unit #4

economic dispatch problem

the main aim in the economic dispatch problem is to minimize the total cost of generating real
power (production cost) at various stations while satisfying the loads and the losses in the
trransmission links. for simp-licity, we consider only thermal plants. when hydro plants are
considered in conjunction with thermal plants , there is limitation of availability of water over a
period of time whihc must be used to save maximum fule at the thermal plants.

input-out curve of a generating unit (see fig.5.1)

Operating
or fuel
cost

C i ( PGi )

MWmin MWmax
Output power MW
P G i

fig.5.1 input-output curve of generating unit

the total generating cost includes fuel, labour and maintenance costs . for simplicity , only fuel
cost is considered to be variable. fuel cost is meaningful in case of thermal
and nuclear stations, but for hydro stations , it is not meaningful . fig. 5.1 specifies the input
F (P ) C (P )
energy rate i Gi or cost of fule used rs per hr as i Gi . the input output curve of fig. 5.1
can be determined experimentally . mwmin minimum loading limit below which it is uneconomical
to oeprate the unit and mwmax is the maxuimum output limit.the curve of fig. 5.1 has
discontinuities that is not shown due to stream valve openings. the curve of fig,. 5.1 can be
approximated by the following function

C i ( PGi ) = ai + bi PGi + ci PGi2 Rs / hr

the slope of the fuel cost curve is called incremental cost (ic) is expressed in rs/mwh

dC i
ICi = = bi + 2ci PGi
dPGi

ic is a linear function of real generator power.

optimal operation:
let us assume that it is known a priori whihc generators are to run to meet a particular load
demand on teh stations. given a station with k generators committed and active power load pd
given , the real power geenration pgi for each generation has to be allocated so as to minimise the
total cost

K
C = ∑ C i ( PGi ) Rs / h
i =1

subject to the inequality constraint

PGi ,min ≤ PGi ≤ PGi ,max , i = 1,2,..K

∑P
i =1
Gi , max > PD
by considering spinning reserves. it is assumed that c is largely dependent on the real power
generation pgi and is insensitive to reactive power generation qgi.
PGj j ≠ i
in general , ci (pgi) is a nonlinear function and ci is independednt of . the problem is
seprable nonlinear programming problem.

∑P
i =1
Gi = PD

consider the augmented cost function

− K 
C = C − λ ∑ PGi − D 
 i =1 

where λ is the lagrangian multiplier.

minimization of the augmented function is achieved by putting

−
dC
=0
dPGi

or,

dC i
= λ, i = 1,2,..K
dPGi

that is, the incremental cost functions are all equal.

dC1 dC 2 dC K
= = ..... = =λ
dPG1 dPG 2 dPGK
the optimal loading of generators corresponds to the equal incremental cost of all the generators.
the last equation is called the corrdination equation numbering k whihc are to be solved
simultaneously with the load demand

∑P
i =1
Gi = PD

since ic curves are not linear in general, λ has to be determined by an iterative process.

1. choose a trial value of λ . i.e ic=ico.

2. solve for pgi , i=1,2,..k

∑P
i =1
Gi − PD < ε
3. if

for a small postive value of ε previopusly assumed, then teh solution is reached. otherwise,

K 
∑ PGi − PD  < 0
4. increment ic ( λ ) by ∆IC if  i =1  or decrement ic by ∆IC if
K 
∑ PGi − PD  > 0
 i =1  and repeat step 2. this is justified because p can be assumed to be a
gi
monotonically increasing function of ic.

example :-

two generators of 100 mw each with incremental characteristics:

dC1
= 2 + 0.012 P1
dP1
dC2
= 1.5 + 0.015 P2
dP2
minimum load on each unit is 10 mw. total load is 150 mw. determine economic
operating schedule.

solution: choose λ=2.

then
dC1 dC2
= = λ = 2.8
dP1 dP2
solving for p1 and p2 gives p1=66.6 mw, p2=86.6 mw
p1+p2=153.2 mw
try λ=2.75. then p1=62.5 mw, p2=83.4 mw and p1+p2=145.9 mw
try λ=2.78. then p1=65 mw, p2=85.4 mw and p1+p2=150.4 mw.
final answer for λ is between 2.77 and 2.78. p1=65 mw, p2=85 mw is near optimum.

economic load dispatch with losses

the above analysis does not take into account loses from source to load. taking the total losses
into account and denoting it as pl, the equations are written as

K
C = ∑ C i ( PGi ) Rs / h
i =1

∑P
i =1
Gi = PD + PL

the augmented equation becomes

− K 
C = C − λ ∑ PGi − PD − PL 
 i =1 
−
dC dC i ∂P
= −λ +λ L = 0
dPGi dPGi ∂PGi

that is,

dC i ∂P
+λ L =λ
dPGi ∂PGi

i =1,2…k

the above equations are called coordination equations.

assuming the losses to be of the form

PL = ∑∑ PGi Bij PGj

i j

we assume in the above model the following: 1.load current at any bus remains constant.fraction
of total equivalent load current 2. generator bus voltage magnitudes and angles are constant.
3.power factor of each source is constant.
then,

∂PL
= 2∑ Bij PGj
∂PGi j

if we assume that the incremental costs are linear

dC i
= a ii PGi + bi
dPGi

using the last two equations, the coordination equation can be written as

aii PGi + bi + λ ∑ 2 Bij PGj = λ

j

solving for pgi

bi
1− − ∑ 2 Bij PGj
λ j ≠i
PGi =
aii
+ 2 Bii
λ
since λ is not known, we use the iterative procedure as follows to find pgi.

