1. **State the problem:** We need to find the internal rate of return (TIR) that satisfies the equation: $$0 = -350000 + \frac{100000}{(1+TIR)^1} + \frac{150000}{(1+TIR)^2} + \frac{50000}{(1+TIR)^3} + \frac{100000}{(1+TIR)^4} + \frac{100000}{(1+TIR)^5}$$ 2. **Formula and explanation:** The internal rate of return (TIR) is the discount rate that makes the net present value (NPV) of all cash flows equal to zero. The formula for NPV is: $$NPV = \sum_{t=0}^n \frac{C_t}{(1+TIR)^t}$$ where $C_t$ is the cash flow at time $t$. Here, the initial investment is $-350000$ at $t=0$, and subsequent cash inflows occur at $t=1$ to $t=5$. 3. **Rewrite the equation:** $$0 = -350000 + \frac{100000}{(1+TIR)} + \frac{150000}{(1+TIR)^2} + \frac{50000}{(1+TIR)^3} + \frac{100000}{(1+TIR)^4} + \frac{100000}{(1+TIR)^5}$$ 4. **Solving for TIR:** This equation is nonlinear and cannot be solved algebraically in closed form. We use numerical methods such as the Newton-Raphson method or trial and error to approximate $TIR$. 5. **Numerical approximation (trial):** - Try $TIR=0.10$ (10%): $$NPV = -350000 + \frac{100000}{1.1} + \frac{150000}{1.1^2} + \frac{50000}{1.1^3} + \frac{100000}{1.1^4} + \frac{100000}{1.1^5}$$ Calculate each term: $$\frac{100000}{1.1} = 90909.09$$ $$\frac{150000}{1.21} = 123966.94$$ $$\frac{50000}{1.331} = 37546.50$$ $$\frac{100000}{1.4641} = 68292.57$$ $$\frac{100000}{1.61051} = 62092.13$$ Sum of inflows: $$90909.09 + 123966.94 + 37546.50 + 68292.57 + 62092.13 = 383807.23$$ NPV: $$-350000 + 383807.23 = 33807.23 > 0$$ - Try $TIR=0.15$ (15%): Calculate denominators: $$1.15, 1.3225, 1.5209, 1.7490, 2.0114$$ Calculate inflows: $$\frac{100000}{1.15} = 86956.52$$ $$\frac{150000}{1.3225} = 113441.80$$ $$\frac{50000}{1.5209} = 32874.32$$ $$\frac{100000}{1.7490} = 57164.18$$ $$\frac{100000}{2.0114} = 49716.05$$ Sum inflows: $$86956.52 + 113441.80 + 32874.32 + 57164.18 + 49716.05 = 340152.87$$ NPV: $$-350000 + 340152.87 = -9847.13 < 0$$ 6. **Conclusion:** Since NPV is positive at 10% and negative at 15%, the TIR lies between 10% and 15%. Using linear interpolation: $$TIR \approx 10\% + \frac{33807.23}{33807.23 + 9847.13} \times (15\% - 10\%) = 10\% + \frac{33807.23}{43654.36} \times 5\% \approx 10\% + 3.87\% = 13.87\%$$ **Final answer:** The internal rate of return (TIR) is approximately **13.87%**.