Question

Given the following two equipment with different service​ lives, assume repeatability and determine which one is more economical. Use a MARR of 6​%per year.

	Eagle	Condor
Capital investment	​$220,000	​$600,000
Net annual revenue	​$110,000	​$130,000
Salvage value	​$0	​$0
Useful life	5 years	10 years

Click the icon to view the interest and annuity table for discrete compounding when i equals=6​% per year.

A.

Condor

B.

They are equivalent

C.

Eagle

Answer 1

Ans: Eagle

Explanation:

PW of Eagle = -220,000 + 110,000(P/A, 6%, 10) - 220,000(P/F, 6%, 5)

                     = -220,000 + 110,000(7.360) - 220,000(0.7473)

                     = -220,000 + 809,600 - 164,406

                     = $425,194

PW of Condor = -600,000 + 130,000(P/A, 6%, 10)

                        = -600,000 + 130,000(7.360)

                        = -600,000 + 956,800

                        = $356,800

Since the PW of Eagle is greate than Condor, therefore, Eagle is more economical.