--- title: "IRR with unrecovered balance, graphical and tabular" subject: "Engineering Economics" chapter: 3 key: "semester_6/engineering_economics:irr-unrecovered-balance-250000" source: https://www.bctengineer.com/q/semester_6/engineering_economics/irr-unrecovered-balance-250000 site: BCT Engineering --- # IRR with unrecovered balance, graphical and tabular Engineering Economics · Chapter 3: Basic Methodologies of Engineering Economic Analysis - Asked in: 2065 Kartik, 8 marks - Video: [Find IRR | Engineering economics | Unrecovered investment balance table and graphic form](https://www.youtube.com/watch?v=O5FoZWr4dQU) — Sudip Dhakal - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/engineering_economics/irr-unrecovered-balance-250000 - Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-3 ## Question Find IRR and show the unrecovered investment balance in the graphical and tabular form. Investment (First) Cost = Rs. 2,50,000 Revenues/Year = Rs. 1,00,000 Expenses/Year = Rs. 30,000 Salvage Value = Rs. 50,000 Useful life = 5 years ## Short answer IRR ≈ 16.48% by interpolation between 16% and 17%. Unrecovered balance table shows Rs 2,50,000 initial, decreasing annually to zero at year 5 with net annual cash flow Rs 70,000 and salvage Rs 50,000. ## Answer IRR is the interest rate i* at which PW of the net cash flows is zero; at i* the unrecovered investment balance falls to exactly zero at the end of the life. #### Cash flows Net annual cash flow = 100,000 - 30,000 = Rs. 70,000. Salvage Rs. 50,000 comes in year 5. ```text 70k 70k 70k 70k 70k+50k | | | | | 0 -----1-----2-----3-----4------5---> year | v 250,000 ``` #### Find IRR PW(i) = -250,000 + 70,000 (P/A, i, 5) + 50,000 (P/F, i, 5) = 0 Trial values: - i = 15%: (P/A) = 3.3522, (P/F) = 0.4972, PW = 9,510 - i = 18%: (P/A) = 3.1272, (P/F) = 0.4371, PW = -9,243 - i = 16%: (P/A) = 3.2743, (P/F) = 0.4761, PW = 3,006 - i = 17%: PW = -3,240 Interpolating between 16% and 17%: IRR = 16% + 3,006 / (3,006 - (-3,240)) × 1% = 16.48% #### Unrecovered investment balance (tabular), i* = 16.48% Balance at end of year t = balance(t-1) × (1 + i*) - net cash flow(t) | Year | Balance at start | Interest on balance | Cash flow received | Balance at end | |---|---|---|---|---| | 1 | 250,000 | 41,191 | 70,000 | 221,191 | | 2 | 221,191 | 36,444 | 70,000 | 187,635 | | 3 | 187,635 | 30,915 | 70,000 | 148,550 | | 4 | 148,550 | 24,475 | 70,000 | 103,025 | | 5 | 103,025 | 16,975 | 120,000 | 0 | The balance becomes zero at the end of year 5, which confirms the IRR. #### Graphical form (bar length is proportional to the unrecovered balance) ```text EOY 0: ############################## 250,000 EOY 1: ########################### 221,191 EOY 2: ####################### 187,635 EOY 3: ################## 148,550 EOY 4: ############ 103,025 EOY 5: 0 ``` **Answer: IRR ≈ 16.48% per year.** The project is acceptable whenever MARR is below this value. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What defines the IRR i*? A. Rate at which PW of net cash flows is zero B. Rate at which payback is 5 years C. Rate equal to MARR D. Interest on the loan Answer: A. IRR is the rate that makes the PW of net cash flows zero. (Key point: IRR definition) 2. What net annual cash flow is used? A. Rs 70,000 (100,000 - 30,000) B. Rs 1,00,000 C. Rs 1,20,000 every year D. Rs 30,000 Answer: A. Revenue minus expenses = 70,000, with salvage 50,000 added in year 5. (Key point: Net annual CF) 3. What is the IRR found? A. About 9.5% B. About 18% C. About 15% D. About 16.48% Answer: D. Interpolating between 16% (+3,006) and 17% (-3,240) gives 16.48%. (Key point: IRR result) 4. What does the balance reaching zero at end of year 5 show? A. MARR is 0% B. The IRR is correct C. Salvage was ignored D. The project failed Answer: B. At the true IRR the unrecovered balance is exactly zero at the end of life. 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