Engineering Economics

Present / Annual / Future Worth Comparison

Compares mutually exclusive alternatives by present worth (PW), annual worth (AW), or future worth (FW) at the MARR, from signed cash flows (costs negative) with uniform-series shorthand and salvage values. Automatically switches to AW when lives are unequal and no common analysis period is given. Use for: 'which machine should the company buy', PW/AW/FW of a cash flow series, equivalent uniform annual cost comparisons. Not for: rate-of-return problems (irr), single factor lookups (tvm-factors).

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

Inputs

Units

Load a sample problem:

alternativesThe mutually exclusive alternatives to compare

alternatives 1

nameShort label for the alternative, e.g. "Machine A"

cashflowsIndividual cash flows. SIGN CONVENTION (critical): costs are NEGATIVE numbers, receipts POSITIVE — a purchase price at t = 0 must be negative.

cashflows 1

periodPeriod number t (0 = now; end-of-period convention)amountSIGNED cash flow: costs/disbursements NEGATIVE, receipts/incomes POSITIVE. A first cost of $50,000 is -50000.

annuitiesUniform-series shorthand: the amount repeats every period from periodStart to periodEnd inclusive (signs as above)

annuities 1

period startFirst period of the uniform series (inclusive, ≥ 1)period endLast period of the uniform series (inclusive)amountUniform end-of-period amount, SIGNED: recurring costs NEGATIVE, revenues POSITIVE

life (optional)Service life in periods; defaults to the latest period with a cash flowsalvage (optional)Salvage/market value RECEIVED at the end of life (a positive number adds a receipt at t = life)

marr percentMinimum attractive rate of return (MARR) in percent per period, e.g. 10methodComparison measure: PW = present worth, AW = annual (equivalent uniform) worth, FW = future worth. With unequal lives and no analysisPeriod the solver switches to AW automatically.analysis period (optional)Common analysis period in periods (study period). Required to compare unequal-lived alternatives by PW or FW.

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.name Machine Acashflows
1.period 0amount -5000
annuities
1.period start 1period end 5amount 1500
2.name Machine Bcashflows
1.period 0amount -8000
annuities
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

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
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
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"

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.name Pump Acashflows
1.period 0amount -9000
annuities
1.period start 1period end 3amount 4000
life 3
2.name Pump Bcashflows
1.period 0amount -15000
annuities
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

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
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
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
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"

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.name Projectcashflows
1.period 0amount -1000
annuities
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

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
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
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"

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.