Chapter 3 Economic Dispatch of Thermal Unitssmart.tabrizu.ac.ir/Files/Content/operation_wood.pdf · The Economic Dispatch Problem Consider a system that consists of N thermal-generating

EEL 6266 Power System Operation and Control

Chapter 3

Economic Dispatch of Thermal Units

© 2002, 2004 Florida State University EEL 6266 Power System Operation and Control 2

The Economic Dispatch Problem

� Consider a system that consists of N thermal-generating units serving an aggregated electrical load, Pload

� input to each unit: cost rate of fuel consumed, Fi

� output of each unit: electrical power generated, Pi

� total cost rate, FT, is thesum of the individualunit costs

� essential constraint:the sum of the outputpowers must equalthe load demand

� the problem is to minimize FT

T G1 P1F1

T G2 P2F2

T GN PNFN

Pload



� The mathematical statement of the problem is a constrained optimization with the following functions:� objective function:

� equality constraint:

� note that any transmission losses are neglected and any operating limits are not explicitly stated when formulating this problem

� Problem may be solved using the Lagrange function

( )�=

=N

iiiT PFF

1

�=

−==N

iiload PP

1

0φ

( ) ��

��

� −+=+= ��==

N

iiload

N

iiiT PPPFFL

11

λφλ



� Principles� the Lagrange function establishes the necessary conditions for

finding an extrema of an objective function with constraints

� taking the first derivatives of the Lagrange function with respect to the independent variables allows us to find the extreme value when the derivatives are set to zero� there are NF + Nλ derivatives, one for each independent variable

and one for each equality constraint

� the derivatives of the Lagrange function with respect to the Lagrange multiplier λ merely gives back the constraint equation

� the NF partial derivatives result in

( )0

d

d =−=∂∂ λ

i

ii

i P

PF

P

L



� Example� determine the economic operating point for the three

generating units when delivering a total of 850 MW� input-output curves

� unit 1: coal-fired steam unit:

� unit 2: oil-fired steam unit:

� unit 3: oil-fired steam unit:

� fuel costs� coal: $ 3.30 / MBtu� oil: $ 3.00 / MBtu

� the individual unit cost rate functions

2333

2222

2111

00482.097.77800194.085.7310

00142.02.7510

PPHPPH

PPH

++=++=

++=

( ) ( )( ) ( )( ) ( ) 2

333333

2222222

2111111

01446.070.232340.3

00582.055.239300.3

004686.076.2316833.3

PPPHPF

PPPHPF

PPPHPF

++=×=

++=×=

++=×=



� Example� the conditions for an optimal dispatch

� solving for λ yields

� then solving for the generator power values

850

02892.070.23dd

01164.055.23dd

009372.076.23dd

321

333

222

111

=++=+==+==+=

PPP

PPF

PPF

PPF

λλλ

41.27

85002892.0

70.2301164.0

55.23009372.0

76.23

=

=−+−+−

λ

λλλ

4.12802892.0)70.2341.27(

8.33101164.0)55.2341.27(

8.389009372.0)76.2341.27(

3

2

1

=−==−==−=

P

P

P



� In addition to the cost function and the equality constraint� each generation unit must satisfy two inequalities

� the power output must be greater than or equal to the minimum power permitted:� minimum heat generation for stable fuel burning and temperature

� the power output must be less than or equal to the maximum power permitted:� maximum shaft torque without permanent deformation

� maximum stator currents without overheating the conductor

� then the necessary conditions are expanded slightly

min,ii PP ≥

max,ii PP ≤

max,

min,

max,min,

dd

dd

:dd

iiii

iiii

iiiiii

PPPF

PPPF

PPPPPF

=∀≤=∀≥

≤≤∀=

λλλ



� Example� reconsider the previous example with the following generator

limits and the price of coal decreased to $2.70 / MBtu� generator limits

� unit 1: 150 ≤ P1 ≤ 600 MW

� unit 2: 100 ≤ P2 ≤ 400 MW

� unit 3: 50 ≤ P3 ≤ 200 MW

� new fuel cost rate function for unit 1:

� solving for λ yields

( ) ( ) 2111111 003834.044.1913777.2 PPPHPF ++=×=

8.383.1099.701

82.24

85002892.0

70.2301164.0

55.23007668.0

44.19

321 ====

=−+−+−

PPP

λ

λλλ



� Example� this solution meets the constraint of generation meeting the

850 MW load demand, but units 1 and 3 are not within limit� let unit 1 be set to its maximum output and unit 3 to its minimum

output. The dispatch becomes:P1 = 600 MW P2 = 200 MW P3 = 50 MW

� hence, λ must equal the incremental cost of unit 2 since it is the only unit not at either limit

� next compute the incremental costs for units 1 and 3

( ) 88.2520001164.055.23d

d

2002

2

2

=+===P

P

Fλ

( )

( ) 15.255002892.07.23d

d

04.24600007668.044.19d

d

503

3

6001

1

3

1

=+=

=+=

=

=

P

P

P

F

P

F



� Example� note that the incremental cost for unit 1 is less than λ indicating

that it is at its maximum

� however, the incremental cost for unit 3 is not greater than λ so it should not be forced to its minimum

� rework with units 2 and 3 incremental cost equal to λ

� note that this dispatch meets the necessary conditions

� incremental cost of electricity = 2.567 cents / kilowatt-hour

0.680.182600

67.25

25085002892.0

70.2301164.0

55.23

321

1

====

=−=−+−

PPP

P

λ

λλ


� Consider a similar system, which now has a transmission network that connects the generating units to the load� the economic dispatch problem is slightly more complicated

� the constraint equation must include the network losses, Ploss

� the objective function, FT

is the same as before

� the constraint equation must be expanded as:

Network Losses

T G1 P1F1

T G2 P2F2

T GN PNFN

Pload

�=

−+==N

iilossload PPP

1

0φ

Transmissionnetwork withlosses, Ploss


Network Losses

� The same math procedure is followed to establish the necessary conditions for a minimum-cost operating solution� Lagrange function and its derivatives w.r.t. the input power:

� the transmission network loss is a function of the impedances and the currents flowing in the network� for convenience, the currents may be considered functions of the

input and load powers

� it is more difficult to solve this set of equations

( )

λλλ

λφλ

=∂

∂+→=��

��

�

∂∂−−=

∂∂

��

��

� −++=+= ��==

i

loss

i

i

i

loss

i

i

i

N

iilossload

N

iiiT

P

P

P

F

P

P

P

F

P

L

PPPPFFL

d

d01

d

d

11


Network Losses

� Example� repeat the first example, but include a simplified loss

expression for the transmission network

� the incremental cost functions and the constraint function are formed as:

� this is no longer a set of linear equations as before

( )[ ]( )[ ]( )[ ]

23

22

21321

3333

2222

1111

00012.000009.000003.0850

00012.02102892.070.23dd

00009.02101164.055.23dd

00003.021009372.076.23dd

1dd

PPPPPPP

PPPF

PPPF

PPPF

P

P

P

F

loss

i

loss

i

i

++==−++

−=+=−=+=−=+=

��

��

�

∂∂−=

λλλ

λ

23

22

21 00012.000009.000003.0 PPPPloss ++=


Network Losses

� Example� a new iterative solution procedure

step 1 pick starting values for P1, P2, and P3 that sum to the load

step 2 calculate ∂Ploss/∂Pi and the total losses Ploss

step 3 calculate λ that causes P1, P2, & P3 to sum to Pload & Ploss

step 4 compare P1, P2, & P3 of step 3 to the values used in step 2; if there is significant change to any value, go back to step 2, otherwise, the procedure is done

� pick generation values

� find the incremental losses

MW150MW300MW400 321 === PPP

( )( )( )( )( )( ) 0360.015000012.02

0540.030000009.02

0240.040000003.02

3

2

1

==∂∂==∂∂==∂∂

PP

PP

PP

loss

loss

loss


Network Losses

� Example� total losses

� solve for λ

� in matrix form

( ) ( ) ( ) MW6.151501012300109400103 252525 =×+×+×= −−−lossP

( )( )( )

( ) 06.15850

036.0102892.070.23

054.0101164.055.23

024.01009372.076.23

321

3

2

1

=+−++−=+−=+−=+

PPP

P

P

P

λλλ

54.2891.13149.29620.437

6.86570.2355.2376.23

0111964.002892.000946.0001164.00976.000009372.0

321

3

2

1

====

��

��

�

=��

��

�

⋅��

��

�

−−

−

λλ

PPP

PPP


Network Losses

� Example� since the values of P1, P2, and P3 are quite different from the

starting values, we return to step 2� find the incremental losses and total losses

� solve for λ in matrix form

( )( )( )( )( )( ) 0317.09.13100012.02

0534.05.29600009.02

0262.02.43700003.02

3

2

1

==∂∂==∂∂==∂∂

PP

PP

PP

loss

loss

loss

( ) ( ) ( ) MW73.159.13110125.2961092.437103 252525 =×+×+×= −−−lossP

55.2833.13638.29803.431

73.86570.2355.2376.23

01119683.002892.0009466.0001164.009738.000009372.0

321

3

2

1

====

��

��

�

=��

��

�

⋅��

��

�

−−

−

λλ

PPP

PPP


Network Losses

� Example� summarization of the iteration process

iteration P1 P2 P3 losses λcount (MW) (MW) (MW) (MW) ($/MWh)

1 400.00 300.00 150.00 15.60 28.542 437.20 296.49 131.91 15.73 28.553 431.03 298.38 136.33 15.82 28.554 432.45 297.92 135.45 15.80 28.555 432.11 298.06 135.63 15.81 28.556 432.19 298.02 135.59 15.80 28.557 432.17 298.03 135.60 15.80 28.55


Chapter 3

Numerical Methods for Economic Dispatch


The Lambda-Iteration Method

� The solution to the optimal dispatch can be approached by graphical methods� plot the incremental cost characteristics for each generator

� the operating points must have minimum cost and satisfy load� that is, find an incremental cost rate, λ that meets the demand PR

� graphically:

Σ

dF1dP1

($/MWh)

dF2dP2

($/MWh)

dF3dP3

($/MWh)

P1 (MW) P2 (MW) P3 (MW)

PR = P1 + P2 + P3

λ



� An iterative process� assume an incremental cost

rate λ and find the sum of the power outputs for this rate� the first estimate will be

incorrect

� if the total power output is too low, increase the λ value, or if too high, decrease the λ value� with two solutions, a closer

value of total power can be extrapolated or interpolated

� the steps are repeated until the desired output is reached

λ[1]λ[3]λ[2]

[1]

[3]

[2]

λ0

error (e)

solution: (|e| < tolerance)

R

N

ii PPe −=�

=1

Lambda projection



� This procedure can be adopted for a computer implementation� the implementation of the power

output calculation is rather independent of the solution method� each generator output could be

solved by a different method� as an iterative procedure, a stopping

criterion must be established� two general stopping rules are

appropriate for this application� total output power is within a

specified tolerance of the load demand

� iteration loop count exceeds a maximum value

start set λ

calculate Pi

for i = 1 to N

calculate

�=

−=N

iiload PP

1

ε

first iteration?

|ε| ≤ tolerance?

project λ

print scheduleend



� Example� consider the use of cubic functions to represent the input-

output characteristics of generating plants

� for three generating units, find the optimum schedule for a 2500 MW load demand using the lambda-iteration method � generator characteristics:

� assume that the fuel cost to be $1/MBtu

� set the value of λ on the second iteration at 10% above or below the starting value depending on the sign of the error

( ) ( ) MWin MBtu/h 32 PDPCPBPAH +++=

Unit 1 749.55 6.95 9.68×10-4 1.27×10-7 320 800Unit 2 1285.0 7.051 7.375×10-4 6.453×10-8 300 1200Unit 3 1531.0 6.531 1.04×10-3 9.98 ×10-8 275 1100

A B C D Pmax Pmin



� Example� initial iteration: λstart= 8.0

� incremental cost functions

� find the roots of the three incremental cost functions at λ = 8.0� P1 = (–5575.6, 494.3), P2 = (–8215.9, 596.7), P3 = (–7593.4, 646.2)

� use only the positive values within the range of the generator upper and lower output limits

� calculate the error

� with a positive error, set second λ at 10% above λstart: λ[2] = 8.8( ) ( ) ( ) MW/h9.7622.6467.5963.4942500 =−−−=e

( ) ( )( ) ( )( ) ( ) 2

38

33

33

22

82

422

21

71

411

1098.931004.12531.6dd

10453.6310375.72051.7dd

1027.131068.9295.6dd

PPPF

PPPF

PPPF

−−

−−

−−

×+×+==

×+×+==

×+×+==

λλλ



� Example� second iteration: λ[2] = 8.8

� find the roots of the three incremental cost functions at λ = 8.8� P1 = (–5904, 822.5), P2 = (–8662, 1043.0), P3 = (–7906, 958.6)

� calculate the error

� error out of tolerance

� project λ

� continue with third iteration

( ) ( ) ( ) MW/h0.3246.95810435.8222500 −=−−−=e

[ ][ ] [ ]

[ ] [ ][ ]( ) [ ] ( ) 5615.88.80.324

0.3249.762

0.88.82221

123 =+−

+−=+

−−= λλλλ e

ee



� Example� results of all iterations

� Issues� under some initial starting points, the lambda-iteration

approach exhibits an oscillatory behavior, resulting in a non-converging solution� try the example again with a starting point of λstart= 10.0

1 8.0 1737.2 494.3 596.7 646.22 8.8 2824.1 822.5 1043.0 958.63 8.5615 2510.2 728.1 914.3 867.84 8.5537 2499.9 725.0 910.1 864.8

Iteration λλλλ Total Generation P1 P2 P3


The Gradient Method

� Suppose that the cost function is more complex

� example:

� the lambda search technique requires the solution of the generator output power for a given incremental cost� possible with a quadratic function or piecewise linear function

� hard for complicated functions; we need a more basic method

� The gradient search method uses the principle that the minimum is found by taking steps in a downward direction� from any starting point, x[0], one

finds the direction of steepest descent by computing the negative gradient of F at x[0]:

5

4

35.2

210)( a

aP

eaPaPaaPF−

+++=

[ ]( )��

�

�

��

�

�

∂

∂−=∇−

ndxF

dxFxF �

10


The Gradient Method

� to move in the direction of maximum descent from x[0] to x[1]:

� α is a scalar that when properly selected guarantees that the process converges

� the best value of α must be determined by experiment

� for the economic dispatch problem, the gradient technique is applied directly to the Lagrange function

� the gradient function is:

� this formulation does not enforce the constraint function

( )]0[]0[]1[ xfxx ∇−= α

( )

( ) ( )

( ) ( )��

�

�

��

�

�

−

−

−

=

��

�

�

��

�

�

∂∂∂∂

∂∂

=∇

�

��

−+=

�

��

=

==

N

iiload

NNNN

N

iiload

N

iii

PP

PFP

PFP

L

PL

PL

L

PPPFL

1

1111

11

dd

dd

λ

λ

λ

λ

��


The Gradient Method

� Example� solve the economic dispatch for a total load of 800 MW using

these generator cost functions

� use α = 100% and starting from

� λ is initially set to the average of the incremental costs of thegenerators at their starting generation values:

( )( )( ) 2

3333

22222

21111

01446.070.23234

00582.055.23930

004686.076.231683

PPPF

PPPF

PPPF

++=++=

++=

MW300and MW,200MW,300 ]0[3

]0[2

]0[1 === PPP

( ) 27.28

)300(02892.070.23

)200(01164.055.23

)300(009372.076.23

3

1

d

d3

1

]0[

131[0] =

��

�

�

��

�

�

+++

++== �

=iii PF

Pλ


% Example 3Egendata = [ 1683 23.76 0.004686

930 23.55 0.00582234 23.70 0.01446 ];

power = [ 300, 200, 300 ];alpha = 1.00, Pload = 800;% find lambda0n = length( gendata );lambda0 = 0;for i = 1 : n

lambda0 = lambda0 + gendata(i,2) + 2 * gendata(i,3) * power(i);endlambda0 = lambda0 / 3clear x0x0 = power, x0(n+1) = lambda0;% calculate the gradientfor kk = 1 : 10

disp(kk)clear gradient gradient = [];Pgen = 0, cost = 0;for i = 1 : n

gradient(i) = gendata(i,2) + 2 * gendata(i,3) * x0(i) - x0(n+1);Pgen = Pgen + x0(i);cost = cost + gendata(i,1) + gendata(i,2) * x0(i) + ge ndata(i,3) * x0(i) * x0(i);

endgradient(n+1) = Pload - Pgen;disp( [x0, Pgen, cost/1000] )x1 = x0 - gradient * alpha;x0 = x1;

end

The Gradient Method

� Example� Matlab program to perform the

gradient search method


The Gradient Method

� Example� the progress of the gradient search is shown in the table below

� note that there is no convergence to a solution

1 28.28 800.0 300.0 200.0 300.0 23,7512 28.28 800.0 301.7 202.4 295.9 23,7263 28.28 800.1 303.4 204.8 291.9 23,7044 28.35 800.2 305.1 207.1 288.1 23,6855 28.57 800.7 306.8 209.5 284.4 23,6766 29.23 801.8 308.7 212.1 281.0 23,6877 31.06 805.0 311.3 215.3 278.4 23,7578 36.08 813.7 315.7 220.3 277.7 23,9839 49.79 837.4 325.1 230.3 282.1 24,63210 87.19 901.9 348.0 253.8 300.0 26,449

Iteration λλλλ Total Generation P1 P2 P3 Cost


The Gradient Method

� A simple variation� realize that one of the generators is always a dependent

variable and remove it from the problem� for example, picking P3, then

� then the total cost function becomes

� this function stands by itself as a function of two variables with no load-generation balance constraint� the cost can be minimized

by a gradient method such as:

� note that the gradient goes to zerowhen the incremental cost at generator 3 is equal to that at generators 1 and 2

213 800 PPP −−=

( ) ( ) ( )2132211 800 PPFPFPFC −−++=

��

�

�

��

�

�

−

−=

��

�

�

��

�

�

=∇

2

3

2

2

1

3

1

1

2

1

d

d

d

dd

d

d

d

d

dd

d

P

F

P

FP

F

P

F

CP

CP

C


The Gradient Method

� A simple variation� the gradient steps are performed in like manner as before

and

� Example� rework the previous example with the reduced gradient

� α is set to 20.00

α⋅∇−= Cxx ]0[]1[

��

��

�=2

1

P

Px

( ) ( )( )( ) ( )( ) ��

��

�

−−−−+−−−−+

=

��

�

�

��

�

�

−

−=∇

212

211

2

3

2

2

1

3

1

1

80001446.0270.2300582.0255.23

80001446.0270.23004686.0276.23

d

d

d

ddd

dd

PPP

PPP

P

F

P

FP

F

P

F

C


% Example 3Fgendata = [ 1683 23.76 0.004686

930 23.55 0.00582234 23.70 0.01446 ];

power = [ 300, 200, 300 ];alpha = 20.00;Pload = 800;% form lambda0n = length( gendata );clear x0x0 = power(1:n-1);% calculate the gradientfor kk = 1 : 10

disp(kk)clear gradient gradient = [];Pn = Pload;for i = 1 : n - 1

Pn = Pn - x0(i);endcost = gendata(n,1) + gendata(n,2) * Pn + gendata(n,3) * Pn * Pn;for i = 1 : n - 1

gradient(i) = gendata(i,2) + 2 * gendata(i,3) * x0(i) - gendata(n,2) - 2 * gendata(n,3) * Pn;cost = cost + gendata(i,1) + gendata(i,2) * x0(i) + ge ndata(i,3) * x0(i) * x0(i);

enddisp( [x0, Pn, 800, cost/1000] )x1 = x0 - gradient * alpha;x0 = x1;

end

The Gradient Method


simplified gradient search method


The Gradient Method

� Example� the progress of the simplified gradient search is shown in the

table below

� note that there is a solution convergence by the 6th iteration

1 800.0 300.0 200.0 300.0 23,7512 800.0 416.1 330.0 54.0 23,2693 800.0 368.1 287.4 144.5 23,2044 800.0 381.5 307.1 111.4 23,1945 800.0 373.3 303.0 123.7 23,1936 800.0 373.6 307.0 119.3 23,1927 800.0 371.4 307.6 121.0 23,1928 800.0 370.6 309.0 120.4 23,1929 800.0 369.6 309.7 120.7 23,19210 800.0 368.9 310.4 120.6 23,192

Iteration Total Generation P1 P2 P3 Cost


Newton’s Method

� The solution process can be taken one step further� observe that the aim is to always drive the gradient to zero

� since this is just a vector function, Newton’s method finds the correction that exactly drives the gradient to zero

� Review of Newton’s method� suppose it is desired to drive the function g(x) to zero

� the first two terms of the Taylor’s series suggest the following

� the objective function g(x) is defined as:

� then the Jacobian is:

0=∇ xL

( ) ( ) ( )[ ] 0=∆′+=∆+ xxgxgxxg

( )( )

( )��

�

�

��

�

�=

nn

n

xxg

xxgxg

,,

,,

1

11

�

�

�

( )��

�

�

��

�

�

∂∂∂∂

∂∂∂∂=′

nnn

n

xgxg

xgxgxg

�

��

�

1

111


Newton’s Method

� the adjustment at each iteration step is

� if the function g is the gradient vector ∇Lx, then

� For economic dispatch problems:

and

� note that in general, one Newton step solves for a correction that is closer to the minimum than would the gradient method

( )[ ] ( )xgxgx 1−′−=∆

LLx

x x ∆��

��

� ∇∂∂−=∆

−1

( ) �

��

−+= ��==

N

iiload

N

iii PPPFL

11

λ

��

�

�

��

�

�

=∇∂∂

�

��

�

�

2

2

1

2

22

2

12

221

2

21

2

dd

d

dd

d

d

d

dd

ddd

d

d

d

x

L

x

L

x

L

xx

Lxx

L

x

L

Lx x

λλ


Newton’s Method

� Example� solve the previous economic dispatch problem example using

the Newton’s method� the gradient function is the same

as in the first example� let the initial value of

λ be equal to zero

� the Hessian matrix takes the following form:

� the initial generation valuesare also the same as in the first example

[ ]

��

�

�

��

�

�

−−−

−

−

−

=

0111

1d

d00

10d

d0

100d

d

23

32

22

22

21

12

P

FP

FP

F

H


% Example 3Ggendata = [ 1683 23.76 0.004686

930 23.55 0.00582234 23.70 0.01446 ];

power = [ 300, 200, 300 ];Pload = 800;% form Hn = length( gendata );H = zeros(n+1,n+1);for i = 1 : n

H(i,i) = gendata(i,3) * 2;H(i,n+1) = -1, H(n+1,i) = -1; end

x0 = zeros(n+1,1);x0(1:n,1) = transpose( power );% calculate the gradient and Hessian matricesfor kk = 1 : 10

disp(kk)gradient = zeros(n+1,1);gradient(n+1,1) = Pload;for i = 1 : n

gradient(i,1) = gendata(i,2) + 2 * gendata(i,3) * x0( i,1) - x0(n+1,1);gradient(n+1,1) = gradient(n+1,1) - x0(i,1); end

dx = H \ gradient;cost = 0;for i = 1 : n

cost = cost + gendata(i,1) + gendata(i,2) * x0(i) + ge ndata(i,3) * x0(i) * x0(i); enddisp( [x0', cost/1000] )x0 = x0 - dx;

end

Newton’s Method


Newton’s method


Newton’s Method

� Example� the progress of the gradient search is shown in the table below

� note the quick convergence to a solution

� compare with the solution of the previous example

1 0.00 800.0 300.0 200.0 300.0 23,7512 27.19 800.0 366.3 313.0 120.7 23,1923 27.19 800.0 366.3 313.0 120.7 23,1924 27.19 800.0 366.3 313.0 120.7 23,192

Iteration λλλλ Total Generation P1 P2 P3 Cost


Chapter 3

Economic Dispatch Using Dynamic Programming


Piecewise Linear Cost Functions

� Common practice� many utilities prefer to represent their generator cost functions

as single- or multiple-segment, linear cost functions

� Typical examples:

PmaxPmin

F(P)

PmaxPmin

dF(P)/dP

PmaxPmin

F(P)

