Engineering Economics Worked Solutions
Engineering Economics Worked
Solutions
Step-by-step solutions to the companion original practice set
Original independent resource. All explanations, examples, and practice questions are newly
written for general study. This is not official course material and contains no instructor-provided
assessments, slides, or proprietary content.
Solution conventions
Dollar values are rounded to the nearest cent unless a decision only requires a comparison. Small
differences may occur from factor-table rounding.
P1 — Single-payment compounding
Problem. A technician deposits $8,500 today in an account earning 6.2% effective annually. What
amount will be available at the end of 7 years?
Method and result. F = P(1+i)^n = 8,500(1.062)^7 = $12,950.62
P2 — Present worth
Problem. A manufacturing upgrade will produce a one-time net benefit of $42,000 at the end of
year 5. At a MARR of 9%, what is the maximum amount that could be spent today?
Method and result. P = F/(1+i)^n = 42,000/(1.09)^5 = $27,297.12
P3 — Uniform annual series
Problem. At 7% effective annual interest, how much must be deposited at the end of each year for
6 years to accumulate $30,000 immediately after the sixth deposit?
Method and result. A = F(A/F,7%,6) = 30,000[.07/((1.07)^6−1)] = $4,193.87
P4 — Capital recovery
Problem. A test fixture costs $26,000, has a $3,000 salvage value after 5 years, and has no other
costs. At 10%, find its equivalent uniform annual cost (EUAC).
Method and result. EUAC = P(A/P,i,n) − S(A/F,i,n) = 26,000(A/P,10%,5) − 3,000(A/F,10%,5) =
$6,367.34
P5 — Arithmetic gradient
Problem. A process-improvement program saves $4,000 in year 1, and the savings increase by
$900 each year through year 6. What is the present worth of the savings at 8%?
Method and result. Cash flows: 4,000; 4,900; 5,800; 6,700; 7,600; 8,500. PW = Σ CF_t/(1.08)^t =
$27,962.47
Original independent study resource | Not affiliated with or endorsed by any university or instructor
Engineering Economics Worked
Solutions
Step-by-step solutions to the companion original practice set
Original independent resource. All explanations, examples, and practice questions are newly
written for general study. This is not official course material and contains no instructor-provided
assessments, slides, or proprietary content.
Solution conventions
Dollar values are rounded to the nearest cent unless a decision only requires a comparison. Small
differences may occur from factor-table rounding.
P1 — Single-payment compounding
Problem. A technician deposits $8,500 today in an account earning 6.2% effective annually. What
amount will be available at the end of 7 years?
Method and result. F = P(1+i)^n = 8,500(1.062)^7 = $12,950.62
P2 — Present worth
Problem. A manufacturing upgrade will produce a one-time net benefit of $42,000 at the end of
year 5. At a MARR of 9%, what is the maximum amount that could be spent today?
Method and result. P = F/(1+i)^n = 42,000/(1.09)^5 = $27,297.12
P3 — Uniform annual series
Problem. At 7% effective annual interest, how much must be deposited at the end of each year for
6 years to accumulate $30,000 immediately after the sixth deposit?
Method and result. A = F(A/F,7%,6) = 30,000[.07/((1.07)^6−1)] = $4,193.87
P4 — Capital recovery
Problem. A test fixture costs $26,000, has a $3,000 salvage value after 5 years, and has no other
costs. At 10%, find its equivalent uniform annual cost (EUAC).
Method and result. EUAC = P(A/P,i,n) − S(A/F,i,n) = 26,000(A/P,10%,5) − 3,000(A/F,10%,5) =
$6,367.34
P5 — Arithmetic gradient
Problem. A process-improvement program saves $4,000 in year 1, and the savings increase by
$900 each year through year 6. What is the present worth of the savings at 8%?
Method and result. Cash flows: 4,000; 4,900; 5,800; 6,700; 7,600; 8,500. PW = Σ CF_t/(1.08)^t =
$27,962.47
Original independent study resource | Not affiliated with or endorsed by any university or instructor
