PSOC.pptx

1
• POWER SYSTEM OPERATION AND
CONTROL (PSOC)

ELECTRIC POWER SYSTEM OPERATION
 Operational objectives of a power system have been to provide a continuous
quality service with minimum cost to the user. These objectives are:
 First Objective: Supplying the energy user with quality service, i.e., at
acceptable voltage and frequency
 Second Objective: Meeting the first objective with acceptable impact upon the
environment.
 Third Objective: Meeting the first and second objectives continuously, i.e.,
with adequate security and reliability.
 Fourth Objective: Meeting the first, second, and third objectives with
optimum economy, i.e., minimum cost to the energy user.
 The term “continuous service” can be translated to mean “secure and reliable
service”. 2

INTEGRATED OBJECTIVES
Interrelated objectives of operation of a power system
 The direction of the arrows indicates the priority in which the objectives are
implemented
Economically constrained operation of a power system.
3

ELECTRIC POWER SYSTEM OPERATIONS
 Task division:
 Operations planning
 Operations control
 Operations accounting
Interrelated tasks of planned scheduling operation 4

OPERATION PLANNING
 The facilities of a large power system consist of many generating units,
transmission lines, transformers, circuit breakers, DC/DC converters & DC/ AC
converters which are to scheduled for orderly operation & maintenance.
 The energy resources of a large power system consist of hydro, nuclear, fossil
power and renewable energy sources such as wind farm, photovoltaic and
micro turbines.
 These facilities are to be managed and utilized to satisfy load demand of a
power system.
 The load demand of a power system is cyclic in nature and has a daily peak
demand over a week period, weekly peak demand over a month period, and
monthly peak demand over a year period.
 Overall objectives of planned scheduling operation are to manage facilities and
optimize resources for satisfying the peak demand of each load cycle, such that
the total cost of operation is minimized. 5

OPERATION CONTROL
 The primary functions of operations control are satisfying the instantaneous load
on a second-to-second and minute-to-minute basis.
 Some of the functions are:
 Economic Dispatch Calculation (EDC)
 Load Frequency Control
 On-Line Load Flow
 Operating Reserve Calculation (ORC)
6

OPERATION CONTROL Contd…
 Economic Dispatch Calculation:
 Economic dispatch calculation of a power system determines the loading of
each generator on a minute-by-minute basis so as to minimize the operating
costs.
 Load Frequency Control (LFC):
 This function is also referred to as governor response.
 As the load demand of the power system increases, the speed of generators will
decrease and this will reduce the system frequency.
 Similarly, as system load demand decreases, the speed of the system generators
would increase and this will increase the system frequency.
 The power system frequency control must be maintained for the power system
grid to remain stable. 7

OPERATION CONTROL Contd…
 Online Load Flow (OLF):
 This function generally utilizes the output of network topology.
 It is the real time network model, and the bus injections from state estimation
for purpose of security monitoring, security analysis and penalty factor
calculations.
 This function performs “if then condition” to determine the possible system
states (voltages) in face of system outages such as loss of a line due to weather
condition or sudden loss of a generator.
 Operating Reserve Calculation:
 The objective of operating reserve calculation is to calculate the actual reserve
carried by each unit and to check whether or not there is a sufficient reserve in
a system.
 The operating reserve consists of spinning reserve (synchronized), non-
spinning reserve (non-synchronized), and interruptible load. 8

UNIT-1:ECONOMIC LOAD DISPATCH(ELD)
 Difference between LFA & ELD:
 In Load Flow Analysis (LFA), for a particular load, generation is fixed at all
generators except slack bus.
 In Economic Load Dispatch, for a particular load, generation is not fixed for all
the generators but they operated under certain limits.
 Scheduling:
 It is the process of allocation of generation among different generating units.
 Economic Scheduling:
 It is a cost effective mode of generation in such a way that the overall cost of
generation should be minimum.
9

UNIT-1: SYSTEM CONSTRAINTS
 The economic power system operation needs to satisfy the following types of
constraints.
 Equality constraints: The sum of real power generation of all the various units
must always be equal to the total real power demand on the system.
 Inequality constraints: These are classified as two types
1. According to the nature:
i) Hard-type and
ii) Soft-type
10

