Internal Rate of Return vs MARR

0 = - P + t \sum \frac{A _{t}}{( 1 + i ^{*} ) ^{t}}, accept if i^{*} \geq i_{MARR}

IRR is the i that drives NPW to zero. For a single conventional investment, accept if IRR ≥ MARR. Multiple sign changes can create multiple IRRs — then use NPW.

$i *$: IRR
$M A R R$: Minimum attractive rate of return

Step-by-step solved example

Invest $1000, recover $400/yr for 3 years. IRR is nearest?

IRR is the i that makes the cash-flow NPW = 0.

1. NPW at 10%
400×2.487−1000=−5 ≈ 0.
2. Decision
i*≈10%. Accept if MARR ≤ 10%.

Answer: IRR ≈ 10%

10 practice questions

0/10 correct

1.Accept a conventional project when

2.Multiple IRRs can occur if

3.IRR is independent of

4.Incremental IRR is used for

5.If NPW(10%)>0 and NPW(20%)<0, IRR is

6.ERR (external rate) assumes

7.Simple payback ignores

8.i*=0 means

9.For independent projects with capital plenty, choose

10.Descartes’ rule of signs bounds