Present Value & Future Value Calculator
(Engineering Economics)
Instructions:

Select the value for which you want to solve:

Future Value (FV)

Present Value (PV)

Uniform Series (A)


Enter at least one of the following inputs:

Future Value (FV)

Present Value (PV)

Uniform Series (A)

Gradient Series (G)


Enter both of the additional inputs:

Interest Rate per Period

Number of interest periods


Click the Calculate button.

The results will be displayed at the bottom of the table.
Methodology, Equations, and Examples:
Engineering economics is a specialized field of economics that strategically applies economic principles to the decisionmaking process within engineering projects. It encompasses a comprehensive analysis of the costs and benefits of different alternatives, with the ultimate aim of evaluating their financial viability. The primary objective is to identify the most economically efficient solution that maximizes benefits while minimizing costs.
One of the key pillars of engineering economics is recognizing and considering the time value of money. This concept acknowledges that the worth of money fluctuates over time due to factors such as inflation and the potential for interest or investment returns. Consequently, a dollar received or spent in the future holds a different value than one received or spent today. Engineering economists employ various tools and formulas to precisely assess the financial impact of time, including the future value formula, present value formula, compounding periods, present value (PV), future value (FV), present value calculator, and future value calculator.
The future value formula is employed to determine the projected value of an investment or cash flow at a specific time in the future, accounting for compounding periods. This formula considers the initial investment or present value (PV), the interest rate, and the number of compounding periods to calculate the future value (FV) of the investment. By utilizing the future value formula, engineers can estimate the potential growth of an investment over time.
Conversely, the present value formula calculates the current worth of an expected future cash flow or investment. It considers the future value (FV), the interest rate, and the number of compounding periods to compute the present value (PV). This formula is crucial in determining the current value of future benefits or costs associated with engineering projects.
Engineers frequently employ present and future value calculators to simplify these calculations. These tools enable precise assessments of the worth today (present value) or the projected value in the future (future value) of cash flows or investments. Engineers can make informed decisions based on accurate financial evaluations using such calculators.
Engineering economics combines economic principles with engineering project decisionmaking. It involves assessing costs and benefits, considering the time value of money, and utilizing formulas such as the future value formula and present value formula. Furthermore, using compounding periods, present value calculators, and future value calculators enables engineers to evaluate the financial viability of alternatives and determine the optimal solution for their projects.
The terms used in this calculator are defined as follows:
A  Annual Worth or Equivalent Annual Cost
F  Future Value
G  Gradient (or Gradient Series)
i  Interest Rate (or Discount Rate)
n  Number of Time Periods
P  Present Value
Here are the formulas used in the Engineering Economics Calculator, along with examples for each:
Compound Amount (F/P, i, n):
The compound amount formula (F/P) calculates the future value of a present sum of money after compounding at a given interest rate for a specific number of periods.
To find F, given P: (F/P, i, n) F = P(1+i)^n
Example: If you invest $5,000 in a savings account with an annual interest rate of 5%, how much will you have after 10 years?
F/P = $5,000 x (1 + 0.05)^10 = $8,144.47
Present Worth (P/F, i, n):
The present worth formula (P/F) computes the current value of a future sum of money that will be received or paid at a future date, discounted back to the present at a given interest rate.
To find P, given F: (P/F, i, n) P = F(1+i)^(n)
Example: If you expect to receive $10,000 3 years from now, and the discount rate is 4%, what is the present value of that amount?
P/F = $10,000 x (1 + 0.04)^3 = $8,889.96
Series Compound Amount (F/A, i, n):
The series compound amount (F/A) formula determines the future value of equal cash flows invested or received at regular intervals over a specific period at a given interest rate.
To find F, given A: (F/A, i, n) F = A [ ( (1+i)^n  1) / i ]
Example: If you invest $500 at the end of each year for the next 8 years with an interest rate of 8%, how much will you have at the end of 8 years?
F/A = $500 x [((1 + 0.08)^8  1) / 0.08 ] = $5,318.31
Series Present Worth (P/A, i, n):
The series present worth formula (P/A) calculates the equivalent present value of equal cash flows received or paid at regular intervals over a specific period at a given interest rate.
To find P, given A: (P/A, i, n) P = A [ ((1+i)^n  1) / (i (1+i)^n ) ]
Example: You will receive $1,000 at the end of each year for the next 5 years, and the interest rate is 7%. What is the present value of this cash flow series?
P/A = $1,000 x [ (( (1 + 0.07)^51) / (0.07 (1 + 0.07)^5)] = $4,100.20
Sinking Fund (A/F,i,n):
The sinking fund factor (A/F) calculates the regular payments (or deposits) needed to accumulate a specified amount of money in a fund at a future time, with a given interest rate. It is commonly used to plan for the replacement or upgrade of an asset.
To find A, given F: (A/F, i, n) A = F x i / [ (1+i)^n  1 ]
Example: You want to accumulate $10,000 in a sinking fund over 5 years with an annual interest rate of 6%. Using the sinking fund formula, you can calculate the regular payments required:
A/F = $10,000 x 0.06 / [ (1 + 0.06)^5  1] = $1,773.96
Capital Recovery (A/P,i,n):
The capital recovery factor (A/P) calculates the equal periodic payments required to recover the initial investment and cover the interest costs over a specific period.
To find A, given P: (A/P, i, n) A = P [ i (1+i)^n / ((1+i)^n  1) ]
Example: You need to recover an initial investment of $50,000 over 10 years with an annual interest rate of 8%. Using the capital recovery formula, you can calculate the equal periodic payments required:
A/P = $50,000 x [0.08 (1+0.08)^10 / ((1+0.08)^10  1) ] = $7,451.47
Compound Gradient (F/G):
The compound gradient factor (F/G) calculates the future value of a series of increasing or decreasing cash flows that compound at a given interest rate.
To Find F, given G: (F/G, i, n) F = G x [ (1+i)^n  1  ni ] / i^2
Example: Suppose you have a series of cash flows that increase by $2,000 every year for 8 years, and the interest rate is 5%. Using the compound gradient formula, you can calculate the future value of the cash flows:
F/G = $2,000 x [ (1 + 0.05)^8  1  8 x 0.05 ] / 0.05^2 = $61,964.36
Discount Gradient (P/G):
The discount gradient factor (P/G) is used in engineering economics to calculate the present worth of a series of cash flows that change over time. It considers both the time value of money and the gradient, which represents the rate of change of the cash flows.
To find P, given G: (P/G, i, n) P = G x [ ((1+i)^n  in  1) / ((i^2) (1+i)^n ) ]
Example: Suppose you have a series of cash flows that increase by $2,000 every year for 8 years, and the interest rate is 5%. Using the compound gradient formula, you can calculate the present value of the cash flows:
P/G = $2,000 x [ (1+.05)^8  0.05 x 8 1) / (( 0.05^2 x (1+0.05)^8) ] = $41,939.91
Arithmetic Gradient Uniform Series (A/G):
The Discount Gradient (A/G) formula in engineering economics is used to calculate the present worth or future worth of a series of equal annual cash flows that change by a constant percentage or gradient over time.
To find A, given G: (A/G, i, n) A = G x [ ((1+i)^n  in 1) / (i (1+i)^n  i) ]
Example: Suppose you are considering a project that will generate an increasing annual cash flow with a constant gradient of $2,000 per year for a duration of 8 years. The interest rate for the project is 5% per year.
A = $2,000 x [ ((1+ 0.05)^8  0.05 x 8  1) / (0.05 x (1 + 0.05)^9  0.05) ] = $6,489.02
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