How to Calculate the Running Cost of an Electrical Device

A 2.0 kW electric heater runs for 30 minutes. Your electricity tariff is 30 cents per kilowatt-hour. Will running it cost 60 cents?

Not quite. The heater draws 2.0 kW for half an hour, so it uses 1.0 kWh of electrical energy. At 30 cents per kWh, the running cost is 30 cents.

The important distinction is that power tells you how quickly a device uses energy, while energy tells you how much it uses altogether. Your electricity bill charges for the second quantity.

To calculate the cost of running an electrical device, you need three things:

• Its electrical input power.
• How long it operates.
• The electricity tariff, usually quoted in cents per kilowatt-hour (c/kWh).

There is one more complication worth understanding: the power printed on an appliance does not always tell you how much energy it actually uses during a complete operating cycle.

01The basic calculation: energy multiplied by electricity price

The running cost is calculated in two stages.

First, calculate the electrical energy consumed:

\[
E_{\mathrm{in}}=P_{\mathrm{in}}t
\]

Then multiply that energy by the electricity tariff:

\[
C=E_{\mathrm{in}}r
\]

Here:

• \(E_{\mathrm{in}}\) is the electrical energy drawn from the supply, measured in kilowatt-hours (kWh).
• \(P_{\mathrm{in}}\) is the electrical input power, measured in kilowatts (kW).
• \(t\) is the operating time, measured in hours (h).
• \(C\) is the running cost, measured in dollars.
• \(r\) is the electricity tariff, measured in dollars per kilowatt-hour.

Notice that the first equation uses input power, not necessarily the useful power produced by the device. We will return to this distinction shortly.

Getting the units right

Electricity tariffs are usually quoted in cents per kilowatt-hour, so your calculations need to use kilowatts and hours.

Remember these conversions:

For example, a device rated at 1500 W has an input power of:

\[
P_{\mathrm{in}}=\frac{1500}{1000}
=1.5\text{ kW}
\]

If it operates for 45 minutes:

\[
t=\frac{45}{60}=0.75\text{ h}
\]

You can now calculate its energy consumption directly in kWh.

A useful way to picture the relationship is to compare power with speed. A car travelling at 60 km/h for two hours covers 120 km. Similarly, a device drawing 2 kW for half an hour consumes 1 kWh.

The analogy is about rate multiplied by time. Electrical energy is not literally distance, but the mathematical relationship is the same.

02Which power should you use? Understanding efficiency

Here is a common trap when calculating the cost of running electrical devices.

Imagine an electric motor that produces 450 W of useful mechanical power. You are told that its efficiency is 75%.

Should you calculate the electricity cost using 450 W?

No. The motor needs more electrical power than the mechanical power it delivers.

Some of the electrical energy is transformed into unwanted forms, such as thermal energy and sound. The electrical input must therefore be greater than the useful mechanical output.

The efficiency equation

Efficiency describes the proportion of input energy converted into useful output energy.

\[
\eta=
\frac{E_{\mathrm{useful}}}
{E_{\mathrm{in}}}
\]

It can also be expressed using power:

\[
\eta=
\frac{P_{\mathrm{useful}}}
{P_{\mathrm{in}}}
\]

Here, \(\eta\) is efficiency, expressed as a decimal or percentage.

Rearranging the power equation gives:

\[
P_{\mathrm{in}}=
\frac{P_{\mathrm{useful}}}{\eta}
\]

For example, a motor producing 450 W of useful mechanical power at 75% efficiency requires:

\[
P_{\mathrm{in}}=\frac{450}{0.75}
\]

\[
P_{\mathrm{in}}=600\text{ W}
\]

It draws 600 W of electrical power, even though its useful mechanical output is only 450 W.

The important distinction:

• If the question gives electrical input power, use it directly to calculate electricity consumption.
• If the question gives useful output power, use the efficiency to calculate electrical input power first.

Do not divide an electrical input rating by efficiency. That would count the losses twice.

03What if the device does not operate at constant power?

So far, we have assumed that the device draws a constant amount of electrical power throughout its operation.

That is a useful starting point, but real appliances do not always behave this way.

Consider a refrigerator. Its compressor switches on and off as the temperature changes. When the compressor operates, the refrigerator draws considerably more power than when the compressor is idle.

A heater controlled by a thermostat behaves similarly.

