1. The problem statement, all variables and given/known data I have attached the question. 2. Relevant equations - 3. The attempt at a solution Well i first converted the 110W (j/s) into kWh which is 0.11kWh. so each tube used 0.11kWh an hour. now because there is 12 in each classroom and 6in each faculty room and 200 class rooms and 400 faculty rooms. That means there are 2400 tubes classroom tubes and 2400 faculty room tubes. that means for the 240 days it is a total power usage of: (0.11 x 2400 x 24 x 240) x 2 = 3041280 kWh. the cost then is 3041280 x 0.082 = $249384.96 When the lights are turned off during the unoccupied 4hours the total power usage then becomes: (0.11 x 2400 x 20 x 240) x 2 = 253400kWh, so the cost is 253400 x 0.082 = 207820.80. which means the campus will save $41564.16 Is this correct??? the only thing that worries me is the numbers. $249384.96 is really high but then again it is an awful lot of fluorescent tubes.