Engineering economics helps engineers compare design choices using money, time, and risk. This cheat sheet covers the time value of money formulas used to move cash flows to the present, future, or equal annual values. Students need these tools to evaluate projects, equipment purchases, loans, savings plans, and replacement decisions.
The focus is on clear formulas, correct notation, and choosing the right method for a cash flow diagram.
Key Facts
- Future value of one present amount is F = P(1 + i)^n, where P is present value, i is interest rate per period, and n is number of periods.
- Present value of one future amount is P = F / (1 + i)^n.
- Future value of a uniform series is F = A[((1 + i)^n - 1) / i].
- Present value of a uniform series is P = A[((1 + i)^n - 1) / (i(1 + i)^n)].
- Capital recovery converts present cost to annual cost using A = P[i(1 + i)^n / ((1 + i)^n - 1)].
- The effective annual interest rate for m compounding periods per year is i_eff = (1 + r/m)^m - 1.
- Net present worth is NPW = present value of benefits - present value of costs, and a project is acceptable when NPW >= 0.
- Benefit-cost ratio is B/C = present value of benefits / present value of costs, and a project is usually acceptable when B/C >= 1.
Vocabulary
- Time Value of Money
- The principle that money available now is worth more than the same amount received later because it can earn interest.
- Present Value
- The equivalent value today of a future amount or series of cash flows.
- Future Value
- The value at a later date of money invested or borrowed today after interest is applied.
- Uniform Series
- A set of equal cash flows that occur at regular time intervals.
- Discount Rate
- The interest rate used to convert future cash flows into present value.
- Net Present Worth
- The total present value of all benefits minus the total present value of all costs for a project.
Common Mistakes to Avoid
- Mixing time periods and interest rates is wrong because i and n must use the same period, such as monthly interest with number of months.
- Using the future value formula when a present value is needed gives the opposite cash flow movement and usually makes the answer too large.
- Forgetting the sign convention makes project comparisons confusing because costs and benefits must be assigned consistent positive and negative signs.
- Treating annual payments as if they occur at time zero is wrong because ordinary uniform series formulas assume payments occur at the end of each period.
- Comparing alternatives with different lifetimes without adjustment is misleading because projects should be compared using a common study period or annual worth.
Practice Questions
- 1 A project requires 10,500 in 4 years. At i = 6% per year, what is the net present worth?
- 2 What equal annual payment A is equivalent to borrowing $12,000 for 5 years at 7% interest per year?
- 3 A machine costs 6,500 per year for 5 years. At i = 8%, should the machine be accepted using present worth?
- 4 Two projects have the same net present worth, but one has higher first cost and lower yearly operating cost. Explain what non-cost factors an engineer should consider before choosing.
Understanding Engineering Economics & TVM Reference
A cash flow diagram is the starting point for nearly every engineering economy problem. Draw a horizontal timeline with one mark for each period. Put money received above the line and money paid below it.
The first cost of a machine usually occurs at time zero. Operating costs, maintenance costs, fuel savings, and revenue occur later. A salvage value is money recovered when the machine is sold at the end of its life.
This picture prevents a common mistake, which is using the right factor at the wrong time. It also makes it easier to see whether payments are single amounts, equal yearly amounts, or changing amounts.
Interest only has meaning when its period matches the cash flow period. A rate quoted per year should be used with yearly cash flows. A monthly loan payment needs a monthly rate and a number of months.
If a bank states a nominal annual rate but compounds monthly, the actual yearly growth is higher than the stated rate. The effective annual rate measures that full yearly growth. Students should carefully read words such as monthly, quarterly, annually, compounded, and end of year.
A payment at the beginning of each period differs from a payment at the end. Most standard uniform series factors assume end of period payments, so beginning payments must be shifted by one period.
Present worth gives every cash flow a common comparison date, usually today. This is useful when alternatives have different patterns of spending and earning. One pump may be cheap to buy but expensive to run.
Another may cost more at first yet use less electricity each year. Converting all amounts to present worth reveals the total economic effect under the chosen interest rate. Annual worth does a similar job by converting each alternative into an equal yearly amount.
It is especially helpful for projects with different service lives, such as a truck lasting five years compared with one lasting eight years. The comparison must use a consistent study period and include replacement when needed.
Engineering decisions involve estimates, not perfect predictions. Energy prices, repair costs, project life, and resale value may change. A sensitivity check tests whether a decision changes when one assumption moves.
For example, a design may be preferred only if electricity costs remain high. That finding matters more than a single calculated answer. Net present worth and benefit cost ratio are useful decision rules, but they do not replace judgment.
Safety requirements, environmental effects, reliability, and legal limits may rule out an option that appears cheapest. Ignore sunk costs because money already spent cannot be recovered by a new decision.
Focus on future cash flows that actually differ between the choices. Clear assumptions, correct timing, and sensible estimates make the calculations useful.