Engineering Economic Analysis

1

254 INCREMENTALANALYSIS

Alternative B

4000 =639(PIA, i, 20)

4000

(PIA, i,20) = _ 639 =6.259 i=15%

Alternative C

5000 =700(PI A, i, 20)

. 5000
(PIA, i,20)=- 700 =7.143

The rate of return is between 12 and 15%:

(

7.469 -7.143

)

i=12% + 7.469 - 6.259 (3%)=12.8%

At this point, we would reject any alternative that fails to meet the MARR criterion of 6%. All

three alternatives exceed the MARR in this example.

Next, we arrange the alternatives in order of increasing initial cost. Then we can examine the

increments between the alternatives.

Initial cost

Uniform annual benefit

Rate of return

A

$2000

410

20%

B

$4000

639

15%

C

$5000

700

12.8%

Incremental cost

Incremental uniform annual benefit

Increment

B-A

$2000

229

Incremental rate of return:

2000 =229(P I A, i, 20)

2000 ~IRR

(PIA,i,20) = 229 9.6%

TheB-A increment is satisfactory; therefore,Bis preferred overA.BecauseBis preferred to

A,the next incremental comparison is of C withB. (IfAwas preferred toB,then the next

incremental comparison would have been of C withA.)

Incremental cost

JncremenJal'l,.uMofIIJ,an,nualb~nefi!

Increment C-B

$1000

pI

• ..-- -...-- - - ...--- -

--- - -----