PmaxPmin

dF(P)/dP


Piecewise Linear Cost Functions

� Piecewise linear cost functions can not be used with gradient based optimization methods� like the lambda-iteration

� such methods will always land on Pmin or Pmax

� A table-based method resolves this problem� technique

� for all units running, begin to raise the output of the unit with the lowest incremental cost segment� if this unit hits the right-hand end of a segment or hits Pmax, find

the unit with the next lowest incremental cost segment and beginto raise its output

� eventually, the total of all units outputs equals the total load� the last unit is adjusted to have a generation, which is partially

loaded for one segment


Dynamic Programming

� A wide variety of control and dynamic optimization problems use dynamic programming (DP) to find solutions� can greatly reduce the computation effort in finding optimal

trajectories or control policies

� DP applications have been developed for� economic dispatch

� hydro-thermal economic-scheduling

� unit commitment

� methods are based on the calculus of variations� but, applications are not difficult to implement or program

� principles are introduced by presenting examples of one-dimensional problems


Dynamic Programming

� Example� consider the cost of transporting a unit shipment from location

A to location N� there are many short paths that connect many stops along the

way, which offers numerous parallel routes from getting from A to N

� each path has an associated cost� e.g., distance and level of difficulty results in fuel costs

� the total cost is the sum of the path costs of the selected route from the originating location to the terminating location

� the problem is to find the minimum cost route


Dynamic Programming

A C

B

D

F

E

G

I

H

J M

L

K

N

1-D Dynamic Programming Example

3

2

5

11

8

4

9

6

6

3

82

11

5

9

4

5

8

67

3

9

4

3


Dynamic Programming

� There are various stages traversed� starting at A, the minimum cost path to N is ACEILN

� starting at C, the least cost path to N is CEILN

� starting at E, the least cost path to N is EILN

� starting at I, the least cost path to N is ILN

� starting at L, the least cost path to N is LN

� Obtaining the optimal route� the choice of the route is made in sequence

� Theory of optimality� the optimal sequence is called the optimal policy

� any sub-sequence is called a sub-policy

� the optimal policy contains only optimal sub-policies


Dynamic Programming� Example (continued)

� divide up the field of paths into stages (I, II, III, IV, V)� at the terminus of each stage, there is a set of nodes (stops), {Xi}

� at stage III, the stops are [{X3} = {H, I, J, K}]� a set of costs can be found for crossing a stage, {VIII (X2, X3)}

� a cost is dependent on the starting and terminating nodes of a stage, VIII(E, H) = 3, VIII(F, I) = 11

� the minimum cost for traversing from stage I to stage i and arrive at some particular node (stop), Xi, is defined as fI(Xi)� the minimum costs from stage I to stage II for nodes {B, C, D}

are: fI(B) = VI(A, B) = 5, fI(C) = VI(A, C) = 2, fI(D) = VI(A, D) = 3

� the minimum cost from stage I to stage III for node {E} is:fII(E) = min [ fI(X1) + VII(X1, E) ] = min[ 5 + 11, 2 + 8, 3 + inf. ]

{ X1} X1 = B = C = DfII(E) = 10 via ACE


Dynamic Programming

A0

C2

B5

D3

F6

E10

G9

I12

H13

J11

M18

L15

K13

N19

1-D dynamic programming example: cost at each node

3

2

5

11

8

4

9

6

6

3

82

11

5

9

4

5

8

67

3

9

4

3

I II III IV V


Dynamic Programming

� at each stage, the minimum cost should be recorded for all the terminus nodes (stops)� use the minimum cost of the terminus of the previous stage

� identify the minimum cost path for each of the terminating nodesof the current stage

(X1) fI(X1) path (X2) fII(X2) path (X3) fIII (X3)path (X4) fIV(X4)path (X5) fV(X5) path

B 5 A E 10 AC H 13 ACE L 15 ACEI N 19 ACEIL

C 2 A F 6 AC I 12 ACE M 18 ADGK

D 3 A G 9 AD J 11 ACF

K 13 ADG


Dynamic Programming

� Economic dispatch� when the heat-rate curves exhibit nonconvex characteristics

it is not possible to use an equalincremental cost method � multiple values of MW output

exist for a given value of incremental cost

� dynamic programming findsoptimal dispatch under such circumstances� the DP solution is accomplished

as an allocation problem

� the approach generates a set of outputsfor an entire set of load values

PmaxPmin

H(P)

PmaxPmin

dH(P)/dP

A nonconvex heat rate curveand its corresponding

incremental heat rate curve


� Example� consider a three-generator system serving a 310 MW demand

� the generator I/O characteristics are not smooth nor convex

� the demand does not fit the data exactly, interpolate is needed between the available closest values, 300 and 325 MW

Dynamic Programming

0 ∞ ∞ ∞50 810 750 80675 1355 1155 1108.5

100 1460 1360 1411125 1772.5 1655 1704.5150 2085 1950 1998175 2427.5 ∞ 2358200 2760 ∞ ∞225 ∞ ∞ ∞

Power Levels (MW) Costs ($/hour)P1 = P2 = P3 F1 F2 F3


Dynamic Programming� Example

� the minimum cost function for scheduling units 1 and 2:

� let P2 cover its allowable range for demands of 100 to 350 MW( ) ( ) ( )22212 PFPDFDf +−=

50 810 ∞ ∞ ∞ ∞ ∞ ∞ -75 1355 ∞ ∞ ∞ ∞ ∞ ∞ -

100 1460 1560 ∞ ∞ ∞ ∞ 1560 50125 1772.5 2105 1965 ∞ ∞ ∞ 1965 75150 2085 2210 2510 2170 ∞ ∞ 2170 100175 2427.5 2522.5 2615 2715 2465 ∞ 2465 125200 2760 2835 2927.5 2820 3010 2760 2760 150225 ∞ 3177.5 3240 3132.5 3115 3305 3115 125250 ∞ 3510 2582.5 3445 3427.5 3410 3410 150275 ∞ ∞ 3915 3787.5 3740 3722.5 3722.5 150300 ∞ ∞ ∞ 4120 4082.5 4035 4035 150325 ∞ ∞ ∞ ∞ 4415 4377.5 4377.5 150350 ∞ ∞ ∞ ∞ ∞ 4710 4710 150

D F1(D) P2 = 50 75 100 125 150 (MW) f2 P2*

(MW) ($/h) F2(P2)= 750 1155 1360 1655 1950 ($/h) ($/h) (MW)


Dynamic Programming� Example

� the minimum cost function for scheduling units 1, 2 and 3:

� let P3 cover its allowable range for the demand( ) ( ) ( )33323 PFPDfDf +−=

100 1560 ∞ ∞ ∞ ∞ ∞ ∞ ∞ -125 1965 ∞ ∞ ∞ ∞ ∞ ∞ ∞ -150 2070 2366 ∞ ∞ ∞ ∞ ∞ 2366 50175 2465 2771 2668.5 ∞ ∞ ∞ ∞ 2668.5 75200 2760 2976 3073.5 2971 ∞ ∞ ∞ 2971 100225 3115 3271 3278.5 3376 3264.5 ∞ ∞ 3264.5 125250 3410 3566 3573.5 3581 3669.5 3558 ∞ 3558 150275 3722.5 3921 3868.5 3876 3874.5 3963 3918 3868.5 75300 4035 4216 4223.5 4171 4169.5 4168 4323 4168 150325 4377.5 4528.5 4518.5 4526 4464.5 4463 4528 4463 150350 4710 4841 4831 4821 4819.5 4758 4823 4758 150375 ∞ 5183.5 5143.5 5133.5 5114.5 5113 5118 5113 150400 ∞ 5516 5486 5446 5427 5408 5473 5408 150425 ∞ ∞ 5818.5 5788.5 5739.5 5720.5 5768 5720.5 150

D f2(D) P3 = 50 75 100 125 150 175 (MW) f3 P3*

(MW) ($/h) F3(P3) = 806 1108.5 1411 1704.5 1998 2358 ($/h) ($/h) (MW)


Dynamic Programming

� Example� the results show:

� generator #2 is the marginal unit� it picks up all of the additional demand increase between 300 MW

and 325 MW

� P1 = 50 MW, P2 = 110 MW, P3 = 150 MW, and Ptotal = 310 MW

� the cost is easily determined using interpolation

� F1 = $ 810, F2 = $1478, F3 = $ 1998, and Ftotal = $4286

300 4168 150 100 50325 4463 150 125 50

D Cost P 1 P2 P3

( ) 14781360100110100125

13601655)110(2 =+−⋅

−−=F


Ramp Rate Constraints

� Generators are usually under automatic generation control (AGC)� a small change in load and a new dispatch causes the AGC to

change the outputs of appropriate units

� generators must be able to move to the new generation value within a short period of time

� large steam units have a prescribed “maximum rate limit”,∆P/ ∆t (MW per minute)� the AGC must allocate the change in generation to other units,

so that the load change can be accommodated quickly enough

� the new dispatch may be at the most economic values, but the control action may not be acceptable if the ramp rate for any of the units are violated



� To produce an acceptable dispatch to the control system, the ramp rate limits are added to the economic dispatch formulation� requires a short-range load forecast to determine the most

likely load and load-ramping requirements of the units� system load is given to be supplied at time increments

t = 1 … tmax with loading levels of Ptload

� the N generators on-line supply the load at each time increment

� each unit must obey a rate limit such that

tload

N

i

ti PP =�

=1

maxmax

1

iii

it

it

i

PPP

PPP

∆≤∆≤∆−

∆+=+



� The units are scheduled to minimize the cost to deliver the power over the time period

� constraints

and

� the optimization problem can be solved with dynamic programming

max1

1 ttPP tload

N

i

ti �=∀=�

=

maxmax

1

iii

it

it

i

PPP

PPP

∆≤∆≤∆−

∆+=+

( )��==

=N

i

tii

t

t

total PFF11

max


Chapter 4

Transmission System Effects


Overview

� The network’s incremental power losses causes a bias in the optimal economic scheduling of the generators� the total real power loss increases the total generation demand

� the power flow forms the basis for the development of loss factors for predicting the real power loss

� the generation schedule may need to be adjusted by shifting generation to reduce flows on transmission circuits because they would otherwise become overloaded� difficult to include the last effect into optimum dispatching

� the power flow must be solved to check for violations


The Power Flow Problem

� Power flow is the name given to a network solution that shows currents, voltages, and active & reactive power flows at every bus within the system� assumptions

� balanced system - positive sequence solution only

� simple generation and load models

� non-linear problem� relates active and reactive power consumption and generation

with voltage magnitudes

� uses� design procedures (system planning)

� study unique operating problems

� provide accurate calculations of loss penalty factors


The Power Flow Problem

� The power flow problem consist of a given transmission network� lines are represented by a pi-equivalent circuit� transformers are represented by a series impedance circuit� generators and loads represent the boundary conditions

� loads are given as active and reactive power consumptions� generators are usually described by the active power production

and the terminal voltage

� The general power flow equation� one set of equations for each bus in the network

( ) ( )( )

( ) ( )( )∑

∑

=−

=−

−−−=

−+−=

N

kkiikkiikkiiinj

N

kkiikkiikkiiinj

BGEEQ

BGEEP

1

1

cossin

sincos

θθθθ

θθθθ


The AC Power Flow

� Formulation of the power flow� building the bus-admittance matrix

� modeling the transmission network of complex impedances as related to the system buses� includes line, transformer, and shunt element impedances

� general construction rule� if a branch exists between nodes i and j,

and

where j is defined for all branches connected to i

∑+=

−=

jijiii

ijij

yyY

yY

0


The AC Power Flow

� define bus characteristics based on typical information available

Power flow bus specifications

Bus Type Active Power, P

Reactive Power, Q

Voltage Magn., |E|

Voltage Angle, θθθθ

Comments

Constant Power Load, Constant Power Bus

Scheduled

Scheduled

Calculated

Calculated

Standard load representation

Load / Shunt Element, Constant Impedance

Calculated

Calculated

Calculated

Calculated

Only impedance value is given

On-Load Tap Changer, Voltage

Controlled Bus

Calculated

Calculated

Scheduled

Calculated

Secondary side of OLTC

transformers

Generator/Synchronous Condenser, Voltage

Controlled Bus

Scheduled

Calculated

Scheduled

Calculated

Standard generator

representation

Reference / Swing Generator, Slack Bus

Calculated

Calculated

Scheduled

Scheduled

Must adjust net power to hold

voltage constant


The AC Power Flow

� Non-linear system solution method� the ac power flow problem

is cast into a root finding problem

� common solution techniques� Gauss-Seidel

� first AC power flow method developed for digital computers

� method has linear convergence

� governing equation:

start

end

printresults

calculateline flows

∆Emax≤ε

solve for Ei

new=f(Pj,Ej)j=1…N

save maximumvoltage change

do for alli=1…N

i≠ref

select initial voltages

Gauss-SeidelPF Method

yesno

loop

[ ]( )T1

0

0

NxxF LM =

−−−= ∑∑

>

−

<∗−

kjjkj

kjjkj

k

kk

kkk EYEY

E

jQP

YE ]1[][

]1[

[sch][sch]][ 1 ξξ

ξξ


The AC Power Flow

� Non-linear system solution techniques� common solution methods

� Newton-Raphson method� instead of treating each bus individually in each iteration, the

correction is found for the whole system

� Newton’s method is based on the idea of driving the error of a function to zero by making a correction on all the independent variables (bus voltage magnitude and angle)

• setting up the equation: f(x) = K

• pick a starting point x0: f(x0) + ε = K

• use Taylor expansion about x0: f(x0) + (df(x0)/dx) ∆x + ε = K

• setting the error to zero: ∆x = [df(x0)/dx]–1 [K – f(x0)]

� Newton’s method is an iterative process for a non-linear system of equations, but it possesses quadratic convergence to a solution

• the set of first order partial differential eq. is called the Jacobian


The AC Power Flow

� Newton-Raphson method� the set of power equations for each bus in the network

� the injected powers on the left-hand side are the knowns for a load bus

� the power mismatch or error is the difference between the left-hand and right-hand sides with a particular guess of voltages

( ) ( )( )

( ) ( )( )∑

∑

=−

=−

−−−=

−+−=

N

kkiikkiikkiiinj

N

kkiikkiikkiiinj

BGEEQ

BGEEP

1

1

cossin

sincos

θθθθ

θθθθ

( ) ( )( )

( ) ( )( )∑

∑

=−

=−

−−−−=∆

−+−−=∆

N

kkiikkiikkiiinji

N

kkiikkiikkiiinji

BGEEQQ

BGEEPP

1

][][][][][][][

1

][][][][][][][

cossin

sincos

ζζζζζζζ

ζζζζζζζ

θθθθ

θθθθ


The AC Power Flow

� Newton-Raphson method� the incremental correction is defined as

� the matrix representation of the incremental correction

∑∑

∑∑

==

==

∆∂∂+∆

∂∂=∆

∆∂∂+∆

∂∂=∆

N

kk

k

iN

kk

k

ii

N

kk

k

iN

kk

k

ii

EE

QQQ

EE

PPP

1

][

1

][][

1

][

1

][][

ζζζ

ζζζ

θθ

θθ

∆∆

∆∆

∂∂∂∂∂∂∂∂∂∂∂∂∂∂∂∂

∂∂∂∂∂∂∂∂∂∂∂∂∂∂∂∂

=

∆∆

∆∆

−

−

−−−−−

−

−

−

n

n

nnnnnn

nnnnnn

nn

nn

n

n

E

E

EQEQQQ

EQEQQQ

EPEPPP

EPEPPP

Q

Q

P

P

1

2

1

111

1111111

2122212

1112111

1

2

1

M

L

L

MMOMM

L

L

M

θθ

θθθθ

θθθθ


The AC Power Flow

� Newton-Raphson method� deriving the Jacobian terms

� off-diagonal terms

( ) ( )( )

( ) ( )( )

( ) ( )( )

( ) ( )( )kiikkiikkikk

i

kiikkiikkik

i

kiikkiikkikk

i

kiikkiikkik

i

BGEEEE

Q

BGEEQ

BGEEEE

P

BGEEP

θθθθ

θθθθθ

θθθθ

θθθθθ

−−−=∂

∂

−+−−=∂∂

−+−=∂

∂

−−−=∂∂

cossin

sincos

sincos

cossin


The AC Power Flow

� Newton-Raphson method� deriving the Jacobian terms

� diagonal terms

2

2

2

2

iiiiii

i

iiiii

i

iiiiii

i

iiiii

i

EBQEE

Q

EGPQ

EGPEE

P

EBQP

−=∂

∂

−=∂∂

+=∂

∂

−−=∂∂

θ

θ


The AC Power Flow

� Newton-Raphson method� solving the incremental equation

� gaussian elimination is often used to solve directly for the changes in voltage magnitudes and angles instead of finding the matrix inverse of the Jacobian explicitly

� the changes in voltage magnitudes and angles are added to the values that were used at the beginning of the iteration

[ ]

∆∆

∆∆

=

∆∆

∆∆

−

−

−−

n

n

nn

nn

Q

Q

P

P

J

EE

EE 1

2

1

1

11

2

1

MM

θθ


The AC Power Flow

� Newton-Raphson method� the solution process runs

according to the flowchart� note that the Jacobian

matrix is sparse

� the matrix algebra is carried out using Gaussianelimination or one of theother various numericalmethods

start

end

printresults

calculateline flows∆Pmax≤ε

∆Qmax≤ε

select initial voltage values

Newton-RaphsonPF Method

solve for ∆|E| and ∆θusing the Jacobian

and power mismatch

update voltagesθi

ξ+1= θiξ +∆θi

|Ei|ξ+1= |Ei|ξ+∆|Ei|

do for alli=1…N

i≠ref

calculate all ∆P & ∆Qform the Jacobian matrixfind max power mismatch


The AC Power Flow

� Example� three bus system

� solve the power flow Unit 165 MW1.02 pu

100 MW

Three Bus Network100 MVA base

Unit 21.05 pu

Bus 1

0.01 + j0.2

0.03 + j0.25

0.02 + j0.4

Bus 2

Bus 3


The AC Power Flow

� Non-linear system solution techniques� common solution methods

� fast-decoupled power flow� the Newton-Raphson is the most robust algorithm used in practice

� the Jacobian matrix must be recalculated for each iteration

� the set of linear equations must be resolved for each iteration

� a faster method was sought after

� simplifications in the Jacobian� in high voltage systems, the branch impedance is primarily

reactive, X >> R, X/R > 20• Bik >> Gik

� bus angles values are relatively close in value, |θi – θj| < 5°• cos(θi – θj) ≈ 1

• sin (θi – θj) ≈ 0


The AC Power Flow

� Fast Decoupled Power Flow� simplifications

� several of the off-diagonal terms tend towards zero

� the remaining off-diagonal terms can be reduced

( ) ( )( )

( ) ( )( ) 0sincos

0sincos

→=−+−−=∂∂

→=−+−=∂

∂

kiikkiikkik

i

kiikkiikkikk

i

BGEEQ

BGEEEE

P

θθθθθ

θθθθ

( ) ( )( )

( ) ( )( ) ikkikiikkiikkikk

i

ikkikiikkiikkik

i

BEEBGEEEE

Q

BEEBGEEP

−≈−−−=∂

∂

−≈−−−=∂∂

θθθθ

θθθθθ

cossin

cossin


The AC Power Flow


� several of the diagonal terms also tend towards zero

� the remaining diagonal terms can also be reduced

0

0

2

2

→=−=∂∂

→=+=∂

∂

iiiii

i

iiiiii

i

EGPQ

EGPEE

P

θ

22

22

2

iiiiiiiii

i

iiiiiiii

i

iiii

EBEBQEE

Q

EBEBQP

EBQ

−≈−=∂

∂

−≈−−=∂∂

<<

θ


The AC Power Flow


� taking into account the zero terms, the incremental corrections can be rewritten as

� substituting in the reduced terms

∑

∑

=

=

∆∂

∂=∆

∆∂∂=∆

N

k k

k

kk

ii

N

kk

k

ii

E

E

EE

QQ

PP

1

][

][

1

][][

ζζ

ζζ θθ

∑

∑

=

=

∆−=∆

∆−=∆

N

k k

k

ikkii

N

kkikkii

E

EBEEQ

BEEP

1

][

][

1

][][

ζζ

ζζ θ


The AC Power Flow

� Fast Decoupled Power Flow� further simplifications

� divide the terms by |Ei|

� assume |Ek| is approximately equal to one

� the equations in matrix form

∑

∑

=

=

∆−=∆

∆−=∆

N

kkik

i

i

N

kkik

i

i

EBE

Q

BE

P

1

][][

1

][][

ζζ

ζζ

θ

∆∆

−−−−

=

∆∆

−−−−

=

∆

∆

∆

∆

MOMM

L

L

MMOMM

L

L

M

][2

][1

2221

1211][

2

][1

2221

1211

2

][2

1

][1

2

][2

1

][1

ζ

ζ

ζ

ζ

ζ

ζ

ζ

ζ

θθ

E

E

BB

BB

BB

BB

EQ

EQ

EP

EP


The AC Power Flow

� Fast Decoupled Power Flow� resulting equations in general form

� where the terms in B´ and B" come from the susceptances of the bus admittance matrix terms

[ ]

[ ]

∆∆

′′=

∆∆

∆∆

′=

∆∆

MM

MM

2

1

22

11

2

1

22

11

E

E

BEQ

EQ

BEP

EP

θθ


The AC Power Flow

� Fast Decoupled Power Flow� advantages

� B´ and B" are constant� calculated once

� only B" may need to change, resulting from a generation VAR limit violation

� about 1/4 the number of terms found in the Jacobian

� disadvantages� solution convergence failure

� when underlying assumptions do not holdi.e., X/R > 20 and |θi – θj| < 5°


“DC” Power Flow

� A significant simplification of the power flow analysis� drop the Q-V equations altogether in the fast-decoupled

approach

� results in a completely linear, non-iterative power flow algorithm

� simply assume that all voltage magnitudes, |Ei|, equal 1.0 pu

� the system equation becomes

� the terms of B´ are as described for the fast decoupled method

[ ]

∆

∆∆

′=

∆

∆∆

NN

B

P

P

P

θ

θθ

MM

2

1

2

1


“DC” Power Flow

� The DC power flow is only good for calculating the MW flows on transmission lines and transformers� does not solve the problem of MVAR and MVA flows

� the power flowing on each branch (line or transformer) is:

� the power injection at each node (bus) is:

( )kiki

ki xP θθ −= 1

( )∑∑==

−′==N

kkiki

N

kkii BPP

11

θθ


“DC” Power Flow

� Example� solve for the MW flows

� the system equation

� all units are expressedin per units

� the solution Unit 165 MW

Flows Calculated by the DC Power Flow

Unit 235 MW

Bus 1 Bus 2

Bus 3

Unit 165 MW

100 MW

60 MW

5 MW

Three Bus Network

Unit 235 MW

40 MW

Bus 1

j0.2

j0.25j0.4

Bus 2

Bus 3

100 MW

0.0

0.90.5

0.55.7

3

2

1

2

1

=

=

−−

θθθ

P

P

−=

−

=

1.0

02.0

1

65.0

1765.01177.0

1177.02118.0

2

1

θθ


Chapter 4

Transmission System Effects


Transmission Losses

� An illustration using a simple system� consider a two generator system

� the generating units are identical

� production costs are modeledusing a quadratic equation

� the losses on the transmission line are proportional to the square of the power flow

� let both units be loaded to 250 MW� the load would be under served

by 12.5 MW

500 MW

Ploss = 0.0002 P12

P1Min = 70 MWMax = 400 MW

P2Min = 70 MWMax = 400 MW

( ) ( ) ( ) 22211 002.07400 iiii PPPFPFPF ++===

487.5 MW

Ploss = 12.5 MW

P1250 MW

P2250 MW


Transmission Losses

� solution using the Lagrange equation

then

� solution: P1 = 178.88, P2 = 327.50, Ploss = 6.38 MWcost = F1(P1) + F2(P2) = 4623.15

( ) ( ) ( )

( )

00002.0500

0004.00.7

00004.01004.00.7

0002.0

500

212

1

22

111

21

212211

=−−+=∂∂

=−+=∂∂

=−−+=∂∂

=

−−+++=

PPPL

PP

L

PPP

L

PP

PPPPFPFL

loss

loss

λ

λ

λ

λ


Transmission Losses

� if the optimal dispatch is ignored and generator 1 is set to supply all the losses, then� P1 = 263.93 and Ploss = 13.93 MW

