Present worth analysis: a worked example with two alternatives

The present worth method

1. Draw the cash-flow table of each alternative: first cost in year 0, yearly costs or savings, salvage value at the end. Costs are negative, receipts positive, and by convention every cash flow happens at the end of its year.
2. Discount each cash flow with the single-payment present worth factor (P/F, i, n) = 1 ÷ (1 + i)^n.
3. Add the present values to get the PW of each alternative.
4. Choose the alternative with the highest PW. If all cash flows are costs, both PWs are negative and you pick the less negative one.

For a uniform series the shortcut is the (P/A) factor: PW = −P − AOC × (P/A, i, n) + S × (P/F, i, n), where P is the first cost, AOC the annual operating cost and S the salvage value.

Working through the example

Machine A costs $40,000, $9,000 a year to run and is worth $5,000 after 5 years. Machine B costs $55,000, only $5,000 a year to run and is worth $8,000 at the end. At a 10% MARR, (P/A, 10%, 5) = 3.7908 and (P/F, 10%, 5) = 0.6209.

PW of A = −40,000 − 9,000 × 3.7908 + 5,000 × 0.6209 = −$71,012. PW of B = −55,000 − 5,000 × 3.7908 + 8,000 × 0.6209 = −$68,987. B has the higher (less negative) present worth, so Machine B is the better choice by $2,026 in today's dollars. Its extra $15,000 up front is more than repaid by $4,000 a year of lower running costs and a higher salvage value.

Worked example: present worth of two machines at a 10% MARR

First cost, operating cost, salvage, life and MARR are the yellow inputs. Cash flows, factors, present values and the choice are formulas.

Worked example: present worth of two machines at a 10% MARR

	A	B	C	D	E	F
1	Input	Machine A	Machine B
2	First cost, $	40,000	55,000
3	Operating cost per year, $	9,000	5,000
4	Salvage value at the end, $	5,000	8,000
5	Life, years (equal)	5	5
6	MARR	10%
7
8	Year	Cash flow A	Cash flow B	(P/F, i, n)	PV of A	PV of B
9	0	-40,000	-55,000	1.0000	-40,000	-55,000
10	1	-9,000	-5,000	0.9091	-8,182	-4,545
11	2	-9,000	-5,000	0.8264	-7,438	-4,132
12	3	-9,000	-5,000	0.7513	-6,762	-3,757
13	4	-9,000	-5,000	0.6830	-6,147	-3,415
14	5	-4,000	3,000	0.6209	-2,484	1,863
15	Present worth (sum of PVs)				-71,012	-68,987
16	Same PW with NPV: year 0 + NPV(MARR, years 1–5)				-71,012	-68,987
17	Same PW with PV factors: −P − AOC × (P/A) + S × (P/F)				-71,012	-68,987
18	Choose (highest PW = lowest present cost)				Machine B
19	Advantage of the choice in today's dollars				2,026

The machines and figures are fictional teaching numbers.

Present worth in Excel: three ways

• Factor table: one row per year, =1/(1+$B$6)^A9 for the factor and cash flow × factor for the present value, then SUM. This is the clearest layout for homework and reports.
• NPV: =B9+NPV(B6,B10:B14). Excel's NPV discounts its first value by one year, so year 0 stays outside the function.
• PV for the uniform series: =-B2+PV(B6,B5,B3)+B4/(1+B6)^B5. PV returns a negative number for a positive payment, which matches the cost sign here.

All three give the same result. Printed interest tables round the factors to four decimals, so a result from tables can differ from Excel by a few dollars.

How the MARR changes the decision

A higher MARR shrinks future cash flows more, which favours the alternative with the lower first cost. At 14% Machine B still wins, but at 15% Machine A has the higher PW by about $100. The switch happens at the rate of return on the extra investment: B costs $15,000 more and returns $4,000 a year plus $3,000 more salvage, an incremental rate of return of about 14.7%.

This is the link between present worth and rate-of-return analysis. If the incremental ROR is above the MARR, the more expensive alternative is worth its extra cost, and the PW comparison at the MARR always gives the same answer. Choosing the alternative with the highest ROR of its own, instead, can pick the wrong one.

When the lives are not equal

Present worth needs the same study period for every alternative. If one machine lasts 4 years and the other 7, comparing PWs over their own lives compares 4 years of service with 7. Either use a common study period, such as the least common multiple of the lives with repeated purchases, or switch to annual worth, the equivalent uniform annual cost, which you can compare directly. For investments judged on returns rather than costs, the same discounting logic is behind NPV and IRR.

Common mistakes

• Choosing the lower first cost. Machine A is $15,000 cheaper to buy but $2,026 more expensive over its life.
• Picking the larger negative number. With cost-only alternatives, the better one has the PW closest to zero.
• Putting year 0 inside NPV. NPV discounts its first value by a year. Add year 0 outside the function.
• Forgetting the salvage value. It is a receipt in the last year and must be discounted like every other cash flow.
• Comparing PWs of unequal lives. Use a common study period or annual worth.
• Mixing sign conventions. Keep costs negative and receipts positive throughout, and check the sign Excel's PV returns.