EE/Econ 458
Introduction to Linear Programming
J. McCalley
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Security Constrained Economic Dispatch*
SUBJECT TO constraints on
Gen resource & demand response resource
Demand response resource
External asynchronous constraints
Stored energy reserve constraints
Energy transaction constraints
OPF constraints
Reliability 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-292011.
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Security Constrained Economic Dispatch
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 buy
MAXIMIZE 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
The SCED algorithm uses a Linear
Programming (LP) solver, and so we will study
LP to understand this optimization method.
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Some preliminaries
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
And so any of the following problems are equivalent:
min  f ( x)
s.t. h( x )  c
g ( x)b
min  f ( x)
s.t. h( x )  c
 g ( x)  b
max  f ( x)
s.t. h( x )  c
g ( x)b
max  f ( x)
s.t. h( x )  c
 g ( x)  b
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Linear Program – Standard Form
Problem P:
min  f ( x)
s.t. h( x )  c
g ( x)b
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.
 Then Problem P is a linear program.
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Linear Program – A Convex Program
Problem P:
min  f ( x)
s.t. h( x )  c
g ( x)b
min f ( x)
Feasible set
subject to
xS
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.
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.
S
 A linear program is a convex programming problem.
 If we find a locally optimal solution, it will be
globally optimal.
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Example 1
max f ( x)  3x1  x2
s.t. x1  x2  16
 x1  x2  4
Ignore inequality.
Form Lagrangian: F ( x,  )  3x1  x2   (x1  x2  16)
Apply first-order
conditions:
F
 3  0    3
x1
F
1   0   1
x2
What happened?
F
 x1  x2  16  0

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Example 1
max f ( x)  3x1  x2
s.t. x1  x2  16
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?
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Example 1
max f ( x)  3x1  x2
s.t. x1  x2  16
 x1  x2  4
Form Lagrangian:
Observe g(x)<b
implies to add
this term.
F ( x,  )  3x1  x2   ( x1  x2  16)   ( x1  x2  4)
F
 3    0
x1
F
1     0
x2
KKT Conditions
F
 x1  x2  1 6  0

F
  x1  x2  4  0

 ( x1  x2  4)  0
 0
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Example 1
max f ( x)  3x1  x2
s.t. x1  x2  16
 x1  x2  4
0
0

1

 1
0  1  1  x1   3
0  1 1   x2    1

1 0 0      16 
   
1 0 0     4 
The feasible region associated
with the inequality constraint
is below the dotted line
The feasible region for the
problem is below the dotted
line and on the thick one
(the equality constraint).
The feasible region for the
problem includes everything
on the thick line to the right of
the intersection point (red dot).
 x1   6 
 x  10
 2   
   2
   
  1 
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Example
max f ( x)  3x1  x2
s.t. x1  x2  16
 x1  x2  4
0
0

1

 1
0  1  1  x1   3
0  1 1   x2    1

1 0 0      16 
   
1 0 0     4 
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)
 x1   6 
 x  10
 2   
   2
   
  1 
 f ( x)  3x1  x2  3(6)  10  28
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Example 1
max f ( x)  3x1  x2
s.t. x1  x2  16
 x1  x2  4
Definition: Any constraint
comprising the feasible region
boundary is an active constraint.
Observation: The solution
occurred at a point where two
active constraints intersect.
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Example 2
Resource allocation: Optimize an objective through allocating
resources to activities subject to constraints on resources.
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.
Material
X
Y
1
40 min
30 min
Person
2
24 min
32 min
3
20 min
24 min
Objective: Maximize profits
Resource: Time of each person
Activities: Producing materials X and Y.
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Example 2
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.
Material
X
Y
1
40 min
30 min
max f ( x, y)  5x  8 y
Person
2
3
24 min
20 min
32 min
24 min
Produce only X
Subject to
40x  30y  480
(person 1)
24x  32y  480
(person 2)
20x  24y  480
x0
y0
(person 3)
40 x  480  x  12
24 x  480  x  20
20 x  480  x  24
f ( x, y)  5x  8 y  5(12)  8(0)  60
Produce only Y
30y  480  y  16
32y  480  y  15
24y  480  y  20
f ( x, y)  5x  8 y  5(0)  8(15)  120
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Example 2
Conclusion: Producing only Y is better than producing only X,
and if we have only these two options, we will produce only Y.
max f ( x, y)  5x  8 y
Subject to
40x  30y  480
(person 1)
24x  32y  480
(person 2)
20x  24y  480
x0
(person 3)
y0
Where on the plot are the
solutions we just found?
Can we increase profits beyond $120 by producing some of each?
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Example 2
Plot the contours of increasing objective function.
f ( x, y)  5x  8 y
f  5x
f  5x  8 y  y 
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The contour f=60 passes
through the point (12,0).
The contour f=120 passes
through the point (0,15).
Is it possible for any point to
be better than the red one?
No. All other points on the f=120 contour occur above the feasible
region; all other contours touching the feasible region are below 120.
Optimal solution is again at a point where 2 active constraints intersect.
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Example 3
Same constraints but slightly different objective.
max f ( x, y)  9 x  8 y
Subject to
40x  30y  480
(person 1)
24x  32y  480
(person 2)
20x  24y  480
x0
(person 3)
y0
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.
Optimal solution is again at a point where 2 active constraints intersect.
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Example 4
Same constraints but slightly different objective.
max f ( x, y)  15x  8 y
Subject to
40x  30y  480
(person 1)
24x  32y  480
(person 2)
20x  24y  480
x0
(person 3)
y0
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.
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Conclusion
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!
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A consideration
How many corner points
does our problem have?
(0,0), (12,0),
(1.7143, 13.7143), (0,15)
But what about (0,16),
(0,20), (20,0), and (24,0)
and the two more outside
the plot?
The first set are feasible
Revised
solution
corner points.
strategy: Search the
The second set are
feasible corner points!
infeasible corner points.
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Why is our revision important?
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:
5
5!
5 * 4 * 3 * 2 *1


2 2!(5  2)! (2 *1)(3 * 2 *1)  10
 
With 40 constraints,
there would be 780
corner points to check.
Some problems have
millions of
constraints!
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An approach
5
2
4
2
6
2
100
90
3
2
7
2
80
70
60
8
2
2
9
1
10
2
50
40
30
20
5
10
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An approach
5
2
4
2
6
2
100
90
3
2
7
2
80
70
60
8
2
2
9
1
10
2
50
40
30
20
5
10
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An approach
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.
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High-Level Version of Simplex Method
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.
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