When several alternatives are being considered, each will have different cash flows over time. We assign the notation as the investment and and to indicate the income and the cost for alternative k at time t. is the life of alternative k and K is the number of alternatives. In the typical problem, the alternatives are mutually exclusive; only one may be chosen.

Alternatives may be compared with the present worth method or the annual worth method. For the present worth method, the NPW of the cash flows for alternative k is computed using the MARR.

We use the argument to emphasize that the present worth is computed over periods.

When all alternatives have the same life, the one with the greatest NPW is selected as the best.

When the alternatives have different lives, we must compare their present worths over a common time horizon called the study period. There are two ways to establish a study period. The simplest is to truncate the longer lived alternatives at the termination of the alternative with the smallest life. We must then include a salvage value for the truncated alternatives for the study period. The second method is to use a study period equal to the least common multiple of the lives. We use the latter method because it does not require estimates of the salvage values for time periods other than the project life. When we have two or more alternatives let n be the least common multiple of the lives.

One way to calculate the NPW over the study period is to first compute the NAW of each alternative and then compute the present worth for n periods.

The best alternative is the one with the greatest value of . An equivalent and sometimes easier method is to chose the alternative with the greatest value of .

These concepts are illustrated with the two alternatives below. The least common multiple of the two lives (6 and 9) is 18 years. Comparing the two on the basis of a single life using the NPW values in I2 and I17 respectively indicates that Machine A should be selected, but this is incorrect. Comparing their NAW values in cells I3 and I18 we see that Machine B should be selected. The NPW values over the study period in cells I4 and I19 also indicate that Machine B is best. This is the correct decision.