� total cost = F1(263.93) + F2(250)= 4661.84

� optimum dispatch tends toward supplying the losses from the unitclose to the load, resulting in a lower value of losses

� the best economics are not necessarily attained at minimum losses� the minimum loss solution:

P1 = 102.08 and P2 = 400 MWPloss = 2.08 MW

� total cost = 4655.43

500 MW

Ploss = 13.93 MW

P1263.93 MW

P2250 MW

500 MW

Ploss = 2.08 MW

P1102.08 MW

P2400 MW (at limit)


Transmission Losses

� Derivation of the penalty factor from incremental losses� start with the Lagrange equation for the economic dispatch

� then

� rearranging the equation

( ) ( )

maxmin1

121

1

0min

,,,

iiii

NiP

N

iiNlossload

N

iii

PPPP

LL

PPPPPPPFL

i

≤≤∀=∂∂→

−++=

=∀

==∑∑

K

Kλ

( ) λ=

∂∂−

−

i

ii

i

loss

P

PF

P

P

d

d1

1

( )01

d

d =

∂∂−−=

∂∂

i

loss

i

ii

i P

P

P

PF

P

L λ


Transmission Losses

� the incremental loss for bus i is defined as

� the penalty factor for bus i is given as

� if the losses increase for an increase in power from bus i, the incremental cost is positive and the penalty factor is greater than unity

� the rearranged minimizing equations become

which are called the coordination equations

1

1−

∂∂−=

∂∂

i

lossi

i

loss

P

PPf

P

P

( )maxmind

diii

i

iii PPP

P

PFPf ≤≤∀= λ


Transmission Losses

� Behaviors of the penalty factor on the coordination equations� when transmission losses are ignored, the penalty factor takes

on the value of unity, making the coordination equation the same as the incremental cost equation

� a penalty factor greater than one (Pfi > 1), representing increased losses for increased generation at bus i, acts on the coordination equation as if the original incremental cost function has been slightly increased� graphically being moved upward

� a Pfi < 1 acts on the coordination equation as if the incremental cost has been slightly decreased� graphically being moved downward


Transmission Losses

� Graphical comparison of the coordination equations with and without accounting for the losses

1

1

d

d

P

F

Pf1 = 1.05

1

1

d

d

P

F1

11 d

d

P

FPf

1P1P′1P ′′

2

2

d

d

P

F

Pf2 = 1.00

2

2

d

d

P

F2

22 d

d

P

FPf

2P2P′ 2P ′′

3

3

d

d

P

F

Pf3 = 0.95

3

3

d

d

P

F3

33 d

d

P

FPf

3P3P′ 3P ′′

λ′λ ′′

factorspenalty withDispatch

losses ignoring Dispatch

=′′=′

i

i

P

P

no penalty factors

with penalty factors


Transmission Losses

� The B matrix loss formula� simplified, practical method for loss and incremental cost

calculations

� basic formula

� where P is the vector of all generator bus net power injections� [B] is a square loss factor matrix of the same dimension as P� B0 is a loss factor vector of the same length as P� B00 is a loss factor constant

� alternative form of the equation

[ ] 000 BBBP TTloss ++= PPP

000 BPBPBPPi

iii j

jijiloss ++= ∑∑∑


Transmission Losses

� using the B coefficients in the economic dispatch equations� equality constraint

� incremental cost equations

� the presence of the incremental losses couples together the coordination equations� makes the solution process

more difficult

� iterative solution algorithm

start

endprint

results

solve coordination equations for Pi for i=1…N

select starting valueof Pi, i = 1…N

Economic Dispatch

with updatedpenalty factors

calculate Ploss using B matrixfind demand PD = Pload + Ploss

calculate penalty factor eqs.for Pfi for i = 1…N

pick starting λ

check demand|ΣPi–PD|<ε?

adjust λ

compare new Pi to Pi of last iteration, save max. change

check solutionmax|Pi–Pi′|<δ?

given: total load, Pload

001

01 11

BPBPBPPPN

iii

N

i

N

jjijiload

N

ii ++++−= ∑∑∑∑

== ==

φ

−−−=

∂∂

∑=

01

21d

di

N

jjij

i

i

i

BPBP

F

P

L λ


Transmission Losses

� Example� loss coefficients

� cost functions

� Pload = 210 MW

040357.0

0189.0

00342.0

0766.0

0294.000901.000507.0

00901.00521.000953.0

00507.000953.00676.0

][

00

0

=

−−

=

−

−=

B

BB

( )( )( ) 1800.4500741.0833.100.240

1505.3700889.0333.100.200

2000.5000533.0669.111.213

32

3333

22

2222

12

1111

≤≤++=≤≤++=≤≤++=

PPPPF

PPPPF

PPPPF


Transmission Losses

� Example� iteration results

1 17.8 227.8 12.8019 50.00 85.34 92.492 11.4 221.4 12.7929 74.59 71.15 75.693 9.0 219.0 12.8098 73.47 70.14 75.394 8.8 218.8 12.8156 73.67 69.98 75.185 8.8 218.8 12.8189 73.65 69.98 75.186 8.8 218.8 12.8206 73.65 69.98 75.18

iteration Ploss PD λ P1 P2 P3


Transmission Losses

� Reference bus versus load center based penalty factors� B matrix assumption

� all load currents conform to an equivalent total load current

� equivalent load current is the negative sum of all generation

� important concepts� an incremental loss is the change in losses

for an incremental change in generation output

� the incremental loss for generator bus i assumes that all other generator outputs remain fixed

� implied principles when using the B matrix� an incremental increase in generator output is

matched by an equivalent increment in load

� alternative approach is to use a reference generator bus

ijP

PP

PP

j

ii

lossloss

≠∀=∆

∆∂

∂=∆

0


Transmission Losses

� Reference generator bus based penalty factors� reference bus always moves for a change in generation

� load demand stays constant

� a change in reference generation would be the negative of the generation change� flows in the system changes as a result of any generator

adjustments

� the change in flows is apt to cause a change in losses

refold

refnew

ref

iold

inew

i

PPP

PPP

∆+=

∆+=

lossiref PPP ∆+∆−=∆


Transmission Losses

� Beta factors� ratio of the negative power change in the reference bus to a

change in generator i:

� economic dispatch can now be defined as follows� economic dispatch is reached when an incremental power shift

from any generator to the reference results in no change in net production cost for any arbitrarily small power change

� this implies

( )i

loss

i

lossi

i

refiiiref P

P

P

PP

P

PPP

∂∂−=

∆∆−∆=

∆∆

−=∆−=∆ 1ββ

( ) ( ) ε≤∆=∆∂

∂=∆= ∑∑ iii

iiii PP

P

PFCPFC 0

( ) ( ) ( ) ( )0

d

d

d

d

d

d

d

d =∆−∆=∆+∆=∆ iref

refrefii

i

iiref

ref

refrefi

i

iii P

P

PFP

P

PFP

P

PFP

P

PFC β


Transmission Losses

� Beta factors� economic dispatch using beta factors

� rewriting the equations

� this is very similar to the coordination equation where the reciprocal of beta replaces the penalty factor

� first order gradient solution method� pick a generation value for the reference bus

� set all other generation according to the equation above

� check for total demand

� readjust the reference as needed until a solution is reached

( ) ( ) ( ) ( )i

ref

refrefi

i

ii

ii

ref

refrefii

i

ii PP

PFP

P

PFP

P

PFP

P

PF ∆=∆→∆=∆d

d

d

d1

d

d

d

d

ββ

( ) ( )∑≠

∆

−=∆

refii

ref

refrefi

i

ii PP

PF

P

PFC

d

d

d

d β


Transmission Losses

� Finding the reference-bus-based penalty factors� computed directly from the N-R power flow equations

� seek to find the ratio of power change at the reference bus whenthere is a change, ∆Pi at the i-th generator

� likewise, find the ratio when there is a reactive power change, ∆Qi at the i-th generator

� the terms ∂Pref∂θi and ∂Pref∂Ei are derived by differentiating the power equation for the reference bus

∑∑∑∑ ∆∂

∂∂∂

+∆∂∂

∂∂

=∆∂∂

+∆∂

∂=∆

ii

i

i

i

ref

ii

i

i

i

ref

ii

i

ref

ii

i

refref P

P

E

E

PP

P

PE

E

PPP

θθ

θθ

∑∑∑∑ ∆∂∂

∂∂

+∆∂∂

∂∂

=∆∂∂

+∆∂

∂=∆

ii

i

i

i

ref

ii

i

i

i

ref

ii

i

ref

ii

i

refref Q

Q

E

E

PQ

Q

PE

E

PPP

θθ

θθ


Transmission Losses� the remaining terms are taken from the inverse Jacobian matrix

� the resulting equation is

or

[ ]

[ ]

∂∂∂∂

∂∂∂∂∂∂∂∂

=

∂∂∂∂

∂∂∂∂∂∂∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

=

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

−

−

Nref

Nref

ref

ref

ref

ref

T

Nref

Nref

ref

ref

ref

ref

N

ref

N

refrefrefrefref

N

ref

N

refrefrefrefref

EP

P

EP

P

EP

P

J

QP

PP

QP

PP

QP

PP

JE

PP

E

PP

E

PP

Q

P

P

P

Q

P

P

P

Q

P

P

P

θ

θ

θ

θθθ

MM

L

L

2

2

1

1

1

2

2

1

1

1

2211

2211

Note: in practice, Gaussian elimination is employed tofind the reference-bus penaltyfactors.


Chapter 5

Unit Commitment


Load Demand Cycles

� Human activity follows cycles� systems supplying services will also experience usage cycles

� transportation, communication, and electric power systems

� electric power consumption follows a daily, weekly, and seasonal cycles� high power usage during the day and evening hours

� industrial and commercial operations and lighting loads

� lower usage on the weekends

� higher usage during the summer and winter� greater temperature extremes

� Load cycles create economic problems for power generation� it is quite expensive to continuously run all generation, which

is needed to meet the peak power demands


Load Demand Cycles

� Definition� commitment means to turn-on a given generation unit

� have the prime mover operating the unit at synchronous speed

� synchronize and connect the unit to the network grid

� Economics� savings are gained by decommitting some of the generation

units when they are not need to meet the current load demand

� the engineering problem is committing enough units to meet current and future load demands while minimizing starting and operating costs


Load Demand Cycles

� Example� consider the cost for operating three generation units

� what combination of units is best to supply a 550 MW load?

Unit 1: F1(P1) = 561 + 7.92 P1 + 0.001562 P12 150 ≤≤≤≤ P1 ≤≤≤≤ 600

Unit 2: F2(P2) = 310 + 7.85 P2 + 0.00194 P22 100 ≤≤≤≤ P2 ≤≤≤≤ 400

Unit 3: F3(P3) = 93.6 + 9.56 P3 + 0.005784 P32 50 ≤≤≤≤ P3 ≤≤≤≤ 200

Uni

t 1

Uni

t 2

Off Off

Uni

t 3

OffOff Off OnOff On OffOff On OnOn Off OffOn Off OnOn On OffOn On On

Max

Gen

0200400600600800

10001200

Min

Gen

050

100150150200250300

P1

Infeasible

0550500295267

40000

255233

1500

500

50

05389491130302787

376000

24402244

InfeasibleInfeasible

P2

P3

F1

F2

16580

5860

586

F3

54185389549754715617

Fto

tal


Load Demand Cycles

� Example� notes:

� the least expensive way to supply the generation is not with allthree units running or with any combination involving two units

� the optimal commitment is to only run unit #1, the most economic unit

� by only running the most economic unit, the load can be suppliedby that unit operating closer to its best efficiency

� if another unit is committed, both unit #1 and the other unit will be loaded further from their best efficiency points, resulting in ahigher net cost


Load Demand Cycles

� Daily load patterns� consider the load demand with a simple peak-valley pattern

� in order to optimize the operation of the system� units must be shut down as load goes down

� then the units must be recommitted as load goes back up

� simple approach to the solution is a simple priority list scheme

9 AM9 PM 3 AM 3 PM3 PM

500 MW

1000 MW

1500 MW1150 MW Peak Load

450 MW Min. Load

Time of day

Tot

al L

oadi

ng


� Example� use a brute force technique to obtain a “shut-down rule” for the range of

loads from 1200 to 500 MW in steps of 50 MW

� when load is above 1000 MW, run all three units

� loading between 600 MW and 1000 MW, run units #1 & #2

� loading below 600 MW, only run unit #1

Load Demand Cycles

9 AM9 PM 3 AM 3 PM3 PM

500 MW

1000 MW

1500 MW

600 MW

Time of day

Tot

al L

oadi

ng

Unit #1

Unit #2400 MW

Unit #3200 MW

1200 on on on1150 on on on1100 on on on1050 on on on1000 on on off

950 on on off900 on on off850 on on off800 on on off750 on on off700 on on off650 on on off600 on off off550 on off off500 on off off

Load Unit 1 Unit 2 Unit 3


Constraints in Unit Commitment

� Primary constraints� enough units are committed to supply the load economically

� Spinning reserve constraints� spinning reserve definition

� the total amount of on-line, synchronized generation power committed less the current loading and power losses supplied� protects the network from an unexpected loss of one or more

generation units

� typical spinning reserve rules� the reserve is a given percentage of the forecasted demand� must be capable of making up the loss of the most heavily

loaded generation unit� reserves must be spread around the system to avoid transmission

limitations (bottling) and permit parts of the system to run as “islands”



� Example� consider a power system consisting of two isolated regions

� transmission tie-lines join the regions and may transfer power up to a maximum of 550 MW in either direction

� five units have been committed to supply 3090 MW of loading

Uni

t C

apac

ity

(MW

)

Uni

t Out

put

(MW

)

Reg

iona

l G

ener

atio

n(M

W)

Uni

t

Reg

ion

Spi

nnin

g R

eser

ve

(MW

)

Reg

iona

l Lo

ad (M

W)

Inte

rcha

nge

(MW

)

1000

800

800

1200

600

4400

900

420

420

1040

310

3090

1740

1350

3090

1

2

3

4

5

Western

Eastern

Total

100

380

380

160

290

1310

1900

1190

3090

160(in)

160(out)

Eastern region

Units 4 & 5

Western region

Units 1, 2 & 3

550 MWmaximum



� Example� verify the allocation of spinning reserves in the system

� western region� generation of largest unit: 900 MW

� available spinning reserve• local: 760 MW; tie-line capacity: 390 MW; eastern region: 450 MW

• total: 1150 MW - load can be supplied

� eastern region� generation of largest unit: 1040 MW

� available spinning reserve• local: 450 MW; tie-line capacity: 550 MW; western region: 700 MW

• total: 1000 MW - load can not be completely supplied

� lack 40 MW of spinning reserve in the eastern region� commit 40 MW of new generation within the eastern region



� Thermal unit constraints� a thermal unit can undergo only gradual temperature changes

� results in a time period of several hours to bring a unit on-line

� minimum up time: it should not be turned off immediately

� minimum down time: once decommitted, the minimum time before a unit can be recommitted

� crew constraint: at a multiple unit plant, there is usually onlyenough personnel to start one unit at at time

� a certain amount of energy is expended to bring a unit on-line� to slowly bring up the temperature and pressure

� this energy does not result in any power delivered from the unit

� the energy cost is brought into the unit commitment problem as a start-up cost



� Start-up costs� the start-up cost can vary from a maximum

cold-start value to a much smaller warm-start value� warm unit: a recently turned-off unit with latent heat

that is near the normal operating temperature

� two approaches available to treating a thermal unit during its down time� allow the boiler to cool down and then heat it back up to

operating temperature in time for a scheduled turn-on

� provide enough fuel to supply sufficient energy to the boiler tojust maintain the operating temperature


Constraints in Unit Commitment� Start-up cost comparison

� cooling

� allowing the unit to cool down

� start-up cost function:

� banking

� input sufficient energy into the boiler to just maintain the operating temperature

� banking cost function:0 1 2 3 4 5 6 7 hr

0

Cfixed

Cstart-up

banking

cooling

break-evenpoint

Time-dependent start-up costs

fixedfuel

t

coldcold CFeHCdownshut

+

−=−−

α1

fixeddownshutfuelbankbank CtFHC += −


Unit Commitment Solution Methods

� Typical utility situation involving the commitment problem� must establish a loading pattern for M periods

� have N generation units available to commit and dispatch� the M load levels and operating limits on the N units are such

that any one unit can supply the load demand and any combination of units can also supply the loads

� Commitment by enumeration� a brute force method

� total combinations to investigate: 2N – 1

� for the total period of M intervals, the maximum number of possible combinations is: (2N – 1)M

� example: for a 24-hour period made up of 1-hr intervals,a 5 unit network become 6.2 × 1035 combinations



� Priority-List Methods� consist of a simple shut-down rule

� obtained by an exhaustive enumeration of all unit combinations at each load level

� or obtained by noting the full-load average production cost of each unit� the full-load average production cost is the net heat rate at full load

multiplied by the fuel cost

� various enhancements can be made to the priority-list scheme by the grouping of units to ensure that various constraints are met



� Typical shut-down rules� at each hour when load is dropping, determine whether

dropping the next unit on the list leaves sufficient generation to supply the load plus the spinning-reserve requirements� if the supply is not sufficient, keep the unit committed

� determine the number of hours before the unit is needed again� if the time is less than the minimum shut-down time for the unit,

keep it committed

� perform a cost comparison� the sum of the hourly production costs for the next number of

hours with the next unit to be dropped being committed

� and the sum of the restart costs for the next unit based on the minimum cost of cooling the unit or banking the unit



� Example� construct a priority list for the units in the first example using

the same cost equations

� the full-load average production costs

� a strict priority order for these units: [ 2, 1, 3 ]

Unit 1: F1(P1) = 561 + 7.92 P1 + 0.001562 P12 150 ≤≤≤≤ P1 ≤≤≤≤ 600

Unit 2: F2(P2) = 310 + 7.85 P2 + 0.00194 P22 100 ≤≤≤≤ P2 ≤≤≤≤ 400

Unit 3: F3(P3) = 93.6 + 9.56 P3 + 0.005784 P32 50 ≤≤≤≤ P3 ≤≤≤≤ 200

Unit 1: F1(600) 600 = 9.7922Unit 2: F2(400) 400 = 9.4010 Unit 3: F3(200) 200 = 11.1848



� Example� the commitment scheme

� ignoring minimum up/down times and start-up costs

� notes� this scheme does not completely parallel the shut-down

sequence described in the first example� there unit 2 was shut down at 600 MW leaving unit 1

� here unit 1 is shut down at 400 MW leaving unit 2

� why the differences? where is the problem?

1 + 2 + 3 300 12001 + 2 250 1000

2 100 400

Combination Min MW Max MW


Chapter 5

Unit Commitment



� Dynamic programming� chief advantage over enumeration schemes is the reduction in

the dimensionality of the problem� in a strict priority order scheme, there are only N combinations

to try for an N unit system

� a strict priority list would result in a theoretically correct dispatch and commitment only if� the no-load costs are zero

� unit input-output characteristics are linear

� there are no other limits, constraints, or restrictions

� start-up costs are a fixed amount



� Dynamic programming� the following assumptions are made in this implementation of

the DP approach� a stateconsists of an array of units

� with specified units operating and the rest decommitted (off-line)

� a feasible state is one in which the committed units can supply the required load and meets the minimum capacity for each period

� start-up costs are independent of the off-line or down-time� i.e., it is a fixed amount w.r.t. time

� no unit shutting-down costs

� a strict priority order will be used within each interval

� a specified minimum amount of capacity must be operating within each interval



� The forward DP approach� runs forward in time from the initial hour to the final hour

� the problem could run from the final hour back to the initial hour

� the forward approach can handle a unit’s start-up costs that are a function of the time it has been off-line (temperature dependent)� the forward approach can readily account for the system’s history

� initial conditions are easier to specified when going forward

� the minimum cost function for hour K with combination I:

� Fcost(K, I) = least total cost to arrive at state (K, I)

� Pcost(K, I) = production cost for state (K, I)

� Scost(K–1, L: K, I) = transition cost from state (K–1, L) to (K, I)

( ) ( ) ( ) ( )[ ]LKFIKLKSIKPIKFL

,1,:,1,min, costcostcost}{

cost −+−+=



� The forward DP approach� state (K, I) is the I th commitment combination in hour K

� a strategyis the transition or path from one state at a given hour to a state at the next hour� X is defined as the number of states to search each period

� N is defined as the number of strategies to be saved at each step� these variable allow control of the computational effort

� for complete enumeration, the maximum value of X or N is 2N – 1

� for a simple priority-list ordering, the upper bound on X is n, the number of units

� reducing N means that information is discarded about the highest cost schedules at each interval and saving only the lowest Npaths or strategies� there is no assurance that the theoretical optimum will be found



� The forward DP approach� restricted

searchpaths� N = 3

� X = 5

IntervalK – 1

IntervalK

IntervalK + 1

X X X

N

N



� Example� consider a system with 4 units to serve an 8 hour load pattern

Incremental No-load Full-load Min. TimeUnit Pmax Pmin Heat Rate Cost Ave. Cost (h)

(MW) (MW) (Btu / kWh) ($ / h) ($ / mWh) Up Down

1 80 25 10440 213.00 23.54 4 22 250 60 9000 585.62 20.34 5 33 300 75 8730 684.74 19.74 5 44 60 20 11900 252.00 28.00 1 1

Initial Condition Start-up CostsUnit off(-) / on(+) Hot Cold Cold start

(h) ($) ($) (h)

1 -5 150 350 42 8 170 400 53 8 500 1100 54 -6 0 0.02 0

Hour Load (MW)

1 4502 5303 6004 5405 4006 2807 2908 500



� Example� to simplify the generator cost function, a straight line

incremental curve is used� the units in this example have linear F(P) functions:

� the units must operate within their limits

PdP

dFFPF loadno

+= −)(

F(P)

Fno-load

Pmin PmaxP

No-load IncrementalUnit Pmax Pmin Cost Cost

(MW) (MW) ($ / h) ($ / MWh)

1 80 25 213.00 20.882 250 60 585.62 18.003 300 75 684.74 17.464 60 20 252.00 23.80



� Case 1: Strict priority-list ordering� the only states examined each hour

consist of the listed four:� state 5: unit 3, state 12: 3 + 2

state 14: 3 + 2 + 1, state 15: all four

� all possible commitments start from state 12 (initial condition)

� minimum unit up and down times are ignored

� in hour 1:� possible states that meet load demand (450 MW): 12, 14, & 15

State Unit status Capacity

5 0 0 1 0 300 MW12 0 1 1 0 550 MW14 1 1 1 0 630 MW15 1 1 1 1 690 MW

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( )

( ) ( ) ( )( ) ( ) ( )[ ] 38.1021102.035036.9861

15,1:12,015,115,1

36.98612080.2330046.1710500.182588.2036.1735

203001052515,1

costcostcost

1321cost

=++=+=

=++++=+++=

SPF

FFFFP Economic Dispatch Eq.

DP State Transition Eq.



� Case 1� in hour 1:

� minimum at state 12 (9208)

� in hour 2:� possible states that meet load demand (530 MW): 12, 14, & 15

K Pcost Scost Fcost

15 9861 350 1021114 9493 350 984312 9208 0 9208

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( )

( ) ( ) ( )

( ) 20859

102110

98430

9208350

min11301

15,2:,115,215,2

113012080.2330046.1718500.182588.201735

203001852515,2

costcostcost

1321cost

=

+++

+=

+==++++=

+++=

LSPF

FFFFP

DP State Transition Eq.