UNIT-1: SYSTEM CONSTRAINTS Contd…
2. According to power system parameters the inequality constraints are.
i) Output power,
ii)Voltage,
iii)Spare Capacity,
iv) Transformer tap position,
v) Transmission line and
vi) Security constraints.
 System Variables:
i) Control variables(PG & QG),
ii) Disturbance variables (PD & QD)
iii) State variables (V & δ) 11

UNIT-1: STEAM UNIT
 A typical boiler-turbine-generator unit is shown in Figure 1.
 This unit consists of a single boiler that generates steam to drive a single turbine-
generator set.
 The electrical output of this set is connected not only to the electric power
system, but also to the auxiliary power system in the power plant.
 A typical steam turbine unit may require 2-6% of the gross output of the unit for
the auxiliary power requirements.
 The necessary to drive boiler feed pumps, fans, condenser circulating water
pumps, and so on.
12
Fig. 1 Boiler-turbine-generator unit

UNIT-1: CHARACTERISTICS OF STEAM UNIT
 Input-Output Characteristics:
 Figure 2 shows the input-output characteristic of a steam unit in idealized form.
 The input to the unit shown on the ordinate may be either in terms of heat
energy requirements [millions of Btu per hour (MBtu/hr)] or in terms of total
cost per hour (Rs/ hr).
 The output is normally the net electrical output of the unit. The characteristic
shown is idealized in that it is presented as a smooth, convex curve.
13
Fig. 2 Input-output curve of a steam turbine generator

 Cost Curves:
 To convert the input-output curves into cost curves, the fuel input per hour is
multiplied with the cost of the fuel(expressed in Rs./million kCal).
 Cost Curves=(kCal*10^6/hr)*(Rs./million kCal)=Rs./hr
 Incremental Fuel Cost Curve (IFC):
 The IFC is defined as the ratio of a small in the input to the corresponding small
change in the output and it is expressed in Rs./MWh.
14
Fig. 3 Incremental heat (cost) rate characteristic

 Incremental Fuel Cost Curve (IFC):
 Mathematically, the IFC curve expression can be obtained from the expression
of the cost curve.
 cost curve expression is, (2nd degree polynomial)
 The IFC is, (linear approximation)
 Heat Rate Curve :
 The Thermal unit is most efficient at a minimum heat rate
15
2
1
2 i
i
i i i G i
G
C a P b P d
  
( ) i i
i
i
i i G i G
G
dc IC a P b P
dP
  
Fig. 4 Net heat rate characteristic of a steam turbine generator unit

UNIT-1: OPTIMIZATION PROBLEM
 MATHEMATICAL FORMULATION (Neglecting The Transmission Losses):
 An optimization problem consists of :
1. Objective function.
2. Constraint equations
 Assumptions:
1. Each unit does not violate the inequality constraints
2. Let the Transmission losses are neglected (PL =0)
3. Cost of ith unit is,
 The objective function is minimize the overall cost(CT) of production of electrical
energy , let n be the number of units in the system and Ci be the cost of ith unit .
 Objective: Min CT = (1) 16
2
1
2 i
i
i i i G i
G
C a P b P d
  
1
( )
i
n
i G
i
C P



UNIT-1: OPTIMIZATION PROBLEM Contd…
 The cost is to be minimized subject to the equality constraints.
Subject: (2)
 This is a constrained optimization problem that may be attacked formally using
advanced calculus methods that involve the Lagrange function.
 In order to establish the necessary conditions for an extreme value of the
objective function, add the constraint function to the objective function after
the constraint function has been multiplied by an undetermined multiplier.
 This is known as the Lagrange function and is shown in Eq. 3.
(3)
 Take the first derivative of the Lagrange function with respect to each of the
independent variables and set the derivatives equal to zero (variables are N+1)
(4)
 Condition for optimality is (5)
 Eq.(5) is called an approximate co-ordination equation because losses are neglected.17
1
0
i
n
G D
i
P P

 

'
1
[ ]
i
n
G D
T
i
P P
C C 

  

'
0
i
G
C
P
 

i
i
G
C
P

 


UNIT-1: COMPUTATIONAL METHODS
 Different types of computational methods for solving the optimization problem.
1. Analytical method when the no of units are small (either 2 or 3)
2. Graphical method
3. Using a digital computer method or λ-iterative method for more no of units.
 Algorithm for λ-iterative method :
i) Guess the initial value of λ0 with the use of cost curve equations.
ii) Calculate PGi according to equation (5).
iii) Calculate
iv) Check whether
v) If , set a new value of λ, i.e., and repeat from step (ii)
vi) If , set a new value of λ, i.e., and repeat from step (ii)
vii) Stop . 18
1
i
n
G
i
P


