riaology
Learning path 6 · Topic 29 of 30

Give unequal lives a fair comparison

A three-year asset and a five-year asset cannot be compared using their own raw life-cycle PW as if they provide equal service.

Change a scenario. Watch the consequence.
AlternativeLifeEAC ($/year)15-year cost PW ($)
A35,719.0343,499.42
B55,420.7641,230.73
EAC(B) minus EAC(A) · $/yearEACA=5719.033233,EACB=5420.759693,ΔEAC=−298.27354EAC_A=5719.033233,\quad EAC_B=5420.759693,\quad \Delta EAC=-298.27354

Choose B for equal required service under repeatability. The common repeated study period is LCM(3,5)=15 years.

Follow the reasoning

Assume identical replacement cycles and continuing service. A costs $10,000 with $1,000 salvage; B costs $18,000 with $2,000 salvage.

Cash-flow dates and signs are part of the answer. Model values retain precision; displays round money to cents. Illustrative assumptions exclude taxes and inflation unless stated.

01

What the model means

Under repeatability and continuing equal service, compare per-cycle annual costs at the same MARR, or repeat full cycles to the least common multiple of lives. LCM(3,5)=15, not simply 3+5. Replacements and salvage at each cycle boundary must be included.

02

Build the calculation

For a specified nonrepeatable study period, use the stated horizon and estimate terminal market values, replacements, alternative service, or reinvestment consistently. Investing a short-lived project’s proceeds to a later date assumes no missing required service; it is not automatically valid for a machine comparison.

EAC=I(A/P,i,N)+E−S(A/F,i,N),PWcommon=EAC(P/A,i,H)EAC=I(A/P,i,N)+E-S(A/F,i,N),\qquad PW_{common}=EAC(P/A,i,H)
Practice with feedback

Set up. Calculate. Explain.

Money: enter dollars

Rebuild the starting scenario

MARR: 10 % per year; Operating cost of 3-year A: 2000 $/year; Operating cost of 5-year B: 1000 $/year. Calculate the EAC(B) minus EAC(A).

Setup hint

Assume identical replacement cycles and continuing service. A costs $10,000 with $1,000 salvage; B costs $18,000 with $2,000 salvage.

Open the visual solutionEACA=5719.033233,EACB=5420.759693,ΔEAC=−298.27354EAC_A=5719.033233,\quad EAC_B=5420.759693,\quad \Delta EAC=-298.27354
  1. Assume identical replacement cycles and continuing service. A costs $10,000 with $1,000 salvage; B costs $18,000 with $2,000 salvage.
  2. Compute each cycle EAC at the same MARR, or repeat full cycles to 15 years and compare cost PW.
  3. For a fixed nonrepeatable study horizon, include terminal market values, replacements or service gaps explicitly. Never compare raw 3-year and 5-year PW as if horizons were equal.
AlternativeLifeEAC ($/year)15-year cost PW ($)
A35,719.0343,499.42
B55,420.7641,230.73
EAC(B) minus EAC(A) · $/yearEACA=5719.033233,EACB=5420.759693,ΔEAC=−298.27354EAC_A=5719.033233,\quad EAC_B=5420.759693,\quad \Delta EAC=-298.27354

Choose B for equal required service under repeatability. The common repeated study period is LCM(3,5)=15 years.

Visual solution

Assume identical replacement cycles and continuing service. A costs $10,000 with $1,000 salvage; B costs $18,000 with $2,000 salvage.

Cash-flow dates and signs are part of the answer. Model values retain precision; displays round money to cents. Illustrative assumptions exclude taxes and inflation unless stated.

Checks allow 0.1% numerical rounding (minimum $0.05 or 0.001 in the stated nonmonetary unit). Interpolated factors and rates are approximations; keep table precision consistent.

Money: enter dollars

Transfer the idea to a new scenario

Use A: $10,000, life 3, salvage $1,000, operating $2,200/year; B: $18,000, life 5, salvage $2,000, operating $900/year. At 8%, find EAC(B)−EAC(A).

Setup hint

Assume identical replacement cycles and continuing service. A costs $10,000 with $1,000 salvage; B costs $18,000 with $2,000 salvage.

Open the visual solutionEACA=5772.301626,EACB=5067.303273,ΔEAC=−704.998353EAC_A=5772.301626,\quad EAC_B=5067.303273,\quad \Delta EAC=-704.998353
  1. Assume identical replacement cycles and continuing service. A costs $10,000 with $1,000 salvage; B costs $18,000 with $2,000 salvage.
  2. Compute each cycle EAC at the same MARR, or repeat full cycles to 15 years and compare cost PW.
  3. For a fixed nonrepeatable study horizon, include terminal market values, replacements or service gaps explicitly. Never compare raw 3-year and 5-year PW as if horizons were equal.
AlternativeLifeEAC ($/year)15-year cost PW ($)
A35,772.3049,407.89
B55,067.3043,373.47
EAC(B) minus EAC(A) · $/yearEACA=5772.301626,EACB=5067.303273,ΔEAC=−704.998353EAC_A=5772.301626,\quad EAC_B=5067.303273,\quad \Delta EAC=-704.998353

Choose B for equal required service under repeatability. The common repeated study period is LCM(3,5)=15 years.

Visual solution

Assume identical replacement cycles and continuing service. A costs $10,000 with $1,000 salvage; B costs $18,000 with $2,000 salvage.

Cash-flow dates and signs are part of the answer. Model values retain precision; displays round money to cents. Illustrative assumptions exclude taxes and inflation unless stated.

Checks allow 0.1% numerical rounding (minimum $0.05 or 0.001 in the stated nonmonetary unit). Interpolated factors and rates are approximations; keep table precision consistent.

Explain the distinction

When is comparing each asset’s cycle EAC defensible?

Can you defend your setup?

Can you place each amount on the timeline, match the interest period, calculate the equivalent quantity, and state the decision or limitation?

← PreviousNext →

Completion is saved on this browser.