12.4: Time Value of Money - Present Worth

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Time Value of Money - Present Worth

Present Worth (P), often referred to as Present Value (PV), represents the equivalent current value at time zero (t=0) of a single cash amount or cash flow stream received or paid in the future (F).

While Future Worth calculates how money grows forward over time through compound interest, Present Worth works in reverse: it discounts a future cash flow back to today's terms. Because money has earning potential, a dollar received in the future is worth less than a dollar held today. Discounting allows engineers to evaluate future costs and revenues on an equal, side-by-side baseline at the present moment.

Engineers routinely use Present Worth calculations to evaluate:

Capital Justification: Determining the maximum initial amount a company should spend today to purchase equipment that will generate specific cost savings in the future.
Comparing Alternatives: Converting future maintenance costs, operational expenses, or equipment salvage values back to so competing design proposals can be compared fairly.
Present Funding Requirements: Calculating the exact lump sum that must be set aside today at a known interest rate to cover a guaranteed future expense.

The Present Worth Mathematical Model (Lump Sum)

Starting with the Future Worth: \[ F = P(1 + i)^N \]

Solving for the Present Worth: \[ P = \frac{F}{(1 + i)^N} \] or \[ P = F(1 + i)^{-N} \]

Where:

= Present worth (equivalent lump-sum value at time )
= Future worth (lump-sum amount at period )
= Interest rate (or discount rate) per compounding period
= Number of compounding periods

The term $\frac{1}{(1 + i)^N}$ or $(1 + i)^{-N}$ is formally called the Single-Payment Present Worth Factor (or Discount Factor).

In engineering economics factor notation, it is written as $(P/F, i, N)$, which reads: "Find $P$, given $F$, at interest rate $i$, for $N$ periods."

Engineering Example: Major Overhaul Funding Ahead (Lump Sum)

A municipality is designing a wastewater treatment facility.

Engineers know that a specialized high-volume pump will require a major overhaul costing in years.

If the city can invest funds today in a municipal bond account yielding annual compound interest, how much money () must be deposited today to cover this future overhaul?

Variables:

Future Worth () =
Interest Rate () = per year
Number of Periods () = years

Apply the Present Worth Formula:

\[ P = \frac{F}{(1 + i)^N} = \frac{\$30,000}{(1 + 0.06)^8} = \frac{\$30,000}{1.593848} = \$18,822.37 \]

3. Engineering Interpretation:

Depositing today at interest will grow to exactly in 8 years.

The Present Worth of the future overhaul is .

Spending any amount less than today to eliminate that future expense would be a good financial decision.

Yearly Expense or Cost (Not Lump Sum)

If the problem were not a lump sum but a yearly expense or cost then we need to use a different formula.

The Present Worth of a Uniform Series is calculated using the Uniform Series Present Worth Factor $(P/A, i, N)$:

\[ P = A \left[ \frac{(1 + i)^N - 1}{i(1 + i)^N} \right] \]

Engineering Example: Yearly Expense or Cost

Example:
$P$ = Maximum initial purchase price (Present Worth)
$A$ = Annual savings = $15,000
$i$ = Interest rate per period = 8% or 0.08 (which we call the discount rate)
$N$ = Useful life = 6 years

Step-by-Step Calculation

1. Calculate $(1 + i)^N$:
\[ (1 + 0.08)^6 = 1.08^6 \approx 1.586874 \]

2. Calculate the $(P/A, 8\%, 6)$ factor:
\[ (P/A, 8\%, 6) = \frac{1.586874 - 1}{0.08 \times 1.586874} = \frac{0.586874}{0.126950} \approx 4.622880 \]

3. Calculate Present Worth ($P$):
\[ P = \$15,000 \times 4.622880 = \$69,343.19 \]

The Present Worth =$69,343.19

The Maximum initial purchase price is $69,343.19

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Engineering Example: Major Overhaul Funding Ahead (Lump Sum)

Yearly Expense or Cost (Not Lump Sum)

Engineering Example: Yearly Expense or Cost