1
i
n
G D
i
P P



1
i
n
G D
i
P P


 1 0

 
  
1
i
n
G D
i
P P



1 0

 
 

FLOWCHART WITHOUT LOSSES
19
Increase λ by Δλ
i.e., (λ = λ + Δλ )
Set PGi=PGi(Max)
START
Read n, ai, bi, di, ԑ, PGi(Min), PGi(Max), and Δλ
Choose a suitable value of λ
Set generator count i=1
Compute PGi
Is
PGi > PGi(Max)
Is
PGi < PGi(Min)
Set PGi=PGi(Min)
Yes
Yes
Increment i=i+1
1
A
B
Compute ΔΡ= | ΣΡGi-ΡD|
No
Yes
Check
if i=n?
1
Yes
A
Check
if
ΔΡ<Ԑ
Print power
generations of
all units and
compute cost of
generation
No
Check
if
ΣΡGi> Ρ D
B
Decrease λ by Δλ
i.e., (λ = λ -Δλ )
No
Yes
No
No

UNIT-1: OPTIMAL LOAD SHEDDING INCLUDING
TRANSMISSION LOSSES
 The mathematical formulation is now stated as:
 Objective: Min CT = (1)
 The cost is to be minimized subject to the equality constraints.
 Subject: (2)
 Lagrange function is (3)
 The minimum point is obtained when,
(4)
 Therefore the condition for optimality is (5)
 Eqn(5) is modified as (6)
 Eqn(6) is called exact co-ordination equation because losses are co-ordinate the
ITL with IFC .
20
1
( )
i
n
i G
i
C P


1
0
i
n
G L D
i
P P P

  

'
1
[ ]
i
n
G L D
T
i
P P P
C C 

   

'
(1- ) 0; i=1......n
i i i
i L
g
G G G
C P
C
P P P

 
   
  
i i
i L
G G
C P
P P
 
 
 
 
i
i
i
G
C L
P

  


UNIT-1: OPTIMAL LOAD SHEDDING Contd…
 The term is called the penalty factor of plant i
 The minimum operation cost is obtained when the product of the incremental
fuel cost and the penalty factor of all units is the same, when losses are taken.
 The approximate expression for loss PL is given by, (7)
 Where, Bij is called loss coefficients and the unit is MW-1
 The incremental transmission loss (ITL) is given by, (8)
 The incremental fuel cost (IFC) is given by, (9)
 Substitute Eqn’s (8)&(9) in Eqn(5);we get, (10)
 To solve this allocation problem use λ-iterative method with losses considered.
21
1
1-
i
i
L
G
L
P
P



1 1
i j
n n
L G ij G
i j
P P B P
 
 
1
2 j
i
n
L
ij G
G j
P B P
P 
 
 
2 i
i
i
i G i
G
C a P b
P
  

 
1( )
1 2
2
j
i
n
i
ij G
j j i
G
i
ii
b
B P
P
a B


 
 




UNIT-1: OPTIMAL LOAD SHEDDING Contd…
 Algorithm for λ-iterative method when losses are considered:
i) Assume a suitable value of λ0 . This value should be more than the largest
intercept of the incremental cost characteristics of the various generators.
ii) Calculate generations (PGi) based on approximate co-ordination Eqn (5).
iii) Calculate generations (PGi) at all buses using Eqn (10), and check if the
difference in power at all generations (PGi) between two consecutive
iterations is less than pre-specified(Ԑ) value. If not repeat from step (ii).
iv) Calculate loss value using Eqn (7), and calculate
v) Check whether change in power ΔΡ ≤ Ԑ, stop the process and calculate the
cost of generations with their values of powers. Otherwise go to next step.
vi) If ,set a new value of λ, i.e., and repeat from step (ii)
vii) If ,set a new value of λ, i.e., and repeat from step (ii)
viii) Stop the process.
22
1
i
n
G L D
i
P P P P

   

1 0

 
  
1 0

 
 
