The math of the MIT/Cambridge Electric Light Co. case is not that complicated. It's simple subtraction, multiplication and division, says MIT attorney John DeTore, citing the analysis submitted for MIT by Elaine Saunders, senior analyst at LaCapra Associates and a former rates manager for a major electric utility. Saunders was calculating the out of pocket loss to CELCo due to MIT's cogeneration plant.

CELCo President Russell Wright had testified that CELCo would need a rate increase of "more than six percent" to make up for the loss from the MIT plant. Since CELCo revenues were \$118 million in 1994, that six percent would be equal to more than \$7 million. But that six percent shrinks to 2/10ths of one percent, Saunders says. DeTore explains it this way:

In round figures, DeTore says, start with the fact that MIT's total business with CELCo was a little over \$9 million in 1994.

Subtract the \$4.4 million of that cost that was due to fuel cost that CELCo no longer will have, and you're down to \$4.6 million.

MIT is still buying about \$2 million a year of power from CELCO for the 25% of its campus that is not connected to the cogeneration plant. Some of that would be fuel cost, and about \$1.6 million in revenue to CELCo would be the base rate.

Subtract the \$1.6 million base rate of continuing revenue from the \$4.6 million and you're down to \$3 million.

Subtract the approximately \$950,000 which CELCo estimates is the market value of the 19.5 megawatts freed up by MIT which can be sold elsewhere, and you're down to \$2,050,000.

Subtract the approximately \$1,605,000 which CELCo estimates will be its new base revenues from growth of electricity use on the system, and you're down to \$445,000 in net lost revenues before tax savings.

Subtract the 38% tax savings (\$170,000) from declaring that \$445,000 loss and you're down to net lost revenues after tax savings of \$275,000 a year (2/10ths of one percent of CELCo revenues).

CELCo claimed that the MIT plant would cause them to lose 25 times that.

"Two tenths of one percent is what it comes down to, at the most! It will probably be even less than that. Two-tenths of one percent--that's when you've got \$100 in your pocket and you lose two dimes!" said DeTore.