If you multiply the maximum rated power by the entire time the appliance is switched on, you may overestimate its actual electricity consumption.

Calculating energy when power changes

When power changes, calculate the energy consumed during each interval separately.

For a device operating at different constant power levels:

\[
E_{\mathrm{total}}
=P_1t_1+P_2t_2+\cdots
\]

Each power value must be multiplied by the time for which that particular power applies.

Then add the energy from all intervals.

For example, suppose an appliance operates:

• At 1000 W for 20 minutes.
• At 500 W for 40 minutes.

Convert the powers into kilowatts:

\[
P_1=1.0\text{ kW}
\]

\[
P_2=0.5\text{ kW}
\]

Convert the times into hours:

\[
t_1=\frac{20}{60}=\frac13\text{ h}
\]

\[
t_2=\frac{40}{60}=\frac23\text{ h}
\]

Calculate the energy for each interval:

\[
E_1=1.0\times\frac13
=0.333\text{ kWh}
\]

\[
E_2=0.5\times\frac23
=0.333\text{ kWh}
\]

Total energy:

\[
E_{\mathrm{total}}=0.333+0.333
\]

\[
E_{\mathrm{total}}=0.667\text{ kWh}
\]

If you had incorrectly assumed that the appliance operated at 1000 W for the entire hour, you would have calculated 1.0 kWh.

That is a substantial overestimate.

The general principle is that electrical energy is the accumulated power over time. On a power-time graph, the area beneath the graph represents the energy transferred.

04How an electricity meter measures actual consumption

What happens if you don’t know the actual power being drawn by an appliance?

You can estimate its consumption from its rating, but an energy meter can provide a more direct measurement.

This is particularly useful for appliances that switch between operating modes.

Setting up a measurement

A simple experiment uses a commercially manufactured plug-in energy meter.

The apparatus is connected in this order:

1. The energy meter plugs into a suitable wall socket.
2. The appliance plugs into the energy meter.
3. The energy meter displays the electrical quantities being measured, which may include instantaneous power and accumulated energy in kWh.

The experiment begins with a simple question:

Does the appliance actually consume its rated power throughout its entire operating cycle?

Suppose an appliance is labelled with an electrical input power of 1.5 kW.

If it operates for 30 minutes at this power continuously, the predicted energy consumption is:

\[
E_{\mathrm{predicted}}=1.5\times0.5
\]

\[
E_{\mathrm{predicted}}=0.75\text{ kWh}
\]

That is our prediction.

Now imagine that the energy meter records an increase of only 0.62 kWh during the same 30-minute operating cycle.

The measured energy is lower than the prediction.

What does this tell us?

It tells us that the average electrical input power during the cycle was lower than 1.5 kW.

We can calculate that average power using:

\[
P_{\mathrm{avg}}=\frac{E_{\mathrm{measured}}}{t}
\]

Substitute:

\[
P_{\mathrm{avg}}=\frac{0.62}{0.5}
\]

\[
P_{\mathrm{avg}}=1.24\text{ kW}
\]

The average input power was 1.24 kW.

Why was the prediction different?

The original calculation assumed that the appliance operated at 1.5 kW for the entire 30 minutes.

That assumption may not be correct.

The appliance might:

• Switch between high-power and low-power operating modes.
• Reduce its power once it reaches a target temperature.
• Spend part of the cycle in standby.
• Draw different amounts of power during different stages of operation.

The energy meter improves the measurement because it accumulates the actual electrical energy consumed throughout the interval rather than assuming a constant power.

The difference between the prediction and observation does not, by itself, establish that the appliance is inefficient or faulty.

The meter measures electrical input energy. It does not tell us how much useful output energy the appliance produces.

To calculate efficiency, we would need an additional measurement of useful energy output.

For safety, use only an appropriately rated, commercially manufactured energy meter. Do not open appliances or modify household mains wiring.

Calculating the cost from a measured meter reading

Suppose the appliance in our experiment operates at a tariff of 29 c/kWh.

Its measured energy consumption was 0.62 kWh.

Convert the tariff:

\[
r=0.29\text{ dollars/kWh}
\]

Calculate the cost:

\[
C=0.62\times0.29
\]

\[
C=0.1798
\]

The measured operating cycle costs approximately 18 cents.

This is often a better estimate than using the maximum rated power, particularly for appliances with variable operating cycles.