0
P
 
0
P
 

FLOWCHART WITH LOSSES
23
Increase λ by Δλ
i.e., (λ = λ + Δλ )
Set PGi=PGi(Max)
START
Read n, ai, bi, di, ԑ, PGi(Min), PGi(Max), and Δλ
Choose a suitable value of λ
Set generator count i=1
Compute PGi
Is
PGi >
PGi(Max)
Set PGi=PGi(Min)
Yes
Yes
Increment i=i+1
1
B
D
Compute Transmission loss,
PL and check in power
change, ΔΡ= | ΣΡGi- ΡL-ΡD|
No
Yes
Check
if i=n?
Yes
Check
if
ΔΡ<Ԑ
Print power
generations of
all units and
compute cost of
generation
No
Is
ΔΡ>0
Decrease λ by Δλ
i.e., (λ = λ -Δλ )
No
Yes
No
No
Is
PGi <
PGi(Min)
Determine PGi corresponding to IPC
Set iteration count k=1
C
Check if
|Pgi
k –Pgi
k-1|
< Ԑ
Yes
Increment
iteration count k,
k=k+1
A
C
D B A 1

UNIT-2: HYDROTHERMAL SHEDULING
 The hydrothermal co-ordination is classified into :
i) Long-Term Co-ordination
ii) Short-Term Co-ordination
 Long-Range Hydro-Scheduling:
 The long-range hydro-scheduling problem involves the long-range forecasting
of water availability and the scheduling of reservoir water releases (i.e.,
“drawdown”) for an interval of time that depends on the reservoir capacities.
 Typical long-range scheduling goes anywhere from 1 week to 1 year or several
years.
 For hydro schemes with a capacity of impounding water over several seasons,
the long-range problem involves meteorological and statistical analyses.
24

UNIT-2: HYDROTHERMAL SHEDULING Contd…
 Short-Range Hydro-Scheduling:
 Short-range hydro-scheduling (1 day to 1 week) involves the hour-by-hour
scheduling of all generation on a system to achieve minimum production cost
for the given time period.
 In such a scheduling problem, the load, hydraulic inflows, and unit
availabilities are assumed known.
 A set of starting conditions (e.g., reservoir levels) is given, and the optimal
hourly schedule that minimizes a desired objective, while meeting hydraulic
steam, and electric system constraints, is sought .
25

UNIT-2: HYDROTHERMAL SHEDULING Contd…
 The factors on which the economic operation of a combined hydro-thermal system
depends on:
i) Load cycle.
ii) Incremental fuel costs (IFC) of thermal power stations.
iii) Expected water inflow in hydro-power stations.
iv) Water head that is a function of water storage in hydro-power stations.
v) Hydro-power generation.
vi) Incremental transmission loss (ITL).
 The few important methods for short-term hydro-thermal co-ordination:
i) Constant hydro-generation method.
ii) Constant thermal-generation method.
iii) Maximum hydro-efficiency method.
iv) Kirchmayer’s method. 26

UNIT-2: SHORT RANGE HYDRO SHEDULING
 Kirchmayer’s method:
 In this method equivalent cost of water is used.
 Let there be α thermal power stations and (n- α) hydro power stations in a
power system.
 Let γj be the equivalent cost in Rupees of one cubic meter of water, and wj be
the water used in cubic meters per hour in power generation in jth hydro station.
 Let ci be the cost of power generation in Rs/hr in thermal ith power station.
 Then the total cost of power generation would be.
 Object: Min CT= (1)
 In this total cost CT is minimized subject to the equality constraint.
 Subject: (2)
27
1 1
( ) Rs/hr
i
n
i T j j
i j
C P w



  

 
1 1
0
i j
n
T H L D
i j
P P P P


  
   
 

UNIT-2: SHORT RANGE HYDRO SHEDULING
Contd…
 The optimal operating state is determined by the Lagrange method.
 The augmented cost function is, (3)
 Carrying out the differentiation of Eqn(3), we get conditions for optimality as
 W.r.t, Thermal power generation the condition is, (4)
 W.r.t, Hdro power generation the condition is, (5)
 Solution of Eqn’s(4-5), yields the economically optimum thermal and hydro
power generations.
 If transmission losses are neglected the Eqn’s(4-5) reduced to
28
*
1 1
[ ]
i j
n
T H L D
T
i j
P P P P
C C



  
    
 
i i
i L
T T
C P
P P
 
 
 
 
j j
j L
j
H H
w P
P P
  
 
 
 
j i
j i
j
H T
w c
P P
 
 
 
 

