Find internal rates of return for signed cash flows within a bounded search range. Follow the calculation and compare NPV at a given MARR.
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Worked examples
These are the solver's own reference problems — the answers come from published sources or independent hand computation, never from the solver itself. The solution below is the live solver output for each.
End-of-period discrete compounding; IRR is the rate i solving NPV(i) = 0 with all flows reinvested/financed at i itself.
Solution steps
Set up the rate-of-return equation
The IRR is the interest rate that makes the discounted receipts exactly balance the discounted costs.
NPV(i) = Σ At/(1 + i)^t = 0
Net cash flows: -1000 at t = 0; 600 at t = 1; 600 at t = 2
Count sign changes (Descartes' rule of signs)
Each sign change in the ordered net cash-flow sequence admits at most one more real root — a conventional investment (one change) has a unique IRR, while non-conventional flows may have several.
Sign changes in the flow sequence: 1
Solve NPV(i) = 0
The evenly spaced net cash flows form a quadratic in the discount factor. Solving it algebraically also finds repeated roots where NPV touches zero without crossing it.
find i* with NPV(i*) = 0
IRR = 13.066 percent
= 13.066
Compare with the MARR
A conventional investment is acceptable when IRR ≥ MARR — equivalently, when its NPV at the MARR is non-negative.
accept if NPV at the MARR ≥ 0 (for conventional investments: IRR ≥ MARR)
NPV at MARR = 41.3223 ≥ 0 (IRR 13.066 percent vs MARR 10 percent) — accept the project
= 41.3223
Results
Internal rate of return (percent)
13.066
Net present value at the MARR
41.3223
Where this answer was checked
source
NPV quadratic solved by hand in x = 1/(1+i) — 1000 = 600/(1+i) + 600/(1+i)²
exact two-flow case: −1000 now, +1210 at t = 2 → IRR = 10% exactly
Given
cashflows
1.period 0amount -1000
2.period 2amount 1210
Assumptions
End-of-period discrete compounding; IRR is the rate i solving NPV(i) = 0 with all flows reinvested/financed at i itself.
Solution steps
Set up the rate-of-return equation
The IRR is the interest rate that makes the discounted receipts exactly balance the discounted costs.
NPV(i) = Σ At/(1 + i)^t = 0
Net cash flows: -1000 at t = 0; 1210 at t = 2
Count sign changes (Descartes' rule of signs)
Each sign change in the ordered net cash-flow sequence admits at most one more real root — a conventional investment (one change) has a unique IRR, while non-conventional flows may have several.
Sign changes in the flow sequence: 1
Solve NPV(i) = 0
NPV(i) is scanned across the rate range for sign changes and each bracket is refined with Brent's method (bisection-safe, superlinear).
Lorie–Savage pump: [−1600, +10000, −10000] → two IRRs, 25% and 400%
Given
cashflows
1.period 0amount -1600
2.period 1amount 10000
3.period 2amount -10000
Assumptions
End-of-period discrete compounding; IRR is the rate i solving NPV(i) = 0 with all flows reinvested/financed at i itself.
Solution steps
Set up the rate-of-return equation
The IRR is the interest rate that makes the discounted receipts exactly balance the discounted costs.
NPV(i) = Σ At/(1 + i)^t = 0
Net cash flows: -1600 at t = 0; 10000 at t = 1; -10000 at t = 2
Count sign changes (Descartes' rule of signs)
Each sign change in the ordered net cash-flow sequence admits at most one more real root — a conventional investment (one change) has a unique IRR, while non-conventional flows may have several.
Sign changes in the flow sequence: 2
Solve NPV(i) = 0
The evenly spaced net cash flows form a quadratic in the discount factor. Solving it algebraically also finds repeated roots where NPV touches zero without crossing it.
find i* with NPV(i*) = 0
NPV(i) = 0 at 2 rates: 25 percent, 400 percent — no single IRR governs
= 25
Results
Internal rate of return, first root (percent) — multiple roots exist
25
Additional IRR root 2 (percent)
400
Notes
The cash flows change sign more than once — multiple internal rates of return are possible (Descartes' rule of signs). When several IRRs exist, none of them is a reliable decision measure by itself; decide using present worth at the MARR instead.
Where this answer was checked
source
Classic multiple-IRR fixture (Lorie–Savage oil-pump problem), quadratic solved by hand — −1600 + 10000/(1+i) − 10000/(1+i)² = 0
verified by
hand-recomputed
derivation
Check i = 0.25: −1600 + 8000 − 6400 = 0 ✓. Check i = 4: −1600 + 2000 − 400 = 0 ✓. In x = 1/(1+i): −10000x² + 10000x − 1600 = 0 → 25x² − 25x + 4 = 0 → x = (25 ± √(625 − 400))/50 = (25 ± 15)/50 → x = 0.8 (i = 0.25) or x = 0.2 (i = 4). Two sign changes → two roots; the solver must report both and warn.