05Electricity tariffs are not always constant

There is one final complication.

The electricity tariff does not necessarily remain the same throughout the day.

Some electricity plans use time-of-use pricing, where electricity costs different amounts depending on when it is consumed.

For example, an illustrative tariff might be:

This means that two appliances can consume exactly the same amount of electrical energy but have different running costs if they operate at different times.

It also means that a single average electricity price cannot always be used.

When the tariff changes, calculate the cost of the energy consumed during each tariff period separately:

\[
C_{\mathrm{total}}
=C_1+C_2+\cdots
\]

For example, suppose a 2 kW appliance operates for:

• One hour during off-peak pricing.
• One hour during peak pricing.

Its energy consumption is:

\[
E_{\mathrm{off}}=2\times1=2\text{ kWh}
\]

\[
E_{\mathrm{peak}}=2\times1=2\text{ kWh}
\]

The total energy consumption is:

\[
E_{\mathrm{total}}=4\text{ kWh}
\]

The cost is:

\[
C_{\mathrm{total}}
=(2\times0.20)+(2\times0.50)
\]

\[
C_{\mathrm{total}}=0.40+1.00
\]

The total running cost is $1.40.

You cannot simply multiply the total energy by whichever tariff looks most convenient.

You must match each amount of energy to the tariff that applied when it was consumed.

06A reliable method for HSC questions

When you encounter a question asking you to calculate the cost of running an electrical device, work through these decisions in order.

If power changes, calculate energy for each interval and add the results.

If the tariff changes, calculate the cost for each interval and add the results.

If you have a measured energy reading, use that instead of estimating energy from a constant rated power.

What this calculation includes, and what it doesn’t

The calculations in this article determine the energy-use component of an electricity bill.

An actual household bill may also include:

• Daily supply charges.
• Other usage charges or tariff arrangements.
• Different peak, shoulder, and off-peak rates.
• Applicable taxes and adjustments.

The worked examples assume that the quoted tariff is the applicable usage rate, with no additional charges included.

They also assume that the electrical energy is drawn from the grid. If a household uses solar generation or battery storage, the amount of energy purchased from the grid may differ from the appliance’s total electrical energy consumption.

For the underlying physics of electrical energy transfer, including the relationship between power, voltage, current, and time, see Electrical Energy in HSC Physics: Using E = Pt.

07Questions and solutions

The following questions progress from straightforward running-cost calculations to problems involving changing power, efficiency, actual meter readings, and time-of-use tariffs.

All electricity prices are illustrative.

Question 1

A portable electric heater has an electrical input power of 2.4 kW.

It operates for 35 minutes. The electricity tariff is 28 c/kWh.

Calculate:

a) The electrical energy consumed.

b) The cost of operating the heater.

Solution 1

The heater consumes 1.4 kWh of electrical energy, giving a running cost of approximately $0.39.

Part a: Calculate the electrical energy.

First, convert the operating time into hours:

\[
t=\frac{35}{60}
\]

\[
t=0.5833\text{ h}
\]

Use:

\[
E_{\mathrm{in}}=P_{\mathrm{in}}t
\]

Substitute:

\[
E_{\mathrm{in}}=2.4\times0.5833
\]

\[
E_{\mathrm{in}}=1.4\text{ kWh}
\]

Part b: Calculate the running cost.

Convert the tariff into dollars:

\[
r=28\text{ c/kWh}
=0.28\text{ dollars/kWh}
\]

Calculate:

\[
C=E_{\mathrm{in}}r
\]

\[
C=1.4\times0.28
\]

\[
C=0.392
\]

Therefore, the running cost is approximately $0.39.

The important point is that the heater’s power rating describes its energy consumption per unit time. Its total energy consumption depends on how long it operates.

Question 2

A thermostat-controlled electric heater operates in a repeating cycle.

During each five-minute cycle:

• It draws 1.2 kW for two minutes.
• It draws 8 W in standby for the remaining three minutes.

The heater completes eight cycles during each operating session.

It is used for four sessions per week.

The electricity tariff is 31 c/kWh.

Calculate the total electrical energy consumed during one week and the weekly running cost.

Solution 2

The heater consumes 1.2928 kWh per week, giving a weekly running cost of approximately $0.40.

The key is to account for both operating states. Standby power is small, but it is not zero.

