Exercise set

Engineering Lifecycle Cost EAC, Replacement, and Maintenance Economics Exercises

Solved engineering economics exercises for lifecycle cost, equivalent annual cost, replacement timing, maintenance cost and downtime economics.

Branch: Industrial and Management Engineering
Content: Exercise set
Updated: Jul 03, 2026
Revision: v1.0.0 · reviewed

These exercises practise lifecycle-cost decisions for engineered assets: installed cost, recurring energy and maintenance, equivalent annual cost, unequal lives, downtime exposure, replacement timing, residual value and maintenance economics.

The goal is to compare alternatives with the same service boundary. A lower capital cost can lose after energy, repair, downtime, residual value or replacement cycle are included.

Assume simplified financial factors when provided. Real lifecycle-cost decisions should also check inflation basis, taxes, warranty, service contracts, permitting delay, downtime consequence, reliability evidence, maintainability and residual-value risk.

Release Evidence Notes

Lifecycle evidence should state service duty, analysis horizon, discount rate, asset life, residual value, included costs, excluded costs and operating assumptions.

EAC evidence is useful when alternatives have unequal lives. It should not be used if the replacement cycle or service duty is not repeatable.

Maintenance economics should connect cost to technical mechanisms: failure modes, inspection interval, spare readiness, repair access and validation after repair.

Downtime cost should be stated separately from repair cost. Lost production, service penalties, safety exposure and restart losses can dominate repair invoices.

Engineering Boundary Notes

This page covers lifecycle cost and replacement economics. Investment NPV, payback, IRR and capital rationing belong in the investment appraisal exercise set. Tax, escalation and benefit ramp-up belong in the cash-flow evidence exercise set.

Scenario Map

Scenario	Exercises	Primary check	Engineering decision
Lifecycle cost	1-5	capital, energy, maintenance, present worth and residual value	Choose the lower cost alternative for equal service.
Equivalent annual cost	6-9	EAC, unequal lives and replacement cycle	Compare repeatable alternatives.
Maintenance and downtime	10-14	repair cost, downtime exposure and preventive action economics	Justify maintenance or redesign.
Replacement release	15-18	defender/challenger cost, sensitivity and hard gates	Keep, replace or defer.

Exercise 1: Six-Year Lifecycle Cost

Two pump packages have:

Cost item	A	B
installed cost	$180000$	$240000$
annual energy	$42000$	$28000$
annual maintenance	$12000$	$8000$

Compare six-year lifecycle cost without discounting.

Solution

For A:

LCC_A=180000+6(42000+12000)=504000

For B:

LCC_B=240000+6(28000+8000)=456000

B is lower by:

48000

Engineering Comment

The higher-capital option wins because recurring savings exceed the extra capital over six years.

Plausibility Check

B costs $60000$ more upfront but saves $18000$ per year, or $108000$ over six years.

Exercise 2: Present Worth of Maintenance Cost

Annual maintenance cost is:

C=14000

for eight years. Use:

(P/A,8\%,8)=5.7466

Find present worth.

Solution

Present worth:

PV=C(P/A)=14000(5.7466)=80452

Engineering Comment

Recurring maintenance can be a large lifecycle component. It should be based on observed work orders when available.

Plausibility Check

Eight undiscounted years are $112000$ ; discounting reduces present worth to about $80000$ .

Exercise 3: Energy Cost Present Worth

Annual energy cost difference between two alternatives is:

\Delta C=9000

over ten years. Use:

(P/A,7\%,10)=7.0236

Find present value of the energy saving.

Solution

Present value:

PV=9000(7.0236)=63212

Engineering Comment

Energy savings should be tied to measured load profile, operating hours and utility tariff.

Plausibility Check

Ten undiscounted years are $90000$ ; discounted present value near $63000$ is plausible.

Exercise 4: Residual Value Present Worth

An asset residual value in year 8 is:

R=30000

Use:

(P/F,8\%,8)=0.5403

Find present value.

Solution

Present value:

PV_R=30000(0.5403)=16209

Engineering Comment

Residual value is often uncertain. Treat it separately so the decision does not hide resale risk.

Plausibility Check

Eight-year discounting cuts the future value nearly in half.

Exercise 5: Net Lifecycle Present Cost

Installed cost is:

180000

maintenance present worth is:

80452

energy present worth is:

63212

residual present value is:

16209

Find net lifecycle present cost.

Solution

Net cost:

NPC=180000+80452+63212-16209=307455

Engineering Comment

Residual value is subtracted because it offsets cost. Keep sign convention explicit.

Plausibility Check

Costs sum to $323664$ before residual, then the residual reduces the total.

Exercise 6: Equivalent Annual Cost from Present Cost

Present lifecycle cost is:

P=307455

Use:

(A/P,8\%,8)=0.1740

Find EAC.

Solution

Equivalent annual cost:

EAC=P(A/P)=307455(0.1740)=53497

Engineering Comment

EAC converts a present cost into an annualized service cost. It is useful for comparing repeatable alternatives.

Plausibility Check

Annualizing over eight years at interest gives more than simple division by eight.

Exercise 7: Unequal-Life EAC Comparison

Alternative A has present cost:

210000

and life 4 years. Alternative B has present cost:

315000

and life 7 years. Use:

(A/P,10\%,4)=0.3155,\quad (A/P,10\%,7)=0.2054

Compare EAC.

Solution

For A:

EAC_A=210000(0.3155)=66255

For B:

EAC_B=315000(0.2054)=64701

B has lower EAC.

Engineering Comment

B costs more upfront but spreads cost over a longer economic life.

Plausibility Check

The EAC values are close, which fits the trade between higher cost and longer life.

