AWA -30,000 5,000 AG, 10, 5 20,949.50

Copyright © 2025 Pearson Education, Inc.Solutions to Chapter 6 Problems

6-1 *Students may realize that the scenario described in the problem has the cost become a revenue (an unlikely situation).

A: In-house treatment

0 1 2 3 4 5

G = $5,000

A = $30,000

B: Outside treatment

0 1 2 3 4 5

$15,000

AW(A) = -$30,000 + 5,000 (A/G, 10%, 5) = −$20,949.50

AW(B) = −$15,000 (F/P, 10%, 1) = −$16,500

Outside treatment is the most economical alternative. 1 / 4

Copyright © 2025 Pearson Education, Inc.6-2 Present Worth Method, MARR = 10% per year

PW D1 (10%) = −$600,000 − $780,000(P/A,10%,8) = −$4,761,222

PW D2 (10%) = −$760,000 − $728,000(P/A,10%,8) = −$4,643,807

PW D3 (10%) = −$1,240,000 − $630,000(P/A,10%,8) = −$4,600,987

PW D4 (10%) = −$1,600,000 − $574,000(P/A,10%,8) = −$4,662,233

Select Design D3 to minimize the present worth of costs.

Future Worth Method, MARR = 10% per year

FWD1 (10%) = −$600,000(F/P,10%,8) − $780,000(F/A,10%,8) = −$10,206,162

FWD2 (10%) = −$760,000(F/P,10%,8) − $728,000(F/A,10%,8) = −$9,954,471

FWD3 (10%) = −$1,240,000(F/P,10%,8) − $630,000(F/A,10%,8) = −$9,862,681

FWD4 (10%) = −$1,600,000 (F/P,10%,8) – $574,000(F/A,10%,8) = −$9,993,967

Select Design D3 to minimize the future worth of costs.

Annual Worth Method, MARR = 1% per year

AWD1 (10%) = −$600,000 (A/P,10%,8) – $780,000 = −$892,440

AWD2 (10%) = −$760,000 (A/P,10%,8) – $728,000 = −$870,424

AWD3 (10%) = −$1,240,000 (A/P,10%,8) – $630,000 = −$862,376

AWD4 (10%) = −$1,600,000 (A/P,10%,8) – $574,000 = −$873,840

Select Design D3 to minimize the annual worth of costs. 2 / 4

Copyright © 2025 Pearson Education, Inc.6-3 The lives of the alternatives are unequal. Let us assume that repeatability does not apply, so the PW can be compared over the different time periods.

PW A = −$200,000 + $60,000 (P/A, 7%, 5) = $46,012

PW B = −$180,000 + $55,000 (P/A, 7%, 6) = $82,157.50

Choose Venture B to maximize present worth.

• / 4

Copyright © 2025 Pearson Education, Inc.6-4 Assume all units are produced and sold each year.

AW A(20%) = −$30,000(A/P,20%,10) + 15,000($3.50 − $1.00) − $15,000 = $15,345

AWB(20%) = −$60,000(A/P,20%,10) + 20,000($4.40 − $1.40) − $30,000 + $20,000(A/F,20%,10)

= $16,460

AW C(20%) = −$50,000(A/P,20%,10) + 18,000($4.10 − $1.15) − $25,000 + $15,000(A/F,20%,10)

= $15,753

Select Design B to maximize the annual worth.

• / 4