� Case 1: DP diagram

� total cost: 73,439� priority order list, up-times and down-times neglected

151 1 1 1

690 MW

141 1 1 0

630 MW

120 1 1 0

550 MW

50 0 1 0

300 MW

stateunit statuscapacity

hour 1450 MW

hour 2530 MW

hour 3600 MW

hour 4540 MW

hour 5400 MW

hour 6280 MW

hour 7290 MW

hour 8500 MW

hour 0

986110211

12

94939843

12

92089208

12

1130120860

12

1093320492

12

1064819857

12

1341033618

12

1226532472

12

557457182

12

1148143953

14

1111343585

14

1082843300

14

896452615

12

859352244

12

830851609

12

686958828

12

649058449

12

619257800

12

704364976

5

666564597

5

636663949

5

1076174442

5

1039374074

5

1010873439

5

574862930

5

350 350 350 350 350

350 350 350

0

350 350

0

0 0

0

0 0

0

750

750

400

0

750

750

400



� Case 2:� complete enumeration (2.56 × 109 possibilities)

� fortunately, most are not feasible because they do not supply sufficient capacity

� in this case, the true optimal commitment is found� the only difference in the two trajectories occurs in hour 3

� it is less expensive to turn on the less efficient peaking unit #4 for three hours than to start up the more efficient unit #1 for that same time period

� only minor improvement to the total cost� case 1: 73,439

� case 2: 73,274



� Case 2: DP diagram

151 1 1 1

690 MW

141 1 1 0

630 MW

120 1 1 0

550 MW

50 0 1 0

300 MW

stateunit statuscapacity

hour 1450 MW

hour 2530 MW

hour 3600 MW

hour 4540 MW

hour 0

986110211

12

94939843

12

92089208

12

1130120860

12

1093320492

12

1064819857

12

1341033618

12

1226532472

12

1148143953

14

1111343585

14

1082843135

13

350 350

350 35013

0 1 1 1610 MW

95769576

12

1101620221

12

1245032507

12

1111343503

13

350

350



� Lagrange Relaxation� dual variables and dual optimization

� consider the classical constrained optimization problem� primal problem: minimize f(x1,…,xn), subject to ω(x1,…,xn) = 0

� the Lagrangian function: L(x1,…,xn) = f(x1,…,xn) + λ ω(x1,…,xn)

� define a dual function

� then the “dual problem” is to find

� the solution involves two separate optimization problems� in the case of convex functions this procedure is guaranteed to

solve the problem

( ) ( )λλλ

qq0

max≥

∗ =

( ) ( )λλ ,,min 21, 21

xxLqxx

=



� Example� minimize f(x1,x2) = 0.25x1

2 + x22

subject to ω(x1,x2) = 5 –x1 – x2

� the Lagrangian function:

� the dual function:

� the dual problem: ( ) ( ) ( )2

05max 25

0

=

=−=∂∂→=

≥

∗

λ

λλλ

λλλ

qqq

( ) ( )

( ) λλλ

λλλλ

52

&2,,min

245

2121, 21

+−=

==→=

q

xxxxLqxx

( ) ( )2122

2121 525.0,, xxxxxxL −−++= λλ



� Iterative form of Lagrange relaxation method� the optimization may contain non-linear or non-convex

functions

� iterative process based on incremental improvements of λ is required to solve the problem� select a arbitrary starting λ� solve the dual problem such that q(λ) becomes larger

� update λ using a gradient adjustment:

� find closeness to the solution by comparing the gap between the “primal” function and the dual function� primal function: J* = min L

� relative duality gap: , in practice the gap never reaches zero

( ) αλλ

λλ

+= qtt

d

d

( )*

**

q

qJ −



� Lagrange relaxation for unit commitment� loading constraint

� unit limits

� unit minimum up-time and down time constraints

� the objective function

TtUPPN

i

ti

ti

tload K10

1

=∀=−∑=

TtNiPUPPU iti

tii

ti KK 1 and 1minmin ==∀≤≤

( )[ ] ( )ti

ti

T

t

N

i

titiupstart

tii UPFUSPF ,

1 1, =+∑∑

= =−



� Formation of the Lagrange function� in a similar way to the economic dispatch problem

� unit commitment requires that the minimization of the Lagrange function subject to all the constraints� the cost function and the unit constraints are each separated over

the set of units� what is done with one unit does not affect the cost of running

another unit as far as the cost function, unit limits, and the up-time and down-time constraints are concerned

� the loading constraint is a coupling constraint across all the units

� the Lagrange relaxation procedure solves the unit commitment by temporarily ignoring the coupling constraint

( ) ( ) ∑ ∑= =

−+=T

t

N

i

ti

ti

tload

tti

ti UPPUPFUPL

1 1

,,, λλ



� The dual procedure attempts to reach the constrained optimum by maximizing the Lagrangian with respect to the Lagrange multiplier

� done in two basic steps� Step 1: find a value for each λt which moves q(λ) towards a

larger value

� Step 2: assuming that λt found in Step 1 is fixed, find the minimum of L by adjusting the values of Pt and Ut

� minimizing L

( ) ( ) ( ) ( )λλλλλ

,,minwheremax,

UPLqqqti

ti

t UP==∗

( )[ ]

( )[ ]{ } ∑∑∑

∑∑∑∑∑

== =

= === =

+−+=

−++=

T

t

tload

tN

i

T

t

ti

ti

ttiti

tii

T

t

N

i

ti

ti

tT

t

tload

tT

t

N

i

titi

tii

PUPUSPF

UPPUSPFL

11 1,

1 111 1,

λλ

λλ



� separation of the units from one another; the inside term can now be solved independently for each generating unit

� the minimum of the Lagrangian is found by solving for the minimum for each generating unit over all time periods

subject to the up-time and down-time constraints and

� this is easily solved as a two-state dynamic programming problem of one variable

( )[ ]∑=

−+T

t

ti

ti

ttiti

tii UPUSPF

1, λ

( ) ( )[ ]{ }∑∑==

−+=T

t

ti

ti

ttiti

tii

N

i

UPUSPFq1

,1

minmin λλ

TtPUPPU iti

tii

ti K1minmin =∀≤≤

Ui = 1

Ui = 0t = 1 t = 2 t = 3 t = 4

Si Si Si



� Minimizing the function with respect to Pit

� at the Uit = 0 state, the minimization is trivial and equals zero

� at the Uit = 1 state, the minimization w.r.t. Pi

t is:

� there are three cases to be considered for Piopt and the limits

� if Piopt ≤ Pi

min then min [Fi(Pi) – λt Pit] = Fi(Pi

min) – λt Pimin

� if Pimin ≤ Pi

opt ≤ Pimax then min [Fi(Pi) – λt Pi

t] = Fi(Piopt) – λt Pi

opt

� if Pimax ≤ Pi

opt then min [Fi(Pi) – λt Pit] = Fi(Pi

max) – λt Pimax

� the two-state DP is solved to minimize the cost of each unit� for Ui

t = 0 the minimum is zero; therefore the only way to have a lower cost is to have [Fi(Pi) – λt Pi

t] < 0

( )[ ]( )[ ] ( ) ( ) tt

iiti

ttiit

i

ti

tiit

i

ti

tii

PFP

PFP

PPFP

PPF

λλλ

λ

=→=+=+

+

d

d0

d

d

d

d

min



� Adjusting λ� λ must be carefully adjusted to maximize q(λ)

� various techniques use a mixture of heuristic strategies and gradient search methods to achieve a rapid solution

� for the unit commitment problem λ is a vector of λt’s to be adjusted each hour

� simple technique� gradient component: where

� heuristic component:� α = 0.01 when is positive

� α = 0.002 when is negative

( )

( ) ∑=

−=

+=

N

i

ti

ti

tload

t

tt

UPPq

q

1d

d

d

d

λλ

αλλ

λλ

( )λλ

qdd

( )λλ

qdd



� The relative duality gap� used as a measure of the closeness to the solution

� for large real-size power-systems unit-commitment calculations, the duality gap becomes quite small as the dual optimization proceeds� the larger the commitment problem, the smaller the gap

� the convergence is unstable at the end� some units are being switched in and out

� the process never comes to a definite end

� there is no guarantee that when the dual solution process stops, it will be at a feasible solution

� the gap equation:( )*

**

q

qJ −



� Lagrange relaxation algorithm using dual optimization

start

pick starting λt

for t = 1 … Tk = 0

J*–q*

q*

loop foreach unit i

two-state dynamic program with Tstages and solvefor Pi

t and Uit

solve for thedual value q*(λt)

solve the economic dispatch for each hourusing committed units

calculate theprimal value (λt)

update λt

for all t

end

make finaladjustmentsto scheduleto achievefeasibility

gap < thresholdgap > threshold



� Example� find the unit commitment for a three-generator system over a

four-hour segment of time� assume no start-up cost and no minimum up-time or down-time

� start with all λt values set to zero

� execute an economic dispatch for each hour when there is sufficient generation committed for that hour� if there is not enough generation, arbitrarily set the cost to 10,000

� the primal value J* is the total generation cost summed over all t

( )( )( ) 20050005.06100

4001000025.08300

600100002.010500

32

3333

22

2222

12

1111

<<++=<<++=<<++=

PPPPF

PPPPF

PPPPF( )MWt

loadPt

1 1702 520 3 11004 330



� Iteration 1:

� q(λ) = 0.0 J* = 40,000 (J* – q*) / q* = undefined� Iteration 2:

� dynamic programming for unit #3

Hour λ u1 u2 u3 P1 P2 P3 dq(λ)/dλ P1edc P2

edc P3edc

1 0.0 0 0 0 0 0 0 170 0 0 02 0.0 0 0 0 0 0 0 520 0 0 03 0.0 0 0 0 0 0 0 1100 0 0 04 0.0 0 0 0 0 0 0 330 0 0 0

U3 = 1

U3 = 0t = 1 t = 2 t = 3 t = 4

3.3247.5

P34 = P3

min

11.0-700.0

P33 = P3

max

5.2152.5

P32 = P3

min

1.7327.5

P31 = P3

min

λ =F(P) – λP =



� Iteration 2:

� q(λ) = 14,982 J* = 40,000 (J* – q*) / q* = 1.67


edc P3edc

1 1.7 0 0 0 0 0 0 170 0 0 02 5.2 0 0 0 0 0 0 520 0 0 03 11.0 0 1 1 0 400 200 500 0 0 04 3.3 0 0 0 0 0 0 330 0 0 0



� Iteration 3:

� q(λ) = 18,344 J* = 36,024 (J* – q*) / q* = 0.965


edc P3edc

1 3.4 0 0 0 0 0 0 170 0 0 02 10.4 0 1 1 0 400 200 -80 0 320 2003 16.0 1 1 1 600 400 200 -100 500 400 2004 6.6 0 0 0 0 0 0 330 0 0 0



� Iteration 4:

� q(λ) = 19,214 J* = 28,906 (J* – q*) / q* = 0.502


edc P3edc

1 5.1 0 0 0 0 0 0 170 0 0 02 10.24 0 1 1 0 400 200 -80 0 320 2003 15.8 1 1 1 600 400 200 -100 500 400 2004 9.9 0 1 1 0 380 200 -250 0 130 200



� Iteration 5:

� q(λ) = 19,532 J* = 36,024 (J* – q*) / q* = 0.844


edc P3edc

1 6.8 0 0 0 0 0 0 170 0 0 02 10.08 0 1 1 0 400 200 -80 0 320 2003 15.6 1 1 1 600 400 200 -100 500 400 2004 9.4 0 0 1 0 0 200 130 0 0 200



� Iteration 6:

� q(λ) = 19,442 J* = 20,170 (J* – q*) / q* = 0.037� Remarks

� the commitment schedule does not change significantly with further iterations� however, the solution is not stable (oscillation of unit 2)

� the duality gap does reduce� after 10 iterations, the gap reduces to 0.027� good stopping criteria would be when the gap reaches 0.05


edc P3edc

1 8.5 0 0 1 0 0 200 -30 0 0 1702 9.92 0 1 1 0 384 200 -64 0 320 2003 15.4 1 1 1 600 400 200 -100 500 400 2004 10.7 0 1 1 0 400 200 -270 0 130 200


Chapter 6

Generation with Limited Energy Supply



� Economic operation requires that expenditures for fuel be minimized over a period of time� condition #1: no limitation on fuel supply

� economic dispatch using only the present conditions as the data

� condition #2: energy resources available at a particular plant is a limiting factor in operations� economic dispatch calculations must account for what has

happened before and what will happen in the future

� examples include limited fuel supplies, fix cost fuels, take-or-pay contracts, surplus fuels, etc.



� Take-or-pay fuel supply contracts� consider a system with N classical thermal plants and one

turbine generator fueled under a take-or-pay agreement� the utility agrees to use a minimum amount of fuel over a

contracted time period in order to purchase at a bulk price

� if the utility fails to use the minimum amount, it agrees to paythe minimum charge for the minimum amount

� while this unit’s cumulative fuel consumption is below the minimum, the system is schedule to minimize the total cost, subject to the constraint that the total fuel consumption for the period for this unit is equal to a specified amount� once the min. fuel amount is used, the unit is scheduled normally

� as a simplification, we will let the maximum fuel consumption isequal to the minimum amount



� Take-or-pay fuel supply contracts� consider an N + 1 unit system with operation over jmax time

intervals, and let:� Pij = unit i power output at time j

� Fij = cost for unit i for interval j

� qTj = fuel input for unit T at time j

� FTj = cost for unit T for interval j

� Pload j = total load at time j

� nj = number of hours in interval j

� then the problem is defined as

1

2

3

T

P1

P2

PN

F1

F2

FN

PT FT

Pload

00min1

,111 1

maxmaxmax

=−−==−=+

∑∑∑∑ ∑

==== =Tj

N

iijjloadjtotal

j

jTjj

j

jTjj

j

j

N

iijj PPPqqnFnFn ψφ



� Take-or-pay fuel supply contracts� ignoring for the moment the generator limits, the term

is a constant because the total fuel to be used at plant T is fixed� the total fuel cost of that fuel is constant and is dropped from the

Lagrange function

� the independent variables are the powers Pij and PTj

� for any given time period, j = k

∑=

max

1

j

jjTjFn

( )

−+

−−+= ∑∑ ∑∑∑== ===

total

j

jjTj

j

j

N

ijTjijloadj

N

iji

j

jj qqnPPPFnL

maxmaxmax

11 1,

11

γλ

kkT

kTk

kTk

ki

kik

ki P

qn

P

LNi

P

Fn

P

L λγλ −==∂∂=−==

∂∂

d

d0and1

d

d0 K



� Take-or-pay fuel supply� γ is referred to as a pseudo-

price or a shadow price� revalues the fuel price of

a limited fuel supply foreconomic dispatching

� discrete load patterns� solving the fuel limited

dispatch requires dividingthe pattern into intervals

Load

(M

W)

time

start

select value for γ

for load Pload jcalculate the

economic dispatch

jTj

Tjj

jij

ijj

P

qn

NiP

Fn

λγ

λ

=∂∂

== K1,d

d

[ ] total

j

jjTj qqn −=∑

=

max

1

ε

tolerancek≤ε endTrueFalse

loopover all jintervals



� Example� find the optimal dispatch for a gas-fired steam plant

� HT(PT) = 300 + 6.0PT + 0.0025PT2 [MBtu/h]

50 ≤ PT ≤ 400

� Frate = 2.0 $/ccf (1 ccf = 103 ft3)

� Hrate = 1100 Btu/ft3

� the plant must burn 40×106 ft3 of gas

� composite of remaining generation� FS(PS) = 120 + 5.1PS + 0.0012PS

2 [$/h]50 ≤ PS ≤ 500

� load pattern� 0h - 4h: 400 MW, 4h - 8h: 650 MW, 8h - 12h: 800 MW

12h - 16h: 500 MW, 16h - 20h: 200 MW, 20h - 24h: 300 MW



� Example� ignoring the gas constraint, the optimum economic schedule is

� operating cost of the composite unit for the 24h period: $ 52,128

� total gas consumed: 21.8×106 ft3 (at a cost of $ 80k / 40×106 ft3)

� total cost: $ 132,128

1 350 502 500 1503 500 3004 450 505 150 506 250 50

period PS PT



� Example� consider now the gas constraint

� using the gamma search method, the γ value ranges from 0.5 to 0.875 with a final value of 0.8742 $/ccf

� the optimum economic schedule is

� operating cost of the composite unit for the 24h period: $ 34,937

� total gas consumed: 40×106 ft3 (at a cost of $ 80k)

� total cost: $ 114,937

1 197.3 202.62 353.2 296.83 446.7 353.34 259.7 240.35 72.6 127.46 135.0 165.0

period PS PT



� Composite generation production cost functions� composite production cost curves are a useful technique to

mix fuel-constrained and non-fuel-constrained generation� combines N non-constrained units into an equivalent generator

� if one of the units hits a limit, its output is held constant

� a simple procedure for generating FS(PS) consists of adjusting λ from λmin to λmax, where

( ) ( ) ( ) ( )

λ====

+++=+++=

N

N

NS

NNSS

P

F

P

F

P

F

PPPP

PFPFPFPF

d

d

d

d

d

d

2

2

1

1

21

2211

K

K

K

==

== Ni

P

FNi

P

F

i

i

i

iKK 1,

d

dmax&1,

d

dmin maxmin λλ



� Curve finding� at each increment of λ,

calculate the total fuel consumption costs and the total power output for all the units� these data points represent

points on the FS(PS) curve

� the composite curve canbe piecewise linear orsmooth (using curve fitting)

start

set λα = λmin

calculate Piα such

that dFi /dPi = λiα

for i = 1 … N

if unit i hitsa limit, setPi

α at limit

∑=

=N

iiS PP

1

fit curve to points of (PS, FS)

( ) ( )∑=

=N

iiiSS PFPF

1

λλλ αα ∆+=+1

max1 λλα ≥+

end

TrueFalse



� Example� consider the three generation

units with the following cost functions

� combine the units into an equivalent composite generating unit and findthe equivalent cost function

( )

( )

( )20050

00723.0955.11117

400100

002716.099.10434

600150

001562.092.7561

3

23333

2

22222

1

21111

≤≤++=

≤≤++=

≤≤++=

P

PPPF

P

PPPF

P

PPPF



1 8.3886 300 4077 41382 8.7115 403 4961 49243 9.0344 507 5878 57994 9.3574 610 6829 67625 9.6803 714 7813 78126 10.0032 750 8168 82057 11.6178 766 8349 83758 11.9407 825 9049 90459 12.2636 885 9768 9744

10 12.5866 944 10507 1047111 12.9095 1019 11470 1143712 13.2324 1088 12369 1236113 13.5553 1111 12669 1266814 13.8782 1133 12975 1298015 14.2021 1155 13288 1329516 14.5241 1178 13609 1361517 14.8470 1200 13937 13939

step λ Ps Fs Fs (fitted)

PS equivalent unit output (MW)

FS

equi

vale

nt u

nit o

pera

ting

cost

($/

h)

Equivalent unit input / output curveLambda Steps for Composite Cost Curve

Equivalent unit input / output function

( )1200300

0041168.07151.465.2352 2.

≤≤

++=−

S

SSSapproxS

P

PPPF



� Finding a solution by the gradient search technique� consider the following equations produced by the

minimization of the Lagrange function

then combining the equations yields:� for an optimal dispatch, γ is constant

for all intervals j, j = 1 … jmax

� this fact is used to form a search process by refining γ� qTj is treated as a vector containing jmax terms

� γ indicates the gradient of the objective function with respect to qTj

kkT

kTkk

ki

kik P

qnNi

P

Fn λγλ ===

d

dand1

d

dK

kT

kT

ki

ki

Pq

PF

dd

dd

=γ



� Two gradient methods� a simple search approach

� does not require aninitial feasible schedule

� does not require aninitial feasible fuelusage schedule

� but the approach doesnot insure optimality

start

[ ] total

j

jjTj qqn −=∑

=

max

1

ε

tolerancek≤ε endTrueFalse

from arbitrary PS schedulecompute FS(PS) and dFS/dPS

assume feasible schedule forPS and PT for all j = 1,…,jmax

calculate γj for j = 1,…,jmax

find j* with minimum γjand decrease fuel useqTj = qTj – ∆qTj for j = j*

find j* with maximum γjand increase fuel useqTj = qTj + ∆qTj for j = j*

0>εTrueFalse

*

*

*

*

*

dd

dd

jT

jT

jS

jS

j

Pq

PF

=γ



� Two gradient methods� a relaxation technique

� requires an initial feasible schedule

� requires an initialfeasible fuel usageschedule

� approach insures optimality

start

end

from a feasible schedulecompute FS(PS) and dFS/dPS

ε≤∆ totalFTrueFalse

assume feasible schedule such that

jS

Sjtotal

j

jjTj P

Fnqqn λ==∑

= d

d&

max

1

calculate

jS

j

jjtotal FnF ∑

=

=max

1

calculate γj for j = 1,…,jmax

select j+ and j– such that γj +

is maximum for j = j+ and γj

– is minimum for j = j–

adjust q in j+ and j–,qT j

+ = qT j + ∆qj/nj, j = j+

qT j– = qT j – ∆qj/nj, j = j–

adjust PT j+, PT j

–

calculate

−+ ==∆+∆=∆

jjjSjjjStotal FFF

calculate new valuesof γj for j+ and j–



� Example� find the optimal dispatch for a gas-fired steam plant

� HT(PT) = 300 + 6.0PT + 0.0025PT2 [MBtu/h]

50 ≤ PT ≤ 400

� Frate = 2.0 $/ccf (1 ccf = 103 ft3)

� Hrate = 1100 Btu/ft3

� the plant must burn 40×106 ft3 of gas

� composite of remaining generation� FS(PS) = 120 + 5.1PS + 0.0012PS

2 [$/h]50 ≤ PS ≤ 500

� load pattern� 0h - 4h: 400 MW, 4h - 8h: 650 MW, 8h - 12h: 800 MW

12h - 16h: 500 MW, 16h - 20h: 200 MW, 20h - 24h: 300 MW



� Example� initial dispatch

� total fuel consumption: 21.84 M ft3

� maximum γ : 1.0877, minimum γ : 0.9240, average γ : 1.0058

� increase unit 4 by 137.2 MW

1 350 50 2.205 1.04542 500 150 4.568 1.02673 500 300 8.455 0.92404 450 50 2.205 1.08775 150 50 2.205 0.96106 250 50 2.205 1.0032

period PS PT qTj γj



� Example� iteration #1




1 350 50 2.205 1.04542 500 150 4.568 1.02673 500 300 8.455 0.92404 312.8 187.2 5.493 0.92795 150 50 2.205 0.96106 250 50 2.205 1.0032








1 242.1 157.9 4.762 0.92042 500 150 4.568 1.02673 500 300 8.455 0.92404 312.8 187.2 5.493 0.92795 150 50 2.205 0.96106 250 50 2.205 1.0032








1 242.1 157.9 4.762 0.92042 360.6 289.4 8.167 0.88113 500 300 8.455 0.92404 312.8 187.2 5.493 0.92795 150 50 2.205 0.96106 250 50 2.205 1.0032








1 199.1 200.9 5.841 0.87602 353.7 296.3 8.355 0.87463 446.0 354.0 9.954 0.87354 260.0 240.0 6.851 0.87455 71.6 128.4 4.042 0.87316 135.3 164.7 4.932 0.8745






� maximum γ : 0.8746, minimum γ : 0.8731, average γ : 0. 8739

� increase unit 5 by -0.8 MW

1 197.4 202.6 5.884 0.87432 353.7 296.3 8.355 0.87463 446.0 354.0 9.954 0.87354 260.0 240.0 6.851 0.87455 71.6 128.4 4.042 0.87316 135.3 164.7 4.932 0.8745







� end search

1 197.4 202.6 5.884 0.87432 353.7 296.3 8.355 0.87463 446.0 354.0 9.954 0.87354 260.0 240.0 6.851 0.87455 72.4 127.6 4.022 0.87406 135.3 164.7 4.932 0.8745



Chapter 6




� Hard limits and slack variables� take account of the limits on the take-or-pay generating unit

� this may be added to the Lagrangian by using two new functions and two new variables, called slack variables

� where S1j and S2j are the slack variables

� they may take on any real value including zero

� the new Lagrangian

maxmin TTT PPP ≤≤

( )

