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CivilSolve

Engineering Economics

Present / Annual / Future Worth Comparison

Compare alternatives by present, annual, or future worth at a given MARR. Enter signed cash flows, uniform series, and salvage values.

Direct solving is free and needs no account. Have a word problem instead? Submit it as text.

Inputs

Load a sample problem:
alternativesThe mutually exclusive alternatives to compare
alternatives 1
cashflowsIndividual cash flows. SIGN CONVENTION (critical): costs are NEGATIVE numbers, receipts POSITIVE — a purchase price at t = 0 must be negative.
cashflows 1
annuitiesUniform-series shorthand: the amount repeats every period from periodStart to periodEnd inclusive (signs as above)
annuities 1

Worked examples

These are the solver's own reference problems — the answers come from published sources or independent hand computation, never from the solver itself. The solution below is the live solver output for each.

Example 1

PW comparison, equal 5-year lives at 10%: A beats B

Given

alternatives
  1. 1.name Machine Acashflows
    1. 1.period 0amount -5000
    annuities
    1. 1.period start 1period end 5amount 1500
  2. 2.name Machine Bcashflows
    1. 1.period 0amount -8000
    annuities
    1. 1.period start 1period end 5amount 2200
marr percent
10
method
PW

Assumptions

  • End-of-period cash-flow convention with a constant MARR per period; amounts are currency-agnostic (costs negative, receipts positive).

Solution steps

  1. Present worth of alternative 1

    Each cash flow of "Machine A" is discounted to t = 0 with a (P/F) factor at the MARR; uniform series are equivalent to a (P/A) factor times the series amount.

    PW = Σ At·(1 + i)^−t at i = MARR

    Alternative 1 flows: -5000 at t = 0; 1500 at t = 1; 1500 at t = 2; 1500 at t = 3; 1500 at t = 4; 1500 at t = 5 ⇒ PW = 686.18 at i = 10 percent

    = 686.18

  2. Present worth of alternative 2

    Each cash flow of "Machine B" is discounted to t = 0 with a (P/F) factor at the MARR; uniform series are equivalent to a (P/A) factor times the series amount.

    PW = Σ At·(1 + i)^−t at i = MARR

    Alternative 2 flows: -8000 at t = 0; 2200 at t = 1; 2200 at t = 2; 2200 at t = 3; 2200 at t = 4; 2200 at t = 5 ⇒ PW = 339.731 at i = 10 percent

    = 339.731

  3. Select the preferred alternative

    The alternative with the highest present worth at the MARR is preferred ("Machine A"). A negative best value means NO alternative earns the MARR — doing nothing beats all of them if that is an option.

    Best: alternative 1 with 686.18 (at MARR = 10 percent)

Results

Present worth of "Machine A"

686.18

Present worth of "Machine B"

339.731

Preferred alternative (1-based index): "Machine A"

1

Where this answer was checked
source
Uniform-series present worth closed form (interest-table route) — PW = −first cost + A·(P/A, 10%, 5)
verified by
hand-recomputed
derivation
(P/A,10%,5): 1.1^5 = 1.61051; (1 − 1/1.61051)/0.1 = (1 − 0.6209213)/0.1 = 3.790787. A: PW = −5000 + 1500·3.790787 = −5000 + 5686.18 = 686.18. B: PW = −8000 + 2200·3.790787 = −8000 + 8339.73 = 339.73. A preferred.

Example 2

unequal lives (3 yr vs 6 yr), PW requested → auto-switch to AW

Given

alternatives
  1. 1.name Pump Acashflows
    1. 1.period 0amount -9000
    annuities
    1. 1.period start 1period end 3amount 4000
    life 3
  2. 2.name Pump Bcashflows
    1. 1.period 0amount -15000
    annuities
    1. 1.period start 1period end 6amount 3800
    life 6
marr percent
10
method
PW

Assumptions

  • End-of-period cash-flow convention with a constant MARR per period; amounts are currency-agnostic (costs negative, receipts positive).
  • Annual Worth comparison assumes each alternative can be repeated with identical cash flows to any common multiple of the lives (repeatability assumption).

Solution steps

  1. Present worth of alternative 1

    Each cash flow of "Pump A" is discounted to t = 0 with a (P/F) factor at the MARR; uniform series are equivalent to a (P/A) factor times the series amount.