Step 1: Calculate the total time spent operating at full power.

Each cycle has two minutes of full-power operation.

There are eight cycles:

\[
t_{\mathrm{on}}=8\times2
\]

\[
t_{\mathrm{on}}=16\text{ min}
\]

Convert to hours:

\[
t_{\mathrm{on}}=\frac{16}{60}
=0.2667\text{ h}
\]

Step 2: Calculate the energy consumed at full power.

\[
E_{\mathrm{on}}=P_{\mathrm{on}}t_{\mathrm{on}}
\]

\[
E_{\mathrm{on}}=1.2\times0.2667
\]

\[
E_{\mathrm{on}}=0.320\text{ kWh}
\]

Step 3: Calculate the total standby time.

Each cycle has three minutes of standby:

\[
t_{\mathrm{standby}}=8\times3
\]

\[
t_{\mathrm{standby}}=24\text{ min}
\]

Convert to hours:

\[
t_{\mathrm{standby}}=\frac{24}{60}
=0.4\text{ h}
\]

Convert standby power into kilowatts:

\[
P_{\mathrm{standby}}=\frac{8}{1000}
\]

\[
P_{\mathrm{standby}}=0.008\text{ kW}
\]

Step 4: Calculate standby energy consumption.

\[
E_{\mathrm{standby}}
=P_{\mathrm{standby}}t_{\mathrm{standby}}
\]

\[
E_{\mathrm{standby}}=0.008\times0.4
\]

\[
E_{\mathrm{standby}}=0.0032\text{ kWh}
\]

Step 5: Calculate the total energy per session.

\[
E_{\mathrm{session}}
=E_{\mathrm{on}}+E_{\mathrm{standby}}
\]

\[
E_{\mathrm{session}}
=0.320+0.0032
\]

\[
E_{\mathrm{session}}=0.3232\text{ kWh}
\]

Step 6: Calculate the weekly energy consumption.

There are four sessions per week:

\[
E_{\mathrm{week}}=4\times0.3232
\]

\[
E_{\mathrm{week}}=1.2928\text{ kWh}
\]

Step 7: Calculate the weekly cost.

Convert the tariff:

\[
r=0.31\text{ dollars/kWh}
\]

Then:

\[
C=1.2928\times0.31
\]

\[
C=0.400768
\]

Therefore, the weekly running cost is approximately $0.40.

A tempting shortcut would be to multiply 1.2 kW by the entire 40-minute session. That would assume the heater draws full power continuously, which contradicts the operating cycle.

This question illustrates why you must pay attention to the power drawn during each interval rather than automatically using the rated power for the entire operating time.

Question 3

An electric water pump is driven by a motor that produces 420 W of useful mechanical output power.

The motor has an efficiency of 70%.

The pump operates for 50 minutes per day, six days per week.

The electricity tariff is 32 c/kWh.

Calculate:

a) The electrical input power of the motor.

b) The total electrical energy consumed during one week.

c) The weekly running cost.

Solution 3

The motor draws 600 W of electrical input power and consumes 3.0 kWh per week. Its weekly running cost is $0.96.

Part a: Calculate the electrical input power.

The given 420 W is useful mechanical output power, not electrical input power.

Use the efficiency relationship:

\[
\eta=
\frac{P_{\mathrm{useful}}}{P_{\mathrm{in}}}
\]

Rearrange:

\[
P_{\mathrm{in}}=
\frac{P_{\mathrm{useful}}}{\eta}
\]

Substitute:

\[
P_{\mathrm{in}}=\frac{420}{0.70}
\]

\[
P_{\mathrm{in}}=600\text{ W}
\]

Therefore, the motor draws 600 W of electrical power.

Part b: Calculate the weekly electrical energy consumption.

First, calculate the total operating time:

\[
t=50\times6
\]

\[
t=300\text{ min}
\]

Convert to hours:

\[
t=\frac{300}{60}=5\text{ h}
\]

Convert the electrical input power into kilowatts:

\[
P_{\mathrm{in}}=0.60\text{ kW}
\]

Now calculate energy:

\[
E_{\mathrm{in}}=P_{\mathrm{in}}t
\]

\[
E_{\mathrm{in}}=0.60\times5
\]

\[
E_{\mathrm{in}}=3.0\text{ kWh}
\]

Part c: Calculate the weekly running cost.