( ) ( )∑∑

∑∑ ∑∑∑

==

== ===

+−++−+

−+

−−+=

maxmax

maxmaxmax

1

22min2

1

21max1

11 1,

11

j

jjjTTj

j

jjTjTj

total

j

jjTj

j

j

N

ijTjijloadj

N

iji

j

jj

SPPSPP

qqnPPPFnL

αα

γλ

22min2

21max1 and jjTTjjTjTj SPPSPP +−=+−= ψψ



� Hard limits and slack variables� α1j and α2j are Lagrange multipliers

� the first partial derivatives for the kth interval are

� when the constrained variable PTk is within bounds, α1j = α2j = 0 and S1j and S2j are nonzero; when PTk is limited, either S1j or S2jis zero and the associated Lagrange multiplier is nonzero

kkk

kkk

kkkkT

kTk

kT

kki

kik

ki

SL

SL

P

qn

P

L

P

Fn

P

L

222

111

21

20

20

d

d0

d

d0

αα

αα

ααλγ

λ

==∂∂

==∂∂

−+−==∂∂

−==∂∂



� Hard limits and slack variables� consider some interval k where PTk = Pmax

� S1k = 0 and α1k ≠ 0, then

� if

the value of α1k takes on the value just sufficient to make the equality true

0d

d1 =++−

kT

kTkkk P

qnγαλ

kT

kTkk P

qn

d

dγλ >



� Example� reconsider the 6-interval 24-hour fuel scheduling

� maximum generation on PT is reduced to 300 MW� in the original optimal schedule, PT = 353.3 in the third time-

period

� when the limit is reduced to 300 MW, the gas-fired unit burns more fuel in other time periods to meet the 40 M ft3 gas consumption constraint



� Example� resulting optimal schedule with Ptmax = 300 MW

� shadow price, γ : 0.8603

� total cost: $ 122,985

1 183.4 216.6 5.54 5.54 0.2 350.0 300.0 5.94 5.86 0.083 500.0 300.0 6.30 5.86 0.444 245.4 254.6 5.69 5.69 0.5 59.5 140.5 5.24 5.24 0.6 121.4 178.6 5.39 5.39 0.

period PS PT λj γj (∂qTj /∂PTj) α1j



� Fuel scheduling� the major elements in the fuel supply and delivery system

� raw-fuel suppliers� coal, oil, and gas companies

� long-term contracts with min./max. limits on fuel per time-period

� prices may change subject to renegotiation provisions in contract

� transportation� railroads, river barges, gas-pipeline companies

� represent problems in scheduling of fuel deliveries

� fuel storage farms - fuel inventory� coal piles, oil storage tanks, underground gas storage facilities

� keeping proper inventory levels to forestall fuel shortages• load levels exceed forecast, suppliers or shippers unable to deliver

� generation stations



� Fuel scheduling� fuel scheduling problem

� the total time period is broken into discrete time increments

� constraint functions� some constrain functions will span two or more time steps

� solution technique� linear programming

� procedure that minimizes a linear objective function

� variables are also subject to linear constraints

� any non-linear functions (objective or constrain equations) must be approximated by linearization about an operating point or by piecewise linear functions



� Linear programming� optimization function

� linear constraints

� and the variables may having specified upper and lower limits

NN xcxcxcZ +++= K2211

MNMNMM

NN

NN

bxaxaxa

bxaxaxa

bxaxaxa

≤+++

≤+++≤+++

K

M

K

K

2211

22222121

11212111

maxmin

max22

min2

max11

min1

NNN xxx

xxx

xxx

≤≤

≤≤

≤≤

M



� One of many Linear programming (LP) techniques� upper-bound linear programming algorithm

� variable limits are handled implicitly

� requires a slack variable with each constraint

� the slack variables� used primarily in inequality constraints

� equals the difference or slack between a constraint and its limit

� transforms an inequality constraint into an equality constraint

� if x1 = 3 and x2 = 2, then xS = 4

� for a “greater than or equal to” constraint, the bounds on the slack variable are changed

� equality constraints may contain a slack variable with bounds at 0

∞<≤=++→≤+

SS xxxx

xx

02864

2864

21

21



� Linear programming� linear equations structure (canonical form)

� the objective function and constraints are arranged in a matrix

� there is at least one variable in each constraint whose coefficient is zero in all the other constraints� this set of variables is called the basis variables

� the remaining variables are called non-basis variables

=

−

+

01

1

1

1

2

1

2

1

2

1

21

121

22221

11211

NSN

S

S

NN

NNN

N

N

b

b

b

Z

x

x

x

x

x

x

ccc

aaa

aaa

aaa

MMOM

L

L

MOMM

L

L



� Linear programming� the solution procedure centers on conducting pivot operations

� a pivot exchanges a non-basis variable for a basis variable

� pivoting� consider the operation for pivoting variable x1 for xS2

� that is equivalent to pivoting column 1 with row 2

� steps� multiply the elements in row 2 by: 1 / a21

� for each row i (i ≠ 2) subtract from row i the product of row 2 and ai1

21222122 1 abbNjaaa jj =′==′ K

Njaccc

iNibabb

Njaaaa

jjj

iii

jiijij

K

K

K

1

2,1

1

21

21

21

=′−=′≠=′−=′

=′−=′



� Linear programming� the result of the pivoting operation

� x1 is now one of the basis variables

′′

′′

=

−′′

′′

+

′′′′

′′′′

z

b

b

b

Z

x

x

x

c

a

a

a

x

x

x

cc

aa

aa

aa

NSN

S

S

S

NS

S

S

NN

NN

N

N

MMOMM

L

L

MOMM

L

L

2

1

2

1

2

2

22

21

2

1

2

12

222

112

1

1

1

0

0

1

0



� Linear programming� the dual upper-bounding algorithm proceeds in simple steps

� variables that are in the basis set are exchanged for variables in the non-basis set using the pivoting operation

� the non-basis variables are held equal to either their upper or lower bounds

� the basis variables are allowed to take any value without respect to their bounds

� the solution terminates when all the basis variables are within their respective limits



� The LP algorithm� starting steps

� for each variable that has a non-zero coefficient cj in the objective function (the cost row) must be set according to the following rule� if cj > 0, set xj = xj

min; else if cj < 0, set xj = xjmax

� if cj = 0, xj can be set to any value; for convenience it is set to xjmin

� add a slack variable xSj to each constraint in the problem� using the values of xj assigned in the above step, set the sleck

variables to make each constraint equal to its limit

� variable exchange� row selection: find the basis variable with the greatest violation

� this is the row to pivot on

� identified as row R



� The LP algorithm� variable exchange using R with the worst violation

� column selection 1: use if constrain row R is below its minimum� pick column S so that cS / (-aR,S) is minimum for all S that satisfy:

• S is not in the current basis

• aR,S is not equal to zero

• if xS is at its minimum, then aR,S< 0 and cS ≥ 0

• if xS is at its maximum, then aR,S> 0 and cS ≤ 0

� column selection 2: use if constrain row R is above its maximum� pick column S so thatcS / (aR,S) is minimum for all S that satisfy:

• S is not in the current basis

• aR,S is not equal to zero

• if xS is at its minimum, thenaR,S> 0 andcS ≤ 0

• if xS is at its maximum, thenaR,S< 0 andcS ≥ 0



� The LP algorithm� variable exchange using row R and column S

� pivot at the selected row R and column S

� the pivot column’s variable, xS, goes into the basis

� if no column fits the column selection criteria, the problem hasan infeasible solution� no values for x1…xN will satisfy all constraints simultaneously

� setting the variables after pivoting� all non-basis variables, except for xS hold there current values

� the most violated variable xR is set to the limit that was violated

� all non-basis variables are determined, set each basis variable to whatever value is required to make the constraints balance� all basis variables may change, many may now violate the limits



� The LP algorithm

end

search the basis variablesfor the one with the worstviolation, designate it as R

most violated variableis above its maximum

Yes

pick column S using the column selection procedure for most

violated variable above its maximum

foundcolumn S?

foundcolumn S?

violation type

any violation?

pivot on selected row and column

pick column S using the column selection procedure for most

violated variable below its minimum

infeasiblesolution

start

infeasiblesolution

Yes

NoNo

No

Yes

most violated variableis below its minimum



� LP example� minimize: Z = 2x1 + x2

� subject to: x1 + x2 = 20 constraint #1 –1.4x1 + x2 ≤ 2 constraint #22 ≤ x1 ≤ 12 limit #12 ≤ x2 ≤ 16 limit #2

–1.4x1 + x2 ≤ 2

x1 + x2 = 20

x2 = 2

x2 = 16

x1 = 2 x1 = 12

cost contours



� LP example� problem cast into canonic form:

� variable 3 is the worst-violated basis variable, R = 1� exceeds its maximum of 0

� using column selection procedure, pivot at column 2, S = 2� s = 1: a11 > 0 x1 = x1

min c1 > 0 then c1 / a11 = 2/1 = 2

� s = 2: a12 > 0 x2 = x2min c2 > 0 then c2 / a12 = 1/1 = 1

constraint #1: x1 + x2 + x3 = 20constraint #2: –1.4x1 + x2 + x4 = 2cost: 2x1 + x2 –Z = 0

0 ≤ x3 ≤ 00 ≤ x4 ≤ ∞

minimum: 2 2 0 0present value: 2 2 16 2.8 6maximum: 12 16 0 ∞



� LP example� pivoting results in:

� variable 4 is the worst-violated basis variable, R = 2� exceeds its minimum of 0

� using column selection procedure, pivot at column 1, S = 1� s = 1: a21 < 0 x1 = x1

min c1 > 0 then c1 / –a21 = 1/2.4 = 0.4166

� s = 3: a22 < 0 x3 = x3max c2 < 0 then x3 is not eligible

constraint #1: x1 + x2 + x3 = 20constraint #2: –2.4x1 – x3 + x4 = –18cost: x1 – x3 –Z = –20

0 ≤ x3 ≤ 00 ≤ x4 ≤ ∞

minimum: 2 2 0 0present value: 2 18 0 –13.2 22maximum: 12 16 0 ∞



� LP example� pivoting results in:

� no violations among the basis variables

� algorithm stops at the optimum

constraint #1: + x2 +.583x3 +.417 x4 = 12.5constraint #2: x1 +.417x3 +.417 x4 = 7.5cost: –1.417x3 +.417 x4 –Z = –27.5

0 ≤ x3 ≤ 00 ≤ x4 ≤ ∞

minimum: 2 2 0 0present value: 7.5 12.5 0 0 27.5maximum: 12 16 0 ∞



� Fuel scheduling by LP� two coal-burning generating units

� both must remain online for a 3 week period

� the combined output is to supply the following load � week 1: 1200 MW

� week 2: 1500 MW

� week 3: 800 MW

� one coal supplier is under contract to supply 40,000 tons per week to be split between the two plants� there are existing inventories at the start of the 3-week period

� find the following� the operation schedule for each plant for each week

� the coal delivery amounts to be made each week



� Fuel scheduling by LP� data

� coal� heat value: 23 Mbtu/ton

� cost: $30/ton or $1.3/Mbtu

� inventories: plant #1 = 70,000 tons, plant #2 = 70,000 tons

� each plant has a maximum coal storage of 200,000 ton

� generators� plant #1: H1(P1) = 380.0 + 8.267 P1 150 ≤ P1 ≤ 600 MW

q1(P1) = 16.52 + 0.3594 P1 [tons/h]F1(P1) = 495.65 + 10.78 P1 [$/h]

� plant #2: H2(P2) = 583.3 + 8.167 P2 400 ≤ P2 ≤ 1000 MWq2(P2) = 25.36 + 0.3551 P2 [tons/h]F2(P2) = 760.8 + 10.65 P2 [$/h]



� Fuel scheduling by LP� solution process

� assume that the units are operated at a constant rate during each week� coal deliveries are made at the beginning of each week

� the problem is set up for 1-week time intervals

� then� plant #1: q1(P1) = 2775.4 + 60. 4 P1 [tons/wk]

F1(P1) = 83,269.2 + 1811 P1 [$/wk]

� plant #2: q2(P2) = 4260.5 + 59.7 P2 [tons/wk]F2(P2) = 127,814.4 + 1789 P2 [$/wk]

� objective function� minimize: Z = F1[P1(1)] + F2[P2(1)] + F1[P1(2)] + F2[P2(2)]

+ F1[P1(3)] + F2[P2(3)]



� Fuel scheduling by LP� constraints

� power delivery� P1(1) + P2(1) = 1200

� P1(2) + P2(2) = 1500

� P1(3) + P2(3) = 800

� coal deliveries� D1(1) + D2(1) = 44,000

� D1(2) + D2(2) = 44,000

� D1(3) + D2(3) = 44,000

� volume of coal� V1(1) + D1(1) –q1(1) = V1(2)

� V2(1) + D2(1) –q2(1) = V2(2)

� V1(2) + D1(2) –q1(2) = V1(3)

� V2(2) + D2(2) –q2(2) = V2(3)

� V1(3) + D1(3) –q1(3) = V1(4)

� V2(3) + D2(3) –q2(3) = V2(4)



� Fuel scheduling by LPvariables

X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13 X14 X15 X16 X17 X18 constantsconstraint

1234567 89

101112

variableminimumvariable

maximum

D1(

1)

P1(

1)

D2(

1)

P2(

1)

V1(

2)

D1(

2)

P1(

2)

V2(

2)

D2(

2)

P2(

2)

V1(

3)

D1(

3)

P1(

3)

V2(

3)

D2(

3)

P2(

3)

V1(

4)

V2(

4)

4000

00

600

150

4000

00

1000

400

2000

000

4000

00

600

150

2000

000

4000

00

1000

400

2000

000

4000

00

600

150

2000

000

4000

00

1000

400

2000

000

2000

000

1 1 12001 1 400001 -60.4 -1 2775.4-V1(1)

1 -59.7 -1 4260.5-V2(1)1 1 1500

1 1 400001 1 -60.4 -1 2775.4

1 1 -59.7 -1 4260.51 1 800

1 1 400001 1 -60.4 -1 2775.4

1 1 -59.7 -1 4260.5


Chapter 8

Production Cost Models


Introduction

� Production cost models� are computational models designed to calculate information

for long-range system planning:� generation system production costs

� energy import requirements

� availability of energy for sale to other systems

� fuel consumption

� employees models of expected load patterns and simulated operation of the system’s generation � uncertainty of load forecasts

� reliability of generating units

� expected need for emergency energy and capacity supplies

� uses statistical computational methods for solving problems


Introduction

� Stochastic production cost models� used for long range studies

� the risk of sudden, random, generating unit failures and random deviations for the mean forecasted load are treated as probability distributions

� load modeling considers the behaviors of the “expected load” patterns that cover periods of weeks, months, and/or years� the load duration cover expresses the time period that the

loading is expected to equal or exceed a given power value

� generating unit modeling includes fuel costs usually expressed over a monthly basis� generating unit scheduled maintenance outages

may involve time periods from days to years


Introduction

� Types of production cost studies

Load ModelInterval under Consideration

Economic Dispatch Procedure

total energy or load duration

load duration or load cycles

seasons or years

months or weeks

block loading (w/o regard to incr. costs)

incremental loading

load durationor load cycles

load cycles

months, weeks, or days

weeks or days

incremental loadingwith forced outages

incremental loading with losses

Long-Range Planning

Operation Planning

Weekly Schedules

�

�

�

�

�

�

� �


Load-duration Curves

� Representation of future loads in which the impact of capacity limitations will be studied

� Building the load-duration curve� consider the expected load

pattern

� build a histogram of load for a given time period and find the load density function, p(x)

� integrate the load densityfunction to obtain the load distribution function, Pn(x)

Time (h)

Load

(M

W)

Load L (MW)

p(L)

Pro

babi

lity

that

load

= L

MW

probability densityfunction

Load L (MW)

P(L

) P

roba

bilit

ylo

ad ≥

LM

W cumulativedistribution

function

1.0

0.0( ) ( )∫∞

−=x

d1 xxpxPn



� Building the load duration curve� multiply the probability by the

period length to show the number of hours that load equals or exceeds a given power level, L

� common convention has theload on the vertical axis

� Block-Loading� simulates the economic dispatch

procedure with this type of load model

� generating units are ordered by cost� units are assumed to be fully loaded or

loaded up to the limitation of the load-duration curveLo

ad L

(MW

)

Hours load equalsor exceed L MW

040

500

1000

1500

8 12



� Block-Loading� example of the Niagara-Mohawk system

2-Mile PointMohawk #1Mohawk #2

8.18.58.7

800300200

UnitIncremental Cost ($/h)

Maximum Capacity (MW)

Rio Bravo #1Rio Bravo #2Rio Bravo #3

9.29.69.7

752520

(8) gas turbines

9.9 400

Load

L(M

W)

Hours load equals or exceed L MW0

500

1000

1500

2-Mile Point (800 MW)

Mohawk #1 (300 MW)

Mohawk #2 (200 MW)

Rio Bravo #1, #2, & #3 (120 MW)

gas turbines (280 MW out of 400 MW)



� Example� consider a two generating unit

system that will serve the following expected load pattern:

� construct a load-duration curvein tabular and graphic form

1008040

206020

20004800800

x-Load (MW)

Duration(h)

Energy (MWh)

Totals: 100 7600

02040

00

20

100100100

x-Load (MW)

ExactDuration (h)

T Pn(k), Hours that Load Equals

or Exceeds x

6080

100

06020

808020

100+ 0 0

100

50

0100806040200

x-Load (MW)

TP

n(x)

, Hou

rs th

at lo

ad e

qual

s or

exc

eeds

x



� Example� the two generating units have the following characteristics

� the fuel cost rate for each unit is a linear function of the power output

� block-load the two units onto the load-duration curve� unit #1 is used first because of its lower average cost per MWh

0800

16080080

1.01.02.0

UnitPower Output(MW)

Fuel Input

(MBtu/h)

2

1

Fuel Cost

($/MBtu)

Fuel Cost Rate($/h)

Incremental Fuel Cost ($/MWh)

40 400 2.0

160800160800

8.0

16.0

0.05

0.10

Unit Force Outage Rate

(per unit)



� Example� block-loaded

� unit #1 is on-line for 100 h� 80 MW output for 80 h

� 40 MW output for 20 h

� unit #2 is on-line for 20 h� 20 MW output for 20 h

� summary of results

100

50

0100806040200

x-Load (MW)

TP

n(x)

, Hou

rs th

at lo

ad e

qual

s or

exc

eeds

x

Unit 1

Unit 2

4080

20

2080

20

8006400

400

UnitLoad(MW)

Duration (h)

2

1

Energy (MWh)

Fuel Used (MBtu)

Fuel Cost ($)

Total: 7600

960064000

4800

9600

Subtotal: 7200 736006400073600

96008320078400


Forced Outages

� Forced outage of a generator unit� the time that the unit is not available due

to a failure of some sort� represents a random event

� taken out of the total time that the unit should be available for service

� the forced outage rate is the ratio of forced outage time over the total time available� schedule outage times for maintenance are excluded in

both the total time available and the forced outage time

� Forced outage rates for all generating units must be accounted for in the expected production costs


Forced Outages

� Example� reconsider the previous example, but now including the effects

of forced outages

� evaluate by load levels� Load = 40 MW; duration 20 h

� Unit 1: on-line for 20 h, operates for 0.95 × 20 = 19 houtput: 40 MW, energy delivered: 19 × 40 = 760 MWh

� Unit 2: on-line for 1 h, operates for 0.90 × 1 = 0.9 houtput: 40 MW, energy delivered: 0.9 × 40 = 36 MWh

� load energy = 800 MWhgeneration = 796 MWhunserved energy = 4 MWhshortage = 40 MW for 0.1 h


Forced Outages

� Example� Load = 80 MW; duration 60 h


� Unit 2: on-line for 3 h, operates for 0.90 × 3 = 2.7 houtput: 40 MW, energy delivered: 2.7 × 40 = 108 MWh

� load energy = 4800 MWhgeneration = 4668 MWhunserved energy = 132 MWhshortage = 80 MW for 0.3 h (24 MWh) and

40 MW for 2.7 h (108 MWh)


Forced Outages

� Example� Load = 100 MW; duration 20 h


� Unit 2: on-line for 20 h, operates as follows• Unit 1 is on-line and operating for 19 h

Unit 2: on-line for 0.90 × 19 = 17.1 houtput: 20 MW, energy delivered: 17.1 × 20 = 342 MWhshortage: 20 MW for 1.9 h

• Unit 1 is supposedly on-line, but not operating 1 hUnit 2: on-line for 0.90 × 1 = 0.9 h

output: 40 MW, energy delivered: 0.9 × 40 = 36 MWhshortage: 100 MW for 0.1 h and 60 MW for 0.9 h

� load energy = 2000 MWh; generation = 1898 MWhunserved energy = 102 MWh

• 100 MW for 0.1 h = 10 MWh; 60 MW for 0.9 h = 54 MWh;20 MW for 1.9 h = 38 MWh


Forced Outages

� Comments� it was necessary to make an arbitrary assumption that the

second unit will be on-line for any load level that equals or exceeds the capacity of the first unit

� the enumeration of the possible states is not complete� need to separate the periods when there is excess capability,

exact matching of generation and load, and shortages

� when there is an exact matching of generation and load, it is referred to as a “zero-MW shortage”

� there are two such periods in the example� 40 MW loading, 20 h duration, unit 2 on: 0.05 × 0.9 × 20 = 0.9 h

� 80 MW loading, 60 h duration, unit 1 on: 0.95 × 0.1 × 60 = 5.7 h

� total zero-reserve expected duration: 6.6 h


Forced Outages

� Summary of all possible states

20

Load(MW)

1

Duration(h)

EventNo.

