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ARCHIVED - Step 5: Calculate Your Savings

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Before deciding to go forward with an energy investment or retrofit, senior management usually wants to know the payback, or how many years it will take a measure to pay for itself.

If you are looking to install new equipment or adopt new measures, simple payback will indicate the amount of time needed for energy savings to equal the purchase price. For example, if a new energy-efficient measure costs $10,000 and will save you $1,000 in energy costs each year, its simple payback is 10 years.

If your old equipment is at the end of its life cycle, incremental payback is the amount of time needed to pay for the difference between an efficient and less-efficient replacement unit. For example, an energy-efficient replacement model costs $700 and a less efficient model costs $500. If you can save $100 a year by buying the efficient model, the incremental payback of the efficient replacement model is two years compared with buying the less efficient replacement model.

Although more detailed formulas such as rate of return or return on investment, both incorporating life cycle costing, are better measures for investment decisions, the template on page 35 can help you calculate approximate savings for each type of lighting upgrade in your facility. There is also an on-line simple-payback calculator at oee.nrcan.gc.ca/eii/tools.cfm.

Calculate Your Lighting Paybacks
A. Number of new units = ________ units
B. Purchase and installation costs per unit = $ ________
    Multiply A . B      Total cost = $ ________
C. Number of new units* = ________ units
D. Old wattage – New wattage ÷ 1000 = ________ kW saved
E. Usage in hours per day = ________ hours per day
F. Usage in days per week = ________ days per week
G. Usage in weeks per year = ________ weeks per year 52
H. Average local cost per kWh of electricity = $ ________ (including demand charges)
Multiply C through H
Annual savings
= $ ________
Simple payback = Total cost ÷ Annual savings =________ years

*This assumes that the number of new lights is the same as the number of old lights.

For example, a hotel has fifty 100-watt incandescent lamps that stay on 24 hours each day. The lamps are to be replaced with 25-watt compact fluorescents that cost $25 each, including installation. The average local cost of electricity is $0.07 per kWh, including demand charges. Calculations are as follows:

Total costs = 50 . $25 = $1,250
Annual savings = 50 units x 0.075 kW saved x 24 hours per day x 7 days per week x 52 weeks per year x $0.07 per kWh = $2,293.20 per year
Simple payback = $1,250 ÷ $2,293.20/yr = 0.5 years

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