deposits) is: $$FV = D \times \frac{(1 + i_m)^{72} - 1}{i_m}$$ We want $FV = PV = 519319.52$. Calculate: $$\frac{(1 + 0.009867)^{72} - 1}{0.009867} = \frac{(1.009867)^{72} - 1}{0.009867}$$ Calculate $(1.009867)^{72}$: $$\ln(1.009867) \approx 0.00982$$ $$0.00982 \times 72 = 0.707$$ $$e^{0.707} \approx 2.028$$ So: $$\frac{2.028 - 1}{0.009867} = \frac{1.028}{0.009867} \approx 104.2$$ 6. **Solve for $D$:** $$D = \frac{PV}{104.2} = \frac{519319.52}{104.2} \approx 4981.5$$ **Final answer:** The monthly deposit needed is approximately **4982 pesos**.