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Engineering economics

IEs often have to evaluate alternatives:

• Which one of several available machines should be purchased to provide a needed function in the production system?
• Will investment in a new production process be worth the anticipated savings?

Methods for making such decisions are part of the area of industrial engineering called engineering economics or engineering economy. Despite its name, this area has more in common with finance than with micro or macroeconomics. The methods of engineering economics help the IE pick the most economic decision, that is, the decision that costs the least money, over the long run.

Consider the following example. An office manager is deciding which copier to purchase or lease. Two alternatives are:

• Copier A, which can be purchased for $10,000. Maintenance and repairs are estimated to be $600 per year and the copier will last 5 years.
• Copier B, which can be leased at $3000 per year, including all maintenance and repairs

More realistic engineering economic studies may involve comparisons that cannot be immediately expressed in dollars. To keep this example simple, I will assume that the copiers have similar features, that the same number of copies will be made on either copier, and that the supplies will cost the same.

More realistic engineering economic studies also look at the tax implications of each alternative. Large engineering investments may invovle tax investment credits, depreciation, and other tax considerations. To keep this example simple, I will ignore any tax effects.

The two alternatives can be displayed in cash flow diagrams:

The horizontal lines represent time, with now shown as time 0, and the 5 years into the future shown by the other numbers. The arrows represent the dollar amounts. In this example, all the arrows point downward, indicating that the dollar amounts are costs or outflows of money; revenues would be shown by upward pointing arrows.

In drawing these diagrams, I assumed that the $10,000 cost of Copier A would be paid now, and that the annual lease amounts for Copier B are due at the start of each year; five payments will pay for the five years of the lease. This diagram shows that we are correctly comparing 5 years of service from the two alternatives; more sophisticated analysis is needed if the life of the copier and the lease period don't match.

A simple, but incorrect, analysis would be to say that Copier A costs $10,000 plus 5 payments of $750, for a total of $13,750, while Copier B costs 5 payments of $3000, for a total of $15,000. Copier A costs less and should be chosen.

That simple analysis is wrong because it ignores the fact that the purchase of Copier A means that the company gives up any earnings it could have made by investing the $10,000. The company could have, for example, bought equipment to produce products and made money, it could have invested in equipment to save money, or it could just have put the $10,000 in the bank and earned interest on the money.

The correct analysis takes account of the time value of money, that is, the fact that $1 has the same value as $(1+i) one year from now, where i is the rate of return the company can earn on its money.

For example, if i=12%, $1 now has the same value as $1.12 one year from now, and has the same value as $(1.12)² or $1.25 two years from now, and so forth, using compound interest.

We can reverse the calculation and compute that $1 one year from now has the same value as $(1/1.12) = $0.89 now, $1 two years from now has the same value as $(1/1.12²) = $0.90 now, and so forth.

To keep the example simple, I am ignoring the effect of inflation.

Now we can compare the two alternatives by converting each dollar at future times to its equivalent value now, called its present value. The present value of Copier A turns out to be $12,703.58, and the present value of Copier B is $12,112.05. Since these present values represent costs, we want to pick the copier with the smaller present value, which is Copier B.

The choice between the copiers depends on the value of i, the interest rate. If I use i=6%, the present worth for Copier A is $13,159.27 and the present worth for Copier B is $13,395.32. We would choose Copier A with that interest rate.

I can plot the present worths of the two copiers as a function of i, the interest rate, as shown below.

The two lines on the graph cross at 7.59%. For a particular value of i, we want to pick the copier with the lower present value. For interest rates below 7.59%, Copier A is the better choice; for interest rates above that value, Copier B is the better value.

This simple example ignored many complications of more sophisticated engineering economics analysis, some of which I have mentioned above. In addition to the points I have already mentioned, this example assumed perfect certainty about future costs; more sophisticated analysis uses probability to express uncertainty about the future.