Transcript A Simple Unit Commitment Problem

A Simple
Unit Commitment
Problem
Valentín Petrov, James Nicolaisen
18 / Oct / 1999
NSF meeting
Economic Dispatch (Covered last time)
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• With a given set of units running, how of the load much should be
generated at each to cover the load and losses? This is the question of
Economic dispatch.
• The solution is for the current state of the network and does not
typically consider future time periods.
Deciding which units to “commit”
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• When should the generating units (G) controlled by the GENCO be
run for most economic operation?
– Concern must be given to environmental effects
• How does one define “economic operation”? Profit maximizing? Cost
minimizing? Depends on the market you’re in.
Problem Setup
• Last meeting we discussed the economic
dispatch problem
• Now we will see how the unit commitment
fits into the general picture
• Unit commitment is bound to the economic
dispatch
• Use similar optimization methods
What is Unit Commitment (1)
• We have a few generators (units)
• Also we have some forecasted load
• Besides the cost of running the units we
have additional costs and constraints
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start-up cost
shut-down cost
spinning reserve
ramp-up time... and more
What is Unit Commitment (2)
• It turns out that we cannot just flip the switch
of certain units on and use them!
• We need to think ahead, and based on the
forecasted load and unit constraints, determine
which units to turn on (commit) and which
ones to keep down
• Minimize cost, cheap units play first
• Expensive ones run only when demand is high
How Do We Solve the Problem
• If a unit is on, we designate this with 1 and
respectively, the off unit is 0
• So, somehow we decide that for the next hour
we will have "0 1 1 0 1" if we have five units
• Based on that, we solve the economic dispatch
problem for unit 2, 3 and 5
• We start turning on U2, U3, U5
• When the next hour comes, we have them up
and running
To Come Up With Unit Commitment
• The question is, _how_ do we come up with
this unit commitment "0 1 1 0 1" ?
• One very simplistic way: if we have very few
units, go over all combinations from hour to
hour
• For each combination at a given hour, solve
the economic dispatch
• For each hour, pick the combination giving the
lowest cost!
Lagrange Relaxation (1)
• Min f = (0.25 x21+15)U1 + (0.255 x22+15)U2
• subject to:
– W = 5 – x1U1 - x2U2
– 0 < x1 < 10
– 0 < x2 < 10
• U may be only 0 or 1
Lagrange Relaxation (2)
• L = (0.25 x21+15)U1 + (0.255 x22+15)U2 +
l(5 – x1U1 - x2U2)
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Pick a value for l and keep it fixed
Minimize for U1 and U2 separately
0 = d/dx1(0.25x21 + 15 - x1l1)
0 = d/dx2(0.255x22 + 15 - x2l1)
Lagrange Relaxation (3)
• 0 = d/dx1(0.25x21 + 15 - x1l1)
– if the value of x1 satisfying the above falls outside
the 0 < x1 < 10, we force x1 to the limit.
– If the term in the brackets is > 0, set U1 to 0,
otherwise keep it 1
• 0 = d/dx2(0.255x22 + 15 - x2l1)
– same as above
Lagrange Relaxation (4)
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Now assume the variables x1, x2, U1, U2 fixed
Try to maximize L by moving l1 around
dL/dl = (5 – x1U1 - x2U2)
l2 = l1 + dL/dl (a)
– if dL/dl > 0, a = 0.2
– if dL/dl < 0, a = 0.005
• After we found l2, repeat the whole process
starting at step 1