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Give step-by-step solution with explanation and final answer: What is the present worth of the cash flows in table below? Use an interest rate of 6% compounded annually. ? Cash Flows Table Year Cash Flow 1 15,000 2 5,000 3 4,000 4 3,000 5 2,000 6-10 10,000 (annually)

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Give step-by-step solution with explanation and final answer: What is the present worth of the cash flows in table below? Use an interest rate of 6% compounded annually. ? Cash Flows Table Year Cash Flow 1 15,000 2 5,000 3 4,000 4 3,000 5 2,000 6-10 10,000 (annually)

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Asked by: mohit
Created at: 2025-09-19 18:10:43

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# Present Worth Calculation of Cash Flows We are given a series of cash flows over 10 years and an annual interest rate of 6%. We need to find the present worth (PW) of these cash flows. ## Cash Flow Table | Year | Cash Flow | |--------|-----------| | 1 | 15,000 | | 2 | 5,000 | | 3 | 4,000 | | 4 | 3,000 | | 5 | 2,000 | | 6-10 | 10,000/year | ## Step 1: Formula for Present Value The present value (PV) of a future cash flow at year \( n \) is: \[ PV = \frac{F}{(1 + i)^n} \] where: - \( F \) = Future cash flow - \( i \) = interest rate (6% = 0.06) - \( n \) = year ## Step 2: Calculate Present Worth for Years 1–5 Let’s calculate each year individually. ### Year 1 \[ PW_1 = \frac{15,000}{(1.06)^1} = 15,000 \div 1.06 = 14,150.94 \] ### Year 2 \[ PW_2 = \frac{5,000}{(1.06)^2} = 5,000 \div 1.1236 = 4,448.42 \] ### Year 3 \[ PW_3 = \frac{4,000}{(1.06)^3} = 4,000 \div 1.191016 = 3,358.55 \] ### Year 4 \[ PW_4 = \frac{3,000}{(1.06)^4} = 3,000 \div 1.262477 = 2,377.78 \] ### Year 5 \[ PW_5 = \frac{2,000}{(1.06)^5} = 2,000 \div 1.338225 = 1,494.64 \] ## Step 3: Present Worth of Years 6–10 (Uniform Series) This is a uniform annual series of \( 10,000 \) per year for 5 years, starting at year 6. First, calculate the present worth at year 5 using the Present Worth Factor for a Uniform Series (\( P/A, i, n \)): \[ PW_{6-10,5} = 10,000 \times \frac{1 - (1 + i)^{-5}}{i} \] Where \( n = 5 \), \( i = 0.06 \): Calculate the factor: \[ (1 + 0.06)^{-5} = (1.338225)^{-1} = 0.747258 \] \[ \frac{1 - 0.747258}{0.06} = \frac{0.252742}{0.06} = 4.21237 \] \[ PW_{6-10,5} = 10,000 \times 4.21237 = 42,123.70 \] Discount this amount back to present (year 0) by moving it 5 years back: \[ PW_{6-10,0} = \frac{42,123.70}{(1.06)^5} = 42,123.70 \div 1.338225 = 31,480.52 \] ## Step 4: Sum All Present Worths \[ PW_{total} = PW_1 + PW_2 + PW_3 + PW_4 + PW_5 + PW_{6-10,0} \] \[ PW_{total} = 14,150.94 + 4,448.42 + 3,358.55 + 2,377.78 + 1,494.64 + 31,480.52 \] \[ PW_{total} = 57,310.85 \] ## Final Answer \[ \boxed{57,311} \] *(rounded to the nearest dollar)* --- **Summary Table** | Component | Present Worth (\$) | |------------------------|-------------------| | Year 1 | 14,151 | | Year 2 | 4,448 | | Year 3 | 3,359 | | Year 4 | 2,378 | | Year 5 | 1,495 | | Years 6–10 (total) | 31,481 | | **Total** | **57,311** |

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