Post on 28-Dec-2015
EE/Econ 458Introduction to Linear Programming
J. McCalley
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Security Constrained Economic Dispatch*
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Gen resource & demand response resourceDemand response resourceExternal asynchronous constraintsStored energy reserve constraintsEnergy transaction constraintsOPF constraintsReliability constraints
- System reserves- transmission - watchlist transmission flowgates- contingencies
*From Chapter 6 of the MISO Energy & Operating Reserve Markets, Business Practices Manual, BPM-002-r10, JUN-29-2011.
SUBJECT TO constraints on
Security Constrained Economic Dispatch
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The SCED algorithm uses a Linear Programming (LP) solver, and so we will study LP to understand this optimization method.
This problem may be simplified to the following
GIVEN:•A number of gens make price-quantity offers to sell•A number of LSEs make price-quantity bids to buyMAXIMIZE social surplus U(P)-C(P), where U(P) is the composite utility (value) function for consumers, C(P) is the composite cost function for suppliers, and P is real power injection vector of network nodes (+ for gen, - for load).SUBJECT TO constraints:
- Pmin<P<Pmax
- Sum of generation = sum of demand- Flows on each circuit <= maximum flow for circuit
Some preliminaries
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1. Maximization of a function is equivalent to minimization of the negated function,
maximize f(x) is the same as minimize (-f(x))
2. An inequality constraint may be equivalently written as g(x)>b or -g(x)<-b
bxg
c)x(h s.t.
xf
)(
)(min
And so any of the following problems are equivalent:
bxg
c)x(h s.t.
xf
)(
)(max
bxg
c)x(h s.t.
xf
)(
)(min
bxg
c)x(h s.t.
xf
)(
)(max
Linear Program – Standard Form
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Requirements: f(x) is a linear function in x, i.e., a1x1+a2x2+…+anx3
All equalities hj(x)=cj are linear functions in x.All inequalities gk(x)=bk are linear functions in x.
bxg
c)x(h s.t.
xf
)(
)(min
Then Problem P is a linear program.
Problem P:
Linear Program – A Convex Program
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bxg
c)x(h s.t.
xf
)(
)(minProblem P:
A linear program is a convex programming problem.
Recall: If f(x) is a convex function, and if S is a convex set, then the above problem is a convex programming problem. f(x) is linear, therefore a convex function. If S is formed by inequalities, it is polyhedral, therefore convex.
S
Sx
xf
subject to
)(min
Feasible set
One linear equality reduces feasible set S to the line. More than one reduces feasible set S to their intersections. S is convex in both cases.
If we find a locally optimal solution, it will be globally optimal.
Example 1
7
4
16
3)(max
21
2
21
xx
xx s.t.
xxxf
1
)16x(3),( 2121 xxxxF Ignore inequality. Form Lagrangian: Apply first-order conditions:
016
101
303
21
2
1
xxF
x
F
x
F
What happened?
Example 1
8
16
3)(max
2
21
xx s.t.
xxxf
1
We can push f(x) as far negative as we like and there will always be an intersection point with the equality constraint.
This problem is unbounded. This can happen with linear programs.But what about the inequality constraints?
Example 1
9
4
16
3)(max
21
2
21
xx
xx s.t.
xxxf
1
Form Lagrangian: )4()16(3),( 21221 xxxxxxxF 1
Observe g(x)<b implies to add this term.
KKT Conditions04
016
01
03
21
21
2
1
xxF
xxF
x
F
x
F
0
0)4( 21
xx
Example 1
10
4
16
3)(max
21
2
21
xx
xx s.t.
xxxf
1
4
16
1
3
0011
0011
1100
1100
2
1
x
x
1
2
10
6
2
1
x
x
The feasible region associated with the inequality constraint is below the dotted line
The feasible region for the problem includes everything on the thick line to the right of the intersection point (red dot).
The feasible region for the problem is below the dotted line and on the thick one (the equality constraint).
Example
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4
16
3)(max
21
2
21
xx
xx s.t.
xxxf
1
4
16
1
3
0011
0011
1100
1100
2
1
x
x
1
2
10
6
2
1
x
x
Minimize the objective:Choose solution that has smallest value of f(x) but is in the feasible region. Choose lowest contour touching the feasible (red) line.
This is shown by the dashed line at the end of its animation.
Graphical analysis reveals the solution is: x1=6, x2=10 (red dot) 2810)6(33)( 21 xxxf
Example 1
12
4
16
3)(max
21
2
21
xx
xx s.t.
xxxf
1
Definition: Any constraint comprising the feasible region boundary is an active constraint.
Observation: The solution occurred at a point where two active constraints intersect.
Example 2
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Resource allocation: Optimize an objective through allocating resources to activities subject to constraints on resources.