UNIT-2: LONG RANGE HYDRO SHEDULING
 To mathematically formulate the optimal scheduling problem in a hydro-
thermal system.
 Few assumptions are to be made for a certain period of operation T (several
years)
i) The storage of a hydro reservoir at the beginning and at the end of period of operation T are
specified.
ii) After accounting for the irrigation purpose, water inflow to the reservoir and load demand on
the system are known deterministically as functions of time with certainties.
 The optimization problem here is to determine the water discharge rate q(t) in
m3/sec, so as to minimize the cost of thermal generation.
 Object: Min CT= (1)
29
0
( )
t
i
c t dt


UNIT-2: LONG RANGE HYDRO SHEDULING Contd…
 Subjected to three constraints:
i) The real power balance equation: (2)
ii) Water availability equation: (3)
iii) The real power hydro-generation: (4)
 Solution of problem-discretization principle:
 This problem is solved by dividing the total time interval T into M subintervals
each of time, ΔT=T/M.
 To simplify the analysis, assume that during each subinterval all the variables
remain fixed.
 The problem is therefore, redefined as
Objective: (5)
30
( ) ( ) ( ) ( ) 0
T H L D
P t P t P t P t
   
0 0
'( ) '(0) ( ) ( ) 0
T T
i
W T W W t dt q t dt
   
 
( ) ( '( ), ( ))
H
P t f W t q t

1
C ( )
M
m m m
i
T T
m
Min C P

 

 Subject to the operating constraints are redined as:
i) Power balance equation is, (6)
ii) Water availability equation is, (7)
iii) Hydro generation in any subinterval can be expressed as:
(8)
 where, h0=9.81x10-3h‫׳‬0, h‫׳‬0 is the height of the storage tank, 0.5(wm-wm-1) is the
average additional height due to storage of water, e is the water head correction
factor and ρ is the no-load discharge of water.
 The sum of discharges during (M-1) intervals will give the desired available
discharge and one of the discharges is taken as dependent variable.
31
0
m m m m
T H L D
P P P P
   
1
0
m
m m m
i q
w w w

   
1
0{1 0.5 ( )}( )
m m m m
H
p h e w w q 

   

 Usually q1 is chosen as dependent variable and hence Eqn(7) corresponding to
water availability can be rewritten as
(9)
 The problem of economic hydro-thermal scheduling is handled by making use
of Lagrangian multiplier.
 The augmented cost function is given as
(10)
 The Lagrangian multipliers can be obtained by differentiating the augmented
function w.r.t, dependent variables (PT
m,PH
m ,wm and q1) and equating it to zero.
 For minimization of the augmented function, differentiate the augmented
function w.r.t, independent variable (qm) and obtain the gradient vector which
should be zero. 32
1 0
1 2
M M
M m m
i
m m
w q
q w w  
   
 
* 1
1 2
1
0
3
( ) ( )
{ [1 0.5 ( )( )]}
m m m m m m m
m m m
T i
T H L D
m m m m m
H
c c P P P P w w w q
P h e w w q
 
 


        
    

Contd…
 The coordination equations are given as:
 Where, α is a positive scalar value with a range of 0.4-0.8.
*
1 1 1
(1 ) 0 or (11)
m m
T L T L
m m m
m m m m m
T T T T T
P P
c c
c
P P P P P
  
 
 
      
    
*
3 1 1 1 3
(1 ) 0 or (12)
m m
L L
m m m m m
m m m
H H T
P P
c
P P P
    
 
      
  
* 1 1 1
0 0
2 2 3 3
0( )
[0.5 ( )] [0.5 ( )] 0 (13)
m m m m
m m
m m orM
c h e q h e q
w
     
  

       

* 1 1 1
0 1
0
2 3
1
[1 0.5 (2 2 )] 0 (14)
i
c h e w w q
q
  
       

*
1
0
2 3
1
[1 0.5 (2 2 )] (15)
m m m
m m
i
m
m
c h e w w q
q
  


      

*
1
(1 ) (16)
m m
new old
m
m
c
q q
q



  


Contd…
 Algorithm for Long-Term Co-ordination:
i) Assume initial set of independent variables, qm for all sub-intervals except the
first sub-interval.
ii) Obtain the values of dependent variables PT
m,wm , PH
m and q1 using Eqn’s (6),
(7), (8), and (9) respectively.
iii) Obtain the Lagrangian multipliers λ1
m, λ3
m, λ2
1 and λ2
m using Eqn’s(11), (12),
(14), and (13) respectively.
iv) Obtain the gradient vector using equation (15) and check whether all its
elements are close to zero within a specified tolerance, if so the optimal value is
reached; if not, go to the next step.
v) Obtain the new values of control variables using the equation (16), then go to
step (ii) and repeat the process.