Convert the tariff:

\[
r=0.32\text{ dollars/kWh}
\]

Then:

\[
C=3.0\times0.32
\]

\[
C=0.96
\]

The weekly running cost is $0.96.

The tempting mistake is to use 420 W directly in the energy equation.

That would calculate the useful mechanical output energy, not the electrical energy purchased from the supply.

Because the motor is only 70% efficient, the electrical input must be greater than the useful mechanical output.

Question 4

An electrical appliance has a rated electrical input power of 1.5 kW.

A plug-in energy meter is connected between the wall socket and the appliance.

During a 30-minute operating cycle, the meter records an increase in cumulative energy consumption of 0.62 kWh.

The electricity tariff is 29 c/kWh.

Calculate:

a) The average electrical input power during the cycle.

b) The actual running cost of the cycle.

c) Explain why the measured energy consumption does not establish the appliance’s efficiency.

Solution 4

The appliance’s average input power is 1.24 kW, and the measured running cost is approximately 18 cents. Its efficiency cannot be determined from the information provided.

Part a: Calculate the average input power.

We know that:

\[
E_{\mathrm{in}}=P_{\mathrm{avg}}t
\]

Rearrange:

\[
P_{\mathrm{avg}}=\frac{E_{\mathrm{in}}}{t}
\]

Convert the operating time:

\[
t=\frac{30}{60}=0.5\text{ h}
\]

Substitute:

\[
P_{\mathrm{avg}}=\frac{0.62}{0.5}
\]

\[
P_{\mathrm{avg}}=1.24\text{ kW}
\]

Therefore, the average electrical input power was 1.24 kW.

This is lower than the appliance’s rated 1.5 kW input power.

The difference may be explained by changing operating conditions or power consumption during the cycle.

Part b: Calculate the running cost.

Convert the tariff:

\[
r=0.29\text{ dollars/kWh}
\]

Calculate:

\[
C=E_{\mathrm{in}}r
\]

\[
C=0.62\times0.29
\]

\[
C=0.1798
\]

The actual running cost is approximately 18 cents.

Part c: Explain why efficiency cannot be calculated.

Efficiency requires both useful output energy and total input energy:

\[
\eta=
\frac{E_{\mathrm{useful}}}{E_{\mathrm{in}}}
\]

The energy meter measures the electrical input energy, which is 0.62 kWh.

However, the question does not provide the useful energy output.

Therefore, efficiency cannot be calculated.

It would be incorrect to compare the measured energy with the energy predicted from the rated power and call the ratio efficiency. That comparison tells us how the actual input consumption differs from the constant-power estimate, not how effectively the appliance converts energy into useful output.

Question 5

An industrial motor is used to complete a task requiring 1.20 kWh of useful mechanical energy.

The motor can operate under two different programs. Its efficiency is constant at 80% in both programs.

The electricity supplier uses the following illustrative time-of-use tariff:

The motor can operate under either of the following schedules:

Plan A

• From 4:00 pm to 5:00 pm, electrical input power is 1.0 kW.
• From 5:00 pm to 6:00 pm, electrical input power is 0.5 kW.

Plan B

• From 4:00 pm to 5:00 pm, electrical input power is 0.5 kW.
• From 5:00 pm to 6:00 pm, electrical input power is 1.0 kW.

Assume that power remains constant within each hour, the motor has no additional standby consumption, and all electrical energy is purchased from the grid.

Answer the following:

a) Calculate the total electrical input energy consumed under each plan.

b) Calculate the useful mechanical energy produced under each plan. Verify that both plans complete the required task.

c) Calculate the running cost of each plan.

d) Which plan is cheaper? Explain why two operating schedules can consume the same total electrical energy and produce the same useful output but have different running costs.

e) An energy meter displays only the final cumulative energy consumption for each run. Could this reading alone identify which plan is cheaper? Explain.

Solution 5

Plan B consumes the same total electrical energy as Plan A but costs more. Plan A costs $0.45, while Plan B costs $0.60.

The important distinction is that equal total energy consumption does not guarantee equal running costs when electricity tariffs change during the operating period.

Part a: Calculate the total electrical input energy.

