Capacity Expansion - Operations Research

Capacity ExpansionOperations Research

Anthony Papavasiliou

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Outline

1 Screening Curves

2 Stochastic Programming Formulation

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Load and Wind in Belgium, 2013

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Load Duration Curve

Load duration curve is obtained by sorting load time series indescending order

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Horizontal Stratification of Load

Load duration curve describes number of hours in the year that loadwas greater than or equal to a given level (e.g. net load was ≥ 10000MW for 2000 hours)

Step-wise approximation:

Base load: 0-7086 MW, lasts for 8760 hours (whole year)

Medium load: 7086-9004 MW, lasts for 7500 hours

Peak load: 9004-11169 MW, lasts for 1500 hours

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Technological Options

Technology Fuel cost ($/MWh) Inv cost ($/MWh)Coal 25 16Gas 80 5

Nuclear 6.5 32Oil 160 2

Fuel/variable cost: proportional to energy produced

Investment/fixed cost: proportional to built capacity

Discounted investment cost: hourly cash flow required for 1 MWof investment

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Optimal Investment Problem

Optimal investment problem: find mix of technologies that can servedemand at minimum total (fixed + variable) cost

The optimal investment problem can be solved graphically withscreening curves

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Screening Curves

Screening curve: Total hourly cost as a function of the fraction oftime that a technology is producing

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Logic of Graphical Solution

Total cost of using 1 MW of a technology depends on amount oftime it produces

Each horizontal slice of load can be allocated to an optimaltechnology, depending on its duration (which technology shouldserve base load? peak load?)

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Optimal Solution

Fraction of time each technology should be functioning:

For oil: 2 + 160 · f ≤ 5 + 80 · f ⇔ f ≤ 0.0375⇒ 0-328 hoursFor gas: f > 0.0375 and 5 + 80 · f ≤ 16 + 25 · f ⇔ f ≤ 0.2⇒328-1752For coal: f > 0.2 and 16 + 25 · f ≤ 32 + 6.5 · f ⇔ f ≤ 0.8649⇒1752-7576 hoursFor nuclear: 0.8649 ≤ f ≤ 1⇒ 7576-8760 hours

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Optimal Solution

Recall,

Base load: 0-7086 MW, lasts for 8760 hours (whole year)

Medium load: 7086-9004 MW, lasts for 7500 hours

Peak load: 9004-11169 MW, lasts for 1500 hours

From previous slide,

Base-load is assigned to nuclear: 7086 MW

Medium load is assigned to coal: 1918 MW

No load is assigned to oil: 0 MW

Peak load is assigned to gas: 2165 MW

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Outline

1 Screening Curves

2 Stochastic Programming Formulation

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Increasing Wind PenetrationWhich load duration curve corresponds to 10x wind power?

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Scenarios

Duration (hours) Level (MW) Level (MW)Ref 10x wind

Base load 8760 7086 3919Medium load 7000 9004 7329

Peak load 1500 11169 10315

Low wind: 10%

High wind: 90%

Goal determine optimal expansion plan

Optimal refers here to the expansion plan that minimizes theexpected total cost.

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Stochastic Program Vs Expected Value Problem

How do we compute each load duration curve?

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Screening Curve Solution

Duration (hours) Level (MW) TechnologyBlock 1 8760 3919 NuclearBlock 2 7176 7086 CoalBlock 3 7000 7329 CoalBlock 4 2050 9004 CoalBlock 5 1500 10315 GasBlock 6 150 11169 Oil

Table: Optimal assignment of capacity for the 6-block load duration curve.

Duration (hours) Level (MW) TechnologyBase load 8760 4235 Nuclear

Medium load 7000 7496 CoalPeak load 1500 10401 Gas

Table: Optimal assignment of capacity for the expected load duration curve.

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Investment and Fixed Cost

SP inv EV inv SP fixed cost EV fixed cost(MW) (MW) ($/h) ($/h)

Coal 5,085 3,261 81,360 52,176Gas 1,311 2,905 6,555 14,525

Nuclear 3,919 4235 125,408 135,520Oil 854 0 1,708 0

Total 11,169 10,401 215,031 202,221

Why are the investment plans different?

Why does the EV solution have a lower fixed cost?

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Merit Order Dispatch

Merit order dispatch rule: In order of increasing variable cost,assigns technologies to load blocks of decreasing duration, untileither all load blocks are satisfied or all generating capacity isexhausted

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Merit Order Dispatch

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Variable Cost

SP var cost EV var cost($/h) ($/h)

Block 1 25,473 25,473Block 2 64,858 60,070Block 3 4,854 4,854Block 4 9,799 29,209Block 5 17,960 17,959Block 6 2,340 13,268

Total 125,285 150,834

The EV solution is expensive in serving block 4 (served largely by gasinstead of coal) and block 6 (why?)

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Value of the Stochastic Solution

Value of the stochastic solution (VSS): Cost difference ofstochastic programming solution and expected value solution whenthe two are compared against the "true" model of uncertainty

Stochastic program: 125,285 (variable) + 215,031 (fixed) =340,316 $/h

Expected value problem: 150,834 (variable) + 202,221 (fixed) =353,055 $/h

VSS = 12,739 $/h

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Multiple Periods

Orange area: sub-structure that recurs as we move backwards⇒ dynamic programming

Block separability: some decisions do not influence the futurestate of the system, only the payoff of each period (which onematters for the future, ’Invest’ or ’Operate’?)

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Math Programming Formulation of 2-Stage Problem

minx,y≥0

n∑i=1

(Ii · xi +m∑

j=1

Ci · Tj · yij)

s.t.n∑

i=1

yij = Dj , j = 1, . . . ,m

m∑j=1

yij ≤ xi , i = 1, . . .n

Ii ,Ci : fixed/variable cost of technology i

Dj ,Tj : height/width of load block j

yij : capacity of i allocated to j

xi : capacity of i

Where is the uncertainty?23 / 25

Towards a Dynamic Programming Algorithm

In order to solve multi-stage problem via dynamic programming, wewould like to express cost of 2-stage problem as a function ofinvestment x

Consider the following LP, with fixed x :

f (x) = miny≥0

n∑i=1

(Ii · xi +m∑

j=1

Ci · Tj · yij)

s.t.n∑

i=1

yij = Dj

m∑j=1

yij ≤ xi

Show that f (x) is a piecewise linear function of x

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Additional Reading

§1.3 BL: capacity expansion planning problem

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