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The demand for electricity varies greatly during the day. Because large amounts

ID: 394667 • Letter: T

Question

The demand for electricity varies greatly during the day. Because large amounts of electricity cannot be stored economically, electric power companies cannot manufacture electricity and hold it in inventory until it is needed. Instead, power companies must balance the production of power with the demand for power in real time. One of the greatest uncertainties in forecasting the demand for electricity is the weather. Most power companies employ meteorologists who constantly monitor weather patterns and update computer models that predict the demand for power over a rolling, seven-day planning horizon. This forecasted seven-day window of demand is referred to as the company’s load profile. Typically it is updated every hour.

Every power company has a baseload demand that is relatively constant. To satisfy this baseload demand, a power company uses its most economical, low-cost power generating assets and keeps them running continuously. To meet additional demands for power above the baseload, a power company must dispatch (or turn on) other generators. These other generators are sometimes called “peakers,” as they help the power company meet the highest demands or peak loads. It costs differing amounts of money to bring different types of peakers online. And because different peakers use different types of fuel (e.g., coal, gas, biomass), their operating costs per megawatt (MW) generated also differ. Thus, dispatchers for a power company continually have to decide which generator to bring online or turn off to meet their load profile with the least cost.

The Old Dominion Power (ODP) Company provides electrical power throughout Virginia and the Carolinas. Suppose ODP’s peak-load profile (that is the estimated load above baseload) in MWs is currently estimated as follows:

DAY

1

2

3

4

5

6

7

Load (MW)

4,300

3,700

3,900

4,000

4,700

4,800

3,600

ODP currently has three peaking generators offline that are available to help meet this load. The generators have the following operating characteristics:

Generator Location

Startup Cost

Cost per Day

Max MW Capacity per Day

New River

$800

$200 + $5 per MW

2,100

Galax

$1,000

$300 + $4 per MW

1,900

James River

$700

$250 + $7 per MW

3,000

To get an offline generator up and running, a startup cost must be paid. Once a generator is running, it can continue to run indefinitely without having to pay this startup cost again. However, if the generator is turned off at any point, the setup cost must be paid again to get it back up and running. Each day that a generator runs, there is both a fixed and variable cost that must be paid. For example, any day that the New River generator is online, it incurs a fixed cost of $200 plus $5 per MW generated. So even if this generator is not producing any MWs, it still costs $200 per day to keep it running (so as to avoid a restart). When they are running, each generator can supply up to the maximum daily MWs listed in the final column of the table.

1) Formulate a mathematical programming model for ODP’s power dispatching problem.

2) Implement your model in a spreadsheet and solve it.

3) What is the optimal solution?

4) Suppose ODP can buy power sometimes from a competitor. How much should ODP be willing to pay to acquire 300 MW of power on day 1? Explain your answer.

5) What concerns, if any, would you have about implementing this plan?

DAY

1

2

3

4

5

6

7

Load (MW)

4,300

3,700

3,900

4,000

4,700

4,800

3,600

Explanation / Answer

Decision Variables:

Xij = megawatts generated at plant i on day j

Rij = 1, if generator at plant i is running on day j; 0, otherwise.

Sij = 1, if generator at plant i is started-up on day j; 0, otherwise.

Where,

i = 1, 2, 3 represents the plants at New River, Galax, and James River respectively

j = 1, 2, … 7 represents the seven days of week

Objective function:

Total cost = (variable Cost of MW produced + Fixed cost of running plant + Start-up cost)

MIN Z = 5 (X11 + X12 + … + X17) + 4 (X21 + X22 + … + X27) + 7 (X31 + X32 + … + X37) +

200 (R11 + R12 + … + R17) + 300 (R21 + R22 + … + R27) + 250 (R31 + R32 + … + R37) +

800 (S11 + S12 + … + S17) + 1000 (S21 + S22 + … + S27) + 700 (S31 + S32 + … + S37)

Subject To:

Daily demand constraint:

X11 + X21 + X31 ³ 4300

X12 + X22 + X32 ³ 3700

X13 + X23 + X33 ³ 3900

X14 + X24 + X34 ³ 4000

X15 + X25 + X35 ³ 4700

X16 + X26 + X36 ³ 4800

X17 + X27 + X37 ³ 3600

Plant start-up and running constraint:

MW produced at plant <= (capacity of plant) x (running condition of plant)

For day 1, if plant New river is online, then X11 <= 2100R11 or X11 - 2100 R11 £ 0

X1i - 2100 R1i £ 0 , i = 1, 2, …, 7

X2i - 1900 R2i £ 0 , i = 1, 2, …, 7

X3i - 3000 R3i £ 0 , i = 1, 2, …, 7

If the plant i is stopped and restarted again, then following constraint is applied:

-R1,i-1 + R1i - S1i £ 0 , i = 2, 3, …, 7

-R2,i-1 + R2i - S2i £ 0 , i = 2, 3, …, 7

-R3,i-1 + R3i - S3i £ 0 , i = 2, 3, …, 7

Xij ³ 0, Rij, Sij binary

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