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Engineering economics helps engineers choose between project alternatives using both technical performance and financial value. A design that works well may still be a poor choice if its costs arrive too early, its benefits arrive too late, or another option creates more value. The central idea is that money has a time value, so a dollar today is worth more than a dollar received in the future.

This matters in decisions about equipment, energy systems, construction projects, manufacturing upgrades, and product design.

Key Facts

  • Future worth with compound interest: F = P(1 + i)^n
  • Present worth of a future amount: P = F/(1 + i)^n
  • Net present value: NPV = present value of benefits - present value of costs
  • Choose the alternative with the largest positive NPV when projects have the same study period and risk level.
  • Simple payback period = initial investment / annual net cash inflow
  • Example: If a project costs 10000nowandreturns10000 now and returns 3000 per year for 4 years at 8%, NPV = -10000 + 3000(P/A, 8%, 4) = -10000 + 3000(3.312) = -$64, so it is slightly unattractive by NPV.

Vocabulary

Time Value of Money
The principle that money available today is worth more than the same amount of money received in the future because it can earn interest.
Present Worth
The value today of a future cost or benefit after discounting it using an interest rate.
Future Worth
The value at a future date of money invested or borrowed today after interest has accumulated.
Net Present Value
The total present value of all benefits minus the total present value of all costs for a project.
Payback Period
The time required for a project’s cash inflows to recover its initial investment.

Common Mistakes to Avoid

  • Adding future dollars directly to present dollars, because cash flows at different times must be converted to a common point in time before comparison.
  • Using the wrong sign for costs and benefits, because costs should reduce value and revenues or savings should increase value in an NPV calculation.
  • Choosing the shortest payback automatically, because payback ignores cash flows after recovery and often ignores the time value of money.
  • Comparing alternatives over unequal study periods without adjustment, because different lifetimes can bias the result unless repeated lives, annual worth, or a common study period is used.

Practice Questions

  1. 1 A machine costs 8000todayandwillsave8000 today and will save 2500 per year for 4 years. Using an interest rate of 6%, calculate the NPV. Use (P/A, 6%, 4) = 3.465.
  2. 2 You deposit $5000 in an account earning 7% annual compound interest. How much will it be worth after 5 years? Use F = P(1 + i)^n.
  3. 3 Two design alternatives have the same initial cost. Alternative A has large savings in the first two years, while Alternative B has larger savings in years 8 through 10. Explain how a higher interest rate affects which alternative is more attractive.