Capacity Expansion - Operations Research
Transcript of Capacity Expansion - Operations Research
Load Duration Curve
Load duration curve is obtained by sorting load time series indescending order
4 / 25
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
5 / 25
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
6 / 25
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
7 / 25
Screening Curves
Screening curve: Total hourly cost as a function of the fraction oftime that a technology is producing
8 / 25
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?)
9 / 25
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
10 / 25
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
11 / 25
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.
14 / 25
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.
16 / 25
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?
17 / 25
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
18 / 25
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?)
20 / 25
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
21 / 25
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’?)
22 / 25
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
24 / 25