1. assume an initial value of λ. this value should be more than ai,i=1,2,…k

2. calculate generations based on equal incremental production.
3. calculate the generation pgi using the last equation with values of pgj ,j ≠ i on the rhs
taken from step 2.
4. check of the generations pgi from successive iterations differ within a prespecified value. if
not go to step 3.if so , continue.
K 
∑ PGi − PD − PL 
5. check whether  i =1  is smaller than a presepecified value in
magnitude. if so, stop and calculate the cost of generation with the values of
powers.
6. otherwise, update λ and go to step 3.

example 1;-

the 2 generator roblem with the same data as before, but with losses. the loss coefficients are
given by

b11= 0.0015; b12= -0.0005; b22= 0.005

assume
dC1
= 0.01P1 + 2
dP1

dC 2
= 0.01P2 + 1.5
dP2
the problem is to determine operating schedule for λ =2.6.

using the data given, the above equation for pgi yields

0.230769 + 0.001PG 2
PG1 =
0.00684615

0.423077 + 0.001PG1
PG 2 =
0.00884615

for equal incremental cost of production,

λ − 2 2.6 − 2
PG1 = = = 60 MW
0.01 0.01
similarly,

1.1
PG 2 = = 110 MW
0.01
using these values of generation, for the case of including losses, we can get using the above
equations,

0.230769 + 0.001PG 2
G1 =
p 0.00684615 =49.97mw

0.423077 + 0.001PG1
PG 2 =
0.00884615 =54.61 mw

iterating further, we get,

pg1=41.38mw; pg2=52.50mw

one more iteration yieds

pg1=41.37mw; pg2=52.50mw
which verifies convergence to solution.

losses at this generation can be calculated using the b –coefficients as pl=7.28mw

load= pg1+pg2-pl=86.59 mw.

example #2
suppose however we are given the load as 160 mw, an assuming the same data as before,
we have

b11= 0.0015; b12= -0.0005; b22= 0.005

assume

dC1
= 0.01PG1 + 2
dP1

dC2
= 0.01PG 2 + 1.5
dP2

we now proceed as follows:

for equal incremental cost

0.01PG1 + 2 = 0.01PG 2 + 1.5

pg1+pg2=160

solving

pg1=55mw; pg2=105mw
λ = 0.01(105) + 1.5 = 2.55

substituting the value of λ in the coordination equation we have

 2
1− + 2(0.0005) PG 2
2.55 0.215686 + 0.001PG 2
PG1 = =
0.01 0.00692156
+ 2(0.0015)
2.55
1.5
1− + 2(0.0005) PG1
2.55 0.4117648 + 0.001PG1
PG 2 = =
0.01 0.00892156
+ 2(0.0025)
2.55

starting with
PG 2 = 105, successive iteration of the above two equations give the following.

PG1 = 46.33
PG 2 = 51.34

PG1 = 38.56 PG1 = 38.45

PG 2 = 50.47 PG 2 = 50.46

adding
PG1 + PG 2 = 88.91 << 160

so, put λ = 5 (say)

the coordination equations for this new value of λ become

0.6 + 0.001PG 2
PG1 =
0.005
0.7 + 0.001PG1
PG 2 =
0.007
starting with pg2=350, w have the following values through iteration of the above two coordination
equations.

PG1 = 190 PG1 = 145.42 PG1 = 144.14

PG 2 = 127.14 PG 2 = 120.7 PG 2 = 120.59
it is noted in the above pg1+pg2 >> 160. so, choose λ =3.75 (say), the coordination
equations are given by

0.46667 + 0.001PG 2
PG1 =
0.0056667
0.6 + 0.001PG1
PG 2 =
0.0076667

iteration of the above coordination equations result in the following.

PG1 = 122.05 PG1 = 98.973 PG1 = 98.44 PG1 = 98.42

PG 2 = 94.18 PG 2 = 91.17 PG 2 = 91.10 PG 2 = 91.09

calculating the losses,

PL = 0.0015(98.42) 2 + 0.0025(91.09) 2 − 2(98.42)(91.09)(0.0005) = 26.31MW

PD = PG1 + PG 2 − PL = 189.51 − 26.31 = 163.2 MW
since pd > 160 mw, we continue our iteration. choose λ=3.69. the coordination equations become

0.4579946 + 0.001PG 2
PG1 =
0.005710027
0.593496 + 0.001PG1
PG 2 =
0.007710027
starting with pg2= 219, w get through successive iterations the following.

PG1 = 118.56 PG1 = 96.40

PG 2 = 92.35 PG 2 = 89.48

PG1 = 95.88

PG 2 = 89.41
pg1=95.86; pg2=89.41

calculating losses,
pl=0.0015(95.86)2+0.0025(89.41)2-2(95.86)(89.41)(0.0005)=25.19 mw

pd=95.86+89.41-25.19=160.08 mw

PD − 160 = 0.08 ≤ 0.1 , we can stop the iterations here.

since

cost of generation

integrating incremental cost of production, we have

0.01PG12
C1 = + 2 PG1 = 0.005 PG12 + 2 PG1
2 rs/hr

0.01PG 2 2
C2 = + 1.5 PG 2 = 0.005 PG 2 2 + 1.5 PG 2
2 rs/hr

total cost of generation c = c1+c2.

for optimal values pg1= 95.86 mw, pg2=89.41 mw, the total cost of generation becomes

c= c1+c2= rs. 411.74/hr