Let's solve the problem step by step. --- ### **Given:** - **Interest rate (discount rate), \( r = 5\% \) per annum** - **Project duration = 8 years** #### **Cashflows:** | Timing | Amount | |----------------------------|---------------| | Initial outlay | 300,000 (t=0) | | Further cost | 200,000 (t=1) | | Advertising income | 10,000 (t=1) | | Regular expenditure | 20,000 p.a. (t=2 to t=8, continuous) | | Regular income | 130,000 p.a. (t=2 to t=8, continuous) | | Decommissioning expense | 10,000 (t=8) | --- ## **(a) Net Present Value (NPV)** ### **Step 1: List all cashflows** - \( t = 0 \): Outflow = \(-300,000\) - \( t = 1 \): Outflow = \(-200,000\), Inflow = \(+10,000\) (Net: \(-190,000\)) - \( t = 2 \) to \( t = 8 \) (continuous): Net cashflow per year = \(130,000 - 20,000 = 110,000\) (inflow) - \( t = 8 \): Outflow = \(-10,000\) ### **Step 2: Present Value Calculations** #### **1. Initial Outlay (\(t=0\)):** \[ PV_0 = -300,000 \] #### **2. Net outflow at \( t=1 \):** \[ PV_1 = -190,000 \times e^{-0.05 \times 1} = -190,000 \times 0.95123 = -180,733.7 \] #### **3. Continuous net inflow from \( t=2 \) to \( t=8 \):** Present Value of continuous cashflow, \( PV_{ann} \): \[ PV_{ann} = \int_{2}^{8} 110,000 \cdot e^{-0.05 t} dt \] \[ = 110,000 \left[ \frac{-1}{0.05} e^{-0.05 t} \right]_{2}^{8} \] \[ = -2,200,000 \left( e^{-0.4} - e^{-0.1} \right) \] Calculate \( e^{-0.4} = 0.67032 \), \( e^{-0.1} = 0.90484 \): \[ PV_{ann} = -2,200,000 (0.67032 - 0.90484) \] \[ = -2,200,000 \times (-0.23452) \] \[ = 515,944 \] #### **4. Decommissioning expense at \( t=8 \):** \[ PV_8 = -10,000 \times e^{-0.05 \times 8} = -10,000 \times e^{-0.4} = -10,000 \times 0.67032 = -6,703 \] --- ### **Step 3: Sum all Present Values** \[ NPV = PV_0 + PV_1 + PV_{ann} + PV_8 \] \[ NPV = (-300,000) + (-180,733.7) + 515,944 + (-6,703) \] \[ NPV = 515,944 - 300,000 - 180,733.7 - 6,703 \] \[ NPV = 515,944 - 487,436.7 \] \[ NPV = 28,507.3 \] --- #### **Final Answer for (a):** \[ \boxed{NPV = \$28,507 \text{ (rounded to nearest dollar)}} \] --- ## **(b) Discounted Payback Period** **Discounted payback period:** Time when the sum of discounted net cash flows becomes positive. ### **Step 1: Calculate cumulative discounted cashflows by year:** #### - **At t=0**: \(-300,000\) #### - **At t=1**: \(-190,000 \times e^{-0.05 \times 1} = -180,733.7\) - Cumulative: \(-300,000 - 180,733.7 = -480,733.7\) #### - **From t=2 onwards:** Net annual cashflow = +110,000 (continuous) - PV of inflows from t=2 to t=T: \(110,000 \int_2^T e^{-0.05 t} dt\) We want to find T such that cumulative PV = 0: \[ -300,000 - 180,733.7 + 110,000 \int_2^T e^{-0.05 t} dt = 0 \] Let’s denote S = cumulative outflow up to t=2: \[ S = -300,000 - 180,733.7 = -480,733.7 \] PV of continuous inflow from t=2 to t=T: \[ PV_{inflow} = 110,000 \left[ \frac{-1}{0.05} (e^{-0.05 T} - e^{-0.05 \times 2}) \right] \] \[ = -2,200,000 \left( e^{-0.05 T} - 0.90484 \right) \] Set cumulative PV = 0: \[ -480,733.7 - 2,200,000 (e^{-0.05 T} - 0.90484) = 0 \] \[ -480,733.7 = 2,200,000 (e^{-0.05 T} - 0.90484) \] \[ \frac{-480,733.7}{2,200,000} = e^{-0.05 T} - 0.90484 \] \[ -0.218515 = e^{-0.05 T} - 0.90484 \] \[ e^{-0.05 T} = 0.90484 - 0.218515 = 0.686325 \] \[ -0.05 T = \ln(0.686325) \] \[ -0.05 T = -0.376684 \] \[ T = \frac{-0.376684}{-0.05} = 7.53 \] So, the discounted payback period is **about 7.53 years**. --- ### **Final Answer for (b):** \[ \boxed{7.53 \text{ years}} \] --- ## **(c) Approximate Internal Rate of Return (IRR)** IRR is the rate \( r \) that sets NPV to zero. Since the NPV at 5% is +\$28,507, let's try a slightly higher rate to approximate IRR using interpolation. Let’s try \( r = 6\% \): #### **Repeat calculations with \( r = 6\% \):** - \( e^{-0.06 \times 1} = 0.94176 \) - \( e^{-0.06 \times 2} = 0.88692 \) - \( e^{-0.06 \times 8} = 0.61878 \) ##### **1. PV of t=1 outflow:** \[ PV_1 = -190,000 \times 0.94176 = -178,934.4 \] ##### **2. PV of continuous net inflow (t=2 to t=8):** \[ PV_{ann} = 110,000 \left[ \frac{-1}{0.06} (e^{-0.48} - e^{-0.12}) \right] \] \[ e^{-0.48} = 0.619 \] \[ e^{-0.12} = 0.8869 \] \[ PV_{ann} = -1,833,333 \times (0.619 - 0.8869) = -1,833,333 \times (-0.2679) = 491,236 \] ##### **3. PV of decommissioning at t=8:** \[ PV_8 = -10,000 \times 0.61878 = -6,188 \] ##### **4. NPV at 6%:** \[ NPV_{6\%} = -300,000 - 178,934.4 + 491,236 - 6,188 \] \[ = 491,236 - 485,122.4 \] \[ = 6,113.6 \] Now, interpolate between 5% (NPV = 28,507) and 6% (NPV = 6,113.6): \[ IRR \approx 5\% + \left( \frac{28,507}{28,507 - 6,113.6} \right) \times (6\% - 5\%) \] \[ = 5\% + \left( \frac{28,507}{22,393.4} \right) \times 1\% \] \[ = 5\% + 1.273 \% \] \[ = 6.27\% \] --- ### **Final Answer for (c):** \[ \boxed{IRR \approx 6.3\%} \] --- ## **Summary of Answers:** **(a) NPV at 5% =** \$28,507 **(b) Discounted payback period =** 7.53 years **(c) Approximate IRR =** 6.3%