Unit 1 Unit 2

40

1

Status Power (MW) Status Power (MW)

Combined Event

Duration (h) Consequence

2 13 04 05 16 17 08 09 110 111 012 0

6080

20100

101010101010

4040

40

8080

8080

00

00

00

00

000

0

0

0

40

40

20

17.11.90.90.1

51.35.72.70.3

17.11.90.90.1

Load satisfied

Load satisfied

0 MW shortage

40 MW shortage

Load satisfied

0 MW shortage

40 MW shortage

80 MW shortage

Load satisfied

20 MW shortage

60 MW shortage

100 MW shortage


Forced Outages

� Summary of generation cost results

� unserved load

100

81

95.0

72.9

6840

522

UnitScheduled

Time On-line (h)

Expected Operating Time (h)

2

1

Expected Generated

Energy (MWh)

Expected Fuel Used

(MBtu)

Expected Production

Cost ($)

Total: 7362

69920

10008

69920

20016

8993679928

6.61.9

038

Unserved Demand (MW)

Duration of Shortage (h)

200

UnservedEnergy (MWh)

Total: 23812.6

12.66.0

Duration of Given Shortage or More (h)

2.80.9

1125460

40 4.11.3

0.30.1

2410100

80 0.40.1


Chapter 8

Production Cost Models


Probabilistic Programming

� Production cost analysis� operating experience indicates that the force outage rate of

thermal-generating units tends to increase with unit size� the frequent long-duration outages of the more efficient base-load

units require running the less efficient, more expensive plants at higher than expected output and the importation of emergency energy

� two measures of system unreliability due to random forced generator failures:� time period when the load is greater than available generation

capacity

� the expected levels of power and energy that must be imported tosatisfy the load demand� sensitivity indicators of the need to add generator or tie-line capacity



� Probabilistic production cost computations� mathematical methods that make use of probabilistic models

� load models

� energy and capacity models for generators� generator models represent the unavailability of basic energy

resources, random forced outages, and the effects of energy sale/purchase contracts

� tie-line models represent the expected cost of emergency energy, also called the cost of unsupplied energy

� techniques for the convolution of the load distributions with the capacity-probability density functions� numerical convolution of discrete functions

� analytical convolution of continuous functions



� Probabilistic production cost computations� the unserved load distribution method

� the individual unit probability capacity density functions are convolved one by one with the load distribution function� each convolution result is an unserved load distribution, which will

be used for the next convolution

� the overall result is a series of unserved load distributions

� the sequence of convolutions is determined by a fixed economic loading criterion that sets the order of the generator units

� the area under the load distributions represent needed energy� a unit’s energy production is found from the difference in the

energy from the unserved load distribution prior to convolution and the energy from the new unserved load distribution after the convolution



� Probabilistic production cost computations� the expected cost method

� the individual unit probability capacity density functions are convolved with one another� the sequence of convolutions is determined by a fixed economic

criterion that sets the order of the generator units

� the convolutions of the generating units produces a set of available capacity distributions, each with an associated cost curve as a function of total power generated

� the resultant expected cost curve is convolved with the load distribution function� this produces the expected value of production cost to serve the

given load forecast



� The unserved load method� load is modeled as a load-duration curve

� a probability distribution that is expressed in hours that the load equals or exceeds a given power value� monotonically decreasing function with increasing load

� can be treated as cumulative probability density function, Pn(x) or can be expressed in hours, T Pn(x)

� for numerical analysis� treat it in terms of regular discrete steps in a recursive algorithm

� procedural steps� if there is a segment of capacity with a total of C MW available

for scheduling� let q be the probability that C MW are unavailable

� and p = 1 –q (that is the available probability of the segment)



� The unserved load method� procedural steps

� then after the segment is scheduled� the probability of needing x MW or more is now P′n(x)

� because the occurrence of loads and unexpected unit outages are statistically independent events, the new probability distribution function is a combination of the two mutually exclusive events

• q Pn(x) is the probability that new capacity C MW is unavailable and needing x MW or more

• p Pn(x + C) is the probability that C is available and needing (x + C) MW or more

( ) ( ) ( )Cxpxqx nnn ++=′ PPP




� first, the generation requirements for any generating segment are determined by the knowledge of the distribution T Pn(x) that exists prior to the dispatch of the particular generating segment� the value of T Pn(0) determines the required hours of operation of a

new unit, and the area under the distribution T Pn(x) between zero and the rating of the unit determines the energy production

� if a generation segment is not perfectly reliable, there will be a residual distribution of demands that cannot be served by this particular segment because of forced outages




� representing the forced outage when the unit is needed for T Pn(0) hours� on average it is only available for (1 –q) T Pn(0) hours

� energy required by the load distribution that the unit could serve

� the unit can only generate (1 –q) E because of its expected unavailability

� suppose that the unit has a linear input-output cost function

� the expected production cost for this period

( ) ( )∫ ∑+=

∆==C C

xnn xxPTxxPTE

00

d

( ) iii PFFPF ⋅+= 10

[ ] ( ) ( ) ( )EqFTqFFE ni −+−= 10P1 10




� there is a residual of unserved demands due to the forced outages of the unit� the new distribution of the probabilities of unserved load

� the process may be repeated until all units have been scheduled� a residual distribution remains that gives the final distribution of

unserved demand

( ) ( ) ( ) ( )CxTqxqTxT nnn +−+=′ P1PP



� Example� use the data of the two generating unit system and the

associated expected load pattern

� first analyze the expected value of production cost to serve the given load forecast by ignoring the forced outages

02040

00

20

100100100

x-Load (MW)

ExactDuration (h)

T Pn(k), Hours that Load Equals or

Exceeds x

6080

100

06020

808020

100+ 0 0

0800

16080080

1.01.02.0

UnitPower Output(MW)

Fuel Input (MBtu/h)

2

1

Fuel Cost ($/MBtu)

Fuel Cost Rate($/h)

Incremental Fuel Cost ($/MWh)

40 400 2.0

160800160800

8.0

16.0

0.05

0.10

Unit Force Outage Rate

(per unit)



� Example� graphical form of the convolution

x-Unsupplied Load (MW)

T Pn(x) (h)

100

50

100500-100 -50

100

50

100500-100 -50

100

50

100500-100 -50x-Unsupplied Load (MW)

x-Unsupplied Load (MW) T Pn(x) (h)

T Pn(x) (h)

original load duration (distribution) curve

load duration curveafter first unit dispatch

load duration curveafter second unit dispatch



� Example� tabular form of the convolution

� shows the load probability for unserved loads of 0 to 100+ MW

02040

100100100

x-Load (MW)

T Pn(x)(h)

6080

100

808020

100+ 0

80200

T P′n(x) = T Pn(x + 80)

(h)

0000

000

T P″n(x) = T P′n(x + 40)

(h)

0000

Energy (for x ≥ 0) 7600 MWh 400 MWh 0 MWh



� Example� repeat, now considering the effects of the forced outage rates

� for the first dispatched unit, q1 = 0.05� the recursive equation to obtain P′n(x)

� the original and resultant unserved load distributions

( ) ( ) ( )80P95.0P05.0P ++=′ xTxTxT nnn

02040

100100100

x-Load (MW)

T Pn(x)(h)

6080

100

808020

100+ 0

76 + 5 = 8119 + 5 = 240 + 5 = 5

T P′n(x) (h)

0 + 4 = 40 + 4 = 40 + 1 = 1

0Energy (for x ≥ 0) 7600 MWh 760 MWh

80200

T Pn(x + 80)(h)

0000


T Pn(x) (h)

load duration curveafter first unit dispatch

100

50

100500-100 -50



� Example� the production cost for unit 1

� may be loaded to 80 MW for 80 h and 40 MW for 20 h (max)

� but the unit is only available for 19 h at 40 MW and for 76 h at80 MW

� cost:� 40 MW, 19 h, E = 760 MWh, fuel = 9120 MBtu, cost = $ 9120

� 80 MW, 76 h, E = 6080 MWh, fuel = 60.8 Gbtu, cost = $ 60800



� Example� for the second dispatched unit, q2 = 0.10

� the recursive equation to obtain P′n(x)

� the first resultant and second resultant unserved load distributions

( ) ( ) ( )40P90.0P10.0P +′+′=′′′ xTxTxT nnn

02040

81245

x-Load (MW)

T P′n(x)(h)

6080

100

441

100+ 0

12.66.04.1

T P″n(x) (h)

1.30.40.10

Energy (for x ≥ 0) 760 MWh 238 MWh


load duration curveafter second unit dispatch

T Pn(x) (h)

100

50

100500-100 -50



� Example� the production cost for unit 2

� the unit is required a total of 81 h � at 0 MW for 81 - 24 = 57 h

� at 40 MW for 5 h

� at 20 MW for 24 - 5 = 19 h

� cost:� 0 MW, 51.3 h, E = 0 MWh, fuel = 4104 MBtu, cost = $ 8208

� 20 MW, 17.1 h, E = 342 MWh, fuel = 4104 MBtu, cost = $ 8208

� 40 MW, 4.5 h, E = 180 MWh, fuel = 1800 MBtu, cost = $ 3600

� total: 72.9 h, E = 522 MWh, fuel = 10008 MBtu, cost = $20016



� Example� final unserved energy distribution

� there remains an expected requirement to supply 100 MW� probability for needing this capacity is 0.1% (0.1 h out of 100 h)

� the cost of supplying the remaining 238 MWh of unsupplied load energy� based on an estimate of the cost of emergency energy supply

� assume that emergency energy cost is $ 24 / MWh

� the resulting cost is then:238 MWh × $ 24 / MWh = $ 5712

� difference between the two examples� ignoring force outages results in a 1.95 % underestimation of fuel

consumption and a complete neglect of emergency supply costs, and an 8.1 % underestimate of the total production costs


Chapter 9

Control of Generation


Generation Control

� Optimal dispatch and scheduling of generation establishes the best operation point with respect to economics

� The operating point must be implemented via generation control� local generator control for each individual generator

� energy control center for the control of a large utility and theflow of power across interconnections to other utilities

� regional control over several utilities and the Independent Power Produces, IPP’s� ISO - Independent System Operator

� RTO - Regional Transmission System Operator


� Many generators supply power to the transmission system� consumer loads are constantly

changing the power level

� some control means needed to allocate the load changes to the generators� a governor on each unit

maintains mechanical speed (electrical frequency)

� supplementary control acts to allocate generation� influences the power output

� control signal usually originates at a remote control center

Overview of Control Problem

generationcontrolsystem

powersystem

turbine-generator unit

turbine-generator unit

measurements of generator electric

power output

measurement of system frequency

measurements of tie-line flows to

neighboring systems

controlsignals


� Definition of important terms� ω = rotational speed

� α = rotational acceleration

� δ = phase angle of a rotating machine

� Tnet = net accelerating torque in a machine

� Tmech= mechanical torque exerted on the machine by the turbine

� Telec= electrical torque exerted on the machine by the generator

� Pnet = net accelerating power

� Pmech= mechanical power input

� Pelec= electrical power output

� I = moment of inertia for the machine

� M = angular momentum of the machine• all quantities, except for the phase angle, are expressed in per unit on the

machine base and/or the standard system frequency base

• steady-state and nominal values have a “0” subscript added

Generator Model

generatorturbine electricalenergy

mechanicalenergy

TmechTelec


Generator Model

� Basic relationships� acceleration principle:

� momentum principle:

� power equation:

� Phase angle deviation� general shaft equation:

� deviation from nominal:

� relationship to torque:

� deviation of power:

� the resulting swing eq.:

αITnet =IM ω=

( ) ααωω MITP netnet ===

( ) t

ttt

dtd αωδ

αωδδαωω=∆=∆

++=+= 221

000

( )( )δωω

δα

∆==−=

∆==−=

2

2

2

2

dd

dd

tnetelecmechnet

telecmechnet

ITPPP

IITTT

( ) ( )ωδ ∆=∆=− ttelecmech MMPP dd

dd

2

2


� Laplace transform of the dynamic power equation

� Block diagram model

( )( )ω

ωωωω

∆=∆−∆∆=∆−∆

∆−∆=∆−∆==∆−∆=∆−∆+−=

=

sMPP

MPP

PPTTTP

TTTTTTT

TT

elecmech

telecmech

elecmechelecmechnetnet

elecmechelecmechelecmechnet

elecmech

dd

00

00

Generator Model

∆Pmech

∆Pelec

∆ωsM

1


Load Model� Electrical loads consist of a variety of devices

� purely resistive devices

� power electronics� motor loads

� motor loads dominate the mix of loads

� Motors exhibit a variable power-frequency characteristic� model of the effect of a frequency change on net load drawn

� D is expressed as a percentage change in load per percentage change in frequency on the motor’s power base

� the value of D must be converted to the system power base for system studies

ω∆⋅=∆ DP freqL )(


Load Model

� Block diagram modeling� basic frequency dependent load

� rotating mass and load as seen by prime mover output

� the net change in the electrical power load, Pelec, is

� where ∆PL is the non-frequency-sensitive load change

D

∆Pmech

∆Pload

∆ωsM

1

∆Pmech

∆Pload

∆ωDsM +

1

D∆ω ∆PL(freq)

ω∆+∆=∆ DPP Lelec


Load Model

� Example� consider an isolated power system on a 1000 MVA base

� 600 MVA generator� M = 7.6 pu MW/pu frequency/sec on the machine base

� 400 MVA load� the load changes by 2% for a 1% change in frequency

� suppose that the load increases by 10 MVA� what is the transient response of the system frequency

� M = 7.6 (600/1000) = 4.56 pu and D = 2.0 (400/1000) = 0.80 pu

( ) ( )

( ) 0125.00125.08.0

01.0

8.0

01.0

8.056.4

101.001.0

175.056.4

8.0

−=−=∆

+=∆→=∆

−−

tt

L

eet

sss

ssP

ω

ω


Load Model

� Example� the final value of ∆ω is -0.0125 pu

� a drop of 0.75 Hz on a 60-Hz system


Prime-mover Model

� The prime mover drives the generating unit� steam turbine

� hydroturbine

� Modeling must account for control system characteristics� e.g., boiler and steam supply

� Model of the non-reheat turbine� relates the steam valve position

to the output power� “charging time”

time constant, TCH� per unit change in valve

position from nominal, ∆Pvalve

prime moversteam

steam valve

rotatingshaft

∆PmechCHTs+1

1∆Pvalve

∆Pvalve

CHTs+1

1∆Pvalve

∆Pload

∆ωDsM +

1∆Pmech

Prime-mover model

Prime-mover-generator-load model


Governor Model

� The governor compensates for changes in the shaft speed� changes in load will eventually lead to a change in shaft speed

� change in shaft speed is also seen as a change in system frequency

� simplest type of control is the isochronous governor

steam

steam valverotating

shaft

tachometer

∫ dt∆Pvalve ωrefKG –1

∆Pvalve

+ = open valve

– = close valve ∆ω

ω

prime mover


� The isochronous governor � to force frequency errors to zero requires the use of an

integration of the speed error

� the isochronous governor can not be used when two or more generators are electrically connected to the same system� fighting between generator governors for system frequency� problems with load distribution between generators

� A load reference control provides settings for both the frequency and the desired output power� a new input, the load reference signal, controls the desired

power output

� feedback loop contains a gain R that determines a speed-droop characteristic

Governor Model


Governor Model

� Load reference control� the speed-droop function handles the load sharing between

generators� there will always be a unique frequency at which the system

loading will be shared among the generators

� the gain R is equivalent to the per unit change in frequency for a 1.0 p.u. change in power: pu

PR

∆∆= ω

steam

steam valverotating

shaft

tachometer

∫ dt∆Pvalve

ωrefKG

∆Pvalve

+ = open valve

– = close valve ∆ω

ω

prime mover

RPref


Governor Model

� Simplification of the block diagram model

s

Kg

ω

+

_

∆Pvalve

PrefR+

_Σωref

RKT

gg

1=

_Σ

_Σ

∆ω

( ) 11

1−+ RKs g

ω+

∆Pvalve

Pref

+

_Σωref

_Σ

∆ωR

1

gTs+1

1

ω+

∆Pvalve

Pref

+

_Σωref

_Σ

∆ωR

1

where


� Speed-droop characteristic

� Allocation of unit outputs with governor droop

Governor Characteristics

P2

frequency

f0

p.u. output power1.00.5

frequency

f0

Unit #2 output power

P1′P1 P2′

f ′

Unit #1 output power


� Speed-changer settings

Governor Characteristics

load reference for nominal speed at no-load

frequency

f0

p.u. output power1.00.5

load reference for nominal speed at 50% loading

load reference for nominal speed at 50% loading


Complete Generator Model

� Block diagram of governor, prime mover, rotating mass, and loads

� transfer function of generator

GTs+1

1 ∆Pvalve

∆ω

∆Pmech

∆PL

+

_

+

+

++

+−

=∆∆

DsMTsTsR

DsMsP

s

CHG

L 11

11

111

1

)(

)(ω

+

_

Σ

R

1

PrefCHTs+1

1

DsM +1

Σ

governor prime moverrotating mass

& load

loading input

load referenceset point


Complete Generator Model

� Steady state behaviors� final value of the transfer function

� using Laplace method

� for several generators connected within the system

[ ]DR

P

DR

DPss LL

st +∆−=

+

−

∆=∆=∆ −→∞= 10 111

1

)(lim ωω

DRR

P

GG

L

+++∆−=∆

L21

11ω


Chapter 9



Tie-line Model

� The power flow across a tie-line can be modeled using a linear load flow approach� steady-state or nominal flow quantity:

� deviation from the nominal tie-line flow

� where ∆δ must be in radians for ∆Ptie to be in per unit

� using the relationship for speed and ∆δ:

then

� where T = 377 /Xtie for a 60-Hz system

( )21flowtie

1 δδ −=tieX

P

( ) ( )[ ] ( )( )

( )21flowtie

21212211flowtieflowtie

1

11

δδ

δδδδδδδδ

∆−∆=∆

∆−∆+−=∆+−∆+=∆+

tie

tietie

XP

XXPP

ωωδ ∆⋅=∆s

0

( ) ( )21210

flowtie ωωωωω ∆−∆=∆−∆=∆s

T

XsP

tie


Tie-line Model

� Simplified control for two interconnected areas

11

1

GTs+

∆ω1

∆Pmech1 +

_

+

_

Σ

1

1

R

Pref-111

1

CHTs+ 11

1

DsM +Σ

load referenceset points

21

1

GTs+

∆ω2

∆Pmech2

∆PL2

+

_

+

_Σ

2

1

R

Pref-221

1

CHTs+ 22

1

DsM +Σ

governors prime movers

∆PL1

Σ

_

+

+

_s

T∆Ptie


Tie-line Model

� Consider two areas each with a generator� the two areas are connected with a single transmission line

� the line flow appears as a load in one area and an equal but negative load in the other area

� the flow is dictated by the relative phase angle across the line, which is determined by the relative speeds deviations

� let there be a load change ∆PL1 in area 1

� to analyze the steady-state frequency deviation, the tie-flow deviation and generator outputs must be examined


Tie-line Model

� after the transients have decayed, the frequency will be constant and equal to the same value in both areas

� applying substitutions yields

( ) ( )

22mech2tie2mech

11mech11tie1mech

2121 0

d

d

d

d

RPDPP

RPDPPP

tt

L

ωω

ωω

ωωωωω

∆−=∆⋅∆=∆+∆

∆−=∆⋅∆=∆−∆−∆

=∆=∆→∆=∆=∆

−−

−−

+∆=∆+

+∆=∆−∆−

22tie

111tie

1

1

RDP

RDPP L

ω

ω


Tie-line Model

� additional simplification yields

� Notes� the new tie flow is determined by the net change in load and

generation in each area, but not influenced by the tie stiffness

� the tie stiffness determines the phase angle across the tie

2121

221

tie

2121

1

11

1

11

DDRR

RDP

P

DDRR

P

L

L

+++

+∆−

=∆

+++

∆−=∆ω


Tie-line Model

� Example� consider two areas, each with a generator, motor loads, and a

single tie-line connecting the two areas.� area 1: R1 = 0.01 pu, D1 = 0.8 pu, Base MVA = 500

� area 2: R2 = 0.02 pu, D2 = 1.0 pu, Base MVA = 500

� a load change of 100 MW (0.2 pu) occurs in area 1

� find the new steady-state frequency and net tie flow change

( )( )

( ) MW6.33pu06719.0102.0

1001318.0

1

Hz92.5960001318.060

001318.00.18.0

02.01

01.01

2.011

22

2121

1

==

+−=

+∆=∆

=−+=

−=+++

−=+++

∆−=∆

DR

P

f

DDRR

P

tie

new

L

ω

ω


Tie-line Model

� Example� change in the prime movers

� change in the motor loads

� change in tie flow

� change in apparent loading)MW100(200.0

)MW6.33(0672.0

)MW659.0(001318.00.1001318.0

)MW527.0(001054.08.0001318.0

)MW94.32(0659.002.0001318.0

)MW88.65(1318.001.0001318.0

)(111

)(22

2)(2

1)(1

22

11

=∆+∆−∆=∆

=∆−∆=∆

−−=⋅−=∆=∆

−−=⋅−=∆=∆

==∆−=∆

==∆−=∆

tiemotorLmechL

motorLmechtie

motorL

motorL

mech

mech

PPPP

PPP

DP

DP

RP

RP

ωω

ω

ω


Generation Control

� Automatic Generation Control (AGC) is a control system that has three major objectives� hold the system frequency at or very close to a specified

nominal value (e.g., 60 Hz)

� maintain the correct value of interchange power between control areas� enforce contracts for shipping or receiving power along the tie-

lines to neighboring utilities

� maintain each generating unit’s operating point at the most economic value


Generation Control

� Supplementary control action� assume for the moment that a single generating unit supplies

load to an isolated power system� a load change will produce a frequency change

� magnitude depends on the droop characteristics of the governor and the frequency characteristics of the system load

� a supplementary control must act to restore the frequency to thenominal value (60 Hz)� accomplished through an integral control loop to the governor

� the control action forces the frequency error to zero by adjusting the speed reference set point


Generation Control

� Supplementary control action

GTs+1

1

∆ω

∆Pmech1 ++

_

Σ

1

1

R

PrefCHTs+1

1

DsM +1

Σgeneratorloadingset point

governor prime mover

∆PL_

s

K

rotating massesand frequency-dependent loads

∆ω

speed (frequency)droop control

frequencyrestoration control


Generation Control

� Tie-line control� two utilities will interconnect their systems for several reasons

� buy and sell power with neighboring systems whose operating costs make the transactions profitable

� improvement to overall reliability for events like the sudden loss of a generating unit

� provide a common frequency reference for frequency restoration

� define tie flows and tie flow changes� total actual net interchange:Pnet int

� + for power leaving the system; – for power entering the system

� scheduled or desired value of interchange: Pnet int sched

� change in tie flow: ∆ Pnet int = Pnet int– Pnet int sched


Generation Control

� Tie-line control� interconnections present a challenging control problem

� consider a two-area system� both areas have equal load and generator characteristics

(R1 = R2, D1 = D2)

� assume that area 1 sends 100 MW to area 2 under an interchange agreement between the system operators of the two areas

� let area 2 experience a sudden load increase of 30 MW, then bothareas see a 15 MW increase in generation (because R1 = R2) and the tie flow increases from 100 to 115 MW

• the 30 MW load increase is satisfied by a 15 MW increase in generation #2 plus a 15 MW increase in tie flow into area 2

� this is fine, except that area 1 contracted to sell only 100 MW• generation costs have increased without anyone to bill

� a control scheme is needed to hold the system to the contract


Generation Control

� Tie-line control� such a control system must use two pieces of information

� the system frequency

� the net power flow across the tie line

� a few possible network conditions include� if the frequency decreases and net interchange power leaving the

system increases, a load increase has occurred outside the area

� if the frequency decreases and net interchange power leaving thesystem decreases, a load increase has occurred inside the area

� define a control area� part of an interconnected system within which the load and

generation will be controlled� all tie-lines that cross a control area boundary must be metered


Generation Control

� Tie-line control� control scheme actions

+ +

– –

– +

+ –

∆ω ∆Pnet int

– 0

+ 0

0 +

0 –

∆PL1 ∆PL2

decrease Pgen in area 1

increase Pgen in area 1

increase Pgen in area 2

decrease Pgen in area 2

Resulting control action

1 2∆Pnet int

∆PL1 = Load change in area 1∆PL2 = Load change in area 2


Generation Control

� Tie-line control� for the first row of the control response table, it is required that

� the required change is known as the area control error (ACE)

� the equations for the ACE for each area

� B1 and B2 are the frequency bias factors, and are set accordingly

02

11

=∆

∆=∆

gen

Lgen

P

PP

ωω

∆−∆−=

∆−∆−=

22intnet2

11intnet1

BPACE

BPACE

22

2

11

1

1

1

DRB

DRB

+=

+=


Generation Control

� Tie-line control� combining the ACE functions with the tie flow equations

results in

011

111

1

111

11

1

2121

12

221

21

22

1

2

1

2121

11

121

21

22

1

1

=+++

∆−

+−

+++

+∆

=

∆=+++

∆−

+−

+++

+∆

=

DDRR

PD

RDDRR

DR

P

ACE

PDD

RR

PD

RDDRR

DR

P

ACE

L

L

LL

L


Generation Control

� Tie-line bias supplementary control for two areas

11

1

GTs+

∆ω1

∆Pmech1

+

_

+

_

Σ

1

1

R

11

1

CHTs+ 11

1

DsM +Σ

21

1

GTs+

∆ω2

∆Pmech2

∆PL2

+

_

+

_Σ

2

1

R

21

1

CHTs+ 22

1

DsM +Σ

governors prime movers

∆PL1

Σ

_

+

+

_s

T∆Ptie

s

K

s

K

1B

2B

Σ

Σ

_

_

_

+

∆Pnet int 1

–∆Pnet int 2

ACE2

ACE1


Chapter 9



Generator Allocation

� A typical control area contains many generators� the individual outputs must be set according to economics

� the solution of the economic dispatch must be coupled to the generation control system

� the input consist of the total generation required for the area� in order to satisfy the load demand and maintain contractual power

flows across the tie lines

� the output is the power distribution across the outputs of all the generators within the control area

� continuously varying system load demand� a particular total generation value will not exist for a very long

time



� Economic generator control� it is impossible to simply specify a total generation, calculate

the economic dispatch schedule, and give the control system the output schedule for each generator� unless such a calculation can be made very quickly

� for digital control system it is desirable to perform the economic dispatch calculation at intervals of 1 to 15 minutes

� independent of the calculation schedule� the allocation of generation must be made instantly whenever

the required area total generation changes

� the allocation control of generation must run continuously

� a rule must be provided to indicate the generation allocation for values of total generation other than that used in the economic dispatch



� The allocation of individual generators over a range of total generation values� accomplished using base points and participation factors

� for period k, the economic dispatch sets the base-point generation values for the total generation value measured at the start of the period

� the base-point generation for the ith unit, Pi baseis the most economic output for the particular total generation value

� the participation factor, pfi, sets the rate of change of the ithunit’s power output with respect to a change in total generation

� the base points and participation factors are used as follows

( ) ( ) ( )( ) ( ) ( )∑

∈

−=∆

∆⋅+=

genallbaseactualtotal

totalbasescheluded

ii

iii

kPtPtP

tPpfkPtP



� Base points and participation factors� participation factors are determined from a generator’s cost

function� assume that both the first and second

derivatives exist for the cost function

� the change in the system’s incremental cost as a function of the change in power output on the ith generator

� the change in system incremental cost equaling the unit’s incremental cost is true for all generating units