    PW = Σ At·(1 + i)^−t at i = MARR

    Alternative 1 flows: -9000 at t = 0; 4000 at t = 1; 4000 at t = 2; 4000 at t = 3 ⇒ PW = 947.408 at i = 10 percent

    = 947.408

  2. Present worth of alternative 2

    Each cash flow of "Pump B" is discounted to t = 0 with a (P/F) factor at the MARR; uniform series are equivalent to a (P/A) factor times the series amount.

    PW = Σ At·(1 + i)^−t at i = MARR

    Alternative 2: discounting all flows at i = 10 percent gives PW = 1549.99

    = 1549.99

  3. Convert to annual worth

    AW spreads each present worth over the alternative's own life with the capital-recovery factor (A/P) — this is what makes unequal lives comparable.

    AW = PW × (A/P, i, n)

    alternative 1: 947.408 × 0.40211 (n = 3) = 380.967; alternative 2: 1549.99 × 0.22961 (n = 6) = 355.889

    = 355.889

  4. Select the preferred alternative

    The alternative with the highest annual worth at the MARR is preferred ("Pump A"). A negative best value means NO alternative earns the MARR — doing nothing beats all of them if that is an option.

    Best: alternative 1 with 380.967 (at MARR = 10 percent)

Results

Annual worth of "Pump A"

380.967

Annual worth of "Pump B"

355.889

Preferred alternative (1-based index): "Pump A"

1

Notes

  • Alternatives have unequal lives and no analysis period was given — PW values over different horizons are not comparable, so the comparison was switched to Annual Worth (AW), which assumes each alternative repeats identically.
Where this answer was checked
source
Annual-worth closed form for unequal-lived alternatives — AW = PW·(A/P, 10%, n) with n = each alternative's own life
verified by
hand-recomputed
derivation
A (3 yr): (P/A,10%,3): 1.1^3 = 1.331; (1 − 0.7513148)/0.1 = 2.486852; PW = −9000 + 4000·2.486852 = 947.41; (A/P,10%,3) = 1/2.486852 = 0.402115; AW = 947.41·0.402115 = 380.97. B (6 yr): 1.1^6 = 1.771561; (P/A,10%,6) = (1 − 0.5644739)/0.1 = 4.355261; PW = −15000 + 3800·4.355261 = 1549.99; (A/P,10%,6) = 0.2296074; AW = 1549.99·0.2296074 = 355.89. A preferred.

Example 3

FW with salvage at 8%: independent two-route check gives 1377.31

Given

alternatives
  1. 1.name Projectcashflows
    1. 1.period 0amount -1000
    annuities
    1. 1.period start 1period end 5amount 400
    life 5salvage 500
marr percent
8
method
FW

Assumptions

  • End-of-period cash-flow convention with a constant MARR per period; amounts are currency-agnostic (costs negative, receipts positive).

Solution steps

  1. Present worth of alternative 1

    Each cash flow of "Project" is discounted to t = 0 with a (P/F) factor at the MARR; uniform series are equivalent to a (P/A) factor times the series amount.

    PW = Σ At·(1 + i)^−t at i = MARR

    Alternative 1: discounting all flows at i = 8 percent gives PW = 937.376

    = 937.376

  2. Convert to future worth

    FW carries each present worth to the end of the analysis horizon with the (F/P) factor.

    FW = PW × (F/P, i, n)

    alternative 1: 937.376 × 1.4693 (n = 5) = 1377.31

    = 1377.31

  3. Select the preferred alternative

    The alternative with the highest future worth at the MARR is preferred ("Project"). A negative best value means NO alternative earns the MARR — doing nothing beats all of them if that is an option.

    Best: alternative 1 with 1377.31 (at MARR = 8 percent)

Results

Future worth of "Project"

1377.31

Preferred alternative (1-based index): "Project"

1

Where this answer was checked
source
Future-worth closed forms, verified by two independent factor routes — FW = PW·(F/P, 8%, 5) vs FW = −P·(F/P) + A·(F/A) + salvage
verified by
hand-recomputed
derivation
Route 1: (P/A,8%,5): 1.08^5 = 1.4693281; (1 − 0.6805832)/0.08 = 3.992710; PW = −1000 + 400·3.992710 + 500·0.6805832 = −1000 + 1597.08 + 340.29 = 937.38; FW = 937.38·1.4693281 = 1377.31. Route 2 (no PW): (F/A,8%,5) = 0.4693281/0.08 = 5.866601; FW = −1000·1.4693281 + 400·5.866601 + 500 = −1469.33 + 2346.64 + 500 = 1377.31. Both routes agree.