For Plan A, the motor operates at 1.0 kW during the first hour:

\[
E_{A1}=1.0\times1
\]

\[
E_{A1}=1.0\text{ kWh}
\]

During the second hour, it operates at 0.5 kW:

\[
E_{A2}=0.5\times1
\]

\[
E_{A2}=0.5\text{ kWh}
\]

Therefore:

\[
E_A=1.0+0.5
\]

\[
E_A=1.5\text{ kWh}
\]

For Plan B, the power levels are reversed.

First hour:

\[
E_{B1}=0.5\times1
=0.5\text{ kWh}
\]

Second hour:

\[
E_{B2}=1.0\times1
=1.0\text{ kWh}
\]

Therefore:

\[
E_B=0.5+1.0
\]

\[
E_B=1.5\text{ kWh}
\]

Both plans consume exactly 1.5 kWh of electrical energy.

Part b: Calculate the useful mechanical energy.

The motor has an efficiency of 80%.

Convert this into a decimal:

\[
\eta=0.80
\]

Use:

\[
E_{\mathrm{useful}}=\eta E_{\mathrm{in}}
\]

For Plan A:

\[
E_{\mathrm{useful,A}}
=0.80\times1.5
\]

\[
E_{\mathrm{useful,A}}=1.20\text{ kWh}
\]

For Plan B:

\[
E_{\mathrm{useful,B}}
=0.80\times1.5
\]

\[
E_{\mathrm{useful,B}}=1.20\text{ kWh}
\]

Both plans produce exactly the required 1.20 kWh of useful mechanical energy.

Neither plan wastes more energy than the other. Both have the same efficiency and consume the same electrical energy.

Part c: Calculate the running cost of each plan.

For Plan A:

During the first hour, 1.0 kWh is consumed at 20 c/kWh.

\[
C_{A1}=1.0\times0.20
\]

\[
C_{A1}=0.20
\]

During the second hour, 0.5 kWh is consumed at 50 c/kWh.

\[
C_{A2}=0.5\times0.50
\]

\[
C_{A2}=0.25
\]

Therefore:

\[
C_A=0.20+0.25
\]

\[
C_A=0.45
\]

Plan A costs $0.45.

Now calculate Plan B.

During the first hour:

\[
C_{B1}=0.5\times0.20
\]

\[
C_{B1}=0.10
\]

During the second hour:

\[
C_{B2}=1.0\times0.50
\]

\[
C_{B2}=0.50
\]

Therefore:

\[
C_B=0.10+0.50
\]

\[
C_B=0.60
\]

Plan B costs $0.60.

Part d: Which plan is cheaper?

Plan A is cheaper by 15 cents.

\[
\Delta C=0.60-0.45
\]

\[
\Delta C=0.15
\]

Although both plans consume 1.5 kWh and produce 1.20 kWh of useful mechanical energy, they distribute their energy consumption differently across the two tariff periods.

Plan A uses more electricity during the cheaper period.

Plan B uses more electricity during the expensive period.

This is why their running costs differ.

A tempting approach would be to calculate the average tariff:

\[
r_{\mathrm{average}}=
\frac{20+50}{2}
\]

\[
r_{\mathrm{average}}=35\text{ c/kWh}
\]

Then multiply this by 1.5 kWh.

However, this would give 52.5 cents for both plans, which is incorrect.

The problem is that the energy consumption is not distributed equally between the two tariff periods. The correct average tariff must account for how much energy is consumed at each price.

Part e: Can the final cumulative meter reading identify the cheaper plan?

No. A final cumulative reading of 1.50 kWh alone cannot distinguish the two plans.

The meter records the same total electrical energy consumption in both cases.

However, it does not reveal how much of that energy was consumed during the cheaper period and how much was consumed during the more expensive period.

To determine the actual running cost, we need energy consumption data for each tariff interval.

This is the important conceptual result: total energy determines the cost only when the applicable tariff is constant.

When electricity prices vary with time, the timing of energy consumption becomes part of the calculation.

08What to understand next

Calculating electricity costs brings together three ideas that are easy to confuse:

• Power describes the rate of electrical energy consumption.
• Energy describes the total amount consumed.
• Efficiency describes how much of the input energy becomes useful output.

Once you can distinguish these quantities, you can make much better comparisons between electrical devices.

The next useful step is understanding power-time graphs and energy efficiency together. They allow you to predict how much electricity a device will consume, even when its power changes throughout an operating cycle.