( ) iiii PPF ∆⋅′′≅∆=∆ 0systemλλ

NN

PF

PF

PF

∆=′′

∆∆=′′

∆∆=′′

∆ λλλL,, 2

21

1dF

i/dP

i=

Fi′

Pi0 Pi

λ0

∆λ

∆Pi

Relationship of ∆λ and ∆Pi



� Base points and participation factors� the total change in generation must equal the change in the

total system demand, and is the sum of all the individual unit changes

� the participation factor for each generating unit is then found as

∑∈ ′′

⋅∆=

∆++∆+∆=∆

genall

21

1

i i

ND

F

PPPP

λ

L

∑∈ ′′

′′=

∆∆=

genall

1

1

i i

i

D

ii

F

F

P

Ppf



� Example� Consider a three generator system

� the cost functions for the three generators

� an economic dispatch has been conducted for a total load demand of 850 MW� the system’s incremental cost is $ 9.148 / MWh

� the dispatch is: P1 = 393.2 MW, P2 = 334.6 MW, & P3 = 122.2 MW

� calculate the participation factors for the current dispatch, and calculate the dispatch for a new total load of 900 MW

( )( )( ) 2

3333

22222

21111

00482.097.778

00194.085.7310

001562.092.7561

PPPF

PPPF

PPPF

++=

++=

++=



� Example� participation factors

� new dispatch

( )( ) ( ) ( )( )

( )15.0

57.681

00964.0

38.057.681

00388.0

47.057.681

10.320

00964.000388.0003124.0

003124.0

13

3

12

2

111

11

1

==∆∆=

==∆∆=

==++

=∆∆=

−

−

−−−

−

D

D

D

P

Ppf

P

Ppf

P

Ppf

( )( )( )( )( )( ) 7.1295015.02.122

6.3535038.06.334

7.4165047.02.393

50850900

3

2

1base11

=+==+=

=+=∆⋅+==−=∆

P

P

PpfPP

P

D

D


Generator Control

� Automatic generator control implementation� the AGC schemes are usually centrally located at a control

center� system measurements, taken at the major substations, other

information and data are telemetered to the control center� unit megawatt power output for each committed generating unit

� megawatt power flow over each tie line to neighboring systems

� system frequency

� control actions are determined in a digital computer

� control signals are transmitted to the generation units at remote generation stations over the same communication channels� raise / lower pulse signals change a generating unit’s load

reference point up or down


Generator Control

� Automatic generator control implementation� the basic reset control loop for a generating unit

consists of an integrator with gain K� the integrator insures that the steady-state

control error goes to zero

� the scheduled power value is the control input� a function of the system frequency deviation,

net interchange error, and the unit’s deviation from its scheduled economic output


� Automatic generator control implementation� the basic generating unit’s power output control loop

� implementation via telemetry

GiTs+1

1 ∆Pmech

output+

_ΣPi sch

CHiTs+1

1

s

Kload ref set point governor and prime mover

Generator Control

∆Pmech

outputPi sch

telemetryremotestation

telemetrymasterstation

controllogic

raise/lower request raise/lower request

at control center

governorand

prime mover

controllogic

at generating plant


Generator Control

� The AGC calculation� the input to the AGC combines the inputs of the various tie-

flows errors with the frequency deviation� the resultang is the area control error

+

_Σ

fstandard

+_

ΣPtie-2+

_Σ

Ptie-1

_

Σ

Psch. net interchange

2B

fmeasured

Ptie-n

+

+

PACE


Generator Control

� The AGC calculation� the control must also drive the generating units to obey the

economic dispatch in addition to pushing the frequency and tie flow errors to zero� the sum of the unit output errors is added to the ACE to form a

composite error signal

� the generation allocation calculation is placed between the ACE and the governor control / unit control loop


Generator Control

� Automatic generator control implementation� a typical layout

Σ ε unit i

ε unit 1

Unit Control Logic

Pbase-1

s

K

B

Psch int

ACE

∆Pgen1fmeas+

Σ+

pf1

pf1

pf1

Pbase-2

+

Σ+

Pbase-3

+

Σ+

+

_Σ raise/lower

control logic

governorand

prime mover

_Σ raise/lower

control logic

governorand

prime mover

+

+

_Σ raise/lower

control logic

governorand

prime mover

Σ

ΣΣ

Σ

Σ

∆Pgen2

∆Pgen3

_

+

fstd

Σ Σ Pbase

Generation Allocation Logic

+

++

__

_

_

Ptie-j

Ptie-i

Ptie-k

++

+ _

+

+ +


Generator Control

� Automatic generator control implementation� good design requirements

� the ACE signal should be kept moderate in size� the ACE is influenced by random load variations� the standard deviation of the ACE should be small

� the ACE should not be allowed to drift� the integral of the ACE should span an appropriate,

but small time period� drift has the effect of creating system time errors or

inadvertent interchange errors

� the control action should be kept to a minimum� many errors are simple random load changes that

should not cause any control action� chasing these random variations only wears out the

unit’s speed-changing hardware


Chapter 10

Interchange of Power and Energy


Economy Interchange

� Having interconnections between various electric power utility systems is beneficial� increases reliability

� protects against unserved load due to loss of generation� reduces the cost of installing additional generation

� more flexible operation� system frequency is more stable due to the increase in rotational

inertia from the summation of all generating units� take advantage of sharing spinning reserve responsibilities

� lower operating cost� take advantage of lower cost generation

� utilities with excess energy can sell production to utilities with high load demands


Economy Interchange

� Improving operating economics� opportunities may arise whenever two power systems are

operating with different incremental costs

� if a sufficient difference in cost exists, then it pays to have both systems exchange power at an equitable price� consider utility A is generating at a lower incremental cost than

utility B

� utility B buys the next megawatt for its load from utility A instead of generating it� this saves money in supplying the incremental load

� utility A benefits economically from the sell of power to utility B as long as utility B is willing to pay a price that is greater than the cost of production of that block of power by utility A


Inter-utility Economy Evaluation

� Example� consider two utilities connected with tie-lines

Utility Unit Generating Unit Generator Limits LoadArea Num. Cost Function Pmin(MW) Pmax(MW) Demand

1 1122 + 15.84P1 + 0.003124P12 150 600

A 2 620 + 15.70P2 + 0.003880P22 100 400 700

3 156 + 15.94P3 + 0.009640P32 50 200

subtotals: 300 1200

4 950 + 13.41P4 + 0.002641P42 140 590

B 5 560.5 + 14.17P5 + 0.003496P52 110 440 1100

6 560.5 + 14.17P6 + 0.003496P62 110 440

subtotals: 360 1470

1 2

3 4

5

6



� Example� first, examine the behaviors when each utility performs an

economic dispatch on its own area without regards to the neighboring area

� results of individual economic dispatches for each area� area A: P1 = 322.7 MW F1 = 6,559.07 λ = 17.856

P2 = 277.9 MW F2 = 5,282.50P3 = 99.4 MW F3 = 1,835.68

FA = 13,677.25

� area B: P4 = 524.7 MW F4 = 8,715.00 λ = 16.185P5 = 287.7 MW F5 = 4,927.13P6 = 287.7 MW F6 = 4,927.13

FB = 18,569.26

� total operating cost = 32,246.50



� Example� now, examine the behaviors when both utility work together

and perform a single economic dispatch for the whole system

� results of the economic dispatch� area A: P1 = 184.0 MW

P2 = 166.2 MWP3 = 54.4 MW

operating cost = 8,530.93

� area B: P4 = 590.0 MWP5 = 402.7 MWP6 = 402.7 MW

operating cost = 23,453.89

� total operating cost = 31,984.82 , λ = 16.990interchange power = 295.4 MW from area B to area A



� Example� area A costs: without interchange: 13,677.21

with interchange: 8,530.93savings: 5,146.28

� area B costs: without interchange: 18,569.23with interchange: 23,453.89increased cost: 4,884.66

� combined net savings: 261.62

� conclusions:� area A can argue that area B had a net operating cost increase of

4884.66 and therefore area A ought to pay area B this cost

� area B can argue that area A had a net decrease in operating cost of 5146.28 and therefore area A ought to pay area B this savings



� The problem with the previous approach is that there is no agreement conserning a mutually acceptable fair price� a common practice is to share the savings equally between the

two parties and set the price for such

� Two power system could operate for less money when interconnected than when operated separately� the dispatch is obtained for the interconnected system by

assuming that all necessary information is available to everyone

� such openness of communication is unrealistic� communication limits and business confidentiality prevents

everyone having all the information



� The simplest way to coordinate the operations between two power systems is to apply comparisons of the incremental costs� process steps

� assume there is no interchange, and each system runs an economic dispatch for its own system

� communicate the incremental cost values and determine who has the lower incremental cost� the one with the lower cost runs a series of dispatches with

increased total demand

� the one with the higher cost runs a series of dispatches with decreased total demand

� compare for equal values of interchange powers the incremental costs for each system, select the interchange with equal costs



� Example� rework the first example by carrying out the steps for

incremental cost comparison between the two utilities� conduct the series of dispatches in steps of 50 MW

steps utility area A: utility area B: Tie flow1 700 MW 17.856 1100 MW 16.185 0 MW2 650 17.710 1150 16.291 503 600 17.563 1200 16.395 1004 550 17.416 1250 16.501 1505 500 17.270 1300 16.656 2006 450 17.123 1350 16.831 2507 400 16.976 1400 17.006 3008 350 16.816 1450 17.181 350



� Example� note that at step 6, area A’s incremental cost is just slightly

above area B’s incremental cost, and that this relationship changes at step 7� for minimum operating cost, the two systems ought to be

interchanging power between 250 MW and 300 MW

� the procedure can then be repeated with smaller steps between 250 and 300 MW as needed


Multiple-Utility Interchanges

� Regional interconnectivity� most power systems are interconnected with all their

immediate neighboring systems� one system may buy and sell interchange power simultaneously

with several neighbors� the price for interchange must be set while taking account of the

other interchanges

� prices are derived from the incremental costs of the various utility systems

� a group of utility power systems may form a power pool� the price might be set by the power and energy pricing policies and

be established by a centralized pool control center

� when there is no centralized center, the order in which transaction agreements are executed is very important in costing the interchange



� Regional interconnectivity� multiple neighboring utilities may engage in “wheeling”

� simply stated, wheeling is the transmission of power from one utility, through one or more intermediate utility systems, to a third utility system

� the intermediate systems’ AGCs keep the net interchange to specified values, regardless of the power being passed through

� the transmitted power changes the transmission losses incurred in the intermediate systems� the increased losses represent an unfair burden on the intermediate

systems, if the utilities are not part of the interchange agreement

� the increased losses are supplied by the intermediate systems’ generation

� utilities may impose a wheeling access charge for such power



� Example� consider a three utility region with interchange agreements

� data for utility areas A & B are the same as in previous examples� area A generation & costs: area B generation & costs:

400 MW @ $ 8452 /h 1100 MW @ $ 18569 /h500 MW @ $ 10165 /h 1400 MW @ $ 23532 /h700 MW @ $ 13677 /h800 MW @ $ 15478 /h

� data for utility area C are as follows: (w/ 550 MW load demand)� area C generation, incremental cost, and total cost

450 MW @ $ 18.125 / MWh & $ 8220 /h550 MW @ $ 18.400 / MWh & $ 10042 /h

� let there be two interchange agreements:� BA: area A buys 300 MW from area B

� AC: area A sells 100 MW to area C



� Example� examine the costs under a split-savings pricing policy

� consider the impact with agreement BA first, then AC

Area AGen.(MW)

Area ACost($/h)

Area BGen.(MW)

Area BCost($/h)

Base Case

Area CCost($/h)

Area CGen.(MW)

With Agreement BA

Savings / Costs

With Agreement AC

Savings / Costs

Split Savings Price

Split Savings Price

700 13677 1100 18569 550 10042

400 8452 1400 23532 550 10042

450 8220500 10165 1400 23532

5225 $/h 4963 $/h 0 $/h

5094 $/h from area A to area B

1822 $/h0 $/h1712 $/h

1767 $/h from area C to area A

Payment Summary pays a net 3327 $/h receives 5094 $/h pays 1767 $/h



� Example� examine the costs under a split-savings pricing policy

� consider the impact with agreement AC first, then BA

Area AGen.(MW)

Area ACost($/h)

Area BGen.(MW)

Area BCost($/h)

Base Case

Area CCost($/h)

Area CGen.(MW)

With Agreement AC

Savings / Costs

With Agreement BA

Savings / Costs

Split Savings Price

Split Savings Price

700 13677 1100 18569 550 10042

800 15478 1100 18569 450 8220

450 8220500 10165 1400 23532

1822 $/h1800 $/h 0 $/h

5138 $/h from area A to area B

5313 $/h 0 $/h4963 $/h

1811 $/h from area C to area A

Payment Summary pays a net 3327 $/h receives 5138 $/h pays 1811 $/h



� Example� comparison and results

� for utilities B and C, the payments are different depending on the order in which the agreements were conducted� it is in the seller’s best interest to sell when the buyer’s incremental

cost is high

� conversely, it is in the buyer’s best interest to purchase when the seller’s incremental cost is low

� utility A sees no difference based on the order of the agreements

� when several two-party agreements are made, the pricing must follow the proper sequence� selling utilities receive more than its incremental production costs

no matter which transaction is initiated first

� the rates paid are different and depend on the order of evaluation


Chapter 10

Interchange of Power and Energy


Power Pools

� When one system is interconnected with many neighbors, all the transaction setup processes with every neighbor become time consuming� individual transactions rarely results in the optimum

production cost

� one solution is for several utilities to from a power pool� the power pool is administered from a central location

� the pool avoids the problem of different prices depending on the order of agreements

� the pool incorporates a central dispatch office


Power Pools

� The energy-broker system� the interconnected power systems deal through a broker

� the broker sets up sales and purchases of energy

� the broker can observe all the buy and sell offers at one time

� by accepting quotations to sell and purchase, the central brokercreates an orderly marketplace where prices, supply, and demand are known simultaneously

� the “bulletin board” is the simplest form of a broker scheme� utility members post offers to buy or sell power and energy at

regular intervals

� members are free to access the bulletin board at all times� members finding attractive offers are free to contact those posting

the offers and make direct arrangements for the transactions


Power Pools

� The energy-broker system� more complex broker systems make arrangements to match

buyers to sellers directly� the broker may also set the transaction prices

� in one scheme, members send the broker hourly buy and sell offers for energy� each member provides information each hour

• the incremental cost and megawatt-hours it is willing to sell

• the decremental cost and megawatt-hours it is willing to buy

� the broker matches buy and sell offers according to certain rules• the lowest cost seller is matched with the highest cost buyer

• repeat until all offers are processed

• price is set to compensate the seller for the incremental generation cost and split the savings of the buyer equally with the seller


Power Pools

� The energy-broker system� price setting formula between energy buyer and seller:

� Fs′ = incremental cost of the selling utility ($/MWh)

� Fb′ = incremental cost of the buying utility ($/MWh)

� Fc = cost rate of the transaction ($/MWh)

� the transaction cost rate is the average of the seller’s incremental cost and the purchaser’s decremental cost

( ) ( )sb21

sb21

sc FFFFFF ′+′=′−′+′=


Power Pools

� Example� four power systems

has submitted the following buy/sell offers to the broker

� the maximum poolsavings that arepossible are:

Fsaving= ($ 1750 + $ 6750) – ($ 2500 + $ 3000)= $8500 – $ 5500 = $ 3000

� the broker sets up the following transactions:� A sells 100 MWh to D

� B sells 50 MWh to D and sells 50 MWh to C

Utilities Incremental Energy Seller’s TotalSelling Cost for Sale Cost IncreaseEnergy ($/MWh) (MWh) ($)

A 25 100 2500B 30 100 3000

C 35 50 1750D 45 150 6750

Utilities Decremental Energy Buyer’s TotalBuying Cost to Buy Cost DecreaseEnergy ($/MWh) (MWh) ($)


Power Pools

� Example� the broker’s transactions:

� the rates and total payments under a split-savings arrangement

Savings Total TransactionTransaction Computation Savings ($)

A sells 100 MWh to D (100 MWh)($45 – $25) / MWh 2000B sells 50 MWh to D ( 50 MWh)($45 – $30) / MWh 750B sells 50 MWh to C ( 50 MWh)($35 – $30) / MWh 250

Total: 3000

Price TotalTransaction ($/MWh) Cost ($)

A sells 100 MWh to D ($45 + $25) / 2 = 35.0 3500B sells 50 MWh to D ($45 + $30) / 2 = 37.5 1875B sells 50 MWh to C ($35 + $30) / 2 = 32.5 1625

Total: 3000


Power Pools

� Example� results

� utility A receives $ 3500 /h from utility D� utility A receives $ 1000 /h above its costs

� utility D saves $ 1375 /h

� utility B receives $ 3500 /h from utilities D and C� utility B receives $ 500 /h above its costs

� utility C saves $ 125 /h

� note that each participant benefits from the transactions

� the chief advantage of a broker system is its simplicity


Power Pools

� Clearing price policy� economist have encouraged a policy in which the broker

pricing scheme sets one single “clearing price” for energy in each time period� the market-determined price level should be based on all the

participants’ needs and willingness to buy and sell

� the market-determined price removes the absolute need to quoting cost-based prices

� utilities are free to quote offers at whatever price level, but would be obligated to deliver or purchase energy quoted at the market clearing price

� the transactions market is similar to the stock exchange

� in times of shortage, price levels could rise dramatically and uncontrollably


Power Pools

� Allocating pool savings� all saving-allocating methods implemented by central pool

dispatching are based on the premise that no member should have higher generation production expenses than in could achieve by dispatching its own generation to meet its own load� the split-savings method meets this criterion

� under a central economic dispatching pool, transactions can be modeled as if the power were sold to the pool by the selling utilities and then bought from the pool by the buying utilities� under this model, allowances are made for the cost and impact of

using some utilities’ transmission networks over other’s in carrying out the pool transactions


Power Pools

� Example� reconsider the previous example of four utilities

� energy interchanges are scheduled by a central dispatching scheme with 10% of the gross system’s savings are set aside to compensate those utilities that provide transmission facilities to the pool

� table of increase, decreased costs, and the net saves:

Utilities Incremental Energy Seller’s TotalSelling Cost for Sale Cost IncreaseEnergy ($/MWh) (MWh) ($)

A 25 100 2500B 30 100 3000

C 35 50 1750D 45 150 6750

Utilities Decremental Energy Buyer’s TotalBuying Cost to Buy Cost DecreaseEnergy ($/MWh) (MWh) ($)

Pool Savings: 3000Savings Withheld for Compensation: 300

Net Savings: 2700


Power Pools

� Example� weighted average incremental costs are computed for selling

and buying power� single clearing prices for buying & selling

� seller’s weighted average incremental cost

� buyer’s weighted average incremental cost

∑∑ ×

=i

ii

E

EFF

( ) ( )/MWh50.27$

MWh100MWh100

MWh100/MWh30$MWh100/MWh25$S =

+×+×=F

( ) ( )/MWh50.42$

MWh150MWh50

MWh150/MWh45$MWh50/MWh35$B =

+×+×=F


Power Pools

� Example� calculation of individual utility savings

� area A sells 100 MWh to the pool

� area B sells 100 MWh to the pool

� area C buys 50 MWh from the pool

� area D buys 150 MWh from the pool

50.787$9.02

/MWh25$/MWh50.42$MWh1001 =×

−×=S

50.562$9.02

/MWh30$/MWh50.42$MWh1002 =×

−×=S

75.168$9.02

/MWh50.27$/MWh35$MWh503 =×

−×=S

25.1181$9.02

/MWh50.27$/MWh45$MWh1504 =×

−×=S


Power Pools

� Example� total net savings for the four utilities

$ 787.50 + $ 562.50 + $ 168.75 + $ 1181.25 = $ 2700.00 net

� the total transfer for this hour are then:� C buys 50 MWh for ($42.5 × 50 – 168.75) = $ 1956.25

D buys 150 MWh for ($42.5 × 150 – 1181.25) = $ 5193.75subtotal $ 7150.00

� A sells 100 MWh for ($27.5 × 100 + 787.50) = $ 3537.50B sells 100 MWh for ($27.5 × 100 + 562.50) = $ 3312.50

subtotal $ 6850.00

� total transmission compensation charge: + $ 300.00total $ 7150.00


Power Pools

� Example� the $ 300 that is set aside for transmission compensation is

split up among the four utility systems according to agreements reflecting the transmission network contributions to the pool’s network


Power Pools

� Wheeling� wheeling occurs on the AC interconnections for those

networks containing more than two utilities (or parties) whenever transactions take place� parties represent non-utility organizations

� consider a system with six interconnected control areas� suppose areas A and C negotiate the sell of 100 MW by A to C

� ignoring losses, A increases net generation by 100 MW and C decreases net generation by 100 MW

50 MW

50 MW

50 MW

50 MW

25 MW

25 MW

25 M

W

A

B

C

D

E

F

A

B

C

D

E

F


Power Pools

� Wheeling� executing a power flow study before and after the agreement

shows the change in flows across the whole system� note that not all the transaction flows over the direct

interconnection between the two systems

� the other systems are wheeling some amount of the transaction

� the wheeled power is referred to as loop flows � the transmission system provides the parallel paths


Chapter 11

Power System Security


System Security

� The main concern in power system operation� minimize the operating cost

� An important operating factor� maintain system security

� System security� practices designed to keep the system

operating when components fail� example: a generator must be taken off-line

because of auxiliary equipment failure� maintaining proper amounts of spinning reserve,

the remaining generators can make up the deficit

� system is served without too low a frequency drop or the need to shed any load


Security Preparedness

� The timing of initiating events that cause components to fail are unpredictable� the system must be operated so that any credible event

will not put the system in a dangerous condition

� power equipment is designed to operate within known limits� protection devices automatically switch out

equipment that exceeds the specified limits

� if an event leaves the system operating with limits violated, the event may be followed by a series of further actions that switchother equipment out of service� if the process of cascading failures continues, large parts or the

entire system may completely collapse; i.e.; a system blackout


System Security

� Three major functions of system security� system monitoring

� contingency analysis

� security-constrained optimal power flow analysis

� System monitoring� provides the operators with up-to-date information

on the condition of the power systems� critical quantities are measured

� voltages, currents, power flows, and the state of circuit breakers and switches

� frequency, generator outputs, and transformer tap positions

� the measurements are sent to the control central � via the telemetry system


System Security

� System monitoring� computers collect the telemetric data, processes and stores

them, and displays information for the operators� the computes check incoming information against

pre-selected limits and annunciate alarms in the event of overloads or out-of-limit voltages

� state estimation combines the telemetric system data with the system’s network model to produce the “best estimate” of the current power system condition or state� measurements, metering devices, and the communication

system contain sources of noise and random errors

� combining monitoring functions with supervisory control functions forms the Supervisory Control And Data Acquisition (SCADA) system


System Security

� Contingency analysis� allows the system to be operated defensively

� many problems in power systems can cause serious trouble within a rapid time period and the human operator can not respond fast enough� cascading failures

� models possible system troubles before they arise� using a model of the power system, a computer algorithm

predicts future operating states and gives alarms to any potential overloads or out-of-voltage limits


System Security

� Security constrained optimal power flow� analysis provides a solution to the optimal dispatch of

generation with a large number of system constraints� other adjustments are made to the dispatch schedule so that a

security analysis results in no contingency violations

� the power system may operate in one of four states� optimal dispatch - state of the system prior to any contingency

� optimal with respect to economic operation, but may not be secure

� post contingency - state after contingency has occurred� this condition has a security violation: line overload, voltage limit

� secure dispatch - state at no contingency with dispatch schedule corrections to account for security violations

� secure post contingency - state after a contingency with no resulting security violations


Security States

� Illustration of states� optimal dispatch

� ignoring losses

� maximum line loading of 400 MW

� post contingency� consider that a transmission line has opened because of a failure

500 MW700 MW

1200 MW

unit #1unit #2

250 MW

250 MW

500 MW700 MW

1200 MW

unit #1unit #2

0 MW

500 MW(overload)


Security States

� Illustration of states� the overload line may cause a cascaded outage

� secure dispatch� unit #1 generation set to maximum secure line loading of 400 MW

