1. Module E3-b
Economic Dispatch
(continued)

2. Continuing from the previous two gen example,
let’s consider what happens if the demand is
increased from PT=400 MW to PT= 550 MW.
First, we need to solve it with only equality constraints.
The LaGrangian function is:

3. KKT conditions are

4. P1 is within limits, but P2 exceeds its 200 MW
upper limit. So…...

5. Constraint associated with the complementarity cdt:
is binding: P2-200=0 (an equality constraint!
==> P2=200.)
Therefore the mu2 multiplier will not be zero
(but all other mu multipliers will still be zero,
since the corresponding inside term will NOT
be zero).

6. So the new LaGrangian function is:

7. KKT conditions become

8. Rewrite the equations in preparation
for putting into matrix form.

9. All constraints are satisfied therefore this is the solution.

10. What is the meaning of mu?
Recall that we interpreted lambda as
the cost of “increasing” the RHS of the
equality constraint by 1 MW for an hour.
Let’s see if mu is the cost of increasing the
RHS of the corresponding inequality constraint
by 1 MW for an hour.
So let’s let the upper bound of P2 be 201 MW.

11. Get total cost corresponding to P2max=200.

12. Refer back to our solution without any
inequality constraints. We obtained
And this also violates P2max<201 MW.
So we can use the set-up we already developed
for bringing in this constraint, except we
simply change from 200 to 201.

14. Get total costs corresponding to this solution.

15. The total costs changed from
Which is almost the same as the mu2
multiplier.
Implication is that if we increase P2max
to 201 MW, we can save $3.34

16. A fundamental EDC concept:
Consider the first KKT condition:
If all gens are regulating, then
IC1 = IC2 = … = ICi = lambda

17. At the optimal (min cost) solution, all regulating generators
(those not at their limits)
will have the same incremental cost,
IC=dC/dP
which will be lambda.

18. This gives rise to the graphical solution method.