- 2065 Kartik · 8 marks
IRR with unrecovered balance, graphical and tabular
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.
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
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.
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)
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.
✓ Answer checked by Claude Opus 5.5 (high effort)
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