UNIT-3: UNIT COMMITMENT
 The total load of the power system is not constant but varies throughout the day
and reaches a different peak value from one day to another.
 Therefore, it is not advisable to run all available units all the time.
 So, it is necessary to decide in advance which generators are to
startup/shutdown, and for how long.
 The computational procedure for making such decision is called unit
commitment.
 Unit commitment means to ‘commit’ a generating unit to ‘turn it on’
 In the case of ELD all the available units should be turned on for all the time.
 In the case of UC only a best of available units to be turned on to supply the
forecast load of the system over a future time period.

 Need for UC:
 The plant commitment and unit ordering schedules.
 Weekly patterns can be developed from daily schedules, likewise monthly and
annual schedules.
 A great deal of money can be saved by turning off the units when they are not
needed for the time.
 Constraints in UC:
i) Spinning reserve:
1. low/high frequency
2. islands
3. fast/slow responding units

ii) Thermal unit constraints:
1. Minimum up/down time
2. crew constraints
3. start-up cost: Two approaches are there
 Start-up cost when cooling=Cc(1-e-t/α) F+Cf
 Start-up cost when banking=Ct × t × F+Cf
 Where Cc is cold start cost(MBtu), F is fuel cost, Cf is fixed cost, α is thermal
time constant for the unit, Ct is cost (Mbtu/h) of maintaining unit at operating
temperature, and t is time (h) the unit was cooled
iii) Hydro unit constraints:
1. Must run constraint
2. Fuel constraints

 Cost function formulation:
1. Running cost.
2. Start-up cost.
3. Shut-down cost.
 The total expression for the cost function is given as:
 Unit commitment solution methods:
1. Priority list (PL) schemes.
2. Dynamic programming (DP) method.
3. Lagrange’s relaxation (LR) method.
1 1 1 1 1
t ij i t i t
t
y
N k x
T
T ij G sc i sd i
t i j i i
F C P C C
 
    
 
  
 
 
   

 Priority list method:
 The simplest unit commitment solution method consists of creating a priority
list of units.
 In this method a simple shut-down rule or priority-list scheme could be obtained
after an exhaustive enumeration of all unit combinations at each load level.
 The priority list could be obtained by noting the full-load average production
cost of each unit.
 where the full-load average production cost is simply the net heat rate at full
load multiplied by the fuel cost.
 The most efficient unit (least average production cost) is loaded first to be
followed by the less efficient units in order as the load decrease.

 Dynamic programming approach:
 Assumptions in the DP approach:
1. A state consists of an array of units with specified units operating and the rest off-line.
2. The start-up cost of a unit is independent of the time it has been off-line.
3. There are no costs for shutting down a unit.
4. There is a strict priority order, and in each interval a specified minimum amount of
capacity must be operating.
 Mathematical representation:
 is called recursive relation.
Where, FN(x) be the min cost in Rs,/hr of generation of ‘x’ MW by N no. of units.
and, fN(y) be the cost of generation of ‘y’ MW by Nth unit.
and, FN-1(x-y) be the min cost of generation of ‘x-y’ MW by remaining N-1 units.
1
( ) min{ ( ) ( )}
N N N
y
F x f y F x y

  

 Procedure for preparing the UC table using the DP approach:
Step 1: Start arbitrarily with considerations of any two units.
Step 2: Arrange the combined output of the two units in the form of discrete
load levels.
Step 3: Determine the most economical combination of the two for all the load
levels.
Step 4: Obtain most economical cost curve in discrete form for the two units
and that can be treated as the cost curve of a single equivalent unit.
Step 5: Add the third unit and repeat the procedure to find the cost curve of the
three combined units.
Step 6: Repeat the process till all available units are exhausted.
 Advantage of DP approach:
 The main advantage is that having obtained the optimal way of loading ‘K’ units, it
is quite easy to determine the optimal way of loading (K+1) units.

UNIT-4: LOAD FREQUENCY CONTROL
 SINGLE AREA CASE:

PSOC.pptx

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