� post contingency� consider the same contingency, resulting in no violations

400 MW800 MW

1200 MW

unit #1unit #2

200 MW

200 MW

400 MW800 MW

1200 MW

unit #1unit #2

0 MW

400 MW


Security States

� Remarks� post-contingency line overloading is avoided by adjusting the

generation on unit #1 and unit #2 before an event� essence of security corrections

� programming tools that make control adjustments to the pre-contingency operation to prevent violations in the post-contingency condition are called “security-constrained optimal power flows” or SCOPF


Factors Affecting Security

� As a consequence of several major widespread blackouts in interconnected system, operation priorities have evolved� operate the system so that power is delivered reliably

� within the constraints placed by reliability considerations, operate the system most economically

� It is highly uneconomical to build a system with sufficient redundancy to eliminate the possibility of dropping load� systems are designed and operated to has sufficient

redundancy to withstand all major failure events� no guarantee for 100% reliability


Factors Affecting Security

� Within the design and economic limitations, the system operators try to maximize the reliability of the system at any given time� usually a power system is never operated with all the

equipment in service at one time� occurrence of failures

� maintenance

� the operators are concern with possible events that cause trouble on a power systems� focus on two major types of events: transmission line outages

and generation unit failures

� line outages and unit failures cause changes in power flows and bus voltages, which impact the remaining system equipment


Contingency Analysis

� Detecting network problems� consider the 6-bus network

� a typical constraint of the system is the maximum loading that atransmission line can carry� expressed in MVA

� the flow on a line may increase due to a contingency in the system� line outage

� generating unit failure

� other operating factors that are impacted include:� the bus voltages

� generation output to make-up the load capacity• the slack generator may pickup all of the lost generation

• or the generators may share in picking up the lost generation



� Example� base case

2.9 MW12.3 MVAR

2.9 MW5.7 MVAR

70 MW70 MVAR

70 MW70 MVAR

43.8 MW60.7 MVAR

42.8 MW57.9 MVAR

25.7 MW16.0 MVAR

1.6 MW9.7 MVAR

19.1 MW23.2 MVAR

18.0 MW26.1 MVAR

60.0 MW89.6 MVAR

15.0 MW18.0 MVAR

27.8 MW12.8 MVAR

15.5 MW15.4 MVAR

26.2 MW12.4 MVAR

1.6 MW3.9 MVAR

50.0 MW74.4 MVAR

33.1 MW46.1 MVAR

28.7 MW15.4 MVAR

43.6 MW20.1 MVAR

4.0 MW2.8 MVAR

34.5 MW13.5 MVAR

4.1 MW4.9 MVAR

42.5 MW19.9 MVAR

70 MW70 MVAR

107.9 MW16.0 MVAR

35.6 MW11.3 MVAR

31.6 MW45.1 MVAR

Bus 2241.5 kV / –3.7º

Bus 3246.1 kV / –4.3º

Bus 6231.0 kV / –5.9º

Bus 5226.7 kV / –5.3º

Bus 1241.5 kV / 0º

Bus 4227.6 kV

/ –4.2º



� Example� show the impact on the system if the line 3-5 is taken out of

service due to a permanent transmission line fault

� notes: � the flow on line 3-6 has increased to 54.9 MW

� most of the other transmission lines experience a change in reactive power flow

� the bus voltages at the load buses are impacted

� the voltage magnitude at bus 5 is about 5% below normal



� Example� line outage

5.1 MW10.7 MVAR

5.1 MW4.1 MVAR

70 MW70 MVAR

70 MW70 MVAR

54.9 MW64.6 MVAR

53.6 MW60.3 MVAR

22.3 MW19.9 MVAR

5.7 MW15.3 MVAR

0 MW0 MVAR

0 MW0 MVAR

60.0 MW68.7 MVAR

19.9 MW25.6 MVAR

26.0 MW11.8 MVAR

20.9 MW24.7 MVAR

22.9 MW16.2 MVAR

5.9 MW10.2 MVAR

50.0 MW91.2 MVAR

37.3 MW49.1 MVAR

26.8 MW14.6 MVAR

43.2 MW22.7 MVAR

7.4 MW7.2 MVAR

37.0 MW22.0 MVAR

7.6 MW0.0 MVAR

42.1 MW22.3 MVAR

70 MW70 MVAR

108.5 MW29.8 MVAR

38.5 MW21.7 MVAR

35.5 MW47.6 MVAR

Bus 2241.5 kV / –3.4º

Bus 3246.1 kV / –3.0º

Bus 6229.8 kV / –5.2º

Bus 5219.3 kV / –5.5º

Bus 1241.5 kV / 0º

Bus 4226.4 kV

/ –4.1º



� Example� show the impact on the system when generating unit at bus 3

is taken off-line due to a steam pipe rupture� the loss of generation is pickup completely by the slack bus

generator at node 1



� Example� unit failure

21.2 MW28.6 MVAR

20.5 MW31.4 MVAR

70 MW70 MVAR

70 MW70 MVAR

21.1 MW25.2 MVAR

20.8 MW25.8 MVAR

40.3 MW41.8 MVAR

8.9 MW2.6 MVAR

0.6 MW6.1 MVAR

0.7 MW10.5 MVAR

0 MW0 MVAR

17.1 MW29.7 MVAR

52.4 MW28.1 MVAR

18.3 MW29.4 MVAR

42.9 MW44.2 MVAR

8.8 MW2.4 MVAR

50.0 MW74.4 MVAR

20.0 MW57.6 MVAR

55.7 MW25.9 MVAR

64.1 MW19.4 MVAR

10.0 MW8.3 MVAR

52.5 MW18.9 MVAR

10.3 MW1.5 MVAR

62.0 MW15.3 MVAR

70 MW70 MVAR

175.0 MW16.6 MVAR

55.2 MW23.1 MVAR

18.2 MW56.2 MVAR

Bus 2241.5 kV / –7.0º

Bus 3221.9 kV / –9.1º

Bus 6214.7 kV / –10.1º

Bus 5216.8 kV / –8.4º

Bus 1241.5 kV / 0º

Bus 4226.5 kV

/ –6.5º


Chapter 11

Power System Security


Security Analysis

� To be useful to the system operators in any way, a security analysis study must be executed very quickly� on the order of 5 to 10 minutes

� Three approaches� fast algorithms using linear approximating models of the

power system that can cover all possible cases

� traditional algorithms for a selection of only the most important cases

� multiple processors / vector processors modifications of traditional algorithms to gain speed to cover a larger selectionof cases


Sensitivity Factors

� Linear sensitivity factors� useful for reaching an approximate analysis of the effect of

each outage� limitations and attributes to the linear (DC) power flow method

� only branch active power flows (MW) are calculated with approximately 5% accuracy

� show the approximate change in line flows for changes in generation and network configuration

� two types of sensitivity factors� generation shift factors:

� change on line l w.r.t. gen. i

� line outage distribution factors:� change on line l w.r.t. line k

i

lil P

fa

∆∆=,

0,k

lkl f

fd

∆=


Sensitivity Factors

� Derivation of sensitivity factors� begin with the linear load flow model

θθθθ = [X] P

� the incremental changes of the bus voltage angles for perturbations of power injections

∆θ∆θ∆θ∆θ = [X] ∆∆∆∆P

� first consider the generation shift sensitivity for the generator on bus i� set the perturbation on bus i to +1 and the perturbation on all

other buses to zero� an equal but opposite perturbation (–1) must occur on the

reference bus (that is, the reference bus absorbs any changes and all other generation buses remain fixed)


Sensitivity Factors

� Derivation of sensitivity factors� the change in bus phase angles are found using matrix

calculations

� this is equivalent to a 1 pu power increase at bus i with a compensating 1 pu power decrease at the reference bus

� the ∆θ values are equal to the derivative of the bus angle with respect to a change in power injection at bus i

� the sensitivity factor for the change in power of line l with respect to a change in generation at bus i is

� line l is connected between buses n and m

−+

=∆row refat

row at

1

1][ iXθ

( ) ( )minili

m

i

n

lmn

lii

lil XX

xPPxxPP

fa −=

−=

−== 1

d

d

d

d11

d

d

d

d,

θθθθ


Sensitivity Factors

� Derivation of sensitivity factors� a line outage is modeled by adding two power injections, one

at each end of the line to be dropped� the line is not actually dropped, but the effective power change

due to the loss of line flow is added as power injections� suppose line k from bus n to bus m takes an outage

� if and , then line k looks like an outagenmn PP~=∆ nmm PP

~−=∆

Lines to remainder of network




Bus n Bus m Bus n Bus m Bus n Bus m

Line k

Pnm

Line k Line k

Pnm = 0 ∆Pn ∆Pm

Flows in lines before line outage Resulting flows in lines after line outage Equivalent flow in lines after line outage

nmP~


Sensitivity Factors

� Derivation of sensitivity factors� beginning with the incremental change in power flows

∆θ∆θ∆θ∆θ = [X] ∆∆∆∆P

� then a change in injected powers at nodes n and m produces

� and define the following� : angles and power flow across line k before the outage

� : the incremental changes resulting from the outage

� : angles and power flow after the outage

mmmnmnm

mnmnnnn

m

n

PXPX

PXPX

P

PP

∆+∆=∆∆+∆=∆

→

∆

∆=∆

θθ

M

M

nmmn P∆∆∆ ,, θθnmmn P,,θθ

nmmn P~

,~

,~ θθ


Sensitivity Factors

� Derivation of sensitivity factors� the outage model

� the injection power needed to model the outage

( )

( ) ( )

( ) ( ) ( ) nnmmmnnk

nmmnk

mnk

nm

mmnmmmnnmnnn

mmmnnn

mnk

mnnm

PXXXx

Pxx

P

PXXPXX

xPPP

∆−++=∆−∆+−=

∆−=∆∆−=∆∆+=∆+=

−=∆−=∆=

2111~

&

~&

~

~~1~

θθθθ

θθθθθθθθ

θθ

( )nm

nmmmnnk

n PXXX

x

P

−+−=∆

21

1

1


Sensitivity Factors

� Derivation of sensitivity factors� define a new sensitivity factor consisting of the change in

phase angle anywhere in the system to the original power Pnm

flow before the outage

( ) ( )

( )( )

( )( )nmmmnnk

iminknmi

nmmmnnk

nmiminki

nm

nmnnmmk

imnm

nmmmnnk

ini

mimnini

nm

inmi

XXXx

XXx

XXXx

PXXx

PXXX

x

XPXXX

x

X

PXPX

P

22

21

1

1

21

1

1

,

,

−+−−=→

−+−−=∆

−+−−

−+−=∆

∆+∆=∆

∆=

δθ

θ

θ

θδ


Sensitivity Factors

� Derivation of sensitivity factors� if either bus n or bus m is the reference bus, only one injection

is made� for m = reference:

� for n = reference:

� if bus i is the reference bus, then δi,nm = 0 since the reference bus angle is constant

� the line outage distribution factor is expressed as

nnk

inknmi Xx

Xx

−=,δ

mmk

imknmi Xx

Xx

−−=,δ

( )( )nmjnmi

lnm

j

nm

i

lk

jil

k

lkl xPPxf

x

f

fd ,,00,

111

δδθθ

θθ−=

∆−∆=

∆−∆=∆=


Sensitivity Factors

� Derivation of sensitivity factors� if i, j, m, nor n is a reference bus

� Linear sensitivity factors� the generator shift and line outage distribution factors are

linear, allowing the use of superposition to extend the factors� consider the impact on the generator shift factor of line l and bus

i after line k has been taken out of service

� this is a compensated generation shift sensitivity factor

( ) ( )( )

( )( )nmmmnnk

jmimjninl

k

nmmmnnk

jmjnkimink

lkl XXXx

XXXXxx

XXXx

XXxXXx

xd

22

1, −+−

+−−=

−+−−−−

=

( ) ( ) iikkliliikkliill

iikkkkliill

PadaPadPaf

PaffdPaf

∆+=∆+∆=∆∆=∆∆+∆=∆

,,,,,,

,,,


Sensitivity Factors

� Shift factors and generator participation factors� superposition allows one to analyze the effects of

simultaneous changes on several generating buses� the loss of generation on bus i may be compensated by governor

action on the remaining units in the interconnected system

� the remaining units pick up in proportion to their maximum MW rating:where Pk

max = max. MW rating for unit k

� testing for the flow for overload on line l, with all remaining units taking a proportion of the lost generating unit

∑≠

=

ikkk

jij P

P

,

max

max

,γ

∑≠

∆−∆+=ij

iijjliilll PaPaff ,,,0ˆ γ


Sensitivity Factors

� Example� using the [X] matrix for the 6-bus network and line data, find

the generator shift factors and the line distribution factors

[ ]

=

16328.008927.005920.012895.008129.00

08927.012215.005422.009077.006435.00

05920.005422.010088.005897.006298.00

12895.009077.005897.016590.008051.00

08129.006435.006298.008051.009412.00

000000

X

l1, line 1-2

0.20

0.30

0.25

0.10

0.30

0.20

0.30

0.40

0.10

0.26

0.20

l2, line 1-4

l3, line 1-5

l4, line 2-3

l5, line 2-4

l6, line 2-5

l7, line 2-6

l8, line 3-5

l9, line 3-6

l10, line 4-5

l11, line 5-6

Line X (pu) Line X (pu) Line X (pu)


Sensitivity Factors

� Example� generation shift factors

� the factor for a unit 3 outage on line 9 is:

� the complete table of factors:

– 0.29– 0.40

– 0.340.22

– 0.03

– 0.30

– 0.13– 0.08

0.370.29

– 0.24

l = 1, line 1-200000

0

00000

– 0.47

– 0.210.050.310.10

– 0.31

– 0.060.00

– 0.010.060.06

l = 2, line 1-4l = 3, line 1-5l = 4, line 2-3l = 5, line 2-4l = 6, line 2-5l = 7, line 2-6l = 8, line 3-5l = 9, line 3-6

l = 10, line 4-5l = 11, line 5-6

i = 1bus 1

i = 2bus 2

i = 3bus 3

( ) ( ) 37.012895.016590.010.0

11

d

d3,63,3

93

93,9 =−=−== XX

xP

fa


Sensitivity Factors

� Example� line outage distribution factors

� the factor for a line 4 outage on line 9 is:

0.460.54

0.18– 0.17

0.33

-------

– 0.180.29

– 0.030.200.21

– 0.11– 0.03

0.15

0.160.22

-------

0.120.13

– 0.62– 0.38

0.51

– 0.500.61

– 0.110.12

0.23-------

– 0.13– 0.39– 0.02

0.140.15

– 0.21– 0.06

0.270.230.30

-------

– 0.230.24

– 0.030.270.27

– 0.12– 0.04

0.160.470.17

-------

0.360.140.64

– 0.17

0.24

– 0.14– 0.04

0.18– 0.40

0.19

-------

– 0.400.150.60

– 0.200.27

0.010.00

– 0.02– 0.53– 0.02

-------

0.42– 0.02

0.470.58

– 0.03

0.01– 0.33

0.320.17

– 0.67

-------– 0.18

– 0.020.190.200.31

0.130.04

– 0.170.13

– 0.19

-------– 0.15

0.56– 0.42

0.44– 0.26

l = 1, line 1-20.590.41

– 0.10– 0.59– 0.19

-------

0.110.010.01

– 0.12– 0.12

0.64

0.36– 0.03

0.76– 0.06

-------

0.03– 0.24

0.00– 0.04– 0.04

l = 2, line 1-4l = 3, line 1-5l = 4, line 2-3l = 5, line 2-4l = 6, line 2-5l = 7, line 2-6l = 8, line 3-5l = 9, line 3-6

l = 10, line 4-5l = 11, line 5-6

k = 1line 1-2

k = 2line 1-4

k = 3line 1-5

k = 4line 2-3

k = 5line 2-4

k = 6line 2-5

k = 7line 2-6

k = 8line 3-5

k = 9line 3-6

k = 10line 4-5

k = 11line 5-6

( )( )

( )( ) 62.0

08051.0216590.009412.025.0

12895.016590.008129.008051.010.025.0

2 2333224

633362329

4

4,9 −=⋅−+−

+−−=

−+−

+−−=

XXXx

XXXXxx

d


Sensitivity Factors

� Example� consider the generator outage on bus 3 with all the pickup of

lost generation coming from the generator on bus 1� calculate the post outage flow on line 2, from bus 1 to bus 4

� base-case flow on line 2, 1-4 = 43.6 MW� base-case generation at bus 3 = 60.0 MW� generation shift factor, a2,3 = – 0.29� new flow: 43.6 MW + (– 0.29)(– 60.0 MW) = 61.0 MW

� consider the line 8 (3-5) outage� calculate the post outage flow on line 9 (3-6)

� base-case flow on line 8 = 19.1 MW� base-case flow on line 9 = 43.8 MW� line outage distribution factor, d9,8 = 0.60� new flow: 43.8 MW + (0.60)(19.1 MW) = 55.26 MW


Contingency Selection

� AC power flow methods� the approximate methods cannot solve for the apparent or

reactive power flow changes

� full ac power flow algorithms are very accurate but are also relatively slow

� because of the way power systems are designed and operated, only a few outages will actually cause trouble� only a few power flow solutions will conclude that an overload

or voltage violation exist

� a solution to the time problem is to find a way of selecting those contingencies that are likely to result in an violation� these cases are studied in detailed

� the remaining cases will go unanalyzed



� AC power flow methods� selecting the problematic cases from the full list of cases is not

an exact procedure

� two sources of errors can arise� placing too many cases on the short list

� conservative approach

� leads to longer run times

� skipping cases� a case that would have shown a problem is not placed on the list

� possibility of having that outage take place and cause trouble without the operators being warned



� Contingency selection� concentric relaxation

� bounding

� Concentric relaxation� an outage has a limited geographical effect

� the loss of a transmission line does not cause much effect at far distances (i.e., 1000 km)

� the power system can be divided into two parts� the affected region

� the unaffected region

� the regions are found by segmenting the system into layers� layers represent the buses that are n-nodes away from the outage



� Concentric relaxation� example of layering the outage effects

� some arbitrary number of layers is chosen and all buses included in that layer and all inner layers are solved with the outage in place� the buses in the outer layers are kept at constant voltage and

phase angle

� selecting the optimum layer is not trivial

p ri

jl m

Layer 0

outaged line

Layer 1

q ks

Layer 2



� Bounding� one problem of the concentric relaxation is overcome by using

adjustable region around the outage to solve for the overloads

� bounding accomplishes this by using three flexible subsystems defined as follow:� the subsystem immediately surrounding the outage, N1

� the external subsystem that is not solved in detail, N2

� the set of boundary buses that separate N1 and N2, N3

� the bounding method makes certain assumptions about the phase angle spread across the lines in region N2� given the injections in N1 and the maximum phase angle

appearing across any two buses in region N3



� Bounding� given a transmission line in N2 with flow fpq

0

� there is a maximum amount that the flow on pq can shift� it can increase to its upper limit, or decrease to its lower limit

� the power shift is translated into a maximum change in phase angle difference

p q i jk m

N2N3N1

∆Pm∆Pk

( ) ( )[ ]−+ −−=∆ pqpqpqpqpq fffff 00max ,ofsmaller

( ) ( ) pqpqqpqppq

pq xfx

f maxmax1 ∆=∆−∆→∆−∆=∆ θθθθ



� Bounding� the maximum change in the phase angle difference across pq

defines an upper limit for angular difference in region N3

� where i and j are any pair of buses in N3

� the right-hand side of the equation provides an upper limit to the maximum change in angular spread across any circuit in N2, providing a limit to the maximum change in their flow

� a completely safe N2 region� where the maximum shifted phase angle difference is small

enough to be less than all of the biased flow limits

� accomplished by expanding the N1 region to encompass the overloaded circuits

jipqpqjiqp xf θθθθθθ ∆−∆<∆→∆−∆<∆−∆ max



� Bounding� expansion rule:

� all circuits in N2 are safe from overload if the value ofthe upper limit defined by the difference in the change of angles is less than the smallest value of the biased line flow over all line pairs pq, where p and q corresponds to the buses at the ends of circuits in N2:

� if this condition fails, N1 and N3 are expended to cover the violations

� calculate a new upper limit for region N3:

� rerun the test over the newly defined region N2

jipqpq xf θθ ∆−∆<∆ max

ji θθ ∆−∆



� Example� using the 6-bus system, study the outage of line 9 (3-6)

� system data and line upper limits

l1, line 1-2

0.50

0.40

0.20

0.40

0.20

0.30

0.20

0.20

0.60

0.20

0.30

l2, line 1-4

l3, line 1-5

l4, line 2-3

l5, line 2-4

l6, line 2-5

l7, line 2-6

l8, line 3-5

l9, line 3-6

l10, line 4-5

l11, line 5-6

LineMW Limit

(pu)fpq

0

(pu)xpq (pu)

0.20

0.30

0.25

0.10

0.30

0.20

0.30

0.40

0.10

0.26

0.20

∆fpqmax

0.084

0.069

0.182

0.075

0.038

0.047

0.197

0.159

-------

0.031

0.052

0.416

0.331

0.018

0.325

0.162

0.253

0.003

0.041

0.449

0.169

0.248

∆fpqmax xpq

0.0168

0.0207

0.0455

0.0075

0.0114

0.0094

0.0591

0.0636

-------

0.00806

0.0104


N2

N1 & N3


� Example

1.8 MW 1.8 MW

70 MW

70 MW

44.9 MW

44.9 MW

24.8 MW

0.3 MW

16.9 MW

16.9 MW

60.0 MW

16.2 MW

25.3 MW

16.2 MW

24.8 MW

0.3 MW

50.0 MW

32.5 MW

25.3 MW

41.6 MW

4.1 MW

33.1 MW

4.1 MW

41.6 MW

70 MW

100 MW

33.1 MW

32.5 MW

Bus 2 Bus 3

Bus 6

Bus 5Bus 1

Bus 4



� Example� to find the maximum change in N3, line 3-6 is examined

� checking for violations� the smallest value of the biased line flows is 0.0075 for line 2-4,

which is less than the maximum change0.0075 < 0.1113

� the criterion fails and we need to expand region N3

( )( )( )

( )1113.00536.00577.0

0536.0449.002876.0

003433.0

2

0577.0449.002876.0

003695.0

2

63

36663336

366663366

36663336

363633363

=+=∆−∆

−=−=−+−

−=∆

==−+−

−=∆

θθ

θ

θ

XXXx

PXXx

XXXx

PXXx



� Example

N2 N3

N1

1.8 MW 1.8 MW

70 MW

70 MW

44.9 MW

44.9 MW

24.8 MW

0.3 MW

16.9 MW

16.9 MW

60.0 MW

16.2 MW

25.3 MW

16.2 MW

24.8 MW

0.3 MW

50.0 MW

32.5 MW

25.3 MW

41.6 MW

4.1 MW

33.1 MW

4.1 MW

41.6 MW

70 MW

100 MW

33.1 MW

32.5 MW

Bus 2 Bus 3

Bus 6

Bus 5Bus 1

Bus 4



� Example� to find the maximum change in N3, line 2-5 is examined

� checking for violations� again, the smallest value of the biased line flows is 0.0075 for

line 2-4, which is greater than the maximum change0.0075 > 0.003564

� the criterion is satisfied

( )( )( )

( )003564.0

2

2

52

36663325

365653255

36663325

362623252

=∆−∆−+−

−=∆

−+−−=∆

θθ

θ

θ

XXXx

PXXx

XXXx

PXXx



� Example

l1, line 1-2

0.50

0.40

0.20

0.40

0.20

0.30

0.20

0.20

0.60

0.20

0.30

l2, line 1-4

l3, line 1-5

l4, line 2-3

l5, line 2-4

l6, line 2-5

l7, line 2-6

l8, line 3-5

l9, line 3-6

l10, line 4-5

l11, line 5-6

LineMW Limit

(pu)fpq

0

(pu)

0.416

0.322

– 0.220

0.316

0.148

0.257

0.191

0.032

-------

0.308

0.508

0.416

0.331

0.018

0.325

0.162

0.253

0.003

0.041

0.449

0.169

0.248

fpq3-6 out

(pu)

overloaded

overloaded

overloaded

Chapter 3 Economic Dispatch of Thermal Unitssmart.tabrizu.ac.ir/Files/Content/operation_wood.pdf · The Economic Dispatch Problem Consider a system that consists of N thermal-generating

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