Objective: Maximize profitsResource: Time of each personActivities: Producing materials X and Y.
Material
Person1 2 3
X 40 min 24 min 20 minY 30 min 32 min 24 min
Three people work 8 hours/day (480 min) making materials X and Y. Their company makes $5 profit per unit X, $8 profit per unit Y. The times required for each person’s contribution towards making a unit are below.Note: A unit of material requires contributions from all three people,
i.e., no individual may make either material on their own.
Example 2
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Material
Person1 2 3
X 40 min 24 min 20 minY 30 min 32 min 24 min
yxyxf 85),(max
4803040 yx
4803224 yx
4802420 yx
(person 1)
(person 2)
(person 3)
Subject to
0
0
y
x
1248040 xx2048024 xx2448020 xx
Produce only X
60)0(8)12(585),( yxyxfProduce only Y
1648030 yy1548032 yy2048024 yy
120)15(8)0(585),( yxyxf
Three people work 8 hours/day (480 min) making materials X and Y. Their company makes $5 profit per unit X, $8 profit per unit Y. The times required for each person’s contribution towards making a unit are below.
Example 2
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Conclusion: Producing only Y is better than producing only X, and if we have only these two options, we will produce only Y.
yxyxf 85),(max
4803040 yx
4803224 yx
4802420 yx
(person 1)
(person 2)
(person 3)
Subject to
0
0
y
x
Where on the plot are the solutions we just found?
Can we increase profits beyond $120 by producing some of each?
Example 2
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Plot the contours of increasing objective function.
yxyxf 85),(
The contour f=60 passes through the point (12,0).The contour f=120 passes through the point (0,15).
No. All other points on the f=120 contour occur above the feasible region; all other contours touching the feasible region are below 120.
8
585
xfyyxf
Is it possible for any point to be better than the red one?
Optimal solution is again at a point where 2 active constraints intersect.
Example 3
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Same constraints but slightly different objective.
4803040 yx
4803224 yx
4802420 yx
(person 1)
(person 2)
(person 3)
Subject to
0
0
y
x
Which of the points is optimal?
The blue one is optimal since there must be a contour between f=120 and f=130 that just touches it, and any contour with higher f will not touch the feasible region.
yxyxf 89),(max
Optimal solution is again at a point where 2 active constraints intersect.
Example 4
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Same constraints but slightly different objective.
4803040 yx
4803224 yx
4802420 yx
(person 1)
(person 2)
(person 3)
Subject to
0
0
y
x
Which of the points is optimal?
The yellow one is optimal since there must the contour f=180 just touches it, and any contour with higher f will not touch the feasible region.Optimal solution is again at a point where 2 active constraints intersect.
yxyxf 815),(max
Conclusion
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The optimal solution was always at a point where two active constraints intersect (corner points). This always happens in an LP.
The solution to an LP, if one exists, is always at a corner point.
Solution strategy: Search the corner points!
A consideration
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How many corner points does our problem have?
But what about (0,16), (0,20), (20,0), and (24,0) and the two more outside the plot?
(0,0), (12,0), (1.7143, 13.7143), (0,15)
The first set are feasible corner points.The second set are infeasible corner points.
Revised solution strategy: Search the feasible corner points!
Why is our revision important?
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Assuming none of our constraints are parallel, the number of corner points is obtained as a combination of 5 distinct things (constraints) taken 2 at a time. In our case:
With 40 constraints, there would be 780 corner points to check.
Some problems have millions of constraints!
10)1*2*3)(1*2(
1*2*3*4*5
)!25(!2
!5
2
5
An approach
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5 102
An approach
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An approach
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1. Pick a corner point at random. 2. Move to an adjacent corner point that is better.
a. If there are two that are better, move to the one that is best.b. If there are no better adjacent corner points, the current
corner point is the solution to the problem.
Optimality (stopping) condition: If a corner point feasible solution is equal to or better than all its adjacent corner point feasible solutions, then it is equal to or better than all other corner point feasible solutions, i.e., it is optimal. Main ideas of proof: 1.If objective function monotonically increases (decreases) in some direction within the decision-vector space, then each adjacent corner point will become progressively better in the direction of objective function increase (decrease) such that the last corner point must have two adjacent corner points that are worse. 2.The monotonicity of objective function increase (decrease) is guaranteed by its linearity.
High-Level Version of Simplex Method
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1. Initialization: Start at a corner point solution.2. Iterative step: Move to a better adjacent corner point
feasible solution. 3. Optimality test: Determine if the current feasible
corner point is optimal using our optimality test (if none of its adjacent feasible corner points are better, then the current feasible corner point is optimal).
a. If the current feasible corner point is optimal, the solution has been found, and the method terminates.
b. If the current feasible corner point is not optimal, then go to 2.