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.
Copyright © 2025 Pearson Education, Inc.
,6-2 Present Worth Method, MARR = 10% per year
PWD1 (10%) = −$600,000 − $780,000(P/A,10%,8) = −$4,761,222
PWD2 (10%) = −$760,000 − $728,000(P/A,10%,8) = −$4,643,807
PWD3 (10%) = −$1,240,000 − $630,000(P/A,10%,8) = −$4,600,987
PWD4 (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.
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.
PWA = −$200,000 + $60,000 (P/A, 7%, 5) = $46,012
PWB = −$180,000 + $55,000 (P/A, 7%, 6) = $82,157.50
Choose Venture B to maximize present worth.
Copyright © 2025 Pearson Education, Inc.
, 6-4 Assume all units are produced and sold each year.
AWA(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
AWC(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.
Copyright © 2025 Pearson Education, Inc.
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.
Copyright © 2025 Pearson Education, Inc.
,6-2 Present Worth Method, MARR = 10% per year
PWD1 (10%) = −$600,000 − $780,000(P/A,10%,8) = −$4,761,222
PWD2 (10%) = −$760,000 − $728,000(P/A,10%,8) = −$4,643,807
PWD3 (10%) = −$1,240,000 − $630,000(P/A,10%,8) = −$4,600,987
PWD4 (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.
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.
PWA = −$200,000 + $60,000 (P/A, 7%, 5) = $46,012
PWB = −$180,000 + $55,000 (P/A, 7%, 6) = $82,157.50
Choose Venture B to maximize present worth.
Copyright © 2025 Pearson Education, Inc.
, 6-4 Assume all units are produced and sold each year.
AWA(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
AWC(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.
Copyright © 2025 Pearson Education, Inc.