Exercise 8: Annualized Salvage Credit

Future salvage present value is:

PV_R=16209

Use:

(A/P,8\%,8)=0.1740

Find annualized salvage credit.

Solution

Annual credit:

A_R=16209(0.1740)=2820

Engineering Comment

Annualized salvage should be shown as a credit, not mixed into operating savings without explanation.

Plausibility Check

The annual value is much smaller than the present salvage value spread over eight years.

Exercise 9: EAC Difference

Two alternatives have:

EAC_A=53497,\qquad EAC_B=48700

Find annual saving from B.

Solution

Annual saving:

S=53497-48700=4797

Engineering Comment

A small EAC difference should be checked against uncertainty in energy, maintenance and residual assumptions.

Plausibility Check

The difference is under ten percent of either EAC, so sensitivity matters.

Exercise 10: Downtime Cost per Failure

Downtime loss is:

1800\ \text{per h}

Average outage duration is:

6\ \text{h}

Repair invoice cost is:

3500

Find cost per failure.

Solution

Downtime cost:

C_d=1800(6)=10800

Total cost:

C_f=10800+3500=14300

Engineering Comment

Downtime dominates the repair invoice. Maintenance economics should include lost production or service impact.

Plausibility Check

Six hours at $1800$ per hour is over ten thousand, plus repair cost.

Exercise 11: Annual Failure Cost

Expected failures per year:

n=4

Cost per failure:

C_f=14300

Find annual failure cost.

Solution

Annual cost:

C_a=nC_f=4(14300)=57200

Engineering Comment

This annual cost can justify preventive work if the failure mechanism is controllable.

Plausibility Check

Four events at about fourteen thousand each gives about fifty-seven thousand.

Exercise 12: Preventive Action Benefit

A preventive action costs:

24000\ \text{per year}

and reduces annual failure cost from:

57200

to:

26000

Find net annual benefit.

Solution

Avoided failure cost:

B=57200-26000=31200

Net benefit:

B_n=31200-24000=7200

Engineering Comment

The action is positive on this screen, but only if it truly reduces the failure modes driving downtime.

Plausibility Check

The avoided cost is slightly above the preventive cost, leaving a modest benefit.

Exercise 13: Maintenance Interval Cost Screen

Short interval maintenance costs:

40000\ \text{per year}

and expected failure cost:

18000

Long interval maintenance costs:

24000

and expected failure cost:

42000

Choose lower expected annual cost.

Solution

Short interval:

C_s=40000+18000=58000

Long interval:

C_l=24000+42000=66000

Short interval is lower by:

8000

Engineering Comment

Economic interval choice should still obey safety and reliability gates. Cost minimum cannot justify unacceptable risk.

Plausibility Check

The long interval saves maintenance cost but loses more in failures.

Exercise 14: Availability Loss Cost

An asset loses:

30\ \text{h/year}

of availability. Production margin is:

2500\ \text{per h}

Find annual availability loss.

Solution

Loss:

C=30(2500)=75000

Engineering Comment

Availability improvements should be evaluated against actual constrained production value, not average revenue if the asset is not a bottleneck.

Plausibility Check

Thirty hours at twenty-five hundred per hour is seventy-five thousand.

Exercise 15: Defender Annual Cost

Keeping an existing asset costs:

32000

maintenance per year and expected downtime cost:

45000

Find defender annual cost.

Solution

Defender cost:

C_D=32000+45000=77000

Engineering Comment

The defender should be compared with the challenger on equivalent service, not sunk historical cost.

Plausibility Check

The two annual cost components add directly.

Exercise 16: Challenger Annual Cost

A replacement has annualized capital cost:

51000

maintenance cost:

14000

and downtime cost:

9000

Find annual cost.

Solution

Challenger cost:

C_C=51000+14000+9000=74000

Engineering Comment

The challenger is only slightly cheaper than the defender, so uncertainty and implementation risk matter.

Plausibility Check

Capital dominates, but lower downtime and maintenance keep total below the defender.

Exercise 17: Replacement Margin

Defender annual cost:

C_D=77000

Challenger annual cost:

C_C=74000

Find replacement margin.

Solution

Margin:

M=C_D-C_C=77000-74000=3000

Engineering Comment

A $3000$ annual margin is narrow. Installation risk, downtime during replacement and residual value should be reviewed.

Plausibility Check

The costs are close, so the margin is small.

Exercise 18: Lifecycle Replacement Release Gate

A replacement review has:

Gate	Requirement	Current result
challenger annual cost	below defender	pass
annual margin	at least $10000$	$3000$
downtime model validated	required	pass
residual value evidence	required	open

Decide whether to approve replacement.

Solution

The challenger is lower cost, but margin and residual evidence fail:

3000<10000

The replacement should not be approved for full release.

Engineering Comment

Replacement can be staged or sent for more evidence. A narrow economic advantage is not enough when residual value is unresolved.

Plausibility Check

Two hard gates fail, so the release decision should be hold or restrict.

Validation Package Checklist

A strong lifecycle-cost solution should check:

whether alternatives deliver the same service duty;
whether capital, energy, maintenance, downtime and residual value are included consistently;
whether EAC is used only for repeatable unequal-life comparisons;
whether downtime cost reflects constrained value;
whether maintenance benefits are tied to failure modes;
whether defender cost excludes sunk cost;
whether residual value and replacement downtime are evidenced;
whether failed economic gates block release.

Common Release Mistakes

Common mistakes include choosing low capital cost while ignoring lifecycle cost, annualizing unequal lives incorrectly, mixing salvage signs, omitting downtime cost, using maintenance savings without failure evidence, comparing defender sunk cost with challenger future cost, accepting a narrow EAC margin without sensitivity, and approving replacement when residual or implementation evidence is open.

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